Feynman Lectures on the Strong Interactions
FFeynman Lectures on the Strong Interactions
Richard P. Feynman
Lauritsen Laboratory, California Institute of Technology, Pasadena, California 91125 revised by James M. Cline
McGill University, Department of Physics, 3600 University St., Montr´eal, Qu´ebec H3A2T8, Canada
These twenty-two lectures, with exercises, comprise the extent of what was meant to be a full-yeargraduate-level course on the strong interactions and QCD, given at Caltech in 1987-88. The coursewas cut short by the illness that led to Feynman’s death. Several of the lectures were finalizedin collaboration with Feynman for an anticipated monograph based on the course. The others,while retaining Feynman’s idiosyncrasies, are revised similarly to those he was able to check. Hisdistinctive approach and manner of presentation are manifest throughout. Near the end he suggestsa novel, nonperturbative formulation of quantum field theory in D dimensions. Supplementarymaterial is provided in appendices and ancillary files, including verbatim transcriptions of threelectures and the corresponding audiotaped recordings. Contents
Preface
1. The quark model (10-15-87)
2. Other phenomenological models (10-20-87)
3. Deep inelastic scattering; electron-positronannihilation (10-22-87)
4. Quantum Chromodynamics (10-27-87)
5. QCD Conventions (10-29-87)
6. Geometry of color space ∗ (11-3,5-87)
7. Semiclassical QCD ∗ (11-10-87)
8. Quantization of QCD ∗ (11-12-87)
9. Hamiltonian formulation of QCD ∗ (11-17-87)
10. Perturbation Theory (11-19-87)
11. Scattering processes (11-24-87)
12. Gauge fixing the path integral ∗ (12-1-87)
13. Quark confinement ∗ (12-3-87)
14. Interlude (1-5-88)
15. Scale dependence (1-5-88)
16. The renormalization group (1-7-88) g
17. Renormalization: applications (1-12-88)
18. Renormalization, continued (1-14-88)
19. Renormalization (conclusion); Lattice QCD(1-19-88)
20. Dimensional regularization, continued(1-21-88) D dimensions 54
21. Physics in D dimensions, conclusion(1-26-88) Q
22. Final lecture (1-28-88)
A. Transcription: Scale dependence (1-5-88) a r X i v : . [ h e p - ph ] J un B. Transcription: Renormalization:applications (1-12-88) C. Transcription: Renormalization, continued(1-14-88) D. Revision examples E. Hadron masses and quark wave functions F. Tables of hadrons G. Rules for amplitudes and observables Preface
During the last year of my Ph.D. at Caltech in 1987-88,I was looking for a course to TA that would not take toomuch time from finishing my dissertation. I had heardthat Feynman did not assign homework in his courses,and in my naivet´e asked him if I could be his teachingassistant for a new course that had been announced, onquantum chromodynamics. After checking my creden-tials with my supervisor John Preskill, he agreed. Onlyafterwards did I realize that the TA in Feynman’s courseswas generally the person who did the transcription of thenotes to create the monograph that would follow. Thiswas not the easy job I had bargained for, and I persuadedSteven Frautschi to assign several other TAs to the courseto divide the labor. We took turns rewriting the lecturesinto publishable form, which Feynman would revise be-fore considering final. Little did I suspect that I was onlypostponing my task by ∼
30 years.Unfortunately most of those corrected drafts becamedispersed with the other TAs, who have left physics. Inmy possession are seven lectures that I prepared for pub-lication, at least some of which were revised by Feynman.(These are denoted by an asterisk ∗ in the section head-ings.) As for the rest, I report what is in my class notes,trying to convey their intent as best I can. Based uponthe rather extensive revisions he made to some of my firstdrafts, the sections he did not check are unlikely to dojustice to all of his intended meanings. Certain parts callfor elaboration, but I abstain from restoring longer expla-nations where I have no record of what Feynman actuallysaid. These fully revised lectures can be found in sections6, 7, 8, 9, 12. I was able to supplement my notes in someplaces with his own (mostly very sketchy) lecture notes,that are available from the Caltech Archives, Folder 41.7of the Feynman Papers.For the lectures of Jan. 5, 12 and 14, 1988, I was ableto refer to tape recordings that were kindly provided byArun K. Gupta, one of the former TAs. I have placed ver-batim transcriptions of these lectures in the appendix, asa supplement to the more conventional versions in themain document. The quality of the recordings makes itimpossible to reproduce every word, and ellipses indicatewords or passages that I could not make out. This is especially the case toward the end of long explanations,where Feynman’s voice would tend to diminish greatly,whereas at the beginning he might almost be shouting.These recordings are available alongside the lectures assupplementary material. I have preserved as much aspossible his original words to convey the style of deliv-ery, which was considerably more colorful and colloquialthan the tone he adopted in the drafts to be published.The reader who compares these “raw” versions with therevised ones will understand why it was sometimes chal-lenging to correctly capture Feynman’s intended mean-ings.One thing you may notice, and that struck me as aneducator now myself, is that Feynman was never in a rushto explain anything (although at times he would speakvery fast), nor did he eschew repeating himself, perhapsin several different ways, to try to get his point across.And of course there was his bent for telling stories, whichI had forgotten about in the context of this course, sinceI had omitted them from my written notes. The “inter-lude,” section 14, which were Feynman’s remarks at thestart of the new term, is kept in the main body of thetext; it has a few interesting stories, and shows that hewould make time to help a high school student with hisgeometry.I have the impression that in some places Feynman hadnot prepared carefully and was working things out onthe spot, sometimes getting them not quite right, and attimes seemingly meandering through the material. Thiswas apparent for example in the early lectures on QCD,where in subsequent class sessions he came back and re-vised previous equations to correct the details. It is inter-esting that no notes corresponding to the QCD lecturesappear in the Caltech Archives folder, suggesting he wasspeaking extemporaneously. There is also repetition ofalready introduced material. Perhaps this was a deliber-ate pedagogical strategy, since it gave the students timeto digest the concepts and to see it being derived fromscratch. It is also possible that his terminal illness wasinterfering with his ability to prepare as well as he mighthave liked to. These detours would have been smoothedover in the version destined for publication, had therebeen time for him to revise the notes.Although there were no homework assignments, therewere some recommended problems that are included inthe lectures. Moreover about a month before the endof the first term, when students were starting to thinkabout the upcoming final exams, Feynman decided thateach of them should do an original research project re-lating to QCD. I recall that many were dumbfoundedwhen this announcement was made. Such an unexpecteddemand made by a lesser instructor would have createdsome outcry, but to a decree from the great man nobodyobjected, and everyone somehow managed to carry outthe task: it was a privilege. Feynman of course gradedthe projects himself, and he comments on them in theinterlude section.One may wonder what the specific content of the un-finished part of the course might have been. Feynmanannounces at the beginning of the second term that itwill be half on perturbative methods followed by non-perturbative. At that time he was interested in QCD in1+1 dimensions, as an exactly solvable model that mightshed light on the real theory. He started working with afew graduate students on this subject, including SandipTrivedi.His private course notes reveal a different direction;around 20 of the 60 pages are devoted to reformulatingvector spaces and calculus in arbitrary noninteger dimen-sions, which he discusses in lectures 20-21. His intent wasto combine this with Schwinger’s functional formulationof field theory, presented in lectures 21-22, to overcomethe difficulty of defining the path integral in nonintegerdimensions. Also in those notes is some material on chi-ral symmetry breaking by the axial anomaly and thetavacua in QCD, that he did not have time to present. Nodoubt the students would have been exposed to his ideasfor deepening our understanding of the strong interac-tions, had he lived until the end of the course.Feynman was an inspiring teacher, presenting every-thing in an incisive and fascinating way, that obviouslyhad his own mark on it. He reinvented the subject as washis wont, even if he was not the first to discover, for ex-ample, the Fadeev-Popov procedure for gauge fixing thepath integral. In the final meetings, he was too weak tostand at the board, and he delivered the lectures whileseated. He died less than three weeks following the lastlecture. His passion for transmitting the excitement ofphysics to a new generation never waned.Sorry this took so long, professor. James M. ClineMontr´eal, 2020
1. THE QUARK MODEL (10-15-87)
We begin our exploration of the strong interactionswith a survey of the hadronic particles, interpreted fromthe quark model perspective. The spin-1 / / u, d, s have charge+2 / , − / , − / state ( J =3 / : (cid:26) ddu ↑↑↑ , J z = 3 / ddu √ ( ↑↑↓ + ↑↓↑ + ↓↑↑ ) , J z = 1 / ( J = 1 / N : ddu √ ( − ↑↑↓ + ↑↓↑ + ↓↑↑ ) , J z = 1 / : uds √ ( − ↑↑↓ + ↑↓↑ + ↓↑↑ ) (1.2)For ∆ and N , the coefficients of the ↑↓↑ + ↓↑↑ spinterms had to be equal, since they are symmetric underinterchange of the first two quarks, which have identicalflavors ( dd ). However this is not a constraint for the uds baryons, so there must exist an additional state Λ Λ : uds √ ( ↑↓↑ − ↓↑↑ ) (1.3)that has isospin 0. The fact that the mass eigenstatesare also eigenstates of isospin indicates that u and d areapproximately degenerate, compared to the scale of thehadron masses.Similarly the mesons can be arranged into multiplets,as we illustrate for the J = 1 − vector mesons in fig. 2.The wave functions are given by ω : √ ( u ¯ u + d ¯ d ) ↑↑ ρ + : u ¯ d ↑↑ ; ρ : √ ( u ¯ u − d ¯ d ) ↑↑ ; ρ − : d ¯ u ↑↑ K ∗− : s ¯ u ↑↑ ; K ∗ : s ¯ d ↑↑ ; K ∗ + : u ¯ s ↑↑ ; K ∗ : d ¯ s ↑↑ ; φ : s ¯ s ↑↑ (1.4)It is interesting to notice that the ω and ρ are very closeto each other in mass. What do we learn about the stronginteractions from this near-degeneracy? Apparently, thestrong interactions conserve isospin.It is also interesting to observe that φ decays muchfaster into KK than into pions. This is an example ofZweig’s rule (OZI suppression), that can be pictureddiagrammatically by the statement that ππ is favored over KK φ One might wonder whether OZI suppression in thisexample is somehow related to the degeneracy of the φ - ω system. In fact there is a connection: if ω had some s ¯ s content rather than being purely made from u ¯ u and d ¯ d , which would spoil the degeneracy, then by the samemixing φ would also have light quark content, allowing fordecays into pions without going through the annihilationdiagram.The pseudoscalar mesons ( J P = 0 − ) have a differentflavor structure from the vector mesons, apart from thesimilarities between the two isotriplets ρ and π ,[ π + , π − , π ] : √ ( ↑↓ − ↓↑ ) (cid:104) u ¯ d, d ¯ u, √ ( u ¯ u − d ¯ d ) (cid:105) (1.5)In this case there is mixing between the isosinglets, η (546) ∼ ( u ¯ u + d ¯ d ) − . s ¯ sη (cid:48) (960) ∼ ( u ¯ u + d ¯ d ) + 0 . s ¯ s (1.6) Σ Λ J P = 1/2 + states Ν + Ν Ξ Ξ − Σ − Σ + charge mass (1320)(1190)(938) s = −1s = −2s = 0 Ω − Ξ ′− Ξ ′0 Σ ′− Σ ′+ Σ ′0 ∆ ∆ − ∆ + (duu) ∆ ++ charge (sss)(ssd) (ssu) J P = 3/2 + states (suu)(sdd) (sdu)(ddd) (ddu) (uuu) (1236)(cid:10)(1385)(1532)(1672) s = 0s = −1s = −2s = −3mass FIG. 1: The baryon octet (left) and decuplet (right). Masses indicated in MeV/ c . Why isn’t η purely ( u ¯ u + d ¯ d ), in analogy to ω , whichwould have made it approximately degenerate with thepions? This has to do with chiral symmetry breaking,which is specific to QCD and not accounted for by thequark model.An interesting prediction of the quark model is elec-tromagnetic matrix elements, that determine the baryonmagnetic moments. We consider those of the proton andthe neutron, where the proton wave function is | p (cid:105) = ( uud ) √ ( − ↑↑↓ + ↑↓↑ + ↓↑↑ ) (1.7)The magnetic moment is given by (cid:28) p (cid:12)(cid:12)(cid:12)(cid:12) q (cid:126) m σ z (cid:12)(cid:12)(cid:12)(cid:12) p (cid:29) (1.8)where q is the charge operator acting on the quarks, and m = m p / qσ z | p (cid:105) = uud √ (cid:16) e − ↑↑↓ + ↑↓↑ − ↓↑↑ )+ 2 e − ↑↑↓ − ↑↓↑ + ↓↑↑ ) − e ↑↑↓ + ↑↓↑ + ↓↑↑ ) (cid:17) (1.9)we find (cid:28) p (cid:12)(cid:12)(cid:12)(cid:12) q (cid:126) m σ z (cid:12)(cid:12)(cid:12)(cid:12) p (cid:29) = 3 µ N (1.10) ω ρ φ ρ −∗− Κ ∗0 Κ ρ +− = 1 ∗0 Κ (770)(1070)(895)(895) s = 1s = 0s = −1J P states Κ ∗+ FIG. 2: The vector meson nonet. where µ N = e (cid:126) / m p is the nuclear magneton. The anal-ogous calculation for the neutron (see eq. (1.2)) gives − µ N . These predictions are compared to the measuredvalues in the table 1. These predictions can be corrected, as shown in thethird row of the table, by taking a more realistic value ofthe constituent u and d quark masses, m q = 1085 / ∼ =362 MeV instead of m p / Further improvement mightarise from taking into account isospin breaking; the u and d masses are not exactly the same. We must certainlytake SU(3) flavor breaking into account for the s quark,whose constituent mass is m s = 1617 / qσ z m | Λ (cid:105) = sud √ qσ z m ( ↑↑↓ − ↑↓↑ ) → − m s qσ z m | Σ + (cid:105) = sud √ qσ z m (2 ↓↑↑ − ↑↓↑ − ↑↑↓ ) → (cid:104) + m s + m q + m q (cid:105) + (cid:104) − m s (cid:105) = m s + m q (1.11) p n Λ Σ + Σ − Ξ − µ th . − − − µ exp . . − . − . ± .
03 2 . ± . − . ± . − . ± . µ corr . . − . − .
58 2 . − . − . In class, RPF only presented the n and p values, and omittedthe “corrected” predictions. I have restored these and some ofthe related discussion from his private notes. It is not explained in his notes where the number 1085 comesfrom; probably it is a consequence of taking the spin-spin inter-actions into account in the baryon mass calculation. qσ z m | Σ − (cid:105) = sdd √ qσ z m (2 ↓↑↑ − ↑↓↑ − ↑↑↓ ) → (cid:104) + m s − m q − m q (cid:105) + (cid:104) − m s (cid:105) = m s − m q qσ z m | Ξ − (cid:105) = dss √ qσ z m (2 ↓↑↑ − ↑↓↑ − ↑↑↓ ) → m q − m s (1.12)These are in rather good agreement with the data, exceptfor the Ξ − .The nonrelativistic quark model can also be used topredict the axial vector current matrix elements, σ z γ .In the quark model we find that σ z γ gives +1 for u and − d , leading to the prediction g A = [1 + 1 + 1] + [ −
1] = 5 / which is high compared to the experi-mental value 1 . ± . Exercise.
What kind of baryon states do you expectwhen there is one unit of internal angular momentum?
2. OTHER PHENOMENOLOGICAL MODELS(10-20-87)
There are complementary phenomenological modelsfor describing the strong interactions, which we brieflyreview here. The first is the relativistic string, whichwas inspired by the observed Regge trajectories. Theseare plots of the spin versus mass squared of hadronicresonances, as in fig. 3. One considers only families ofresonances having the same parity, requiring J to jumpby ∆ J = 2. Empirically the trajectory is linear, whichwas not predicted by the quark model. However the rel-ativistic string, illustrated in fig. 4, gets the correct re-lation. In the simplest version of the model, the masses m J FIG. 3: A Regge trajectory. This calculation, also taken from his written notes, seems tobe based on unstated insights from the extended quark modelanalysis that takes into account the small components of theDirac spinors. This is worked out in the next lecture. c c(a) (b) (c)
FIG. 4: String model for mesons (a) and baryons (b,c). of quarks or antiquarks on the ends of the string are ne-glected, and one cares only about the constant tension T = energy/length of the strings, which represent fluxtubes of the strong interaction field. One finds that m (the energy squared) is proportional to the angular mo-mentum, with the endpoints of the string moving at thespeed of light. To explain the linear Regge trajectoriesof baryons in this picture, one could imagine flux tubeconfigurations as in fig. 4(b,c). Configuration (c) wouldobviously lead to the same prediction of linear Regge tra-jectories as for mesons.In the bag model (fig. 5(a)), quarks in a hadron “pushaway the vacuum” and move around in this evacuatedregion with nearly zero mass. It takes energy to makethe hole, which is interpreted as the hadron mass. Anapplication is to the decay ρ → e + e − (fig. 5(b)), wherewe recall that ρ = ( u ¯ u − d ¯ d ) / √
2. One must know thewave function of the q ¯ q bound state at the origin, ψ (0), toestimate the amplitude. Although the bag model gives areasonable estimate for ψ (0), the bag nevertheless turnsout to be too stable to get the rate right. One needs tomake it more dynamical than in the bag model picture,so that it shrinks more easily when the q and ¯ q are closeto each other. And the bag should turn into a flux tubewhen the pair is well-separated.The parton model is useful for describing high-energyprocesses, including inelastic scattering of electrons onnucleons. An example is shown in fig. 6, for the caseof eN → eN π . Let us think about the partons in theinitial state nucleon, in a reference frame where it ismoving to the right with 4-momentum p µ = ( E, p, , E ∼
10 GeV for example, and the virtual photon 4-momentum is q µ = (0 , (cid:126)Q ). A parton in the nucleus will e + e − u(d)(a) u(d) (b) γ FIG. 5: (a) Bag model for baryons. (b) Electromagnetic decayof ρ meson. − eN N γπ FIG. 6: High-energy electron-nucleon scattering have momentum components p (cid:107) = xp, p ⊥ ∼
300 MeV (2.1)parallel and perpendicular to the beam, respectively,where x is the momentum fraction of the parton that in-teracts with the virtual photon. In this frame, its momen-tum just gets reversed after scattering, px → − px = Q/ q = 0. In an arbitrary framewe can write the momentum fraction as x = Q pQ = − q − p · q . (2.2)The momentum distribution of partons in the nucleoncan be thought of as coming from their respective wavefunctions, written in momentum space. Naively, wewould expect the probability to find a quark with mo-mentum in the interval [5 , .
5] GeV in a 10 GeV protonto be the same as for the interval [10 ,
11] GeV in a 20GeV proton. Each parton has its own probability distri-bution u ( x ) , d ( x ) , s ( x ) : quarks¯ u ( x ) , ¯ d ( x ) , ¯ s ( x ) : antiquarks g ( x ) : gluon (2.3)For example u ( x ) is the probability density for finding a u quark with momentum fraction x . Of course these defi-nitions depend upon which hadron the parton belongs to.If we define the above functions as belonging to the pro-ton, then the amplitude for the photoproduction processis proportional to u ( x )+ d ( x )+ s ( x )+ ¯ u ( x )+ · · · , ( ep → eN π ) (2.4)for scattering on protons, whereas it is d ( x )+ u ( x )+ s ( x )+ ¯ u ( x )+ · · · , ( en → eN π ) (2.5)for scattering on neutrons since u ( x ) in a proton must beequal to d ( x ) in a neutron. The amplitudes will of coursealso depend upon Q . Challenge.
Compute the width for φ → e + e − .
3. DEEP INELASTIC SCATTERING;ELECTRON-POSITRON ANNIHILATION(10-22-87)
In the last lecture we saw that the electron-proton scat-tering cross section is proportional to a function F ep ( x ) = ( u + ¯ u ) + ( d + ¯ d ) + ( s + ¯ s ) (3.1)where u ( x ) is the probability density for finding a u quarkwith momentum fraction x in the proton. The momen-tum distribution functions are subject to constraints1 = (cid:18) protoncharge (cid:19) = (cid:90) (cid:2) ( u − ¯ u ) − ( d − ¯ d ) − ( s − ¯ s ) (cid:3) dx (cid:18) neutroncharge (cid:19) = (cid:90) (cid:2) ( d − ¯ d ) − ( u − ¯ u ) − ( s − ¯ s ) (cid:3) dx (cid:90) ( s − ¯ s ) dx (3.2)where again we assumed the neutron is related to theproton by interchange u ↔ d . From these it follows that (cid:90) ( u − ¯ u ) dx = 2 (cid:90) ( d − ¯ d ) dx = 1 (cid:90) ( s − ¯ s ) dx = = 0 (3.3)It has been shown that as x →
0, the distribution func-tions have the behavior u = ¯ u = d = ¯ d ∼ . x (3.4)Likewise, s = ¯ s scales as 1 /x . This behavior can be un-derstood as coming from brehmsstrahlung of soft gluons,which have a distribution of 1 /x . These (virtual) gluonsdecay into soft quark-antiquark pairs, qqg explaining why all flavors have the same 1 /x dependenceat low x , regardless of whether they are particles or an-tiparticles: gluons can decay into all flavors equally. Thegluons are known to comprise a significant fraction of thetotal partons, (cid:90) dx g ( x ) ≈ .
44 (3.5)So far, we have taken for granted that we know thecharges of the quarks. One experiment that constrainsthe charges is annihilation of electrons and positrons,fig. 7. Denoting the cross section for annihilation into e + e − by σ el , we can express that for annihilation into u ¯ u e − e + e + e − e − e + q q γ γ (a) (b) FIG. 7: Electron-positron annihilation into e + e − (a) and q ¯ q (b). e + e − jetjet FIG. 8: Electron-positron annihilation into hadronic jets. as σ u ¯ u ∼ σ el · σ d ¯ d or σ s ¯ s we get σ el ·
3. Forthe inclusive cross section to produce hadrons, we addthe three flavors together to obtain σ e + e − → hadrons σ el = 43 + 13 + 13 = 2 (3.7)assuming the energy is below the c quark threshold.Above this threshold, 2 → /
3, and above the b quarkthreshold 10 / → / w = ln x FIG. 9: Distribution of pions with momentum fraction x injets. RPF has apparently ignored the t -channel contribution to σ el here. Or course what we really see is not quarks in the finalstate, but rather jets (fig. 8), primarily K ’s and π ’s, withsmaller admixtures of nucleons and antinucleons. Onecan define probability distribution functions for hadronsin the jets in analogy to those of the partons, for example π ( x ), where now the momentum fraction is defined as x = energy of π energy of e + e − (3.8)Like for the quarks, these distributions go like 1 /x atsmall x . Denote the components of the π momentumparallel to and transverse to the average jet momentumas p (cid:107) and p ⊥ . Then dxx = dp (cid:107) E = dp (cid:107) (cid:113) m π + p ⊥ + p (cid:107) (3.9)This shows that at small x the distribution of particleswith momentum fraction x is flat as a function of w =ln x , fig. 9. Lorentz transforming to a frame where theaverage momentum of the two jets does not add to zerocauses the distribution to be translated to the right orleft in w . More generally, we can define fragmentationfunctions D qh ( x ) that denote the probability distributionfor producing a hadron of type h and momentum fraction x from a jet originating from a quark of flavor q . Thesecan be measured in deep inelastic scattering experiments.The formation of two jets from the breaking of a stringof strong interaction flux is in some ways analogousto a simpler problem, the spontaneous emission of anelectron-positron pair from a constant electric field. Solv-ing the Dirac equation in this background, I find that theprobability of pair production is exp( − e π m + (cid:126)p (cid:126)E ). Nowimagine that the electric field is created by two chargedplates that are moving together with velocity v . Thepairs should be produced with net total momentum. Butlocally they are created from a uniform field, so how dothey know they should have net momentum?A related problem concerns the wave function ofquarks in a stationary proton versus a moving proton.The wave function is not a relativistic invariant, nor evensomething that transforms nicely. Exercise.
From the Schr¨odinger equation, how does thesolution for the wave function ψ transform when the po-tential changes by the Galilean transformation V ( (cid:126)r ) → V ( (cid:126)r − (cid:126)vt )? (Answer: ψ → e im ( (cid:126)v · (cid:126)r − (cid:126)v t ) ψ ( (cid:126)r − (cid:126)vt, t ).)Similarly, the wave function for positronium, ψ ( (cid:126)x , (cid:126)x , t ), depending on the positions of its con-stituents, changes in a complicated way, as can beunderstood by Lorentz transforming and noting that inthe new frame, the events that were ( t, (cid:126)x ) and ( t, (cid:126)x )in the original frame are no longer simultaneous; see fig.10. We have to evolve one particle forward and the otherbackward in time to find the new wave function. Hence ψ in the new frame, call it ψ (cid:48) , is not just a function ofthe original ψ , but rather ψ (cid:48) = f ( ψ, H ), depending alsoon the Hamiltonian H of the system. t x ´xt FIG. 10: Relativity of simultaneity for positronium con-situents in the rest frame versus a boosted frame. qaq e + e − (E, −p)(E, p)E E FIG. 11: Hadronization after e + e − → q ¯ q production, showingthe breaking of the QCD string. We can say something more quantitative about the dis-tributions of transverse and longitudinal momenta how-ever. The center of mass energy of the e + e is 2 E ∼ = 2 | (cid:126)p | .The produced quarks carry less momentum because ofthe mass produced in the QCD string during hadroniza-tion, E = (cid:113) p (cid:107) + p ⊥ + m = xp (cid:112) p ⊥ + m ) / ( xp ) (recalling that p (cid:107) = xp ). Hence E − p (cid:107) ∼ xp . (3.10)A measure of momentum loss is given by the sum overthe different kinds of hadronic particles h produced, | ∆ (cid:126)p | = (cid:88) h (cid:90) ( E − p (cid:107) ) dp (cid:107) E C h d p ⊥ = (cid:88) h (cid:90) d p ⊥ C h ( p (cid:107) − E ) (cid:12)(cid:12)(cid:12) p = (cid:88) h (cid:90) d p ⊥ C h (cid:113) p ⊥ + m h (3.11)where C h depends on the particle. As shown, one can dothe integral over p (cid:107) exactly. This gives us a way of mea-suring the string tension T , since the loss of momentum This equation which holds at large xp does not seem to be neededfor what follows. E L a s y m p t o t e FIG. 12: Regge trajectory for the flexible bag model. is given by Newton’s law, (cid:88) L + R | ∆ (cid:126)p | = T × a (3.12)where the sum is over both left and right sides of thediagram (the two jets) and a is the time it takes to breakthe string, as shown in fig. 11.The theory of the string tension is not very quantita-tive. It comes from Regge trajectories, but these are notknown for highly stretched strings, and moreover in thereal situation there are quarks at the ends of the strings,that have not been taken into account. Exercise.
The “flexible bag” model takes the quarkmass to depend on the quark separation in a hadron,with Hamiltonian H = m ( r ) ddt (cid:126)r · ddt (cid:126)r + V ( r ), V ( r ) = kr , m ( r ) = µr . The quark mass represents the inertia of thegluon field (bag). Prove that E ∼ aL + b + c/L + . . . forlarge angular momentum L . Find E for L = 0 and forlarge L . [RPF shows graphically his result in fig. 12.] Problem.
What happens for the relativistic treatmentof the string? We must formulate a relativistic equation.String theory! Solution:
The proper tension T is the energy per unitlength. The radial variable goes from 0 to a , so the ve-locity varies as v ( r ) = r/a along the string, which rotatesat angular frequency ω = 1 /a . The differential force act-ing on an element of the string is dF = ωµvdr where µ = T / √ − v . Therefore the total energy and angularmomenta are E = 2 (cid:90) a T dr √ − v = 2 T a (cid:90) dv √ − v = πT aJ = 2 (cid:90) a µvr dr = 2 T a (cid:90) v dv √ − v = π T a and we understand the Regge trajectory behavior, E =2 πT J . Comparing to data, 2 πT = 1 .
05 GeV , giving T = 0 .
167 GeV . RPF in his private notes devotes four pages to working this out,first classically for circular orbits, then quantum mechanicallyusing exponential and Gaussian variational anz¨atze for the wavefunction. This was around the time of the first string revolution. I repro-duce the following answer from RPF’s private notes.
References . Quark model references:O. W. Greenberg, “Spin and Unitary Spin In-dependence in a Paraquark Model of Baryonsand Mesons,” Phys. Rev. Lett. , 598 (1964).doi:10.1103/PhysRevLett.13.598O. W. Greenberg and M. Resnikoff, “Symmetric QuarkModel of Baryon Resonances,” Phys. Rev. , 1844(1967). doi:10.1103/PhysRev.163.1844R. P. Feynman, M. Kislinger and F. Ravndal, “Currentmatrix elements from a relativistic quark model,” Phys.Rev. D , 2706 (1971). doi:10.1103/PhysRevD.3.2706Isgur MIT bag model references:A. Chodos, R. L. Jaffe, K. Johnson, C. B. Thornand V. F. Weisskopf, “A New Extended Modelof Hadrons,” Phys. Rev. D , 3471 (1974).doi:10.1103/PhysRevD.9.3471T. A. DeGrand, R. L. Jaffe, K. Johnson andJ. E. Kiskis, “Masses and Other Parameters of theLight Hadrons,” Phys. Rev. D , 2060 (1975).doi:10.1103/PhysRevD.12.2060
4. QUANTUM CHROMODYNAMICS (10-27-87)
We will denote color indices by a, b, · · · = r, b, g andflavor by f so that the noninteracting part of the quarkLagrangian is (cid:88) f ¯ ψ f ¯ a ( i /∂ − µ f ) ψ fa (4.1)while the interaction term (for a single flavor) is¯ ψ ¯ a A µ ¯ ab γ µ ψ b (4.2)and the gluon kinetic term is tr( ∂ µ A b ¯ aν − ∂ ν A b ¯ aµ ) ≡ tr( E b ¯ aµν ) (4.3)(Here only the noninteracting part of the field strengthis used.) The gauge transformations are given by ψ (cid:48) = Λ ψA (cid:48) µ = Λ † A µ Λ + Λ † i∂ µ Λ E µν = ∂ µ A ν − ∂ ν A µ − [ A µ , A ν ] → Λ † E µν Λ = E (cid:48) µν (4.4) These were originally given at the end of lecture 5, but logicallythey belong here since they pertain to quark models. No specific references are given, but probably RPF had in mindIsgur’s papers from 1978-1979 on the quark model. RPF omits coupling constants and numerical factors in eqs. (4.2-4.4), but restores them in (4.5). RPF uses E µν and F µν interchangeably for the field strength. The fully gauge invariant QCD Lagrangian is L QCD = (cid:88) f ¯ ψ f [ i /∂ − /A − µ f ] ψ f + 12 g tr E µν E µν (4.5)The quark-gluon interaction can also be written as L q − int = tr A µ J µ (4.6)with the current J µ = (cid:88) f J µ,f ; J ¯ baµ,f = ¯ ψ ¯ bf γ µ ψ af (4.7)Notice that there are 6 × × × Exercises.
1. If B ( x ) is a 3 × B (cid:48) = Λ † B Λ, show that ∂ µ B does not transformhomogeneously in this way, but ∂ µ B − i [ A µ , B ] does. I.e., ∂ µ − i [ A µ , ] ≡ D µ is the covariant derivative forfields transforming in the octet representation.2. Prove that [ D µ , D ν ] B = i [ E µν , B ].3. If A µ → A µ + δA µ then δE µν = D µ δA ν − D ν δA µ .4. By varying A µ in L , show that D µ E µν = g J ν ,where J is the quark current.5. Derive the equation of motion of the quarks,( i /∂ − µ f − /A ) ψ = 0 .
6. From this, show that D µ J µ,f = 0; also ∂ µ tr J µ,f = 0.7. Show that D µ J µ = (cid:80) f D µ J µ,f =. Hint: E µν )that D µ E νσ + D ν E σµ + D σ E µν = 0. Note thatif ˜ E µν ≡ (cid:15) µνστ E στ then this can be written as D µ ˜ E µν = 0.9. Define color electric and magnetic fields E x = E xt etc. and B z = E xy etc. (check the signs). Then rewritethe field equations in terms of these quantities. Let E =( E x , E y , E z ) etc . Show that D · B = 0 D × E + D t B = 0 D · E = g ρ where ρ = J t D × B − D t E = g J (4.8)where D = (cid:154) − i [ A , ] and D t = ∂ t − i [ A t , ].0Define the matrices λ = λ = − i i λ = − λ = λ = − i i λ = λ = − i − i λ = (cid:113) − λ = (cid:113) (4.9)Our convention is that tr( λ i λ j ) = 2 δ ij . We define A µ = A aµ λ a . There is no A µ term because this would corre-spond to an extra U(1) force. Let (cid:126)A µ = ( A µ , . . . , A µ ). Exercises, continued.
10. Show that Λ † ∂ µ Λ is trace-less, i.e., it has no λ component.The structure constants f ijk are defined through (cid:2) ( λ i ) , ( λ j ) (cid:3) = if ijk ( λ k ) (4.10)One can show that f ijk is totally antisymmetric. We alsodefine the anticommutator { λ i , λ j } = 2 d ijk λ k (4.11)Unlike f ijk , we can find the extra generator k = 0amongst those on the right-hand side.
11. Show that ∂ µ A iν − ∂ ν A iµ − f ijk A jµ A kν = E iµν . Wecan rewrite this as ∂ µ (cid:126)A ν − ∂ ν (cid:126)A µ − (cid:126)A µ × (cid:126)A ν = (cid:126)E µν bydefining a cross product in color space as( (cid:126)C × (cid:126)D ) i = f ijk C j D k (4.12)Similarly define the dot product (cid:126)C • (cid:126)D = (cid:88) i C i D i (4.13)12. Prove that (cid:126)C • ( (cid:126)C × (cid:126)D ) = 0and (cid:126)A × ( (cid:126)B × (cid:126)C ) + (cid:126)B × ( (cid:126)C × (cid:126)A ) + (cid:126)C × ( (cid:126)A × (cid:126)B ) = 0However (you don’t need to prove this), the familiaridentity (cid:126)A × ( (cid:126)B × (cid:126)C ) = (cid:126)B ( (cid:126)A • (cid:126)C ) − (cid:126)C ( (cid:126)A • (cid:126)B ) is only truefor SU(2) and not for general SU( N ).13. Show that C = D µ B = ⇒ (cid:126)C = ∂ µ (cid:126)B − (cid:126)A µ • (cid:126)B (4.14)14. Rewrite L using component notation. This seems to be a notational innovation of RPF.
Consider successive transformations ψ (cid:48) = ΛΨ, ψ (cid:48)(cid:48) = M ψ (cid:48) . Then ψ (cid:48)(cid:48) = N ψ ≡ M Λ ψ obviously. This is anexample of the group multiplication law for the color ro-tations. For many purposes we may be interested in in-finitesimal rotations, Λ = 1 + i a . Under this, the gaugefield transforms as A (cid:48) µ = (1 − i a ) A µ (1 + i a ) + (1 − i a ) ∂ µ (1 + i a )= A µ − i [ a , A µ ] + i∂ µ a = A µ + iD µ a (4.15)So it is always possible to impose temporal gauge, A =0, since this only requires solving a first order differen-tial equation. In the following however we will discuss adifficulty that arises when charges are present.What happens to a quark’s color as it is transportedthrough a gluon field? The transformation between twosets of color axes separated by a distance ∆ x µ can bewritten as 1 + iA µ ∆ x µ . Now suppose that every set ofaxes is changed locally by a rotation Λ( x ). Then the newtransformation relating the two sets of axes isΛ † ( x + ∆ x )(1 + iA µ ∆ x µ )Λ( x ) ≡ iA (cid:48) µ ∆ x µ (4.16)Therefore A (cid:48) µ = Λ † A µ Λ + Λ † i∂ µ Λ (4.17)which is the finite version of (4.15).
It would be very satisfying if we could justify some ofthe phenomenological approaches I considered earlier, us-ing QCD as a starting point. Heavy quarkonium systems,being approximately nonrelativistic, are the simplest sys-tems to consider, and can be described by a potential ofthe form V ∼ αr + br + . . . (4.18)where I have omitted the spin-spin and spin-orbit in-teractions. (In the complete Hamiltonian there is alsoan annihilation term H A that can cause transitions like u ¯ u ↔ s ¯ s , that give rise to η - η (cid:48) mixing.) The terms writ-ten describe the linearly confining potential representingthe mass of the string connecting the quark to the an-tiquark, and the Coulomb-like interaction, which mightrather be something like e − µr /r .One can also make predictions for relativistic systemslike the vector mesons; see S. Godfrey, N. Isgur, Phys.Rev. D32, 189 (1985). Then it is advantageous to useharmonic oscillator wave functions as a basis for com-puting matrix elements of the Hamiltonian to get a goodapproximate solution and compare to the data. Not only1can one compute the mass spectrum, but also strong in-teraction decay amplitudes, such as for φ → K ¯ K . Butall of this still relies on making a reasonable guess forthe form of the potential, and it would be preferable toderive these interactions directly from QCD.Let us recall how the analogous calculation works inQED, for the potential between a proton and an electron.We start with the fundamental interactions, L = F µν F µν + ¯ ψ e /Dψ e + ¯ ψ p /Dψ p (4.19)and from this we would like to derive the nonrelativisticeffective Hamiltonian H NR = p µ + V ( r ) (4.20)The potential can be computed perturbatively, byFourier transforming the amplitude, V ( r ) ∼ ++ ++ . . . . . . (4.21)But for QCD we know that the linear term is a nonper-turbative effect, so a different approach is needed.A better way might be to solve the classical equationsof motion for the gauge field, in the presence of a sourceterm, where the Lagrangian is L = F µν F µν + A µ J µ = ( ∂ µ A ν ∂ µ A ν − ∂ µ A ν ∂ ν A µ ) + A µ J µ → − ( A ν (cid:3) A ν + A ν ∂ µ ∂ ν A µ ) + A µ J µ (4.22)This gives the equation of motion − (cid:3) A µ + ∂ µ ∂ ν A ν + J µ = 0 (4.23)which by taking the divergence implies the current is con-served, ∂ µ J µ = 0. Since ∂ · F = ∂ µ F µν = ∂ µ ( ∂ µ A ν − ∂ ν A µ ), eq. (4.23) can be written in the gauge-invariantform ∂ ν F µν = J µ (4.24)Now we must solve (4.24) when the source is J = e ( δ ( (cid:126)r ) − δ ( (cid:126)r − (cid:126)a )), supposing that the two charges arelocated at the origin and at (cid:126)r = (cid:126)a respectively. In elec-trodynamics this is easy, thanks to the linearity of thetheory. We just superpose the solutions from the twosources, call them (cid:126)E = q r ˆ r, (cid:126)E = q r (cid:48) ˆ r (cid:48) (4.25)where (cid:126)r (cid:48) = (cid:126)r − (cid:126)a . Then we can compute the interactionenergy by integrating the energy density in the fields, E ∼ | (cid:126)E + (cid:126)E | : V ( a ) = (cid:90) d x E int = (cid:90) d x (cid:126)E · (cid:126)E (4.26)This shows how one might be able to compute the quark-antiquark potential without relying on perturbation the-ory; it would require knowing the classical solution forthe gluons fields in the presence of a static source. Therefore we would like to solve for the chromoelectricfield in the presence of a source. However it is no longerpossible write this only in terms of the field strength, aswe could for QED. In the temporal gauge A = 0, theGauss’s law equation is D · E = (cid:154) · E − i ( A · E − E · A ) = g ρ (4.27)where ρ = J = (cid:88) f ¯ ψ bf γ ψ af (4.28)which is a matrix in color space. Before fixing the gauge, D · E = D · ( (cid:154) A − ∂ A − [ A , A ]) → − D · ∂ A (4.29)so another way of writing (4.27) in A = 0 gauge is − D · ∂ A = g ρ (4.30)However we should first verify that there is no obsta-cle to transforming to the A = 0 gauge when externalcharges are present. An issue, as I will show, is whetherone can consider the source to be static. Starting fromsome configuration with A (cid:54) = 0, we would like to con-struct the gauge transformation that makes A (cid:48) = 0, bysolving eq. (4.17) with µ = 0. One can guess that it is atime-ordered exponential,Λ = P exp (cid:18) i (cid:90) tt dt (cid:48) A ( t (cid:48) ) (cid:19) (4.31)and verify that this is a solution, since ∂ Λ = ∂ (cid:18) i (cid:90) tt dt (cid:48) A ( t (cid:48) ) − (cid:90) tt dt A ( t ) (cid:90) t t dt A ( t ) + . . . (cid:19) = A ( t ) − A ( t ) (cid:90) tt dt (cid:48) A ( t (cid:48) ) + . . . (4.32)So there is no difficulty in transforming to the temporalgauge. But we must also consider how the source (4.23)transforms: ρ → Λ † ρ Λ (4.33)Recall that the quark changes its color when it emits agluon; that’s why the charge is a matrix. Does the gaugedependence mean that it makes no sense to ask whatis the potential between two spatially separated chargematrices?Before getting too ambitious and trying to solve withthe source having both a quark and an antiquark, let’sfirst imagine the seemingly easier case of a single quark,even though the solution is not expected to fall off at2large distances. Suppose that a quark starts out beingred. At a later time, after emitting or absorbing a gluon,it is some linear combination of red, green, blue: q ( t ) = r ( t ) b ( t ) g ( t ) (4.34)where | r | + | b | + | g | = 1, say. It gives a color chargematrix of the form ρ ¯ ab = ¯ q ¯ a γ q b . Clearly a nontrivialsolution will have time dependence, associated with thefact that the source is not a color singlet. To avoid this,we would have to include the antiquark contribution, soas to form a gauge-invariant source, ρ = ¯ q a ( x ) (cid:104) P e i (cid:82) x + rx A µ dx µ (cid:105) ab q b ( x + r ) (4.35)which is no longer a matrix, since we have traced overthe color indices. But this trace is not actually presentin the equation of motion (4.27), which has the explicitform D · E = (cid:154) · E − i [ A · , E ]= (cid:154) · ˙ A − i [ A · , ˙ A ] = g ρ (4.36)in A = 0 gauge. In this form it is clear that tr D · E = 0since every term is proportional to an SU(3) generator.Therefore (as we already knew) only the traceless part of ρ can act as a source for the gluons.Let’s rewrite (4.27) in the SU(3) vector notation I in-troduced previously, ∇ i ˙ (cid:126)A i − iA ai ˙ A bi [ T a , T b ] = g (cid:126)ρ . (4.37)Using [ T a , T b ] = if abc T c and rescaling A → g A , thisbecomes ∇ i ˙ (cid:126)A i + g (cid:126)A × i ˙ (cid:126)A i = (cid:126)ρ . (4.38)Now let (cid:126)A i = t (cid:126)E i ; then since (cid:126)E × i (cid:126)E i = 0, we get Gauss’slaw ∇ i (cid:126)E i = (cid:126)ρ . So it looks like we have succeeded in find-ing a class of solutions, that looks like just eight copiesof the Abelian problem. Not so fast! In electrodynamics,there is no difficulty in setting the magnetic field to zerofor a static charge configuration. But in QCD the colorfield sources itself, and now it is no longer obvious thatwe can set B = 0. Since B contains the term [ A, A ],it would vanish for special charge distributions where ρ and ρ (whose generators are diagonal) are the onlynonzero components. But such solutions are not helpfulfor understanding the distinctive properties of QCD, inparticular the confining potential.More generally there could be an integration constant, (cid:126)A i = (cid:126)a i ( (cid:126)x ) + t (cid:126)E i , giving the extra term ∇ i (cid:126)E i + g (cid:126)a × i (cid:126)E i = (cid:126)ρ (4.39)in Gauss’s law. What is the physical significance of (cid:126)a i ?Recall that B z = E xy = ∂ x A y − ∂ y A x − [ A x , A y ] (4.40) so that (cid:126)a i ( x ) gives a time-independent contribution tothe chromomagnetic field, B z = ∂ x a y − ∂ y a x + t [ ∂ x E y − ∂ y E x ] (4.41) − [ a x + tE x , a y + tE y ]= ∂ x a y − ∂ y a x − [ a x , a y ]+ t ( ∂ x E y − ∂ y E x − [ E x , a y ] − [ a x , E y ]) − t [ E x , E y ]The time-dependent terms still vanish for the specialcharge distributions ρ , ρ , while the time-independentone vanishes if in addition a i and a i are curl-free.In electrodynamics, a static electric and magnetic fieldin temporal gauge are described by A i = a i + tE i with ∇ × E = 0 and ∂ t a i = ∂ t E i = 0. We can’t seem to dothat here: E i = − ∂ i A + ∂ A i + [ A i , A ]= ∂ A i in temporal gauge; B x = ∂ y A z − ∂ z A y − [ A y , A z ] (4.42)because the commutator in B generically gives rise totime dependence. This seems to imply that we cannotimpose A = 0 gauge when charges are present. In elec-trodynamics it is more common to express a static solu-tion as (cid:126)E = (cid:126) ∇ A in Coulomb gauge where; (cid:126) ∇ · (cid:126)A = 0.Then (cid:126) ∇ · (cid:126)E ∼ (cid:126) ∇ · ˆ rr = ∂ i x i r = 3 r − (cid:18) (cid:19) x i x i r = 0 (4.43)But we previously showed that it is always possible to goto temporal gauge; why should it matter what gauge wechoose?One reason it could matter is the gauge-covariance ofthe source. Suppose that ρ was initially static in a gaugewhere A (cid:54) = 0. When we transform to temporal gauge, itis no longer static! InsteadΛ † ρ Λ = (cid:16)
P e i (cid:82) t dt (cid:48) A ( t (cid:48) ) (cid:17) † ρ (cid:16) P e i (cid:82) t dt (cid:48) A ( t (cid:48) ) (cid:17) (4.44)which is t -dependent, unlike in the Abelian case. Onemight try to fix the problem by rewriting Gauss’s law interms of gauge invariant quantities on the left-hand sideof the equation. Using D · E = (cid:154) · E − i [ A , E ], it wouldread (cid:154) · E = g ρ + i [ A , E ] (4.45)But that doesn’t work, since E itself is not gauge invari-ant!Eq. (4.35) suggests that it might be possible to finda solution where the charge remains static if we workinstead in an axial gauge with n · A = 0 for some spacelikevector n , for example A z = 0, in the case where thequark and antiquark are separated along the z direction.Then the gauge transformation needed to transform from A z (cid:54) = 0 to A (cid:48) z = 0 isΛ = P e i (cid:82) z dz (cid:48) A z ( z (cid:48) ) (4.46)3If the initial gauge field was static, ˙ A µ = 0, then Λ † ρ Λremains static. In fact, the same argument would haveworked in temporal gauge since then
P e i (cid:82) t dt (cid:48) A ( t (cid:48) ) = P e itA = e itA , which would be consistent, but then (cid:126)E =0. In summary, it seems to be difficult to find the classicalgauge configurations that would explain the origin of thequark-antiquark potential.
