hhep-th/yymmnnnITP–UU–08/05SPIN–08/05
Lectures on instantons
Stefan Vandoren and Peter van Nieuwenhuizen Institute for Theoretical Physics and Spinoza InstituteUtrecht University, 3508 TD Utrecht, The Netherlands [email protected] C.N. Yang Institute for Theoretical PhysicsState University of New York at Stony Brook, NY 11790, USA [email protected]
Abstract
This is a selfcontained set of lecture notes on instantons in (super) Yang-Mills theoryin four dimensions and in quantum mechanics. First the basics are derived from scratch:the regular and singular one-instanton solutions for Yang-Mills theories with gauge groups SU (2) and SU ( N ), their bosonic and fermionic zero modes, the path integral instantonmeasure, and supersymmetric Yang-Mills theories in Euclidean space. Then we discussapplications: the θ -angle of QCD, the solution of the U (1) problem, the way Higgs fieldssolve the large-instanton problem, and tunneling and phase transitions in quantum me-chanics and in nonabelian gauge theories. These lecture notes are an extension of a reviewon Yang-Mills and D-instantons written in 2000 by both authors and A.Belitsky [1]. Contents a r X i v : . [ h e p - t h ] F e b Collective coordinates, the index theorem and fermionic zero modes 25 β function for SYM theories . . . . . . . . . . . . . . . . . . . . 52 N = 4 supersymmetric Yang-Mills theory 57 N = 4 SYM . . . . . . . . . . . . . . . . . . . . . . . . . . . 578.2 Euclidean N = 4 SYM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588.3 Involution in Euclidean space . . . . . . . . . . . . . . . . . . . . . . . . . 61 U (1) problem 8914 Baryon decay 9115 Discussion 94A Winding number 94B ’t Hooft symbols and Euclidean spinors 97C The volume of the gauge orientation moduli space 100D Zero modes and conformal symmetries 106 Introduction
In the last decades enormous progress has been made in understanding nonperturba-tive effects, both in supersymmetric field theories and in superstring theories. By non-perturbative effects we mean effects due to solitons and instantons, whose masses andactions, respectively, are inversely proportional to the square of the coupling constant [2].Typical examples of solitons are the kink, the vortex, and the magnetic monopole in fieldtheory, and some D-branes in supergravity or superstring theories. In supersymmetric fieldtheories these solutions preserve half of the supersymmetry and saturate BPS bounds. Asfor instantons, we have the Yang-Mills (YM) instantons in four dimensions [3, 4, 5], ortunnelling phenomena in quantum mechanics with a double-well potential as described bythe kink, see e.g. [6], and there are various kinds of instantons in string theory, for examplethe D-instantons [7]. Also instantons preserve half the number of supersymmetries in su-persymmetric field theories. Instantons can also be defined in field theories in dimensionshigher than four [8], but we discuss in this chapter mainly the case of four dimensions.Instantons in ordinary (i.e., nongravitational) quantum field theories are by definitionsolutions of the classical field equations in Euclidean space with finite action. Only fora finite classical action S cl is the factor exp[ − h S cl ] in the path integral nonvanishing.We shall consider instantons in nonabelian gauge theories in flat spacetime (there are noinstantons in abelian gauge theories in flat space), both regular instantons (which actuallyhave a singularity at | x | = ∞ ) and singular instantons (which have a singularity at apoint x = x but not at | x | = ∞ ). A singular gauge transformation maps the first intothe second, and vice-versa . Around a given instanton solution, there are the quantumfluctuations. The action contains terms with 2, 3, 4 . . . quantum fields, and one canperform perturbation theory around the instanton. The terms quadratic in quantum fields In gravity there are various definitions of instantons: Einstein spaces with selfdual Weyl tensors,selfdual Riemann tensors, solutions of the Einstein equations with/without finite action etc. Since in gravityspacetime is part of the solution, one usually considers spacetime topologies which are different from thatof R . A selfdual Riemann tensor leads to an Einstein space ( R µν = Λ g µν ) whose Einstein-Hilbert actionis either infinite (if the cosmological constant Λ is nonvanishing), or it only gets contributions from theGibbons-Hawking boundary term [9]. In general, the semiclassical approximation of the Einstein-Hilbertaction is not well defined due to the unboundedness of the action inside the path integral. To cure this,one probably has to discuss gravitational instantons inside a full theory for quantum gravity. For instantonsolutions in flat space but using curvilinear coordinates (for example S , or cylindrical coordinates) see[10]. For the ”regular solution”, A aµ is finite on R ∪ ∞ = S everywhere, but this does not mean that it isregular. It is finite only because one can use two different patches to cover S , and A aµ is regular in eachpatch. If one maps infinity to the origin by a space-inversion transformation ( x µ = y µ /y ), then one findsa singularity at the origin. In this sense the ”regular solution” is singular. We further clarify this issue inthe next section. normalizable . (It isan eigenfunction of the quantum field operator with eigenvalue zero). In a trivial vacuumthere are no zero modes: there are, of course, solutions of the linearized field equations, butthey are not normalizable. We must treat the zero modes in instanton physics separatelyfrom the nonzero modes; for example, they have their own measure in the path integral.The nonzero modes live in the space orthogonal to the zero modes and in this space onecan invert the linearized field equations for the fluctuations and construct propagators, anddo perturbation theory.Instantons describe tunnelling processes in Minkowski space-time from one vacuum attime t to another vacuum at time t . The simplest model which exhibits this phenomenonis a quantum mechanical point particle with a double-well potential having two vacua,or a periodic potential with infinitely many vacua. Classically there is no trajectory fora particle to travel from one vacuum to the other, but quantum mechanically tunnellingoccurs. The tunnelling amplitude can be computed in the WKB approximation, and istypically exponentially suppressed. In the Euclidean picture, after performing a Wickrotation, the potential is turned upside down, and it is possible for a particle to propagatebetween the two vacua, as described by the classical solution to the Euclidean equations ofmotion. The claim is then that the contributions from instantons in Euclidean space yielda good approximation of the path integral in Minkowski space. We shall prove this for thecase of quantum mechanics.Also in YM theories, instantons are known to describe tunnelling processes betweendifferent vacua, labeled by an integer winding number, and lead to the introduction ofthe CP-violating θ -term in the action [11, 12]. It was hoped that instantons could shedsome light on the mechanism of quark confinement. Although this was successfully shownin three-dimensional gauge theories (based on the Georgi-Glashow model) [13], the roleof instantons in relation to confinement in four dimensions is less clear. Together withthe non-perturbative chiral U (1) anomaly in an instanton background, which leads tobaryon number violation and a solution of the U (1) problem [4, 5], instantons are usedin phenomenological applications to QCD and the Standard Model. To avoid confusion,note that the triangle chiral anomalies in perturbative field theories in Minkowski space-time are canceled by choosing suitable multiplets of fermions. There remain, however,chiral anomalies at the non-perturbative level. It is hard to compute the non-perturbativeterms in the effective action which lead to a breakdown of the chiral symmetry by usingmethods in Minkowski space-time. However, by using instantons in Euclidean space, one4an relatively easily determine these terms. The nonperturbative chiral anomalies are dueto fermionic zero modes which appear in the path integral measure (in addition to bosoniczero modes). One must saturate the Grassmann integrals over these zero modes, and thisleads to correlation functions of composite operators with fermion fields which do violatethe chiral U (1) symmetry. The new non-perturbative terms are first computed in Euclideanspace, but then continued to Minkowski space where they give rise to new physical effects[5]. They have the following generic form in the effective action (we suppress here possibleflavor or adjoint indices that the fermions can carry) S eff ∝ e n − π g ( O ( g ) ) + iθ o (¯ λ ¯ λ ) n , (1.1)where 2 n is the number of fermionic zero modes ( n depends on the representation of thefermions and the gauge group). The prefactor is due to the classical instanton action andis clearly non-perturbative. The terms indicated by O ( g ) are due to standard radiativecorrections computed by using Feynman graphs in an instanton background. The term (cid:0) ¯ λ ¯ λ (cid:1) n involving antichiral spinors ¯ λ is produced if one saturates the integration in thepath integral over the fermionic collective coordinates and violates in general the chiralsymmetry. On top of (1.1) we have to add the contributions from anti-instantons, generat-ing ( λλ ) n terms in the effective action, where λ denotes chiral spinors. As we shall discuss,for Majorana spinors in Euclidean space the chiral and anti-chiral spinors are independent,but in Minkowski space-time they are related by complex conjugation, and one needs thesum of instanton and anti-instanton contributions to obtain a hermitean effective action.We shall also apply the results of the general formalism to supersymmetric gauge the-ories, especially to the N = 4 SU ( N ) supersymmetric Yang-Mills (SYM) theory. Here N stands for the number of supersymmetries. Instantons in N = 1 , N = 1 models, one is mainly interested in the calculation ofthe superpotential and the gluino condensate [15, 16]. In some specific models, instantonsalso provide a mechanism for supersymmetry breaking [16], see [17] for a review on theseissues. In the case of N = 2, there are exact results for the prepotential [18] based onlyon general symmetry principles and electric-magnetic duality; the prepotential acquirescontributions from all multi-instanton sectors. These predictions were successfully testedagainst direct field theoretical calculations in the one-instanton sector in [19], and for atwo-instanton background in [20]. More recently, new techniques were developed to per-form multi-instantons calculations in [21]. Finally, the nonperturbative structure of N = 4SYM has been studied thoroughly in the context of the AdS/CFT correspondence [22].SYM instantons in the limit of large number of colors were succesfully shown to reproducethe D-instanton contributions to certain correlation functions, both for single instantons523, 24] and for multi-instantons [25]. Other correlation functions were studied in [26, 27].For a recent review of instantons in supersymmetric gauge theories, see [28].The material is organized as follows. In section 2, we discuss the winding number ofgauge fields, and we present the standard one-instanton solution in SU (2) and in SU ( N ).This already raises the question how to embed SU (2) into SU ( N ), and we discuss thevarious embeddings. In section 3 we discuss instanton solutions in general: we solve theduality condition and find multi-instanton solutions which depend on their position andtheir scale. We concentrate on the one-instanton solutions, and first determine the singularsolutions, but then we make a (singular) gauge transformation and obtain the regularsolutions. In section 4 we start the study of “collective coordinates”, the parameters onwhich the most general instanton solutions depend. We show that the number of collectivecoordinates is given by an index theorem for the Dirac operator in an instanton background.We then give a derivation of this index theorem, and conclude that a k -instanton solutionin SU ( N ) has 4 N k bosonic collective coordinates, 2
N k fermionic collective coordinates forfermions in the adjoint representation, and k fermionic collective coordinates for fermionsin the defining (fundamental, vector) representation. In section 5 we explicitly constructthe zero modes for gauge group SU ( N ) in a one-instanton background. First we constructthe bosonic zero modes; these are associated to the collective coordinates for translations,dilatations and gauge orientations. Next we derive the explicit formula for the generalsolution of the fermionic zero modes of the Dirac equation in a one-instanton background,first for SU (2) and then for SU ( N ).In section 6 we construct the one-instanton measure for the bosonic and fermionic collec-tive coordinates. We explain in detail the normalization of the zero modes since it is crucialfor the construction of the measure. We convert the integration over the coefficients of thebosonic zero modes to an integration over the corresponding bosonic collective coordinatesby the Faddeev-Popov trick, but for fermionic zero modes we do not need this procedurebecause in this case the coefficients of fermionic zero modes are already the fermionic col-lective coordinates. In section 7 we discuss the one-loop determinants in the background ofan instanton, arising from integrating out the quantum fluctuations. We then apply this tosupersymmetric theories, and we use an index theorem to prove that the determinants forall supersymmetric YM theories cancel each other. Furthermore, we compute the completenonperturbative β function for supersymmetric Yang-Mills theories by assuming that themeasure for the zero modes does not depend on the renormalization scale µ . However,since it is not known to which regularization scheme this procedure corresponds, this re-sult cannot be checked by standard perturbative calculations. In section 8 we discuss the N = 4 supersymmetric Yang-Mills theory in Euclidean space and its instantons.The remaining sections contain applications. Section 9 discusses the problem of large6nstantons and its solution in terms of Higgs fields and spontaneous symmetry breaking.Section 10 gives a detailed discussion how instantons can describe tunnelling. In section11 we use a quantum mechanical model with a double-well potential to discuss the phasetransition from a false vacuum to the true vacuum by bubble formation. Section 12 containsthe strong CP problem, the mystery that the θ angle is so small. Section 13 discusses thatinstantons solve the U (1) problem and in section 14 we finally discuss how instantons leadto baryon decay.In a few appendices we set up our conventions and give a detailed derivation of sometechnical results in order to make the material self-contained. In appendix A we providedetails of the calculation of the winding number. In appendix B we discuss the ’t Hoofttensors and the spinor formalism in Euclidean space. In appendix C we calculate thevolume of the moduli space of gauge orientations. Finally, in appendix D we show thatconformal boosts and Lorentz rotations do not lead to additional zero modes. We start with some elementary facts about instantons in SU ( N ) Yang-Mills theories. Theaction, continued to Euclidean space, is S = − g (cid:90) d x tr F µν F µν ; F µν = F aµν T a . (2.1)The generators T a are traceless anti-hermitean N by N matrices satisfying [ T a , T b ] = f abc T c with real structure constants and tr( T a T b ) = − δ ab . For instance, for SU (2) one has T a = − i τ a , where τ a are the Pauli matrices and f abc = (cid:15) abc . Notice that with theseconventions the action is positive. Further conventions are D µ Y = ∂ µ Y + [ A µ , Y ] for anyLie algebra valued field Y , and F µν = ∂ µ A ν − ∂ ν A µ + [ A µ , A ν ], so that F µν = [ D µ , D ν ]. TheEuclidean metric is δ µν = diag(+ , + , + , +). In (2.1), the only appearance of the couplingconstant is in front of the action. The group metric g ab = δ ab is an invariant tensor , sowe may raise and lower indices with δ ab and δ ab . Thus we may also write [ T a , T b ] = f abc T c ,and from now on we shall write group and Lorentz indices either as covariant indices or ascontravariant indices, depending on which way is most convenient.By definition, a Yang-Mills instanton is a solution of the classical Euclidean equationsof motion with finite action. The classical equations of motion read D µ F µν = 0 . (2.2) From tr [ T c , T a T b ] = tr ([ T c , T a ] T b + trT a [ T c , T b ]) it follows that g ab is an invariant tensor: transforming its indicesby an adjoint transformation with parameter λ c yields zero: δg ab = λ c f cad g db + λ c f cbd g ac = 0.
7o find solutions with finite action, we require that the field strength tends to zero atinfinity faster than | x | − ≡ r − , hence the gauge fields asymptotically approach a puregauge A µ | x | →∞ = U − ∂ µ U , (2.3)for some U ∈ SU ( N ). To prove that gauge fields are pure gauge if the curvature F µν vanishes, is easy. Using U ∂ µ U − = − ∂ µ U U − we must solve for U from ∂ µ U = − A µ U ,whose solution is the path-ordered integral U = exp[ − (cid:82) x A µ ( y )d y µ ]. This expression doesnot depend on the path chosen because F µν = 0. (Note, however, that if two gauge fieldconfigurations, say A Iµ and A IIµ , yield the same curvature, F µν ( A I ) = F µν ( A II ), they neednot be gauge equivalent. A simple example proves this. Consider A Iµ = (cid:8) − ByT , BxT , , (cid:9) ; A IIµ = (cid:110) A II = √ BT , A II = √ BT , , (cid:111) , (2.4)where B is a constant and T a are the generators of SU (2) with structure constants f abc = (cid:15) abc . Clearly F ( A I ) = BT and also F ( A II ) = BT while all other components of F µν vanish. To prove that A Iµ cannot be written as U − ( ∂ µ + A IIµ ) U we note that if there wassuch a group element U , it should satisfy U F µν U − = F µν , hence U should commute with T . This implies that U would be given by exp( f ( x ) T ) for some real function f ( x ). Then − ByT = ∂ x f T + e − fT √ B T e − fT which has no solution. One can also calculate aWilson loop W = tr P exp (cid:72) Adl . This expression is gauge invariant, and if one chooses asloop a square in the x − y plane with sides L and L , one finds W I = BL L T ; W II = 2 √ B ( L T + L T ) (2.5)If A Iµ and A IIµ were gauge equivalent, W I should have been equal to W II ).There is actually a way of classifying fields which satisfy the boundary condition in(2.3). It is known from homotopy theory that all gauge fields with vanishing field strengthat infinity can be classified into sectors characterized by an integer number called thePontryagin class, or the winding number, or instanton number, or the topological charge k = − π (cid:90) d x tr F µν ∗ F µν , (2.6)where ∗ F µν = (cid:15) µνρσ F ρσ is the dual field strength, and (cid:15) = 1. Note that it is notnecessary that these gauge fields satisfy the field equations, only that their field strength Another way of satisfying the finite action requirement is to first formulate the theory on a compactified R , by adding and identifying points at infinity. Then the topology is that of the four-sphere, since R ∪ ∞ (cid:39) S . The stereographic map from R ∪ ∞ to S preserves the angles, and is therefore conformal.Also the YM action is conformally invariant, implying that the action and the field equations on R ∪ ∞ are the same as on S (the metric on the sphere is g µν = δ µν (1 + x ) − ). The finiteness requirement issatisfied when the gauge potentials can be smoothly extended from R to S . The action is then finitebecause S is compact and A µ is well-defined on the whole of the four-sphere. r = ∞ . The derivation of this result can be found in AppendixA. As part of the proof, one shows that the integrand in (2.6) is the divergence of a current K µ = − π (cid:15) µνρσ tr A ν (cid:0) ∂ ρ A σ + A ρ A σ (cid:1) . (2.7)The four-dimensional integral in (2.6) then reduces to an integral over a three-sphere atspatial infinity, and one can use (2.3) to show that the integer k counts how many times thisspatial three-sphere covers the gauge group three-sphere S ≈ SU (2) ⊂ SU ( N ). In moremathematical terms, the integer k corresponds to the third homotopy group π ( SU (2)) = Z . So k as defined in (2.6) does not depend on the values of the fields in the interior, butonly on the fields at large | x | . This can also directly be seen: under a small variation A µ → A µ + δA µ one has F µν → F µν + D µ δA ν − D ν δA µ , and partial integration (allowedwhen δA µ is only nonzero in a region in the interior) yields δA ν D µ ∗ F µν which vanishesdue to the Bianchi identity D [ µ F νρ ] = 0. (To prove this Bianchi identity one may use F νρ = [ D ν , D ρ ]. In [ D µ , [ D ν , D ρ ]] + [ D ν , [ D ρ , D µ ]] + [ D ρ , [ D µ , D ν ]] there are then 12 termswhich cancel pairwise.)Since we require instantons to have finite action, they satisfy the above boundary con-ditions at infinity, and hence they are classified by k , which we call the instanton number.Gauge potentials leading to field strengths with different instanton number can not berelated by continuous gauge transformations. This follows from the fact that the instantonnumber is a gauge invariant quantity. In a given topological sector, the field configurationwhich minimizes the action is a solution of the field equations. (It is a priori not obviousthat there exist field configurations that minimize the action, but we shall construct suchsolutions, thereby explicitly proving that they exist). We now show that, in a given topo-logical sector, the solution to the field equations that minimizes the action has either aselfdual or anti-selfdual field strength F µν = ± ∗ F µν = ± (cid:15) µνρσ F ρσ . (2.8)This equation is understood in Euclidean space, where ( ∗ ) = 1. In Minkowski space thereare no real solutions to the selfduality equations since ( ∗ ) = −
1. As seen from (2.6), instantons (with selfdual field strength) have k > whereas anti-instantons(with anti-selfdual field strength) have k <
0. (Recall that trT a T b is negative). Tosee that minimum action solutions are indeed selfdual or anti-selfdual, we perform a tricksimilar to the one used for deriving the BPS bound for solitons: we write the action as thesquare of a sum plus a total derivative term S = − g (cid:90) d x tr F = − g (cid:90) d x tr ( F ∓ ∗ F ) ∓ g (cid:90) d x tr F ∗ F ≥ ∓ g (cid:90) d x tr F ∗ F = 8 π g ( ± k ) . (2.9)9e used that tr ∗ F ∗ F = tr F F and omitted Lorentz indices to simplify the notation. Theequality is satisfied if and only if the field strength is (anti-) selfdual. The value of theaction is then S cl = (8 π /g ) | k | , and has the same value for the instanton as for theanti-instanton. However, we can also add a theta-angle term to the action, which reads S θ = − i θ π (cid:90) d x tr F µν ∗ F µν = iθk = ± iθ | k | . (2.10)The plus or minus sign corresponds to the instanton and anti-instanton respectively, sothe theta-angle distinguishes between them. In Minkowski spacetime this term is the samebecause both d x and F µν ∗ F µν produce a factor i under a Wick rotation. We give a moredetailed treatment of the theta-angle term and its applications in Section 12.It is interesting to note that the energy-momentum tensor for a selfdual (or anti-selfdual)field strength always vanishes T µν = − g tr (cid:8) F µρ F νρ − δ µν F ρσ F ρσ (cid:9) = 0 . (2.11)(Because in Euclidean space T = − g tr ( (cid:126)E − (cid:126)B ), the Euclidean “energy” T need not bepositive definite). This agrees with the observation that the instanton action (cid:82) d x tr F = (cid:82) d x tr ∗ F F is metric independent in curved space. The vanishing of the energy-momentumtensor is consistent with the fact that instantons are topological in nature. It implies thatinstantons do not curve Euclidean space, as follows from the Einstein equations.An explicit construction of finite action solutions of the Euclidean classical equations ofmotion was given by Belavin et al. [3]. We shall derive this solution, and others, in section3, but to get oriented we present it here, and discuss some of its properties. The gaugeconfiguration for one instanton ( k = 1) in SU (2) contains the matrices σ µν or ¯ σ µν . Oneoften writes it in terms of the ’t Hooft η tensors, related to ¯ σ µν by ¯ σ µν = iη aµν τ a where τ a are the generators of SU (2). We discuss these tensors in Appendix B. The regularone-instanton solution reads then A aµ ( x ; x , ρ ) = 2 η aµν ( x − x ) ν ( x − x ) + ρ ,A µ ≡ A aµ (cid:16) τ a i (cid:17) = − ¯ σ µν ( x − x ) ν ( x − x ) + ρ , (2.12)where x and ρ are arbitrary parameters called collective coordinates. They correspondto the position and the size of the instanton. The above expression solves the selfduality Note that T is proportional to tr( F F + F F ), which is equal to minus itself due to the selfdualityrelations F = F , F = − F and F = F . Similarly T vanishes because it is proportional to thetrace of ( F + F + F ) − ( F + F + F ). x = x ,as long as ρ (cid:54) = 0. The real antisymmetric eta-symbols are defined as follows η aµν = (cid:15) aµν µ, ν = 1 , , , η aµ = − η a µ = δ aµ , ¯ η aµν = (cid:15) aµν µ, ν = 1 , , , ¯ η aµ = − ¯ η a µ = − δ aµ . (2.13)The η and ¯ η -tensors are selfdual and anti-selfdual respectively, for fixed index a . Theyform a basis for antisymmetric four by four matrices, and we have listed their propertiesin Appendix B. They are linear combinations of the Euclidean Lorentz generators L µν ,namely η aµν = ( J a + K a ) µν and ¯ η aµν = ( J a − K a ) µν , where J a = (cid:15) abc L bc and K a = L a ,and ( L mn ) µν = δ mµ δ nν − δ mν δ nµ with m, n = 1 ,
4. In this subsection we use η tensors, butin later sections we shall use the matrices σ µν and ¯ σ µν .The field strength corresponding to this gauge potential is (use (B.5)) F aµν = − η aµν ρ [( x − x ) + ρ ] , (2.14)and it is selfdual. Thus (2.12) is a solution of the classical field equations. Far away, A aµ becomes proportional to the inverse radius r so that it contributes a finite amount tothe integral for the winding number which is of the form (cid:82) A ( r dΩ), while F µν becomesproportional to r , yielding a finite action. However, A aµ itself vanishes at r → ∞ , hencewe have a smooth configuration on S . Notice that the special point ρ = 0, correspondingto zero size instantons, leads to zero field strength and corresponds to pure gauge. Strictlyspeaking, this point must therefore be excluded from the instanton moduli space of col-lective coordinates. Finally one can compute the value of the action by integrating thedensity tr F µν F µν = − ρ [( x − x ) + ρ ] . (2.15)Using the integral given at the end of Appendix B, one finds that this solution correspondsto k = 1.One may show by direct calculation that the regular one-anti-instanton solution is alsogiven by (2.12) but with ¯ η aµν . (In the proof one uses that the first formula in (B.5) alsoholds for ¯ η aµν ).We shall also derive the one-instanton solution in the singular gauge. In terms of η symbols it reads A aµ = 2 ρ ¯ η aµν ( x − x ) ν ( x − x ) [( x − x ) + ρ ] = − ¯ η aµν ∂ ν ln (cid:26) ρ ( x − x ) (cid:27) . (2.16)This gauge potential is singular for x = x , where it approaches a pure gauge configurationas we shall show in the next section, A µ x → x = U ∂ µ U − . The gauge transformation U is11ingular and relates the regular gauge instanton (2.12) to the singular one (2.16) at allpoints. The field strength in singular gauge is then (taking the instanton at the origin, x = 0, otherwise replace x → x − x ) F aµν = − ρ ( x + ρ ) (cid:110) ¯ η aµν − η aµρ x ρ x ν x + 2¯ η aνρ x ρ x µ x (cid:111) . (2.17)Notice that despite the presence of the anti-selfdual eta-tensors ¯ η , this field strength isstill selfdual, as can be seen by using the properties of the eta-tensors given in (B.5). Thesingular gauge is frequently used, because, as we will see later, zero modes fall off morerapidly at large x in the singular gauge. One can compute the winding number again insingular gauge. Then one finds that there is no contribution coming from infinity. Instead,all the winding is coming from the singularity at the origin. The singular solution issingular at x , so one would expect that the regular solution is singular at infinity. Thismay seem puzzling since we saw that the regular solution was smooth on S . However, todecide whether a configuration is smooth at r → ∞ , one should first transform the point atinfinity to the origin and then study how the transformed configuration behaves near theorigin. Making the coordinate transformation x µ = y µ /y or x µ = − y µ /y , not forgettingthat a vector field transforms as A (cid:48) µ ( y ) = ( ∂x ν /∂y µ ) A ν ( x ), one finds that the transformedregular k = 1 solution is indeed singular at the origin . In fact, it is equal to the singular k = − ρ replaced by ρ .At first sight it seems that there are five collective coordinates for the k = 1 solution.There are however extra collective coordinates corresponding to the gauge orientation. Onecan act with an SU (2) matrix on the solution (2.12) to obtain another solution, A µ ( x ; x , ρ, (cid:126)θ ) = U − ( (cid:126)θ ) A µ ( x ; x , ρ ) U ( (cid:126)θ ) , U ∈ SU (2) . (2.18)with constant (cid:126)θ . One might think that these configurations should not be considered asa new solution since they are gauge equivalent to the expression given above. This isnot true, however, the reason being that, after we fix the gauge, we still have left a rigid SU (2) symmetry which acts as in (2.18). So in total there are eight collective coordinates,also called moduli. In principle, one could also act with the (space-time) rotation matrices SO (4) on the instanton solution, and construct new solutions. However, these rotations canbe undone by suitably chosen gauge transformations [29]. Actually, the Yang-Mills actionis not only invariant under the Poincar´e algebra (and the gauge algebra), but it is also This coordinate transformation in R can be viewed as a product of two conformal projections, onefrom the plane to the coordinate patch on the sphere S containing the south pole, and the other fromthe other coordinate patch on S with the north pole back to the plane. The transformed metric is g (cid:48) µν ( y ) = δ µν /y , so conformally flat. Then the action for the A (cid:48) µ ( y ) in y -coordinates is again the usualflat space action in (2.1), and the transformed instanton solution is an anti-instanton solution. K µ ). As shown in Appendix D, for theEuclidean conformal group SO (5 , SO (5) consisting of SO (4) rotations anda combination of conformal boosts and translations ( R µ ≡ K µ + ρ P µ ), leaves the instantoninvariant up to gauge transformations. This leads to a 5 parameter instanton moduli space SO (5 , /SO (5), which is the Euclidean version of the five-dimensional anti-de Sitter space AdS . The coordinates on this manifold correspond to the four positions and the size ρ ofthe instanton. On top of that, there are still three gauge orientation collective coordinates,yielding a total of eight moduli for the k = 1 instanton in SU (2).Instantons in SU ( N ) can be obtained by embedding SU (2) instantons into SU ( N ). Forinstance, a particular embedding is given by the following N by N matrix A SU ( N ) µ = (cid:32) A SU (2) µ (cid:33) . (2.19)where the instanton resides in the 2 × SU ( N ) element on the solution (2.19) and obtain a newone. Not all elements of SU ( N ) generate a new solution. There is a stability group thatleaves (2.19) invariant, acting only on the zeros, or commuting trivially with the SU (2)embedding. Such group elements should be divided out, so we consider, for N > A SU ( N ) µ = U (cid:32) A SU (2) µ (cid:33) U † , U ∈ SU ( N ) SU ( N − × U (1) . (2.20)One can now count the number of collective coordinates. From counting the dimension ofthe coset space in (2.20), one finds there are 4 N − SU (2) solution, we find in total 4 N collective coordinates for a one-instanton solution in SU ( N ). It is instructive to work out the example of SU (3). Here weuse the eight Gell-Mann matrices { λ α } , α = 1 , . . . ,
8. The first three λ a , a = 1 , ,
3, form an SU (2) algebra and are used to define the k = 1 instanton by contracting (2.12) or (2.16)with λ a . The generators λ , . . . , λ form two doublets under this SU (2), so they act onthe instanton and can be used to generate new solutions. This yields four more collectivecoordinates. Then there is λ , corresponding to the U (1) factor in (2.20). It commuteswith the SU (2) subgroup spanned by λ a , and so it belongs to the stability group leavingthe instanton invariant. So for SU (3) and k = 1, there are seven gauge orientation zeromodes, which agrees with 4 N − N = 3.The embedding of instanton solutions as a 2 × N × N matrix rep-resentation of SU ( N ) is not the only embedding possible. For example, one can also use13he 3 × T a of SU (2), and put the instanton inside a 3 × N of SU ( N ). This of SU (2) is sometimes called “the other SU (2) in SU (3)”, but itis simply the adjoint representation of SU (2), which is also the defining representation of SO (3), and is given by ( T a ) ij = (cid:15) iaj , T = −
10 1 0 ; T = − ; T = − (2.21)This representation has the same structure constants f abc = (cid:15) abc as the representation T a = τ a / (2 i ), but now tr { T a T b } = − δ ab , four times larger.In fact, going back to the construction of the instanton, we note that any representation T a of SU (2) yields an instanton solution for SU ( N ) as long as it fits inside the N × N matrices of SU ( N ) [30] A µ = 2 η aµν T a x ν x + ρ . (2.22)The of SU (2) with T a = τ a i yields (2.19), but any other representation yields anotherembedding.For SU (3) there are only two possibilities. We can embed the instanton using the of SU (2); this yields (2.19). But we can also use the matrices T a given in (2.21) as the first 3generators of SU (3). For SU ( N ) we can use any spin j representation of SU (2) providedit fits inside the N × N matrices. Since the action and winding number are proportionalto the trace tr T a T b , which is proportional to the quadratic Casimir operator j ( j + 1) timesthe dimension 2 j + 1 of the spin j representation , we see that we get instanton solutionswith winding number k = ± j ( j + 1)(2 j + 1). For j = 1 / k = ±
1. Forthe first few SU ( N ) the results are as follows SU (3) : k = ± k = ± j = 1 / j = 1) SU (4) : k = ± k = ± k = ±
10 ( j = 1 / , , / k = ± j = 1 / SU (5) : k = ± , ± , ± , ±
20 ( j = , , , k = ± , ± j = ⊕ and j = ⊕ . (2.23)All these instanton solutions with winding number | k | > k instantonsembedded as 2 × Use δ ab trT a T b = − trC ( R ) = − (2 j + 1) C ( R ) where the quadratic Casimir operator for the represen-tation R with spin j is given by C ( R ) = − δ ab T ( R ) a T ( R ) b = j ( j + 1). /r . Bringing k instantons together such that they sit all at the samepoint, gives solutions of the kind above. So far apart there is a small positive interaction,but when they are brought together the interaction energy vanishes. Hence, there must bedomains of attraction in between. This already shows that the interaction of instantons isa complicated problem [30]. In fact, one can deform these single-instanton solutions suchthat a multi-instanton solution is obtained in which the single-instantons do not attractor repel each other. In other words, in such a multi-instanton solution the positions, sizesand gauge orientations of the single instantons are collective coordinates.For the general multi-instanton solution, the dependence on all collective coordinatesis in implicit form given by the ADHM construction [31]. For a recent review, see [32].In the next section we will obtain explicit formulas for the dependence on 5 k collectivecoordinates. Explicit formulas for the dependence on all collective coordinates only existfor the k = 2 instanton solution [31, 33, 34, 32] and the k = 3 instanton solution [35].We end this section with some remarks on embeddings into other gauge groups [36]. For k = 1 and gauge group SO ( N ), it is known that there are 4 N − SO (4) = SU (2) × SU (2) generated by η aµν and ¯ η aµν , and putting the instantonin one of the SU (2) groups. The stability group of this instanton is SO ( N − × SU (2),so we obtain (for N > A SO ( N ) µ = U (cid:32) A SU (2) µ (cid:33) U † , U ∈ SO ( N ) SO ( N − × SU (2) . (2.24)The number of collective coordinates of such solutions follows from the dimension of thecoset (which is 4 N − SU (2) instanton, we arriveat 4 N − N = 6, we can usethe isomorphism between SO (6) and SU (4). For both countings, we arrive at 16 moduli.Similarly, we can analyze the symplectic gauge groups U Sp (2 N ). Here we can simplychoose the lower diagonal SU (2) = U Sp (2) embedding inside
U Sp (2 N ) for a k = 1 in-stanton. The stability group of this embedding is now U Sp (2 N − A Sp ( N ) µ = U (cid:32) A SU (2) µ (cid:33) U † , U ∈ U Sp (2 N ) U Sp (2 N − . (2.25)The dimension of U Sp (2 N ) is N (2 N +1), and so the total number of collective coordinatesthat follows from this construction is 5 + (4 N −
1) = 4( N + 1), which is the correct number The dimension of U (2 N ) is 4 N and the generators have the form (cid:32) a + is b − b † a + is (cid:33) where a i is N = 2, we have the isomorphism U Sp (2) = SO (5), which in both countings leadsto 12 collective coordinates.For higher instanton number, not all instantons can be constructed from a properlychosen embedding. There the ADHM formalism must be used. We just mention herethat the total number of collective coordinates is 4 kN, k ( N −
2) and 4 k ( N + 1) for thegauge groups SU ( N ) , SO ( N ) and U Sp (2 N ) respectively. The geometric relation betweeninstanton moduli spaces and quaternionic manifolds (whose dimension is always a multipleof four) can e.g. be found in [37]. Note that we have not shown that all solutions of (2.2) with finite action are given byselfdual (or anti-selfdual) field strengths. In principle there could be configurations whichare extrema of the action, but are neither selfdual nor anti-selfdual . For the gauge group SU (2) this has been a long standing question. The first result was established in [38, 39, 40]where it was shown that for gauge groups SU (2) and SU (3), nonselfdual solutions cannotbe local minima, hence if they exist, they should correspond to saddle points. The existenceof nonselfdual solutions with finite action and gauge group SU (2) was first established in[41], for k = 0, and later for k (cid:54) = 0 in [42]. For gauge group SU (3) some results havebeen obtained in [43, 44]. The situation seems to be quite complicated, and no elegant andsimple framework to address these issues has been found so far. For bigger gauge groups,it is easier to construct non-selfdual (or anti-selfdual) solutions. This becomes clear inthe example of SO (4) = SU (2) × SU (2). If we associate a selfdual instanton to the firstfactor, and an anti-selfdual instanton to the second factor, the total field strength satisfiesthe equations of motion (2.2) but is neither selfdual nor anti-selfdual. Even simpler is theexample of SU (4). By choosing two commuting SU (2) subgroups, we can embed both an antisymmetric and s i is symmetric. Complex symplectic matrices M = (cid:32) A BC D (cid:33) satisfy M T Ω + Ω M =0 where Ω = (cid:32) I − I (cid:33) . The restriction that the unitary generators be also symplectric leads to N + N ( N −
1) constraints ( D + A T = 0 and C − C T = B − B T = 0). It is possible to construct solutions for SU (2) that are not selfdual, but not with finite action. Anexample is A µ = − σ µν x ν r . Its field strength is F µν = σ µν /r + ( x µ σ µρ − x ν σ µρ ) x ρ /r . One can checkthat it satisfies the second order equation of motion (2.2) (both ∂ µ F µν and [ A µ , F µν ] vanish), but thisconfiguration is not selfdual since F µν − ∗ F µν = σ µν r . Because this field strength does not tend to zerofast enough at infinity, the action evaluated on this solution diverges logarithmically. U (2) instanton and an anti-instanton inside SU (4), A SU (4) µ = (cid:32) A + µ A − µ (cid:33) , (2.26)where A ± µ denotes the (anti-) selfdual SU (2) gauge potentials with topological charges k ± .Clearly the total field strength is neither selfdual nor anti-selfdual, but satisfies the secondorder equations of motion. The instanton action is finite and the total topological chargeis k + − k − .From the embedding (2.26) one can generate more solutions by acting on the gaugepotential with a global gauge transformation U ∈ SU (4). In this way, one generates newexact and nonselfdual solutions which are not of the form (2.26).For SU ( N ) gauge groups, one has even more possibilities. One can embed k + instantonsand k − anti-instantons on the (block)-diagonal of SU ( N ), as long as 2( k + + k − ) ≤ N . Ifwe take both k + > k − >