5. QCD CONVENTIONS (10-29-87)
In the previous lectures we may have been a bit care-less with numerical factors and signs. Let’s now try toget all of these right and establish a consistent set of con-ventions. First, we can verify that the quark Lagrangianshould read ¯ ψ ( i /∂ − /A ) ψ to be invariant under the gaugetransformations ψ → Λ ψ, ¯ ψ → ¯ ψ Λ † ,A µ → Λ A µ Λ † + i ( ∂ µ Λ) Λ † (5.1)Second, we carry out the gauge transformations ∂ µ A ν − ∂ ν A µ → Λ( ∂ µ A ν − ∂ ν A µ )Λ † (5.2)+ ( ∂ µ Λ) A ν Λ † − ( ∂ ν Λ) A µ Λ † + Λ A ν ∂ µ Λ † − Λ A µ ∂ ν Λ † + i∂ µ (( ∂ ν Λ)Λ † ) − i∂ ν (( ∂ µ Λ)Λ † )and[ A µ , A ν ] → [Λ A µ Λ † + i ( ∂ µ Λ)Λ † , Λ A ν Λ † + i ( ∂ ν Λ)Λ † ]= Λ[ A µ , A ν ]Λ † + i [Λ A µ Λ † , ( ∂ ν Λ)Λ † ] (5.3)+ i [( ∂ µ Λ)Λ † , Λ A ν Λ † ] − [( ∂ µ Λ)Λ † , ( ∂ ν Λ)Λ † ]Using ( ∂ ν Λ)Λ † = − Λ ∂ ν Λ † , we can show that the termsinvolving derivatives of Λ cancel in the linear combination F µν = ∂ µ A ν − ∂ ν A µ + i [ A µ , A ν ] (5.4)which is therefore the covariant field strength. The chro-moelectric and magnetic fields are E i = F i = ∂ A i − ∂ i A + i [ A , A i ] B x = F yz = ∂ y A z − ∂ z A y + i [ A y , A z ] (5.5)The full Lagrangian is L = 12 g tr F µν F µν + ¯ ψ ( i /∂ − /A − m ) ψ (5.6) This statement was not in my notes; it conveys my impressionthat RPF was explaining from memory the sequence of difficul-ties he encountered when looking for classical solutions, sometime prior to the course. There is no record of these attempts inhis personal notes.
Let’s vary it with respect to A to find the equation ofmotion. The variation of the first term is δ L = 1 g tr F µν δF µν + . . . (5.7)= 1 g tr (cid:16) δA µ ∂ ν F µν + i ([ δA µ , A ν ] + [ A µ , δA ν ]) F µν (cid:17) = 2 g tr (cid:16) δA µ ∂ ν F µν + i [ δA µ , A ν ] F µν (cid:17) (5.8)Then writing A µ = A aµ T a = A aµ ( λ a ), g δ L δA aµ = 2tr (cid:16) T a ∂ ν F µν + i [ T a , A ν ] F µν (cid:17) = ∂ ν F aµν + 2 i A bν F cµν tr (cid:16) [ T a , T b ] T c (cid:17) = ∂ ν F aµν + i [ A ν , F µν ] a (5.9)where we usedtr([ T a , T b ] T c ) = i f abc = tr([ T b , T c ] T a ) (5.10)Therefore ∂ ν F aµν + i [ A ν , F µν ] a = g ¯ ψT a γ µ ψ (5.11) Exercise.
Show that in Coulomb gauge ∂ i A i = 0, theGauss’s law constraint becomes ∇ A + 2 ig [ A i , ∂ i A ] − g [ A i , [ A i , A ]] = gρ (5.12)after rescaling A µ → gA µ . ρ is the matrix charge definedin eq. (4.28).
6. GEOMETRY OF COLOR SPACE ∗ (11-3,5-87) Our discussion of the QCD Lagrangian has been ofa largely algebraic nature to this point, but much intu-ition can be gained by considering the local color sym-metry in geometric terms. At each point in spacetimewe imagine there exists a set of axes in the color space, At this point RPF writes δ L /δA aµ on the left side, but on theright side gives the variation of L with respect to A a ¯ bµ labeled bythe (3 , ¯3) indices, rather than varying with respect to A aµ labeledby the adjoint index a . This gives a result twice as large as itshould be (due to the normalization of the generators), whichRPF recognizes as being wrong and therefore concludes that thegluon kinetic term should really be normalized as (1 / g )tr F .This may be another case of him extemporizing. I have correctedthe derivation here. In the lectures, RPF derives this but I leave it as an exercise.Part of the derivation involves assuming that ∂ A i = 0 in thecommutator [ A i , ∂ A i ], which seems not generally true. This section, which was revised by RPF, combines lectures 6 and7, given on Nov. 3 and 5, 1987. It repeats some material thatwas presented earlier. I retained the redundancies in the interestof historical accuracy. x ), under which a quark transforms as ψ (cid:48) ( x ) = Λ( x ) ψ ( x ) . (6.1)Since one rotation may be followed by another, ψ (cid:48)(cid:48) ( x ) = Λ (cid:48) ( x ) ψ (cid:48) ( x ) = Λ (cid:48) ( x )Λ( x ) ψ ( x ) ≡ Λ (cid:48)(cid:48) ( x ) ψ ( x ) ;they form a group, with Λ (cid:48)(cid:48) = Λ (cid:48) Λ being the group multi-plication law. Requiring that Λ not change the length ofa color vector is equivalent to demanding that Λ † Λ = 1.Thus the Λ’s would represent the group U(3) of unitary3 × e iθ , which would give rise to anadditional long-range interaction like the electromagneticforce. To eliminate this we note thatdet Λ (cid:48)(cid:48) = det Λ (cid:48) det Λrepresents the U(1) transformations (it is Abelian), so weshould make the restrictiondet Λ = 1 , i.e., Λ is a special unitary matrix, hence the group isSU(3).The transformation law for the gluon field has a lessimmediately obvious interpretation than that for thequarks, eq. (6.1). For infinitesimal rotations Λ = 1+ i a , A (cid:48) µ = (1 − i a ) A µ (1 + i a ) + (1 − i a ) ∂ µ (1 + i a )= A µ − i [ a , A µ ] + i∂ µ a = A µ + iD µ a (6.2)where D µ is the covariant derivative. How can this beunderstood geometrically?To answer this, it must first be realized that there is, a priori , no way of telling whether a color frame at point x is parallel to one at x + ∆ x , because the color space iscompletely unrelated to spacetime. An analogy is tryingto choose local tangent frames on a curved space, suchas the surface of a two-sphere, that are “parallel” to eachother. It is not possible to do without defining a law ofparallel transport for vectors, so that we know what itmeans for two vectors at different locations to be parallel.Similarly in QCD one needs a rule for comparing orienta-tions of nearby color frames. This is the function of thegauge field A µ ( x ), in much the same way as the metrictensor (to be more precise, the Christoffel symbol) de-fines parallel transport in the geometry of curved space. Neither RPF nor I noticed the inconsistency with eq. (5.1), whichis the correct version having i∂ µ instead of ∂ µ in the first line. x µ x µ ∆ U U U U x µ −∆ x ν x ν −δ δ FIG. 13: Parallel transport of a quark around a closed loop.
Define the relative orientation between two nearby colorframes, at x and x + ∆ x , to be given by the rotationmatrix U ( x, ∆ x ) = 1 + iA µ ( x )∆ x µ . (6.3)Now suppose that every set of axes is rotated by Λ( x ),depending on the position x . Then the new transforma-tion relating the frames at x and x + ∆ x is U (cid:48) ( x, ∆ x ) = Λ † ( x + ∆ x ) U ( x, ∆ x )Λ( x )= 1 + iA (cid:48) µ ( x )∆ x µ (6.4)It follows that A (cid:48) µ ( x ) = Λ † ( x ) A µ ( x )Λ( x ) + i Λ † ( x ) ∂ µ Λ( x ) (6.5)which shows that our geometric interpretation of A µ ( x )agrees with its previously determined transformation law.If one was to take a quark at x with color vector (cid:126)q andparallel-transport it to x + ∆ x , its color would change to U ( x, ∆ x ) (cid:126)q . Of particular interest is the change in (cid:126)q whentransported around a closed loop, such as the one shownin fig. 13. Let U = U ( x, ∆ x ), U = U ( x + ∆ x, δx ), U = U ( x + ∆ x + δx, − ∆ x ), U = U ( x + δx, − δx ). Thetransformation of (cid:126)q in going around the loop is (cid:126)q (cid:48) = U U U U (cid:126)q = U tot (cid:126)q (6.6)One notices that U tot ( x ) has a simple transformation un-der local SU(3) rotations, U tot ( x ) → Λ † ( x ) U tot ( x )Λ( x ) , which is just how the field strength F µν ( x ) transforms.This is not an accident: if you expand U tot in terms ofthe gauge field as in (6.3), you will find that U tot = 1 + iF µν ( x )∆ x µ δx ν (6.7)plus terms of order (∆ x ) and ( δx ) . Notice that ∆ x µ δx ν is the area of the loop. So F µν ( x ) tells us how much colorrotation a quark suffers under transformations around in-finitesimal loops. It is analogous to the Riemann tensor,which does the same thing for vectors in curved space.With the concept of parallel transport in hand, covari-ant differentiation becomes quite transparent. If ψ ( x )is a quark field, it is not ψ ( x + ∆ x ) − ψ ( x ) that is of5physical interest, because this includes the difference dueto arbitrary orientations of the local color axes. Weshould rather compare ψ ( x + ∆ x ) with ψ ( x ) transportedto x + ∆ x . Therefore define the covariant derivative as∆ x µ D µ ψ ( x ) = ψ ( x + ∆ x ) − U ( x, ∆ x ) ψ ( x )or equivalently D µ ψ ( x ) = ( ∂ µ − iA µ ) ψ ( x ) . (6.8)Similarly for the field strength,∆ x α D α F µν ( x ) = F µν ( x +∆ x ) − U † ( x, ∆ x ) F µν ( x ) U ( x, ∆ x )which implies D α F µν ( x ) = ∂ α F µν ( x ) − i [ A α , F µν ] (6.9)Even as seemingly abstract an equation as the Bianchiidentity can be understood geometrically. This is one ofthe statements you were asked to prove previously, D α F βγ + D γ F αβ + D β F γα = 0 . (6.10)For concreteness, let ( α, β, γ ) = ( x, y, z ). Then the firstterm is D x F yz . In terms of fig. 14, this is the change incolor axis orientation around the top loop minus that ofthe bottom loop. The use of D z rather than ∂ z meansthat the bottom loop was parallel transported to the po-sition of the top loop before making the subtraction. Thecontribution to D z F xy from each link of the cube is de-noted by a line with an arrow that shows the relativesign of the contribution. From the figure, it is easy to seethat when the remaining terms in (6.10) are included,each link will contribute twice, once in each direction.Therefore the sum is zero.The matrix U ( x, ∆ x ) that connects nearby color axescan be used to construct the rotation connecting colorframes that are separated by a finite distance. Choose apath connecting the two points and divide it into smallincrements, labeled by x µi , such that ∆ x µi = x µi +1 − x µi asshown in fig. 15. Then the rotation matrix between x µ and x µf is S ( x , x f ) = (cid:89) i (1 + iA µ ( x i )∆ x µi ) . (6.11)As ∆ x µi →
0, this becomes equivalent to (cid:81) i e iA µ ( x i )∆ x µi .We would like to write it as exp( i (cid:82) A µ dx µ ), which would xyz FIG. 14: Geometric interpretation of the Bianchi identity. x x f FIG. 15: Path connecting two points in spacetime. be true if the A µ were numbers, but since the A µ don’tcommute at different positions, we cannot add the expo-nents. Instead one defines the path ordering operator P e b + b = e b e b , where it is understood that b is farther along the paththan b . Therefore S ( x , x f ) = P exp (cid:18) i (cid:90) x f x dx µ A µ (cid:19) = 1 + i (cid:90) x f x dx µ A µ (6.12) − (cid:90) x f x dx µ (cid:90) x f x dx (cid:48) ν A ν ( x (cid:48) ) A µ ( x ) + · · · Notice that S is by no means unique; it depends uponthe path chosen. Under a gauge transformation however, S ( x , x f ) = Λ † ( x f ) S Λ( x ) (6.13)regardless of the path. Exercise.
Using the definition (6.12), show that P exp (cid:18) − (cid:90) x f x dx µ Λ † ( x ) ∂ µ Λ( x ) (cid:19) = Λ † ( x f )Λ( x )Hint: consider the differential equation satisfied by (6.12)for ∂S/∂x µ .The connection S is useful for making bilocal operatorsgauge invariant. For example, in the full interacting the-ory, the two-point function for the field strength vanishesbecause of gauge invariance, since (cid:104) | F αβ ( x ) F µν ( x f ) | (cid:105) = (cid:104) | Λ † ( x ) F αβ ( x )Λ( x )Λ † ( x f ) F µν ( x f )Λ( x f ) | (cid:105) for an arbitrary Λ( x ). This can only be satisfied if (cid:104) F F (cid:105) = 0. However the functiontr (cid:0) S † ( x , x f ) F αβ ( x ) S ( x , x f ) F µν ( x f ) (cid:1) is gauge invariant, and has a meaningful nonvanishingexpectation value.The statement that (cid:104) F F (cid:105) = 0 for F ’s at distinct pointsimplies some way of defining expectation values withouttampering with the gauge symmetry of the path inte-gral over A µ , to be discussed later on. In practice, it isnecessary to choose a condition that fixes the gauge, byassociating F µν ( x ) with a unique vector potential A µ ( x ).For example, it is always possible to demand that A = 0by transforming A µ → Λ † A µ Λ + i Λ † ∂ µ Λ, whereΛ † A Λ + i Λ † ∂ µ Λ = 06One can show that the solution to this equation isΛ = T exp (cid:18) i (cid:90) t A ( t (cid:48) ) (cid:19) dt (cid:48) where T is the same as P but for a purely timelike path.Another gauge condition that is conceptually useful isto minimize the quantity (cid:90) d x tr( A µ ( x )) (6.14)In this gauge, P exp( i (cid:82) dx A ) does not change much ifthe path is varied only slightly. Therefore it is possible todefine a global orientation for color axes on short enoughdistance scales: a quark that looks red at point x willstill look red after parallel transport if it is not carriedtoo far. Because of this it makes some sense to say thatquarks of the same color repel each other, whereas quarksthat are antisymmetric in their colors attract each other,as will be shown in the next lecture. In an arbitrarygauge it would not be meaningful to say that two bluequarks repel each other, unless they were at the sameposition, since what is blue at one place may not be blueat another.The above choice of gauge is closely related to anothermore familiar one. Under an infinitesimal gauge trans-formation A µ → A µ + D µ α , the change in (6.14) is2 (cid:90) d x tr A µ D µ α = − (cid:90) d x tr( α D µ A µ ) (6.15)If (cid:82) tr( A µ ) is at a minimum, then (6.15) must vanish forall α ( x ). This implies D µ A µ = ∂ µ A µ = 0 (6.16)However the two gauges are not equivalent, because thefirst one asks for the absolute minimum of (cid:82) tr( A µ ),whereas ∂ µ A µ = 0 only requires that (cid:82) tr( A µ ) be ata local minimum. Therefore ∂ µ A µ = 0 may have manysolutions, and it does not uniquely fix the gauge. Thisproblem was first discussed by Gribov in the context ofthe path integral. A synopsis of the popular gauge choices is A t = 0 Weyl ∂ µ A µ = 0 Lorentz (cid:154) · A = 0 Coulomb (6.17) This material appears in my original notes but was omitted fromthe revised version above. ab dci i
RGRG RG
FIG. 16: (a) Left: quark-quark scattering by gluon exchange.(b) Right: same, with particular choice of colors.
In addition, there is an analog to (6.14) due to Mandula,which is to minimize (cid:82) d x A : A . Exercise.
Show that E µν • (cid:101) E µν ≡ (cid:15) µναβ E µν • E αβ couldbe added to the Lagrangian (usually written as the ac-tion θ (cid:82) d x E µν (cid:101) E µν ), but it makes no contribution tothe equations of motion: it is a total derivative.
7. SEMICLASSICAL QCD ∗ (11-10-87) I know you are eager to move on to the quantum theoryof chromodynamics, now that we have studied it at theclassical level, but there always has to be some professordeterring you by saying “before we do that, let’s look atsuch-and-such!” Accordingly, before we quantize QCD Iwant to discuss a somewhat tangential but very impor-tant issue: can we explain the properties of the hadrons,even qualitatively , with the theory of QCD? That is, wewould like to see that we are at least going in roughly theright direction before we invest all our effort in it. Forexample, it would be quite discouraging if at the lowestlevel of analysis QCD predicted that the three quarks ina baryon will want to fly apart.But we shall see that it does work, and we won’t evenhave to do that much work ourselves to see it, if we justremember a few things from quantum electro dynamics.This is because at lowest order in the coupling constant g ,the interaction between two quarks is given by essentiallythe same Feynman diagram as that for electron-electronscattering, fig. 16. The only difference is that in thecase of QCD, each vertex comes with a group theoryfactor λ iab and λ icd to account for the fact that the quarksare changing color when they exchange a gluon of color i . The sum over intermediate gluon colors then gives afactor (cid:126)λ ab • (cid:126)λ cd in the amplitude. Of course we know that fig. 16(a) is not a good approx-imation for the quark-quark scattering in a hadron, be-cause the coupling is large. But we just want to see thatit is going in the right direction , when we do make thisapproximation. Knowing that a single photon exchangegives rise to the Coulomb potential in electrodynamics,we can immediately write the quark-quark potential from These should be accompanied by extra factors of 1 / T a = λ a / V ( r ) = g r (cid:126)λ ab • (cid:126)λ cd (7.1)Now all that remains is to evaluate the (cid:126)λ • (cid:126)λ factor inthe color channels appropriate for baryons and mesons.This could be done by using fancy group theory tech-niques, but I find it useful to take a more simple-mindedapproach at first. It is not necessary to know grouptheory—all one needs to know is that there are threecolors!Let us suppose that quark 1 (on the left) starts out be-ing red, and converts to green by emitting a red-antigreengluon. In order to conserve color, quark 2 must havestarted out being green and turn to red when it absorbsthe gluon. Using the convention(R, G B) T = (red, green, blue) T (7.2)for the components of a color vector, there are two λ matrices contributing to the process shown in fig. 16(b),namely λ and λ , λ = , λ = − i i (7.3)due to the fact that a R ¯ G gluon corresponds to a partic-ular linear combination, λ + iλ . Therefore the contri-bution to (cid:126)λ ab • (cid:126)λ cd from fig. 16(b) is1 · i ( − i ) = 2 (7.4)Now suppose that the colors of the initial state quarkdid not change. This could happen if gluons correspond-ing to the color-diagonal generators λ and λ were ex-changed, λ = − , λ = (cid:113) − . (7.5)For example, the contribution to (cid:126)λ • (cid:126)λ from fig. 17 is1 · √ (cid:18) − √ (cid:19) = −
23 (7.6)
BBRR
FIG. 17: A color-conserving q - q scattering process. From (7.4) and (7.6) we could guess that the generalexpression for (cid:126)λ • (cid:126)λ is (cid:126)λ ab • (cid:126)λ cd = 2 P c.e.ab,cd − δ ab δ cd (7.7)where P c.e. is the color-exchange operator, P c.e.ab,cd = (cid:40) a = d and b = c B BB B : (cid:18) − √ (cid:19) (cid:18) − √ (cid:19) = 43 (7.9)which has only a λ -type gluon. This agrees with (7.7)since 2 P c.e.BB,BB − δ BB δ BB = 4 / −
1. (Of course,it would also be an eigenstate of the identity operator δ ab δ cd with eigenvalue +1.) On such a state, (cid:126)λ • (cid:126)λ | ψ (cid:105) = (cid:0) P c.e. − (cid:1) | ψ (cid:105) = − | ψ (cid:105) . (7.10)It means that in the antisymmetric channel, the quark-quark potential is V ( r ) = − g r . (7.11)The sign is important: since electron-electron scatteringis repulsive, the relative sign here tells us that quarksthat are antisymmetric in their colors attract . This isprecisely what we want: in a baryon the quarks are in acompletely antisymmetric state, which is the only way tomake a color-neutral object out of three color triplets. Sowe understand, in a rough way, why protons, neutrons,∆’s, etc. exist.The mesons can be understood similarly. In this casethe initial state is color symmetric, | ψ (cid:105) = 1 √ (cid:0) R ¯ R + B ¯ B + G ¯ G (cid:1) . (7.12)If we focus on the R ¯ R part, there are three graphs toconsider, fig. 18. They contribute − (cid:126)λ • (cid:126)λ factors (the ex-tra minus sign coming from the coupling of vectors to , B R R,R GG BR RR R R
FIG. 18: Scattering of q -¯ q within a meson. − (1 · i ( − i )) = − , − , − (cid:16) · √ · √ (cid:17) = − (7.13)respectively. Obviously the B ¯ B and G ¯ G parts of | ψ (cid:105) dothe analogous thing, so that the quark-antiquark poten-tial for color-symmetric states is V ( r ) = − g r . (7.14)Graphs like fig. 18 give an attractive potential for elec-trons and positrons in QED, and we see that quarks andantiquarks in a meson must also attract each other.One notices that the q ¯ q force in mesons is twice asstrong as the qq force in baryons. But it is interestingto note that the total force per quark is the same in eachsystem, since 12 quarks (cid:18) (cid:19) = 83 for mesons (7.15)and 13 quarks × (3 pairwise forces ) × (cid:18) (cid:19) = 83 for baryons(7.16) Problem.
Show that the general q ¯ q interaction dueto one-gluon exchange (summed over gluon colors) canbe written as − | s (cid:105)(cid:104) s | where is the identity operator in color space, and | s (cid:105)(cid:104) s | projects onto the color singlet state, | s (cid:105) = √ (cid:0) R ¯ R + B ¯ B + G ¯ G (cid:1) So far, so good: QCD explains why the hadrons exist,even at this crude level of approximation where g wastaken to be small. Now we would like to see if it explainssome more detailed observations, like the nondegeneracyof the ∆ and the neutron:∆ : 1232 MeV; | ψ (cid:105) = ( udd )( ↑↑↑ ) (7.17) N : 935 MeV; | ψ (cid:105) = ( udd ) √ (2 ↓↑↑ − ↑↓↑ − ↑↑↓ )If the masses were coming solely from the constituentmasses of the quarks, these states would be degeneratedue to their identical quark content. The only differ-ence between them seems to be their spin wave func-tions. Therefore their mass splitting must be due tospin-dependent forces. This comes as no surprise sincethe photon-exchange graph gives a spin-spin interactionas well as the Coulomb interaction in QED. Let us recallwhat the sign of the force is in electromagnetism. Specif-ically, if we could make an s -wave from two electrons,
12 21
FIG. 19: An electron spin (2) in the dipole field produced byanother electron (1). would their spins tend to align or anti-align? We drawthe second spin in the magnetic field of the first, withboth of them pointing up; see fig. 19. When the secondspin is beside the first, the interaction energy is positive,since spin 2 would prefer to flip so as to align with the (cid:126)B field. When spin 2 is above spin 1, the interaction energyis negative. To make an s -wave we must average over thepositions of spin 2 relative to spin 1 in a spherically sym-metric way. One can show that at any nonzero radius,the interaction energy is zero when this is done.However, zero is not the correct answer, as we alreadyknow. The problem is that the spins have been idealizedas pointlike objects. If they actually have a small butfinite spatial extent, the magnetic field of spin 1 will looklike fig. 20. Because of the interior region, the integrationover positions of spin 2 give a net positive interactionenergy, and two electrons in an s -wave tend to align. Thisis known as the Fermi interaction; it is the first term inthe expression V spin ( (cid:126)r ) ∝ − π (cid:126)σ · (cid:126)σ δ (3) ( (cid:126)r )+ 1 r (cid:0) (cid:126)σ · (cid:126)σ − (cid:126)σ · ˆ r )( (cid:126)σ · ˆ r ) (cid:1) . (7.18)The second term comes from the exterior region that wediscussed previously; its angular average is zero, as wenoted.From this we can deduce that the color magnetic mo-ment interaction energy for two quarks, in a relative s -state, is positive if the spins are aligned and the colorsare antisymmetric. This is because the sign of the spinforce relative to that of the Coulomb force is determinedby the Lorentz indices of the diagram in fig. 16, so thisrelative sign must be the same for QCD and QED. Hencethe ∆ is heavier than the N —it takes more energy to lineup all the spins.What about the Σ and the Λ? They have identicalquark content, and equal numbers of aligned spins, yetthe Σ is more massive. However, the spin wave functionsare not the same, | Σ (cid:105) = ( sud ) √ (2 ↓↑↑ − ↑↓↑ − ↑↑↓ ) | Λ (cid:105) = ( sud ) √ ( ↑↑↓ − ↑↓↑ ) (7.19)We must remember that the magnetic moment of a quarkis inversely proportional to its mass, so the s quark has9 FIG. 20: Effect of a nonpointlike spin 2 (red) overlapping thecentral region of the magnetic field produced by spin 1. smaller spin interactions. In the limit that its magneticmoment is neglected, the Σ is 4 / / a qq , a qs , a ss between two light quarks,one light and one strange, and two strange quarks, re-spectively. Then the spin-spin coupling for Σ and Λparticles isˆ O = a qq ( σ u · σ d ) + a qs ( σ s · σ u + σ s · σ d ) (7.20)where a qs < a qq , or equivalently, defining a spin-exchangeoperator P s.e.qq (cid:48) ,ˆ O = a qq (2 P s.e.ud −
1) + a qs (2 P s.e.su −
1) + a qs (2 P s.e.sd − , P s.eud (2 ↓↑↑ − ↑↓↑ − ↑↑↓ ) = 2 ↓↑↑ − ↑↑↓ − ↑↓↑ ,P s.esu (2 ↓↑↑ − ↑↓↑ − ↑↑↓ ) = 2 ↑↓↑ − ↓↑↑ − ↑↑↓ ,P s.esd (2 ↓↑↑ − ↑↓↑ − ↑↑↓ ) = 2 ↑↑↓ − ↑↓↑ − ↓↑↑ (7.22)Thereforeˆ O| Σ (cid:105) = [(2 − a qq + ( − − a qs ] | Σ (cid:105) = ( a qq − a sq ) | Σ (cid:105) (7.23)Similarly one finds thatˆ O| Λ (cid:105) = [( − − a qq + (2 − a qs ] | Λ (cid:105) = − | Λ (cid:105) (7.24)Therefore the mass difference is M Σ − M Λ = 4( a qq − a qs ),which is positive since a qq > a qs . Problem.
Find the mass splittings of the rest of thebaryon
12 + octet and
32 + decuplet states. Assuming that a qq − a qs = a qs − a ss prove the Gell-Mann–Okubo formula,2( M Ξ + M N ) = 3 M Λ + M Σ A similar analysis can be done for the mesons, and itis observed that the level splittings of heavy quarkoniumexcitations, such as the ψ and Υ systems, are similar tothose of positronium. However there is an interestingdistinction between the spin forces in quarkonium andthose of positronium. In the latter an extra contributionto the spin-spin interaction arises from the annihilationdiagram, fig. 21. However, at lowest order in g , no suchprocess can occur for quarkonium. This is because the q ¯ q pair in a meson forms a color singlet, which cannotannihilate into a colored object like a gluon. It cannoteven annihilate into a pair of gluons, for it is spin 1 − ,a state not available to two gluons. It requires at leastthree gluons, which means a high power of the couplingconstant, g , which is rather small at the scale of theseparation between a heavy q and ¯ q in the ψ or Υ. Alsothe numerical coefficient of the annihilation amplitude issmall, making the width for disintegration of ψ or Υ intohadrons quite narrow. Hence the OZI rule is understoodfor these particles, at least. References A. De Rujula, H. Georgi and S. L. Glashow, “HadronMasses in a Gauge Theory,” Phys. Rev. D , 147 (1975).doi:10.1103/PhysRevD.12.147T. Appelquist, R. M. Barnett and K. D. Lane, “Charmand Beyond,” Ann. Rev. Nucl. Part. Sci. , 387 (1978).doi:10.1146/annurev.ns.28.120178.002131N. Isgur, “Hadronic structure with QCD: From α to ω (via ψ and Υ),” AIP Conf. Proc. , 1 (1982).doi:10.1063/1.33447In the last reference, Isgur supposed that the potentialbetween quarks in a hadron is V ( r ij ) = kr ij + U ( r ij )where U is a perturbation. One needs to evaluate (cid:104) U (cid:105) and (cid:104) U | x i − x j | (cid:105) expectation values. This gives two pa-rameters to fit the spectrum. He discusses the fact thateven heavy quark systems have sufficiently large wave FIG. 21: Annihilation diagram for positronium, not presentfor quarkonium. The following references (here modernized to include DOIs) anddiscussion of Isgur’s work are present in my original notes, butsomehow got omitted from the revised versions submitted toRPF. V ( r ) = − α/r + br is important. But the whole potentialapproach is approximate for various reasons, includingrelativistic effects and annihilation/creation processes.
8. QUANTIZATION OF QCD ∗ (11-12-87) We now turn to the quantization of QCD. Recall thatthe action for the gluon fields, interacting with a non-dynamical source, is S [ A ] = (cid:90) (cid:18) g (cid:126)F µν • (cid:126)F µν + (cid:126)J µ • (cid:126)A µ (cid:19) d x (8.1)I will assume that you are familiar with the path integralformulation of quantum mechanical amplitudes. If A i and A f are the initial and final configurations of the gluonfield, the transition amplitude for going from A i to A f is K [ A f , A i ] ≡ (cid:90) fi D A e iS [ A ] (8.2)where the integral is supposed to be over all field con-figurations A aµ ( (cid:126)x, t ) such that A aµ ( (cid:126)x, t i ) = A ai,µ ( (cid:126)x ) and A aµ ( (cid:126)x, t f ) = A af,µ ( (cid:126)x ). Formally, the measure is defined asan infinite product over all points in spacetime between t = t i and t = t f , D A = (cid:89) (cid:126)x,t,µ d A µ ( (cid:126)x, t ) (8.3)The eight-dimensional measure d A µ simply means theproduct over the eight components of color, (cid:81) i dA iµ . The action (8.1) is invariant under the local color gaugetransformations of A µ , A (cid:48) µ = Λ † A µ Λ + i Λ † ∂ µ Λ (8.4)This implies that the measure D A is also gauge invariant,since it transforms as dA (cid:48) µ = Λ † dA µ Λ at each point, andthe Jacobian of this transformation is trivial when weconsider d A µ . Problem.
Prove this. Now as you know, this path integral is plagued withinfinities. One rather trivial kind is the infinite volumeof spacetime. Another sort, the ultraviolet divergences,comes from the uncountably infinite dimensional natureof the measure, an integral for each point of spacetime.This kind I want to ignore for the moment—it can be This paragraph, present in my notes, was omitted from the orig-inal revised version. This sentence may seem extraneous, but RPF was correcting mymisconception that the Haar measure for the group manifold wassomehow incorporated. The Jacobian matrix is tr( λ a Λ † λ b Λ). One can show its deter-minant is trivial by considering an infinitesimal transformationto leading order, and using det = exp tr ln. cured by approximating spacetime as a discrete lattice,in some gauge invariant way. This has been discussed byWilson, and we shall describe it later. It is the basis fora numerical method to evaluate the path integral.But in QCD we are still left with another in-finity, due to the gauge symmetry itself. If werepresent the space of functions A µ ( x ) in two di-mensions, we have trajectories of gauge fields thatare related to one another by local color rotations: space of A µ (x) This means that the expectation value of a physically rel-evant operator, that is gauge invariant, will diverge likethe volume of local gauge transformations in functionspace, (cid:90) D A e iS [ A ] ∼ (cid:90) D Λ( (cid:126)x, t ) = ∞ (8.5)Put another way, there are directions in which A canchange ( i.e., by gauge transformations) for which the in-tegrand is invariant and the region of integration is infi-nite.However we can get a finite and meaningful result bydefining a kind of expectation value of any gauge invari-ant functional F [ A ] of the field, and limiting ourselves tocomputing only such quantities, (cid:104) F (cid:105) = lim R→∞ (cid:82) D A F e iS [ A ] (cid:82) D A e iS [ A ] (8.6)where R is a relatively finite region in the space of gaugefields; for example we might limit the range of each A µ to be [ − M, + M ] for some large value of M , the same inthe numerator and denominator, and then take the limitas M → ∞ . (I say “relatively finite” because there arestill infinitely many integration variables, one for eachpoint in spacetime, unless we go to a lattice.) Gaugeinvariance is broken temporarily by this procedure be-cause | A (cid:48) µ | need not be less than M even if | A µ | is, butthis should be no problem in the limit R → ∞ .Having found a finite and gauge-invariant definition ofamplitudes, we are free to choose a gauge. A convenientchoice is A = 0. Recall that it is always possible toreach this gauge from a configuration where A old0 (cid:54) = 0,using the transformation matrixΛ( (cid:126)x, t ) = P exp (cid:18) i (cid:90) t A old0 ( (cid:126)x, t (cid:48) ) (cid:19) dt (cid:48) . (8.7)In general, the path integral measure D A old µ changesby a Jacobian factor when we transform the variables1 A old µ → A new µ , where A new0 = 0. How can this be? Themeasure was supposed to be gauge invariant! But this istrue only for gauge transformations that are independentof A µ . Nevertheless, we will show that for the specialcase of (8.7), the Jacobian is still almost trivial, eventhough Λ( (cid:126)x, t ) depends on A µ . This is because Λ de-pends only on A , and thus for the three space directionsat least, d A (cid:48) m = d A m (we use n, m, etc. for the spatialcomponents of ν, µ , while the time component is 0.) Theexpectation value of an operator can now be written as (cid:82) R old e iS [ A old ,A old0 ] F [ A old , A old0 ] (cid:81) n =1 D A old n D A old0 (cid:82) R old e iS [ A old ,A old0 ] (cid:81) n =1 D A old n D A old0 = (cid:82) R new e iS [ A new , F [ A new , (cid:81) n =1 D A new n D A old0 (cid:82) R new e iS [ A new , (cid:81) n =1 D A new n D A old0 (8.8)by making the gauge transformation (8.7). In the secondexpression, the integrands are independent of A , andthe factors of (cid:82) D A old0 cancel between numerator and de-nominator, using the definition (8.6).If F were not a gauge invariant operator, we couldalways replace it by its gauge-averaged expression,ˆ F [ A ] = (cid:82) d Λ F [ A (cid:48) ] (cid:82) d Λ (8.9)where A (cid:48) is as in (8.4), and d Λ is the invariant groupmeasure, satisfying (cid:90) d Λ f (Λ) = (cid:90) d Λ f (ΛΛ ) (8.10)for any SU(3) matrix Λ and any function f . Problem: If F is not gauge invariant, show that (cid:104) F (cid:105) as defined in (8.6) is the same as (cid:104) ˆ F (cid:105) .Now the path integral is reduced to the simpler expres-sion, Z = (cid:90) e iS [ A ,A ] D A (8.11)where S = 12 g (cid:90) d x (cid:16) (cid:126) E • · (cid:126) E − (cid:126) B • · (cid:126) B (cid:17) . (8.12)Since A = 0, (cid:126) E = − ∂ (cid:126) A = − ˙ (cid:126) A (8.13)and B is just as it was before, (cid:126) B = (cid:154) × (cid:126) A − (cid:126) A ×× (cid:126) A (8.14) RPF had written (cid:126) B = (cid:154) × (cid:126) A − (cid:126) A × (cid:126) A . I have corrected it here andin some subsequent equations to indicate the operation neededfor the spatial indices of the interaction term. The spatial crossproduct introduces a factor of 2 that must be compensated. (for example, B z = ∂ x A y − ∂ y A x − [ A x , A y ]). In termsof the gauge field, the Lagrangian is L = 12 g (cid:90) d x (cid:16) ˙ (cid:126) A • · ˙ (cid:126) A − ( (cid:154) × (cid:126) A − (cid:126) A ×× (cid:126) A ) (cid:17) . (8.15)This is analogous to the Lagrangian for a particle movingin a potential L = m ˙ x ( t ) − V ( x ( t )) (8.16)where (cid:126) A is like the position of the particle, ˙ (cid:126) A : (cid:126) A playsthe role of the kinetic term, and (cid:126) B : (cid:126) B is the potential.This situation is unique to the A = 0 gauge; in othergauges we would have terms like ˙ (cid:126) A : (cid:126) A , and the separationbetween kinetic and potential energy would no longer beso clean.For simplicity we suppressed the source term. In A =0 gauge, it is L source = (cid:90) d x (cid:126) A • · (cid:126) J , (8.17)and one sees that the charge density ρ does not enter.Then the equation of motion for (cid:126) A implied by the fullLagrangian is ¨ (cid:126) A − D × B = g J (8.18)We also have the nondynamical equations D · B = 0 , ˙ B + D × E = 0 (8.19)which are identities, due to the way B and E are definedin terms of A . Recall that we found one further equationby varying the covariant form of the Lagrangian, namelyGauss’s law, D · E = g ρ (8.20) Problems.
Gauss’s law does not seem to arise fromthe path integral formulation in A = 0 gauge. Whathappened to it?What condition must the source J µ obey in order for (cid:82) d x (cid:126)J µ • (cid:126)A µ to be gauge invariant? What is the phys-ical significance of this condition when J µ is the quarkcurrent?Show that the equations of motion of ψ imply that D µ J µ = 0.
9. HAMILTONIAN FORMULATION OF QCD ∗ (11-17-87) Just as in ordinary quantum mechanics the state ofa system is specified by a wavefunction ψ ( x, t ), in thepurely gluonic version of QCD we can characterize phys-ical states by a wave functional Ψ[ A ( (cid:126)x, t )] that dependson the gauge field A . (We continue to work in A = 02gauge.) In this lecture we shall explore the analogy some-what, and discuss the Hamiltonian formalism for evolvingΨ in time.Perhaps the closest analogy to the situation in fieldtheory, where we have infinitely many dynamical vari-ables, would be a lattice of atoms, say, interacting witheach other through some potential V that depends onthe positions (cid:126)q ( (cid:126)n, t ), where (cid:126)n is a lattice vector telling uswhich atom is being referred to. The action is S = (cid:90) dt L = (cid:90) dt (cid:32) m (cid:88) (cid:126)n (cid:12)(cid:12)(cid:12) ˙ (cid:126)q ( (cid:126)n, t ) (cid:12)(cid:12)(cid:12) − V ( (cid:126)q ( (cid:126)n, t )) (cid:33) . (9.1)Now if ψ i ( (cid:126)q ( (cid:126)n )) is the wave function at some initial time t i , then the amplitude for reaching a state ψ f ( (cid:126)q ( (cid:126)n )) at alater time t f is given by an ordinary integral (cid:90) (cid:89) (cid:126)n d q i ( (cid:126)n ) d q f ( (cid:126)n ) (9.2) ψ f ( (cid:126)q f ( (cid:126)n )) K (cid:16) (cid:126)q f ( (cid:126)n ) , t f ; (cid:126)q i ( (cid:126)n ) , t i (cid:17) ψ i ( (cid:126)q i ( (cid:126)n ))and the function K that propagates the initial state isgiven by a path integral K (cid:16) (cid:126)q f ( (cid:126)n ) , t f ; (cid:126)q i ( (cid:126)n ) , t i (cid:17) = (cid:89) (cid:126)n D (cid:126)q ( (cid:126)n, t ) e iS [ (cid:126)q ( (cid:126)n,t )] (9.3)where the integral is over functions (cid:126)q ( (cid:126)n, t ) satisfying (cid:126)q ( (cid:126)n, t i ) = (cid:126)q i ( (cid:126)n ) ,(cid:126)q ( (cid:126)n, t f ) = (cid:126)q f ( (cid:126)n ) . (9.4)In the same way, we can assign to each state in QCD awave functional Ψ[ A ( (cid:126)x, t )] such that | Ψ | is the probabil-ity density for the gauge field to have the value A ( (cid:126)x, t ),for each point in space, at a given time. Here A corre-sponds to (cid:126)q , and the position (cid:126)x corresponds to the latticevector (cid:126)n in the atomic crystal analog. The kernel for timeevolution of Ψ[ A ] was given in (8.2), which is the analogof (9.3).Alternatively, the time evolution of Ψ can be describedin differential rather than integral form—Schr¨odinger’sequation! To do this, we must first find the Hamiltonian.In the finite system (9.1), the canonically conjugate mo-menta are (cid:126)p ( (cid:126)n ) = ∂L∂ ˙ (cid:126)q ( (cid:126)n ) = m ˙ (cid:126)q ( (cid:126)n ) (9.5)They can be represented by (cid:126)p ( (cid:126)n ) = 1 i ∂∂(cid:126)q ( (cid:126)n ) (9.6)(taking (cid:126) = 1) since − i∂/∂(cid:126)q has the same commutationrelation with (cid:126)q as (cid:126)p has canonically. Similarly in QCDthe momentum conjugate to A an ( (cid:126)x ) is p an ( (cid:126)x ) = δLδ ˙ A an ( (cid:126)x ) = ˙ A aN ( (cid:126)x ) = − E an ( (cid:126)x ) (9.7) where we used the Lagrangian (8.15) after rescaling A → g A , and the operation that appears is a functional derivative, δA an ( (cid:126)x ) δA bm ( (cid:126)x (cid:48) ) = δ ab δ nm δ (3) ( (cid:126)x − (cid:126)x (cid:48) ) (9.8)This is the natural generalization of the partial deriva-tive to the case of infinitely many variables, labeled by acontinuous index (cid:126)x . Notice that (cid:126)p is just minus the colorelectric field. It can also be written as p an ( (cid:126)x ) = 1 i δδA an ( (cid:126)x ) (9.9)similarly to (9.6). Now the Hamiltonian can be con-structed. For a discrete system like the lattice, it is H = (cid:32)(cid:88) (cid:126)n (cid:126)p ( (cid:126)n ) · ˙ (cid:126)q ( (cid:126)n ) − L (cid:33) ˙ (cid:126)q = (cid:126)p/m . (9.10)For QCD, one simply replaces the sum with an integral,so that H = (cid:90) d x (cid:16) (cid:126) E + (cid:126) B (cid:17) = (cid:90) d x (cid:32) − (cid:18) δδ(cid:126) A (cid:19) + (cid:126) B (cid:33) . (9.11)Then the Schr¨odinger equation for Ψ[ A ] is1 i ∂∂t Ψ = H Ψ . (9.12)Previously we noted that three of the four Maxwellequations of QCD emerged from the gauge-fixed La-grangian (8.15), and the definitions of E and B , butGauss’s law, D · (cid:126) E = g (cid:126)ρ , did not appear. However, the time derivative of Gauss’slaw can be deduced as follows: ∂∂t (cid:16) D · (cid:126) E (cid:17) = − ∂∂t (cid:16) (cid:154) · ˙ (cid:126) A − i(cid:126) A · × ˙ (cid:126) A (cid:17) = − (cid:154) · ¨ (cid:126) A + i(cid:126) A · × ¨ (cid:126) A = D · ¨ (cid:126) A (9.13)Using the equation of motion for A , eq. (8.18), this be-comes ∂∂t (cid:16) D · (cid:126) E (cid:17) = − D · ( D × (cid:126) B ) − g D · (cid:126) J (9.14)The middle term would vanish trivially if D was the ordi-nary gradient, but since the components of D do not com-mute, more care is required. One finds that D · ( D × (cid:126) B ) is[ D x , D y ] B z plus cyclic permutations.3But this is just i [ F xy , B z ], as you showed in a previousexercise, which vanishes because F xy = B z . Furthermorethe source is covariantly conserved, D µ J µ = 0 , so that D · (cid:126) J = − ∂∂t (cid:126)ρ in A = 0 gauge. (In our conventions, A µ B µ = − A B + A i B i , and ρ = − J . ) Therefore the condition ∂∂t ( D · (cid:126) E ) = g ∂∂t (cid:126)ρ (9.15)is a consequence of the equations of motion. Conse-quently, if the wave functional satisfied (cid:104) D · (cid:126) E − g (cid:126)ρ (cid:105) Ψ[ A ( x )] = 0 (9.16)at some initial time, it would continue to do so forever.Therefore Gauss’s law can be implemented by imposingit as a constraint on the state of the system, Ψ.Notice that (9.16) is a functional differential equation,since E is to be interpreted as − iδ/δ A . Moreover (9.16) isan infinite set of constraints, one at each point in space.One might wonder whether solutions exist, since the con-straint operator (cid:126)C ( (cid:126)x ) ≡ D · (cid:126) E ( (cid:126)x ) − g (cid:126)ρ ( (cid:126)x ) (9.17)does not commute at different positions,[ (cid:126)C ( (cid:126)x ) , (cid:126)C ( (cid:126)x (cid:48) )] (cid:54) = 0 . To be consistent, we require that this new operator alsoannihilates the wave functional. If the commutator isa linear combination of (cid:126)C ’s there is no problem, but ifnot we might generate more and more constraints, to thepoint that no solution existed. It turns out to be nicerto investigate this not with the (cid:126)C ’s directly, but ratherwith their weighted averages, defined byΓ( µ ) ≡ (cid:90) µ i ( (cid:126)x ) C i ( (cid:126)x ) d x . (9.18)One can show that[Γ( µ ) , Γ( ν )] = Γ( λ ) (9.19)where (cid:126)λ = (cid:126)µ × (cid:126)ν (9.20) This choice of the metric signature is not consistent throughoutthe lectures.
Thus the commutators produce no new constraints; in-stead they form a closed algebra. It is the algebra of thecolor group SU(3), for we could define generators of localcolor transformationsˆΓ( µ ) = µ i ( x ) T i in the three-dimensional ( i.e. fundamental ) representa-tion of SU(3), and they would satisfy the same relations(9.19,9.20) as the Γ( µ ). Problem.
Prove eqs. (9.19,9.20).It is therefore not surprising that the Γ operators gen-erate gauge transformation on the state Ψ. That is, e i Γ( µ ) Ψ[ A ] = Ψ[ A (cid:48) ] (9.21)where A (cid:48) is the gauge field obtained from A by trans-forming with the matrixΛ( (cid:126)x ) = e i(cid:126)µ ( (cid:126)x ) • (cid:126)T (9.22)The alert reader may wonder how it is possible to dogauge transformations, since we have already fixed thegauge to A = 0. However the transformation (9.22) istime-independent, so Λ † ∂ Λ = 0 and any such Λ willkeep A = 0. Hence the Gauss’s law constraint on Ψmeans that Ψ must be invariant under the residual gaugetransformations that preserve the A = 0 condition. Wecan prove this directly from the equation itself: let (cid:20)(cid:90) (cid:126)α • (cid:16) D · (cid:126) E − g (cid:126)ρ ( (cid:126)x ) (cid:17) d x (cid:21) Ψ[ A ] = 0 (9.23)for some (cid:126)α ( (cid:126)x ). For simplicity suppose that there are noquarks, so that ρ = 0. Then (9.23) can be rewritten as (cid:20)(cid:90) D (cid:126)α ( (cid:126)x ) : (cid:126) E ( (cid:126)x ) d x (cid:21) Ψ[ A ] = 0where we have integrated by parts. (The reader shouldsatisfy himself that partial integration works for covariantderivatives.) Using the operator form of E = − iδ/δ A ,this becomes (cid:90) D (cid:126)α ( (cid:126)x ) : δ Ψ δ(cid:126) A d x = 0 (9.24)Now the functional version of Taylor’s theorem says thatΨ[ (cid:126) A ( x ) + (cid:126) a ( x )] = Ψ[ (cid:126) A ( x )] + (cid:90) (cid:126) a : δ Ψ[ (cid:126) A ( x )] δ(cid:126) A ( x ) d x (9.25)to first order in (cid:126) a . So for infinitesimal (cid:126)α ( (cid:126)x ), Gauss’s lawis equivalent toΨ[ (cid:126) A ( x ) + D (cid:126)α ( x )] = Ψ[ (cid:126) A ( x )] (9.26)which is just the result claimed, since D (cid:126)α ( x ) is the effectof an infinitesimal gauge transformation.4 p e p e p e p e(a) (b) (c) (d) FIG. 22: Diagrams for the hydrogen atom.
However not all gauge transformations can be built upfrom infinitesimal ones. This is most easily demonstratedfor the SU(2) subgroup of SU(3) generated by (cid:126)σ = ( λ , λ , λ )Imagine that we have cut off the infinite volume of spaceby introducing a large radius R . It is easy to check thatthe matrix Λ = (cid:114) − x R + i (cid:126)x · (cid:126)σR (9.27)is unitary and has determinant 1, yet cannot be written inthe form of exp( i(cid:126)α • (cid:126)λ ). We should therefore require thatΨ[ A ] be invariant under all such “large” gauge transfor-mations, as well as the ordinary ones.
10. PERTURBATION THEORY (11-19-87)
Now we must confront the question, how do we calcu-late anything quantitatively in this theory? Perturbationtheory (P.T.) is not very useful for bound state proper-ties since the coupling g is large. On the other hand,we have some experience with nonperturbative processeseven in nonrelativistic quantum mechanics; the hydro-gen atom is not a perturbative problem—it is an exactnonperturbative solution to Schr¨odinger’s equation. Butit is nevertheless made more accurate by the smallnessof the coupling e . Consider fig. 22(a). Since photon ex-change is relatively infrequent, we can replace the pho-ton exchanges by instantaneous effective interactions de-picted in fig. 22(b). The Schr¨odinger equation takes aninitial state of the proton and the electron and propa-gates them freely via − ( ∂/∂x ) plus interactions V ( x ).This only works if the coupling constant is small; oth-erwise diagrams like fig. 22(c,d) become too importantand we would have to find some other kind of effectiveinteraction potential to represent their effect.But there are some processes for which perturbationtheory in QCD works relatively well, for example p ¯ p col-lisions (fig. 23). At sufficiently high energies, only onegluon might be exchanged, similarly to fig. 21. One canthen predict the dynamics of the jets rather precisely us-ing perturbation theory. Recall the massless scalar field theory with Lagrangian L = 12 (cid:16) ˙ φ − ( ∇ φ ) (cid:17) + g φ − S ( x ) φ The equation of motion is¨ φ − ∇ φ ≡ (cid:3) φ = g φ − S (10.1)Let us ignore interactions for the moment and focus onthe source term. Going to Fourier space, the equation ofmotion becomes( ω − (cid:126)k ) φ k = k φ k = S k (10.2)To solve for φ k , a prescription must be given for treatingthe pole of the propagator, i.e., we must add i(cid:15) , φ k = S k k + i(cid:15) (10.3)Putting back the interaction, we have the rule that theamplitude for three φ particles to interact is g . Then forexample the s -channel scattering diagram is= g k + i(cid:15) (10.4)Let’s now compare this to the case of QCD in A = 0gauge, where the propagator comes from the free La-grangian L = 12 (cid:16) ˙ (cid:126) A • · ˙ (cid:126) A − ( (cid:154) × (cid:126) A ) (cid:17) . (10.5) jetjetp p FIG. 23: Electron-positron annihilation into hadronic jets. Throughout these lectures, RPF avoids writing factors of i forthe vertices and the propagators. Later, he will claim that theonly factors of i that are necessary to keep track of can be as-sociated with loops. I have not checked whether his rules forthe cubic and quartic gluon interactions give rise to the correctsign of interference for the diagrams contributing to four-gluonscattering, but his later claim implies that he did so. A → g A to get the coupling out ofthe propagator and back into the interactions. Includinga classical source (cid:126) S , the equation of motion is¨ (cid:126) A − (cid:154) × ( (cid:154) × (cid:126) A ) = − (cid:126) S (10.6)or equivalently¨ A i − ∇ A i + (cid:154) ( (cid:154) · A i ) = − S i (10.7)Hence, going to Fourier space,( ω − k ) A + k ( k · A ) = S (10.8)(omitting the gauge index i and subscript k for brevity).Next dot k into this equation to get( ω − k ) k · A + k k · A = ω k · A = k · S . (10.9)We can therefore eliminate k · A from eq. (10.8) and find k A = 1 ω ( S − k ( k · S )) . (10.10)In analogy to the scalar field example, one can read offthe propagator P ( k ) = − kk /ω k + i(cid:15) (10.11)in the form of a 3 × kk denotesthe outer product of the spatial momenta components. Now we would like to perturbatively compute pathintegrals involving gauge invariant functionals F [ A , A is set to zero), (cid:104) F (cid:105) = (cid:82) e iS [ A , F [ A , D A (cid:82) e iS [ A , D A (10.12)where the action is S [ A ,
0] = (cid:90) (cid:104) ˙ (cid:126) A : ˙ (cid:126) A − (cid:126) B : (cid:126) B (cid:105) d x + g (cid:90) (cid:126) A : (cid:126) J d x (10.13)with (cid:126) B = (cid:154) × (cid:126) A − ( g/ (cid:126) A ×× (cid:126) A and the current J in = ¯ ψ ( (cid:126)x, t ) γ n λ i ψ ( (cid:126)x, t ) ( n = 1 , , . (10.14)The amplitude for a quark-gluon interaction can bewritten as u p u p a ni = (cid:112) πg (cid:88) n,i (cid:18) ¯ u γ n λ i u (cid:19) a in (10.15) where a in is the combined spin-color polarization of thegluon and the coupling is rescaled so that the Coulombinteraction is g /r instead of g / (4 πr ). Therefore thegluon-exchange diagram giving rise to the potential is p p p p k = πg (cid:0) ¯ u γ n λ i u (cid:1) (cid:32) δ nm − k m k n ω k (cid:33) δ ij (10.16) × (cid:0) ¯ u γ m λ j u (cid:1) This can be put into a simpler form using the conserva-tion of the quark currents, e.g., k µ (¯ u γ µ λ i u ) = ¯ u ( /p − /p ) λ i u = ¯ u ( m − m ) λ i u = 0 (10.17)since the gluon vertex conserves flavors. Therefore k n ¯ u γ n λ i u = ω ¯ u γ λ i u and we can rewrite the right-hand side of (10.16) as πg (cid:0) ¯ u γ µ λ i u (cid:1) (cid:18) − δ µν k (cid:19) δ ij (cid:0) ¯ u γ ν λ j u (cid:1) (10.18)in which the gluon propagator takes a Lorentz covariantform. This is an illustration of the fact that physicalamplitudes are independent of the choice of gauge. Nev-ertheless we must make some choice. Consider the gluonLagrangian with no choice of gauge imposed, L ∼ (cid:2) ∂ µ A ν − ∂ ν A µ − A × µ A ν (cid:3) (10.19)The noninteracting part has the structure k δ µν − k µ k ν in momentum space, which is noninvertible. Hence weneed to fix the gauge to define the gluon propagator.Another interesting observation is that when varyingthe full action, the equation of motion takes the form (cid:3) A ν − ∂ ν ∂ µ A µ = S ν = ⇒ k A ν − k ν ( k · A ) = S ν ( k )= ⇒ k · S (10.20)where the source S ν now includes contributions that arenonlinear in A for nonvanishing background gauge fields,in addition to the quark current contribution. In thiscase, current conservation is more complicated than forQED, where the current comes only from the chargedfermions. RPF writes δ µν for the Minkowski metric tensor, even thoughhe is not working in Euclidean space. Moreover he normallydoes not distinguish between covariant and contravariant Lorentzindices. Consider the lowest order processes contributing togluon propagation (postponing for the moment the issueof gluon loops): = 1 + iX (10.21)The probability associated with this amplitude is 1 + i ( X − X ∗ ) + . . . . This means that i ( X − X ∗ ) is theprobability of not producing a gluon from a gluon, whichis the probability of instead producing a q ¯ q pair: Im = 2 (10.22)However if we try to do the same thing with gluons inthe loop, and using a covariant gauge for the gluon prop-agator, the analogous relation breaks down. To fix it, weneed to add ghost fields,
2= Im + (10.23)The ghost fields (dashed line) do not appear on the right-hand side of eq. (10.23) because ghosts never enter intoa final state. Alternatively, we can work in A = 0 gaugeinstead of covariant gauge, and dispense with the ghosts.Before Faddeev and Popov solved the problem of con-sistently incorporating ghosts at any order of perturba-tion theory, I was trying to figure out how to make itwork beyond one loop, without success. At that time itdid not occur to me to use A = 0 gauge, because of aprejudice based on experience with QED. People origi-nally tried to formulate QED in a nonrelativistic gauge, (cid:154) · A = 0, but nobody understood how to renormalizethis theory in a relativistic way.Another way to try to get around the gauge fixingproblem is to temporarily give the gluons a mass. Thenthe propagator exists, and one can try to take m g → The factor of i associated with the loop will be explained at theend of the lecture. For now we will avoid the ghosts by continuing in A = 0 gauge. The next step is to write the rules forthe gluon self-interaction vertices. We continue to writecombined spin/color polarization vectors a, b, c, . . . cor-responding to a plane wave solution A iν = ( a iν e iq µa x µ + b iν e iq µb x µ + c iν e iq µc x µ , and taking all q i to point inwardstoward the vertex, so that (cid:80) i q µi = 0. Recall that thecubic interaction Lagrangian is g ( ∂ µ A ν − ∂ ν A µ ) A µ × A ν (10.24)By substituting the plane wave solution for the fields, wecan read off the rule for the 3-gluon vertex, g ( q µa a ν − q νb a µ ) • ( b µ × c ν ) + cyclic permutations (10.25)The permutations can be reorganized into the form q a q b q c c λ k b ν j a µ i = (cid:112) πg ( q a − q c ) ν (cid:0) b ν • ( a µ × c µ ) (cid:1) +( q b − q a ) ν (cid:0) c ν • ( b µ × a µ ) (cid:1) +( q c − q b ) ν (cid:0) a ν • ( c µ × b µ ) (cid:1) (10.26)after rescaling the coupling as before. Similarly, for thefour-gluon amplitude, we obtain a d cb = 4 πg a µ × b ν ) • ( c µ × d ν )+ symmetric permutations (10.27)For completeness, the gluon propagator is again= δ mn − k m k n /ω k δ ij (10.28)and the rules for quarks are ji = δ ij /p − m q (10.29)7and u u p p a = (cid:112) πg ¯ u ( (cid:126)a µ • (cid:126)λ γ µ ) u (10.30)We can rewrite the gluon propagator in a morecovariant-looking form by introducing the 4-vector η µ =(1 , , , q µ = ( ω, (cid:126)k ), we have ω = η · q and(0 , (cid:126)k ) = q µ − η µ ( η · q ), so k n k m ω = ( q µ − η µ ( η · q )) ( q ν − η ν ( η · q ))( η · q ) (10.31)Moreover − δ mn = δ µν − η µ η ν . The gluon propagator(ignoring color indices) becomes P µν = − δ µν + q µ η ν + q ν η µ q · η − q µ q ν η ( η · q ) q + i(cid:15) (10.32)which has the property P µν η ν = 0 . All of the η -dependence must drop out of physical amplitudes forthem to be Lorentz invariant. We will see in section 12that the q µ q ν term can be eliminated by an appropriategauge fixing procedure. Thus far all the rules have been unencumbered by anyfactors of i . We can consistently push them into the rulesfor loops, integrating over the internal momentum of theloop, i (cid:90) d p (2 π ) for gluon loops − i (cid:90) d p (2 π ) for quark loops (10.33)These are the only factors of i that one ever needs. The minus sign for fermion loops can be understoodas the result of doing a 360 ◦ rotation, illustrated by tak-ing the two ends of a belt and exchanging their positions RPF aligns (cid:126)p against the flow of fermion number since the spinoris ¯ u , not ¯ v . We will see this again in section 11. The pointof defining final state momenta as being negative is to make theMandelstam variables all look the same, all involving plus signsrather than minus signs. See note 27 The following material appears in my notes as an interruptionof the new subject that has just been introduced, section 11,as though it suddenly occurred to RPF that he had meant todiscuss it earlier. In characteristic showman fashion, he removedhis own belt to do the demonstration. (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) µµ B FIG. 24: Left: magnetic monopole with a charge transportedaround its Dirac string. Right: the charge is transportedaround the monopole within the plane. while keeping the orientations of ends of the belts fixed.Although each fermion by itself undergoes only a 180 ◦ rotation, relative to each other it is 360 ◦ , which as weknow for fermions introduces a relative sign, symbolizedby the kink in the belt.Similarly when we exchange the positions of two fermionsin diagrams such as ,they differ from each other by a minus sign because ofFermi statistics. There is inherently a 360 ◦ rotation oftheir relative orientations. A similar exchange occurswhen there is a fermion loop, leading to the minus signin the diagrammatic rule.An interesting aside illustrates the origin of this signfor a composite system that behaves like a fermion. Thisis the combination of a magnetic monopole of strength µ and an electrically charged scalar, with charge q , sep-arated by some distance, (cid:126)r (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) q µ r vBE Even in the absence of any relative motion between thetwo constituents, this system has angular momentum in Characteristically, RPF does not call this by its common name,dyon. Several pages of his personal notes are devoted to thisproblem. (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) FIG. 25: Interchange of two monopole-charge systems. the (cid:126)r direction, which can be deduced by imagining thatwe try to move the charge with velocity (cid:126)v as shown. Since q is moving in a magnetic field, it experiences a force thatgives a torque on the system, as if it were a gyroscope,showing that it has angular momentum.Since B = (cid:154) × A , B can never have a divergence un-less there is a Dirac string. If µ is quantized such that µq = (cid:126) /
2, then the change in phase of an electron as itmoves around the string is e ie (cid:82) (cid:126)A · (cid:126)dx , which is − µ isquantized properly. This is easiest to see in the right-hand part of fig. 24 by considering the phase change of acharge moving in the plane of the monopole using Green’stheorem, e iq (cid:82) A · dx = e iq Φ = e iq ( 12 4 πµ ) = e iq πµ = e iπ (10.34)where Φ is the flux of B through the upper hemisphere.But we have merely rotated the system by 360 ◦ , so thisphase change shows that it behaves like a particle of spin1 / −
1. It comes from the combinedphase changes of the charges as they move around theopposite monopole by 180 ◦ .
11. SCATTERING PROCESSES (11-24-87)
Let us now consider the scattering of two quarks asshown in fig. 26. This can be measured by doing p - p scattering, since the parton model allows us to relatethe two processes. The parton distribution functions aremeasured by deep inelastic scattering experiments.Most of the time the scattering does not producejets, but these are the observables we are interested in.The scattered quarks determine the directions of thejets. This part of the problem—how quarks hadronizeinto jets—can be understood from the observations of e + e − → [2 hadron jets] through the electromagnetic pro-cess shown in fig. 27.Of course it is also possible to get jets originating fromgluons produced in the QCD scattering process. This The unconventional choice of momentum labels is deliberate; seenote 30. p p p p timeq qq FIG. 26: Quark-quark scattering. e − q qe + FIG. 27: Electromagnetic production of jets. has to be taken into account, but for simplicity we willstart with the quark production process. Our goal is tomeasure α s through scattering. Historically this analysishelped to design the experiments observing these pro-cesses, and QCD helped to tune the phenomenologicalmodels needed to make the predictions.Recall the gluon propagator (10.32) in η · A = 0 gauge.The terms involving η drop out of the amplitude becauseof the conservation of the external quark currents. Thenthe amplitude becomes T = [4 π ] g (cid:18) ¯ u γ µ λ i u (cid:19) q (cid:18) ¯ u γ µ λ i u (cid:19) (11.1)Conventionally we express it in terms of the Mandel-stam variables. Neglect quark masses, since we want highenough energies to get clean jets. Then s = ( p + p ) = 4 E in c.m. frame t = ( p + p ) = − (momentum transfer) u = ( p + p ) (11.2)which have the property s + t + u = (cid:80) i =1 m i (left as anexercise for the reader to prove). In the center-of-massframe, (E, p cos θ , p sin θ ) (E, p, 0) θ (E, −p, 0) the momentum transfer is given by t = (0 , p (1 − cos θ ) , p sin θ ) = − p (1 − cos θ ) = q (11.3)9Next we need | T | , which depends upon the polariza-tions of the quarks. If these are not measured, then weare only interested in | T | = (cid:88) init . spins& colors (cid:88) final spins& colors | T | (cid:16) no . of spins &colors in init . state (cid:17) (11.4)If we measure the final state polarizations but the incom-ing beams are unpolarized, then we should omit the sumover final state spins.In eq. (11.1) we have implicitly assumed that thespinors u i are products of spin and color factors, whichwe can write as u = U α , etc . (11.5)Then | T | = (cid:18) πg q (cid:19) (cid:20) ¯ U ¯ α γ µ λ i U α (cid:21) (cid:20) ¯ U ¯ α γ µ λ i U α (cid:21) × (cid:18)(cid:20) ¯ U ¯ α γ ν λ j U α (cid:21) (cid:20) ¯ U ¯ α γ ν λ j U α (cid:21)(cid:19) ∗ → (cid:88) spins (cid:18) πg q (cid:19) (11.6) × (cid:0) ¯ U γ µ U (cid:1) (cid:0) ¯ U γ µ U (cid:1) (cid:16) (cid:0) ¯ U γ ν U (cid:1) (cid:0) ¯ U γ ν U (cid:1) (cid:17) ∗ × (cid:18) ¯ α λ i α (cid:19) (cid:18) ¯ α λ i α (cid:19) (cid:18) ¯ α λ j α (cid:19) ∗ (cid:18) ¯ α λ j α (cid:19) ∗ The sum on spins gives, for example, u ¯ u = /p + m (11.7)leading to the tracestr (cid:16) γ ν ( /p + m ) γ µ ( /p + m ) (cid:17) tr (cid:16) ( /p + m ) γ µ ( /p + m ) γ ν (cid:17) Similarly, the sum on colors gives (cid:88) colors (cid:18) ¯ α λ i α ¯ α λ j α (cid:19) = tr λ i λ j δ ij (11.8)Recall thattr γ ν /p γ µ /p = 4 (cid:16) p ν p µ + p µ p ν − ( p · p ) δ µν (cid:17) tr /p γ µ /p γ ν = 4 (cid:16) p ν p µ + p µ p ν − ( p · p ) δ µν (cid:17) neglecting masses. The total color factor is δ ij δ ij N −
14 (11.9)where we indicated the more general result for SU(N) atthe end, and we have not yet included the 1 /N from averaging over the quark colors. Putting everything to-gether, | T | = (cid:18) (4 πg ) N q (cid:19) · ( N − × [2( p · p )( p · p ) + 2( p · p )( p · p )]= N − N (cid:18) (cid:19) (4 πg ) t ( s + u ) (11.10)and the differential cross section is dσdt = | T | π s (11.11)where dt = 2 E d cos θ .Similar results can be found for quark-gluon ( QG ) andgluon-gluon ( GG ) scattering. At low momentum trans-fer, and apart from an overall proportionality constant,their relative squared matrix elements go as QQ → QQ = 89 s t QG → QG = 2 s t GG → GG = 92 s t (11.12)We see that the formulas simplify at small t .
12. GAUGE FIXING THE PATH INTEGRAL ∗ (12-1-87) We now have a complete set of rules for calculatingamplitudes perturbatively in the gauge A t = 0. (It is notperfectly complete because we have not yet specified howto deal with the infinities arising from loop diagrams; thiswill be the subject of some later chapters.) Ordinarily Iwould not complicate matters by introducing an addi-tional formalism that gives the same answers in the end,but Faddeev and Popov (Phys. Lett. B25 (1967) pp. 29-30) have invented another way of fixing the gauge that isso elegant and useful that it deserves mention. It allowsone to evaluate the path integral with an arbitrary gaugecondition, of which A t = 0 is just a special case.Before deriving the method, it will be useful to know atechnique that allows the path integral in A t = 0 gauge, (cid:90) e iS [ A , D A ( x , t ) , (12.1)to be rewritten as an integral over all four of the A µ , withan extra A t -dependent term in the Lagrangian. Notice RPF had written | T | / (8 π s ); I have restored the missing factorof 1 /
2. For same-flavor quarks, there should be yet another factorof 1 / (cid:90) e iS [ A ,φ ] D A , (12.2)for any function φ . One way to see this is to gauge trans-form A in (12.1) to A (cid:48) = Λ † A Λ + Λ † (cid:154) Λ A (cid:48) t = φ = Λ † ˙Λ , (12.3)where Λ, which is determined by solving ˙Λ = φ Λ, doesnot depend upon A , hence D A is invariant. Eq. (12.2)can also be written as (cid:90) e iS [ A ,A t ] D A δ [ A t − φ ] (12.4)Since it does not depend on φ , we can functionally inte-grate over φ with some weight, sayexp (cid:18) iµ g (cid:90) d x φ (cid:19) (12.5)and change the path integral by only an overall multi-plicative factor. This factor has no effect on an expecta-tion value of a gauge-invariant functional, (cid:104) F (cid:105) = (cid:82) D φ e ic (cid:82) φ (cid:82) e iS [ A ,A t ] F ( A , A t ) D A D A t δ [ A t − φ ] (cid:82) D φ e ic (cid:82) φ (cid:82) e iS [ A ,A t ] D A D A t δ [ A t − φ ](12.6)since it cancels between numerator and denominator.Now the φ integral is trivial because of the delta func-tional, and the path integral is (cid:90) e iS [ A ,A t ]+ iµ g (cid:82) A t d x D A (12.7)Because of the new term in the action, the gluon propa-gator now exists, even though A t is no longer fixed to bezero. Exercise.
Show that the propagator for (12.7) is P µν ( k ) = 1 k (cid:18) − δ µν + k µ η ν k · η + k ν η µ k · η − k µ k ν η ( k · η ) (cid:19) − k µ k ν η µ ( k · η ) (12.8)where η µ = (1 , , , µ = − k , which is impossible because µ is just aconstant, not a Fourier transform variable. But the samething can be accomplished by using e i µ g (cid:82) d x ( ∂ µ φ ) (12.9)instead of (12.5) as the weight factor. Then we get (12.8),but with µ → − k µ , and µ can be chosen so that thelast two terms in (12.8) cancel. This is the justification A t W(A) = ε W(A) = A ε FIG. 28: Gauge orbits sliced by a gauge condition W . for saying, in a previous lecture, that the k µ k ν term inthe propagator was irrelevant.We are aiming for an expression similar to (12.7), butfor some arbitrary gauge condition, not necessarily A t =0. For this the Faddeev-Popov procedure will be needed,which since I didn’t invent it myself, I claim is extremelysubtle! Suppose we wanted the gauge condition to be ∂ µ A µ = 0. Then the path integral must look somethinglike (cid:90) e iS [ A ] δ [ ∂ µ A µ ] D A . (12.10)But this is not quite right, even though it would be rightfor the A t = 0 case. Consider the space of all gauge fieldconfigurations, represented schematically by plotting A t along one axis and A along the other. Let W ( A ) = 0be the desired gauge condition, represented by a surfacein the function space, that cuts across the trajectories ofgauge-equivalent A µ ( x )’s, called “gauge orbits” (see fig.28). Previously we integrated over the line A t = 0. Thedelta functional δ [ A t ] can be interpreted as the limit ofa less singular constraint, which is to integrate over thestrip between A t = 0 and A t = (cid:15) , divide by (cid:15) , and take (cid:15) →
0. However if we try to do the same thing for thesurface W ( A ) = 0, the gauge orbits will not necessarilycross the strip at the same angle everywhere, and thesimple constraint δ [ W ( A )] will weight some orbits toomuch, some too little, as one moves along the surface.An extra factor is needed to compensate for the varyinglength of the orbits crossing the strip.Call this compensating factor ∆( A ). It will be shownthat ∆( A ) − = (cid:90) D g δ [ W ( A g )] (12.11)where A gµ = Λ † ( g ) A µ Λ( g ) + Λ † ( g ) i∂ µ Λ( g ) , (12.12)is the gauge-transformed A µ and Λ( g ) is the matrix rep-resentation of the abstract group element g . D g standsfor the invariant group measure at each point in space-1time. It has the property (cid:90) D g δ [ W ( A g )] = (cid:90) D g δ [ W ( A hg )]= ∆( A h ) − (12.13)for any h in SU(3), so we see that ∆( A ) is gauge invariant.To define the path integral, insert a factor of 1 = (cid:82) ∆( A ) (cid:82) D g δ [ W ( A g )] into (cid:82) e iS [ A ] D A . The expecta-tion value of F [ A ] is then (cid:104) F (cid:105) = (cid:82) e iS [ A ] F D A (cid:82) e iS [ A ] D A (12.14)= (cid:82) e iS [ A ] F ∆( A ) δ [ W ( A g )] D A D g (cid:82) e iS [ A ] ∆( A ) δ [ W ( A g )] D A D g . The nice thing about these integrals is that they don’tdepend on g (assuming, as usual, that F is gauge invari-ant). Make the change of variables A → A g − . Becauseeach factor in the integrals is gauge invariant except forthe delta functional, the g -dependence disappears, andthe (infinite) integrals (cid:82) D g cancel between numeratorand denominator. We are therefore left with the pathintegral Z = (cid:90) e iS [ A ] ∆( A ) δ [ W ( A )] D A , (12.15)as claimed.Eq. (12.15) is the desired generalization of (12.1) for A t = 0 gauge, but it is not in a very useful form for ex-plicit computations. ∆( A ) is some horribly complicatedfunctional which in general nobody knows how to com-pute. Fortunately, it is not necessary to know ∆( A ) forall values of A , but only where W ( A ) = 0, and there it can be determined. Take W ( A ) = ∂ µ A µ , for example.We must evaluate (cid:90) δ [ ∂ µ A gµ ] D g (cid:12)(cid:12)(cid:12) ∂ µ A µ =0 . (12.16)We first assume that there is a unique solution to ∂ µ A gµ =0, such that g is the identity when ∂ µ A µ is already zero.Therefore we can focus on infinitesimal gauge transfor-mations, (cid:126)A gµ = (cid:126)A µ + D µ (cid:126)α = (cid:126)A µ + ( ∂ µ − A × µ ) (cid:126)α . (12.17)Then (12.16) becomes (cid:90) δ [ ∂ µ A µ + ∂ µ D µ α ] D α (cid:12)(cid:12)(cid:12) ∂ µ A µ =0 . (12.18)Recall that for finite-dimensional integrals, (cid:90) d n x δ ( n ) ( M ab x b ) = 1 | det M | . In the present case, eq. (12.18), we get a functional de-terminant, Det − ( ∂ µ D µ ) , which depends on A µ through D µ . Therefore the pathintegral (12.5) is (cid:90) e iS [ A ] δ [ ∂ µ A µ ] Det( ∂ µ D µ ) D A (12.19)in the gauge ∂ µ A µ = 0.The next step is to reexpress the determinant so thatit looks like a new term in the action. If instead ofDet( ∂ µ D µ ) we had Det − / ( ∂ µ D µ ), we could use thefunctional generalization of the formula (cid:90) d n x e − x a M ab x b = (2 π ) n/ √ det M .
However, there is an analogous formula for anticommut-ing variables that does what is needed, (cid:90) D P D ¯ P e ¯ P MP = det
M . (12.20)Here P is an anticommuting function, { P ( x ) , P ( x (cid:48) ) } =0, ¯ P is its complex conjugate, and M is a differentialoperator. (The reader who is unfamiliar with this typeof integral should work through the following exercise.) Exercise.
Complex anticommuting variables are de-fined to satisfy θ = ¯ θ = { θ, ¯ θ } = 0. The complete tableof integrals for such variables is, by definition, (cid:90) dθ = (cid:90) dθ ¯ θ = 0; (cid:90) dθ θ = 1(and similarly for the complex conjugates). If there are2 N variables θ i , ¯ θ i , i = 1 , . . . , N , then they all anti-commute with each other. By the above rule, the onlynonvanishing integrals over all the θ i , ¯ θ i are (cid:90) d n θ d n ¯ θ (cid:2) ¯ θ θ . . . ¯ θ N θ N (cid:3) = 1 , and those integrals that differ from it by a permutationof the variables in the integrand. (If the permutation isodd, the integral will be − (cid:90) d n θ d n ¯ θ e ¯ θ a M ab θ b = det M .
Applying this technique to the path integral gives (cid:90) e iS [ A ] δ [ ∂ µ A µ ] e i (cid:82) ¯ P ( x ) ∂ µ D µ P ( x ) d x D P D ¯ P , (12.21)where the P ’s transform in the octet representation ofSU(3), since that is the representation on which D µ actedin (12.18).The new fields P , ¯ P are called Faddeev-Popov ghosts to underscore the fact that they are not physical fieldslike quarks or gluons, but only a mathematical conve-nience. Because they are anticommuting, they behavelike fermions in the sense that ghost loops contribute a2factor of − ∂ µ ¯ P D µ P = ∂ µ ¯ P ∂ µ P − ∂ µ ¯ P · ( A µ × P ) . (12.22)The Feynman rules for ghosts are seen to be − a b = 1 k δ ab ;coupling : p p c a µ c = p µ (cid:126) ¯ c • ( (cid:126)a µ × (cid:126)c )where (cid:126)c and (cid:126) ¯ c are octet color vectors, just like (cid:126)a , . . . , (cid:126)a . In addition, the gauge field propagator is sim-ply a, µ b, ν = δ µν δ ab k in this gauge, since (cid:90) d x ( ∂ µ A ν − ∂ ν A µ ) = (cid:90) d x (cid:0) ( ∂ µ A ν ) − ( ∂ µ A µ ) (cid:1) after integrating by parts, and ∂ µ A µ = 0.Now we are almost done, but the expression (12.21)is still not easy to use since we don’t know how to doGaussian integrals with a constraint. This is where thetrick introduced at the beginning of the chapter comesin. If instead of ∂ µ A µ = 0 one used ∂ µ A µ = f , ∆( A )would be the same as before, and (12.21) would be (cid:90) e iS [ A ] ∆( A ) δ [ ∂ µ A µ − f ] D A . (12.23)This expression does not really depend on f because itwas obtained from the same starting point, (cid:82) e iS [ A ] D A ,for any f . So again it is permissible to integrate over f with a weight factor exp( i (cid:82) f d x ). The final result is (cid:90) e iS [ A ]+ i g (cid:82) ( ∂ µ A µ ) d x + i (cid:82) ∂ µ ¯ P D µ P d x D A D ¯ P D P . (12.24)The second term in the action cancels the similar termin S [ A ] so that the gluon propagator is still δ µν /k , butthere is no longer any restriction on A µ in the integral. In deriving this we assumed that for a given vectorfield A µ the gauge transformation g needed to arrangethat the divergence of the new field is zero, ∂ µ A gµ = 0, The following paragraph was added to my revision of the lectureby RPF. is unique. This was found to be false by Gribov. Thus ∂ µ A µ = 0 does not completely specify the gauge. ThusFaddeev’s argument looks imperfect—first there are sev-eral places in g where there are contributions to (12.15).In addition our ghost gives Det( ∂ µ D µ ) but our analy-sis from (12.18) wants the absolute value | Det( ∂ µ D µ ) | .There is much confusion, but I think (from some studiesI made some time ago) the final integral is really cor-rect. At any rate no error would be expected in per-turbation theory because for configurations with small A µ the gauge that makes ∂ µ A µ = 0 is unique. Gribov’sambiguity appears only for sufficiently large A . We shallsee examples of it later. Exercise.
Rewrite (12.24) for a general gauge condi-tion, W ( A ).