0, the solution is clearly not selfdual or anti-selfdual andthe instanton action, including the theta-angle, is given by S = 8 π g ( k + + k − ) + iθ ( k + − k − ) . (2.27)In a supersymmetric theory, these solutions will not preserve any supersymmetry. This isinteresting in the context of the AdS/CFT correspondence that relates N = 4 SYM theoryto type IIB superstrings. In [45], it is shown that these non-selfdual Yang-Mills instantonsare related to non-extremal (non BPS) D-instantons in IIB supergravity. To find explicit instanton solutions, we solve the selfduality (or anti-selfdualty) equations F µν = ∗ F µν where ∗ F µν = (cid:15) µνρσ F ρσ with µ, ν = 1 , (cid:15) = (cid:15) = 1. Since D µ ∗ F µν vanishes identically due to the Bianchi identity, we then have a solution of the field equa-tions, D µ F µν = 0. The main idea is to make a suitable ansatz, and then to check that ityields solutions. The ansatz is (we restrict ourselves for the moment to the gauge group SU (2)) A µ ( x ) = α σ µν ∂ ν ln φ ( x ) , (3.1)where α is a real constant to be fixed and σ µν is the 2 × σ µν a lot,we first discuss their properties in some detail, and then we shall come back below (3.22)to the construction of instanton solutions. 17 .1 Lorentz and spinor algebra In Euclidean space, a suitable 4 × γ µ = (cid:32) − i ( σ µ ) αβ (cid:48) i (¯ σ µ ) α (cid:48) β (cid:33) , σ µ = ( (cid:126)τ , iI )¯ σ µ = ( (cid:126)τ , − iI ) , (3.2)where (cid:126)τ are the Pauli matrices. We use slashes instead of dots on the spinor indices toindicate that we are in Euclidean space. All four Dirac matrices are hermitian, and satisfy { γ µ , γ ν } = 2 δ µν . The matrix γ is diagonal γ ≡ γ γ γ γ = (cid:32) I − I (cid:33) , (3.3)and chiral spinors correspond to projections with (1 ± γ ) which yield the upper or lowertwo components of a nonchiral four-component spinor.Since we are in Euclidean space, it does not matter whether we write the index µ as acontravariant or covariant index. In Minkowski space this representation (with γ replacedby γ where γ = iγ , so that ( γ k ) = +1 but ( γ ) = −
1) is used for two-component spinorformalism. Four-component spinors are then decomposed into two-component spinors as ψ = (cid:0) λ α ¯ χ ˙ α (cid:1) , and this explains the position of the spinor indices on σ µ and ¯ σ µ in (3.2).The Euclidean Lorentz generators ( SO (4) generators) acting on 4-component spinors are M µν = ( γ µ γ ν − γ ν γ µ ) and satisfy the Euclidean Lorentz algebra[ M µν , M ρσ ] = δ νρ M µσ − δ νσ M µρ − δ µρ M νσ + δ µσ M νρ . (3.4)However, this representation is reducible: the upper and lower components of ψ formseparate representations M µν = 12 (cid:32) ( σ µν ) αβ
00 (¯ σ µν ) α (cid:48) β (cid:48) (cid:33) . (3.5)In terms of σ µ and ¯ σ µ we then find the following two inequivalent spinor representationsof SO (4) : M µν = σ µν and M µν = ¯ σ µν , where σ µν = ( σ µ ¯ σ ν − σ ν ¯ σ µ ) ; ¯ σ µν = (¯ σ µ σ ν − ¯ σ ν σ µ ) . (3.6)(It is customary not to include the factor in M µν = σ µν into the definition of σ µν ).The matrices σ µν and ¯ σ µν satisfy some properties which we shall need repeatedly. Firstof all, they are anti-selfdual and selfdual, respectively σ µν = − (cid:15) µνρσ σ ρσ ; ¯ σ µν = (cid:15) µνρσ ¯ σ ρσ . (3.7)18his follows most easily by noting that the matrices γ µ satisfy γ [ µ γ ν ] = − (cid:15) µνρσ γ ρ γ σ γ where γ µν ≡ γ [ µ γ ν ] = ( γ µ γ ν − γ ν γ µ ). For example, γ γ = − γ γ γ because (cid:15) = +1.From this (anti)-selfduality one derives another useful property (cid:15) µνρσ σ στ = δ µτ σ νρ − δ ντ σ µρ + δ ρτ σ µν . (3.8)It is easiest to prove (3.8) by substituting (3.7) into the left-hand side, and decomposingthe product of two (cid:15) -tensors into a sum of products of Kronecker tensors. Another proofis based on the “Schouten identity” which is the observation that a totally antisymmetrictensor with 5 indices vanishes in 4 dimensions (because there are always at least two indicesequal). Writing the left-hand side of (3.8) as (cid:15) µνρα δ βτ σ αβ and using the Schouten identity (cid:15) µνρα δ βτ = (cid:15) βνρα δ µτ + (cid:15) µβρα δ ν τ + (cid:15) µνβα δ ρτ + (cid:15) µνρβ δ ατ , (3.9)the identity (3.7) can be used to prove the property (3.8) (the last term in (3.9) yieldsminus the contribution of the term on the left-hand side). In a similar way one may prove (cid:15) µνρσ ¯ σ στ = − δ µτ ¯ σ νρ + δ ντ ¯ σ µρ − δ ρτ ¯ σ µν . (3.10)The extra overall minus sign is due to the extra minus sign in the selfduality relation in(3.7).Further identities are the commutator of two Lorentz generators, and the anticommu-tator which is proportional to the unit matrix in spinor space[ σ µν , σ ρσ ] = 2 δ νρ σ µσ + three more terms , { σ µν , σ ρσ } = 2( δ µσ δ νρ − δ µρ δ νσ ) + 2 (cid:15) µνρσ . (3.11)One easy way to prove or check these identities is to use 4 × { γ γ , γ γ } = 2 γ and { γ , γ } = 0 but { γ , γ } = − γ , γ ] = − γ . Because γ = (cid:32) I − I (cid:33) , it is clear that the ¯ σ µν satisfy the same commutation and anticommuta-tion relations, but with a different sign for the (cid:15) symbol. In particular, { ¯ σ µν , ¯ σ ρσ } = 2( δ µσ δ νρ − δ µρ δ νσ ) − (cid:15) µνρσ . (3.12)All these identities can also be derived using two-component spinor formalism for vectors.For example, a vector v µ is written as v αα (cid:48) ≡ v µ σ µαα (cid:48) , and then one may use such identitesas δ µν ∼ δ αα (cid:48) ββ (cid:48) ∼ δ αβ δ α (cid:48) β (cid:48) ; δ µν ∼ (cid:15) αβ (cid:15) α (cid:48) β (cid:48) . (3.13)If one never introduces any vector indices at all but only uses spinor indices, this spinorformalism turns about all identities into trivialities, but we prefer to also keep vector indices19round. The other extreme is to expand σ µν and ¯ σ µν into Pauli matrices τ a as σ µν = i ¯ η aµν τ a and ¯ σ µν = iη aµν τ a where η aµν and ¯ η aµν are constructed from (cid:15) aij and δ ai tensors, as in (2.13).A whole calculus of these “’t Hooft-tensors” can be set-up, and is often used. We discussit in appendix B. We shall not limit ourselves to one of these extremes; proofs are giveneither by using 2-component spinors or 4 × A µ in (3.1) merits a short discussion. A Lie-algebravalued gauge field A µ has indices i, j for a representation R of an SU ( N ) group. For SU (2)the generators in the defining representation are the Pauli matrices τ a divided by 2 i , hence A µ = ( A µ ) ij = A aµ (cid:0) τ a i (cid:1) i j . The ansatz for the instanton can then be written as( A µ ) ij = ( σ µν ) ij x ν f ( x ) . (3.14)The indices µ, ν are Lorentz indices, but the indices i, j are SU (2) indices. Hence thematrix ( σ µν ) ij carries simultaneously spacetime indices and internal SU (2) indices. Thematrices σ µν are indeed proportional to τ a , σ µν = i ¯ η aµν τ a , as one may check for specificvalues of µ and ν , using( σ µν ) ij = (cid:110) ( σ µ ) iβ (cid:48) (¯ σ ν ) β (cid:48) j − ( σ ν ) iβ (cid:48) (¯ σ µ ) β (cid:48) j (cid:111) ( σ µ ) iβ (cid:48) = { (cid:126)τ , i } , (¯ σ µ ) β (cid:48) j = { (cid:126)τ , − i } . (3.15)The matrices η aµν and ¯ η aµν are actually invariant tensors of a particular SU (2) group. Thereare three groups SU (2): the gauge group SU (2) g and the rotation group SO (4) = SU (2) L × SU (2) R generated by η aµν and ¯ η aµν . The tensor η aµν is invariant under the combined SU (2) g gauge transformations acting on the index a generated by (cid:15) abc , and the SU (2) L Lorentztransformations generated by η bρσ . Indeed, under infinitesimal variations with parameter λ a we find, using (B.5), δη a = (cid:15) abc η b λ cg + λ cL [ η c , η a ] = 0 if λ ag = λ aL . (3.16)Furthermore, η aµν is separately invariant under the SU (2) R subgroup of the Lorentz groupgenerated by ¯ η bρσ ; this follows from [ η a , ¯ η b ] = 0. In fact, η a = L a + (cid:15) abc L bc and ¯ η a = − L a + (cid:15) abc L bc from which [ η a , ¯ η b ] = 0 easily follows. Spinor indices are raised and lowered by (cid:15) -tensors following the northwest-southeastconvention: v α (cid:48) = (cid:15) α (cid:48) β (cid:48) v β (cid:48) and v α = (cid:15) αβ v β . So (¯ σ µ ) β (cid:48) α = (cid:15) β (cid:48) δ (cid:48) (cid:15) αγ (¯ σ µ ) δ (cid:48) γ . There are variousdefinitions of these (cid:15) tensors in the literature; we define (cid:15) αβ = − (cid:15) α (cid:48) β (cid:48) . (3.17) The 4 × L a and L bc have entries ( L a ) µν = δ aµ δ ν and ( L bc ) µν = δ bµ δ cν − δ cµ δ bν . Theyform the defining representation of the Euclidean Lorentz algebra. (cid:15) αβ = (cid:15) αβ but also (cid:15) α (cid:48) β (cid:48) = (cid:15) α (cid:48) β (cid:48) because one needs two (cid:15) tensors toraise or lower both indices of an (cid:15) tensor. We fix the overall sign by (cid:15) αβ = (cid:15) ij where (cid:15) = 1.A crucial relation in the spinor formalism which we shall frequently use is¯ σ µ,α (cid:48) i = σ µ,i α (cid:48) , (3.18)where we recall that σ µ,i α (cid:48) = σ j β (cid:48) µ (cid:15) ji (cid:15) β (cid:48) α (cid:48) .Using 2-component spinor indices for vectors,(¯ σ µ ) α (cid:48) α A µ ≡ A α (cid:48) α and (¯ σ ν ) β (cid:48) j x ν ≡ x β (cid:48) j , (3.19)the ansatz for the instanton solution in (3.1) with spinor indices for A µ becomes(¯ σ µ ) α (cid:48) α ( A µ ) ij ≡ A α (cid:48) αij = (¯ σ µ ) α (cid:48) α ( σ µν ) ij x ν f ( x )= (cid:110) δ β (cid:48) α (cid:48) δ iα x β (cid:48) j − (cid:15) α (cid:48) β (cid:48) (cid:15) αj x iβ (cid:48) (cid:111) f ( x ) = (cid:8) δ iα x α (cid:48) j + (cid:15) αj x iα (cid:48) (cid:9) f ( x ) . (3.20)The trace over ( ij ) clearly vanishes, and this fixes the relative sign. We worked out thematrix (¯ σ µ ) α (cid:48) α ( σ µν ) ij using ¯ σ µα (cid:48) α σ iβ (cid:48) µ = 2 δ β (cid:48) α (cid:48) δ iα , (3.21)and ¯ σ µα (cid:48) α (¯ σ µ ) β (cid:48) j = ¯ σ µα (cid:48) α ( σ µ ) jβ (cid:48) = ¯ σ µα (cid:48) α σ kγ (cid:48) µ (cid:15) kj (cid:15) γ (cid:48) β (cid:48) = 2 (cid:15) α (cid:48) β (cid:48) (cid:15) αj . (3.22) Let us now come back to the construction of instanton solutions. Substituting the ansatzfor A µ in (3.1) into the definition of F µν yields with (3.11) F µν = ασ νρ ∂ µ ∂ ρ ln φ − ασ µρ ∂ ν ∂ ρ ln φ + α [ σ µρ , σ νσ ]( ∂ ρ ln φ )( ∂ σ ln φ )= ( ασ νρ ∂ µ ∂ ρ ln φ − µ ↔ ν ) + 2 α ( σ µσ ∂ ν ln φ ∂ σ ln φ − µ ↔ ν ) − α σ µν ( ∂ ln φ ) . (3.23)We want to solve the equation F µν = ∗ F µν . The dual of F µν can be written as an expressionwithout any (cid:15) tensor by using the identities for (cid:15) µνρσ σ στ and (cid:15) µνρσ σ ρσ in (3.7) and (3.8).One finds ∗ F µν = (cid:15) µνρσ F ρσ = α(cid:15) µνρσ σ σα ∂ ρ ∂ α ln φ + 2 α (cid:15) µνρσ σ ρβ ∂ σ ln φ ∂ β ln φ − α (cid:15) µνρσ σ ρσ ( ∂ ln φ ) = σ νρ ( α∂ ρ ∂ µ ln φ − α ∂ ρ ln φ ∂ µ ln φ ) − µ ↔ ν + σ µν ( α∂ ln φ ) . (3.24)21quating F µν to ∗ F µν yields two equations for φ , namely one for the terms with σ νρ andthe other for the terms with σ µν α∂ µ ∂ ρ ln φ − α ∂ µ ln φ ∂ ρ ln φ = α∂ µ ∂ ρ ln φ − α ∂ µ ln φ ∂ ρ ln φ , − α ( ∂ ln φ ) = α∂ ln φ . (3.25)The first equation is identically satisfied (for that reason we equated F µν to + ∗ F µν ), whilethe second equation can be rewritten as ∂ ln φ + 2 α ( ∂ ln φ ) = 0. For α = it simplifiesto ∂ φ/φ = 0. (Setting α = is not a restriction because rescaling φ → φ / α achieves thesame result).Setting φ = x yields for x (cid:54) = 0 a solution: ∂ φ/φ = x ∂ µ ( − x µ /x ) = 0. However, thisis also a solution at x = 0 because ∂ x − is proportional to a delta function (note that thedimensions match) and x δ ( x ) = 0 ∂ x = − π δ ( x ) . (3.26)To check the coefficient, we integrate over a small ball, which includes the point x = 0; weobtain then (cid:82) ∂ x d x = (cid:82) r d r dΩ µ ∂ µ x = (cid:82) r dΩ µ ( − x µ /x ) = − π . (The surface of asphere in 4 dimensions is 2 π ).We have thus found a selfdual solution A µ ( x ) = σ µν ∂ ν ln (cid:20) ρ ( x − a ) (cid:21) . (3.27)We have added unity to φ in order that A µ ( x ) vanishes for large | x | . A more generalsolution is given by A µ ( x ) = σ µν ∂ ν ln φ with φ = 1 + k (cid:88) i =1 ρ i ( x − a i ) , (3.28)which also solves ∂ φ/φ = 0. These are a class of k -instanton solutions, parameterized by5 k collective coordinates. In particular for k = 1 we find the one-instanton solution A sing µ ( x ) = σ µν ∂ ν ln (cid:20) ρ ( x − a ) (cid:21) , = − σ µν ρ ( x − a ) ν ( x − a ) (( x − a ) + ρ ) ( k = 1 , singular) . (3.29)This solution is clearly singular at x = a , but one can remove the singularity at x = a bya singular gauge transformation (which maps the singularity to x = ∞ ). To determinethis gauge transformation we first study the structure of the singularity. Near x = a thesingular solution becomes A sing µ ( x ) ≈ − σ µν ( x − a ) ν ( x − a ) , (3.30)22hich is a pure gauge field with U ( x − a ) in (A.9) U − ∂ µ U = − σ µν ( x − a ) ν ( x − a ) ; U ( x ) = x + ix k σ k √ x = i ¯ σ µ x µ / √ x . (3.31)Note that U is unitary, and U − equals − iσ µ x µ / √ x , which follows from the property σ ρ ¯ σ µ + σ µ ¯ σ ρ = 2 δ ρµ .From (3.29) and (3.30) it follows that we can write A sing µ as A sing µ ( x ) = ρ ( x − a ) + ρ U − ∂ µ U . (3.32)It is now clear that an opposite gauge tranformation removes the singularity at x = 0 A reg µ ( x ) = U ( ∂ µ + A sing µ ) U − = ∂ µ U U − (cid:18) − ρ ( x − a ) + ρ (cid:19) = ( U ∂ µ U − ) ( x − a ) ( x − a ) + ρ . (3.33)The expressions U − ∂ µ U and U ∂ µ U − are closely related; in fact, one finds by directevaluation U ∂ µ U − = − ¯ σ µν ( x − a ) ν ( x − a ) . (3.34)Thus the regular one-instanton solution is given by A reg µ = − ¯ σ µν ( x − a ) ν ( x − a ) + ρ ( k = 1 , regular) . (3.35)Of course, the singular and the regular solution are both selfdual, because self-duality isa gauge-invariant property, but the field strengths differ by a gauge transformation. Setting a = 0 for simplicity, one finds for the field strengths in the regular and singular gauge F reg µν = 2¯ σ µν ρ [ x + ρ ] ( k = 1 , regular) ,F sing µν = U − F reg µν U = − ix ρ σ ρ √ x σ µν ρ ( x + ρ ) ix σ ¯ σ σ √ x ( k = 1 , singular) . (3.36)It is clear that F reg µν is selfdual because ¯ σ µν is selfdual, but also F sing µν is selfdual as is clearfrom acting with (cid:15) µνρσ on ¯ σ µν . Using some further identities which follow from the results for [ γ µν , γ ρ ] and { γ µν , γ ρ } ¯ σ µ σ νρ = δ µν ¯ σ ρ − δ µρ ¯ σ ν − (cid:15) µνρσ ¯ σ σ ; ¯ σ µν ¯ σ ρ = δ νρ ¯ σ µ − δ µρ ¯ σ ν − (cid:15) µνρσ ¯ σ σ ,σ µν σ ρ = δ νρ σ µ − δ µρ σ ν + (cid:15) µνρσ σ σ ; σ ρ ¯ σ µν = δ ρµ σ ν − δ ρν σ µ + (cid:15) ρµνσ σ σ , one finds for the k = 1 singular solution F sing µν = 2 ρ ( x + ρ ) (cid:16) − x µ x ρ x σ ρν + 2 x ν x ρ x σ ρµ + σ µν (cid:17) In this form the selfduality is no longer manifest. S = − g (cid:90) tr F µν F µν d x = − g (cid:90) tr F µν ∗ F µν d x = 8 π g . (3.37)The same result is obtained by direct evaluation of this integral.The anti-instanton (the solution with k = −
1) is closely related to the instanton solution.Recall that we derived the instanton solution by making the ansatz A µ = ασ µν ∂ ν ln φ ,evaluating F µν and ∗ F µν in terms of σ µν matrices, and then setting F µν = ∗ F µν . For theanti-instanton solution we make the ansatz A µ = β ¯ σ µν ∂ ν ln φ . The expression for F µν isunchanged (except that A µ contains ¯ σ µν instead of σ µν ), but the ¯ σ µν are selfdual insteadof anti-selfdual, hence the expression for (cid:15) µνρσ ¯ σ στ has opposite signs from (cid:15) µνρσ σ στ . Theequation with ∂ µ ∂ ρ ln φ again cancels if F µν = − ∗ F µν , which leads to opposite windingnumber ( k = − ∂ ln φ + 2 β ( ∂ ln φ ) = 0, hence β = andagain φ = 1 + (cid:80) Ni =1 ρ i ( x − a i ) . This yields for the singular-gauge anti-instanton solution A sing µ = − ¯ σ µν ρ ( x − a ) ν ( x − a ) [( x − a ) + ρ ] , ( k = − , singular) (3.38)Setting again temporarily a = 0, we find near x = 0 A sing µ ≈ − ¯ σ µν x ν /x = U ∂ µ U − , (3.39)with the same U = i ¯ σ µ x µ / √ x as before. Similarly as for the instanton, we have A sing µ = U ∂ µ U − ρ x + ρ A reg µ = U − ( ∂ µ + A sing µ ) U = ∂ µ U − U (cid:18) ρ x + ρ − (cid:19) = U − ∂ µ U (cid:18) x x + ρ (cid:19) . (3.40)Using the expression for U − ∂ µ U in (3.31) one finds A reg µ = − σ µν ( x − a ) ν ( x − a ) + ρ ( k = − , regular) (3.41)The curvatures for the anti-instanton solution are obtained by interchanging σ µν and ¯ σ µν in the instanton solution F reg µν = 2 σ µν ρ [( x − a ) + ρ ] ( k = − , regular) . (3.42)So, the only difference between the instanton and anti-instanton solutions is the exchangebetween σ µν and ¯ σ µν in F µν and A µ . For the instanton solution, F reg µν and A reg µ depend24n ¯ σ µν , but A sing µ depends on σ µν , and F sing µν also depends on σ µν (setting a = 0 again fornotational simplicity), F sing µν = U F reg µν U − = ix ρ ¯ σ ρ √ x σ µν ρ ( x + ρ ) − ix σ σ σ √ x ( k = − , singular) . (3.43)If one evaluates the product of the σ matrices as in footnote 11, one finds an expressionfor F sing µν in which the anti-selfduality is no longer manifest. We found in section 2 one-instanton solutions ( k = 1) in SU ( N ) with 4 N parameters.The question arises whether these are all the solutions. To find this out, one can considersmall deformations of the solution, A µ + δA µ , and study when they preserve selfduality.Expanding to first order in the deformation, and using that the variation of a curvature isthe covariant derivative of the variation of the gauge field, this leads to the condition D µ δA ν − D ν δA µ = ∗ ( D µ δA ν − D ν δA µ ) , (4.1)where the covariant derivative depends only on the classical solution but not on δA µ . Inaddition we require that the new solution is not related to the old one by a gauge trans-formation. This can be achieved by requiring that the small deformations are orthogonalto any small gauge transformation D µ Λ, for any function Λ, i.e. (cid:90) d x tr { ( D µ Λ) δA µ } = 0 . (4.2)This certainly rules out deformations of the form δA µ = D µ Λ. After partial integrationthe orthogonality requirement leads to the usual gauge condition in the background fieldformalism D µ δA µ = 0 . (4.3)At this point the reader may start feeling uneasy because the conditions (4.1) and (4.2)may seem too strong. First of all, the deformation should be a solution but need not be(anti-) selfdual. Furthermore, the field equation for the fluctuations consists of the sumof a classical piece and a piece from the gauge fixing term, so that, requiring each part tovanish separately may seem too restrictive. However, one can prove the following generalresult [46]. Arbitrary solutions of the fluctuations around an (anti-) instanton which aresquare-integrable so that they do not change the winding number, are themselves also(anti-) selfdual and transversal. To prove this property, note that the field equations25or the fluctuations read D µ F µν ( A + δA ) + D ν ( D µ δA µ ) = 0. The second term comesfrom the gauge-fixing term. Taking the D ν derivative, the first term vanishes while thesecond term yields D ( D µ δA µ ) = 0, hence D µ δA µ on-shell. The terms in the classicalaction which are quadractic in the fluctuations can be written as − ( f µν − ∗ f µν ) where f µν = D µ δA ν − D ν δA µ . The minimum of the action yields a solution, hence f µν = ∗ f µν on-shell. Thus imposing (4.1) and (4.3) is not too restrictive.The requirement that δA µ be square integrable is due to the fact that the inner productof zero modes δA µ will later give us the metric or moduli space, which in turn will giveus the integration measure of the moduli space. Also, for the index theorem which will beused to determine the number of zero modes, one needs the L norm for fluctuations. It isremarkable that the zero modes which satisfy the differential equations in (4.1) and (4.3)are all square integrable.In references [46, 36] the solutions of (4.1) subject to the condition (4.3) were studiedusing the Atiyah-Singer index theorem. Index theory turns out to be a useful tool whencounting the number of solutions to a certain linear differential equation of the form ˆ DT =0, where ˆ D is some differential operator and T is a tensor. We will elaborate on this inthe next subsection and also when studying fermionic collective coordinates. The ultimateresult of [36] is that there are 4 N k solutions, leading indeed to 4 N collective coordinatesfor k = 1 [46]. An assumption required to apply index theorems is that the space has tobe compact. One must therefore compactify Euclidean space to a four-sphere S , as wasalready discussed in footnote 4. In this section we will make more precise statements about the number of solutions to theselfduality equations by relating it to the index of the Dirac operator. The problem isto determine the number of solutions to the (anti-)selfduality equations with topologicalcharge k . For definiteness we consider anti-instantons, so we look for deformations whichsatisfy an anti-selfduality equation.As explained in the last subsection, we study deformations of a given classical solution A cl µ + δA µ . Let us define φ µ ≡ δA µ and f µν ≡ D µ φ ν − D ν φ µ . The covariant derivative herecontains only A cl µ . The constraints can then be written as¯ σ µν D µ φ ν = 0 ; D µ φ µ = 0 , (4.4)which are 3 + 1 relations. Indeed, more explicitly, (¯ σ µν ) αα (cid:48) D µ φ ν are 3 Lie-algebra valuedexpressions because α, α (cid:48) = 1 , σ µν = 0. To prove the first relation, multiply by ¯ σ ρσ and take the trace. Since the trace of [¯ σ ρσ , ¯ σ µν ] vanishes, while { ¯ σ ρσ , ¯ σ µν } = 2( δ µσ δ ρν − νσ δ ρµ ) − (cid:15) ρσµν , one finds D σ φ ρ − D ρ φ σ − (cid:15) µνρσ D µ φ ν = 0, which is the anti-selfdualitycondition (3 relations). Both equations can be written as one simple equation as follows:¯ σ µ σ ν D µ φ ν = 0 , (4.5)because ¯ σ µ σ ν = δ µν + ¯ σ µν , and the spinor structures of δ µν and ¯ σ µν are independent.Introducing two-component spinor notation with¯/ D = ¯ σ µ,α (cid:48) β D µ = ¯ D α (cid:48) β ; σ αβ (cid:48) ν φ ν = Φ αβ (cid:48) , (4.6)the deformations of an anti-instanton can be written as follows¯/ D Φ = ¯ D α (cid:48) β Φ βγ (cid:48) = 0 . (4.7)Note that Φ βγ (cid:48) is in the adjoint representation, so (4.7) stands for ∂ α (cid:48) β Φ βγ (cid:48) + [ A α (cid:48) β , Φ βγ (cid:48) ]= 0. Using the explicit representation of the matrices σ µ in (B.9), we can represent thequaternion Φ by Φ = (cid:32) a b ∗ b − a ∗ (cid:33) , (4.8)with a and b complex adjoint-valued functions. Then (4.7) reduces to two spinor equations,one for λ = (cid:16) ab (cid:17) ; ¯/ Dλ = 0 , (4.9)and one for iσ λ ∗ = (cid:32) b ∗ − a ∗ (cid:33) . Conversely, for each spinor solution λ to the Dirac equa-tion, one may show that also iσ λ ∗ is a solution. (Use (¯ σ µ ) ∗ = − σ ¯ σ µ σ ). Indeed, if λ yields a deformation ( δA , δA , δA , δA ), then iσ λ ∗ corresponds to the deformation( δA (cid:48) , δA (cid:48) , δA (cid:48) , δA (cid:48) ) with δA (cid:48) = − δA , δA (cid:48) = δA , δA (cid:48) = δA and δA (cid:48) = − δA . They arenot related by a Lorentz transformation because the coordinates x µ are not transformed.Thus given λ , we obtain two linearly independent deformations of the (anti-) instanton.As we already stressed, the spinors λ are in the adjoint representation. We shall discussother representations later.Given a solution λ of the spinor equation, one can still construct two other solutions ofthe deformation of the anti-instanton, which differ by a factor i Φ (1) = (cid:32) a b ∗ b − a ∗ (cid:33) , Φ (2) = (cid:32) ia − ib ∗ ib ia ∗ (cid:33) . (4.10)The reason we do not count iλ as a different solution for the spinors but treat Φ (1) andΦ (2) as independent has to do with reality properties: δA aµ should be real, and Φ (1) andΦ (2) yield different variations δA µ . Namely, a = φ + iφ and b = φ + iφ , soΦ (1) : δA = φ , δA = φ , δA = φ , δA = φ , (2) : δA = φ , δA = − φ , δA = − φ , δA = φ . (4.11)It may seem miraculous that we find a second solution without any hard work, but closerinspection reveals that no miracle is at work: under the substitutions δA → δA , δA →− δA , δA → δA , δA → − δA , one of the anti-selfduality equations is exchanged with thegauge condition, and the other two duality equations get interchanged. Also for solitonsthis way of counting zero modes is encountered: for example for vortices one complexfermion zero mode corresponds to two real bosonic zero modes [47].In fact, because Φ (2) = Φ (1) iσ , one might wonder whether Φ (3) = Φ (1) ( − iσ ) andΦ (4) = Φ (1) ( − iσ ) yield further solutions. One obtainsΦ (3) = (cid:32) − ib ∗ − iaia ∗ − ib (cid:33) ; Φ (4) = (cid:32) b ∗ − a − a ∗ − b (cid:33) (4.12)which are just the Φ (1) constructed from σ λ ∗ and iσ λ ∗ . So there are no further inde-pendent solutions [46]. Therefore, the number of solutions for Φ is twice the number ofsolutions for a single two-component adjoint spinor. So, the problem of counting the num-ber of bosonic collective coordinates is now translated to the computation of the Diracindex, which we discuss next. Both motivated by the counting of bosonic collective coordinates, as discussed in the lastsubsection, and by the interest of coupling Yang-Mills theory to fermions, we study theDirac equation in the background of an anti-instanton. We start with a massless four-component complex (Dirac) fermion ψ , in an arbitrary representation (adjoint, fundamen-tal, etc) of an arbitrary gauge group γ µ D µ ψ = / Dψ = 0 . (4.13)We recall that a Dirac spinor can be decomposed into its chiral and anti-chiral components ψ = (cid:32) λ α ¯ χ α (cid:48) (cid:33) ; λ ≡ (cid:0) γ (cid:1) ψ , ¯ χ ≡ (cid:0) − γ (cid:1) ψ . (4.14)We use the Euclidean representation for the Clifford algebra discussed before γ µ = (cid:32) − iσ µ αβ (cid:48) i ¯ σ µα (cid:48) β (cid:33) , γ = γ γ γ γ = (cid:32) − (cid:33) . (4.15)In Euclidean space the Lorentz group decomposes according to SO (4) = SU (2) × SU (2).The spinor indices α and α (cid:48) correspond to the doublet representations of these two SU (2)28actors. As opposed to the case of Minkowski space, λ α and ¯ χ (cid:48) α are not in complex-conjugaterepresentations. The Dirac equation then becomes (cid:54) ¯ Dλ = 0 , (cid:54) D ¯ χ = 0 , (4.16)where (cid:54) D and (cid:54) ¯ D are two-by-two matrixes, see (4.6), and λ and ¯ χ are independent complextwo-component spinors. We now show that in the presence of an anti-instanton, (4.16)has zero modes for λ , but not for ¯ χ . Conversely, in the background of an instanton, (cid:54) D has zero modes, but (cid:54) ¯ D has not. A zero mode is by definition a solution of the linearizedfield equations for the quantum fluctuations which is normalizable . The fermionic fieldsare treated as quantum fields (there are no background fermionic fields), so normalizablesolutions of (4.16) are zero modes.The argument goes as follows. Given a zero mode ¯ χ for / D , it also satisfies (cid:54) ¯ D (cid:54) D ¯ χ = 0.In other words, ker / D ⊂ ker (cid:8) (cid:54) ¯ D (cid:54) D (cid:9) where ker denotes the kernel. Next we evaluate (cid:54) ¯ D (cid:54) D = ¯ σ µ σ ν D µ D ν = D + 12 ¯ σ µν F µν , (4.17)where we have used ¯ σ µ σ ν + ¯ σ ν σ µ = 2 δ µν , and ¯ σ µν was defined in (3.6). But notice that theanti-instanton field strength is anti-selfdual whereas the tensor ¯ σ µν is selfdual, so the secondterm vanishes. From this it follows that ¯ χ satisfies D ¯ χ = 0. Now we can multiply D ¯ χ with its conjugate ¯ χ ∗ and integrate to get, after partial integration and assuming that thefields go to zero at infinity , (cid:82) d x | D µ ¯ χ | = 0. From this it follows that ¯ χ is covariantlyconstant, D µ ¯ χ = 0, and so F µν ¯ χ = 0. Since F µν ¯ χ = F aµν T a ¯ χ , with T a the generators ofthe gauge group SU (2) in a representation R , we conclude that F aµν ( x ) T a ¯ χ ( x ) must vanishat all points x . Since F aµν is proportional to η aµν (or ¯ η aµν ), and η aµν η bµν is proportional to δ ab , we find that T a ¯ χ ( x ) vanishes for all a and all x . Then D µ ¯ χ = 0 reduces to ∂ µ ¯ χ = 0,and this implies that ¯ χ = 0. We conclude that / D ¯ χ has no square-integrable solutions.Stated differently, − D is a positive definite operator and has no zero modes. Note thatthis result is independent of the representation of the fermion.For the λ -equation, we have (cid:54) D (cid:54) ¯ Dλ = 0, i.e. ker (cid:54) ¯ D ⊂ ker (cid:8) (cid:54) D (cid:54) ¯ D (cid:9) , and we obtain (cid:54) D (cid:54) ¯ D = D + 12 σ µν F µν . (4.18)This time the second term does not vanish in the presence of an anti-instanton, so zeromodes cannot be ruled out. In fact, there do exist fermionic zero modes, because weshall construct them. Knowing that (cid:54) D has no zero modes, one easily concludes thatker (cid:54) ¯ D = ker (cid:8) (cid:54) D (cid:54) ¯ D (cid:9) and ker (cid:54) D = ker (cid:8) (cid:54) ¯ D (cid:54) D (cid:9) = 0. Normalizability of zero modes requires that ¯ χ tends to zero faster than 1 /r (usually like 1 /r orsometimes 1 /r ). Then the boundary term with ¯ χ ∗ D µ ¯ χ indeed vanishes. Dλ = im ¯ χ and / D ¯ χ = − imλ , anditeration yields (cid:54) ¯ D (cid:54) D ¯ χ = m ¯ χ . The crucial observation is that m is positive, while (cid:54) ¯ D (cid:54) D isnegative definite. Hence, no zero modes exist for massive spinors.Now we can count the number of solutions using index theorems. The index of the Diracoperator is defined as Ind (cid:54) ¯ D = dim ker (cid:54) D (cid:54) ¯ D − dim ker (cid:54) ¯ D (cid:54) D . (4.19)This index will give us the number of zero modes, since the second term is zero and sinceany renormalizable solution of (cid:54) D (cid:54) ¯ Dλ = 0 satisfies (cid:54) ¯ Dλ = 0 as we have shown. There areseveral ways to compute its value. We begin by writing the index as followsInd (cid:54) ¯ D = lim M → T r (cid:26) M − (cid:54) D (cid:54) ¯ D + M − M − (cid:54) ¯ D (cid:54) D + M (cid:27) , (4.20)where M is an arbitrary parameter. The trace Tr stands for a sum over group indicesand spinor indices, and includes an integration over space-time. We shall discuss that thisexpression (before taking the limit) is independent of M . This implies that the operators (cid:54) D (cid:54) ¯ D and (cid:54) ¯ D (cid:54) D not only have the same spectrum but also the same density of states fornon-zero eigenvalues . That they have the same non-zero eigenvalues is clear: if ψ is aneigenfunction of (cid:54) ¯ D (cid:54) D , then (cid:54) Dψ is an eigenfunction of (cid:54) D (cid:54) ¯ D with the same nonvanishingeigenvalue and (cid:54) Dψ does not vanish. Conversely, if ψ is an eigenfunction of (cid:54) D (cid:54) ¯ D withnonzero eigenvalue, then (cid:54) ¯ Dψ does not vanish and is an eigenfunction of (cid:54) ¯ D (cid:54) D with the samenonvanishing eigenvalue.To show that (4.20) is independent of M , we rewrite the index in terms of four-dimensional Dirac matrices, I ( M ) ≡ Ind (cid:54) ¯ D = T r (cid:26) M − (cid:54) D + M γ (cid:27) , (4.21)where now / D = D µ γ µ . / D × = (cid:32) D × ¯/ D × (cid:33) (4.22)We rewrote the trace of the two terms in (4.20) over a two-dimensional spinor space as thetrace of one term over a four-dimensional spinor space. It has been argued that indepen- One can also (as is customary in the literature) place the system in a large box to discretize thespectrum, and let the boundary conditions for the eigenfunctions of (cid:54) D (cid:54) ¯ D determine the boundary conditionsfor the eigenfunctions of (cid:54) ¯ D (cid:54) D , and vice-versa, such that the non-zero eigenvalues are the same. Such atreatment for the kink has been worked out in detail in [47]. However, in the limit of infinite volume, thedensities of states can become different, as we shall discuss. M follows by taking the M -derivative (see [46]), ∂∂M I ( M ) = − T r (cid:26) (cid:54) D ( − (cid:54) D + M ) γ (cid:27) . (4.23)Using that γ anticommutes with (cid:54) D and that the trace is cyclic, we find − T r (cid:54) D (cid:54) Dγ A = T r (cid:54) Dγ (cid:54) DA = T r (cid:54) D (cid:54) Dγ A , (4.24)where A = ( − (cid:54) D + M ) . Hence T r ( (cid:54) D γ /A ) would seem to vanish and this would provethat I ( M ) is independent of M . The problem with this proof is that one can give acounter example: one can repeat all the steps for the supersymmetric kink, and this wouldthen imply that the densities for chiral and anti-chiral fermion modes are equal. However,one can directly calculate these densities for the supersymmetric kink, and one then findsthat they are different [48] ∆ ρ ( k ) = − mk + m , (4.25)where m is the mass of the fluctuating fields far away from the kink. Applied to the caseof instantons, the situation was considered in [46]. In [49, 50, 51] it was noted that theproof of [46] was incomplete. Cyclicity of the trace (on which the proof in [46] that ∆ ρ ( k )vanishes is based), breaks down due to the presence of massless fluctuating fields . Onecan directly compute I ( M ), using a more detailed index theorem [50, 51], and then findsthat the densities of chiral and antichiral fermionic modes in an instanton background areequal, ∆ ρ ( k ) = 0 for instantons . (4.26)Given that the density of states of the operator (cid:54) D (cid:54) ¯ D in (4.20) is the same as the densityof states of the operator (cid:54) ¯ D (cid:54) D , there is a pairwise cancellation in (4.20) coming from the sumover eigenstates with non-zero eigenvalues, both for the discrete and continuous spectrum.So the only contribution is coming from the zero modes, for which the first term in (4.20)simply gives one for each zero mode, and the second term vanishes because there are nozero modes. The result is then clearly an integer, namely dim (cid:8) ker (cid:54) ¯ D (cid:9) . Since I ( M ) isindependent of M , one can evaluate it in the large M limit instead of the small M limit.The calculation is then identical to the calculation of the chiral anomaly, which we nowreview. One would expect that at the regularized level the trace is cyclic but one may expect that one shouldalso regularize infrared aspects of the problem. Consider for example quantum mechanics for a harmonicoscillator with mass term m q . Define a = (cid:112) m q + ip/ √ m and a † = (cid:112) m q − ip/ √ m . For m tending tozero, the vacuum is annihilated by p + O ( m ) but the vacuum becomes non-normalizable when m vanishes.Still, at finite m , tr [ p, q ] = 0. γ . It can be writtenas Ind (cid:54) ¯ D = tr (cid:90) d x < x | M − / D ¯/ D + M − M − ¯/ D / D + M | x > , (4.27)where tr denotes the trace over group indices and spinor indices. Because γ = diag(+1 , +1 , − , − | x > of the position operator. The operators D µ depend on the opera-tors ˆ x µ and the operators ˆ p µ . When ˆ x reaches | x > , it becomes a c -number x . Similarlyˆ p µ | x > = − ¯ hi ∂∂x µ | x > . The latter statement follows by contracting with a complete set ofmomentum eigenstates, using that < k | ˆ p µ = ¯ hk µ < k | because ˆ p µ is hermitian < k | ˆ p µ | x > = ¯ hk µ < k | x > = ¯ hk µ e − ikx (2 π ) = − ¯ hi ∂∂x µ e − ikx (2 π ) = − ¯ hi ∂∂x µ < k | x > = < k | − ¯ hi ∂∂x µ | x > . (4.28)So, from now on we will replace the operators D µ (ˆ x, ˆ p x ) by D µ ( x, − ¯ hi ∂∂x ). These ∂∂x act onthe x in | x > and do not act on | k > .Let us now insert a complete set of eigenstates of ¯/ D / D and / D ¯/ D , respectively. The indexbecomes thenInd ¯/ D = tr (cid:88) m,n (cid:90) d x < x | n L >< n L | O L | m L >< m L | x > − same with L ↔ R , (4.29)where O L is the first operator in (4.27) and O R the second. As we already discussed theeigenfunctions < x | n L > = ϕ ( L ) n ( x ) and < x | n R > = ϕ ( R ) n ( x ) have the same nonvanishingeigenvalues λ n and the same densities.So the eigenfunctions with nonzero eigenvalues do not contribute to the index. (Notethat it does not make sense to look for eigenfunctions of / D or ¯/ D because these operatorschange the helicity of the spinors). There are in general a finite number of zero modes inthe L sector but none in the R sector. HenceInd ¯/ D = (cid:90) d x (cid:32)(cid:88) n ϕ ( L ) n ( x ) ϕ ( L ) m ( x ) ∗ − (cid:88) m ϕ ( R ) n ( x ) ϕ ( R ) m ( x ) ∗ (cid:33) M δ mn λ n + M + (cid:88) α (cid:90) d xϕ ( L ) α ( x ) ϕ ( L ) α ( x ) ∗ = n ( L ) , (4.30)where ϕ ( L ) α ( x ) are the (square-integrable) zero modes, and n ( L ) is the number of these. Asum over spinor indices is taken in (4.30). 32o actually compute the index (namely, to compute the integer n ( L ) ), we use momentumeigenstates instead of eigenfunctions of / D ¯/ D and ¯/ D / D Ind ¯/ D = (cid:90) d x (cid:90) d k (cid:90) d k (cid:48) tr < x | k (cid:48) >< k (cid:48) | M − / D + M γ | k >< k | x > , (4.31)where we recall that γ = (cid:32) − (cid:33) and / D = (cid:32) − i / Di ¯/ D (cid:33) .As we have discussed, the operator D µ = ∂∂x µ + [ A µ ( x ) , · ] acts on the coordinates x in | x > but not on the k in | k >< k | , and the trace tr sums over the group indices and thespinor indices of γ µ in / D = γ µ D µ and γ . Using < k | x > = e − ikx / (2 π ) and pulling theseplane waves to the left, the derivatives ∂ µ act on the c -numbers x in e − ikx and are replacedby ∂ µ − ik µ . The matrix element < k (cid:48) | M ( − / D + M ) − γ | k > is equal to < k (cid:48) | k > timesthe operator [ M / ( − / D + M )] γ and < k (cid:48) | k > = δ ( k − k (cid:48) ). When the plane wave e − ikx hasbeen pulled all the way to the left, the plane waves e ik (cid:48) x and e − ikx in < x | k (cid:48) > = e ik (cid:48) x / (2 π ) and < k | x > = e − ikx / (2 π ) cancel each other, and one is left withInd ¯/ D = (cid:90) d x (cid:90) d k (2 π ) tr (cid:26) M − ( − i / k + / D ) + M γ (cid:27) . (4.32)The denominator can be written as 1( k + M ) − ( − ik · D + D µ D µ + γ µ γ ν F µν ) , (4.33)and we can exhibit the M dependence by rescaling k µ = M κ µ , yieldingInd ¯/ D = (cid:90) d x M (cid:90) d κ (2 π ) tr κ + 1) − (cid:16) − iκ µ D µ M + D µ D µ M + γ µ γ ν F µν M (cid:17) γ . (4.34)Expanding the denominator, only terms due to expanding two, three or four times cancontribute in the limit M → ∞ , but only the terms with at least four Dirac matricescan contribute to the trace due to the matrix γ . Thus we only need retain the square of γ µ γ ν F µν , and the index becomesInd ¯/ D = (cid:90) d x (cid:90) d κ (2 π ) κ + 1) tr (cid:0) F µν γ µν F ρσ γ ρσ γ (cid:1) = (cid:90) d x π (2 π ) (cid:90) ∞ r d r ( r + 1) (tr T a T b )(tr γ µν γ ρσ γ ) F aµν F bρσ = (cid:90) d x π (tr T a T b )( (cid:15) µνρσ F aµν F bρσ ) , (4.35)33here we used that (cid:82) dΩ µ = 2 π and (cid:82) ∞ r d r ( r +1) = . Note that both a trace over groupindices and a trace over spinor indices has been taken. The result for the index is twicethe product of the winding number in (2.6) and a group theory factorInd ¯/ D = 2 (cid:18) π (cid:90) d xF aµν ∗ F bµν (cid:19) tr T a T b . (4.36)For a representation R of SU ( N ) for the fermions, we define tr T Ra T Rb = − δ ab T ( R ). Bydefinition one has T ( R ) = for the fundamental representation, and then T ( R ) = N forthe adjoint representation . Hence, finally,Ind ¯/ D = | k | for the fundamental representation,= 2 N | k | for the adjoint representation . (4.37)(For an anti-instanton, k is negative. The factor 2 corresponds to our earlier observationthat iσ λ ∗ is also a zero mode if λ is a zero mode.) Furthermore, as shown in the lastsubsection, an (anti-) instanton in SU ( N ) has twice as many bosonic collective coordinatesas there are fermionic zero modes in the adjoint representation. This proves that there are4 N k bosonic collective coordinates for an instanton with winding number k and gaugegroup SU ( N ). In two later sections we will show how to set up and do (one-loop) perturbation theoryaround an (anti-) instanton. This will require the reduction of the path integral measureover instanton field configurations to an integral over the moduli space of collective coor-dinates. In order to achieve this we need to know the explicit form of the bosonic andfermionic zero modes. This is the content of this section. We follow closely [52]. To compute T ( R ) for the adjoint representation, write the carrier space for the adjoint representationof SU ( N ) as u i ¯ v j − N δ ij ( u k ¯ v k ). Then, for i (cid:54) = j , T adj a u i ¯ v j = ( T ( f ) a ) ii (cid:48) u i (cid:48) ¯ v j + ( T f ∗ a ) jj (cid:48) u i ¯ v j (cid:48) . For a diagonalgenerator A of the fundamental representation of SU ( N ) with entries ( iα , . . . , iα N ) with real α j one has Au i = iα i u i and Au i ¯ v j = ( iα i − iα j ) u i ¯ v j , so A ( u i ¯ v j − N δ ij u k ¯ v k ) = ( iα i − iα j )( u i ¯ v j − N δ ij u k ¯ v k ) and (cid:80) α i = 0. Hence tr A = − (cid:80) Ni =1 ( α i ) for the fundamental representation, but tr A = − (cid:80) i,j ( α i − α j ) for the adjoint representation. The latter sum can also be written as N (cid:88) i,j =1 ( α i − α j ) = (cid:16)(cid:88) α i (cid:17) N − (cid:16)(cid:88) α