13. QUARK CONFINEMENT ∗ (12-3-87) The utility of being able to quantize QCD in a varietyof different gauges is that some gauges are particularlyconvenient for certain applications. In electrodynamics,the Coulomb gauge has the virtue that fields satisfyingthe gauge condition (cid:154) · A = 0 (13.1)represent truly physical, transverse degrees of freedom. Italso has some peculiarities. The Hamiltonian has a non-local, instantaneous interaction (action at a distance),which must combine with the interactions of the trans-verse photons so that the net force propagates at thespeed of light.Interactions that look instantaneous are well suited toSchr¨odinger’s equation, which requires the potential be-tween particles at equal times. It would be quite awk-ward to explicitly describe finite-velocity forces in theSchr¨odinger equation because the potential for one par-ticle at a time t would depend on the positions of theothers at the retarded times, and one would need thepast histories of all the particles to propagate the sys-tem forward in time. In what follows we will derive theCoulomb gauge path integral for QCD in the Hamilto-nian form (which is closely related to the Schr¨odingerpicture) and see some indications of quark confinement.Write the gauge field as A iµ = ( φ i , A i ) . (13.2)Then in the gauge (13.1), the vacuum amplitude is X = (cid:90) e iS [ A ] Det( (cid:154) · D ) δ [ (cid:154) · A ] D A D φ , (13.3) In my course notes, I have some elaboration of this point: “RPFconjectures that the sign of the Det changes only if the gauge con-dition does not uniquely determine A µ . In this case, he believesthat integrating over the different solutions gives a compensatingerror that makes up for using Det instead of | Det | .” D = ( (cid:154) − A × ) is the covariant derivative in theadjoint representation. The constraint in (13.3) impliesthat if A is expanded in plane waves, C ( k ) e i k · x , then k · C = 0. The color electric and magnetic fields are(omitting the gauge index for brevity) E = ( − (cid:154) φ + ˙ A + A × φ ) = ˙ A − D φ (13.4) B = (cid:154) × A − A ×× A (13.5)and the action is S [ A ] = 12 g (cid:90) ( E − B ) d x + (cid:90) ( ρφ − J · A ) d x (13.6)in the presence of an external source J µ = ( ρ, J ). Noticethat exp( i g (cid:82) E ) can be rewritten as e i g (cid:82) E d x = c (cid:90) e − ig (cid:82) (cid:144) d x + i (cid:82) (cid:144) · E d x D (cid:144) (13.7)by completing the square. Then the amplitude becomes X = (cid:90) e − ig (cid:82) (cid:144) d x + i (cid:82) (cid:144) · ( ˙ A − D φ ) d x + i g (cid:82) B d x × e i (cid:82) ( ρφ − J · A ) d x Det( (cid:154) · D ) D A T D φ D (cid:144) (13.8)where δ [ (cid:154) · A ] has been eliminated by integrating onlyover the transverse part of A , denoted by A T , and theintegrand is evaluated at (cid:154) · A = 0.Now any vector field can be split into a longitudinaland a transverse part, (cid:144) = (cid:144) transverse + (cid:144) longitudinal ≡ P + (cid:154) f (13.9)where (cid:154) · P = 0, by definition. Then X = (cid:90) exp (cid:34) − ig (cid:90) (cid:0) P · P + 2 P · (cid:154) f + ( (cid:154) f ) (cid:1) d x + i (cid:90) P · (cid:16) ˙ A − (cid:154) φ + A × φ (cid:17) d x + i (cid:90) ( ρφ − J · A ) d x + i (cid:90) (cid:154) f · ( ˙ A − D φ ) d x + i g (cid:90) B · B d x (cid:35) × Det( (cid:154) · D ) D A T D φ D P D f . (13.10)The underlined terms can be eliminated by integrating byparts and using the fact that (cid:154) · P = (cid:154) · A = 0. An overallconstant, Det( (cid:154) ), has been omitted from the functionalmeasure D f . Grouping the remaining φ -dependent fac-tors together and integrating over φ gives a delta func-tional, δ [ P × A + D · (cid:154) f + ρ ] (13.11) See note [25]. which says that f must satisfy f = − ( D · (cid:154) ) − ( ρ + P × A ) . (13.12)This is reminiscent of the analogous equation in electro-dynamics, ˜ f = − ∇ ρ , whose solution is˜ f = 14 π (cid:90) ρ ( R (cid:48) ) | R − R (cid:48) | d R (cid:48) , (13.13)but the complexity of the QCD version (13.12) preventsus from obtaining such a nice closed-form solution for f .There is a fortunate simplification from the constraint(13.11) however. The delta functional produces a factorof Det − ( D · (cid:154) ) (13.14)when the integral over f is performed (recall that δ ( ax ) = δ ( x ) /a ). If it was Det − ( (cid:154) · D ) instead, it would can-cel the Faddeev-Popov determinant in (13.10). However, (cid:154) · D is the same as D · (cid:154) in the present case, since D · (cid:154) g = ( (cid:154) − A × ) · (cid:154) g = ∇ g − A × • (cid:154) g and (cid:154) · D = (cid:154) · ( (cid:154) − A × ) g = ∇ g − ( (cid:154) · A ) × g − A × • (cid:154) g ;the (cid:154) · A term vanishes in the path integral (13.10) whichcan therefore be written as X = (cid:90) e i (cid:82) P · ˙ A −H ( P , A )] d x D A T D P ; (13.15) H = 12 g (cid:0) ( ∇ f ) + P · P + B · B + J · A (cid:1) (13.16)This is the Hamiltonian form of the path integral. Letus recall from ordinary quantum mechanics the connec-tion between it and the Lagrangian form, X L = (cid:90) e iS D Q ( t ) (13.17)where Q is the particle coordinate, S = (cid:82) Ldt , and theLagrangian is L = m Q − V ( Q ) (13.18)for a particle of mass m . From L one derives the canon-ical momentum P = ∂L∂ ˙ Q = m ˙ Q (13.19)and Hamiltonian H = ( P ˙ Q − L ) (cid:12)(cid:12)(cid:12) ˙ Q = P/m = P m + V ( Q ) . (13.20)4The Hamiltonian path integral is given by X H = (cid:90) e i (cid:82) [ P ( t ) ˙ Q ( t ) − H ( P,Q )] dt D P ( t ) D Q ( t )= (cid:90) e i (cid:82) [ P ( t ) ˙ Q ( t ) − P m − V ( Q )] dt D P D Q (13.21)This is seen to be the same as X L after completing thesquare and integrating over P . However, it would dono good to carry out the integral over P in the case ofQCD, (13.16), because it would introduce a complicatedfunctional determinant depending on A T , due to the way P enters (cid:154) f , (12.12).We now concentrate on the interaction Hamiltonianfrom (13.16), H I = 12 g (cid:90) ( (cid:154) f ) d x = − g (cid:90) f ∇ f d x = − g (cid:90) (cid:2) ρ + P × • A (cid:3) D · (cid:154) ∇ D · (cid:154) (cid:2) ρ + P × • A (cid:3) d x (13.22)You will recall that ρ is the charge density of the externalquark field. This suggests interpreting P × • A as the colorcharge density of the transverse gluons. To see that thisinterpretation makes sense, recall that a complex scalarfield φ = ( φ + iφ ) has the charge density ρ = 12 i (cid:16) φ ∗ ˙ φ − ˙ φ ∗ φ ) (cid:17) = 12 i ( φ ∗ π ∗ − πφ )= ( φ π − φ π ) (13.23)where π = ∂ L /∂ ˙ φ is the canonically conjugate momen-tum. Eq. (13.23) is like the cross product of the vectors( φ , φ ) and ( π , π ).Now we would like to deduce the potential betweenquarks from the operator ( D · (cid:154) ) − ∇ ( D · (cid:154) ) − in (13.22).Although D · (cid:154) cannot be inverted in closed form, it canbe expanded in powers of the gluon field, which corre-sponds to weak coupling. We get1 D · (cid:154) ∇ D · (cid:154) = 1 ∇ + 2 1 ∇ A × • (cid:154) ∇ + 3 1 ∇ A × • (cid:154) ∇ A × • (cid:154) ∇ + . . . (13.24)The first term corresponds to the Coulomb potential, asin (13.13). If the following terms are evaluated to 2nd order weget a strong attraction. In β = 11 − n f it contributes The remainder of this section was added to my submitted draftby RPF. The beta function has not yet been introduced; this will come inthe next lecture. +12 units—most of the confining effect. Thus the in-stantaneous Coulomb interaction probably rises with dis-tance. The charge densities of transverse (“real”) glu-ons and quarks are ( ρ + P × A ) and their interaction via H I , via ( D · (cid:154) ) − ∼ = ∇ − , to second order in H I , makesa vacuum polarization of the normal sign, contributing − − n f to β .This leads to an interesting model of a string connect-ing heavy quarks. Because of the rapid rise (with r ) of the force between charges, unbalanced color charges ρ + P × A at any distance are intolerable. Suppose westart with a red quark at r = 0 and say anti-red far awayto the right. By creating a dipole gluon ¯ RG within thisrange where opposite colors are tolerable we cancel red-ness at larger transverse distances. But then the G endof the gluon is unbalanced, so another gluon dipole ¯ GB forms (the energy for which coming from the decrease in G energy). This continues until we meet the final quark.Thus we have a state of superpositions of color arrange-ments, (cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:1)(cid:1) (cid:0)(cid:0)(cid:1)(cid:1) (cid:0)(cid:0)(cid:1)(cid:1) qgluonq gluon gluonR RRR G G B B In the transverse region away from the string, all the col-ors cancel and there is not much energy in the Coulomb-like quantum chromostatic field.We shall have a more general and precise discussion ofthese strings later on in the course, and will see if this isa useful viewpoint of a string. (Such pictures have alsobeen discussed by Greenstreet. )
14. INTERLUDE (1-5-88) [From the audio tape. RPF begins with some remarksabout the research project.]
First, last term we addedan assignment to write something up; I’ve given some ofthem back already and here are the rest. I’ve written alot on them but it doesn’t mean that I’ve corrected every-thing. Mainly I’m trying to suggest some ways of lookingat things, which is not meant as a criticism necessarily;well it could be a criticism, but very often a great deal ofwhat I’m writing is not a criticism. I think this is a veryuseful way of teaching, and I would like to do the samething this term, at the end of this term, having anotherpaper to hand in on the first day of exam week at the endof the term. This seems to be a slip since the equations imply a linear poten-tial between charges, leading to a constant force, as is usuallyunderstood. It can be derived using dimensional analysis, byevaluating the Green’s function ∇ − at large separations. RPF means Greenberg, who wrote several papers about quarkconfinement. And this paper this term will have to be better thanthe paper of the last term. You remember that I decidedon that rather late; I only gave you a few weeks. So Isaid it doesn’t have to be too good. So everybody passed.On some of the papers I was rather surprised and disap-pointed and thought that you were unable or didn’t havethe time to do your subject justice, and I made some re-marks that you should talk to me about it or somethinglike that; that’s only one or two so don’t worry. Some-times a person isn’t prepared or doesn’t have the back-ground or the focus needed, and it’s worth finding outearlier rather than later.The other possibility is that you’ve got some kind ofa block somewhere, a misunderstanding of what it’s allabout. And it’s very surprising [noise] . . . it looks verysimple to me, but it’s only a block. I know that becauseI’ve had that experience once myself, and I understand it.I also used to do tutoring, and I’ve discovered many blockslike this . . . and straightened the person out who had thewrong idea. I’ll give you an example. My own block oc-curred when I studied solid geometry in high schoool. Iwas pretty good at math as you might guess, and I thoughtI’d have a good time in solid geometry. The class startedand I didn’t understand anything. The guy would askquestions, and I couldn’t figure out what the hell the an-swer was, and moreover something that I thought wasusually pretty dopey would come out as the right answer.It would proceed like this and I was getting foggier andfoggier for about a week. All of a sudden, thank God ,it suddenly hit me. I understood. They were drawingthings like this, parallelograms, overlapping. Then some-times lines that would come out [connecting the paral-lelograms] ; the question was were they perpendicular toanother line like that . . . [laughter] . I was looking at everyone of the diagrams as though it were a plane diagram,but it was a three-dimensional picture. So when I finallyfigure it out the teacher told me “that’s why we call itsolid geometry, you idiot!” [laughter] . I couldn’t figureout the theorems and equations and relations because Iwas looking at the diagrams flat. That’s a block.Another example was a student that a lady came to meand told me that her son was very good in math but wasneeding some help in geometry, he was in high school.So the first thing I did was to ask him some questions,like if this is a rectangle and this is the diagonal, howfar is it from here to here? It’s the same, he says; andthen a few more questions, showing that he had a first-rate intuition about geometry; there was no problem aboutunderstanding what geometry was about. But to makea long story short, he had a double block. One: theyhad in his course in high school that you write what theycall proofs. You write something and then you write thereason. Given: . . . and so on and so on. The way hethought was, and he didn’t understand, was how you knewwhat to write on each line. He thought there was somelogical process to know what to write on each line. And Ihad to explain to him, no, what you did was you first hadto figure out how to prove it. And then after you figured out how to prove it you wrote the proof down. Okay, thatwas stupid, but that’s the way they did it.Another block was, he didn’t realize the rule that thetheorems that you were allowed to use in proving, thethings you could write on the right-hand side as rea-sons, always had to be statements that were earlier thanthe thing you were trying to prove. For example, no,you can’t use the Pythagorean theorem here because youhaven’t gotten there yet. Sounds dumb, but it’s just ar-tificial conventions of human beings [long section whereRPF is speaking away from the microphone] .Same thing happens in algebra, where people have greatdifficulty because it’s not realized and it’s not explained tothem that x is used in two ways to represent a number;when you have a problem like this [writing on board] .The problem is x is some special number and you haveto find it, and when you have a statement like this [morewriting] that’s an entirely different use; there it means it’strue no matter what x is.So these blocks may be the cause of some . . . if youlook at your papers and find some remark . . . let’s talkabout it . . . a chance to maybe figure out if you have somedifficulty. Alright? Most of you have no such difficulty.Is everything alright? Do you want me to give moreproblems during the year, or is it okay if we do these finalexam papers, because they’re very useful, I think . . . If youhave some objections, come and tell me because I knowit’s been . . .
15. SCALE DEPENDENCE (1-5-88)
So far in this course we have emphasized perturba-tion theory as the main tool for making theoretical pre-dictions, but it is important to keep in mind that thepath integral is not limited to this treatment. For exam-ple, lattice gauge theory, which we will discuss in somedetail, is a way of computing observables nonperturba-tively. And I will bring up some other possible ideas forgoing beyond perturbation theory. That will occupy usin the second part of this term. For the first part, we willcontinue to explore aspects of perturbation theory.I would like to spend some time in this lecture on thetopic of running coupling constants, which must be han-dled carefully in order to avoid confusion. In fact, there isquite a great potential for confusion in this subject, andan apparent complexity, because of the lack of agreementabout the best conventions for carrying out the renormal-ization of the couplings, and also some misguided sug-gestions that the running coupling should be defined interms of some particular processes.
But before we discuss the running, we should try tounderstand what are the most efficient ways of experi-mentally determining the values of the couplings in the6QCD Lagrangian,1 g tr F F + (cid:88) i ¯ q i ( i /∂ + m i ) q i (15.1)where I have summed over the quark flavors, i = u, d, s, c, b (and presumably t , although it has not yetbeen observed). The subscript 0 means that these arethe bare couplings, that would coincide with the physicalvalues if we were to make predictions only at tree level,but which of course will differ once we start to includeloops. So we have at least six Lagrangian parameters, g , m u , m d , m s , m c , m b (15.2)In principle, we could determine all of these by mea-suring six independent observables, since generically eachone would constrain different combinations of the param-eters. But in practice we usually focus on one thing at atime, and try to choose an observable that is most sensi-tive to the quantity of interest. For example, to compute m b , we could initially estimate it as approximately halfthe mass of the Υ meson (bound state of b ¯ b ), since weknow that Υ gets most of its mass from the quarks andnot the gluons. This would give m b ∼ m c as being about half of the mass ofthe ψ . Of course the Υ and ψ do get some of their massfrom the gluons, and so we could try to improve on theseestimates by doing a bound-state calculation to take thatinto account—but that would depend on other parame-ters, namely g .To determine g , we could try to compute the massof a particle like the proton, which is believed to getmost of its mass from the gluons, since m u , m d (cid:28) m p .However we then encounter the problem that we don’tknow how to compute m p ; it is very far from being aperturbative calculation, and we would have to rely onthe lattice, which for the present looks hopeless, althoughmaybe someday it will be feasible.Instead, to measure g it is more practical to look at aprocess involving higher energies, much higher than thequark masses, so that the measurable quantity dependsvery weakly on the m i ; in this way we can disentanglethe dependences and determine g independently. At ar-bitrarily high energy E (cid:29) m i this should become an in-creasingly good approximation since the amplitudes willdepend only on the ratio m i /E .Let me start with a process that at first looks like itwill not help us, since it does not seem to depend on theQCD coupling g at all: electroproduction of quarks: e − e + qq q Q e (15.3)Of course the real physical process does depend on g because what we actually observe is not the quarks, but rather e + e − → hadrons, and the hadronization processdepends strongly on g . But this is not the kind of g dependence we are interested in because (as I will ar-gue later in the course) it is more characteristic of thelow-energy scales of the hadron masses and it is not aperturbatively computable process. Instead, we want toignore the details of hadronization and pretend that thequarks are produced freely. This is actually a much bet-ter approximation than one might at first imagine, for thesimple reason that hadronization mainly occurs after thequarks have been produced. Therefore it cannot affectthe production cross section (at least at energies thatare not too close to resonances), which is the quantitythat we can compute perturbatively, and which we canmeasure more or less cleanly despite the complications ofhadronization.Hence the electroproduction process does not dependon g at all, at the leading order in couplings; I am go-ing to take advantage of a subleading effect to suggesta way of measuring g . But first let’s look at the lead-ing contribution, in the high-energy limit where we canignore all the masses. We see that the calculation forproducing quarks is hardly different from that for pro-ducing µ + µ − . The only differences are the charges of thequarks, and the fact that quarks come in three colors.Consider e + e − → u ¯ u ; the charge of u is +2 /
3, so we canrelate the quark and lepton cross sections as σ u ¯ u = 3 (cid:18) (cid:19) σ µ + µ − The factor of 3 is commonly understood as coming fromthe number of quark colors: you can produce either a redquark, or a blue or a green. This is actually a cheat: inreality you can only produce a color-singlet combination1 √ (cid:0) R ¯ R + B ¯ B + G ¯ G (cid:1) ≡ a √ √ a (15.4)In this formula, a represents the amplitude for producinga quark of a definite color, that we assumed could occurwhen we multiplied its cross section by 3. We see thatthe amplitude of the color singlet state that is actuallyproduced is bigger by a factor of √
3, and so this is theproper explanation of the factor of 3 in the cross section.But either way of thinking about it gives the right answer.Similarly for producing d ¯ d , we have σ d ¯ d = 3 (cid:18) − (cid:19) σ µ + µ − , and we can continue this to include all the higher massquarks if the center-of-mass energy is sufficient to pro-duce them.The interesting quantity to measure and compare topredictions is the ratio R = σ total σ µ + µ − (15.5)7whose contributions, at high enough Q , we can read offfrom equations like (15.4,15.5). The experimental datalook like this: µ + µ − σσ total 4/33.72 M M ψ
40 GeV ER =
At each quark threshold, a resonance for the correspond-ing bound state occurs, giving rise to the peaks and dips.These are the hadronization complications that we wouldlike to avoid. In between the thresholds, we observe thesimple behavior that we can understand within our ap-proximations: R is just flat, and it is counting the num-ber of quarks that can be produced, weighted by theirelectric charges (squared).Now this is all very beautiful, but it doesn’t yet helpus to determine the QCD coupling. For that, we shouldconsider a higher-order process, where a gluon gets radi-ated from one of the quarks, or exchanged: e − e + e − e + qq qq g γγ g (15.6)One can show that at high energies, these processes arealso independent of the quark masses, and they lead to amultiplicative correction of the leading order prediction R , R = R (cid:16) α g π + . . . (cid:17) (15.7)where α g = g / π . Then if we can measure R wellenough, the deviation between R and the lowest orderprediction R will determine α g . So this is a possible wayof measuring g . It is not quite as direct as we would like,since it depends on a relatively small correction, of ordera few percent, to the basic quantity that is insensitive to g . Can we do better?A more direct approach would be to observe the gluonthat is emitted. That of couse is impossible, just like forthe quarks, since they are all colored objects. Instead,we observe the jets of hadronized particles emerging fromthose primary particles, (15.8) where I have drawn the momentum vectors of the in-dividual hadrons and enclosed them with an envelopeto indicate the jet. At high energies, the jets are well-defined, with lengths (in momentum space) of order thecenter-of-mass energy E , and widths of order the QCDscale ∼ α g , so it isa more senistive determination than using the ratio R .In the following discussion, I will continue to focus onhigh-energy processes that allow us to neglect the quarkmasses and focus on the coupling g . You may have noticed something peculiar about thediagrams I drew in (15.6): they mix up different ordersof perturbation theory. And of course there are otherloop diagrams not shown, like the self-energy correctionto the quark from a gluon loop. As you probably knowfrom a previous course on electrodynamics, the tree dia-gram with the gluon emission cannot properly be sepa-rated from this self-energy correction: the infrared diver-gence of the soft gluon emission is canceled by a similardivergence in the self-energy diagram, when it interfereswith the tree diagram. So that is one reason we cannotavoid considering the loops. But a more important one,for the present discussion, is that the loops will modifyour predictions at high energies, which I have stressed isthe best regime for comparing predictions to experiment.And as you know, these loop contributions are prob-lematic because they diverge at high virtual momenta,so we need to introduce the procedure for cutting off thedivergences. A typical kind of integral that we have to8deal with (continuing in our approximation of neglectingmasses) is (cid:90) d kk ( p − k ) , (15.9)which by counting powers of momenta in the numeratorand denominator is logarithmically divergent. A naiveway of regularizing this, which later we will see does notquite work in the case of gauge theories, is to modify thepropagators by taking1 p → lim Λ →∞ p − p − Λ (15.10)This will render (15.9) finite, in the intermediate stepbefore taking the limit Λ → ∞ , and yields a divergentterm going as ln Λ. This divergence can be absorbed byredefining the coupling g , before taking the limit.Let’s recall how this works in electrodynamics, forelectron-electron scattering. First consider the low-energy, large distance limit, corresponding to scatteringin the Coulomb potential e /r . The tree-level scatteringamplitude is Q ∼ πe Q (15.11)I want to consider the Q → e becomes the familiar constant value e exp ∼ = (cid:112) / ∼ = 0 . This is the physical value,which is not the same as the bare value e that is neededto cancel the divergences from loops, such as (15.12)One can show that the divergences from this loop (plusthe others not shown) can be canceled by defining e suchthat 1 e = 1 e (Λ) + 13 π ln Λ Q (15.13)When we originally wrote the Lagrangian for QED, wethought that e was a parameter very close in value tothe measured charge; now we see that it is not a fixedparameter, but rather a function of the cutoff. In theend nothing can depend on Λ, since we are taking thelimit Λ → ∞ . What about the dependence on Q ? Wethought that e is supposed to go to 0 .
09 as Q → Notice RPF’s unconventional normalization of the coupling, thatwe have seen before and which he sometimes abandons in favorof the usual one in later lectures.
In reality, we should have something like Q + m e in thelog; then this makes sense as Q →
0. The expression(15.13) is valid for Q (cid:29) m e .Of course there is one peculiarity in QED: we can’t takethe Λ → ∞ limit! There is some scale Λ ∼ m e e π × / at which e diverges, and we cannot make sense out of thetheory beyond this point since e becomes negative. Thisis the Landau pole. In practice, it is such a high scale, fargreater than the mass of the universe, that we don’t care:the residual dependence of amplitudes on Λ is negligible.It is more of an aesthetic shortcoming. But a very nicefeature of QCD is that the analogous correction to 1 /g has the opposite sign, so there is never any Landau pole;instead g goes to zero as Λ → ∞ .In this course I will discuss three different methods ofregularization, which are summarized as follows:1 . Pauli-Villars: 1 k → k − k − Λ As I mentioned, this is not quite right as written; we needto be more careful to avoid spoiling gauge invariance; wewill come to that later.2 . Dimensional regularization: α g becomes dimensionful when d (cid:54) = 4; α g → α g µ d − where we analytically continue amplitudes in the number d of spacetime dimensions, and3 . Lattice regularization:has a dimensionful parameter: a ↔ / Λwhere we have approximated spacetime as a discrete lat-tice, with sites separated by the lattice spacing a , thatplays the role of 1 / Λ relative to the first method. Anyof these approaches leads to equivalent results when wetake the continuum limits, Λ → ∞ , (cid:15) → a →
0. Eachone of them has its own different expression for how thebare couplings depend on the cutoff, but the physical pre-dictions that they make to a given order in perturbationtheory are the same. To fully define the theory we willalso have to discuss gauge fixing.Let me continue to work in the Λ cutoff scheme fornow; we will come back to the other regulator methodsin later lectures. One can show that there is a generalform for the Λ dependence of the bare coupling, whichwe will derive later on,1 g (Λ) = β ln Λ M + β β ln (cid:18) ln Λ M (cid:19) + c + (cid:18) c ln Λ /M + . . . (cid:19) . (15.14)Here · · · denote terms falling with Λ even faster than1 / ln Λ , that do not concern us in the large-Λ limit. The9interesting observation is that the constants β and β arethe same in different regularization methods, if we makethe appropriate identifications ln Λ → /(cid:15) or ln Λ → ln(1 /a ) to translate between them. There is an arbitraryscale M appearing, known as the renormalization scale.Notice that any change M → e − γ M can be absorbedinto a change of the constants c and c (for Λ (cid:29) M ):1 g → g + β γ (cid:124)(cid:123)(cid:122)(cid:125) + β γβ (cid:124)(cid:123)(cid:122)(cid:125) ln Λ M + O (cid:18) ln − Λ M (cid:19) δc δc (15.15)Thus the c i are not universal at all, and they also dependon the method of regularization. Their values get fixed,within the given method, by comparing some predictionin which g appears to its corresponding measured value.The essential fact is that it should not matter which ob-servable we choose; any quantity that is sufficiently sen-sitive to g will suffice. And once that is done, we canuse the same formula to predict other observables, anduse these to test the theory.You may wonder why I bother to write the c term atall, since it is irrelevant as we take the continuum limit.There may be some practical situations where we are notable to take this limit, notably the lattice, where it iscomputationally prohibitive to do so. The best we cando there is to take a to some small value. To get the bestdescription of the data at two different values of a , wewould need to keep the c term, so as to properly comparepredictions at different values of a , to see whether wethink a is small enough to trust our predictions.It is worth remarking on the observation that for QCD, g → → ∞ is the same as the limit M → Q , the amplitude will havefactors going as ln Λ /Q . When we combine these withthe coupling g , the cutoff dependence disappears, andthe log becomes ln M /Q , which diverges as M → Q is larger than the characteristic scale of QCD, λ (cid:39) (1 GeV) , where perturbation theory startsto break down.Related to this, another possible source of confusionis the arbitrariness of the choice of M in (15.14). Ishowed that M is perfectly arbitrary since we can ab-sorb any change of M into a redefinition of the c i con-stants. That is true, but in practice, for a given applica-tion, some choices of M will be more advantageous thanothers. Generally, if we are going to use an observationat a particular scale Q to fix the value of g , then itmakes sense to choose M ∼ Q , to get rid of large loga-rithms ln M /Q , that would also appear in the higher-loop diagrams that we are not including in our calcula-tion. This choice reduces the error coming from these higher-order diagrams. This argument also shows whyit would not make sense to try to take M all the wayto zero: perturbation theory in QCD is breaking downat scales Q (cid:46) λ P . So we should not take M < λ P .It is sometimes convenient to define M by imposing achoice like c = 0 on the arbitrary constant, to make eq.(15.14) look nicer. Then M takes on a physical signifi-cance (within a particular regularization method), thatcan be called the scale where QCD is becoming nonper-turbative. In this convention, M = λ P , which has beendetermined to be around 200 MeV.Because the coupling only runs logarithmically, it takesconsiderable experimental effort to reach energies whereconvergence of the perturbation expansion improves dra-matically. And on the lattice, it takes enormous numer-ical effort. Consider that on a 4D lattice, by cutting thelattice spacing in half, we compound the computationalproblem by a factor of 2 = 16, yet the coupling has de-creased by a factor of only 1 / ln 4 ∼ = 0 .
7. Thus it seemsnearly hopeless to achieve quantitative accuracy on thelattice, given the limitations of computers, and it wouldbe nice to come up with another nonperturbative methodthat is not just brute force.
16. THE RENORMALIZATION GROUP (1-7-88)
Last time we saw that the bare coupling could be ex-pressed in a way that I will rewrite in the form1 g (Λ) = β ln Λ λ P + β β ln (cid:18) ln Λ λ P (cid:19) + const . + O (cid:18) /λ P ) (cid:19) (16.1)where λ P is short for a scale I am calling “ λ Politzer ,” andthe last term, that vanishes as the cutoff goes to infin-ity, is scheme-dependent. If we choose g in this way, fora given method of regularization, physics is unchangedfrom scheme to scheme and as the cutoff varies. If werewrite the second term on the right-hand side of (16.1)as ( β /β ) ln (cid:0) / ( β g ) (cid:1) , the terms that vanish as Λ → ∞ are cleaner and have no ln ln coefficients in the numera-tors of 1 / ln Λ . Now we can differentiate, d (cid:16) g (Λ) (cid:17) d (cid:16) ln Λ λ P (cid:17) = β + β g + β g + . . . (16.2)This is closely related to the beta function, β ( g ) = − dg d ln Λ = β g + β g + . . . (16.3) This statement is justified in eq. (16.4). β and β are scheme-independent, but β , β , . . . are scheme-dependent and they affect only the O (1 / ln Λ ) terms. As I explained in the last lecture, wemust choose the constant in (16.1) in order to define λ P ;so we made a specific choice. We define the constant tobe zero in the MS scheme of dimensional regularization,and we can do likewise in Pauli-Villars. This defines λ P .It is an arbitrary choice which depends on the regulariza-tion method. Making this choice, we can write an exactformula 1 g (Λ) = β ln Λ λ P + β β ln (cid:18) β g (cid:19) (16.4)which is an implicit definition of g (Λ). It satisfies thedifferential equation d (cid:16) g (Λ) (cid:17) d (cid:16) ln Λ λ P (cid:17) = β + β g + β β g + . . . (16.5)= β − β β g (cid:32) β = 11 − n f β = 102 − n f (cid:33) where n f is the number of quark flavors. Notice that β and higher coefficients are all determined by β and β in this definition of g (Λ). For large Λ, the differentchoices one could make for these higher coefficients arenot important.As I mentioned in a previous lecture, in QED we havea problem when we try to do the analogous thing. Let e denote the physical, observed value of the coupling.Then 1 e = 1 e (Λ) + 13 π ln (cid:18) Λ m (cid:19) (16.6)where we have a plus sign instead of minus between thetwo terms. Hence1 e (Λ) = 1 e − π ln (cid:18) Λ m (cid:19) (16.7)As Λ gets large, 1 /e (Λ) becomes negative, and we loseunitarity. This means we can’t push QED to arbitrarilyhigh energies, at least not greater thanΛ m ∼ e . (16.8)Grand unification is one solution to this problem. In my notes I have parenthetically, “(Even just electroweak the-ory seems to fix it.)” perhaps referring to the modification of theQED beta function by the other standard model interactions. g Recall that we could measure g through the ratio R = σ e + e − → had . σ e + e − → µ + µ − ; ( R −
1) = g π + a g + · · · (16.9)at high energy. We wanted something that didn’t dependon the mass of the quarks, another example being qq scat-tering at high energies. In this way we could concentrateon determining g . We would also like our observable tobe dimensionless, such as Q σ since the cross section σ has dimensions of (length) − . To avoid extraneous de-pendences, we can imagine keeping the scattering anglesfixed as we increase the energy scale.In general, our calculation of this quantity will dependon g , Λ and Q ; call it D theory ( g , Λ , Q ). At first weconsider g to be independent of Λ—it’s just a parame-ter. Now we want the physically measured value of theobservable D phys . = D ( g (Λ) , Λ , Q ) (16.10)to be independent of Λ as Λ → ∞ (and of course itmust have a well-defined limiting value). Recall that thecoupling is dimensionless, which you can see from theaction, S = (cid:90) d x g F µν F µν (16.11)= (cid:90) d x g (cid:0) ∂ µ A ν − ∂ ν A µ − A × µ A ν (cid:1) and remembering that [ A ] = 1 /L hence [ F ] = 1 /L .So [ F ] = 1 /L and since the action is dimensionless,[ g ] = 1: the coupling is dimensionless. Therefore D theory ( g , Λ , Q ) can only depend on the dimension-less ratio Λ /Q if we keep g fixed. Hence we can write D theory in the form D theory ( g , ln Λ /Q ). Similarly, ourprediction for the physically observed value must take theform D phys . = D (cid:0) g (Λ) , ln Λ /Q (cid:1) (16.12)and it cannot depend on Λ. This tells us how it dependson Q for large Q .Suppose we ignore the β term in g (Λ), and the g term in ( R − D theory (cid:18) g , ln Λ Q (cid:19) = g π (16.13) D phys . (cid:18) g (Λ) , ln Λ Q (cid:19) = g (Λ) π = 1 πβ ln Λ λ P The subscript on g is to distinguish the coupling g that isconsidered to be independent of Λ from g (Λ). D theory (cid:18) g , ln Λ Q (cid:19) = g π + g (cid:18) β ln Λ Q + a (cid:19) D phys . (cid:18) g (Λ) , ln Λ Q (cid:19) = g (Λ) π (16.14)= 1 πβ (cid:16) ln Λ λ P − ln Λ Q (cid:17) = 1 πβ ln Λ λ P + ln Λ /Q πβ ln Λ /λ p + · · · where · · · represents terms of order g . The next orderterms in D theory must be g π (cid:18) β ln Λ Q + 2 a ln Λ Q + a (cid:19) Terms of the form ( β ln Λ /Q ) n are the leading logs.The fact that you can sum them is all due to demandingthat D phys . does not depend on Λ . If we do the sumthen we get 1 πβ ln Q λ P + a π (cid:16) β ln Q λ P (cid:17) + · · · (16.15)where the first term comes from the leading logs, thesecond from the next-to-leading logs, and so on. Now wecould write( R −
1) = α ( Q ) π + a π α ( Q ) + · · · (16.16)where α ( Q ) = 1 β ln Q λ P (16.17)This shows that α ( Q ) operates like a coupling constantthat depends on energy. We did this by neglecting β . Ifwe keep β , we get1 α ( Q ) = β ln Q λ P + β β ln (cid:18) β α ( Q ) (cid:19) (16.18)Evaluating it at different energies Q in GeV, using λ P =0 . Q n f α .
43 0 .
26 0 .
23 0 .
19 0 .
16 0 .
15 0 .
14 0 .
12 0 . ± . This table, which RPF apparently computed himself, was notgiven in the lecture, but it appears in his private notes for thislecture.
Now we are ready to derive the renormalization groupequations, using the fact that physical quantities cannotdepend on Λ. dD phys . /d Λ = 0 implies that ∂D∂g (cid:12)(cid:12)(cid:12)(cid:12) g = g (Λ) × dg (Λ) d ln(Λ /λ P ) − ∂D∂ ln Q = 0 (16.19)where dg d ln Λ = − g β ( g ) (16.20)= − (cid:2) β g (Λ) + β g (Λ) + β g (Λ) · · · (cid:3) Suppose we have worked out the theoretically predictedvalue D th (cid:18) g , ln Λ Q (cid:19) = b + b g + b g + b g + · · · (16.21)where the b m may depend on t ≡ ln Λ /Q . Let b (cid:48) m = db m /dt . Then the RG equation says that b (cid:48) + b (cid:48) g + b (cid:48) g + b (cid:48) g = (cid:2) β g (Λ) + β g (Λ) + β g (Λ) · · · (cid:3) × (cid:2) b + 2 b g + 3 b g · · · (cid:3) (16.22)This tells us that b (cid:48) = b (cid:48) = 0 ; (16.23)hence b and b are constants, that we can calculate the-oretically; call them c and c . Moreover b (cid:48) = c β = ⇒ b = c β ln Λ Q + c b (cid:48) = ( b β + 2 b β ) = c β + 2 c β ln Λ Q + 2 c β = ⇒ b = c β ln Λ Q + (2 c β + c β ) ln Λ Q + c (16.24) etc. Then D phys . (cid:12)(cid:12) g (Λ) = c + c g (Λ) + (cid:18) c β ln Λ Q + c (cid:19) g (Λ)+ (cid:18) c β ln Λ Q + (2 c β + c β ) ln Λ Q + c (cid:19) g (Λ)+ · · · (16.25)At each order n in perturbation theory, a new constantterm c n arises. But this new constant is typically lessimportant than the preceding terms appearing in the co-efficient of g n , that come with higher powers of logs. Inother words, c determines all the leading logs, c and c all the next-to-leading logs, etc . We can simplify (16.25)2because we know it doesn’t depend on Λ. Organizing itin terms of the c i coefficients we get D phys . = c + c α ( Q ) + c α ( Q ) + c α ( Q ) + · · · (16.26)where α ( Q ) satisfies dα ( Q ) d ln Q = β α + β α + β α + · · · (16.27)If one could measure D at such a high energy that α ( Q )was small and the series converged rapidly, it would pro-vide way to determine λ P ∼
200 MeV.
17. RENORMALIZATION: APPLICATIONS(1-12-88)
Renormalization is a confusing subject, and one factorcontributing to the confusion is the proliferation of dif-ferent conventions. I am guilty of this by my preferrednormalization of the gauge coupling, which is not thesame as that of the rest of the world. For your conve-nience, let me translate some previous results into themore conventional form, where the coupling and its as-sociated fine-structure constant are related as α = g π (17.1)Then the running of the renormalized coupling, (16.18),takes the form4 πα ( Q ) = β ln Q λ P + β β ln 4 πβ α ( Q ) (17.2)Now let’s review what we learned in the previous lec-ture, concerning the utility of this expression, that sumsup the leading logarithmic dependences in the pertur-bation expansion. Namely, we can take an amplitudecomputed at tree level, and replace its α dependence byeq. (17.2), to resum the most important subclass of loopcontributions to all orders, which improves the tree-levelprediction. Furthermore, we can extend this to sublead-ing contributions. Suppose we computed an amplitudeat one-loop order and found a result going as M = α + α (cid:18) ln Λ Q + c (cid:19) (17.3)The term with ln Q is already contributing to the lead-ing logs that we obtain by replacing α → α ( Q ) in thelowest order contribution. Thus the correct way to in-corporate the next-to-leading logs is to replace M → α ( Q ) + c α ( Q ) (17.4)I have avoided some of the complications of renormal-ization so far by only discussing gauge invariant, physical quantities. We could also consider the renormalization ofmore general quantities like Green’s functions (cid:104) A ( x ) · · · A ( x n ) (cid:105) . (17.5)Then it is not sufficient to talk about only the renormal-ization of the couplings, but also the wave function renor-malization, that contributes to anomalous dimensions inthe scaling of such a Green’s function. I am not going todiscuss these kinds of issues, but there are many refer-ences that do so, for example Renormalization by JohnCollins, Cambridge University Press, 1984. If you areonly interested in physical, measurable quantities, thesecomplications can be avoided.One further point pertaining to the previous lecture isabout the dependence of the beta function on the numberof flavors. We have noticed that one of the nice featuresof QCD is its good behavior in the ultraviolet. This as-sumes there aren’t too many flavors of quarks. There arefive or six that we already know about, but there couldbe more—nobody knows why there should only be threefamilies. But probably there are not 17 flavors of quarks,which is the critical number that would ruin the goodUV behavior of QCD. Before embarking on explicit calculations of loop dia-grams, it is enlightening to understand the general struc-ture of divergences of the theory, and you are probablyalready familiar with this, but I would like to review itnevertheless. Consider some rather complicated diagramlike ∼ (cid:90) d k d p ( p , k , p · k + · · · ) × f ( p, k, · · · )(17.6)We count 7 propagators, schematically indicated by thedenominator of (17.5), 8 powers of momentum from theintegration measure, and a numerator f that goes like(momentum) from rationalizing the fermion propaga-tors and counting the 3-gluon interactions. So accordingto power counting, this goes as (momentum) − and istherefore superficially convergent. We call this exponent N d , the superficial degree of divergence. It tells us thata given diagram generically behaves as N d = 0 , log divergence N d = 2 , quadratic divergence N d < , converges (17.7)We call it “superficial” because it is possible to constructexceptional cases in which there is a divergence even if3 N d <
0. This would be the case if the integrals somehowfactorized into a product of one that was highly conver-gent times another that diverged. More typically how-ever, we will see that this N d often over estimates thedegree of divergence, as a consequence of gauge symme-try.Now it would be rather tedious to have to do this kindof counting for every possible diagram that may arise,but fortunately we don’t have to. There is a beautifultopological relation that does it for us, solely in terms ofthe numbers of external lines of different kinds, indepen-dently of how complicated the diagram is, such as thenumber of loops. The relation is easiest to prove for avacuum diagram with no external legs. It is a fact fromtopology that such a diagram satisfies N d = 4. Imagine that we cut an internal quark line in aloop in such a diagram, to add two external quarks. Wethereby remove one integration over loop momenta andone fermion propagator, which reduces N d by 3. On theother hand, imagine adding an external gluon to someline on the diagram. It creates an extra fermion propa-gator, or else an extra gluon propagator with a couplingproportional to momentum in the numerator. Or we con-vert a triple-gluon vertex with dimension 1 into a dimen-sionless 4-gluon vertex. In any case, we reduce N d by 1.Therefore it must be that N d = 4 − N g − N q , (17.9)where N g ( N q ) is the number of external gluon (quark)lines.Another way of deriving (17.9) is to use dimen-sional analysis for the general case. Let’s illustratethis for some diagram with an arbitrary number ofexternal particles, representing a process in whichparticle 1 decays into N − (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) N−1 final particles12 N3
Rate of decays of 1 into N − d Γ = 12 E | T | N (cid:89) i =2 (2 π ) δ ( p i − m i ) d p i (2 π ) × (2 π ) δ (4) (cid:32) p − N (cid:88) i =2 p i (cid:33) (17.10)I have drawn them as though they are all external gaugebosons, but we will also discuss the case when some ofthem are quarks. The shaded blob could contain anynumber of loops and internal lines; it doesn’t matter howcomplicated the diagram is. Next to it I gave the formulafor the differential decay rate, that we know has dimen-sions of mass or energy. Therefore we can determine the dimensions of the amplitude | T | that corresponds to thediagram. By moving the factor 1 /E to the other side ofthe equation, we get[energy ] = [ | T | ] [energy] N − [energy] − hence [ | T | ] = [ E ] − N (17.11)where N is the total number of lines coming out, regard-less of whether they are bosons or fermions.If the amplitude T had the same dimensionality asthe amputated loop diagram, we would conclude that N d = 4 − N , since the coupling g is dimensionless, andso all dimensional factors are associated with momenta.In the case where all external particles were bosons, thiswould be the right answer. If some of them are fermions,it is not right because the external fermions have spinorsassociated with them, that have dimensions, and thismakes the mass dimension of the diagram differ from N d . But that is easy to correct for. You recall thatsumming the exterior product of two spinors over theirpolarizations gives the projection operator (cid:80) u ¯ u = /p + m ;therefore each spinor has dimension 1 /
2. This means wehave to correct the previous result for N d to read N d = 4 − N g − N q − N q (17.12)in agreement with (17.9).I made a remark above, about gauge invariance, orpossibly other symmetries, causing some diagrams to bemore convergent than predicted by our formula for N d .The most famous example is the case of N g = 2, whichare diagrams contributing to the gluon vacuum polariza-tion. Gauge invariance tells us that they should dependon the external gluon momentum q as q δ µν − q µ q ν (17.13)This means that two factors of momentum that wecounted toward the degree of divergence are not loopmomenta; instead they can be brought outside of theintegral, making it more convergent that we naively esti-mated. To remind ourselves of this possibility, we couldadd an extra term −P , so that N d = 4 − N g − N q − P (17.14)where P is the known power of the coefficient in front ofthe integral. Exercise.
Show that the power of couplings g P of anarbitrary diagram is given by P = 2 × ( N g + N q − g for the 3-particle vertices and g for the4-gluon vertex.A fortunate consequence of the formula (17.14) is thatonly a finite number of the different kinds of diagramsare divergent, when we classify them by their numbers of4 N g N q N d P external lines. Therefore we can make a table to illustrateall the possibilities: Table I: Superficial degrees of divergence, N d (before ac-counting for P ), for graphs with N q external quarks and N g external gluons. P is the power of external momentum fac-tors. The tadpole diagrams, with N g = 1, provide anotherexample of our statement that symmetries can make adiagram more convergent than power counting wouldsuggest, ∼ (cid:104) A µ (cid:105) , (17.16)This diagram is naively divergent with N d = 1, but infact it vanishes. One can think of it as an expectationvalue of the gluon field. Such a thing, if nonzero, wouldspoil Lorentz invariance, as well as gauge invariance. Andit would break discrete symmetries like C and P.We already discussed the vacuum polarization dia-gram, the fact that it is proportional to ∼ δ µν q − q µ q ν (17.17)We will derive this result later on. It is an example ofhow one must be careful about regulating the divergencesfrom the loops in a gauge invariant way. The behav-ior (17.17) reduces its N d by 2, so that instead of beingquadratically divergent, it is only logarithmically diver-gent, but if the regularization method failed to respectgauge invariance, it would be afflicted with this more se-vere quadratic divergence.Moreover, symmetry prevents the diagram (17.18)from being linearly divergent as its N d = 1 would sug-gest. The only kind of Lorentz-invariant loop integrandone could write, consistent with this power counting,has the form p · X/ ( p ) , schematically, where p is theloop momentum and X represents an external momen-tum or polarization vector. But p µ / ( p ) is odd under p µ → − p µ , so its integral must vanish, as long as the reg-ularization procedure does not introduce any pathologythat would spoil this reasonable expectation. And we canalso argue that it must vanish by Lorentz symmetry, since (cid:82) d p p µ / ( p ) would define some preferred direction inspacetime if it were nonzero. Therefore the 3-gluon am-plitudes are also only logarithmically divergent. In factgauge invariance provides yet another reason this must beso: we know that the three-gluon interaction comes witha power of external momentum, from its Feynman rule,and this explains why the actual form of the integrandmust be q µ / ( p ) , giving N d = 0.There is one possible caveat to the gauge invarianceargument that should be kept in mind however. Thereis no guarantee that individual diagrams will be gaugeinvariant; only the sum of all diagrams contributing to agiven process at a given order must necessarily be gaugeinvariant. Before we embark on explicit calculations of loop dia-grams, I wanted to discuss the relative advantages ofsome choices of gauge relative to others, when it comes todefining the gluon propagator. In our earlier discussionof gauge fixing, lecture 12, we discussed a particular classof gauges η · A = 0, involving an arbitrary four vector η µ .This is actually not the most convenient one for doingperturbative calculations, even though it was conceptu-ally appealing. For one thing, it spoils Lorentz invariancetemporarily, although these terms must cancel out in theend.A simpler choice would be the Lorentz gauge ∂ µ A µ =0. Let us recall how the Faddeev-Popov procedure wouldwork in this case. The gauge-fixed path integral takesthe form Z = (cid:90) e i (cid:82) ( ∂ µ A ν − ∂ ν A µ − A × µ A ν ) ∆( A ) δ [ ∂ µ A µ ] D A µ , (17.19)omitting for simplicity the quarks. We rewrite the de-terminant ∆( A ) as a path integral over ghost fields. Wetake advantage of the fact that this determinant doesnot change at all if we impose a slightly different choiceof gauge, ∂ µ A µ = f ( x ), with an arbitrary function f .Therefore we are free to do a weighted average over thepath integral, Z → (cid:90) D f e i (cid:82) f d x Z (17.20)Then the delta functional gets rid of (cid:82) D f , and we are leftwith a new term ( ∂ µ A µ ) in the Lagrangian, that allowsthe propagator to be defined. To see this, consider themodified equation of motion for the gauge field, includinga source term, (cid:3) A ν − (cid:24)(cid:24)(cid:24)(cid:24) ∂ ν ∂ µ A µ = S ν (17.21)5and notice that the crossed-out term is removed by thenew gauge-fixing term, and therefore we may invert the (cid:3) operator and solve for the gauge field, A ν = 1 (cid:3) S ν = 1 k S ν , (17.22)which shows that the propagator is simply δ µν /k in thisgauge. That obviously simplifies many perturbative com-putations, compared to the axial gauge propagator.Of course nothing obliges us to choose e i (cid:82) f d x as theweighting factor. One can equally well take e ξ i (cid:82) f d x with some arbitrary number ξ (cid:54) = 0. This yields a moregeneral class of covariant propagators of the form P µν = 1 k (cid:18) δ µν − η k µ k ν k (cid:19) (17.23)where η is related in some simple way to ξ . Exercise.
Find the relation between η and ξ .The choice η = 0 is known as Feynman gauge. Anothervery convenient choice is η = 1, the Landau gauge. It hasthe property of being transverse, k µ P µν ( k ) = 0, whichleads to some simplifications in loop calculations. Forexample it greatly reduces the number of diagrams in theprocess we are going to consider next. Nowadays youcan find computer programs that will do the symbolicalgebra for you, for computing such diagrams. Nev-ertheless, it is much easier to avoid mistakes if you canreduce the number of diagrams.
So far we have made numerous statements and exposi-tions of a rather general nature, without getting into thedetails of computing loop diagrams. I would now like togo over some of those details, just to illustrate the cal-culational techniques. The example I will consider is thescattering of two quarks. To lowest order, as we know, itlooks like qji lk ∼ g q λ Aji λ Alk In the lecture RPF says it reduces the number from 17 to 4.Perhaps he had in mind that the number of terms in the 3-gluonvertex is greatly reduced when taking only the transverse terms.Also the ghosts decouple in Landau gauge. Wolfram’s SMP (Symbolic Manipulation Program), the forerun-ner of Mathematica, was in use at Caltech at this time.
When we go to the next order, there are quite a fewdiagrams, including + ++ + + +− x + + permutationsghost m δ + + + + One simplification we can immediately make is to firstisolate the primitive divergences. For example, considerthe diagrams of the form + + . . .
The interesting part of this calculation is the loop, notthe external currents nor the gluon propagators that con-nect them to the loop. We might as well calculate thediagrams + + . . . by themselves, since it is trivial to take that result andadd to it all the tree-level parts such as the gluon prop-agators and the spinors for the external currents.Now to illustrate the techniques, I am just going tocompute the simplest and dullest of all of these, namelythe quark loop contribution to the vacuum polarization, q qp−qp b ν j a i µ Once you understand the principles, it is just a matter oftedious effort to compute the harder ones. By apply-ing the rules, we can write down the expression for thisdiagram, g (cid:18) − Tr (cid:90) d p (2 π ) b jν /p − /q − m λ j γ ν /p − m λ i γ µ a iµ (cid:19) (17.25) RPF mentions in the transcript that it is not so straightforwardto get a gauge invariant result for diagrams with gluon loops,using Pauli-Villars regularization. a iµ and b jν to be combined spin-colorpolarization vectors for the external gluons, but if youprefer you could replace these by spin polarization vec-tors (cid:15) µ and (cid:15) ν , and consider the gluon colors to be simply i and j . The trace here is a sum over all possible interme-diate states, both spins and colors. Therefore it is reallythe product of two traces, one for the Dirac matrices andone for the color matrices.Next, we should rationalize the quark propagators—multiply numerator and denominator by ( /p + m )—andcarry out the traces. This looks liketr (cid:20) /p − /q − m γ ν /p − m γ µ (cid:21)(cid:124) (cid:123)(cid:122) (cid:125) tr (cid:18) λ i λ j (cid:19)(cid:124) (cid:123)(cid:122) (cid:125) (17.26)tr (cid:20) ( /p − /q + m ) γ ν ( /p + m ) γ µ [( p − q ) − m ][ p − m ] (cid:21) δ ij
2= 2 (cid:104) ( p − q ) µ p ν + ( p − q ) ν p µ − δ µν ( p · ( p − q ) − m ) (cid:105) [( p − q ) − m ][ p − m ] δ ij Notice that the gluon color is conserved by the loop.Now we are left with the integral, that has exactly thesame form as in QED. There is a famous trick for combin-ing the denominators, that I adapted from Schwinger byeliminating a step from his Gaussian integral method, (cid:90) dx [ ax + b (1 − x )] = 1 ab (17.27)This allows us to combine the propagators into the form1( p − p · q + q − m )( p − m ) = (cid:90) dx p − p · qx + q x − m ] (17.28)so that the loop integral becomes (cid:90) dx (cid:90) d p ( p − p · qx + q x − m ) (17.29) × (cid:0) p µ p ν − p µ q ν − p ν q µ − δ µν ( p − p · q − m ) (cid:1) To further simplify it, we wish to complete the square inthe denominator, by shifting the integration variable by p → p + qx : (cid:90) dx (cid:90) d p ( p + q x (1 − x ) − m ) (17.30) × (cid:16) p µ p ν + ( p · q terms) + 2 q µ q ν x (1 − x ) − δ µν ( p − q x (1 − x ) − m + p · q terms ) (cid:17) Once it is in this form, it is not necessary to keep carefultrack of the p · q terms in the numerator, since they areodd in p and integrate to zero. In my notes at this point I have written “stolen from Schwingerby eliminating a step from his Gaussian integral method.”
Everything so far seems perfectly innocuous and stan-dard, but if we want to be careful, you will notice thatI have cheated. The integral is divergent, so how do weknow that the step of shifting the integration variable by p → p + qx is legitimate? It could conceivably change theresult in some unphysical way, unless we have carefullydefined what we mean by this integral. To be rigorous,we must specify exactly how we are going to regularizethe integral, to cut off the ultraviolet divergence.The method I want to use in this lecture is the histor-ical one, invented by Pauli and Villars. As I alluded ear-lier, it turns out to be too simplistic to change the prop-agator as in (15.10). This was attempted by some of theearly workers in the field, and it was found to spoil gaugeinvariance. Instead, one needs to apply this prescriptionto the whole amplitude . The consistent way is to replacethe quark mass in the denominators by m → m + Λ ,and subtract the resulting expression from the originalamplitude. The new integrand obtained in this way hasgood behavior in the ultraviolet, and so the procedure ofshifting the integration variable is perfectly consistent,and moreover it preserves the gauge invariance, as wewill see. And it is also consistent with our assumptionthat integrals like (cid:90) d p p µ p − m = 0 (17.31)should vanish, with the understanding that this is nowa shorthand for the fully regulated expression, where wehave subtracted the corresponding term with m → m +Λ .The statement (17.31) looks trivial, but we can use itto derive a more interesting result, now that we are confi-dent that shifts in the integration variable are legitimate.By shifting p → p − a , we obtain (cid:90) d p p µ ( p − a ) − m = a µ (cid:90) d p ( p − a ) − m And then by differentiating with respect to a ν and setting a µ = 0, we get the useful identity (cid:90) d p δ µν ( p − m ) − p µ p ν ( p − m ) = 0 (17.32)Incidentally, this could also be obtained more directly,using (cid:90) d p ∂∂p ν p µ p − m = 0which should be true for the integral of the derivative ofanything. We could be suspicious of such a statementin the unregulated theory, since the surface term mightfail to vanish, but it rigorously vanishes in the regulatedtheory.7Let us now return to the calculation we started above,the computation of the vacuum polarization diagram.We want to reorganize the numerator of (17.30) so thatit has one term in the same form as (17.32): (cid:90) d p ( p + q x (1 − x ) − m ) (cid:104) p µ p ν − δ µν ( p + q x (1 − x ) − m ) − q µ q ν x (1 − x )+2 δ µν q x (1 − x ) (cid:105) This has the pleasing feature that the first term, whichappears to be quadratically divergent, actually vanishes.We are left with the second term, having the form (17.13)that I said must arise as a consequence of the gauge sym-metry:2( q µ q ν − δ µν q ) (cid:90) x (1 − x ) dx (cid:90) d p ( p + q x (1 − x ) − m ) (17.33)This leaves the logarithmically divergent integral, whoseevaluation I will take up in the next lecture.
18. RENORMALIZATION, CONTINUED(1-14-88)
To remind you, we were computing the one-loop cor-rections to quark-quark scattering, and had noted that itis convenient to factorize the amplitude with the gluonvacuum polarization correction in the form= J iµ g q B q J iµ (18.1)where B represents the simpler diagram withthe gluon lines amputated, and J iµ are the external quarkcurrents, with color indices i . We found that B takes theform B = 2( δ µν q − q µ q ν ) I, (18.2) I = (cid:90) dx x (1 − x ) (cid:90) d p/ (2 π ) [ p − m + q x (1 − x )] Because the currents are conserved, q · J = 0, we cansimplify (18.1) slightly, = 2 J iµ g Iq J iµ (18.3)which has the same form as the tree-level contribution,= J µ g q J µ (18.4)Written in this way, it is clear that the loop contributioncan be expressed as a change in the coupling constant, g → g + 2 g I → g − g I ≡ g (18.5) where the first arrow indicates a result that is consis-tent to the order of perturbation theory at which we areworking, while the second one uses the hindsight of re-summing the leading logs, that we discussed in the lastlecture.Now to do the logarithmically divergent integral I , Iwill continue to use Pauli-Villars regularization, althoughlater on we will introduce the more elegant method ofdimensional regularization. Hence we must subtract from I the similar quantity with the modified propagator1( p − ( m + Λ ) + q x (1 − x )) (18.6)Rather than directly subtracting, there is a nicer way toimplement this, by first thinking of I as a being a functionof m , I ( m ) = (cid:90) dx x (1 − x ) (cid:90) d p/ (2 π ) [ p − m + q x (1 − x )] (18.7)and then differentiating with respect to m . Doing thismakes the integral convergent, even without subtractinganything. If we then integrate I (cid:48) with respect to m , − (cid:90) m +Λ m I (cid:48) ( M ) dM = I ( m ) − I ( m + Λ ) (18.8)the result is the subtraction we originally wanted to carryout. The trick then is to evaluate the convergent integralappearing in I (cid:48) ( M ), and postpone doing the integralover M until afterwards: (cid:90) m +Λ m dM (cid:90) dx x (1 − x ) (cid:90) d p/ (2 π ) [ p + q x (1 − x ) − M ] (cid:124) (cid:123)(cid:122) (cid:125) ≡ (cid:90) d p (2 π ) p − L ) = 132 π iL (18.9)In performing the integral, I have glossed over a few stepsthat I assume you are already familiar with, from a pre-vious course on quantum field theory, notably doing theWick rotation to avoid the poles from the i(cid:15) prescription,which gives the factor of 1 /i .Next we carry out the integral over M , (cid:90) m +Λ m dM π i M − q x (1 − x ) (18.10)= 116 π i ln (cid:18) m + Λ − q x (1 − x ) m − q x (1 − x ) (cid:19) . Since we are ultimately interested in the limit as Λ →∞ , this can be simplified by ignoring the finite terms inthe numerator of the argument of the logarithm. As wediscussed before, the ln Λ divergence gets absorbed intothe tree-level contribution by redefining the bare coupling g .8To further simplify the discussion, I would like to con-sider momenta such that − q (cid:29) m , so that we canignore the quark mass. You might be concerned thatthis could give rise to an infrared divergence from theplaces where x = 0 or 1 when we perform the integralover x , but because the integrand is a log, these singu-larities are integrable and lead to no difficulty. In thisapproximation, we have (cid:90) dx x (1 − x )16 π i (cid:20) ln (cid:18) Λ − q (cid:19) − ln x − ln(1 − x ) (cid:21) = (cid:18)
16 ln (cid:18) Λ − q (cid:19) − (cid:19) (18.11)Remember that − q ≡ Q is positive, since (cid:126)q is the mo-mentum transfer in the electron-electron scattering. Theintegral of the logarithm can be done using integrationby parts.Now we have evaluated the integral I , and we can putit back into the expression (18.3): J Q J g + g π (cid:18) ∆ β (cid:18) ln Λ Q (cid:19) + a (cid:19)(cid:21) (18.12)where ∆ β = − n f and a = 109 n f (18.13)where n f is the number of quark flavors having mass lessthan Q ; otherwise our approximation Q (cid:29) m is notvalid. Here ∆ β is just the quark contribution to β ; thefull β gets an additional contribution of +11 from thegluon loop, that we are not calculating here.We can now see more explicitly how the renormaliza-tion of the bare coupling is derived, which absorbs thedependence on the cutoff arising from the loop diagram.Consider the effective coupling defined in (18.5), g = g − g π β ln Λ Q = 16 π (cid:16) π g (cid:17) − β ln Λ Q . (18.14)The second line makes it clear how g must depend on Λin order that g eff be independent of Λ,16 π g (Λ) ≡ g (Λ) = β ln Λ λ P + c (18.15)where I have inserted a renormalization scale λ P , since g is a Lagrangian parameter that cannot depend on theexternal momentum Q . As I previously mentioned, weare free to choose a convention for defining λ P such thatthe arbitrary constant c vanishes, if so desired. Although we imagined that the value of g has beenfixed in this example by comparing to a particular ob-servable, the scattering cross section for two quarks, it is important to emphasize that once g (Λ) has been deter-mined, it is now valid for the study of any process; wedo not need to define a separate g (Λ) for every differ-ent observable. One way to understand this is from thefact that there is a finite number of primitively divergentdiagrams in the theory, repeated in this table, (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)
20 01 00 10 00N
Degreeofdivergence actualnaive
You recall that the tadpole diagram vanishes. The renor-malizability of the theory implies that all the divergentdiagrams can be related to an effective Lagrangian con-tribution that has the same form as the bare Lagrangian.So for example, all the divergent diagrams with only ex-ternal gluons must correspond to terms in the Lagrangianin this manner:( ∂ µ A ν − ∂ ν A µ ) A × ν A • µ ∂ µ A ν A × ν A • ν A × µ A ν A priori, we would have to renormalize three different pa-rameters to absorb the divergences. But because of gaugeinvariance, we know that they must organize themselvesinto the form tr F µν F µν , that depends only on the singleparameter g . Therefore these divergences are related toeach other in such a way that they can all be absorbed bythe renormalization of the single parameter g . This isperhaps easiest to see in the convention where we keep g out of the field strength definition and put it as a prefac-tor in the Lagrangian, (1 /g )tr F µν F µν . Then the threedivergences indicated above would all contribute to theshift in the effective Lagrangian that goes asln Λ tr F µν F µν (18.16)This of course assumes that the regularization methoddid not spoil gauge invariance. Otherwise we would haveto fudge the results to make this work out.A similar argument applies to the terms with externalquarks. The self-energy diagram requires us to renormal-ize an additional quantity, the quark mass. But the ver-tex correction does not need anything new; gauge sym-metry guarantees that the same renormalization of g asneeded for the gluons also suffices for the coupling toquarks.To make this clearer, let’s rephrase these statementsin the context of the path integral. We could imagine,if we were sufficiently adept, being able to carry out the9integral over quarks for a fixed gauge field background, (cid:90) D A µ e i g (cid:82) F µν F µν (cid:90) D ψ D ¯ ψ e i (cid:82) ¯ ψ ( i /D − m ) ψ = (cid:90) D A µ e i g (cid:82) F µν F µν det (cid:0) i /D − m (cid:1)(cid:124) (cid:123)(cid:122) (cid:125) (18.17) G ( A ) ∼ e i ln Λ (cid:82) F µν F µν giving a functional determinant of the covariant Dirac op-erator, that I am calling G ( A ). Even though we can’t cal-culate G ( A ) exactly, from perturbation theory we knowthat it has a divergent contribution like I have indicated,which has the same form as the tree-level gluon action.We can then rewrite the bare coupling in terms of a renor-malized coupling, which is finite as Λ → ∞ , plus a cor-rection designed to cancel the ln Λ divergences. In theconvention I have chosen above, it would be easier tothink of it as a correction to 1 /g :1 g = 1 g + δ (cid:18) g (cid:19) (18.18)These extra terms labeled as δ (1 /g ) are known as the counterterms .There is an interesting consequence of the running forthe convergence properties of the loops when we start tothink about the higher-order contributions, such as + + + . . . One can think of such diagrams as though they were atone order lower in perturbation theory, but constructedfrom propagators that have already been dressed at oneloop, + + . . . ∼ Q ln Q (18.19)An integral that is normally considered to be logarithmi-cally divergent would instead behave like (cid:90) Λ dp pp ln p ∼ ln(ln Λ ) (18.20)This is still divergent, but more mildly so. And going tohigher order, one could get higher powers of logs in thedenominator, which would make the integral converent, (cid:90) Λ c dp pp ln p ∼ (cid:90) Λ c d ln p ln p ∼ c − α ( Q ) ∼ / ln( Q ) to givethis more convergent behavior. I would like to discuss a different viewpoint of the run-ning coupling, that you may encounter in the literature,and that I consider to be misguided. The idea is to choosesome physical amplitude—suppose for simplicity that attree level it is linear in α —and to consider it as a functionof Q . One could then define a running coupling α ( Q )to be exactly determined by this physical observable. Anexample would be the amplitude for the scattering of twoquarks, to all orders in perturbation theory, (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) ≡ α ( Q ) Q (18.22)However there would be no such thing as perturbationtheory in regard to this particular process, since there isnothing to expand in: α ( Q ) is the exact result. Anotherexample is the correction R − e + e − → hadrons that we discussed previously. It gives rise to adifferent definition of the coupling, call it α ( Q ) = π ( R −
1) (18.23)If we were to compare these two definitions, we wouldfind out that they approximately agree, and a useful wayto compare them would be by differentiating and tryingto reconstruct the beta function. We would find thatboth definitions satisfy equations of the form dαd ln Q = β α + β α + β α + . . . (18.24)and that the first two coefficients β , β agree for bothdefinitions. But beyond that, the remaining coefficientsare in general different between the two definitions, andunrelated to the values of β , β . Contrast that to thedefinition I made, dαd ln Q = β α − β β α = β α + β α + β β α + . . . (18.25)where the higher coefficients are all determined. In ourprocedure, the amplitudes for the two processes have toboth be calculated perturbatively in α . This is howphysics should work: we have a definite theory that isindependent of the process and we predict the observ-able from it. The theory should not be predicated onone particular process or another.Adding to the confusion caused by such proposals isthe misconception that there is a momentum-dependentcoupling constant in the Lagrangian. As I explained be-fore, the Q dependence in α ( Q ) is just a shorthand toremind us how the loop-corrected amplitude depends on Q , that we can deduce by replacing λ P → Q in the def-inition of g (Λ). But the actual coupling that goes intothe Lagrangian is g (Λ), which does not depend on Q .0 Let me finish by giving a preview of how the logarithmsarise in dimensional regularization, that we will discussin more detail in the next lecture. Because the actionmust be a dimensionless quantity, when we continue thedimension of spacetime to some value D = 4 − (cid:15) , thecoupling that was dimensionless in D = 4 is no longer so.For this discussion I will adopt the field normalizationwhere F ∼ ∂A +[ A, A ] and the coupling constant appearsin front of F . Because of this form of F , there is nochoice but to say that the dimensions of A are [ A ] =1 / [ L ] = [ M ], and therefore the action looks like1 g (cid:90) F µν d D x (cid:124) (cid:123)(cid:122) (cid:125) dimensions of M − D = [ M (cid:15) ] (18.26)Therefore to make the action dimensionless, we musthave [ g ] = [ M (cid:15) ] (18.27)To make this explicit, it is convenient to relate g to adimensionless coupling g and a mass scale that I sug-gestively call λ P , g = g λ (cid:15)P (18.28)Now imagine redoing the calculation that led to eq.(18.5) using dimensional regularization. The result takesthe form g + g Q − (cid:15) (cid:18) β (cid:15) (cid:19) → g − β Q − (cid:15) (cid:15) (18.29) ≈ β (cid:15) λ − (cid:15)P − β (cid:15) Q − (cid:15) in which the ln Λ divergences of Pauli-Villars get re-placed by 1 /(cid:15) poles. Here I have resummed the leadinglogs and observed that necessarily g − = 2 β /(cid:15) + const . ,in order to cancel the divergence. Then we notice that2 (cid:15) ( λ − (cid:15)P − Q − (cid:15) ) = ln( Q /λ P ) + O ( (cid:15) ) (18.30)just like the outcome of the cutoff method of regulariza-tion.
19. RENORMALIZATION (CONCLUSION);LATTICE QCD (1-19-88)
I would like to add something to our previous discus-sion concerning the renormalization group equation. Re-call that the definition16 π β g (Λ ) − β β ln 16 π β g (Λ ) = ln Λ λ P (19.1) tells us (in an implicit way) how the coupling g mustdepend on the cutoff Λ. Now imagine some physical pro-cess, whose amplitude M —for definiteness I will pickan example where it is dimensionless—we compute fromthe theory with the cutoff, using the coupling g . M is generically a function of g , some momenta which forsimplicity I will represent by a single scale Q , and thecutoff, of the form M theory (cid:18) g , ln Λ Q (cid:19) (19.2)The explicit Λ dependence goes away, if we replace g → g (Λ) using the coupling defined by eq. (19.1). Inperturbation theory, we would find that M theory = g + g (cid:18) β ln Λ Q + c (cid:19) + . . . = α ( Q ) + c α ( Q ) (19.3)Now since M theory (cid:0) g (Λ) , ln Λ /Q (cid:1) is independent of Λ,we are free to set Λ = Q . Then the physical amplitudeis M phys = M theory ( g ( Q ) ,
0) (19.4)where g ( Q ) is defined by eq. (19.1), or explicitly16 π β g ( Q ) − β β ln 16 π β g ( Q ) = ln Q λ P (19.5)and now the physical amplitude is M phys = g ( Q ) + c g ( Q ) (0 + c ) + . . . (19.6)Comparing to (19.3), we can see that g ( Q )16 π = α ( Q ) (19.7)since M phys = α ( Q ) + c α ( Q ) + . . . (19.8)This makes clear what is the correct interpretation of therunning coupling constant that I was criticizing in theprevious lecture. The Q dependence is not present inany fundamental coupling in the Lagrangian, but ratherit arises from taking advantage of the Λ-independenceof the physical amplitude and using our freedom to setΛ equal to the relevant scale of the process, to get ridof the log. Of course, this is a consequence of solvingthe renormalization group equations, but this point ofview seems to me simpler and more intuitive than theRG equations. The preceding sentences are not in my notes, but seem to be thelogical connection to the previous lecture. Now we will move on to the main subject of this lec-ture, which is a comparison of different kinds of cutoffschemes, including the lattice and dimensional regular-ization. As we have mentioned before, the path inte-gral (cid:90) e iS D A = (cid:90) e i g (cid:82) F µν F µν d x + ... D A (19.9)is meaningless until a UV cutoff is introduced. I findthat the lattice is the most physically satisfying way ofaccomplishing this. We approximate spacetime as a lat-tice, and discretize all the field variables and notions ofdifferentiation; for example( ∂φ ) → (cid:15) (cid:0) φ ( x ) − φ ( x + (cid:15) ) (cid:1) (19.10)K. Wilson invented this technique especially for solvingQCD on the computer. In this framework, the gauge field A µ is a connection that relates the relative orientationsof the color frames at neighboring points on the lattice;these can be different from each other by an arbitrarySU(3) rotation. Hence it is natural to regard A µ as livingon the links connecting neighboring lattice points, ratherthan sitting on the lattice points themselves. Considertwo neighboring points, labeled by 1 and 2, and definethe link variable U = e i (cid:82) A iµ λi dx µ (19.11)where the path is a straight line connecting lattice points1 and 2. The quarks, on the other hand, live on thesites. In this formulation, the U ij become the dynamicalvariables rather than A i .Now we need to formulate the action in terms of thelink variables. Of course it has to be gauge invariant. Aninvariant quantity must involve a closed path in configu-ration space, : U U U U (19.12)= 1 + iF ixy λ i a + O ( A )The small a × a planar region together with this prod-uct is known as a plaquette. U U U U is invariantunder gauge transformations at any of the interior points RPF called it “dimensional renormalization” in his lectures, butwould have adopted the more common terminology for the ver-sion to be published. , ,
2, but to make it also invariant at site 1, we musttake the trace:tr [ U U U U ] = tr 1 + 0 + O ( A ) (19.13)The O ( A ) term is the interesting part. Notice that U = 1 + i (cid:88) b i λ i . . . ¯ U = 1 − i (cid:88) b i λ i . . . tr ¯ U U = tr 1 + (cid:88) b i b i + . . . (19.14)But since U is unitary, it must be that all the termsof O ( b ) cancel out, implying that we need to also keeptrack of such terms in the individual U matrices. It turnsout that the relevant ones are U = 1 + i (cid:88) b i λ i − (cid:88) b i b i + . . . ¯ U = 1 − i (cid:88) b i λ i − (cid:88) b i b i + . . . (19.15)When we keep these second-order terms in the expansionof the U ij variables in the plaquette, and expand theresult to O ( A ), the resulting expression is the finite-difference version of the free part of the gauge kineticterm, F aµν F aµν a (19.16)where µν = xy in the example shown in (19.12). Sum-ming on all plaquettes gives a sum on µ, ν as well as thesum over all locations, resulting in the action S = 2 g (cid:88) plaquettes (cid:0) tr[ U U U U ] − tr 1 (cid:1) , (19.17)where the additive constant is unimportant. We mustbe careful about the relative orientations of the U ’s onthe links, as indicated by the arrows in (19.12), to re-spect gauge invariance. This is not indicated explicitlyin (19.17) but it should be kept in mind. The factor of a in (19.16) represents the integration measure, since (cid:88) F a → (cid:90) d x F (19.18)in the continuum limit. Next we turn to the interaction of gluons with quarks.How do we represent ¯ ψ /Dψ ? Consider (cid:88) links ¯ q ( x ) γ µ [ q ( x + µ ) − q ( x )] (19.19) I have in parentheses the question “take real part of tr[
UUUU ]?”in my notes. The answer is yes; for a nice review from this era,see J. Kogut, 10.1103/RevModPhys.51.659. x + µ = x + a(cid:126)e µ , in terms ofa lattice unit vector (cid:126)e µ that points in the µ direction.But this is not yet gauge covariant, and it is missing theintegration measure factor. Instead, take ia (cid:88) links ¯ q ( x ) γ µ [ U x,x + µ q ( x + µ ) − q ( x )] (19.20)In fact, the second term ¯ q ( x ) γ µ q ( x ) can be dropped, sinceit has no effect on the dynamics, but just contributes anoverall phase to the path integral. Let’s look more closely at the gauge invariance of thegluon kinetic term. One can transform the color axesindependently at each lattice site. Suppose we transformsite 1 by the SU(3) matrix Λ(1), and similarly at site 2by Λ(2). One finds that U changes by U Λ(2)Λ(1) U → Λ † (2) U Λ(1) (19.21)Therefore the action is invariant becausetr (cid:20) . . . U Λ(2)¯Λ(2) (cid:124) (cid:123)(cid:122) (cid:125) U . . . (cid:21) = tr [ . . . U U . . . ]1 (19.22)This reasoning also shows how the quark kinetic term isinvariant.However we must still specify the form of the path inte-gral measure for the link variables. Since U is an SU(3)transformation, a gauge invariant measure is required.This is known to mathematicians as the Haar measure.Since SU(3) is a compact group, (cid:82) dU is finite at a givensite, unlike the usual measure (cid:82) dA . We will not go intothe mathematical details of the Haar measure here, butit is intuitively similar to the more familiar integrationmeasure for rotations (sin β dα dβ dγ for SO(3) in termsof Euler angles).Just like for a Pauli-Villars cutoff Λ, we need to findout how g must vary with the lattice spacing a in orderfor the theory to give results that are independent of a in the continuum limit, a →
0. Technically, this is notso easy to do as for the Λ cutoff, but conceptually, itcan be carried out in the same way. We would haveto work out perturbation theory on the lattice to do itproperly. Less rigorously, one can expect that g ( a ) hasa similar structure to our previous expression for g (Λ) ifwe identify Λ = c/a for some constant. Then the taskbecomes determining the correct value of c . Another wayof thinking about it is in terms of the renormalizationscale we called λ P in eq. (16.1). We could define Λ = 1 /a ,but we would find that some other value of λ P is neededto describe the same physics on the lattice relative to theΛ cutoff. I have elaborated here on what is written in my notes: “becauseit is just a number (?)”
To my mind, the lattice is the most concrete and leastmysterious of all regulators, and it is the most physi-cally sensible. But—and now I’m going to speculate—itseems like we are missing something by having to rely onthese rather ad hoc schemes for defining our theories. It’scomparable to Leibnitz and Newton inventing the inte-gral calculus, which also looks like taking the continuumlimit of a lattice,lim h → (cid:88) m f ( x + mh ) h = (cid:90) f ( x ) dx . (19.23)The value of the integral does not depend on the machin-ery of cutting off the small scales and taking the limit, soone need not be preoccupied with the details of exactlyhow to discretize, like we are doing with all of our differ-ent regulator schemes for the path integral. We know howto do the ordinary integrals directly. Similarly we believethat the path integral has some kind of intrinsic meaningthat does not depend on the cutoff scheme, but the dif-ference is that we can’t avoid that whole discussion, andthe dependence on details of whether we use this kindof cutoff or that kind of cutoff. This makes me dream,or speculate, that maybe there is some way, and we arejust missing it, of evaluating the path integral directly,without having to make this detour into the machineryof renormalization, since we know that the physics hasto be independent of it in the end. Previously we already discussed dimensional regular-ization in a preliminary way. Here I would like to do itin somewhat more detail. The basic observation is thatan integral like (cid:90) d D p ( p − m ) (( p − q ) − m )) converges for D < . (19.24)Hence we define D = 4 − (cid:15) , and take the limit (cid:15) → /(cid:15) . Thisidea was used to great advantage by K. Wilson to under-stand phase transitions in statistical mechanics, and itwas applied to gauge theories by ‘t Hooft and Veltman:K. Wilson, Phys. Rev. D7, 2911 (1973)G. ‘t Hooft and M.J.G. Veltman, Nucl. Phys. B44, 189(1972) Exercise.
Prove the following statements in D dimen-sions:1. g has dimensions of E − D .2. Defining η g and η q as the number of external gluon orquark lines, and L as the number of loops, the power of g for any diagram is 2( L −
1) + η g + η q .3. The dimension, in powers of mass, of the matrixelement (amplitude) | T | is d = D + N − DN/ N = η g + η q .34. The dimension of the integral for a diagram is d =4 − η g − η q − L (4 − D ).Now I would like to consider the defining propertiesof the momentum space integrals in D dimensions, toshow that they can be evaluated rigorously and with noambiguity. There are four basic properties.1. (cid:82) f ( p ) d D p is linear: (cid:90) [ αf + βg ] d D p = (cid:18) α (cid:90) f + β (cid:90) g (cid:19) d D p (19.25)Among other things, this means that we can use Fourierand Laplace transforms in D dimensions.2. Shifts of the integration variable are allowed, (cid:90) f ( p + a ) d D p = (cid:90) f ( p ) d D p (19.26)where a is a constant vector.3. Scaling: (cid:90) f ( αp ) d D p = 1 α D (cid:90) f ( p ) d D p (19.27)4. Normalization: (cid:90) e − p / d D p = (2 π ) D/ (19.28)since (cid:82) du e − u / = √ π .We can define Fourier transforms in the usual way, (cid:90) d D p f ( p ) e πip · x = φ ( x ) (cid:90) d D p φ ( x ) e − πip · x = f ( p ) (19.29)Next let us evaluate some integrals. Notice that (cid:90) e − α p d D p = (cid:32)(cid:114) πα (cid:33) D (19.30)Then we can use the shift property to find that (cid:90) e − α p e ip · x d D p = (cid:114) πα D e − x / α (19.31)Other integrals can be generated from this one by differ-entiating with respect to α . Exercise.
Prove that (cid:90) e − α p ( a · p )( b · p ) d D p = X ( a · p ) (19.32)and find X . Further prove that (cid:90) e − α p p d D p = DX (19.33) Another useful integral is (cid:90) d D p (2 π ) D ( p ) R [ p − C ] M = i ( − R + M (4 π ) D/ C R − M + D/ (19.34) × Γ( R + D/
2) Γ( M − R − D/ D/
2) Γ( M )The reader is invited to derive this one as well.
20. DIMENSIONAL REGULARIZATION,CONTINUED (1-21-88)
Another very useful class of integrals is that wherethe integrand f depends only on the magnitude of p .(Imagine that we have already Wick-rotated to Euclideanspace.) Then (cid:90) f ( p ) d D p = C D (cid:90) f ( ρ ) ρ D − dρ (20.1)To determine C D , we can consider the case where f = e − ρ α/ , since we already know the value of this integralfrom (19.30). Comparing with (20.1), (cid:114) πα D = C D (cid:90) ∞ e − ρ α/ ρ D − dρ (20.2)= C D (cid:90) ∞ e − uα/ u D − du √ u = C D (cid:18) α (cid:19) D (cid:90) ∞ e − uα/ u D − du (cid:124) (cid:123)(cid:122) (cid:125) Γ( D/ C D = 2 π D/ Γ( D/
2) (20.3)
Exercise.
Find a general formula for (cid:82) d D p f ( p , a · p ).Hint: relate it to (cid:82) e − αp / ip · x d D p .To combine denominators, we can use formulas like (cid:90) dx x m − (1 − x ) n − [ ax + b (1 − x )] m + n = Γ( m )Γ( n )Γ( m + n ) 1 a m b n (20.4) In my notes there is a question mark over the = sign and aparenthetical note to check the formula, probably a caution fromRPF. It is correct. (cid:90) d D p (( p − q ) − m ) ( p − m ) (20.5)= (cid:90) dx (cid:90) d D p p − p · qx + q x − m ) = (cid:90) dx (cid:90) d D p (cid:48) p (cid:48) + q x (1 − x ) − m ) = (cid:90) dx C D (cid:90) ∞ dρ ρ D − ρ + q x (1 − x ) − m ) Claim: (cid:90) d D p f ( p )( a · p )( b · p ) = a · bD (cid:90) d D p f ( p ) p (20.6)Similarly, (cid:90) f ( p )( a · p )( a · p ) · · · ( a n · p ) d D p (20.7)= (cid:2) ( a · a )( a · a ) + . . . (cid:3) (cid:90) d D p f ( p ) p n where n is assumed to be even.As we showed before, dimensional regularization re-produces the logarithms that we get from Pauli-Villarsregularization, through the combination of the 1 /(cid:15) poleswith terms like ( p /µ ) (cid:15) . One shortcoming however is inthe definition of chiral theories, since it is not clear howto define γ in D dimensions, nor correspondingly thetotally antisymmetric tensor (cid:15) αβγδ . This of course is nota problem for QCD where parity is conserved.A novel potential use of dimensional regularization,which is not usually considered, is that it could providemore than just a method for evaluating divergent loopintegrals: it is also possible to use it to define a quan-tum field theory nonperturbatively in 4 − (cid:15) dimensions.I spent some time thinking about this, but was not ableto get anything interesting out of it. D dimensions Inspired by dimensional regularization, it is interest-ing to try to formulate a more complete picture of whatphysics would look like in an arbitrary number of dimen-sions. We start by imagining a linear vector space in D dimensions, with vectors x , y , etc. Given any two suchvectors, linearity implies that αx + βy is also a vector (20.8) In the next lecture RPF expands on this, presenting a way toformulate quantum field without recourse to the path integral orcanonical quantization.
We need a scalar product, a number associated with ev-ery pair of vectors, x · y maps (vectors) → R (20.9)It must be linear,( αx + βx (cid:48) ) · y = α x · y + β x (cid:48) · y (20.10)and associative, x · y = y · x (20.11)Now suppose we had four such vectors. We could con-struct the object( a · b )( c · d ) = a µ c ν b µ d ν (20.12)where the indices are no longer numbers taking on dis-crete values, but rather markers telling us which vectorsshould be dotted with each other. It is just an alternativenotation. In this case, what would it mean to write anexpression like n µ = a µ c ν d ν (20.13)in which we have an index that is not contracted? Suchan equation makes sense if we interpret it to mean that x µ n µ = x µ a µ c ν d ν for all x . (20.14)We can also think of n µ as being a linear map from vec-tors into R . And we can generalize this to several uncon-tracted indices, like m µν = a µ b ν + c ν d µ (20.15)which means that m µν maps pairs of vectors into R , etc .Hence there is no need for the indices to take on discretevalues as they would in an integer number of dimensions.Contraction. In addition to operating on two vectors, m µν can be contracted on its own indices, m µµ , which isjust a number in R . For the above example, it is obviously a · b + c · d . However there is a special bilinear mapping, δ : ( x, y ) → x · y (20.16)that we call δ µν in discrete dimensions. Then δ µν y ν = y µ (20.17)is a consistent definition of δ µν . Now δ µµ is a pure num-ber, that we are free to choose. Let us make the definition δ µµ ≡ D (20.18)Obviously, there is no restriction that D should be aninteger.Next let’s consider how calculus should work. We canconsider a nonlinear mapping, F ( x ) : vectors → R , suchas F ( x ) = x · x (1 + ( a · x ) ) (20.19)5What is the derivative of F ? We define D c F ≡ lim (cid:15) → F ( x + (cid:15)c ) − F ( x ) (cid:15) (20.20)where c is some vector. This is the directional derivative,along the c direction. It is a linear function of c , so itmust be of the form c · (something). Therefore we define D c F ≡ c · ∇ F = c µ ∇ µ F (20.21)This specifies the ∇ µ operator, independently of c , since(20.21) must be true for any c . For the example (20.19), c · ∇ F = c · ∇ x · x [1 + ( a · x ) ] − x · x [1 + ( a · x ) ] a · x ( c · ∇ )( a · x )(20.22)as usual. Now( c · ∇ )( x · x ) = lim (cid:15) → ( x + (cid:15)c ) · ( x + (cid:15)c ) − x · x(cid:15) = 2 c · x, (20.23)so ∇ µ ( x · x ) = 2 x µ , as expected, and ( c · ∇ )( a · x ) = a · c ,giving ∇ µ ( a · x ) = a µ and ∇ µ x ν = δ µν .Next we construct the Laplacian, by considering suc-cessive derivatives: a µ b ν ∇ µ ∇ ν F = ( a · ∇ )( b · ∇ ) F (20.24)This allows us to isolate ∇ µ ∇ ν F , which is a tensor T µν ,whose contraction gives the Laplacian. For example, ∇ µ ∇ µ x · x = 2 δ µν ∇ ( x · x ) = 2 δ µµ = 2 D (20.25)So far, all of our results look completely reminiscent oftheir counterparts in integer dimensions. But the conceptof orthogonal subspaces leads to a novelty. Suppose wehave a vector a such that a · a (cid:54) = 0. For any vector x wecan define the part that is orthogonal to a as x (cid:48) = x − a · xa · a a (20.26)so that a · x (cid:48) = 0. Then any vector can be written asa piece proportional to a plus a piece in the orthogonaldirection. Furthermore the dot product of two vectorssplits into x · y = x (cid:48) · y (cid:48) + a · y a · xa · a (20.27)If we take away the external vectors, this gives the defini-tion of the Kronecker delta of lower dimensionality, livingin the subspace orthogonal to a , δ µν = δ µ (cid:48) ν (cid:48) + a µ a ν a · a (20.28)with δ µ (cid:48) ν (cid:48) defined in the orthogonal subspace. Its traceis δ µ (cid:48) µ (cid:48) = D − D is an integer, but does not if D is noninteger. Itmeans that we can construct infinitely many directionsthat are all mutually orthogonal in the noninteger case. Exercise.
Suppose that we have two vector spacesof dimension D and D , and vectors x , y , x , y re-spectively defined in the two spaces. There is a rule forcombining them to make a vector of dimension D , suchthat x = x ⊕ x , y = y ⊕ y = ⇒ x + y = ( x + y ) ⊕ ( x + y ) (20.30)Prove that D = D + D , and (cid:90) F ( p ) d D p = (cid:90) F ( p , p ) d D p d D p (20.31)
21. PHYSICS IN D DIMENSIONS,CONCLUSION (1-26-88)
We have not yet discussed how to extend the Dirac al-gebra to arbitrary dimensions. Let’s consider the gammamatrices. For any vector a µ , we can associate a quantity /a = a µ γ µ (21.1)that has the property { /a, /b } = 2 a · b (21.2)Therefore { γ µ , γ ν } = 2 δ µν and /a = a · a , where δ µµ = D as usual.For the trace properties, I like to define two traces,that are normalized differently from each other. Whenacting on the unit matrix in the Dirac space they givetr[1] = 1; Tr[1] = D , (21.3)so tr[ X ] = (1 /D )Tr[ X ]. They satisfy the usual propertytr[ /A /B ] = tr[ /B /A ] (21.4)from which you can derive that tr[ /a/b ] = a · b and tr[ /a ] = 0.In fact, by using the propertytr[ /a/b/c/d ] = 12 D (cid:0) Tr[ /a/b/c/d ] + Tr[ /d/c/b/a ] (cid:1) (21.5)and its generalization to an arbitrary number of gammamatrices in the products, one can demonstrate that thetrace of any odd number of gamma matrices vanishes.The Dirac equation in D dimensions can be written as i / ∇ ψ − V ( x · x ) γ ψ = Eψ (21.6)for a spherically symmetric potential. It can be solvedexactly; I have carried this out.6 Exercise.
Reduce the problem (21.6) to a conventionalordinary differential equation.It turns out that the sum of two spaces of dimension1 / x · y ) = ( x · x )( y · y ) in 1 dimension . (21.7)But this property does not hold when you construct itfrom two half-dimensional spaces, along the lines of theexercise at the end of the previous lecture. Exercise.