i (cid:17) (cid:16)(cid:88) α j (cid:17) + (cid:16)(cid:88) α j (cid:17) N = 2 (cid:16)(cid:88) α i (cid:17) N . So T ( R adj ) = 2 N T ( R f ). .1 Bosonic zero modes and their normalization In order to construct the bosonic zero modes and discuss perturbation theory, we firstdecompose the fields into a background part and quantum fields A µ = A cl µ ( γ ) + A qu µ . (5.1)Here γ i denote a set of collective coordinates, and, for gauge group SU ( N ), i = 1 , . . . , N k .Before we make the expansion of the action, we should first fix the gauge and introduceghosts, c , and anti-ghosts, b . We choose the background gauge condition D cl µ A qu µ = 0 . (5.2)The gauge-fixing term is then L fix = − g tr( D µ A qu µ ) and the ghost action is L ghost = − b a ( D µ ( A cl µ ) D µ ( A cl µ + A qu µ ) c ) a . The action, expanded through quadratic order in the quan-tum fields, is of the form S = 8 π g | k | + 1 g tr (cid:90) d x (cid:8) A qu µ M µν A qu ν + 2 b M gh c (cid:9) , (5.3)with M gh = D and M µν = (cid:0) D δ µν − D ν D µ + F µν (cid:1) + D µ D ν ≡ M (1) µν + M (2) µν , = D δ µν + 2 F µν , (5.4)where we have dropped the subscript cl. Here, M (1) stands for the quadratic operatorcoming from the classical action, and M (2) is due to the gauge fixing term . (Recall that F µν acts on A qu ν as [ F µν , A qu ν ]). In an expansion as in (5.3), one encounters zero modes (i.e.normalizable eigenfunctions of the operator M µν with zero eigenvalues). They are of theform Z ( i ) µ ≡ ∂A cl µ ∂γ i + D cl µ Λ i , (5.5)where the gauge parameter Λ i is chosen to keep Z µ in the background gauge, so that D cl µ Z ( i ) µ = 0 . (5.6)The first term in (5.5) is a solution of M (1) (i.e. an eigenfunction with zero eigenvalue),as follows from taking the derivative with respect to γ i of the field equation. Namely, δS cl /δA cl µ = 0 for all γ i , so0 = ∂∂γ i δS cl δA cl µ ( x ) = (cid:90) δ S cl δA cl ν ( y ) δA cl µ ( x ) ∂ γ i A cl ν ( y )d y . (5.7) To arrive at this expression for M (1) µν , use that F µν = F cl µν + ( D cl µ A qu ν − D cl ν A qu µ ) + [ A qu µ , A qu ν ] and notethat − g tr2 F cl µν [ A qu µ , A qu ν ] = g tr A qu µ [ F cl µν , A qu ν ]. D µ Λ is also a solution of M (1) , since it is a pure gauge transformation. Thesum of the two terms is also a solution of M (2) , because Λ is chosen such that Z µ is inthe background gauge. As we shall show, the solutions in (5.5) are normalizable, hencethey are zero modes. Due to these zero modes, we cannot integrate over all quantumfluctuations, since the corresponding determinants would vanish and yield divergences inthe path integral. They must therefore be extracted from the quantum fluctuations, in away we will describe in a more general setting in the next subsection. It will turn out tobe important to compute the matrix of inner products U ij ≡ (cid:104) Z ( i ) | Z ( j ) (cid:105) ≡ − g (cid:90) d x tr (cid:8) Z ( i ) µ Z µ ( j ) (cid:9) = 1 g (cid:90) Z ( i ) aµ Z ( j ) aµ d x . (5.8)We put a factor g in front of the usual L inner product because the metric U ij willbe used to construct a measure (det U ij ) / for the zero modes, and this measure is alsoneeded if one considers the quantum mechanics of zero modes γ i ( t ). The action for thesetime-dependent γ i ( t ) is U ij γ i ˙ γ j with the same prefactor g as in the Yang-Mills gaugeaction.We now evaluate this matrix for the anti-instanton. For the four translational zeromodes, one can easily keep the zero mode in the background gauge by choosing Λ i = A cl ν .Indeed, Z ( ν ) µ = ∂A cl µ ∂x ν + D µ A cl ν = − ∂ ν A cl µ + D µ A cl ν = F cl µν , (5.9)which satisfies the background gauge condition. The norms of these zero modes are U µν = 8 π | k | g δ µν = S cl δ µν . (5.10)As indicated, this result actually holds for any k , and arbitrary gauge group.Next we consider the dilatational zero mode corresponding to ρ and limit ourselves to k = −
1. Taking the derivative with respect to ρ leaves the zero mode in the backgroundgauge, so we can set Λ ρ = 0. In the singular gauge of (3.38) we have Z ( ρ ) µ = − ρ ¯ σ µν x ν ( x + ρ ) . (5.11)(To show that (5.6) is satisfied, note that ( ∂/∂x µ ) Z ( ρ ) µ = 0 since ¯ σ µν is antisymmetric, while[ A cl µ , Z ( ρ ) µ ] = 0 since both involve ¯ σ µν x ν ). Using (B.18) and (B.21), one easily computesthat U ρρ = 16 π g = 2 S cl . (5.12) This is also easy to prove by direct calculation: ( D δ µν − D ν D µ + F µν ) D ν Λ is equal to D ν [ D ν , D µ ]Λ + F µν D ν Λ, and this vanishes since [ D ν , D µ ] = F νµ and D ν F νµ = 0. More generally, δS cl /δA cl ν ∼ D cl µ F cl µν is gauge-covariant, hence D cl µ F cl µν ( A ρ + D ρ Λ) − D cl µ F cl µν ( A ρ ) = [ D cl µ F cl µν , Λ] which vanishes on-shell (fieldequations transform into field equations). Hence δ S cl δA cl µ δA cl ρ ( D ρ Λ) = M (1) µν D ν Λ vanishes.
36n regular gauge one finds Z ( ρ ) µ = ρσ µν x ν ( x + ρ ) which has clearly the same norm. This resultcan also be derived from ∂∂ρ A reg µ ( k = −
1) = ∂∂ρ U − ( ∂ µ + A sing µ ( k = − U and the identity U − ¯ σ µν x ν U = − σ µν x ν .The gauge-orientation zero modes can be obtained from (2.18). By expanding U ( θ ) =exp( θ a T a ) infinitesimally in (2.18) we get to lowest order in θ (the case of general θ will bediscussed shortly) ∂A µ ∂θ a = [ A µ , T a ] , (5.13)which is not in the background gauge (the matrices T a are in the fundamental represen-tation). To satisfy (5.6) we have to add appropriate gauge transformations, which differfor different generators of SU ( N ). First, for the SU (2) subgroup corresponding to theinstanton embedding, we add, for the singular gauge,Λ a = − ρ x + ρ T a , (5.14)and find that Z µ ( a ) = D µ (cid:20) x x + ρ T a (cid:21) . (5.15)(using ∂ µ x x + ρ = − ∂ µ ρ x + ρ ). One can now show, using (B.5), that the zero mode (5.15) isin the background gauge, and its norm reads U ab = 4 π g ρ δ ab = δ ab ρ S cl . (5.16) A few details may be helpful. One finds for this zero mode in the singular gauge, using (3.38) and(B.15), Z ( a ) µ = 2 x µ ρ ( x + ρ ) − T a + 2 η bµν (cid:15) bac T c x ν ρ / ( x + ρ ) . It is covariantly transversal: ∂ µ acting on the first term plus the commutator of A sing µ with the second termvanishes upon using (B.5). (The commutator of the first term with A sing µ is proportional to (¯ σ µν x ν ) x µ andvanishes). The norm is due to integrating the sum of the square of the first term and the second term,using (B.21) with n = 1 and m = 4. All terms which contribute to Z ( a ) µ , namely ∂∂θ a A sing µ and ∂ µ Λ a, sing and [ A sing µ , Λ a, sing ] fall off as 1 /r for large | x | , and Z ( a ) µ itself is nonsingular at x = 0.In regular gauge one finds from (3.40) ∂ γ A reg µ = ∂ γ U − ( ∂ µ + A sing µ ) U = U − ∂ γ A sing µ U , and the transversality condition becomes U − D µ ( A sing ) U [ U − ∂ γ A sing µ U + U − D µ ( A sing ) U U − Λ a, sing U ]= D µ ( A reg )[ ∂ γ A reg µ + D µ ( A reg µ ) U − Λ a, sing U ] = 0 . Hence, Λ a, reg = U − Λ a, sing U , and now all contributions to Z ( a ) µ in the regular gauge fall only off as 1 /r .Only their sum Z ( a ) , reg µ falls off as 1 /r , just as Z ( a ) , sing µ . It is clearly simpler to work in the singular gauge,because then all integrals separately converge.
37e need the gauge-orientation zero modes for arbitrary values of θ because this is neededfor the group (Haar) measure. They are obtained as follows. By differentiating U ( θ ) andusing that U − ∂∂θ α U is equal to e αa ( θ ) T a , where the function e αa ( θ ) is called the groupvielbein (with α a curved and a a flat index according to the usual terminology ), oneobtains ∂∂θ α A µ ( θ ) = [ A µ ( θ ) , e αa ( θ ) T a ] (5.17)For Λ ( α ) we take now Λ ( α ) ( θ ) = − ρ x + ρ e αa ( θ ) T a , and then we obtain for the gauge zeromodes at arbitrary θZ µ ( α ) ( θ ) = D µ ( A ( θ )) (cid:18) x x + ρ e αa ( θ ) T a (cid:19) = U − (cid:20) D µ ( A ( θ = 0)) (cid:18) x x + ρ ∂ α U U − (cid:19)(cid:21) U . (5.18)We define ∂ α U U − = f αa ( θ ) T a . Note that tr ∂ α U U − ∂ β U U − = tr ( U − ∂ α U U − ∂ β U ) = e αa e βb tr T a T b = f αa f βb tr T a T b . Hence the left-invariant metric e αa e βb δ ab is equal to theright-invariant metric. There is a geometrical interpretation of these results [53, 54].There are only two differences with the θ = 0 case(i) the factors U ( θ ) and U − ( θ ) in front and at the back; these drop out in the trace(ii) the factors of f αa multiplying T a . Taking the trace one obtains the group metric U αβ ( θ ) = (cid:104) Z ( α ) µ | Z ( β ) µ (cid:105) = e αa ( θ ) e βb ( θ ) U ab ( θ = 0) = e αa ( θ ) e βa ( θ ) ( ρ S cl ) . (5.19)Hence, in the square root of the determinant of U one finds a factor det e αa (becausedet( e αa δ ab e βb ) = (det e αa ) ), and this yields the Haar measure µ ( θ ) = det e αa ( θ )d θ . (5.20)Using this measure one can calculate the group volume V of SU (2), V = (cid:82) (det e αa )d θ ,which is independent of the choice of coordinates θ . (We chose the parametrization U ( θ ) =exp θ a T a , but any other parametrization yields the same result.) The group vielbein is given by e αa ( θ ) T a = T α + 12! [ T α , θ · T ] + 13! [( T α , θ · T ] , θ · T ] + · · · , whereas the adjoint matrix representation M adj ( θ ) is given bye − θ · T T a e θ · T = M adj ( θ ) ab T b = T a + [ T a , θ · T ] + · · · . One has M adj ( θ ) ab = (exp θ c f · c · ) ab . There is a relation between the group vielbein and the adjoint matrix: (cid:0) θ β ∂∂θ β + 1 (cid:1) e αb ( θ ) = ( M adj ( θ )) αb The functions e αa ( θ ) are sometimes called the left-invariant one-forms, while f αa ( θ ) are the right-invariant one-forms.
38e have now calculated all norms. It is fairly easy to prove that there is no mixingbetween the different modes, for example U µ ( ρ ) = U µa = U ( ρ ) a = 0. Thus the matrix U ij for SU (2) is eight by eight, with non-vanishing entries along the block-diagonal U ij = δ µν S cl S cl 12 g αβ ( θ ) ρ S cl × , (5.21)The square root of the determinant is √ U = S ρ (cid:113) det g αβ ( θ ) = 2 π ρ g (cid:113) det g αβ ( θ ) (for SU (2)) . (5.22)Let us now consider the remaining generators of SU ( N ) by first analyzing the exampleof SU (3). For simplicity, we restrict ourselves again to lowest order in θ a . There are sevengauge orientation zero modes, three of which are given by (5.15) by taking for T a thefirst three Gell-Mann matrices λ , λ , λ multiplied by − i . For the other four zero modes,corresponding to λ , . . . , λ , the formula (5.13) still holds, but we have to change the gaugetransformation in order to keep the zero mode in background gauge,Λ k = (cid:34)(cid:115) x x + ρ − (cid:35) T k , k = 4 , , , , (5.23)with T k = ( − i/ λ k . The difference in x -dependence of the gauge transformations (5.14)and (5.23) is due to the change in commutation relations. Namely, (cid:80) a =1 [ λ a , [ λ a , λ β ]] = − (3 / λ β for β = 4 , , ,
7, whereas it is − λ β for β = 1 , ,
3. (These are the values of theCasimir operator of SU (2) on doublets and triplets, respectively). As argued before, thereis no gauge orientation zero mode associated with λ , since it commutes with the SU (2)embedding. The zero modes are then Z µ ( k ) = D µ (cid:34)(cid:115) x x + ρ T k (cid:35) , k = 4 , , , , (5.24)with norms U kl = δ kl ρ S cl , (5.25)and are orthogonal to (5.15), such that U ka = 0. This construction easily generalizes to SU ( N ). One first chooses an SU (2) embedding, and this singles out 3 generators. The These zero modes are given by Z µ ( k ) = ρ x ν / ( √ x ( x + ρ ) / )( δ µν T k + 2 η aµν [ T a , T k ]) in the singulargauge, see (3.38). For the first three zero modes we found instead Z µ ( a ) = ρ x ν / ( x + ρ ) (2 δ µν T a +2 η bµν [ T b , T a ]) with [ T b , T a ] = (cid:15) bac T c . The norm of (5.24) is proportional to tr T k T l + 4tr [ T a , T k ][ T a , T l ] =4tr T k T l , where we used (B.5). N −
2) doublets under this SU (2) and the rest aresinglets. There are no zero modes associated with the singlets, since they commute withthe SU (2) chosen. For the doublets, each associated zero mode has the form as in (5.24),with the same norm ρ S cl . This counting indeed leads to 4 N − ρ S cl ) N − , and so √ U = 2 N +7 ρ (cid:18) πρg (cid:19) N (for SU ( N )) . (5.26)This result is a factor 2 N − smaller than [52, 1], since we chose U ( θ ) = exp θ a T a insteadof exp(2 θ a T a ). This ends the discussion about the (bosonic) zero mode normalization. In this subsection we will explicitly construct the fermionic zero modes (normalizable so-lutions of the Dirac equation) in the background of a single anti-instanton. For an SU (2)adjoint fermion, there are 4 zero modes according to (4.37), and these can be written asfollows [63] λ α = − σ αρσ β (cid:16) ξ β − σ βγ (cid:48) ν ¯ η γ (cid:48) ( x − x ) ν (cid:17) F ρσ . (5.27)The SU (2) indices u and v are carried by ( λ α ) uv and ( F ρσ ) uv .To prove that these spinors are solution of the Dirac equation, use ¯ σ µ σ ρσ = δ µρ ¯ σ σ − δ µσ ¯ σ ρ − (cid:15) µρστ ¯ σ τ . Then (cid:54) ¯ Dλ vanishes since D µ F ρσ vanishes when contracted with η µρ , η µσ or (cid:15) µρστ . Actually, this expression also solves the Dirac equation for higher order k , butthere are then additional solutions, 4 | k | in total for SU (2), see (4.37). The four fermioniccollective coordinates are denoted by ξ α and ¯ η γ (cid:48) , where α, γ (cid:48) = 1 , . They are the fermionic partners of the translational and dilatational col-lective coordinates in the bosonic sector. These solutions take the same form in any gauge,one just takes the corresponding gauge for the field strength. The canonical dimension of ξ and ¯ η is − / /
2, respectively.For SU ( N ) (and always k = −
1) there are a further set of 2 × ( N −
2) zero modes inthe adjoint representation, and their explicit form depends on the gauge chosen. In regulargauge, with color indices u, v = 1 , . . . , N explicitly written, the gauge field is given by(3.41) (setting x = 0, otherwise replace x → x − x ) A µuv = A aµ ( T a ) u v = − σ uµν v x ν x + ρ , σ uµν v = (cid:32) σ αµν β (cid:33) . (5.28) To check that the expression with ¯ η is a solution, one may use that ¯ σ ρ σ µν σ ρ = 0. Note that one maychange the value of x in (5.27) while keeping F µν fixed, because the difference is a solution with ξ β . λ α uv = ρ (cid:112) ( x + ρ ) ( µ u δ αv + (cid:15) αu ¯ µ v ) . (5.29)Here we have introduced Grassmann collective coordinates µ u = ( µ , . . . , µ N − , ,
0) ; (cid:15) αu = , . . . , , (cid:15) αβ (cid:48) , . . . , , with N − β (cid:48) = u , (5.30)and similarly for ¯ µ v and δ αv . Thus the SU ( N ) structure for the fermionic instanton is asfollows λ ∝ (cid:32) µ ¯ µ ξ, ¯ η (cid:33) . (5.31)The canonical dimension of µ and ¯ µ is − /
2. To prove that ( λ α ) uv in (5.29) satisfies theDirac equation ¯ σ µ ( ∂ µ λ + [ A µ , λ ]) = 0, note that the terms ( A µ ) uw µ w and ¯ µ w ( A µ ) wv vanishdue to the index structure of A µ and µ, ¯ µ . Because A µ has only nonzero entries in thelower right block, there cannot be fermionic instantons in the upper left block.In singular gauge, the gauge field is given by (3.38) A µ uv = − ρ x ( x + ρ ) ¯ σ µν uv x ν . (5.32)Notice that the position of the color indices is different from that in regular gauge. This isdue to the natural position of indices on the sigma matrices . The fermionic anti-instantonin singular gauge reads [55] λ αuv = ρ (cid:112) x ( x + ρ ) ( µ u x αv + x αu ¯ µ v ) , (5.33)where for fixed α , the N -component vectors µ u and x αv are given by µ u = ( µ , . . . , µ N − , , , x αv = (cid:16) , . . . , , x µ σ αβ (cid:48) µ (cid:17) with N − β (cid:48) = v . (5.34)Further, x αu = x αv (cid:15) vu and ¯ µ v also has N − SU (2)instanton in the lower-right block of SU ( N ). Notice that the adjoint field λ is indeedtraceless in its color indices. This follows from the observation that µ and ¯ µ only appear at To be very precise, we could have used different Pauli matrices ( τ a ) uv for the internal SU (2) generators.Then we could have defined a matrix (¯ σ µν ) uv by ¯ σ µν = iη aµν τ a , and the SU (2) indices in (5.32) and (5.33)would have appeared in the same position as in (5.29). It is simpler to work with only one kind of Paulimatrices. SU ( N ). In general µ and ¯ µ are independent, but if there is areality condition on λ in Euclidean space, the µ and ¯ µ are related by complex conjugation.We will discuss this in a concrete example when we discuss instantons in N = 4 superYang-Mills theory. We should also mention that while the bosonic collective coordinatesare related to the rigid symmetries of the theory, this is not obviously true for the fermioniccollective coordinates, although, as we will see later, the ξ and ¯ η collective coordinates canbe obtained from ordinary supersymmetry and conformal supersymmetry in super Yang-Mills theories.A similar construction holds for a fermion in the fundamental representation. Now thereis only one fermionic collective coordinate, see (4.37), which we denote by K . The explicitexpression for k = − ( λ α ) u = ρ (cid:112) x ( x + ρ ) x αu K . (5.35)In regular gauge it is given by ( λ α ) u = (cid:15) αu ( x + ρ ) / K . (5.36)The Dirac equation for ( λ α ) u is proportional to − x µ ¯ σ µ,α (cid:48) β (cid:15) βu − ¯ σ µ,α (cid:48) β (cid:15) βv ( σ µν ) uv x ν (5.37)and to show that this vanishes one may use ( σ µν ) uv (cid:15) βv = ( σ µν ) uβ and the symmetry of theLorentz generators ( σ µν ) uβ = ( σ µν ) βu and ¯ σ µ σ µν = 3¯ σ ν . Having determined the bosonic and fermionic zero modes for k = ± SU ( N ) gauge group, we now discuss the measure for the zero mode sector of path integrals.The one-loop corrections due to the nonzero modes, will be discussed in the next section. We now construct the measure on the moduli space of bosonic collective coordinates, andshow that the matrix U plays the role of a Jacobian. We first illustrate the idea for a The color index should again be written as ( λ α ) u (cid:48) because λ reg ,u = ( U ) uv (cid:48) λ sing v (cid:48) with U uv (cid:48) = σ uv (cid:48) µ x µ / √ x . However, we drop these primes. The proof that (5.35) satisfies the Dirac equation uses¯ σ µα (cid:48) β σ βρu x µ x ρ = (cid:15) α (cid:48) u x and (¯ σ µρ ) α (cid:48) v (¯ σ µν ) uv = − (¯ σ µρ ) α (cid:48) v (¯ σ µν ) vu = 3 δ ρν (cid:15) α (cid:48) u . φ A , and action S [ φ ] (for example, thekink in one dimension). We expand around the instanton solution φ A ( x ) = φ A cl ( x, γ ) + φ A qu ( x, γ ) . (6.1)The collective coordinates are denoted by γ and, for notational simplicity, we assume thereis only one. At this point the fields φ A qu can still depend on the collective coordinate, asthey can include zero modes. The action, up to terms quadratic in the quantum fields, is S = S cl + φ A qu M AB ( φ cl ) φ B qu . (6.2)The operator M has zero modes given by Z A = ∂φ A cl ∂γ , (6.3)since, as we explained in (5.7), M AB Z B is just the derivative of the field equation ∂S cl /∂φ A cl with respect to the collective coordinate. More generally, if the operator M is hermitian(or rather self-adjoint ), it has a complete set of eigenfunctions F α with eigenvalues (cid:15) α , M AB F Bα = (cid:15) α F Aα . (6.4)One of the solutions is of course the zero mode Z = F with (cid:15) = 0. Any function can beexpanded into a basis of eigenfunctions, in particular the quantum fields, φ A qu = (cid:88) α ξ α F Aα , (6.5)with coefficients ξ α . The eigenfunctions have norms, determined by their inner product (cid:104) F α | F β (cid:105) = (cid:90) d x F Aα ( x ) F Aβ ( x ) . (6.6)The eigenfunctions can always be chosen orthogonal, such that (cid:104) F α | F β (cid:105) = δ αβ u α . Theaction then becomes S = S cl + (cid:88) α ξ α ξ α (cid:15) α u α . (6.7)If there is a coupling constant in front of the action (6.2), we rescale the inner productwith the coupling, such that (6.7) still holds. This was done in (5.8). The path-integralmeasure is now defined as [d φ ] ≡ ∞ (cid:89) α =0 (cid:114) u α π d ξ α . (6.8) More precisely, if there is an inner product ( φ , φ ) = (cid:82) φ A H AB φ B d x with real φ , φ and with metric H AB , and H AB H BC = δ AC , then one may define φ A H AB = φ B so that ( φ , φ ) = (cid:82) φ A φ A d x . If onefurther defines H BC M CD = M BD , then M AB is hermitian if ( φ , M φ ) = ( M φ , φ ). The need for amatrix to define an inner product is familiar from spinors, but for bosons the metric is in general trivial( H AB = δ AB ).
43e perform the Gaussian integration over the ξ α and get (cid:90) [d φ ] e − S [ φ ] = (cid:90) (cid:114) u π d ξ e − S cl (det (cid:48) M ) − / . (6.9)One sees that if there were no zero modes, the measure in (6.8) produces the correct resultwith the determinant of M . In the case of zero modes, the determinant of M is zero, andthe path integral would be ill-defined. Instead, we must leave out the zero mode in M ,take the amputated determinant (denoted by det (cid:48) ), and integrate over the mode ξ . Byslightly changing some parameters in the action (for example by adding a small mass term)the zero mode turns into a non-zero mode, and then one needs (cid:112) u π d ξ as measure. So,continuity fixes the measure for the zero modes as in (6.8).The next step is to convert the ξ integral to an integral over the collective coordinate γ [56]. This can be done by inserting unity into the path integral. Consider the identity1 = (cid:90) d γ δ ( f ( γ )) ∂f∂γ , (6.10)which holds for any (invertible) function f ( γ ). Taking f ( γ ) = −(cid:104) φ − φ cl ( γ ) | Z (cid:105) , and recallingthat the original field φ is independent of γ , we get1 = (cid:90) d γ (cid:18) u − (cid:28) φ qu (cid:12)(cid:12)(cid:12)(cid:12) ∂Z∂γ (cid:29) (cid:19) δ (cid:16) (cid:104) φ qu | Z (cid:105) (cid:17) = (cid:90) d γ (cid:18) u − (cid:28) φ qu (cid:12)(cid:12)(cid:12)(cid:12) ∂Z∂γ (cid:29) (cid:19) δ (cid:16) ξ u (cid:17) . (6.11)This trick is similar to the Faddeev-Popov trick for gauge fixing. In the semiclassicalapproximation, the term (cid:104) φ qu (cid:12)(cid:12)(cid:12) ∂Z∂γ (cid:69) is subleading and we will neglect it . The integrationover ξ is now trivial and one obtains (cid:90) [d φ ] e − S = (cid:90) d γ (cid:114) u π e − S cl (det (cid:48) M ) − / . (6.12)For a system with more zero modes Z i with norms-squared U ij , the result is (cid:90) [d φ ] e − S = (cid:90) (cid:89) i =1 d γ i √ π (det U ) / e − S cl (det (cid:48) M ) − / . (6.13) It will contribute however to a two-loop contribution. To see this, one first writes this term in theexponential, where it enters without ¯ h , so it is at least a one-loop effect. Then φ qu has a part proportionalto the zero mode, which drops out by means of the delta function insertion. The other part of φ qu isgenuinely quantum and contains a power of ¯ h (which we have suppressed). Therefore, it contributes attwo loops [57] (see also [58] for related matters). One obtains from (6.11) det (cid:104) ∂ γ i A cl µ | Z ( j ) (cid:105) times (det U ab ) − / . The matrix elements (cid:104) ∂ γ i A cl µ | Z ( j ) (cid:105) areequal to (cid:104) Z ( i ) | Z ( j ) (cid:105) = U ij minus (cid:104) D µ Λ ( a ) | Z ( b ) µ (cid:105) . The latter term can be partially integrated, and vanishessince there are no boundary contributions, neither in the singular nor in the regular gauge. (For the regulargauge one needs an explicit calculation to check this statement.) Z , which can be seen as rescalingsof the collective coordinates. More generally, the matrix U ij can be interpreted as a metricon the moduli space of collective coordinates. The measure is then invariant under generalcoordinate transformations on the moduli space.One can repeat the analysis for gauge theories to show that (6.13) also holds for Yang-Mills instantons in singular gauge. For regular gauges, there are some complications dueto the fact that neither of the two terms in (5.5) does fall off fast at infinity, but onlytheir sum is convergent. In singular gauge, each term separately falls off fast at infinity.For this reason, it is more convenient to work in singular gauge. The measure for thebosonic collective coordinates for k = 1 SU ( N ) YM theories, without the determinantfrom integrating out the quantum fluctuations which will be analyzed in the next section,becomes 2 N +2 π N − ( N − N − g N (cid:90) d x d ρρ ρ N . (6.14)This formula contains the square-root of the determinant of U in (5.26), 4 N factors of1 / √ π , and we have also integrated out the gauge orientation zero modes. This may bedone only if we are evaluating gauge invariant correlation functions. The result of thisintegration follows from the volume of the coset spaceVol (cid:26) SU ( N ) SU ( N − × U (1) (cid:27) = 2 N − π N − ( N − N − , (6.15)which is a factor 2 N − larger than in [52, 1], because we have used the normalizationtr( T a T b ) = − δ ab , while in [52, 1] tr( T a T b ) = − δ ab was used. We found in (5.26) anotherfactor 2 − (4 N − , and indeed the result for the total measure in (6.14) is the same as in[52, 1]. The derivation of this formula can be found in Appendix C, which is a detailedversion of [52]. We must also construct the measure on the moduli space of fermionic collective coordinates.Consider (5.27). The fermionic zero modes are linear in the Grassmann parameters ξ α and¯ η α (cid:48) . Thus these ξ α and ¯ η α (cid:48) correspond to the coefficients ξ α in (6.5). One obtains the zeromodes by differentiating λ α in (5.27) w.r.t. ξ α and ¯ η α (cid:48) , and for this reason one often callsthese ξ α and ¯ η α (cid:48) the fermionic collective coordinates. This is not quite correct, becausecollective coordinates appear in the classical solution (the instanton) but we shall usethis terminology nevertheless because it is common practice. We use again the measurein (6.8). There are in this case no factors √ π because of the Grassmann integration,and instead of (det M (cid:48) ) − / we now obtain (det M (cid:48) ) / in (6.9). Because the parameters45 α , ¯ η α , etc. appear linearly in the zero modes, we do not need the Faddeev-Popov trickto convert the integration over zero modes into an integration over collective coordinates.So for fermions the Grassmannian coefficients of the zero modes are at the same time thecollective coordinates .We shall discuss these issues in more detail when we come to supersymmetric gaugetheories, but now we turn to computing the norms of the fermionic zero modes.For the zero modes with ξ in (5.27), one finds Z α ( β ) = ∂λ α ∂ξ β = − σ αµν β F µν . (6.16)The norms of these two zero modes are given by( U ξ ) βγ = − g (cid:90) d x tr (cid:8) Z α ( β ) Z α ( γ ) (cid:9) = 4 S cl δ βγ , (6.17)where we have used the definition in (5.8) and contracted the spinor indices with the usualmetric for spinors. This produces a term in the measure (cid:90) d ξ d ξ (4 S cl ) − . (6.18)The result (6.18) actually holds for any k . We get the square root of the determinantin the denominator for fermions. One really gets the square root of the super determi-nant of the matrix of inner product, but because there is no mixing between bosonic andfermionic moduli, the superdeterminant factorizes into the bosonic determinant divided bythe fermionic determinant.For the ¯ η zero modes, we obtain, using some algebra for the σ -matrices, Z αβ (cid:48) = ∂λ α /∂ ¯ η β (cid:48) = ( σ µν σ ρ ) αβ (cid:48) F µν x ρ , (6.19)and ( U ¯ η ) α (cid:48) β (cid:48) = 8 S cl δ α (cid:48) β (cid:48) ρ , (6.20)so that the corresponding measure is (cid:90) d¯ η d¯ η (8 ρ S cl ) − , (6.21)which only holds for k = 1.Finally we compute the Jacobian for the fermionic “gauge orientation” zero modes.For convenience, we take the solutions in regular gauge (the Jacobian is gauge invariantanyway), and find from (5.29) (cid:0) Z α ( µ w ) (cid:1) uv = ρ (cid:112) ( x + ρ ) δ αv ∆ uw , (cid:0) Z α (¯ µ w ) (cid:1) uv = ρ (cid:112) ( x + ρ ) (cid:15) αu ∆ wv , (6.22) Sometimes one finds in the literature that U ξ = 2 S cl . This is true when one uses the conventions forGrassmann integration (cid:82) d ξ ξ α ξ β = (cid:15) αβ . In our conventions d ξ ≡ d ξ d ξ . N by N matrix ∆ is the unity matrix in the ( N −
2) by ( N −
2) upper diagonalblock, and zero elsewhere. So ∆ restricts the values of u, w and v to up to N − δ αv and (cid:15) αv the index runs over the next two values. Consequently, the norms of Z µ and Z ¯ µ are easily seen to be zero, but the nonvanishing inner product is( U µ ¯ µ ) uv = − g (cid:90) d x tr Z α (¯ µ u ) Z α ( µ v ) = 2 π g ∆ uv , (6.23)where we have used the integral (B.21). It also follows from the index structure that the ξ and ¯ η zero modes are orthogonal to the µ zero modes, so there is no mixing in the Jacobian.Putting everything together, the fermionic part of the measure for N adjoint fermionscoupled to SU ( N ) YM theory, with k = 1, is given by (cid:90) (cid:32) N (cid:89) A =1 d ξ A (cid:33) (cid:18) g π (cid:19) N (cid:32) N (cid:89) A =1 d ¯ η A (cid:33) (cid:18) g π ρ (cid:19) N N (cid:89) A =1 (cid:32) N − (cid:89) u =1 d µ A,u d¯ µ Au (cid:33) (cid:18) g π (cid:19) N ( N − . (6.24)Similarly, one can include fermions in the fundamental representation, for which the Jaco-bian factor is U K ≡ (cid:90) d x Z αu Z αu = π , (6.25)for each species. Here K is the Grassmann collective coordinate of (5.36). Hence in thiscase the fermionic part of the measure is (cid:90) N f (cid:89) A =1 d K A (cid:32)(cid:114) π (cid:33) N f (6.26)for N f fundamental Weyl spinors coupled to SU ( N ) YM theory with k = 1.Note that we did not put a factor g in front of the integral in (6.25), whereas we usedsuch a factor for fermions in the adjoint representation. The reason we do not use such afactor for fermions in the fundamental representation has to do with the action. One finds afactor g in front of the Yang-Mills action, and therefore also, by susy, in front of the Diracaction for gluinos. However, in the matter action the g -dependence has been absorbed bythe gluons, so there is no factor g in front of the matter fermions. The measure of the zeromodes uses the metric of the collective coordinates. In soliton physics (and instantons canbe considered as solitons in one higher dimension) one obtains this metric if one lets thecollective coordinates become time dependent and integrates over d x in the action oneends up with a quantum mechanical action of the L = ( U ij ˙ γ i ˙ γ j + U αβ ˙ ξ α ˙ ξ β + U α (cid:48) β (cid:48) ˙¯ y α (cid:48) ˙¯ y β (cid:48) + U AB ˙ K A ˙ K B (6.27)Since U ij , U αβ and U α (cid:48) β (cid:48) are produced by the Yang-Mills action and its susy partner, while U AB is due to the matter action, there is no g -dependence in (6.26).47 One loop determinants
Having determined the measure on the moduli space of collective coordinates, we now com-pute the determinants that arise by Gaussian integration over the quantum fluctuations.Before doing so, we extend the model by adding real scalar fields and Majorana fermionsin the adjoint representation. The action is S = − g (cid:90) d x tr (cid:8) F µν F µν + ( D µ φ ) ( D µ φ ) − i ¯ λ (cid:54) ¯ Dλ − iλ (cid:54) D ¯ λ (cid:9) . (7.1)Here, λ is a two-component Weyl spinor which we take in the adjoint representation . InMinkowski space there is a reality condition between the two complex 2-component spinors λ and ¯ λ , and as a result ¯ λ ˙ α transforms in the complex conjugate of the representation of λ α , but in Euclidean space this reality condition is dropped. So λ α and ¯ λ α (cid:48) are independentcomplex variabes. For the Grassmann integration this makes no difference. Written withindices the Euclidean Dirac action in (7.1) reads {− iλ α ( σ µ ) αβ (cid:48) D µ ¯ λ β (cid:48) − i ¯ λ α (cid:48) (¯ σ µ ) α (cid:48) β D µ λ β } where λ α = λ β (cid:15) βα and ¯ λ α (cid:48) = (cid:15) α (cid:48) β (cid:48) ¯ λ β (cid:48) . Generalization to fundamental fermions is straight-forward. The anti-instanton solution around which we will expand is A cl µ , φ cl = 0 , λ cl = 0 , ¯ λ cl = 0 , (7.2)where A cl µ is the anti-instanton. This background represents an exact solution to the fieldequations. The bosonic and fermionic zero modes are taken care of by the measure forthe collective coordinates, while in the orthogonal space of nonzero modes, one can definepropagators and vertices, and perform perturbation theory around the (anti-) instanton.After expanding A µ = A cl µ + A qu µ , and similarly for the other fields, we add gauge fixingand ghost terms S gf = − g (cid:90) d x tr (cid:110)(cid:0) D cl µ A qu µ (cid:1) − b D c (cid:111) , (7.3)such that the total gauge field action is given by (5.3). The integration over A µ gives[det (cid:48) ∆ µν ] − / , ∆ µν = − D δ µν − F µν , (7.4)where the prime stands for the amputated determinant, with zero eigenvalues left out. Wehave suppressed the subscript ‘cl’ and Lie algebra indices. Integration over the scalar fieldsresults in [det ∆ φ ] − / , ∆ φ = − D , (7.5) As before λ = (cid:32) λ α ¯ λ ˙ α (cid:33) , but the 4-component Majorana spinor ¯ λ is defined by λ T C both in Minkowskiand in Euclidean space, where C is the charge conjugation matrix, C = (cid:32) (cid:15) αβ (cid:15) ˙ α ˙ β (cid:33) . Then ¯ λ = (cid:0) λ α , − ¯ λ ˙ α (cid:1) and Lorentz (or rather SO (4)) invariance is preserved in Euclidean space because the relation Cγ µ = − γ µ,T C holds in both spaces. In Euclidean space we denote the indices of ¯ λ by α (cid:48) instead of ˙ α . gh ] , ∆ gh = − D . (7.6)For the fermions λ and ¯ λ , we need a bit more explanation. Since neither (cid:54) D nor (cid:54) ¯ D ishermitean(even worse, (cid:54) D maps antichiral spinors into chiral spinors), we cannot evaluatethe determinants in terms of their eigenvalues. But both products∆ − = − (cid:54) D (cid:54) ¯ D = − D − σ µν F µν , ∆ + = − (cid:54) ¯ D (cid:54) D = − D , (7.7)with spinor indices still suppressed, are hermitean. Let us label the nonzero modes by asubscript i . Then we can expand λ in terms of commuting eigenfunctions F i of ∆ − withanticommuting coefficients ξ i , and ¯ λ in terms of eigenfunctions ¯ F i of ∆ + with coefficients ¯ ξ i .We have seen before that both operators have the same spectrum of non-zero eigenvalues (cid:15) i , and the relation between the eigenfunctions is ¯ F i = √ (cid:15) i ¯/ DF i and F i = − √ (cid:15) i / D ¯ F i . (Theminus sign is needed in order that ¯ F i = √ (cid:15) i ¯/ DF i = (cid:15) i ( − ¯/ D / D ) ¯ F i = ¯ F i ). Defining the pathintegral over λ and ¯ λ as the integration over ξ i and ¯ ξ i , one gets the determinant over thenonzero eigenvalues . The result for the integration over the fermions can be written insymmetrized form as [det (cid:48) ∆ − ] / [det ∆ + ] / . (7.9)As stated before, since all the eigenvalues of both ∆ − and ∆ + are the same, the determi-nants are formally equal. This result can also be obtained by writing the spinors in termsof Dirac fermions; the determinant we have to compute is then (cid:2) det (cid:48) ∆ D (cid:3) / , ∆ D = (cid:32) (cid:54) D (cid:54) ¯ D (cid:33) . (7.10) Namely, the action becomes − g tr (cid:90) d x (cid:20) − i (cid:18) ¯ ξ i ¯ F i (cid:54) ¯ Dξ j ( − (cid:54) D ¯ F j ) √ (cid:15) j (cid:19) − iξ j F j (cid:54) D ¯ ξ i (cid:54) ¯ DF i √ (cid:15) i (cid:21) = − i ξ i ξ j √ (cid:15) j (cid:10) ¯ F ai | ¯ F aj (cid:11) + i ξ j ¯ ξ i (cid:10) F aj | F ai (cid:11) √ (cid:15) i . (7.8)Next we use that the norms of ¯ F i and F i are equal: (cid:10) ¯ F i | ¯ F j (cid:11) = 1 √ (cid:15) i (cid:28) (cid:54) ¯ DF i | √ (cid:15) j (cid:54) ¯ DF j (cid:29) = 1 √ (cid:15) i (cid:15) j (cid:10) F i |− (cid:54) D (cid:54) ¯ DF j (cid:11) = (cid:114) (cid:15) j (cid:15) i (cid:104) F i | F j (cid:105) = 1 √ (cid:15) i (cid:15) j (cid:10) − (cid:54) D (cid:54) ¯ DF i | F j (cid:11) = (cid:114) (cid:15) i (cid:15) j (cid:104) F i | F j (cid:105) . Hence, as expected, the F i and ¯ F j for different eigenvalues are orthogonal to each other, and the norms of F i and ¯ F j are the same. Denoting g (cid:82) d x ( F ai ) ∗ F ai = < F ai | F ai > by u i , one finds for the path integral (cid:90) d ¯ ξ i dξ i e iξ i ¯ ξ i u i √ (cid:15) i = iu i √ (cid:15) i . Hence the measure is d ξ i √ u i d¯ ξ j √ u j , and the one-loop determinant is (cid:81) i ( (cid:15) i ) / . + and ∆ − , one would expect that the determinants for the bosons can be expressed in termsof the determinants of ∆ − and ∆ + . For the ghosts and adjoint scalars this is obvious,det ∆ φ = det ∆ gh = [det ∆ + ] / . (7.11)We get det ∆ φ = det( − D ) = det ∆ / and det ∆ gh = det( − D ) = det ∆ / because thespinor space is two-dimensional.For the vector fields, we rewrite the operator ∆ µν in (7.4) in terms of the fermion operator∆ − . Using tr(¯ σ µ σ ν ) = 2 δ µν and tr(¯ σ µ σ ρσ σ ν ) = 2( δ µρ δ σν − δ µσ δ ρν − (cid:15) µρσν ) we obtain thefollowing identity for ∆ µν = − δ µν D − F µν ,∆ µν = tr { ¯ σ µ ∆ − σ ν } = ¯ σ µ α (cid:48) β (cid:16) ∆ − βγ (cid:17) σ γα (cid:48) ν = (¯ σ µα (cid:48) β )(∆ − βγ δ α (cid:48) δ (cid:48) )(¯ σ ν γδ (cid:48) ) (7.12)where (∆ − ) βγ δ α (cid:48) δ (cid:48) is block-diagonal on the basis βα (cid:48) = γδ (cid:48) = (11) , (21) , (12) , (22).∆ − βγ δ α (cid:48) δ (cid:48) = ∆ − ∆ − − ∆ − − ∆ − − ∆ − (7.13)This proves that det (cid:48) ∆ µν = [det (cid:48) ∆ − ] . (7.14)Now we can put everything together. The one-loop determinant for a Yang-Mills system,including the ghosts, coupled to n real adjoint scalars and N Weyl spinors (or Majoranaspinors) also in the adjoint representation is[det (cid:48) ∆ − ] − N / [det∆ + ]
14 (2+