Show that a D = 0 space constructed fromtwo spaces of equal and opposite dimension has nontrivialproperties.Newtonian mechanics is quite straightforward in D di-mensions. We define a time-dependent vector x ( t ) andimpose the principal of least action, with the action S = (cid:90) (cid:18) m (cid:18) dxdt · dxdt (cid:19) − V ( x ) (cid:19) dt (21.8)Here I do not insist on a central potential; for instance V ( x ) could have the form V ( x ) = f ( x · x ) + (cid:88) a i · x g ( x · c ) + . . . , (21.9)for example V ( x ) = ( a · x + x · x ) / (1 + ( x · x ) ). You canprove that orbits stay in the same plane—this is just con-servation of angular momentum—for a spherically sym-metric potential; it’s very dull.One can solve the wave equation in noninteger dimen-sions, ∇ φ = − κ φ ; φ = φ ( x ) (21.10)where ∇ = (cid:154) · (cid:154) , as we discussed in the last lecture.It can be done using the Fourier transform, which worksin D dimensions, with φ = e ik · x as usual. If there is asource s , so that ∇ φ = − κ φ + s (21.11)then by Fourier transforming one can solve˜ φ ( p ) = ˜ S ( p ) p − k (21.12)and also do the inverse Fourier transform to obtain φ ( x )in position space.In quantum chromodynamics we would like to be ableto generalize the path integral to D dimensions, Z = (cid:90) e iS (cid:89) D A ( x ) , S = 12 g (cid:90) E µν · E µν (21.13)with E µν = ∇ µ A ν . . . . Here A ν ( x ) is just a vector fielddefined on the space of vectors, both in D dimensions.Therefore there is no difficulty in defining the action.However the measure is problematic because the space-time is too abstract in D dimensions. In particular, we don’t know how to construct a lattice if D is not an in-teger.This motivates us to consider a different formulation ofquantum field theory. Imagine some superfunctional thatI will denote by { F [ A ] } , that acts on functionals F [ A ] ofthe field in the same way as the usual normalized pathintegral, for the cases of integer dimensions where weknow how to define it: (cid:82) e iS F [ A ( x )] (cid:81) D A (cid:82) e iS (cid:81) D A ≡ { F [ A ] } (21.14)Even though the definition of the left-hand side is notobvious for noninteger dimensions, we can generalize itsknown properties in integer dimensions to obtain a func-tional differential equation, that defines our mysterioussuperfunctional. Namely, we know that the path in-tegral is invariant under a change of variables A µ → A µ + (cid:15)α µ ( x ), that I take to be infinitesimal. Moreover,the measure by itself is invariant under this trivial shift.Recall that the functional derivative δF [ A µ ( x )] /δA µ ( x )is defined by δF = (cid:90) (cid:15)α µ ( x ) δF [ A µ ] δA µ ( x ) d D x (21.15)It follows that (cid:26) δF [ A µ ] δA µ ( x ) (cid:27) + i (cid:26) F δSδA µ ( x ) (cid:27) = 0 (21.16)This equation is equivalent to the path integral, but morefundamental since it extends to the case of nonintegerdimensions, and it can be used as our starting point.It is a statement of Schwinger’s action principle, fromwhich one can derive the usual perturbation expansion,but it can also serve as a nonperturbative definition ofthe theory.One issue we have glossed over is the signature of thespacetime metric. This is a discrete choice in integer di-mensions, and it is not perfectly clear how to deal with itin arbitrary dimensions. Should the fractional differencein the dimension be spacelike or timelike? In a more ex-treme case, what would the world look like if the metricof spacetime was δ µν = − − ? (21.17)It might be advantageous to find some kind of geometricaldescription of events to answer this.Although it is a somewhat different issue than noninte-ger dimensions, one could also think about derivatives offractional order. We know that conventional derivativeshave the form dfdx = lim (cid:15) → f ( x ) − f ( x − (cid:15) ) (cid:15) (21.18) d fdx = lim (cid:15) → f ( x ) − f ( x − (cid:15) ) + f ( x − (cid:15) ) (cid:15) . /
2? I will illus-trate the correct generalization for the 1 / d / fdx / = lim (cid:15) → f ( x ) − f ( x − (cid:15) ) + f ( x − (cid:15) ) − . . .(cid:15) / (21.19)where − /
2, 1 / x ) / . This turns out to be a valid pro-cedure. And it has an inverse: you can find half-orderintegrals as well, in an analogous way.Returning to dimensional regularization, I wanted toshow in slightly more detail how the Q dependence ofthe coupling comes out from the perturbation series inthat method. Here I will set D = 4 + (cid:15) and take (cid:15) → g + g (cid:18) β ln Λ Q + c (cid:19) + . . . = α ( Q ) + α ( Q ) c + . . . (21.20)In dimensional regularization, we must introduce an ar-bitrary mass m to make the coupling dimensionless: g = g m − (cid:15) ≡ α m − (cid:15) (21.21)The renormalization of the coupling constant now be-comes 4 πβ α ( Q ) − β β ln (cid:18) πβ α ( Q ) (cid:19) (21.22)= 4 πβ α + 2 (cid:15) + ln Q m + γ E − ln 4 π where γ E = 0 . . . . is Euler’s constant. Nothing phys-ical depends on m , nor on (cid:15) . By choosing the arbitrarynormalization of m appropriately, we can cancel the 2 /(cid:15) pole (the minimal subtraction MS scheme), or the polealong with the γ E − ln 4 π terms (modified minimal sub-traction, MS scheme). Q We have seen that at high Q , the effective couplingof QCD is supposed to become small. But this by it-self does not guarantee perturbation theory is necessarilyvery good, since the numerical coefficients of the expan-sion might turn out to be large. In particular, there isalways a low-energy effect mixed in with any high-energyprocess because of hadronization of the final state par-ticles. This part of the process is taking place at scaleswhere the coupling is definitely not small and the cal-culation is not perturbative. It might seem like after allour wonderful efforts of using renormalization to improvethe perturbative predictions at high energy, we could getfoiled by these low-energy effects. However all is not lost,because we can separate these two phenomena from eachother in a more or less clean manner. If we can argue that the details of hadronization are independent of thehigh energy scale Q , then this separation can be donequantitatively.Recall that in e + e − collisions that produce q ¯ q , we willsee two hadronic jets, e − e + qq γ or if a gluon is radiated at a large enough angle, we willsee three jets, e − e + qq γ At low Q , the third one is likely to be too soft to appearas a distinct jet. In this case the envelope of momentumvectors would appear to be a cylinder with a small bumpfor the third soft jet At high Q , the cigar gets relatively narrower, and thejets become better defined. The gluon jet starts to be-come more distinct as Q increases, like bringing a pic-ture into focus. And as Q continues to increase, greaternumbers of jets start to appear.We can measure the distributions of hadrons in jets atlow Q , described by what are known as fragmentationfunctions. Although we don’t know the fragmentationfunction for gluons, we can make some educated guess.The important point, related to my claim above, is thatthese functions do not change appreciably with Q , whichleads to the factorization phenomenon that I described.These distributions take the form c dp z E f ( p ⊥ ) d p ⊥ (21.23)where p z is the momentum along the jet axis, and p ⊥ isthe transverse momentum (shown as having a spread of ∼ f ( p ⊥ )is flat, within the jet, you will notice that this distributionis Lorentz invariant.8A useful quantity for characterizing particles in thejets is rapidity. Consider the quantity E − p z , where E = (cid:112) p z + p ⊥ + m is the energy of a particle in thejet. Under a boost in the z direction, this becomes γ ( E + vp z ) − γ ( p z + vE ) = γ (1 − v )( E − p z ): it changesmultiplicatively. Similarly E + p z → γ (1 + v )( E + p z ).Therefore the rapidity w = ln E + p z E − p z (21.24)changes by w → w + ln 1 + v − v (21.25)where v is the boost parameter. It is related to w by v = tanh w (21.26)and in terms of w a boost takes the form E (cid:48) = E cosh w = E √ − v ,p (cid:48) = p sinh w . (21.27)It turns out that the distribution f ( p ⊥ ) is not quiteflat over its region of support, but instead goes as f ( p ⊥ ) ∼ α ( p ⊥ ) d p ⊥ p ⊥ ∼ d ln p ⊥ ln p ⊥ ∼ d ln ln p ⊥ . (21.28)Hence it is nearly constant, the log of a log. This smalldeviation from flatness has been observed and provides aconfirmation of QCD. [The following figures appear without explanation inmy notes.] t α( ) s α (t) α ′ + α t α( ) p π − π n t σ ∼ =
22. FINAL LECTURE (1-28-88)22.1. Schwinger’s formulation of QFT, continued
I would like to come back to the alternative formulationof quantum field theory that I started to discuss last time,eq. (21.16). To understand in more detail how to use it,let’s consider the simpler example of a scalar field theory,where it takes the form (cid:26) ∂F∂φ (cid:27) + i (cid:26) F ∂S∂φ (cid:27) = 0 , (22.1)and now S = (cid:90) (cid:0) ( ∇ φ ) − µ φ − λφ (cid:1) d x . (22.2)Remember that we are free to choose any functional F ;it is instructive to take F [ φ ] = e i (cid:82) σ ( x ) φ ( x ) d x . (22.3)From it, we can generate Green’s functions by takingfunctional derivatives δ n /δσ n . Then, with n = 1, i Σ( x ) ≡ { iσ ( x ) F } = − i (cid:8) F ( −∇ − µ ) φ (cid:9) (22.4)to zeroth order in λ . If we define f ( x ) = { F φ ( x ) } , wesee that i ( ∇ + µ ) f = i Σ( x ) , (22.5)which can be solved to get f = (cid:0) ∇ + µ (cid:1) − Σ( x ) (22.6)This of course is just the propagator acting on Σ. Fouriertransforming to momentum space, it reads f ( p ) = − p − µ Σ( p ) (22.7)To define what happens at the pole, we need to make an i(cid:15) prescription, as usual.Now we can write { φ ( x ) F } = (cid:90) dy (cid:110) I ( x − y ) σ ( y ) e i (cid:82) σ φ d x (cid:111) (22.8)where I ( x − y ) is the propagator in position space. Hence (cid:110) φ ( x ) e i (cid:82) σ φ d x (cid:111) = (cid:90) dy I ( x − y ) σ ( y ) (cid:110) e i (cid:82) σ φ d x (cid:111) (22.9)or δ { F } δσ ( x ) = (cid:90) I ( x − y ) σ ( y ) dy { F } . (22.10)We can integrate this to get { F } = e (cid:82) σ ( x ) I ( x − y ) σ ( y ) dx dy . (22.11)9This is all at the level of free field theory so far. I leave itas an exercise for you to show that eq. (22.10) generalizesto δ { F } δσ ( x ) = (cid:90) I ( x − y ) σ ( y ) dy (cid:34) − λ (cid:18) − i δδσ ( y ) (cid:19) (cid:35) { F } (22.12)in the presence of the interaction. This can be solved per-turbatively, or perhaps if you are clever enough, in somenonperturbative fashion. Obviously, there is no very sen-sitive dependence on the number of spacetime dimensionsin this formulation, so it could serve as a nonperturbativedefinition of the theory in 4 − (cid:15) dimensions. Now I would like to come back to some things that westarted to discuss earlier in the course, during the firstfew lectures. Remember the parton picture of the proton,where the quarks and gluons have various momentumprobability distributions inside the proton, that we de-noted by u ( x ), d ( x ), etc . I had mentioned a conceptualproblem, the fact that even if we knew the wavefunctionof all the constituents for a proton at rest, this is notsufficient for determining u ( x ), d ( x ), . . . . The problemhas to do with how the wave function transforms undera boost, t x x x φ( x,y,z) φ i n f o r m a ti on x,y,z) ψ(′ t xx x In one reference frame, at t = 0, the wave function forthree quarks at respective positions x, y, z is φ ( x, y, z ),while in some boosted frame, at t (cid:48) = 0, it is ψ ( x, y, z ).It is a nontrivial task to get ψ from φ , since we need tosolve the Schr¨odinger equation to propagate the quarksforward in time. As we saw before, this is complicated by the fact that the concept of the wavefunction is notrelativistic.But once we know the distribution functions at highenergies, it turns out that they don’t change very muchas you go to even higher energies. We can determinethese functions by doing proton-electron scattering, e − p The reason that the distributions continue to change athigher energies is that the kinematics are not so simpleas in this diagram: in reality, gluons are radiated, inparticular in the forward direction, where we don’t seethem as distinct jets. In fact as we increase Q , gluons aremore likely to be emitted, which leads to u ( x ) dependingon Q and not just the momentum fraction x . This givesa correction to the naive parton picture, that neglectedsuch effects. Qualitatively, the correction looks like α (Q )xu(x) QCD naive parton prediction You might think that gluon emission should decrease athigh Q because of the running of α ( Q ), but it turnsout that the increase in phase space outweighs this effect.It is similar to the infrared/brehmsstrahlung problem inQED.Consider electroproduction of quarks, e + e − → q ¯ q .Many soft gluons will be emitted, but they all get lumpedin with the quark jets, and are of little consequence a pri-ori . Our concern is how this description may change asa function of Q .Recall the fragmentation functions, D h ( z, Q ), that tellus the distribution of the momentum fraction z carriedby hadron h in the jet. We would like to know how itdepends on Q . Since the main dependence on Q comesfrom the coupling, and this log dependence is also trackedby the cutoff, we can infer that dDd ln Q = δDδg ∂g ∂ ln Λ = δDδα ( Q ) ∂α ( Q ) ∂ ln Q , (22.13)or, defining τ = ln Q , dDdτ = δDδα β α . (22.14)0So we need δD h /δα , which arises at first order in per-turbation theory, and is related to the probability foremitting a gluon (or possibly a q -¯ q pair) that carries awaysome fraction (1 − y ) of the momentum: q 1 q y(1−y) This leads to the evolution equation δDδα = (cid:90) P ( y ) D h ( z/y, Q ) dyy (22.15)where P ( y ) is the probability of emitting a gluon thatleaves the quark with momentum fraction y , if it wasnormalized to be 1 initially. One can show that P ( y ) = 23 π ln Q m (cid:18) y − y (cid:19) + . . . (22.16)where the . . . represent terms with weaker dependenceon Q . Exercise.
Consider e + e − → q ¯ q with the kinematicsindicated below: e − e + q p p q g γ z directionpp k Show that the probability to emit the gluon is propor-tional to P ∼ α (cid:15) + (cid:15) + p z + p z (cid:15) (cid:15) ∼ α ( (cid:15) + (cid:15) ) (22.17)where (cid:15) i is the energy of quark i , and the second expres-sion is the result from integrating over the quark direc-tions.We can write dDdα = τ P · D + . . . (22.18)where P · D denotes the convolution (cid:82) P D dy/y , or defin-ing κ = β ln τ , dDdκ = P ·
D . (22.19)Hence if we evolve from Q to Q , the change in κ is∆ κ = β ln τ τ = β ln (cid:18) ln Q /λ ln Q /λ (cid:19) (22.20) This is a double logarithm, so the change is typicallyquite small. Take for example Q = 6 GeV , Q ∼ = 40 GeV ,Q = 4000 GeV , Q ∼ = 2 × GeV (22.21)This gives ∆ κ ∼ = . To get a change as large as ∆ κ = 1,we would need to go to Q = 400 ,
000 GeV!So it is necessary to vary the energy quite dramaticallyto see any appreciable change in the fragmentation func-tion D h . And only by starting from rather low energieswill we observe much variation at all. Notice that at suchlow energies as 6 GeV, many of our approximations thatwere appropriate for high Q are not very good. Theimportant point is that the details of hadronization areindeed insensitive to Q as long as Q is well above theQCD scale. Acknowledgment.
JC thanks Isabelle Masse forproofreading and for helpful suggestions.Appendices A-C are the verbatim transcriptions fromthe audio tapes of lectures 15, 17 and 18. Appendix Dcontains scans of RPFs hand-written addenda and cor-rections to revised drafts of two lectures. The remainingappendices are material that RPF handed out to the classparticipants, some of them written in his own hand.
Appendix A: Transcription: Scale dependence(1-5-88)
Okay. Now that we’re starting the second term, we’veformulated several times, in many different ways, in dif-ferent kinds of gauges, the rules for perturbation the-ory; also the formulas in terms of path integrals. Oneof the purposes of path integrals is a statement of theequations which is not strictly speaking necessarily simplyperturbation theory. If there was some way to computethe path integrals, and there is for instance numerically,there would be a scheme for making calculations whichwould not rely simply upon the need for perturbations.Because of our limited ability in doing path integrals untilthe present time, we’re only pretty good in perturbationtheory. We got so good at it from working with quan-tum electrodynamics where the coupling constant is verysmall, and therefore we’ve had lots of practice. But pleasedon’t think that we have to do everything by perturbationtheory.During this term we are going to talk in the first halfof it about perturbation theory and what we can learnof quantum chromodynamics from perturbation theory.The second half of the term will be an attempt to under-stand the behavior of this theory, the fact that it confines RPF says electrodynamics but it is clear he meant to say QCD. quarks and so on, in some way by looking at the path in-tegrals, without actually expanding them in perturbationtheory. It will not be mathematically accurate; it will bequalitative. Say this will get big, this will get small, Ithink this will be bigger than that, and therefore this willhappen. You will be very dissatisfied if you want preci-sion. One of our problems as we’re discovering, right,is doing these things with precision. The state we are innow is one where we will have to discuss it in a qualita-tive way. So that’s all I can do, but we’ll do that in thesecond part of the term. In the first part of the term, wewill find out how much we can do by perturbation theoryto test the theory.So far we made only passing reference to the runningof the coupling constant. This subject requires some careto avoid confusion. But I must say that it would not beat all difficult, there would not be any particular problem,and it is a very simple matter; the confusion comes be-cause we can’t calculate anything, and so we try to sayas much as we can without calculating . . . It’s somethinglike the subject of thermodynamics, which appears to bequite complicated, but if you use always the same vari-ables, such as temperature and volume, to represent thesystem, it’s much simpler than if you suddenly say nowwait, let’s suppose I want to plot this on entropy andpressure; it’s the perpetual change of variables from oneto the other that makes the subject so complicated.So there is a certain apparent complexity here, whichis due to our inability, or indefiniteness, in choosing amethod of calculation, or indefiniteness in deciding whatprocess to calculate. That makes it look a little compli-cated. Let me make believe, at first, that we could calcu-late whatever we want. Then the problem would be thefollowing. It would be straightforward. We would startwith our theory, with Lagrangian g F F + ¯ ψ ( i /∂ + m ) ψ . . . Let me put the constant and call it g . And then in thepart that has to do with the quarks, with different flavors,there would be masses for the different flavors. So therewould be a number of constants which are in the theory, g , m u , m d , m s , m c , m b I’m putting a subscript 0 on them, which means those arethe values that we put into the equations; in case there’sany question of what I mean, that’s what I mean: thenumbers that we put into the original Lagrangian to makethe calculation. So this is perfectly definite. There’s the c quark; we may discover one day that there are others [quark flavors] so we might need a few other parameters,but for the future. So at the present time we’ve got The top quark had not yet been discovered at this time, thoughits existence was not doubted. these six numbers, the parameters that we can put intothe theory.Now suppose we start out with this theory, we put someparameters in, and we compute something: the mass ofthe proton, the mass of the pion, and so on. If we com-puted six quantities, and we could compute perfectly andthe theory were right and experiments were available forall those quantities, then we could determine these pa-rameters. Then if I computed a seventh quantity, thatwould be a completely predicted quantity and we begin totest the theory. That’s simple and straightforward andthat’s all there is to it, except . . . First of all, we can’tcompute the mass of the proton, so we can’t determinethese constants. In view of that you also would knowthat it would be very practical to compute some quanti-ties rather than others.For instance, if you wanted to compute the mass of the b quark, you could probably do a pretty good job by tryingto compute the mass of the Upsilon, which would be prettyclose to the mass of two b ’s, m b . We’ve got a verycrude beginning for m b without being able to calculate.We would get m b most likely from masses of the Upsilonand its excited states, and we learn about the interactionstrength somewhere, we correct for the excitation statesthat would make a very accurate value of m b , whichwould presumably be very close to one half of the mass ofthe Upsilon—reasonably close to a half the mass of theUpsilon, in other words to, uh, 5, 10 GeV [speaking tohimself] the mass of the Upsilon is . . . so it has to be 5 [for m b ] . Yes that’s right. In the same way the mass ofthe c could be determined as being about 1.8. Now, if wedid the calculation more elaborately, you see we would bepicking out the light quantities; instead of having the massof the proton and the mass of the pion looking for sometiny deviation which was due to the c quarks, which arehardly affecting either one of them, neither one of them,then to get the [ c quark] mass, that’s not the way to doit.So it’s sensible to try to pick out physical quantitiesthat are more sensitive to particular parameters than toothers. Of course in principle you could compute any oldsix quantities with infinite accuracy and deduce all theparameters. But it would be more practical to choose sixquantities that are more directly sensitive to the masses.I’ve already got rid of m b . . . .Now we want to try to concentrate on things whichdepend—because there’s a great deal of interest in thatquantity—physical quantities that depend upon g , andare not very sensitive to the others. I believe that oneis the mass of the proton, because the mass of the u andthe d . . . we have good evidence are very small, and thatthe mass of the proton is not due to the mass of thequarks inside, I mean at least not directly . . . it has todo with the value of g . It’s hardly sensitive to m s , butwe could imagine that someday we could correct for that. . . However at the present time it is hopeless to computethe mass of the proton from these constants theoretically,and therefore we can’t determine g from the mass of the proton, even though someday we could.Another kind of effects where I look for g are thosephenomena that use high energy and which—so high thatthe masses of these things [quarks] don’t make any differ-ence. And insofar as these [quark masses] are involved,we can presumably compute their effect. In other wordswe look for processes in which we expect there would bea limit, that limit would still exist for this process if themasses of these things went to zero.I’ll give you some examples. This is only to suggestthings to look at that we can calculate that will help usto isolate parameters, in particular g . One interestingexperiment is e + e − → hadrons. e − e + qq q Q e (A.1)
And the idea of that is if you do that at high energies, theelectron and positron annihilate and produce a photon,which you can understand. And then the photon producesa pair of quarks, as a sort of initial disturbance. We havean operator Q f ¯ ψ f γ µ ψ f for each quark flavor, and thisoperator starts by generating a pair of quarks. Now whathappens after that is of course that this [quark] mayberadiates a gluon, the gluon splits into quarks, and theycombine together and they make π ’s and they make K ’sand you get a big splash of junk. The total cross sectionfor doing this is the chance that we got started, so tospeak. You can calculate the probability that we got thesethings started by just figuring that they’re free, becausethe energy is so high. And then after a while they scratchtheir heads and say, “hey, I’m not supposed to be ableto come out, I’ve gotta do something else,” but they’realready there. I don’t know if you feel this intuition verywell, I get it myself but I don’t know how to express it;that at high enough energies, when we start this processand then after that . . .Let’s put it this way: suppose you did this, and youthought one day that you made a ρ . The next time yourealize that you’re not going to see the ρ but the ρ actuallydisintegrates into a pair of pions. Well the fact is, by thetime you got to the ρ , and then it went into π ’s, it isn’tgoing to change the total cross section; whether the ρ does or doesn’t disintegrate doesn’t make any differenceto the total rate. The ultimate things that happen to theseobjects are not much affected by things that happen late,and therefore at low energy, and therefore involving themasses of the quarks and so on. So therefore the totalcross section shouldn’t involve the masses of the quarks.It shouldn’t involve anything in QCD, there’s no couplingconstant at all [the QCD coupling] because this rate to produce this pair of quarks that act like free particles, wecalculate it directly.In the same way we could compute the rate to pro-duce a pair of µ ’s and calculate that pure electrodynamicthing and call it σ µµ . Then I could calculate the prob-ability of producing hadrons here [in the diagram (A.1)with hadronization of the quarks] ; the cross section wouldbe the cross section for producing µ ’s—which also doesn’tdepend much on the mass of µ —we take a very high q sothe masses don’t make a difference; then we would havethe cross section for making, let’s say, u quarks. Thenthe charge is / for a u quark and the cross section goesas σ u ¯ u = 3 (cid:18) (cid:19) σ µ + µ − to produce u quarks; so this would be the probability ofproducing u quarks. We can produce the u quarks in threecolors: red, green and blue, and therefore the thing ismultiplied by 3. Is there any question about that? Yeah? [a question is asked, inaudible] No sir, because this is anelectromagnetic phenomenon I’m talking about, this is a [photon] not a gluon, so this coupling is not QCD. Anyother questions? You had me for a minute [laughter] .This three , I say you can produce any color but that’sa lot of nonsense, you can’t produce any color, becauseyou have to go into a singlet state. So you’re going to gointo the state √ (cid:0) R ¯ R + B ¯ B + G ¯ G (cid:1) = 3 a √ √ a (A.2) Now let’s say the amplitude to go into the red-anti-redstate, what I first calculated over there, let’s call that am-plitude a . Then the amplitude to go into blue-blue wouldalso be a , and the amplitude to go into green-green wouldalso be a , so the amplitude to go into this state wouldbe a times / √ —this is a normalized state—which is √ a . And the rate is a . In other words, three times therate of making R ¯ R . You can fake it if you like, i.e. slop-pily, that there’s the same chance for red quarks or greenquarks or blue quarks, therefore I add them—multiply bythree. Or realize that you don’t produce that state at allbut you produce a superposition. But you come out withthe same answer.So that’s for u quarks. But then we might produce d quarks also. And by the same method of thinking, σ d ¯ d = 3 (cid:18) − (cid:19) σ µ + µ − , etc.And now, we might produce s quarks. Now s quarksare not very heavy, so if an e + - e − experiment is doneat several GeV, or 10 GeV or something like that, thenthat’s another one third [writes on board] . . . And then,if we have enough energy to get above the c , we produce c quarks [writes on board] and then maybe the b ; depends,but if we have a total energy between say 9 and 10, proba-bly 9.65 . . . Then there’s another factor yet; for the ratio of the total cross section to the cross section of the µ , weget a curve µ + µ − σσ total 4/33.72 M M ψ
40 GeV ER = which is an interesting thing, it bobbles around, there arebumps and things for the ρ meson . . . And as you getto high energies, here’s 3.6, it comes here and there’s awonderful resonance . . . and makes the ψ and then someparticles, and then the background comes up here, andover here it starts to make the Υ . . . until we have 30 or40 GeV, and we don’t see the resonance we expected forthe t quarks. But it does show more or less constant aslong as you’re in a region where you’re not at the sameorder as a mass of a new kind of quark that you canmake. And it rises from these various plateaus with thesenumbers [gotten by adding the squared charges from theprevious calculations] . . .At any rate, that’s marvelous , but that doesn’t deter-mine any of those constants—too bad. However, maybethe idea if we do this at high energy, and worry ourselvesabout the b quarks and the c quarks and the interactions,we might be able to get a little accuracy this way. Thereis, of course, an interaction; the trick is . . . that it doesn’tinvolve the masses. e − e + e − e + qq qq g γγ g (A.3) . . . and therefore that should be a good approximation.On the other hand there is the possibility that we cancalculate what happens with the possible emission of agluon . . . It could be that emitted a gluon. Or it could bethat there were interaction forces between these [quarks] by the exchange of a gluon. Now it is not quite as ob-vious that the effects of these things will not depend onthe masses of the particles; but calculations by puttingmasses in show that it really doesn’t. And that there is acorrection that now involves, as you would like, the cou-pling constant g which we discussed. And what happensis, that we get that the same theoretical ratio [as we dis-cussed before] is multiplied by a correction, R = R pure freeparticle (cid:16) α g π + . . . (cid:17) (A.4) which is proportional to the coupling constant. I’m go- ing to use α g . . . strong interactions, put the g just toremind you for gluons . . . plus higher terms. This, then,would be a way, if we could measure accurately enough,to determine the α g , and therefore g . It would be mostsensitive to g . [Question from me: what was that sub-script you put on R , those words?] The words say, “purefree particle” theory. This is the real ratio, corrected tothe first degree for quantum chromodynamics. You canmake a power series expansion in the coupling constant,the first term of which is α g /π . And this then is a wayto determine a quantity which is particularly sensitiveto g and which is presumably not sensitive to the othermasses, although we do have to do a little work to getrid of these things, we make corrections for these things,depending on what region of the graph you want, we cor-rect for that mass . . . and we can do a fairly good job ofcorrecting for the masses . . .So that’s one way of getting a quantity which dependson g and it would be a possible thing. The trouble withit is it’s a correction to an experiment which gives 99— [pauses to think and correct himself, in undertone] nothis is I think this is 5% of the total—95% of the answer,that doesn’t depend on quantum electrodynamics [QCD] at all, and you’ve only got a 5% correction, it’s not veryeasy, you can’t do it very well. So it would have beena nice thing, and it would have been a nice experiment,so if we’re talking ideally that would have been a placeto look for something to calculate and to measure to getsomething that’s practically dependent on g .Alright, now there’s another thing that’s observed. Ac-cording to this model, here, when we knock these twoquarks out, and they’re going very fast, and then theyjust tear out and I don’t know what, radiate gluons, anddo all kinds of things, they fall apart and make whateverstrings there are . . . and what happens is that if we lookat the momenta ——I draw it in a plane because it’s three-dimensional, themomenta, what we get is thousands of hadrons, lotsand lots, most of them pions. Okay. And if we plot-ted the momenta, and I’m only going to plot it in twodimensions instead of three, we find that they’re all dis-tributed in sort of a—at least if q is very large—in a kindof a long ellipse, which is much longer than it is wide; Noise interferes, but I believe he says that α g = g , with theusual factor of 4 π absorbed into his unconventional definition of g . this is . . . the order of, well if you added all the momenta. . . conservation of energy, it’s a rather big number . . . butthey have a certain width, the width is of order of a half aGeV . . . Now, suppose you try to calculate the chance thatthis happens, and that this thing goes off . . . you remem-ber what I do, I have to calculate this correction to thisdiagram [RPF is apparently explaining how the three-jetconfiguration on the right arises from radiation of a gluonfrom one of the quarks as in the previous diagram (A.3)] and I add that to the rate of this [the 2 jet] .Now in this case I could look at this one [the 2-jet] andone would notice that there’s a small chance that whathappens, looking down on the plane, is that geometricallythere’s a momentum like that; you’ve got one quark com-ing out over here in this direction, and this quark startsout this way, if you want, if I employ the virtual diagram.What you see, though, is a quark coming this way, and agluon going that way, the total momentum of which bal-ances this [the other jet] . So if we didn’t see this regionhere, it would seem . . .That can happen, at wide angle, you can ask for wide-angle gluons coming out. What we see experimentallyis that from time to time, it doesn’t look like this [pre-sumably the 2-jet diagram] , it looks more like this [the3-jet diagram] . And if this is sufficiently . . . [probablyreferring to the hardness of the gluon jet] and stuck out. . . then we can interpret it as being this [the 3-jet pic-ture] . Why I have to say that is of course, if it’s notsufficiently obvious, it might be that [the 2-jet picture] with a fluctuation. I mean if these two are close enoughtogether, how can you tell the difference? You can’t. Soyou’d have to take the case where there’s a pretty good an-gle, which turns out to be a low chance. It’s low because,you see, when these two open out . . . [RPF explains on theboard that in this case where the gluon is hard, the extraintermediate quark propagator is carrying large momen-tum, which suppresses the amplitude] . So this happensrarely, but we can see it. And although it does, from thephenomenological point of view does involve soft massesand so on in determining whether we get π ’s or K ’s andhow many, we can at least count how many jets we get,and estimate these momenta fairly well.The only uncertainty is whether we should include aparticular particle here; is that part of this jet or part ofthat jet? So there is some sloppiness in it. The sloppinesswill become less as the energy of the experiment is in-creased. There’s some sloppiness, but we can do a prettygood job of guessing that these things come through this [2-jets] , and by measuring their rate, we get a pretty goodidea of this [3-jet] rate, that has an α g directly, becausethe amplitude for this process has a coupling constant init, and this rate has a coupling constant squared, so thisis a direct measurement of α g . . .Later on in the course therefore we’re going to calculatethat . . . and compare it to experiment to try to determinethe coupling constant. Okay? I’ve therefore illustratedtwo examples, and there are others, . . . that seem to iso-late experimental data and seem to be able to measure, that we can roughly calculate. I say “seem to be able tomeasure” because we have all these little uncertainties,so it’s nothing exact, but a pretty good measurement anda pretty good calculation by which we can determine α g today and prove our claims . . . So that is a perturbativeeffect of something that depends mostly on α g .But what I wanted to explain mainly, the main thing Iwanted to explain, is that there are physical processes, forwhich we can say that we don’t need to know the massesof the other particles. That the process has the samelimiting value, the same probability, whether these areall zero or not. It’s not so easy to correct for them; itdepends on the energy. Let us, since we’ve demonstratedmore or less that there are such processes, let us assumethat there are some physical data, which involve just that [ g ] and not these [the quark masses] , because we makecorrections for this, and there are physical phenomena forwhich we believe, theoretically, the phenomena will stillexist as the masses went to zero, of the quarks, and itdoesn’t involve any other length [scale] .So we could imagine those things, those are specialkinds of data that don’t involve the masses of the quarks,and without saying so, from now on I’m talking aboutthat kind of data. When I talk about a physical quantity,I’m going to suppose it’s that kind of a quantity, okay?Not something like the mass of the π , which depends onthe mass of the u , or the mass of the K , or the differencebetween the Λ and the proton, which is certainly depen-dent on the mass of the s . Those are not the kind ofthing that I want to talk about here. Alright? . . . Thiswill focus our attention.Now we’re ready to go, huh? No! Another compli-cation sets in to . . . The theory doesn’t make sense. Theperturbation theory gives infinities . . . the theory diverges.You all know that. I don’t have to prove it to you; we’lldiscuss it all later in detail. We’re going to go back overall this and do it. I’m describing where we’re going to goand what we’re going to see. So the theory has infinities.In the case of electrodynamics, which you studied, andas you know what this does in that case is that there issome kind scheme for cutting off all the integrals that aredivergent, in other words (cid:90) d kk ( p − k ) . Then this is divergent logarithmically because you’ve gotfour k ’s down here and four k ’s up there. And what peo-ple do is they say that the propagator for instance for aphoton is replaced by subtracting from it what you wouldget if the propagator had a mass and then taking the limitas the mass goes to infinity, p → lim Λ →∞ p − p − Λ (A.5) and obtain results which have logarithms in them; youhave cutoffs, actually, and what do you do with these in-finities? What we do with these infinities is the following. If we discuss for example the scattering of two electronsby a photon at large distances, where the potential is e /r ,or the scattering amplitude is Q ∼ πe Q (A.6) If we were to compute this, at very low Q , and long dis-tances, then this [the coupling e ] is an experimental num-ber. And now we discuss that this experimental numberis not the same as the number e that I would put intothe theory, right away, because there are virtual diagramswhere this may make a pair and the pair may annihilate,and like that, and this has got that kind of a divergencein it, and corrects this [(A.6)] . What we’ve discovered isthat [taking his time to write it from memory]1 e = 1 e + 13 π ln Λ Q Yes. Now what we’re supposed to do is to make Λ go toinfinity and this gets some kind of nonnegative result. Q is the momentum transfer, which is supposed to be small.But what we say is, when we do the theory with a cutoff,we change the theory, because the theory by itself doesn’tmean anything. So we use a particular cutoff, and wetake an e such that the physics is independent—is cor-rect, agrees with experiment. We choose a cutoff Λ andan e so that it agrees with experiment. And now if wechange the cutoff, we change the e so that it continuesto agree with experiment. In this case, we could changethe e —if we change the Λ we could change the e so thatthis . . .It’s very important, the right sign, and I get the rightsign by [long pause while RPF checks that the sign of therunning is correct] . That’s 137 [1 /e ] . As Λ gets verybig, this [the log] can get bigger than 137, so this [1 /e ] would have to go negative which makes no sense. Thistheory really doesn’t work, because it means this wouldhave to go negative; I must be missing a sign. . . In anycase, to encounter this problem you need to put in such alarge cutoff that the logarithm is π × which is morethan 1000. Then we would need Λ ∼ e Q , which is amass greater than the mass of the universe. So there is nopractical problem. So that’s why we repeat all these cal-culations, without ever getting into any trouble, in prac-tice. In practice, we never have to take the Λ so large to Recall that RPF prefers to normalize the gauge couplings as α = 1 /e . get a good accuracy . . . From a theoretical standpoint thatseems satisfactory.Alright. So what the trick is, and the point is, is thatwhen we’re putting our cutoff down, we’re changing thetheory. And when we use different kinds of cutoffs, we’reusing different theories. However, as it turns out thatby putting different g ’s in, we can get the same physicsfrom the different theories. For each theory you have tohave its g , for each Λ there has to be an e . . .But what happens in electrodynamics [QCD] is better,because the sign is the other way around, and this is whathappens. We could ask, for any experimental data . . . wecould ask the following thing. . . . So we first have to mod-ify the theory, to make it work. Really, we have to definea process of calculation. There are several methods. Oneis the method I mentioned before [(A.5)] , . k → k − k − Λ plus some tricks to keep gauge invariance. It can be done,it’s not very good about that without those tricks for gaugeinvariance. . . . [Tape was changed here. RPF is describ-ing dimensional regularization.]2 . dimensional regularization. α becomes dimensionful when N (cid:54) = 4; α = α s (Λ D ) − N . . . as I will show you, physics with a fractional numberof dimensions. So we calculate with dimension N . Andthen we write N (cid:54) = 4 . As it turns out, when the number ofdimensions is N , the coupling constant, which I’m goingto write as α instead of g , the coupling constant α hasa dependence on a scale, an energy scale; that is, it’s nota [dimensionsless] number, it’s not a pure number as itis in four dimensions; it has dimensions. So if I put adimension, say Λ D or something which is some kind of alength, to a certain power, namely N − , then this thingin front [ α s ] will be a constant, as we will vary Λ . . .Third method: we replace space and time—spacetime—by a lattice of points in spacetime. And to define what todo on that lattice, which is analogous to the Lagrangianhere, and I will discuss that, that’s called the latticemodel. . Lattice model.Has a dimensionful parameter: a = 1Λ a The lattice model has a dimension to the lattice, howsmall it is. The dimension a corresponds to an energy—I’m going to talk about energies— / Λ a ; this is the spacingof the lattice, the lattice spacing. Later on in the course,I will discuss both of these methods, not so much this one,but I’ll talk about this one too.Oh by the way! This method is not defined yet, wealso have to say in what gauge we do it in. There’s an axial gauge, there’s a ∂ µ A µ = 0 gauge, which propaga-tor, whatnot. They’re all variations on a theme. All I’mtrying to say is, all these things are mutilations of ourbeautiful scheme. Mutilations which we have to make,because otherwise it’s all meaningless. However, we ex-pect the following. Let’s take the lattice model; it’s theeasiest to understand. Surely, if the number of pointstaken is sufficiently fine, we’ll get a damn good repre-sentation . . . [he compares to numerical algorithms forapproximating differential equations as finite-differenceequations] if we don’t get enough accuracy we make thelattice smaller. For any particular size of lattice, thereare artifacts—errors—if you want to talk about some-thing this big and the lattice is that big . . . The size of thephysical phenomenon here would be something involvingthe reciprocal of the momentum cubed . . . Then the q [mo-mentum] is the reciprocal of the wave number, and wewould like to get our constant [ a ] smaller than the wavenumber, or we’ll never be able to represent that . . . Butin order to get good accuracy, if we want to make it re-ally good, we should make it still smaller, because as weall know, the more fine we have the lattice, the betterthe representation. This is some sort of limit, that if wetook the lattice fine enough, we should get more and moreperfect agreement with everything.However . . . so we would like these Λ ’s—or in thiscase it’s a question of how fast you come to the limit N = 4 , and in this case it’s a question, surely puttingsomething like that [the wrong-sign propagator in Pauli-Villars regularization] is going to change the physics if Λ is not enormous compare to Q . Right? The prop-agator is different—you changed it. But you don’t thinkyou changed it much. So, we have the idea, and it turnsout to be right, that if we have a phenomenon at a cer-tain scale Q , and if we take Λ ’s much bigger than Q , andwe consider different Λ ’s, we can always find a constant [bare coupling] to put in that gives the same physics. Atfirst sight you might not have realized that you’ll have tochange the g . You would have thought that’s going tobe fixed, but the theory with its divergences shows that inthe same way as in electricity there are going to be logs. . .And so what happens is, when we ask ourselves, how tochoose—let’s pick a datum to analyze, any one of them. . . We imagine we could calculate it. Then we ask how tochoose the g —the other constants are involved, but wesuppose that we select the datum as something sensitiveto g . . . [RPF is illustrating this on the board, but Ido not have it copied in my notes] . . . so the physics isalways the same. Or the physical datum comes out alwaysthe same. The datum agrees with the theory . . . . Withone datum we can always make it agree with the theoryby adjusting for a given Λ by picking out the g . Foranother datum, we might not get perfect agreement, ifthe Λ wasn’t large enough, right? Because the theory isa little dopey at low Λ . If it was a lattice, the latticescale is too big. So we really want this in the limit ofvery, very large Λ . We aren’t interested in the formula for a Λ of the same order as Q or Q or something likethat. That’s not the problem, because things will workfor one datum but the other one won’t work, because thetheory isn’t right on the scale of . . . Is there any questionabout this idea? I’ve tried to explain the idea, and I hopethat if you don’t catch on tell me what’s bothering you,and we’ll straighten it out now. [Question from a student, inaudible] That’s right, we’reassuming that this thing is going to work. Yes, this busi-ness of choosing a lattice is analogous to the usual one,choosing a lattice for, say, doing the diffusion equation.And you expect it to become more and more accurate asyou make the lattice finer. The only complication is thatwe have to keep changing the g as we do so, in order tokeep the physics the same. But the presumption is thatwe will be able to do that, that there will be a definitelimit. That’s an assumption which is, well I don’t knowwhether people can claim they have proved it, but it seemsto be true. Okay?Anyway we want the physics to be the same when wechange the Λ , but also, by the way, when we changethe method [of regularization] . Like I said [referring toa previous illustration on board] , various Λ ’s and vari-ous methods; I’ve labeled the Λ ’s differently for differentmethods, but it’s the same idea. And I’m now going totell you the answer . . . and later on I’ll prove it . . . Is thereanother question? Okay, the answer is that the value of g that you have to choose has the following expansion in ln Λ : g = β ln Λ + β β ln (cid:0) ln Λ (cid:1) + c + (cid:16) a ln Λ + . . . (cid:17) (A.7) where the further terms that are small [falling with ln Λ] ;these don’t interest us. We’re supposing that Λ is bigenough . . . β and β are computable and known. c is ar-bitrary; it’s where we have the room to choose, to makethe thing fit the data. That’s the constant we chooseto make it fit the data, so c is arbitrary, you can haveany constant there, and it’s chosen to fit the data. Butthe formula, and what you have to choose, depends onthe method—by the method I mean whether you use thelattice method or dimensional renormalization or whatmethod of ultraviolet . . . you use. a [in (A.7)] also de-pends on the method; β and β do not depend on thecutoff method. [Question about the arguments of the logs not beingdimensionless.] Yes, now that’s ridiculous, isn’t it? Verygood, Λ ’s are energies. And I’ve written logs that aren’tany—ha ha ha . . . Well, the old professor can fix that,we’ll divide by (1 GeV) : g = β ln Λ GeV + β β ln (cid:18) ln Λ GeV (cid:19) + c + (cid:16) a ln Λ + . . . (cid:17) (A.8) But why ? We’d better contemplate that some other joe will come along and use
10 GeV in there. Sowe’d better contemplate what happens if instead of writ-ing with underneath with M underneath, beforesomebody else comes along, he likes to write his thing dif-ferent. He would write this. And I would like to explainto you why it doesn’t make any difference. Because thatmakes it look that it is still more arbitrary, that we haveanother thing . . . we don’t. And I’ll show you why. Wemight have a different constant [ c (cid:48) ] , that’s the clue: g = β ln Λ M + β β ln (cid:18) ln Λ M (cid:19) + c (cid:48) + a (cid:48) ln Λ M + . . . = β ln Λ + β β ln (cid:0) ln Λ − ln M (cid:1)(cid:124) (cid:123)(cid:122) (cid:125) +( c (cid:48) − β ln M ) . . . ln Λ (cid:18) − ln M ln Λ (cid:19) = β ln Λ + β β ln ln Λ + (cid:0) c (cid:48) − β ln M (cid:1) + (cid:18) a (cid:48) − β β (cid:19) + . . . (A.9) You see that all I did was change the constant. If I hadused “1” in here [for M ] , I’ve got a certain constant,when I made them fit the data. If I had used 10 or M ,I’d get a different constant, that’s all. So I’m still okay,right? The M makes it look as if there is another pa-rameter, which adds to the confusion of this damn thing.Because when you hear about choosing M and choosing Λ and choosing g , there’s no choosing M , really: it doesn’tmake any difference. It’s just a question of the definitionof the constant when you go to fit the data.Now the part that I hadn’t finished is here; you no-tice that as Λ goes to infinity . . . [RPF explains that theregularization-dependent term a (cid:48) becomes negligible asthe cutoff is removed.] So I must take my Λ big enoughthat this term doesn’t amount to anything. So every-thing’s okay, and that answers your question about theunits. Is that alright?Sir? [Question from me: Is this equation just the two-loop approximation?] Yes—no, it’s exact! [me again:there aren’t more logs of logs?]