N − n ) . (7.15)This expression simplifies to the ratio of the determinants when N − n = 1. Particularcases are N = 1 n = 0 → (cid:20) det ∆ + det (cid:48) ∆ − (cid:21) / , Consider ¯ σ µ,α (cid:48) β and σ γδ (cid:48) ν as 4 × × µ and ν one has ¯ σ µα (cid:48) β σ βα (cid:48) ν = 2 δ µν , hence det[¯ σ µ,α (cid:48) β ] = 4. = 2 n = 2 → (cid:20) det ∆ + det (cid:48) ∆ − (cid:21) / , N = 4 n = 6 → (cid:20) det ∆ + det (cid:48) ∆ − (cid:21) . (7.16)These cases correspond to supersymmetric Yang-Mills theories with N -extended super-symmetry. Notice that for N = 4, the determinants of ∆ + and ∆ − separately cancel, sothere is no one-loop contribution.For N = 1 , + det (cid:48) ∆ − = exp (cid:32)(cid:88) n ω (+) n − (cid:88) n ω ( − ) n (cid:33) , (7.17)with eigenvalues λ n = exp ω n . The frequencies ω (+) n and ω ( − ) n can be discretized by puttingthe system in a box of size R and imposing suitable boundary conditions on the quantumfields at R (for example, φ ( R ) = 0, or dd R φ ( R ) = 0, or a combination thereof [4]). Theseboundary conditions may be different for different fields. The sums over ω (+) n and ω ( − ) n aredivergent; their difference is still divergent (although less divergent than each sum sepa-rately) but after adding counterterms ∆ S one obtains a finite answer. The problem is thatone can combine the terms in both series in different ways, giving different answers. Bycombining ω (+) n with ω ( − ) n for each fixed n , one would find that the ratio (det ∆ + / det (cid:48) ∆ − )equals unity. However, other values could result by using different ways to regulate thesesums. We have discussed before that for susy instantons the densities of nonzero modesare equal, hence for susy instantons the contributions in (7.16) from the one-loop deter-minants cancel. This makes these models simpler to deal with than non-susy models. Forordinary (nonsusy) Yang-Mills theory, the results for the effective action due to differentregularization schemes differ at most by a local finite counterterm. In the background fieldformalism we are using, this counterterm must be background gauge invariant, and sincewe consider only vacuum expectation values of the effective action, only one candidate ispossible: it is proportional to the gauge action (cid:82) d x tr F and multiplied by the one-loopbeta-function for the various fields which can run in the loop,∆ S ∝ β ( g ) (cid:90) d x tr F ln µ µ . (7.18)51he factor ln ( µ /µ ) parametrizes the freedom in choosing different renormalization schemes.A particular regularization scheme used in [4] is Pauli-Villars regularization. In thiscase ’t Hooft first used x -dependent regulator masses to compute the ratios of the one-loopdeterminants ∆ in the instanton background and ∆ (0) in the trivial vacuum. Then heargued that the difference between using the x -dependent masses and using the more usualconstant masses, was of the form ∆ S given above. The final result for pure YM SU ( N ) inthe | k = 1 | sector is [4, 52] (cid:34) det (cid:48) ∆ − det ∆ (0) − (cid:35) − (cid:34) det ∆ + det ∆ (0)+ (cid:35) / = exp (cid:8) N ln( µρ ) − α (1) − N − α (cid:0) (cid:1)(cid:9) . (7.19)Here we have normalized the determinants against the vacuum, indicated by the superscript(0). From the unregularized zero mode sector one obtains a factor ρ N , see (6.14), andPauli-Villars regularization of the 4 N zero modes yields a factor M NP V . All together oneobtains π g + ln( M P V ρ ) in the exponent for SU (2), where g is the unrenormalizedcoupling constant. Subtracting ln( M P V ρ ) to renormalize at mass scale 1 /ρ , one is leftfor the effective action with π g − ln( ρ/ρ ) ≡ π g ( ρ ) . Replacing ln( ρ/ρ ) by ln( µ/µ ), thisis the correct one-loop renormalization equation for the running of the coupling constant.For supersymmetric theories, the nonzero mode corrections to the effective action cancel,and performing the same renormalization procedure as for the non-supersymmetric case,one now obtains only from the zero modes the correct β function. For N = 4 one finds avanishing β function.The fluctuations of the SU (2) part of the gauge fields and the Faddeev-Popov ghostsyield the term α (1) in (7.19), while the fluctuations of the 2( N −
2) doublets (correspondingto λ , · · · λ for SU (3)) yields the term with α (cid:0) (cid:1) . The numerical values of the function α ( t ) are related to the Riemann zeta function, and take the values α (cid:0) (cid:1) = 0 . α (1) = 0 . ρ , andtherefore changes the ρ -dependence of the integrand of the collective coordinate measure.Combined with (6.14) one correctly reproduces the β -function of SU ( N ) YM theory. Thecalculation of the contribution of the nonzero modes can be simplified by using a so-called O (5) formalism [59] which uses the conformal symmetries of instantons, in addition tothe nonconformal symmetries. One still has to regulate the sums over zero-point energies,and both Pauli-Villars regularization [59] and zeta-function regularization [60] have beenapplied to the O (5) formulation. β function for SYM theories In supersymmetric gauge theories, the contributions to the one-loop partition function bythe nonzero modes in the bosonic and fermionic loops cancel each other [61]. Although this52as only been shown to occur in a gravitational background without winding, we assumehere that still occurs in an instanton background. Actually, all contributions from thenonzero mode sector cancel: higher-loops as well as possible nonperturbative corrections.The zero mode sector can be regularized by Pauli-Villars fields, and since the partitionfunction yields a physical observable, namely the cosmological constant (the sum over zero-point energies), the result for the partition function should not depend on the regularizationparameter M P V (the Pauli-Villars mass). From this observation one can derive a differentialequation for the coupling constant g ( M P V ), which yields the exact β function: it containsall perturbative contributions [62].Before going on we should comment on the fact that from the 3-loop level on the resultfor the β function depends on the regularization scheme chosen. It is sometimes claimedthat therefore higher-loop results for the β function have no meaning. This is incorrect:given a particular scheme, all orders in perturbation theory of β have meaning. In thederivation below of the β function we shall find an all-order result, but it is not (yet?)known which regularization scheme for Feynman graphs would reproduce these results. Sothe all-order expression for β has in principle meaning, but in practice one cannot do muchwith it. One can only say: there must exist a regularization scheme which, if used for thecalculation of higher-loop Feynman graphs, will produce the all-order result for β obtainedbelow.We begin with pure supersymmetric gauge theory. We recall that the measure of thezero modes of a single instanton or anti-instanton ( k = ±
1) for N = 1 susy with gaugegroup SU ( N ) and one Majorana or Weyl fermion in the adjoint representation is given byd M k = ± = e − π g (cid:34) d x d ρρ N +2 (cid:18) ρg (cid:19) N (cid:18) M P V √ π (cid:19) N π N − ( N − N − (cid:35) × (cid:34) d ξ d ξ S cl M P V d¯ η d¯ η ρ S cl M P V (cid:81) N − u =1 d µ u d¯ µ u (cid:0) S cl M P V (cid:1) N − (cid:35) Vol (cid:26) SU ( N ) SU ( N − ⊗ U (1) (cid:27) (7.20)where S cl = 8 π /g . The volume of the gauge group was given in (6.15) but becauseit does not depend on g or M P V it will play no role below. Note that this measure isdimensionless; d x d ρ/ρ is dimensionless, and the remaining ρ and M P V occur only in thecombination ρM P V . Also d ξ/M P V and d ¯ η/ ( ρ M P V ) are dimensionless. The prefactore − π /g is of course the classical action for the one-instanton background, and we have leftout the term with the term with the theta-angle. In the first square brackets we find theproduct of the measure for the bosonic zero modes in (6.14) and factors (cid:113) M PV π for eachbosonic zero mode from the corresponding Pauli-Villars modes. The second expressionin square brackets contains the contribution to the measure from the fermionic zero modes If the one-loop determinant for the bosonic fields is (det M b ) − / and for the fermionic fields det M f , √ M PV for each fermionic zero mode. Clearly, each bosoniczero mode contributes a factor M P V /g and each fermionic zero mode contributes a factor g/ √ M P V .The dependence of d M on M P V and g is thus as followsd M ∝ e − π g ( M P V ) N (cid:18) g (cid:19) N , (7.21)where g depends on M P V , so g = g ( M P V ). So g is the bare coupling constant in theregularized theory, and g and M P V vary such that the renormalized coupling constant g R is kept fixed. Usually one considers the renormalized coupling constant as a function ofthe renormalization mass µ , and then the bare coupling constant g satisfies µ ∂∂µ g = 0.Using dimensional regularization and g = Z g ( g ren ) g ren µ (cid:15)/ with (cid:15) = 4 − n yields then the β function. If one uses Pauli-Villars regularization there are two masses which play arole: the cut-off (regulator) mass M P V and the physical renormalization mass µ . The barecoupling depends on one of them, the renormalized coupling on the other. g = g ( M P V ) g R = g R ( µ ) M P V ∂∂M PV g ( M P V ) = β ( g ) µ∂/∂µ g R ( µ ) = β ( g R ) µ ∂∂µ g ( M P V ) = 0 M P V ∂∂M PV g R ( µ ) = 0 (7.22)Physical quantities depend on µ but not on M P V . If one wants to apply the renormalizationgroup to the measure, one must use the approach based on ( M P V ∂/∂M
P V ) g R ( µ )= 0 because the regularized measure depends on M P V , not on µ . The results for the β function obtained from both schemes differ by a sign, because the logarithms in the theorydepend on ln( M P V /µ ).Equating the derivative of the logarithm of the measure w.r.t. in (7.21) M P V to zeroyields then M P V ∂∂M
P V (cid:18) − π g + 3 N ln M P V − N ln g (cid:19) = 0 . (7.23)Hence M P V ∂∂M
P V g ≡ β = (cid:32) N Ng − π g (cid:33) , (7.24)or, written in terms of α = g π g π β = M P V ∂∂M
P V α = − N α π − αN π . (7.25) then the Pauli-Villars method yields further determinants det( M b + M P V ) +1 / and det( M f + M P V ) − .The zero modes are eigenfunctions of M b and M f with eigenvalue zero, so their Pauli-Villars counterpartsbecome nonzero modes with eigenvalues M P V and M P V . β -function for pure N = 1 supersymmetric Yang-Mills theory. It is straightfor-ward to extend this result to pure N -extended supersymmetry with N Majorana or Weylfermions in the adjoint representation. One finds for SU ( N ) M P V ∂∂M
P V α = − α π N − N N − α π (2 N − N N ) . (7.26)It is clear that for N = 2 there is only a one-loop contribution to β , and for N = 4the β function vanishes altogether. These are well-known properties of pure extended susygauge theories. For N = 1 one finds agreement for one- and two- loops. Beyond two loopsthe result for the beta function becomes scheme dependent, so it becomes then pointlessto investigate whether agreement holds.Let us now add matter. In susy QCD with N f flavours the matter part consists of N f pairs of chiral superfields Q i and ˜ Q i with i = 1 , N f in the N and N ∗ representations of SU ( N ). Each fermion in Q and ˜ Q has one zero mode, see (4.37) and (5.35), while thescalars do not have any zero modes. So the zero mode measure for the matter part isaccording to (6.26)d M (matter) = (cid:18) π (cid:19) N f M P V ) N f N f (cid:89) u =1 d K u d ˜ K u . (7.27)Renormalization leads to a further term in the measure, and thus in the β function. Insusy only the kinetic term ¯ φ e V φ of the matter fields gets a Z factor L = Z ¯ φ ren e V ren φ ren , φ = √ Zφ ren . (7.28)and rather than a factor √ M PV for each fermion with one flavor, we now get in the measurea factor ( ZM P V ) − / for each zero mode. (The Pauli-Villars field operator becomes ZM f + M P V , so the zero modes continue to produce a factor M − / P V in the Pauli-Villars sector, not ( ZM P V ) − / . In the nonzero mode sector one can neglect the dependence on M P V ,and here the Z factors of bosons and fermions cancel due to susy).For the gauge multiplet we factorized out a factor 1 /g in front of the action of all fieldsof the gauge multiplet, so that the fields gA µ = ˜ A µ do not renormalize. (We use here thebackground formalism in which Z g = Z − / A , where Z A is the wave function renormalizationconstant for the background fields.) Thus the renormalization of the gauge multiplet istaken care of by the renormalization of the factor 1 /g in (7.23).From here on we proceed as before. The measure for gauge group SU ( N ) with N f flavors is now given by dM k = ± = e − π /g ( M P V ) N (cid:18) g (cid:19) N (cid:18) ZM P V (cid:19) N f (7.29)55e denote the anomalous dimension by γ i where γ i = µ ∂∂µ ln Z i = − M P V ∂∂M
P V ln Z i = γ (the same for i = 1 , ..., N f ) , (7.30)and obtain g π β = M P V ∂∂M
P V α = − α π (cid:32) N − N f (1 − γ )1 − αN π (cid:33) . (7.31)Expanding in terms of α , the result agrees with the results in the literature for the one-loopand two-loop β functions for N = 1 susy QCD with N f pairs of chiral fields Q i and ˜ Q i .Namely, the one- and two-loop β function for an N = 1 vector multiplet coupled to achiral multiplet in a representation R , including the effects of the Yukawa couplings whosecoupling constant is also g (in fact, the renormalized coupling constant g R ), is given by[64] g π µ ∂∂µ g = α π ( − C ( G ) + T ( R ))+ α π ( − C ( G ) + 2 C ( G ) T ( R ) + 4 C ( R ) T ( R )) (7.32)For N f pairs of chiral matter fields (cid:80) T ( R ) = N f , and C ( G ) = N for SU ( N ). Usingalso that the anomalous dimension γ = µ ∂∂µ ln Z for a complex fermion in the fundamentalrepresentation N of SU ( N ) is equal to − αC ( R ) /π , we indeed find agreement. The β function in (7.31) can be rewritten such that only the numbers of zero modesappear. β ( α ) = − α π (cid:32) n g − n f − (cid:88) g γ g + (cid:88) f γ f (cid:33) . (7.33)Here n g is the number of bosonic zero modes (4 N ) , n f the total (gluino and matter) numberof fermionic zero modes (2 N + 2 N f ), and the sums (cid:80) g and (cid:80) f run over the gluon andfermion zero modes. For gluons and gluinos λ , the anomalous dimension is the same (dueto susy) and proportional to the β function γ g = γ λ = β/α . (7.34)Substitution of this result yields back (7.31). This result does not yet agree with theresults in the literature for the β -function of gauge fields minimally coupled to scalars and Often one defines γ = µ ∂∂µ ln √ Z ; here we follow [62] With the usual normalization γ = µ ∂∂µ ln Z / is equal to γ = − α π C ( R ) [64]. λ but rather λ = g . At the two-loop level onetherefore gets extra contributions which one must add to the results from the literature,and then one gets complete agreement. N = 4 supersymmetric Yang-Mills theory An interesting field theory with instantons is the N = 4 super Yang-Mills theory [65]. Theaction is of course well known in Minkowski space, but instantons require the formulationin N = 4 Euclidean space. Due to absence of a real representation of Dirac matrices infour-dimensional Euclidean space, one cannot straightforwardly define Majorana spinorsin Euclidean space. This complicates the construction of Euclidean Lagrangians for su-persymmetric models [66, 67, 68]. For N = 2 , N = 4 Euclidean model via the dimensional reduction of ten-dimensional N = 1 superYang-Mills theory along the time direction. One can also define a continuous Wick rotationfor the spinors directly in four dimensions [68]. N = 4 SYM
The N = 4 action in Minkowski space-time with the signature η µν = diag( − , + , + , +) isgiven by S = 1 g (cid:90) d x tr (cid:26) F µν F µν − i ¯ λ ˙ αA (cid:54) ¯ D ˙ αβ λ β,A − iλ Aα (cid:54) D α ˙ β ¯ λ A ˙ β + 12 (cid:0) D µ ¯ φ AB (cid:1) (cid:0) D µ φ AB (cid:1) − √ φ AB (cid:8) λ α,A , λ Bα (cid:9) − √ φ AB (cid:8) ¯ λ ˙ αA , ¯ λ ˙ α,B (cid:9) + 18 (cid:2) φ AB , φ CD (cid:3) (cid:2) ¯ φ AB , ¯ φ CD (cid:3)(cid:27) . (8.1)The on-shell N = 4 supermultiplet consists of a real gauge field A µ , four complex Weylspinors λ α,A (equivalently, four Majorana spinors) and an antisymmetric complex scalar φ AB with labels A, B = 1 , . . . , R symmetry group SU (4). The realityconditions on the components of this multiplet are the Majorana conditions (cid:0) λ α,A (cid:1) ∗ = − ¯ λ ˙ αA and ( λ Aα ) ∗ = ¯ λ ˙ α,A and ¯ φ AB ≡ (cid:0) φ AB (cid:1) ∗ = (cid:15) ABCD φ CD . (8.2) Unless specified otherwise, equations which involve complex conjugation of fields will be understoodas not Lie algebra valued, i.e. they hold for the components λ a,α,A , etc. SU (4) transformations. The sigma matrices aredefined by σ µ α ˙ β = (1 , τ i ), ¯ σ µ ˙ αβ = ( − , τ i ) for µ = 0 , , , (cid:16) σ α ˙ βµ (cid:17) ∗ = σ β ˙ αµ = ¯ σ ˙ αβµ = (cid:15) ˙ α ˙ γ (cid:15) βδ ¯ σ µ ˙ γδ , with (cid:15) ˙ α ˙ β = (cid:15) ˙ α ˙ β = − (cid:15) αβ = − (cid:15) αβ . Since φ AB isantisymmetric, one can express it on a basis spanned by the real eta-matrices (see AppendixB) φ AB = 1 √ (cid:8) S i η iAB + iP i ¯ η iAB (cid:9) , ¯ φ AB = 1 √ (cid:8) S i η iAB − iP i ¯ η iAB (cid:9) , (8.3)in terms of real scalars S i and real pseudoscalars P i , i = 1 , ,
3. Because η iAB is selfdualand ¯ η iAB anti-selfdual, η iAB = η iAB and ¯ η iAB = − ¯ η iAB . Then the reality conditions arefulfilled and the kinetic terms for the ( S, P ) fields take the standard form. The action in(8.1) is invariant under the following supersymmetry transformation laws with parameters ζ Aα and ¯ ζ ˙ αA δA µ = − i ¯ ζ ˙ αA ¯ σ µ ˙ αβ λ β,A + i ¯ λ ˙ β,A σ α ˙ βµ ζ Aα ,δφ AB = √ (cid:16) ζ α,A λ Bα − ζ α,B λ Aα + (cid:15) ABCD ¯ ζ ˙ αC ¯ λ ˙ α,D (cid:17) ,δλ α,A = − σ µν αβ F µν ζ β,A − i √ ζ ˙ α,B (cid:54) D α ˙ α φ AB + (cid:2) φ AB , ¯ φ BC (cid:3) ζ α,C , (8.4)which are consistent with the reality conditions. Let us turn now to the discussion of theEuclidean version of this model and discuss the differences with the Minkowski theory. N = 4 SYM
To find out the N = 4 supersymmetric YM model in Euclidean d = (4 ,
0) space, wefollow the same procedure as in [65]. We start with the N = 1 SYM model in d = (9 , SU (4) = SO (6) R -symmetry group, this reduction leads to a model with an internalnon-compact SO (5 , R -symmetry group in Euclidean space. As we will see, the realityconditions on bosons and fermions will both use an internal metric for this non-compactinternal symmetry group.The N = 1 Lagrangian in d = (9 ,
1) dimensions reads L = 1 g tr (cid:8) F MN F MN + ¯ Ψ Γ M D M Ψ (cid:9) , (8.5)with the field strength F MN = ∂ M A N − ∂ N A M + [ A M , A N ] and the Majorana-Weyl spinor Ψ defined by the conditions Γ Ψ = Ψ , Ψ T C − = Ψ † iΓ ≡ ¯ Ψ . (8.6)58ere the hermitean matrix Γ ≡ ∗ Γ is a product of all Dirac matrices, Γ = Γ . . . Γ ,normalized to ( ∗ Γ ) = +1. Furthermore, C − is the charge conjugation matrix, satisfying C − Γ M = − Γ TM C − . The Γ -matrices obey the Clifford algebra (cid:8) Γ M , Γ N (cid:9) = 2 η MN withmetric η MN = diag( − , + , . . . , +). The Lagrangian transforms into a total derivative underthe standard transformation rules δA M = ¯ ζΓ M Ψ , δΨ = − F MN Γ MN ζ , (8.7)with Γ MN = [ Γ M Γ N − Γ M Γ N ]. The susy parameter is a Majorana-Weyl spinor, ¯ ζ = ζ T C − = ζ † iΓ and (cid:63)Γ ζ = ζ . To proceed with the dimensional reduction we choose aparticular representation of the gamma matrices in d = (9 , Γ M = (cid:8) ˆ γ a ⊗ γ , [8] × [8] ⊗ γ µ (cid:9) , Γ = Γ . . . Γ = ˆ γ ⊗ γ , (8.8)where the 8 × γ a and ˆ γ of d = (5 ,