Oh yeah, maybe downhere, there’s log log, this times log log, stuff like that.But always smaller, okay? No, there’s no log log log, no.There’s no log log log. I’ll explain to you why. It may bewrong in that there may be a term—I’m not sure, okay,like ln Λ times ln ln Λ , or something like that, which isstill smaller than this one, but not much. And things likethat, but these are all dropping out as Λ goes to infinity.So, I should say, terms of this order or smaller are goingto drop out, that’s what this curly line means. Okay? Allwe have to do is to take the Λ very large and then wecan do that.Alright. And that’s the formula how then we believe,that if we do that, choose the g so it’s equal to this andadjust the constant c , we can fit the first physical da-tum. And then a second physical datum . . . what with thesame thing, with the same constant, you should get a fit to experiment, provided that we’ve chosen the Λ largeenough that this [the a term] is small enough that every-thing’s okay. Alright. Are there any questions? That’sall there is to it. That’s all there is to what we call theway the theory’s supposed to work.Review. The theory diverges. In electrodynamics that’sreally serious [the Landau pole problem] but we don’tpay any attention to it for practical reasons, e is suffi-ciently small . . . In quantum chromodynamics because ofthe opposite sign, there isn’t any real difficulty [becauseof asymptotic freedom] . We can choose the g in termsof the method that we use to make the cutoff in sucha way that we would expect that the physical data agreewith experiment. That is what the theory is. The theoryis, strictly speaking, not the Lagrangian which we wrotedown or the path integral we wrote down, but the pathintegral plus all this crap about how to make a cutoff,plus this baloney about how we have to choose g . Andthen we should take the limit as Λ goes to infinity to getthe most accurate result. That’s the theory. The theoryof quantum chromodynamics is not defined by the La-grangian alone. To put it another way, you cannot sayto a mathematician “hey, here’s my Lagrangian . . . , fig-ure out the consequences,” because you haven’t told himthe full physics of what you intend to do, which is—because if you give him that he’ll find out that the answeris infinity—it doesn’t make any sense. The true theory isthe Lagrangian plus a cutoff scheme, plus a proposition asto how the g ’s go, so that the results will be independentof Λ , the cutoff scale, as the cutoff scale gets sufficientlyfine. And this has to be the way to do it.Now you can find this out by perturbation theory, ofcourse. You’ll notice that as Λ gets very large, the g ’sare very small. When the g ’s are small we can computeeverything by perturbation theory, and that’s the way youcomputed the β and the β . Well why didn’t we com-pute c ? Because you can’t; it’s an arbitrary constant.And why didn’t we compute a ? Because that depends onthe method. We can compute it for each method, butit doesn’t do us any good. I’ll tell you why. This is asignal that there will be errors of / Λ [1 / ln Λ ] in theend, because we haven’t taken a fine enough lattice spac-ing, using the example of the lattice. Because the latticedoesn’t really represent the continuum. And this is a kindof measurement error. So it’s no use to compute this “ a ”accurately for a given scheme . . .Yes? [Question: Λ is something you choose arbitrar-ily?] No, we try to make it as large as possible. [student:We try to make it as large as possible, but we can choose it to be as large as possible.]
Yes, that’s true. [student:Okay, it seems like we can always choose Λ large enoughto make g small enough that we can use perturbation the-ory] That’s correct [student: and then we can use pertur-bation theory on any problem.]
That’s correct. [student,not satisfied: I have been told that . . . ]
But the seriesdiverges . . . for processes with small Q , small momen-tum transfer, the corrections to the propagator get biggerand bigger [student: even when g is small?] What hap- pens is, when you take this exchange between two quarks. . . you get corrections . . . say a loop of gluons or some-thing, which modify the propagator between two quarks.Now this thing, when you calculate it, involves somethinglike ln Λ /Q , when you calculate it. This interaction willnow will have a term like this with an extra g . Althoughthe g is small, the ln Λ is undoing it, and you get afinite . . . which isn’t small. It gets to be small in its effectif Q were big enough, as we will learn next time, but ifyou ask the question at low Q , it just doesn’t work. Thedivergences of the perturbation theory undo the smallnessof the . . . [bare coupling in the UV] . Okay?Alright. Well the difficulty with the perturbation theoryis not that it doesn’t exist; it’s that you can’t sum it. Wedon’t know how. Sometimes we can sum some terms,but we can’t do a very good job, we have to think aboutit, rather than calculate it, even though with sufficientlylarge Q , the effect of g is small, for smaller Q the effectsare bigger . . . [The following apparently refers to eq. (A.8) or (A.9).] I’ll be putting a mass squared here from time to time,maybe . . . you’ll appreciate that it doesn’t represent anindependent choice of . . . It does—there is a way of mak-ing it look like an independent choice. Obviously, thereis . . . suppose that we finally fix the data and worked itout and determined c . And somebody could find an M sothat this canceled out. And then he could say that thisformula for /g is exactly this. And there’s no constantand the other constant is M . And there’s all these dif-ferent ways of representing the same thing, which causesa tremendous amount of confusion to a lot of people, andI’m sorry for that, because we really . . . calculate it, so wedon’t know the M ; some people come out with . orsomething like that and say that’s what it is, . ,whatever. So they don’t know it well enough, so we don’thave the numbers accurate enough that we can do one oranother of these things once and for all and be done withit, so we have to kind of leave all these balls in the air,as to which way you would prefer to write it—whether tochoose an M and say that it’s the constant [ c ] I want todetermine, or to say the constant I’m going to choose iszero and it’s the M I want to determine. So you’ll heardifferent people saying different things, but you have tounderstand that they’re all equivalent. It always makes ita little easy to do it on the blackboard because I’ve pre-pared the lecture, but then you have stop and think arethey really equivalent . . . or you’ll forget how I did that.Are there any other questions about the idea? As youcan probably see, because of the logarithms, you might ex-pect /g is not really converging, and if we are workingat
100 GeV or something like that, and you wanted tochange Λ to change the logarithm in order to get some-thing that . . . it’s damned hard! . . . And therefore some- RPF meant to say 0 . QCD that heusually denotes as λ P . thing like the lattice . . . They have a technical thing; theycan do some calculations on lattices . . . The lattices reallyaren’t small enough to get a good answer . . . They’re assmall as they can make them and still do the calculation,because of the number of . . . that are available . . . Butwhen they try to make the Λ smaller to be more accu-rate, they need a lot more computer time, because in fourdimensions if you decrease the lattice by one half, youhave sixteen times as many points to compute. Sixteentimes as much work. But even changing the lattice by / is only changing Λ by a factor of two, and ln Λ doesn’tdo much. And so it is very difficult, in fact I would sayvirtually impossible, to make the numerical calculationpractical . . . limitations of computers. To make the nu-merical calculations with greater accuracy . . . So I thinkwe have to study this theory, not only to figure out ananalytic way . . . to understand well enough what happensat short distances . . . so that we have a better way of com-puting that is less sensitive to this brute force scheme thatthey’re now using. During this course I will discuss allthe numerical calculations, and more of these methods,and everything else; this is just an introductory lecture toexplain where we have to go. Appendix B: Transcription: Renormalization:applications (1-12-88) [A student is asking whether RPF is going to explainthe correspondence between dimensional regularizationand the cutoff to which he has been referring so far.]
Yes,I am. I worked it out the other day and it’s very simple;I understood it [RPF says something to the effect that hemight get some details wrong here since he is going bymemory]
The point is that in dimensional analysis [regu-larization] the coupling constant has a dimension, so yourepresent the physical coupling constant as g times somedimension—some energy, which corresponds to our Λ —to a power of 4 minus the number of dimensions. Nowwhen you do an integral over correction terms in pertur-bation theory, all of those integrals converge if the num-ber of dimensions is less than 4, and so the correctionsto the coupling constant—that’s what I’m trying to getstraight . . . Now as the d approaches 4 it turns out—let’ssay we write d = 4 − (cid:15) —then there’s an (cid:15) down here asyou approach . . . It ends up that you’re trying to workout something like (1 /(cid:15) ) × Λ − (cid:15) which gives you /(cid:15) times − (cid:15) ln Λ . . . I was trying to get that straight just before Icame . . . I couldn’t figure it fast enough . . . but there’s adirect correspondence . . .We have been talking about the renormalization of thecoupling constant in a kind of abstract way, and as usualat the beginning of each lecture I have to fix up some mi-nor things . . . in order to make everything I said conso- RPF means the lattice spacing nant with the outside world, all the equations in whichI wrote g , in all those the g should be replaced by g / π ; then α = g π (B.1) This is the real world. This is me. My α was πg times π , which is not good to do. This is the right thing; firstyou substitute this, then you substitute that . . . All theequations are changed with the appropriate positions ofthe π ’s. There was one equation that we chose to define α ( Q ) , and that was /α ( Q ) , which now becomes πα ( Q ) = β ln Q λ P + β β ln 4 πβ α ( Q ) (B.2) The equations I’d written before didn’t have the π ’s. Al-right?Now I will just remind you of what we discovered, thatwhen we did perturbation theory to order g to some pro-cess, that’s all the order we worked out. And this is re-placed ultimately by α ( Q ) . . . that’s in perturbation the-ory; it becomes this if you would sum the leading logs,because you always know they come in common. It can’tbe g , because that’s ln ln Λ . . . nothing depends on the cut-off . . . In other words, when you do first order perturba-tion theory, you simply replace the g ’s by α ( Q ) , and youget a much more accurate result. You’ve included all thehigher order leading logs. If you try to do it to next orderin the coupling constant, it gets a little more complicated.You get g + g (cid:18) ln Λ Q + a (cid:19) (B.3) —I’m not going to try to get my π ’s right—and we knowalready there was one of these [in the argument of the ln] in there, that has to be there, and that’s just to get theright coefficient. And then there will be some constant [ a ] that has to be worked out when you do the second orderperturbation. If that’s the case, then this turns into α ( Q ) + a α ( Q ) (B.4) So the way to do second order perturbation theory, is afteryou do it, take away the logarithmic term and just look atthe constant term, and the constant term is the coefficientof the second order term in α . And it will be a littlemore complicated with the 3rd order term. But we canwork it all out. And it tells you, in other words, fromthe perturbation expansion . . . to write it not in terms of g but in terms of α ( Q ) . It’s the effect of summing theleading logs, that are evidently going to come in, although Apparently RPF made a side-by-side comparison of the two no-tations, but I only copied the “real world” version in my notes. we haven’t worked it out for the higher terms. This is amuch harder . . .So in fact therefore people say that the coupling con-stant is dependent on Q because of the running—that’swhat running means . . . I just want to say it again, itlooks complicated but it’s relatively simple; at first orderyou replace g . . . at second order . . . .There’s many papers and places where you can readabout this process of renormalization; I mentioned a [ver-sion of ?] the renormalization group equations which willlook much simpler than any ones you’ll see anywhere,unless of course I have it physically right and have doneit nicely. The problem is making sure that the quantitythat we’re dealing with depends . . . And that the quantitywe’re dealing with is physical and doesn’t involve some-thing like just a Green’s function, an expectation of φ atone point and φ at another point. Because the φ ’s–wavefunctions, or field operators—also shift their coefficientswith various c ’s and so on. If you deal with a physicalquantity that you measure, you don’t have any of thatstuff. For example, if we have an expectation of vectorpotentials (cid:104) A ( x ) · · · A ( x n ) (cid:105) at one point times another point or something like that,then you have to watch out that these vector potentials arealso changing their definition as we change the couplingconstant. So in the renormalization group equationsfor Green’s functions . . . well it looks much more compli-cated, but it really isn’t that much more complicated, it’sjust the physical ideas were adequately described . . . it’salways a good idea to stick to physical questions . . . Soa lot of . . . book or any paper on renormalization, youfind it enormously more complicated than anything you’veseen; they’ve added all kinds of extra stuff. It also hasa lot about the history where people tried this and thatand did this and did that and proved it this way andproved it that way . . . The subject looks worse than it is.Okay. So try to look it up and . . . that I cheated yousomehow in describing it . . . I did find a nice book calledRenormalization by John Collins, Cambridge UniversityPress, 1984. I can’t read it all, it’s too complicated forme. Now there’s thousands of references on renormal-ization . . .Now that’s the end of last time—I’m always fixing upthe lecture before. Oh, there’s one more [thing to fixup] ; I’ve forgot the β /β here [eq. (B.2)] , and I’d bet-ter tell you where the π ’s are . . . The fact that this ispositive, if it was only about electrodynamics and didn’thave the solution to write the conclusion to be negative;it also suggests, the fact that it’s positive means that ev-erything will work . . . wonderful theory . . . [RPF refers to In my notes I have written that RPF is talking about anomalousdimensions here.
However if the num-ber of flavors is more than something like 17, then we’rein trouble. So most people believe there is not 17 flavorsof quarks. We only know of five so far. It might be six,since people like to fill out the symmetry with t quarks.But there is no theoretical reason to say that there isn’tanother group of three, like u , d , s —( u , d ), ( s , c )–strangeand charm, and you have beauty and truth or something. . . [inaudible, some other word starting with “t,” draw-ing laughter from the class] . And maybe there’s x and y .And maybe there’s w and v —we don’t know; there’s nounderstanding as to why there’s more than one family,or why it stops at three families . . .Okay, well, . . . the problem is how do we actually calcu-late something . . . and perhaps this whole problem . . . andyou’ll see it all coming out . . . So that’s what I’m aboutto do. But even there, before I do a particular prob-lem, I want to do something about guessing where thedivergences are going to come. We all know from do-ing quantum electrodynamics and other field . . . I’m as-suming you’ve taken a course in field theory . . . comesout to be divergent because of . . . And so let’s try to findout when we’re going to get . . . So if you were to take avery complicated diagram to calculate, some crazy thing. . . there’s a quark, quark, gluon, gluon . . . it doesn’t makeany difference . . . something like that . . . (cid:90) d k d p ( p , k + · · · ) × f ( p, k, · · · ) (B.5) And then such an object will end up with two integralsover momenta, one for this loop and one from this loop,four-dimensional integrals . . . Therefore we need to inte-grate over eight variables, and in the end the question isdoes it diverge? The real question is, will there be—whatkind of formula are we going to integrate? There willbe various k ’s, minus this and that, propagators, maybethere’ll be six propagators. I don’t mean there’s a sixthpower of the propagator, I mean there are six of thesekind of things in a row. Perhaps . . . So I’m not going toworry about whether it’s p or k . But there will be fromthe gradients in the couplings, up here [the numerator f ] there will be some k ’s and p ’s. And then the questionis when we go to do these integrals, we will get a diver-gence, a logarithmic divergence or . . . depending on howmany powers are down here and how many powers areup here. If there’s more powers downstairs than thereare upstairs, then it will be a convergent integral.So what we have to do is count how many powers thereare upstairs and downstairs. And that means lookingat all these couplings and seeing if there are gradientsin them, taking two powers for every propagator of glu-ons, one power for every propagator of quarks . . . Now Ithink there are no gradients in the coupling of a gluon to a quark and so on. And we get all . . . and have a bigcounting job. And it will depend on the structure of thediagram. And now, for a miracle. There are very manyrelationships between the way diagrams are constructed,and what kinds of topology you can take. Of course oneof the typical theorems of topology is that . . . I don’t knowwhy that should be relevant for this because it doesn’t haveto be a planar diagram, that Edges + Vertices = Faces + 2 (B.6)
In other words, there’s a relationship between the num-ber of loops, the number of vertices and the number ofpropagators . . . But what I’m trying to say is that thenumber of loops, the number of junctions, the number ofcouplings, and all this stuff are not completely indepen-dent of each other, but they’re related to each other. Andthose relationships turn out to mean that I can make thiscount and tell you the answer in a very nice way, it’s verysimple. The net power of any integral, the number of nu-merator over the denominator—in this case for example,one, two, four and four is eight . . . there are twelve downhere, let’s say there’s one more here, that would be four-teen, so that’s a net of minus two, so that’s convergent. The net power of all the momenta, which is N , has thisproperty: N = 0 , log divergence N = 2 , quadratic divergence N < , converges (B.7) and the wonderful thing is, no matter how complicatedthe diagram, the formula is that N = 4 − N g − N q , (B.8) four minus the number of gluon lines coming in fromthe outside, minus / the number of quark lines comingfrom the outside, period! It doesn’t make any differencehow it’s all structured in there. That’s an entertainingthing; you can play around and try to prove it to yourself. . . by actually counting things up and showing various re-lations of the number of intersections and junctions andthree-point couplings. See, for example, this relation be-tween the number of junctions and the number of lines,because each line has two ends, so you know, take thenumber of lines divided by two, it’s going to tell you howmany junctions there are. At any rate, this ends up as be-ing true, which is most remarkable. Well now I’m goingto prove . . . [Question from student: . . . superficial degree of diver-gence?] Yes, yes, yes, superficial degree of divergence. It In my notes I have written that this assumes a vacuum diagram,no external legs. The answer for the diagram in (B.4) should be − is often called a naive counting divergence because, whatcould happen, is that this whole problem, turns out thereare momenta up here, but they’re not the momenta ofthe integrand that you have to integrate over, but theymight be the momenta of the outside lines—let’s call q the typical momenta coming in, that we’re not integrat-ing over. Then dimensionally, from the point of view ofthe number of energy terms, it’s the same dimension, butthe integration is more convergent. [Student: Or whatcould happen is that we have two integrals, over k and q ; the q integral is very convergent, and the k integralis divergent.] That could happen, but it usually doesn’t. [Another student: but anyway this is a worst case, thiscounting?]
Yes.Let me explain how I did this. A way of looking at it,one way of a direct count, the most obvious way . . . Here’sanother way. This, we could say, is part, it’s a diagramfor some process. It’s a piece of a lot of terms that aregoing to be added together to produce a matrix element T for a process. Now since they’re going to be addedtogether, they all have the same dimension, so the di-mension of this is the same as the dimension of T . Butin the case of T , we have various rate formulas. Let’stake an example. We have a single particle going into. . . I don’t care how, and disintegrating into N − parti-cles, one coming in and N − coming out. Then wesay that the rate at which this happens goes like this: N−1 final particles12 N3
Rate of decays of 1 into N − particles: d Γ = 12 E | T | N (cid:89) i =2 (2 π ) δ ( p i − m i ) d p i × (2 π ) δ (4) (cid:32) p − N (cid:88) i =2 p i (cid:33) (B.9) including the (2 π ) ’s—which have nothing to do with di-mensions, but I’m being accurate for a change—that’svery rare. That’s the exact formula for the rate.There’s a similar formula for the cross section, but that’swith two particles coming in; let’s just take the case ofone.What I’m going to use it for is to determine the dimen-sions of T . By the way, in this case it’s very importantthat the coupling constant has no dimensions. We’re go-ing to have coupling constants, g , g , g , g all over thisthing, and those g , g , g , g ’s is not going to make anydifference to the dimension. So all I have to do is findthe energy dimension of this thing, and thereby obtainthe maximum degree to which it could possibly diverge.Okay, now the rate is one over the lifetime of that object,and which is therefore an energy, [ Rate ] = (cid:20) τ (cid:21) = [ energy ] (B.10) The factors of (2 π ) − for each d p i are missing and multiplied by this energy [moving E to the left-handside] , we get[energy ] = [ | T | ] [energy] N − [ energy ] − Okay? And so we find out that [ | T | ] = [ E ] − N or [ | T | ] = [ E ] − N (B.11) where N is the total number of lines coming out. Andthat’s what this formula [(B.8)] was supposed to be, onlyit’s slightly . . . Because we then can prove—you have towatch out, we talked about the dimension of the inte-gral that we’re going to get when we do this. Well that’snot quite the same, because whenever a quark comes in,there’s a spinor for that quark, and that spinor has a di-mension. So the integral [the loops inside the diagram,with external wave functions removed] is not the sameas the dimension of T . But you have [the dimension of] the integral is four minus the number of gluons minusthe number of quarks, which is what I have there [(B.11] ,that’s T , but for each of the quarks there was a spinor,which has a dimension of / , and so the integral doesn’tquite have the same dimension as the T , I = 4 − N g − N q − N q (B.12) To remind you, that a quark or spinor has a dimensionof / , you remember that if you’re going to sum thisover spins, when you sum this over spins and there arequarks in it, what you do is you say, oh I know, I’ll getsome kind of a matrix element, and then I put a projector (cid:80) u ¯ u = /p + m in it when I sum over spins. That meanswhen you sum over spins, you put an extra energy in. Sothe dimensions of the T when summed over all the spinshas these factors, one of these for every quark. That was | T | , therefore half an energy for each quark, so that’swhere this [last term of (B.12)] comes from.And now finally if you have some good reason to knowthat the final answer . . . for some reason you know, gaugeinvariance might be such a reason, that you know the finalanswer must have zero divergence [in terms of contractingexternal momenta with the amplitude] , for a gluon withmomentum q that it has to come in this way, q δ µν − q µ q ν (B.13) so that it will automatically give that q µ on that is zero.So let’s say we know there must be a factor like this infront. Then of course we know the dimension of the in-tegral is that much smaller. So the thing to do to remindyourself of that is to take − p , N = 4 − N g − N q − p (B.14) where p is the known power of the coefficient in front.What I mean by that is the power of external momenta.If you say well I’m going to take the worst case, then p is zero. We’ll consider p zero; I’m going to talk about thecases where it isn’t . . . Alright?Now let’s—oh, I had noticed something, that I copied. . . scrap of paper; I’m not going to remember all of it,therefore I can’t guarantee it, so you might like to try toprove it. I also got interested in what the order in g isfor a given diagram. So I did all my algebra . . . to findthe order. And the particular way that I worked out theorder was the conventional way, in which the fields thatI usually use are replaced by gA so that the action lookslike ( ∂ µ A ν − ∂ ν A µ − g [ A µ , A ν ]) —that’s not my conven-tional way. That’s the way I want to calculate the order. . . When I define it that way, and I find the order of g ,I find the following rule, × ( loops ) + N g + N q − . . . to give you an exercise . . . I don’t guarantee it, becauseI found it on an envelope without checking it. In case youfind that useful, maybe you could disprove it or prove it;it would be interesting to try.Let’s find out what kind of diagrams diverge. And tobe specific in drawing the diagrams, I’m going to drawthe lowest order in each kind. The lowest order in g ,not the lowest divergence . . . Then we can represent ourdiagram by telling how many gluons there are, and howmany quarks there are. Here’s a little table, and for eachcase I’m going to draw a diagram to illustrate it. Nowif there are no lines coming in, so the diagram has noexternal lines, well we never have to calculate it . . . Sowe start with one gluon. And that’s a thing that lookslike this: the gluon’s coming along . . . you can have aloop of gluons, or you can have a loop of quarks; that’sa typical diagram. I just draw one typical diagram ofthis kind. There would be no electron lines coming infrom the outside, and this number [ N ] would be three . . . N g N q N N , for graphswith N q external quarks and N g external gluons. [Question from student: you mean quarks?] Yes, always In my notes I had added the extra examples.
I mean quarks, not electrons. I say “electrons” and I say“photons,” but I mean gluons when I say “photons,” andI mean quarks when I say “electrons.”Now the next case would be that there were two gluons,and there was something going around. It could be aquark. Or if you prefer, to make it more interesting,make it a gluon, I don’t care. Because these are typical,I’m only illustrating. If there are two gluons coming in,as you know, now the divergence is 2. Well you can keepthis going. Now we’ve got three gluons, with somethinggoing around here, these are just to illustrate the idea,and there is no quarks coming, the divergence is 1. Orthere could be four, [RPF makes sound effects as he drawsthe legs] beep, beep, beep, beep, [and the loop] loo-loo-loo-loo-loo; now we have four of these, and the divergence is0. And now if I put five, then it gets convergent. And Istop now; I’m only interested in the divergences.So I start now, over again, this time putting in somequark lines. Now you can’t just have one quark line be-cause of the conservation of quarks, so the first case youwould get would be something like that, it would be no glu-ons coming in, and two quark lines coming in, and that’s [ N =]
1. And then you could have two quark lines comingin and one gluon line, and you get 0. Now the next thingwould be with four quarks, but that’s already convergent.And that’s the end. Those are the only diagrams—sorrythe only types of diagrams—that will bring divergences.Mind you, this diagram here, this is lower order, it’splus any internal complications, don’t forget. That is, thesame divergence will appear to occur if I draw a diagramlike this. It’s the same divergence. And that, due tothat theorem [eq. (B.12)] . . .Now we discuss [in more detail] the individual terms. . . Well if you have a gluon coming in . . . the vacuumhere . . . ∼ (cid:104) A µ (cid:105) , (B.16) a kind of expectation of the vector potential. The vectorpotential could be in any direction in the vacuum, and itaverages to zero. So this is physically zero, and we nevercalculate it, because of symmetry. The symmetry is this:we can make A → − A , the theory is unchanged; not true,not true . . . a little more subtle way; there is a symme-try in here. You change the sign of . . . something likethat; anyway there’s no direction you can . . . the gluon,so there’s no expectation for the mean gluon field. Sothere’s no term, you don’t have to worry about it now.In fact we never calculate it . . . there’s never any . . .Now ordinarily we would expect this ∼ δ µν q − q µ q ν (B.17) I did not copy the diagram, but the reader can imagine addingloops to the lowest order diagrams. to be a quadratic divergence, from what it says here [in thetable] . It turns out that the conditions of gauge invari-ance in the case of quantum electrodynamics for instance,and also in quantum chromodynamics, means that therehas to be such a factor [ q − q µ q ν ] , and I’m not going toprove it now, I’m just telling you about this, that gaugeinvariance makes that p = 2 [in eq. (B.14)] . And thisturns this to a log divergence. In other words we over-count . . . it’s not as bad as we think. However if we goto calculate it, if we’re not very careful with the cutoff,suppose that the method of cutting off doesn’t guaranteegauge invariance, then we can easily get a quadratic di-vergence . . . we screwed up, okay? But if you do it right,so that you don’t lose the gauge invariance in the cutoffprocess, then you can show that this will only be log.Now the divergence of the first power, (B.18) really never occurs. Because if we would have–how couldwe have it? We could have p dot something, like q orsomething, and then you would have p · q/ ( p ) or some-thing. This would be [in the denominator] sphericallysymmetrical, and this would be [in the numerator] lop-sided. If I change p → − p in the integral I get the samething with the minus sign. The mean value of p µ , a sin-gle p , integrated over all directions, is zero. So for thatreason, this “1,” here and here both [in the 3rd columnof the table] the 1 is really equivalent in the end to 0; it’sonly log divergent. Although it looks like it’s linearly di-vergent, the linear pieces average out, provided that yourcutoff isn’t lopsided, alright? . . . If you have a reasonablecutoff that’s symmetrical, you only get log divergences.And these [the terms with superficial degree of divergence0] are log divergent. And so it turns out that this wholemess, in practice, this whole thing, are all log divergent,if the cutoff has any degree . . . [Question from student: does gauge invariance reducethe degree of divergence of some of the diagrams] There’sthis [the vacuum polarization] [student: but not the 3-gluon diagram?]
Yes actually, this cuts this down [by] AA cross a curl A . So there’s one gradient that comes onthe external line. So it always turns out that this “1,”in order to keep the dimensions right, ends up as someexternal line [momentum] . So if we want the momentumof the incoming particle to be in front . . .Yes. [Another student: So you do get gauge invariantcouplings.]
Something tells us that there will be an ex-tra gradient in front. But it’s also true that the mirrorsymmetry of the . . . the same result. Alright? Another RPF means (cid:154) × (cid:126) A · (cid:126) A ×× (cid:126) A ; see eq. (8.15). question? [Shouldn’t gauge invariance also reduce thelogarithmic divergence in the . . . ] Yes and no. That’s. . . more difficult because there’s no procedure. This isgauge invariant, but it’s coupled directly with A and wedon’t see any extra gradients. So there’s no way to de-crease the apparent power. The only way to make itsmaller is by having momentum come in front. Whenyou have two quarks—talking about the last diagram—two quarks and a gluon, you’re going to imitate a termwith a quark, and a quark or an antiquark, and a gluon,and there’s no gradient, so it comes out that it’s logarith-mic.Now it is possible to choose a gauge, by the right choiceof gauge, you can make any of these damn integrals zero [in terms of the degree of divergence] , at the expense thatthe others change [?] . . . You have to be careful tocompute a complete physical process always. You wantto make absolutely sure to compute something, the to-tal answer of which is gauge invariant. Then you can’treally say—it is possible—you remember all the differ-ent propagators we had for the gluon? Well it dependson what propagator you use, whether you use a propaga-tor with k · η/k or whether you use the propagator with δ µν , or still another propagator which is interesting. Iwanted to mention this before . . . When we were doingthe . . . business, with this gauge for instance; suppose westarted with this gauge ∂ µ A µ = 0 , and we come out andwe have to do (cid:90) e i (cid:82) ( ∂ µ A ν − ∂ ν A µ − A × µ A ν ) ∆( A ) δ [ ∂ µ A µ ] D A µ (B.19) and I’m not going to write the quark business in. Thenwe had in addition the statement that there was a de-terminant ∆( A ) , and this produced ghosts, it was repre-sented by ghosts, and on top of that was a delta functionof ∂ µ A µ . And then we integrate over all A . But I sug-gested that we get exactly the same ∆( A ) here if you tryto make this δ [ ∂ µ A µ − f ] , and that the answer was in-dependent of that [ f ] . And then suggested further thatyou multiply by e − f / or something [times] D f . Andthe result of that was to bring up a e − ( ∂ µ A µ ) / —I’m justoutlining what I did. In order to eat, in the square of this [( ∂ µ A ν − ∂ ν A µ ) ] the divergence pieces, and then I couldshow you that the equation of motion which was (cid:3) A ν − (cid:24)(cid:24)(cid:24)(cid:24) ∂ ν ∂ µ A µ = S ν (B.20) At any rate this term [slashed out] didn’t appear at all,and therefore A ν = 1 (cid:3) S ν = 1 k S ν (B.21) So we got the propagator δ µν /k . Now the interestingthing is what happens if you put a different number here I don’t understand the claim, since the propagator (12.8) (with µ → − k ) has the same power-counting properties as usual. , a ( ∂ · A ) / . And I’lljust leave it as an exercise, because if you put a differentnumber in there, you get a propagator of this form, k (cid:18) δ µν − η k µ k ν k (cid:19) (B.22) where η is not the same as a , I can’t remember exactly; η is something like a/ (1 ± a ) . Ah, it should be when a = 1 this disappears; when a goes to infinity, this should go to1. Because a = ∞ brings us all the way back to here; thisis Gaussian, such a tightly Gaussian, it’s equivalent to adelta function. And it says you calculate everything ex-actly when ∂ µ A µ = 0 . And that’s a propagator like that [ δ µν − k µ k ν /k ] . Now you see that if you take a k µ ofthat, you get zero automatically, because the divergenceof A µ is always exactly zero. Well, this general propa-gator when η = 1 is called Landau’s propagator; when η = 0 it’s called the Feynman propagator. And other η ’sare possible too; I call it to your attention because it’s in-teresting . . . the effects of this going to zero . . . you’ve gotto be careful what propagator you use . . . what sizes youget for the different . . . It’s only when you have a gauge-invariant quantity that you get an answer that does notdepend on the . . . propagator you use. [Some of the lecture was lost during the change oftapes.] . . . Nowadays it’s possible to do all these diagramsand all these calculations on machines, programs for alge-bra . . . programmed specifically for working on these dia-grams and integrals involving quark data, and therefore itgets to be no big deal. You choose a propagator [gauge] ,you turn on the switch, and it does all the 17 diagrams.Whereas by hand, you are happy to discover that by usingLandau’s gauge, you only have four diagrams; rememberI had 17. This would be useful . . . without . . . machines.The reason that four diagrams is better than 17 is mainly,it’s impossible to do anything without making mistakes,when you have too many pieces . . . Alright?I am now going to calculate, at last; let us talk aboutthe scattering of two quarks. To lowest order we alreadyknow that it looks like this q ∼ g q (B.23) and we computed it. I don’t remember what we got for thecolor . . . but it involves a g and a one over q . . . That’sthe lowest order. Obviously this is g . . . now it’s goingto be α when we’re done . . .Okay, now we get the next order, the g . We’re tryingto get the next order in g . . . So what we do—this is thelowest order—next order, we have a lot of possibilities, + ++ + + +− x + + permutationsghost m δ + + + + Could be that you had a quark loop here, coming out thisway. Could be you had a gluon loop coming out here.Could be that you had a gluon loop like that—what?Yes, with the four gluons that couple there; I believethat’s zero when you work it out, but I’m not . . . It’sobviously simple. Then there’s a diagram that lookslike this [vertex correction] plus one on the other side.Then there’s a diagram that looks like this [3-gluonvertex correction] —I wish the blackboard went a little bitfurther, I’ll draw it up here, a diagram that looks likethis [quark self-energy correction and mass counterterm] . . . subtract the effect of this, this thing is divergent,the correction to the quark mass, you just have tosubtract it. Alright now I’ve drawn all the diagramsexcept the mirror image of this one, the mirror imageof this one, this [self-energy] could be here or here orhere . . . Alright? Now any other diagrams you can thinkof ? [Student mentions the ghost.]
Ah yes, I’m sorry,the ghost. Important, important, the ghost, the ghost.Very vital. Ghost, ghost, ghost, ghost. Thank you, yes.Anyone else think of some oversight? . . . Alright. Now the situation is, a lot of people will liketo compute this + + . . . and say, well this is just a gluon going along, so don’t doall this, just do this, + + . . . on the gluon . . . [Student (me): there’s a box diagramwith . . . ]
Just a moment. Just let me finish this part.Just do these [now including the gluon loop contributionto the vacuum polarization] . You can’t, because this de-pends on what kind of gauge propagators you use, it’s not I realize that the box diagram is missing a few minutes later inthe lecture and interrupt RPF. a gauge-invariant process . . . You have to finish it by hav-ing quarks make the gluons . . . that’s why I made it morecomplicated. The answer is that depending on what kindof propagator you use, you get different answers for that.It’s only when you put the whole mess together that youget an answer that’s independent of what propagator youuse. Somebody was going to say something. [Student:there’s a box diagram too, where you’ve got two gluonsconnecting the quark propagators.] Of course, sure thereis. Yes. And crossed, right? I thought we’d find some-thing . . .What we can do is do one as simple as possible, andthen say the rest of them, you know the rules for puttingthis in and the rules for putting that in, they follow thesame kind of . . . the labor is enormous . . . how to sumover the colors, various things . . . If you had to do it, youwould do it. Or find a book that will do it, okay? But inorder to understand the nature of the results that we’regoing to get, I’m going to take only one case. I’m goingto take the dullest and the simplest case, this one, q qp−qp b ν j a i µ . . . The others teach you a little bit, but . . . you’ve prob-ably done this . . . So here we go. Alright? . . . I’m goingto have something going around here; let’s say that themomentum coming in here is q , and this has a momen-tum p let’s say, and then since the momentum coming inis the same as the momentum going out . . . If you don’tknow what I’m doing, then it’s because you didn’t take acourse that’s supposed to be a prerequisite for this, hav-ing to do with perturbation field theory, and you’re goingto have to learn it. Alright? If you have taken such acourse, this will be very boring, and I’m sorry, but I’ll goas fast as I can, and hope that you . . . will stop me andask some questions . . . Alright, now here we would havea quark going around, and we would have a coupling inhere, which is a γ µ in the direction of the polarization ofthis, so let’s say that a is a vector polarization, and let’ssay a i is the color of the gluon. So this would be γ µ andit would be multiplied by a iµ . But there would also be a λ i matrix for the color matrix that this couples with, g (cid:18) − Tr (cid:90) d p (2 π ) b jν /p − /q − m λ j γ ν /p − m λ i γ µ a iµ (cid:19) (B.24) Now the next thing happens is that this damn thing prop-agates, the quark propagates around to here, and the factthat they’re propagating, one over /p minus the mass ofthis quark. Sir: [student: Do you mean to have λ i / Yes sir, I do, I do. Yes, thank you. Then comes this baby There are many other crossed diagrams missing. which I’ll call b , and we can say the polarization is γ ν ,we’re going to be multiplying by b ν and by j . And thenwe’ll also have a λ j , for the color business there, andthen we’ll have—I didn’t leave enough room did I?—wellmaybe I just about did, I have the propagator p daggerminus q minus m —oh, I forgot the 2 again, I shouldn’tdo that. And I have two g ’s for the couplings at eitherend. And those are the kind of g ’s that I was tagging g . . .Of course, that looks like this is second order, but Ididn’t put the rest of these two lines on [the externalquarks] , and when I did that there would be two more g ’s, and this is important. The first diagram which gottwo g ’s and the second diagram has four g ’s . . . Now, Ihave to sum over everything. I have to sum over all thepossibilities for the momentum of the quark loop, and Ihave to sum over the colors, and the spinors. Well, firstthe spin. When you come around and come back, you’vegot all these matrices, and it becomes a trace. So reallyit should have been a matrix trace, a trace for the gammamatrices, tr (cid:20) /p − /q − m γ ν /p − m γ µ (cid:21)(cid:124) (cid:123)(cid:122) (cid:125) tr (cid:18) λ i λ j (cid:19)(cid:124) (cid:123)(cid:122) (cid:125) (B.25)tr (cid:20) ( /p − /q + m ) γ ν ( /p + m ) γ µ [( p − q ) − m ][ p − m ] (cid:21)(cid:124) (cid:123)(cid:122) (cid:125) δ ij (cid:104) ( p − q ) µ p ν + ( p − q ) ν p µ − δ µν ( p · ( p − q ) − m ) (cid:105) / . . . . . . In addition, we’ll have a matrix operator for the color,and then it carries all the way around here and goes tothis color, and then it carries all the way around so weget another kind of a trace, this time on the colors. Sothis is a trace on the gamma matrices, you might say,and then there’s another . . . trace on the color indices, of λ j λ i over 4. And this one is easy, we know the color ofthis—without the / —is δ ij , so that’s δ ij . So the firstthing it tells us of course is that the color of the quarkthat comes out must be the same as the one that wentin. That’s the conservation of color; if it’s a red-antibluegluon here, it will be a red-antiblue gluon there . . .This particular trace [the Dirac trace] , the famous wayof handling that, is to multiply numerator and denomi-nator by /p − /q + m and by p dagger plus m , and then inthe denominator, you’ll have the rationalized thing there, ( p − q ) − m and p − m , and then there will be thisfactor we had before, δ ij over 2. Alright? . . . Any ques-tion up to here? In fact it’s the same as the correspondingcorrection in electrodynamics, except there’s some slightlydifferent number . . . from the colors. This trace here thatI’ve written can be directly simplified; this trace is ex-actly the same as this dot this plus the other way around. . . minus . . . it’s all memory . . .So the net result is that effectively I have to do this in-tegral, an integral that you must have learned about whenyou were doing electricity, QED, so I turn over here to dothe integral. But before I do the integral, I look at it, and notice how it diverges. We can see by counting there’s 2 p ’s in the numerator, there’s 4 p ’s in the denominator,there’s 4 p ’s in d p . That cancels but you’ve still got 2 soit’s quadratically divergent. In exact agreement with theprediction. But, now the trickery, the method for doingthis like this, is to have a whole list of integrals . . . a listlike this—I’ll do the integrals in a minute—but things likethis, (cid:90) d p (2 π ) p − m ) = 116 π im (B.26) or something; I’m not chasing that, alright? This permitsyou to do anything that has powers in the denominator,by integrating over . . . But this [(B.25)] has two differ-ent kinds of powers in the denominator, so there is aninvention for putting that together, which runs like this: (cid:90) dx [ ax + b (1 − x )] = 1 ab (B.27) The integral from 0 to 1 dx . . . and this you can verifydirectly. And therefore if you take the product of twopieces, you can write it as one denominator. I copiedit from Schwinger, actually, I cheated. He had anotherway of doing it which was extremely clever. Involving aGaussian integral. I noticed that I could eliminate onestep . . . and make it look . . . [laughter] . So therefore . . . p − p · q + q − m )( p − m ) = (cid:90) dx p − p · qx + q x − m ] (B.28) Alright? Alright, . . . what we’re going to replace thatthing by, and so we would have this integral from 0 to1 dx , to be done, later, and then we’ll have the integral d p , and then we have something like . . . (cid:90) dx (cid:90) d p ( p − p · qx + q x − m ) (B.29) × (cid:0) p µ p ν − p µ q ν − p ν q µ − δ µν ( p − p · q − m ) (cid:1) Alright. Now we have to do this [ p ] integral. But wecan’t, because it’s divergent. The first thing that’s a goodidea to do, always, is to shift the p , let me shift p , let p → p + qx . Then we have (cid:90) dx (cid:90) d p ( p + q x (1 − x ) − m ) (B.30) × (cid:16) p µ p ν + ( p · q terms ) + 2 q µ q ν x (1 − x ) − δ µν ( p − q x (1 − x ) − m + p · q terms ) (cid:17) Here RPF makes a joke that elicits laughter, something like“Classic, right?” Perhap RPF is saying something like “and make it look like myown idea.”
I just completed the square here. And then there are someterms with p · q , I don’t know what they are . . . they arelinear in p , you’ll see why I don’t care about them in aminute . . . Anything linear in p that I didn’t bother towrite out, the reason is that when I integrate over alldirections of p , the plus and minuses are going to go out,and since I’m going to get nothing from them, I didn’tbother to be careful. Alright?Now I have cheated. Why? We made a mistake. Be-cause, the integral is divergent; I didn’t tell you how tomake it convergent. Therefore this business of shifting p by qx —how do I know that the integral—that you canshift p by qx and get the same result? I must insure thatmy method of renormalization, whatever the hell I’m go-ing to use, has that property that if you shifted the p , itwould be alright, okay? So that’s one thing. The secondis, what about the integral over p at zero? That’s right, ifthe method of renormalization is round, okay, but roundagainst what? Against the shifted p ? If you say it’s roundagainst this p , it’s not round against the shifted p . Youknow what I mean, symmetric in both directions. It’scheating. So only if I made a method of renormaliza-tion, I mean a method of cutting off the integral, specific,and carefully, can I really do these things which I’ve beendoing.Now it happens to turn out that one method that’s beeninvented for this kind of loop of quarks or electrons is this:you subtract the same expression with a larger value forthe mass of the electron. In other words, the method ofrenormalization, I should have at least specified before Imade those steps. And the method that I want to use isgoing to be this one. You take the value and then takewith m replaced by a bigger m , which I like to writeas m → Λ + m , and subtract. That’s the scheme.That scheme, it so happens, will permit the steps whichI did of shifting. It was all right. But it’s very easy toslide off the wagon and make operations which are notquite right until you specify the right way . . . You see,this method of subtraction maintains the gauge invari-ance for this diagram. Because if this were electricity,everything is exactly the same except for a number, andif this is electricity, if this is any mass whatever, this isa gauge invariant integral, and the gauge invariance ismaintained by subtracting the same expression. To showyou that it isn’t necessarily obvious what to do, an earlyworker in the field first proposed or tried to subtract fromthis propagator the propagator with a different mass. Inother words, to use for the propagator p − m → p − m − p − (Λ + m ) (B.31) That doesn’t work. That doesn’t keep the gauge invari-ance. Pauli and Villars pointed out that if you sub-tracted the whole thing, the whole closed . . . [amplitude] then you maintain the gauge invariance. So you seeit’s easy to . . . if you don’t maintain the gauge invari-ance . . . electrodynamics . . . Pauli-Villars . . . and that’sthe way if you do this then all these things are legitimate. Okay?Now I’ll show you a very interesting—is that okay, youhad a question? It’s nerve-wracking, alright. But it’sbeen straightened out, in the case of an electron. I want topoint out, that when we come to the second diagram here,with the gluon, it’s more subtle and more complicated.And the exact way to do it is very hard, and it took alot of finagling around to get it right, when I first triedit. But I was able to guess and push and hammer. Inthe meantime, another method of cutoff was invented,which is called the dimensional renormalization, inventedby Wilson and ’t Hooft . . . for which gauge invariance andcovariance in space in four dimensions are automaticallymaintained in [ d dimensions] , not a chance of losing it,so that it’s a good scheme. We don’t have to have theold-fashioned hammer tricks. Here we know a good trickthat will work, and I wanted just to point out . . . Now Iwould like to point out—I think we could almost see—Now I want to show you something. Suppose we had amethod of renormalization, that we knew was going to besymmetric, and everything is going to be alright. Then,I claim this integral, (cid:90) d p p µ p − m = 0 (B.32) if we could renormalize, to make it finite anyway, wouldhave to be zero because of the asymmetry. Provided wehad a good method to protect the asymmetry, and sub-tracting this thing with a different mass obviously doesthat. That’s it, no problem. Now differentiate both sidesof this . . . Now what I want to prove, let’s see . . . If youdifferentiate inside the integral with respect to [ p sub] ν ,you’ll get (cid:90) d p ∂∂p ν p µ p − m = (cid:90) d p δ µν ( p − m ) − p µ p ν ( p − m ) = 0(B.33) It’s not hard to prove, by the same kind of symmetry, oneway or another, you might not like the way I did it, thatthis is also zero. [Student: what did you differentiate? ] I differentiatedthe integrand with respect to p ν . Isn’t it legitimate todifferentiate the integrand if you’re going to integrate itback anyway? Well, if you don’t like it this way, thenanother way to do it is to shift p , and differentiate withrespect to the shift in p . You start out with some kind ofthing like (cid:90) d p p µ ( p − a ) − m = a µ (cid:90) d p ( p − a ) − m which is just this thing [(B.32)] with p substituted with p − a . And then differentiate both sides of this expressionwith respect to a ν ; then put a µ = 0 . Alright? I reconstructed the previous equation from the tape rather thanmy notes; apparently it was less explicit, leading to this question.
I write this particular thing by putting it all under thesame denominator, and putting this here [arranging thenumerator so that the first term is like in (B.33)] : (cid:90) d p ( p + q x (1 − x ) − m ) (cid:104) p µ p ν − δ µν ( p + q x (1 − x ) − m ) − q µ q ν x (1 − x )+2 δ µν q x (1 − x ) (cid:105) So we have this sort of general statement that it is adisplaceable method of doing things. You can think ofit as a shift of the origin; the method of renormalizingdoesn’t have anything to do with it. The question is, ifthis were automatically zero. And then if we look at thiswe expect that . . . we have just the right combination: wehave p µ p ν and δ µν times the denominator—watch outon that sign! [The q term in the numerator has theopposite sign to that in the denominator.] This part’sokay but this is wrong. Well I’ll fix it; I’ll put plus andI’ll make it minus 2 [in the second line] . Alright, that’snot wrong. Now this thing, times the delta, plus this one,go to zero . . . by the argument about the way to do it . . . inthe integral . . .The result of that is that all of this can be . . . the netresult of the whole thing, altogether, is q µ q ν − δ µν q ) (cid:90) x (1 − x ) dx (cid:90) d p ( p + q x (1 − x ) − m ) (B.34) and I’m close to the end of the hour . . . I wanted tosubtract—this is only a logarithmically divergent thing—and I subtract the upper limit and so on. It’s going tointroduce something that goes like the logarithm of thiscutoff divided by some pole mass which is practically the q of the quark. It’s a little more complicated than thatand I’ll finish it next time. What I’m trying to say isthat we can get to this thing in front, and that is neces-sary for gauge invariance, because q dot [the prefactor] is zero . . . The current is conserved.Now let me, since it’s just a few more minutes to ten,just to remind you . . . I’ll explain that this means thatthe vector potential that is coming in and out . . . finishthe job next time . . . and also next time I’ll start to ex-plain . . . dimensional renormalization, which is so handy,because the old-fashioned way you had to do a lot ofthinking, trickery, to make sure you didn’t screw up. . . invariance. Appendix C: Transcription: Renormalization,continued (1-14-88)
We were doing one of a number of diagrams that haveto do with the correction to scattering of two quarks. Thescattering of these two quarks behaves as— g is the cou-pling constant—something like g over q in the first or-der. Or that’s sometimes called second order because it’sin g . . . First order in g . That’s straightforward. We want to get corrections for it. And the corrections ap-peared to be a large number of diagrams which I wrote. . . and you’ll see that there’s a g here and here and hereand here, = J µ g q B q J µ (C.1) where B = . So this is going to be a correctionwhich will be of the order g . There will be two propaga-tors /q here and /q there, and so on. So it will be /q multiplied by an integral, B = ( δ µν q − q µ q ν ) I, (C.2) I = (cid:90) dx x (1 − x ) (cid:90) d p/ (2 π ) [ p − m + q x (1 − x )] What we’re going to discover is that this is—sorry, notby the integral, but by B , the bubble . . . I left out all theindices, the colors and all that. You can go away anddiscover that the colors are all . . . multiplied by a certainintegral . . . in this integral there is a number of constantswhich I inadvertently, carelessly dropped—twos and π ’sand things which presumably you can calculate . . . Now, the currents [ J µ ] , which I didn’t write here, whichare operating here and here . . . this bubble, when actingon the current, the q —the current is conserved—and the q acting on the current is zero, so that q · J = 0 . Thebubble gives you δ µν , which means that these two currentsare in the same direction, and this q in the bubble eatsone of these q ’s. So this thing turns into = J µ g q I J µ (C.3) So therefore, it’s the same form as this [the tree levelcontribution] = J µ g q J µ (C.4) except we have g times the integral up here instead of g in the correction. So the easy way to think about it is it’sa correction to this coefficient at the top, g → g + g I (C.5) Alright? Are there any questions about that?In getting to this form we did a little hocus-pocus aboutcorrecting, shifting origins and this and that, and talkingabout tricks to get rid of the quadratic divergence whichoriginally arose. However we still have a divergence and I had added the 1 / (2 π ) factor as an afterthought in my notes. we have to talk about how to handle it. And the par-ticular rule that works for log divergences in quark loopsis to subtract—the rule I’m going to use here, now— isthe method of subtracting the same result for differentmasses. But I will describe, perhaps today, but later, themethod called dimensional renormalization, which is tochange this d p to d . p . This is a function of only p ,so d p is something like p dp , with a coefficient that is π or something, that depends on the number of dimen-sions; in three dimensions it’s p dp times some othernumber, π . And in d dimensions d d p is p d − dp timessome coefficient involving Gamma functions of d and soon . . . That’s all there is to it—mainly that’s all there isto dimensional renormalization; to use d = 3 . , and then . , go to the limit. Alright? That’s what it’s about.However it’s very pretty and I must have spent a lot oftime because I enjoy it . . .Anyhow, the way of renormalizing is to subtract—thisis what we’re going to do now p − ( m + Λ ) + q x (1 − x )) (C.6)[from the integrand of I ] . That will make the results con-vergent . . . It would be—another way to make this sub-traction is to consider this integral as a function of m , I ( m ) = (cid:90) dx x (1 − x ) (cid:90) d p/ (2 π ) [ p − m + q x (1 − x )] (C.7) and then consider taking I (cid:48) ( m ) , the derivative with re-spect to m , and integrating that − (cid:90) m +Λ m I (cid:48) ( M ) dM = I ( m ) − I ( m + Λ ) (C.8) Let’s take the derivative with respect to m , but we’regoing to have to have a variable for it, so let’s call it M ; integrate that with respect to this M . . . you cancertainly do this . . . This is a trick that I wanted to use. . . converges . . . Well of course I need to differentiate andthen put it under the integral sign, calculate the integraland then . . .. . . This particular method would produce this final re-sult without all those tricks about shifting and so on; allthose could be perfectly done by . . . and not notice all thatstuff . . . all that stuff that I did for quadratic divergences,it’s also taken care of . . . I did that to show you how incases where there’s some confusion, it is always possi-ble to get an answer. People were very clever to squeezeanswers out of these . . .The logarithmic divergences are always much easier tohandle, much less uncertain than quadratic and higherdivergences. And when we were doing this stuff for ex-ample we found that one of these things would produce alogarithmic divergence directly. So another thing to do isto compute a process that we know produces a logarith-mic divergence, and then have no more trouble, and use gauge invariance to get the terms in front. But with di-mensional renormalization, you don’t need any guessing. . .Okay, we’re going to do it this way [by Pauli-Villarssubtraction] here. So first I’ll differentiate that [(C.7)] and I will find myself needing to do the integral (cid:90) m +Λ m dM (cid:90) dx x (1 − x ) 2 (cid:90) d p/ (2 π ) [ p + q x (1 − x ) − M ] (cid:124) (cid:123)(cid:122) (cid:125)(cid:90) d p (2 π ) p − L ) = 132 π iL (C.9) And now at the end when I’m all done I have to integratewith respect to M . . . Alright, so that’s where I’m at, yousee where I got that: differentiate that with respect to M ,the second power . . . now it’s got the third power. Nowof course this integral will present no divergence becausethere are six p ’s in the denominator and four in the nu-merator; then this integral can be done. We can do itin lots of ways, and I’m not going to bother . . . Anybodywho’s ever done anything in perturbation theory is alwaysgoing to put this very same integral, sooner or later. It’sthe integral of d p/ (2 π ) divided by p minus something,cubed, and it’s equal to [one over] 32 π iL or somethinglike that. Factors of 2 or so I’m not going to . . . I can’tremember and I didn’t bother to look it up . . .This then, this integral here then—I’m keeping theselines this way so that you don’t have to write that overand over and over—this piece becomes (cid:90) m +Λ m dM π i M − q x (1 − x ) (C.10)= 116 π i ln (cid:18) m + Λ − q x (1 − x ) m − q x (1 − x ) (cid:19) Okay, now the idea is that Λ is the cutoff, it’s supposedto be higher than any of this part with the mass, even q .We only want this theory as Λ → ∞ . As Λ → ∞ then,relatively speaking we can drop this [ m − q x (1 − x )] .Of course we get an infinite answer because we have adivergence and that’s where all the trouble began. Whathappens to that infinity? It’s fixed by changing the cou-pling constant g with Λ , by making g a function of Λ , sothat the variation of the first order takes away the . . . ifyou use different Λ ’s you use different g ’s at the end.I’m going to continue this calculation disregarding themasses. Suppose you have a large momentum transfer,and disregard the mass. The purpose of this is only todo the arithmetic; if you want to you can always do itwith [nonvanishing mass] . . . I’m going to disregard thisjust so to take a simple example where q is much largerthan the m . . . It looks like it’s dangerous because when x is small, even if q is large, maybe that [the mass] isimportant, but it turns out in the log it don’t make anydifference, but anyway you do it with m , I don’t want to do it with m . I’m only illustrating, explaining whatcomes up . . . So I’m going to write this as (cid:90) dx x (1 − x )16 π i (cid:20) ln (cid:18) Λ − q (cid:19) − ln x − ln(1 − x ) (cid:21) = (cid:18)
16 ln (cid:18) Λ − q (cid:19) − (cid:19) (C.11) We suppose that − q is positive, actually. It’s going tobe a momentum transfer for the scattering . . .Now I have to integrate this and I’m almost finished,see? So I have to integrate x (1 − x ) . That’s a constant [meaning that ln(Λ /q ) does not depend on x ] . . . that’swell within my power. x − x [integrates to] x / − x / that would be the integrand, and if I put x from 0 to 1—nothe integrand is the differential of that—I get / minus / is / . It’s going to come out to / . Silly now,because I’ve lost the constant . . . so I have the logarithm,I’m just trying to keep things that are relevant . . . Now Ihave to do the log of x times this, which I’ll do by parts.And I get / minus / actually, when I do it by parts Iget some kind of number here, something like / , okay?Coming from integrating those logs, which are easy to do,and I’m sure you’ll . . . Yeah, / . Anyway you get someconstant. And that’s it.And that’s the integral “ I ,” and it will have the Λ there.How are we going to look at the physics? Assuming thephysics is right, and we’re going to get an answer, nowwe put that back: this “ I ” goes back in here [eq. (C.3)] ,right? So, I’m just going to write the coefficient of theterm . . . J Q J g + g π β ln Λ − q (cid:124)(cid:123)(cid:122)(cid:125) + a (C.12) Q where β = − n f and a = 109 n f (C.13) g plus g times the various numbers of π ’s, which I haverecovered by looking at the answer in the book, alright?Times a certain constant [ β ] times the logarithm of Λ over − q plus another constant, where for us, β it turnsout is − / when we put it in this form. Alright? Andthe “ a ” for us is equal to—this is / of that, / minus / , probably something like plus / —highly question-able . . .Now first of all, to remind us all that − q is positive,let’s call it Q . So I’m going to make it like that. This [eq. (C.12)] is all multiplied by /Q , and by the currentsand so forth in the final interaction. That’s what comesout, alright? . . . The reason I wrote it this way is thatthere’s going to be more contributions that come fromthe other diagrams that we haven’t worked out. And I’mgoing to have to add those in when I discuss . . .Oh— [the factor of n f was not written in eq. (C.13)at first] there’s more than one flavor of quark, and eachflavor of quark makes a loop, and each one those is the same as this one. And insofar as Q is large enough toneglect the mass of the quark, insofar and therefore forthe first few, certainly for the u and the d and very likelyfor the s , maybe for the c quark, there will be a certainnumber of flavors that we would use, that contribute tothis formula; it would be the number of flavors whosemass is less than Q . The flavors with masses higher than Q are not much contribution. The flavors that are inbetween, then you just have to do the integral better andso on. So put the number of flavors [in (C.13)] . Alright?It looks like this answer means that the probability ofscattering will depend upon the cutoff Λ —there it is, ex-plicitly there. And therefore our original program, whichwas that we were going to calculate somehow in quantumchromodynamics and make predictions, but we found ourtheory was divergent, and what are we going to do? Wellwe have a . . . But the trick is to arrange that it doesn’tdepend on the cutoff, by supposing that g is chosen, foreach Λ that you choose, you must take a different g . Youmust choose a g which is a function of Λ , chosen so thatthe answer to a physical question does not depend on Λ .We need a formula for that to do that. Well this is veryhard to figure out here by looking at this thing, what kindof a shenanigans, how you’re going to vary g to get ridof this ln . Of course this could also be written as g , andto the same order as I have it here, it’s convenient towrite it this way, g − g π β ln Λ Q = 16 π (cid:16) π g (cid:17) − β ln Λ Q (C.14) There’s one other term I forgot about, this (16 π ) , this isvery much like . . . That’s equivalent to this approximation [Taylor expanding to first nontrivial order] ; we’re onlyworried about this log part now; the “ a ” is something elsewe have to work out a little more accurately. To thisorder, as far as the ln Λ , it’s like this . . . which could alsobe written /g , or I would prefer to put my π here,so perhaps there’s a π here, minus β ln Λ /Q . Nowthis [ g ] , we can suppose depends upon Λ , and we have tomake this [the right-hand side of (C.14)] not depend on Λ .If we make g , if we arrange all the time that π over g , which is by the way g / π . . . old notation where Ihad forgotten the π squared, so you can remember this inconnection with the previous lecture, without seeing thatlousy π squared all the time. This, if we suppose this,which is defined as this, π g (Λ) ≡ g (Λ) = β ln Λ λ P + const . (C.15) is equal to, when we vary the Λ , we make sure that this issome kind of constant, well we should make this is ln Λ plus a constant. And since it’s a constant, we can putanything down here, that we want . . . This will cancel the ln Λ , yes? No, you’ve got to have a β in here [initiallyforgotten in the above formula] .So if we suppose our g ’s are chosen like this, thenwe’ll get an answer that is independent of ln Λ ; that’s the trick. When we do different degrees of convergencewith the cutoff, and we change the cutoff, and we change g that we use appropriately for that cutoff, then we canarrange the whole thing so that it doesn’t make any differ-ence where that cutoff is, that’s the miracle of this theory,and that’s why we have a theory. Because otherwise youwould have predictions that would depend on still anotherparameter which is the cutoff, where we have to writeour theory with explicit formulas for the cutoff . . . Thisway, we don’t; we just have to say that it’s going to hap-pen, and hope that it does, it’s been proved that it does. . . Alright?Well this is explaining the machinery that we’re goingto . . . The thing is we haven’t gone to the next order, g ,and discovered that this isn’t quite enough; that there hasto be a log-log term there. I’ll explain why there’s a log-logterm . . .Yes? [Question about the constant in conjunction with λ P .] I don’t know that constant. It’s arbitrary, you chooseanything you want. Later on we chose, specifically, tomake Politzer’s λ , to make that constant serve for a spe-cial method of cutting off dimensional renormalization,which is not what I’m going to use, so the π , the / andall that stuff is changed, because of the kind of handlingof the integral. And because it’s a different—yes, if youhad decided that the method of cutting off was going to bethe Λ method, then in that Λ , the way I did it, changingthe masses of the quarks, that would be enough . . . thena more convenient λ Politzer might be to make that zero [referring to the constant in (C.15)] . . . . Anything wouldbe alright. Somebody has to make a choice somewhere.. . . There’s a choice to make that constant zero for di-mensional renormalization . . . Let’s take the case of thelattice. You say that the Λ corresponds to the wavelengthof the spacing. So is Λ [equal to] 1 /a or π/a ? One guydoes one way, another one another way; all he does ischange the scale of Λ . So you need to put that as the logof π squared over here, or you can change the definitionof this . . . That’s why I like to put the constant here atthe end . . . Any other questions? [Question: why does this procedure work for all pro-cesses, using the same g (Λ) for all of them?] That hasbeen proved. That we had to assume, that the theory was. . . It’s not obvious at all. It turns out that . . . it’s nottrue only of the calculation at second order, but at thenext order also it’s independent of the process. Only be-yond that does it become dependent on the process. Andthat’s connected to the discovery that the rate of changeof the /g with respect to Λ computed as a series . . . OhI shouldn’t say it that way, I should have used a real pro-cess. The first two terms . . . how we know is somethingI didn’t prove. [Student interjects, I think it is me: Isn’tthe answer to that question the fact that you have a fi-nite number of . . . with primitive divergences?] Yes, we’lldiscuss that. Thank you. Yes, thank you. That’s good,let’s discuss it. He’s got the right answer . . . We worked out the various divergences. (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)
20 01 00 10 00N
Degreeofdivergence actualnaive
This one [the tadpole] vanishes. Then there was diver-gences of this kind . . . two things coming out, never mindwhat’s in here; there would be three gluons, rememberthis little table I made? With the degree of divergence. . . There was a table that said how many glues therewere and how many quarks there were on the outsidelines, from which we calculated as 4 minus this [num-ber of gluon lines] times [minus] 3 / that [the number ofquark lines] . And the result for N here was 2, 1, 0, 1, 0,and everything else convergent. Now using gauge invari-ance you can show always that in this case [ N = 2] therehave to be two q ’s outside, in front . . . and therefore thisis really . . . 0, logarithmic divergence. And this one [thelinear divergence] by symmetry again, a single power ofmomentum . . . this is also a logarithmic divergence, so isthis. So altogether they’re all logarithmic divergences, it’sno big deal. So these are the kinds of things that are di-vergent. This [the vacuum polarization] will always have q µ q ν − q δ µν , something like that. Because of dimensionshowever, you see there must be, the whole thing has di-mension one so there has to be at least one q sticking outin front . . . these things [the two-point function] , insofaras they diverge, at high momentum, they must be num-bers times these two A ’s with q ’s in it. And if you callthe vector potential of this A , and [of that] A , and youmake the Fourier transform back again, in other words,if we hadn’t had those cutoffs and I’ve just got quarks inhere—er gluons in here—then this bubble . . . for having a q in front, which I illustrated here . . . double gradient onthe A . And because of gauge invariance, the result mustbe ( ∂ µ A ν − ∂ ν A µ ) A × ν A • µ ∂ µ A ν A × ν A • ν A × µ A ν Likewise this term [the three-point function] , involvingthree potentials, is equivalent to the effect of some di-rect contact—the divergent piece—is equivalent to somekind of contact which involves three A ’s and one gradient.But because of gauge invariance, the only thing you canwrite that has that property is this kind of thing, to goalong with this one [the kinetic term] . And furthermore,these four, has four A ’s, and they will turn out to be ofthis form. Not only that, but the [coefficients] of every one of these things will be adjusted just right so that thecombination of these things with their coefficients, all thecoefficients will be right, so that this is equal to ln Λ F µν F µν (C.16) times a number, which involves this divergent log . . . Thelog divergent part looks like this. So if I had computedthis one or this one, I would have gotten the same result.The reason it has to have this form is gauge invariance.If I did not destroy the gauge invariance by the cutoffmethod. Now the particular cutoff method I used wasforced to not spoil the gauge invariance . . .Likewise this thing [the quark self-energy] is going toinvolve two ψ ’s. And this one [the vertex correction] isgoing to involve two ψ ’s and an A . This one [the ver-tex correction] corresponds to changing the coupling con-stant here, and this one [the self-energy] corresponds tothis—there is a term ¯ ψmψ . . . corresponds to the idea ofchanging the mass of the quark. Well I have just erasedsomething here that I think I need: when I change Λ , I’llchange g because I’m going to suck that number into theoriginal zero order (1 /g ) F µν F µν which is what I startedwith, and this gives corrections—are going to producecorrections—to this thing times F µν F µν , and I’m goingto say “Oh. I could have started with a g , I’ll make this g change to eat that number.” In other words /g will varyin such a way to eat that number. And that’s what thoseformulas [(C.15)] are for /g . They’re just designed toeat these logs. The gauge invariance enables you to knowthat all of these are going to all go together . . . you justhave to look to higher order divergences, and find out thisnever stops . . . but you have to show that you don’t keepgetting more and more in trouble. Which I’ll show youwhy . . . in a minute.To make this even clearer, if I have to; to look at itanother way. We originally have to do an integral thatlooks like this. Then there’s another term, which I’ll writeas a factor ¯ ψ gradient dagger minus A , which I willwrite like this . . . You integrate this over ψ and also over A , (cid:90) D A µ e i g (cid:82) F µν F µν (cid:90) D ψ D ¯ ψ e i (cid:82) ¯ ψ ( i /D − m ) ψ = (cid:90) D A µ e i g (cid:82) F µν F µν det (cid:0) i /D − m (cid:1)(cid:124) (cid:123)(cid:122) (cid:125) g ( A ) ∼ e i ln Λ f ( A ) (C.17) That’s the kind of thing we’re trying to do. Now we canimagine first that I had just done this integral [over ψ ] completely, it would be nice if we could do it for arbitrary A ; this is some terrifying functional of A . [question fromstudent ] Yes, yes, yes, exactly. This is one over—no, RPF habitually says “dagger” to mean what we call “slash” Evidently RPF wrote g ( A ) first without identifying it as thefunctional determinant. the determinant of the Dirac operator i gradient minus m minus A , with some color terms; yes, that’s just whatit is; it’s the same thing, we can’t work this out either.Now we can expand this by perturbation theory, and try tomake a calculation, and we discover that this is . . . at veryshort distances, high momentum, at very short distancesthere’s some trouble. The trouble comes from too manygradients on top of each other, the propagators from thisthing have delta functions in the origin, delta functionson a line, two of them on top of each other, they stronglydiverge. So for very high frequency A , this function hasto be discussed, this is a little bit wrong. So we had to fixit a little bit. It’s still a determinant, we just fixed it bya cutoff. When we fix it by the cutoff, we discover thatthis thing is e to the i times ln Λ multiplied by anotherfunction of A , of course you can always write it that way.But insofar that this only involves high frequencies, shortdistances, A is at two points very close together.And, if everything has been done right, since this ex-pression here [the determinant] is gauge invariant, withrespect to A , because if I make a gauge transformation of A , and then fix up the ψ so as to get the same answer,this has to be a gauge invariant expression . . . we discov-ered over there that it involved the gradient of A squared.But we know that it’s gauge invariant, and therefore if Idid it completely, I would get the whole string [the threeterms in F ] , I could only get this; this is the only gaugeinvariant expression which starts like that. And that itstarts like that is a statement of the forms that we got—by the way, although we didn’t notice it, but this A , thiskind of propagator, is just—see, if I put an A on each sideof this, this becomes A ν squared and two gradients. Wellthat’s ∂ µ squared. Let’s figure out which way [to con-tract the Lorentz indices] . The other one is q µ A ν q ν A µ . . . Then assume an A which is a plane wave and sub-stitute it in this expression [the gluon kinetic term] ; youwould get this kind of thing back [ δ µν q − q µ q ν ] , so this isin fact the operation producing this kind of combination.So we’re getting the first term right. Because we onlylooked at the two-gluon. If we looked at the three-gluon,we would be surprised to discover that it produces A dot A to the cube [ A × ν A • µ ∂ µ A ν ] , which is just this com-bination. And even more surprised to discover that thecoefficient is the same g exactly. The surprise woulddisappear when we realized that it has to be with a cutoffscheme that [preserved gauge invariance] . . .And in the same way, when we go to integrate over A ,we know that there’s a problem with the meaning of thisthing [the functional measure] ; let’s forget about it. Butwe could imagine some kind of rule: stop the integrationsabove a certain high frequency—unfortunately, that’s notgauge invariant—do a lattice. Well let’s say stop high fre-quencies, forget about . . . one of these days—dimensionalrenormalization. So you cut this off at high frequency;you say wait a minute, what if I cut it off at a differenthigh frequency? Then I could say that the intermediatebetween the medium high and the very high frequencies iswhat I’m integrating over to see what happens if I cut it off at the medium frequency. That will produce a numberof terms that will involve the logarithm of the very highand the medium high frequencies. And the coefficient ofthat log . . . will again have to be gauge invariant and havethe same kind of form . . . [change of tape] . . . it’s just pretty; the only place where there appearto be divergences are just the places you need to makethe simple form to be the same shape as the original one.Sometimes the good way to look at it is that this thing canbe compensated by putting a term like this with a g . TheLagrangian [gets a] correction term, this can be put in byputting a term like that times some q times some number.So the Lagrangian has a correction term, those are calledcounterterms; in other words, if we started with a La-grangian, instead of saying /g exactly, times F µν F µν ,we say we’re going to with a Lagrangian which has al-ready in it counterterms, this thing minus those numbers,which you’re gonna find out what they’re gonna be, times F µν F µν , which are counterterms. That’s the Lagrangianthat I started with, and then we’re going to have a cutoffat Λ . And the cutoff at Λ is going to be equivalent to mak-ing corrections—divergences—well, they’re not divergentbecause we’re cutting them off, which undo these, to getsomething which is independent of Λ . So if I write thereal g [the renormalized value]1 g = 1 g + δ (cid:18) g (cid:19) (C.18) this is equivalent to starting with some constant, call that g , the constant we started with, plus some countertermwhich will depend on Λ , and they’re built in such a way asto compensate the divergences that we get here. The netresult when we’re all finished is just that constant times F µν F µν , independent of the cutoff. And that’s where allthose formulas come from, that talk about the /g beingcorrected by things that depend on Λ . Alright? Any otherquestions? It’s just another way of describing the samething. But the beauty of it is . . . the divergent terms areexactly right to reproduce the form of the Lagrangian, andtherefore by changing the coupling constant we can undothe Λ dependence. Alright?Now I do have to complete the discussion, to discusshigher order calculations . . . I only did this one [the gluonvacuum polarization] to lowest order—I didn’t do every-thing, I didn’t do these loops, but let’s suppose I had; welllet’s say this one, I don’t care, look, it doesn’t make anydifference, quarks or gluons [in the loop] . I’m not go-ing to worry about . . . However, we just noticed when wewere counting divergences that now we’re in trouble, be-cause now after we did all that, then we look and we findsomething like this, and that table of calculations saysthat this will have the same divergence as this and this.By the way, there was a step in here that was veryclever: taking this and putting it [in the denominator of(C.14)] implies something about the higher orders—theleading logs, what that’s all about, is that I have not only taken this [one loop] diagram, + + + . . . but I’ve added this diagram, and this diagram [two loops] and so on, to get the sum of one plus x plus x plus x plus x . . . is equivalent to / (1 − x ) , and when I did this [(C.14)] , I was already predicting the higher terms, butI know where they’re coming from, obviously . . . So I’vedone all the single loops.Now, so this looks as if it’s going to produce anothercontribution to the log, and so on, and the millions ofdiagrams—but it’s not true. Because, I look at it thisway, every time from now on that I see a gluon propa-gator, I really should correct it, or could—can correct it,by putting a loop in there like so, or two loops or threeloops, and so on, + + . . . ∼ Q ln Q (C.19) I want to include all these loops. So what this ought to be,this line now really means the propagation compensated,or corrected by these loops. Now we found out that—Iguess I should have emphasized, that when I make thischoice, substitute that back in here, I find this thing, β times the log of Λ , minus this logarithm; this is the log-arithm of Q . . . And therefore the effective propagationis not really /Q , but is really / ( Q ln Q ) if the loopsare included. That’s slightly more convergent. That goesdown a little bit faster than Q . So that these propagatorsare not strictly speaking /Q , they’re / ( Q ln Q ) . Bythe way, we sometimes write that as α ( Q ) /Q ; we talkedabout that. Anyway it’s one over log, so these divergencesare no longer computed right by just saying things like (cid:82) d p/p with zero powers left over, which is equivalent ofcourse to (cid:82) d ( p ) /p . But the propagators are not /p ,they have logs in them, and I have to tell you how manythere are; the lowest possibility is that it begins with ln p , (cid:90) Λ dp pp ln p ∼ ln(ln Λ ) (C.20) Indeed, there are terms, you might have higher powers oflogs . . . but not worse, you have at least the log. But yousee what this looks like, this differential log over log? Thisis log log, less divergent. So if I integrate this to somehigh frequency, this is still divergent, but it’s ln(ln Λ ) .And when I go to the next order, I’m going to get log log.Of course there can be terms with log squared here, butthose will be convergent: (cid:90) Λ dp pp ln p ∼ (cid:90) Λ d ln p ln p ∼ (C.21) In other words this integral [(C.20)] , this doesn’t quiteconverge, but if I had more logs in there it would converge.So that there are little log-log divergences, no worse.So the next order terms, when you include the correc-tions in them from the lower order, does not produce thesame divergence . . . You might say, well how do I knowthat there are . . . maybe they’re only log squared or logcubed . . . β terms . . . [Question from student: why didyou have just two powers of the momenta in the numera-tor of that integral?] Well it’s a logarithmic divergence. Idon’t care about . . . it’s just to understand the log . . . Thedifferential log over log is log log, that’s it.Now, you say well now I’m going to go on to the third.But it’s no longer true that it goes on to the next one,because the higher ones give more powers of logs downhere. And more powers of logs down here, it converges,it stops. Thank God. Now you say, well, it wasn’t reallylog p , it was that plus ln ln p , and I’ll let you makethe argument that this doesn’t change the divergence; itdoesn’t make it converge any better . . . [Question fromstudent about absence of triple logs] w Because the nextorder produces an integral like this [(C.21)] . . . There’s noway to isolate log-log. This correction is this propagator,which is corrected by one over log. If it’s got a log-log init, it’s additional. In other words the corrections of β log plus β log log. So the correct thing to put in there,if it’s anything, is ln p plus ln ln p at that order. Butthat argument, that doesn’t make any difference comparedto the ln p when you calculated the divergence. Whathappens in the higher orders, you get more logs downhere, but you get this to a higher power . . . there’s noway to isolate this . . . [Question from student] This has to do with the behav-ior in terms of Q . The propagator is ln Q . . . it’s truethat the correct formula has ln Q plus ln ln terms; thatdoesn’t . . . you just get rid of this log and subtract it fromthe . . . and try to get the log log isolated; there’s noth-ing that comes out . . . as you would like it, as you mightimagine it. . . . Q (ln Q + ln ln Q ) . Of course there’s nopropagator that goes inversely to log log. That’s not theway it goes. It goes inversely to the sum. And that wouldbe log log log. You don’t get this kind of a form, becauseyou can’t isolate that piece . . . My argument here is veryheuristic, but it does work and it gives you an idea ofwhy . . . this is the way the thing works out . . . more orless why it works out . . . any other question?In order to add to your confusion . . . different conven-tions . . . different ways of looking at it; some are betterthan others. If you want to read the literature, you haveto read everybody’s ideas. Some are better . . . becausethey get rid of some confusion, so they straighten some-thing out. Now if you learned only the way which is allstraightened out . . . then you have some trouble readingliterature in which something is a little older or some-thing which the guy is using some old-fashioned idea;well I wouldn’t say old-fashioned but less . . . then youcouldn’t understand the subject completely. What I de-scribed, I had tried to prove differently than is general in any textbook—if it’s in some textbook, you don’t thinkit’s original, not at all . . . I did it myself—but I have totell you about what I consider a kind of mistake, okay?Which is very prevalent and it’s all over the place. It’s not the way I’m trying to explain it. Now the way I’mtrying to explain it is that the coupling constant—whenyou make a cutoff, you change the theory. And the an-swers in general appear to depend on the cutoff. But itturns out they depend on two things: the way you cut itoff, Λ , and also the coupling constant you put in. Butby the very wonderful situation we have of renormaliza-tion, that when you adjust the coupling constant correctly,when you change the cutoff, you change the coupling con-stant, you’ll get the same predictions in the long-rangewavelength physics . . . And that’s what I wrote in the be-ginning, was a formula for how you have to change . . . inorder to make the results independent of the cutoff. Youchange the cutoff, you’d better change the [bare coupling] . . .Next. It turned out that in many circumstances whereyou could expect the mass of the quarks to be unimportant. . . that for such processes, the behavior of the processcould be worked out as a perturbation theory in g , c πg + g (cid:18) b ln Λ Q + a (cid:19) + . . . = c α ( Q ) + a α ( Q ) + . . . (C.22) These coefficients depend on the logarithm of the cutoffand the momentum of the operation, the process, somemomentum associated with the process, some definitionof the momentum. And this sum, these logs, could besummed so this was written in the form of α ( Q ) plus a α ( Q ) and so on, and this is for some physical pro-cess. In other words, for some physical process, the cal-culations go like this. For a different process, all the coef-ficients would be different . . . there might be a π , so let’sput one in here . . . anyhow, it would go like this andit would be—so, therefore when you start to work pertur-bation theory there’s a rule . . . work at second order youget the log, but you’ve already eaten that when you madethe substitution. So this is a way I managed to writeit, and it’s perfectly okay, and the formula for how loops. . . describe . . . the type of thing that’s connected . . . whichwe wrote down . . . I’m just repeating . . . thing I found outby . . . or rather not the best way to do it.Now let me tell you the wrong way, what I considernot as good a way. It’s a way that you could have doneit, but it’s got annoyances in it. It works like this. Youstart out to define an α . Now in order to make sure thatthis α is not exactly the same as that α , I’m going toput an underline on it, so you’ll always know which oneI’m talking about. Now define an α ( Q ) by a physicalprocess. I’m going to give you examples . . . For instance, I have the π in (C.22) crossed out in my notes we recalculate the scattering of two quarks, to all orders,exactly. All orders exactly, it’s going to be written as thissuper-duper α over Q , (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) ≡ α ( Q ) Q (C.23) and that’s going to define α . There would be no suchthing as finding a power series expansion of the co-efficient, which is what I would do there [eq. (C.22)] ;. . . that’s alpha, by definition, to all orders. Anotherthing would be, another way for example, when we talkabout e + e − → hadrons, the ratio [ R ] , and remember thatthe formula for that ratio in the first order perturbationtheory is α/π , remember that? Or rather g /π ?I’m not worried about the π ’s in the definition of α ( g ) ,that’s a pain in the ass that I can’t remember. Let me de-fine α ( Q ) ; this is another definition, let’s put two lines,it’s another definition, it’s identically equal, α ( Q ) = π ( R −
1) (C.24)
But anyway, we could do this to all orders, this is definedto all orders. So there’s no such thing as a perturbationexpansion for this to first order and next order and nextorder . . . it’s just a definition. That would be a possibil-ity. As it turns out, that up to the first two orders, theformula for this α and the formula for that α and my α all agree, in terms of the first log and the log-log. All ofthose formulas satisfy dαd ln Q = β α − β β α = β α + β α + β β α + . . . (C.25) And for every one of these definitions, for any one ofthese alpha bar things, the α satisfies exactly the sameequation, up to the fourth order, dαd ln Q = β α + β α + β α + . . . (C.26) but the next order depends on the process. These [the co-efficients of the first two terms] don’t. In other words itdepends on the process needed to define . . . the α . So forpractical purposes up to second order it doesn’t make anydifference, but if you want to make things definite so thatone guy can compare his results to the other, in higherorder, they’re all mixed up, because one guy is using oneway, another is using another way, because of the dif-ference in processes. You say what’s any better, why notuse a definite process? Because . . . Instead of using a def-inite process, I used a definite theory . . . You’ll notice this [ β = β /β ] is a special choice, but it’s definite, and thisinvolves . . . independent of process. And an advantage is,you don’t have to compute this special process . . . you’vegot it done . . . If you want to know some physics then youhave to compute. And that’s saying that you should really calculate the power series for this process, in terms of α ,my α . . . We should naturally expect to do each processseparately as a perturbation expansion, instead of arbi-trarily choosing one . . . one is no better than the other. . . it all adds confusion to . . .They then said that the physical coupling constant de-pends on Q , but there’s no definition, it depends on howyou define it. You could say that the coupling constantdepends on momentum squared, but . . . So, people talkabout this as if it’s a running coupling constant, but youcan’t put that into the Lagrangian, as a running couplingconstant. The only thing you can put into the Lagrangianis something that depends on Λ , not on Q , so I was ratherconfused . . . You see how much confusion . . . in the defi-nitions; Politzer’s λ . . . we saw the equations depend onwhich method of cutoff you use, and how you define theconstant, is it zero or is it Euler’s constant times the logof π . . . And on top of that, on top, I wanted to add theambiguities that slipped in . . . to define alpha and it’s not. . .Now in the electrodynamic world, there was a wonder-ful special process to find the electric charge, which wasunique, which is, let me evaluate the interaction of theparticles when they’re very far apart—the very long wave-length coupling of photons to electrons. In quantum chro-modynamics you can’t find any . . . like that . . . For verylong wavelengths . . . so we have no simple phenomenonwhich . . . Any other questions? Alright then.Someone asked me last time how dimensional renor-malization produces the same results. The answer is moreor less the following. You would make a process in thescattering—first of all we have less than four dimensions.We have something like g (cid:90) F µν d D x (cid:124) (cid:123)(cid:122) (cid:125) dimensions of Energy − D (C.27) and then integrate with respect to D dimensions of space-time, D is not four. Now that means—and what about thedimensions? In F µν as you all know and must have writ-ten . . . there’s the combination of ∂A and A A , and thatmeans that A is an inverse length or an energy. That’sindependent of dimension. And that F µν is an energysquared. And F µν squared is an energy to the fourth. Alength is an inverse energy. So this quantity would havedimensions of energy to the − D . Therefore g is notdimensionless. g has dimensions of energy to the − D , [ g ] = Energy − D = E (cid:15) (C.28) So here, one way is to just say, alright, I know that.Another way is to write g as some other constant timessome particular length to the − D , I’ll call that epsilon: g = c λ (cid:15)P (C.29) . . . Now what happens is, if you do perturbation theory, g plus g times an integral, same way as we did before.Except those integrals because they don’t have d p anymore, they only have a D integral, are less divergent, infact they converge. So you can actually do the integral,and there’s no problem, and you find that the integralvaries as Q − (cid:15) g + g Q − (cid:15) (cid:18) β (cid:15) (cid:19) = 1 g − β Q − (cid:15) (cid:15) (C.30) ≈ β (cid:15) λ − (cid:15)P − β (cid:15) Q − (cid:15) minus (cid:15) because it’s D − . . . so you get this kind ofa term and . . . it’s no problem, everything converges andit’s fine. But as we vary (cid:15) , we discover that the coefficientthat we actually get, the coefficient here, is a certain con-stant [2 β ] over (cid:15) . That is, if I did the calculation withdifferent (cid:15) , I’d get something that varies with (cid:15) this way:the coefficient diverges as (cid:15) → .I can go through all this usual stuff of rearranging thesum and . . . when Q is very large this is small; but thepoint is that aside from constants which will be taken out [RPF writes out the right-hand side of eq. (C.30) at thispoint] . . . and as (cid:15) approaches zero, λ − (cid:15)P − Q − (cid:15) ) = − (cid:15) ln( λ P /Q ) (C.31) The point is, the theory is convergent. It depends on (cid:15) ,and has a very high coefficient as (cid:15) goes to zero, andin the limit produces logarithms, just like the logarithmsyou see, and we have to adjust the coupling constant.So it has the right behavior with (cid:15) . . . the same problem,how does this g depend on (cid:15) ? [Student: in dimensionalregularization, don’t you want to take λ to be close to theenergy scale of the . . . ] λ P . . . In the end yes, it’s better,yes I think that’s the right thing to do . . . So what we’resaying is that the coupling constant has the dimensionsof the physical energies we’re interested in . . . but it turnsout the strength varies inversely as (cid:15) . . . Appendix D: Revision examples
Additions written by RPF to the revision of lecture 12.7Additions written by RPF to the revision of lecture 13.89There were sometimes also subtractions: edits by RPF to the revision of lecture 8.0
Appendix E: Hadron masses and quark wave functions
The following three pages were copied out of an unidentified textbook and handed out at the beginning of thecourse. Handwritten corrections of quark wave functions were added by me.123
Appendix F: Tables of hadrons
These tables of meson and baryons were written by RPF.4 . A - ( ' Q ? - - - A ( 2 I - p[/~775/ -g/- - ~ ( 2 1 7 4 .%77?- _ _ - --- I Appendix G: Rules for amplitudes and observables
These were also hand-written by RPF. Annotations in blue were made by me at the time the course was given.7 ' r i a flo~y- ~ o f l o p y . ..