1) with a = 1 , . . . , γ a = (cid:32) Σ a,AB ¯ Σ aAB (cid:33) , ˆ γ = ˆ γ . . . ˆ γ = (cid:32) − (cid:33) , (8.9)In Euclidean d = (6 ,
0) one defines Σ a,AB = ( η kAB , i ¯ η k,AB )¯ Σ aAB = ( − η kAB , i ¯ η kAB ) (8.10)but in Minkowski space one puts a factor − i in front of the first one. So explicitly Σ a,AB = (cid:8) − iη ,AB , η ,AB , η ,AB , i ¯ η k,AB (cid:9) , ¯ Σ aAB = (cid:8) iη AB , − η AB , − η AB , i ¯ η kAB (cid:9) so that (cid:15) ABCD Σ a CD = − ¯ Σ aAB . The first three matrices ˆ γ , ˆ γ , ˆ γ are symmetric while the latter three matricesˆ γ , ˆ γ , ˆ γ are antisymmetric. Meanwhile γ µ and γ are the usual Dirac matrices of d = (4 , γ , in 6 dimensions with the time direction and thus it is anti-hermitean and has square −
1; all others (as well as all Dirac matrices in d = (4 , d = (9 , d = (5 ,
1) and d = (4 , C + and C − in even dimensions, satisfying C ± Γ µ = ± ( Γ µ ) T C ± , and After partial integrations, the Yang-Mills action transforms into [ ζΓ N ψ a ) D M F a,MN and the Diracaction varies into − ¯ ψ a Γ M ( − Γ P Q D M F P Q ζ ). The sum of these two variations cancels if one uses theBianchi identity D [ M F P Q ] = 0. The variation of A M in the covariant derivative in the Dirac action cancelsseparately due to the 3-spinor identity ( ¯ ψ a Γ M ψ b )(¯ ζΓ M ψ c ) f abc = 0 which holds in 3,4,6 and 10 dimensions. + = C − ∗ Γ . These charge conjugation matrices do not depend on the signature of space-time and obey the relation C − ∗ Γ = ± ( ∗ Γ ) T C − with − sign in d = 10 , d = 4. The transposition depends on the dimension and leads to ( C ± ) T = ± C ± in d = 10, ( C ± ) T = ∓ C ± in d = 6, and finally ( C ± ) T = − C ± for d = 4. Explicitly, the chargeconjugation matrix C − is given by C − ⊗ C − where C − = γ γ = (cid:32) (cid:15) αβ (cid:15) α (cid:48) β (cid:48) (cid:33) , C − = i ˆ γ ˆ γ ˆ γ = (cid:32) δ AB δ AB (cid:33) . (8.11)Upon compactification to Euclidean d = (4 ,
0) space, the 10-dimensional Lorentz group SO (9 ,
1) reduces to SO (4) × SO (5 ,
1) with compact space-time group SO (4) and R -symmetry group SO (5 , Ψ in ten dimensions with16 (complex) nonvanishing components decomposes as follows into 8 and 4 componentchiral-chiral and antichiral-antichiral spinors Ψ = (cid:32) (cid:33) ⊗ (cid:32) λ α,A (cid:33) + (cid:32) (cid:33) ⊗ (cid:32) λ α (cid:48) ,A (cid:33) , (8.12)or more explicitly ψ T = [( λ α , λ α , λ α , λ α , , , , , (0 , , , , , λ α (cid:48) ); (0 , ¯ λ α (cid:48) ); (0 , ¯ λ α (cid:48) ); (0 , ¯ λ α (cid:48) )] . (8.13)Here λ α,A ( α = 1 ,
2) transforms only under the first SU (2) in SO (4) = SU (2) × SU (2), while¯ λ α (cid:48) ,A changes only under the second SU (2). Furthermore, ¯ λ α (cid:48) ,A transforms in the complexconjugate of the SO (5 ,
1) representation of λ α,A . To understand this latter statement,note that the SO (5 ,
1) generators are ˆ M ab = (ˆ γ a ˆ γ b − ˆ γ b ˆ γ a ), and ˆ γ is antihermitian.Furthermore, ˆ γ , ˆ γ , ˆ γ and ˆ γ are purely imaginary. Thus(ˆ γ a ) ∗ = − ˆ S ˆ γ a ˆ S − ; ˆ S = ˆ γ ˆ γ . (8.14)This matrix ˆ S is not the charge conjugation matrix. The Lorentz generators ˆ M ab and ˆ S are block diagonal ˆ M ab = (cid:32) Σ ab
00 ¯ Σ ab (cid:33) ; ˆ S = (cid:32) S S (cid:33) , (8.15)where Σ ab = ( Σ a ¯ Σ b − Σ b ¯ Σ a ) and ¯ Σ ab = ( ¯ Σ a Σ b − ¯ Σ b Σ a ) while S = − η η = η . Itfollows that SΣ a S − = − ( Σ a ) ∗ ⇒ SΣ ab S − = ( Σ ab ) ∗ , ¯ Σ b S − = − ( Σ b ) ∗ ⇒ S ¯ Σ ab S − = ( ¯ Σ ab ) ∗ . (8.16)Thus the two spinor representations of SO (5 ,
1) are each pseudoreal (they are not realsince S is antisymmetric), but they are not equivalent to each other. For SO (6) (cid:39) SU (4),the two spinor representations are of course complex and inequivalent to each other. For SO (3 ,
1) the opposite is the case: there the two spinor representations are complex, andequivalent to each other under complex conjugation, ( σ µν ) ∗ = σ ¯ σ µν σ because ˆ S = γ isoff-diagonal.Substituting these results, the Lagrangian reduces to L N =4 E = 1 g tr (cid:40) F µν F µν − i ¯ λ α (cid:48) A (cid:54) ¯ D α (cid:48) β λ β,A − iλ Aα (cid:54) D αβ (cid:48) ¯ λ β (cid:48) ,A + (cid:0) D µ ¯ φ AB (cid:1) (cid:0) D µ φ AB (cid:1) − √ φ AB (cid:8) λ α,A , λ Bα (cid:9) − √ φ AB (cid:110) ¯ λ α (cid:48) A , ¯ λ α (cid:48) ,B (cid:111) + (cid:2) φ AB , φ CD (cid:3) (cid:2) ¯ φ AB , ¯ φ CD (cid:3) (cid:41) , (8.17)where we still use the definition ¯ φ AB ≡ (cid:15) ABCD φ CD . These scalars come from the ten-dimensional gauge field, and can be grouped into φ AB = √ Σ a AB A a , where A a are the firstsix real components of the ten dimensional gauge field A M . Using η ab = { √ Σ aAB , √ ¯ Σ bAB } with η ab = ( − , +1 , +1 , +1 , +1 , +1) the vector indices are turned into SU (4) indices. Writ-ing the action in terms of the 6 scalars A a , one of these fields, say A , has a different signin the kinetic term, which reflects the SO (5 ,
1) symmetry of the theory. In the basis withthe φ AB fields, we obtain formally the same action for the Minkowski case by reducing ona torus with 6 space coordinates, but the difference hides in the reality conditions whichwe will discuss in the next subsection. The action is invariant under the dimensionallyreduced supersymmetry transformation rules δA µ = − i ¯ ζ α (cid:48) A ¯ σ µ α (cid:48) β λ β,A + i ¯ λ β (cid:48) ,A σ αβ (cid:48) µ ζ Aα ,δφ AB = √ (cid:16) ζ α,A λ Bα − ζ α,B λ Aα + (cid:15) ABCD ¯ ζ α (cid:48) C ¯ λ α (cid:48) ,D (cid:17) ,δλ α,A = − σ µν αβ F µν ζ β,A − i √ ζ α (cid:48) ,B (cid:54) D αα (cid:48) φ AB + (cid:2) φ AB , ¯ φ BC (cid:3) ζ α,C ,δ ¯ λ α (cid:48) ,A = − ¯ σ µν β (cid:48) α (cid:48) F µν ¯ ζ β (cid:48) ,A + i √ ζ α,B (cid:54) ¯ D α (cid:48) α ¯ φ AB + (cid:2) ¯ φ AB , φ BC (cid:3) ¯ ζ α (cid:48) ,C . (8.18)Again, these rules are formally the same as in (8.4). Note that the indices A, B are loweredby complex conjugation, but the spinor indices α and α (cid:48) are lowered by (cid:15) - symbols. The Majorana-Weyl condition (8.6) on Ψ leads in four-dimensional Euclidean space toreality conditions on λ α which are independent of those on ¯ λ α (cid:48) , namely, (cid:0) λ α,A (cid:1) ∗ = − λ β,B (cid:15) βα η BA , (cid:0) ¯ λ α (cid:48) ,A (cid:1) ∗ = − ¯ λ β (cid:48) ,B (cid:15) β (cid:48) α (cid:48) η ,BA . (8.19)61hese reality conditions are consistent and define a symplectic Majorana spinor in Eu-clidean space. The SU (2) × SU (2) covariance of (8.19) is obvious from the pseudoreality ofthe of SU (2), but covariance under SO (5 ,
1) can also be checked (use [ η a , ¯ η b ] = 0). Sincethe first Σ matrix has an extra factor i in order that ( Γ ) = −
1, see (8.8), the realitycondition on φ AB involves η AB (cid:0) φ AB (cid:1) ∗ = η AC φ CD η DB . (8.20)The Euclidean action in (8.17) is hermitean under the reality conditions in (8.19) and(8.20). For the σ -matrices, we have under complex conjugation (cid:16) σ αβ (cid:48) µ (cid:17) ∗ = σ µ αβ (cid:48) , (¯ σ µ α (cid:48) β ) ∗ = ¯ σ α (cid:48) βµ . (8.21)Due to the nature of the Lorentz group the involution cannot change one type of indicesinto another, as opposed to the Minkowskian case. We have seen in previous sections that the instanton measure on the moduli space forpure SU (2) gauge theory with one anti-instanton ( k = −
1) is given by (dropping overallmultiplicative factors of two and π )d M ∝ d x d ρρ g M P V e − π g = (cid:18) d x d ρρ g (cid:19) e − π g +4 ln( ρM PV ) . (9.1)The one-loop corrections coming from the determinants further modify the factor 4 into4 − = , see (7.19), and in addition yield some constants in the exponent. The integralover ρ , the instanton size, is clearly nonsingular for small ρ as long as asymptotic freedomholds , but for large ρ it diverges severely. However, in a Higgs model, the mass term forthe gauge bosons ( L = − A µ g v if there are no instantons) yields further terms of theform − h L cl (Higgs) = − h π v ρ + . . . . (9.2)Thus for spontaneously broken gauge theories the ρ integral acquires a Gaussian cut-off,and yields a finite result. This solves the large- ρ problem for the electroweak interactions.For QCD the situation is more complicated; in fact, the large- ρ problem is presumablyintimately related to confinement. We now give some details. One integrates ρ up to the renormalization scale µ , and instantons with scale ρ yield the prefac-tor exp( − π /g ). The g in this prefactor depends on ρ , not on µ . One finds then that g ( ρ ) =8 π / ( − β ln( ρ Λ)) where β = − C ( G ) + · · · is negative if asymptotic freedom holds. So, if − β + 3 ≥ ρ = 0. SU (2) Higgs doublet is given by L H = D µ ϕ ∗ D µ ϕ + λ ( ϕ ∗ ϕ − v ) D µ ϕ = ∂ µ ϕ + A µ ϕ ; ϕ = (cid:32) ϕ + ϕ (cid:33) ; A µ = A aµ τ a i . (9.3)With < ϕ > = v , the ordinary Higgs effect in Minkowski space gives a mass term L = − A µ v for the vector bosons . We could also discuss other representations for the Higgsfield but the analysis is very similar, and a doublet is of course the most interesting case.In Euclidean space we take for A µ the regular selfdual instanton solution with k = 1 A µ = − ¯ σ µν x ν x + ρ . (9.4)We next solve the ϕ field equation in this instanton background. For general λ , an exactsolution to the coupled equations seems out of reach. We therefore drop the potentialterm and only require that | ϕ | → | v | at large | x | . So this is not an exact solution, butthe first term in an approximate solution. We shall discuss the higher-order terms later.As we shall show, the solution to the equation D µ D µ ϕ = 0 , | ϕ ( | x | → ∞ ) | = v is of theform ϕ = f ( r ) (cid:32) − ix + x x + ix (cid:33) . This clearly looks awkward, and a more covariant way toconstruct the solution is to write ϕ as ϕ = (cid:32) ϕ + − ( ϕ ) ∗ ϕ ( ϕ + ) ∗ (cid:33) (cid:32) (cid:33) , (9.5)and to make the ansatz ϕ = vf ( x ) (cid:16) ¯ σ µ x µ / √ x (cid:17) (cid:32) (cid:33) , (9.6)with f ( x ) → x → ∞ . (Recall that one can always write (cid:0) ϕ + ϕ (cid:1) as √ ( σ + i(cid:126)χ · (cid:126)σ ) (cid:0) (cid:1) and this yields the form of ϕ given in (9.5) up to an inessential factor i ).The function f ( x ) satisfies a second-order differential equation, but we do not analyzethis equation, but present the result and check that it solves D µ D µ ϕ = 0: ϕ = v (cid:115) x x + ρ ¯ σ µ x µ √ x (cid:32) (cid:33) = v (cid:112) x + ρ ¯ σ µ x µ (cid:32) (cid:33) . (9.7) One usually decomposes ϕ into ϕ = √ ( σ − iχ ), see below, with < σ > = v σ . Then v = v σ , andthe mass of the vector boson is m A = gv . This is ’t Hooft’s approach [4]. Note that the field equation for A µ is not restricted due to thebackreaction of the Higgs field. Affleck [72] considered instead the case v = 0, λ arbitrary, in which casethe usual instanton solution together with ϕ = 0 solves the coupled equations. Both approached yieldequivalent results. ¯ σ µ x µ √ x (cid:0) (cid:1) has unit norm. It is straight-forward to check that this expression for ϕ satisfies the field equation. Namely, omittingthe overall factor v and the spinor (cid:0) (cid:1) one finds D µ ϕ = ∂ µ ϕ + A µ ϕ = ¯ σ µ ( x + ρ ) − x µ x ν ¯ σ ν ( x + ρ ) / − (¯ σ µν x ν )(¯ σ ρ x ρ )( x + ρ ) / = ¯ σ µ ρ / ( x + ρ ) / D µ D µ ϕ = ∂ µ D µ ϕ + A µ D µ ϕ = − σ µ x µ ρ − (¯ σ µν x ν )¯ σ µ ρ ( x + ρ ) / = 0 . (9.8)Having found the solution of the field equation of the Higgs scalar in the background ofan instanton, we now substitute it into the action to find the corrections to the classicalaction. The kinetic term only yields a surface integral due to partial integration (cid:90) D µ ϕ † D µ ϕ d x = (cid:90) dΩ µ ( ϕ † D µ ϕ )= lim x →∞ π ( x ) / v (cid:32) (cid:33) T σ ν x ν (cid:112) x + ρ √ x x τ ¯ σ τ ρ ( x + ρ ) / (cid:32) (cid:33) = 2 π v ρ . (9.9)This is the extra term mentioned in (9.2).However, the contribution of the term with the potential is divergent λ (cid:90) ( ϕ ∗ ϕ − v ) d x = λ (cid:90) (cid:18) v ρ x + ρ (cid:19) d x = ∞ . (9.10)The reason for this divergence is clear: we did not solve the full field equation, but rathertook the instanton solution of pure Yang-Mills theory, and solved the field equation for thescalar in this background, omitting the potential term.We enter here the difficult area of “constrained instantons” [72, 73]. There does notexist an exact and stable solution of the coupled field equations, as can be shown asfollows. Suppose there was a solution with ϕ (cid:54) = 0, and a finite but nonvanishing actionfor the scalars. If one replaces A µ ( x ) by aA µ ( ax ) and ϕ ( x ) by ϕ ( ax ) (which preserves theboundary condition | ϕ | → v ) then the action becomes upon also setting ax = yS cl ( a ) = (cid:90) d y (cid:20) − g tr F µν ( y ) + 1 a | D µ ϕ ( y ) | + 1 a λ ( ϕ ∗ ( y ) ϕ ( y ) − v ) (cid:21) . (9.11)Note that all three terms in the action are positive. Replacing A µ ( x ) by aA µ ( ax ) for a nearunity amounts to a particular small variation of A µ , and similarly for ϕ . So one can make64he value of the action slightly smaller by making a slightly larger then unity. This provesthat no solution exists. In fact, if a tends to infinity, we approach the bound S = 8 π /g ,but this bound can never be reached. The expression for aA µ ( ax ) is equal to the instantonsolution with ρ replaced by ρ/a , and for a → ∞ we get a zero-size instanton. That leavesopen the possibility that a local minimum might stil exist, but detailed analysis shows thatthis is not the case. This scaling argument is called Derrick’s theorem [74], and often yieldsvaluable information without having to perform integrals.One can still use an approximate solution to find a large part of the contributions to thepath integral, and this approximate solution is obtained by first inserting a constraint intothe path integral which yields an exact solution, and then to integrate over this constraint.The idea is as follows. There are one or at most a finite number of directions in fieldspace along which the action decreases (“destabilizing directions”, in our SU (2) model thedirections parametrized by a ). Deformations in all other directions increase the action. Theconstraint prevents deformations in the destabilizing directions, and on first minimizes theaction with the constraint present. The solution is called the constrained instanton. It lookslike the instanton for pure Yang-Mills theory at short distances but decays exponentiallyat large distances. It has a particular value of ρ . Finally, one integrates with the measurefor the zero modes over all values of ρ . The expectation is that this should capture mostof the path integral, even though one is not expanding around a solution of the theorywithout constraint. For the SU (2) instanton one may add a term σ (cid:82) d x [tr F − c ρ − ]to the action to constrain deformations in the direction of the gauge zero mode ( ∂/∂ρ ) A cl µ ,and a term σ [ (cid:82) d x ( ϕ ∗ ϕ − v ) − c ρ − ] to freeze deformations in the directions of thematter zero mode ( ∂/∂ρ ) ϕ cl , with ϕ cl given by (9.7). One might fix the values of c and c such that the constraint is satisfied for the instanton solution and ϕ in (9.7). The Lagrangemultipliers σ and σ are then fixed order by order in perturbation theory, by requiringsuitable boundary conditions for the deformations.The result is that one can make an expansion of the full approximate solution in termsof ρv and finds then the following results in the singular gauge [72, 73, 32]:(i) inside a core of radius ρ = m W where m W = gv , the approximate solution given in (9.7)is still valid(ii) far away the solution decays exponentially, A µ ∼ exp( − m W | x | ) and | ϕ − v | ∼ exp( − m H | x | )with m H = 2 √ λv .(iii) the integral over | D µ ϕ | has the same leading term 2 π ρ v + O (cid:16) λ ( vρ ) ln( vρ √ λ ) (cid:17) ,but the potential term is now convergent and yields a result O (cid:16) λ ( vρ ) ln( vρ √ λ ) (cid:17) .Hence, the Higgs effect indeed solves the large ρ problem, and asymptotic freedom solvesthe small ρ problem. Constrained instantons are also relevant for N = 1 ,
10 Instantons as most probable tunnelling paths
Instantons of nonabelian gauge theories can be interpreted as amplitudes for tunnellingbetween vacua in Minkowski space with different winding numbers Q . We shall deter-mine a path in Minkowski spacetime which yields the “most probable barrier tunnellingamplitude”. We follow closely [76], but related work is found in [77, 78].We begin with one particular path A I,µ = { (cid:126)A I ( (cid:126)x, t ) , A I, ( (cid:126)x, t ) } from which we constructa class of paths which all differ by how fast one goes from one configuration at t to thenext at t . Namely, we make a coordinate transformation from t to λ ( t ) in Minkowskispacetime and consider the following collection of paths (cid:126)A ( λ ) I ( (cid:126)x, t ) = (cid:126)A I ( (cid:126)x, λ ( t )) ; A ( λ ) I, ( (cid:126)x, t ) = A I, ( (cid:126)x, λ ( t )) ˙ λ ( t ) (10.1)(Often one works in the temporal gauge A ( λ )0 = 0 because this makes the physical inter-pretation clearer. All our results are, however, gauge invariant). The case λ ( t ) = t yieldsthe original path, but different λ ( t ) yield paths which all run through the same sequenceof 3-geometries (cid:126)A I ( (cid:126)x, t ) , (cid:126)A I ( x, t ) , (cid:126)A I ( (cid:126)x, t ) . . . but at different speeds. The variable λ ( t )can be considered as a kind of collective coordinate which measures a kind of continuouswinding number because we will start with one winding number and end up with anotherwinding number. For t between t and t this continuous winding number is due to anintegral (cid:82) d x (cid:82) tt dt (cid:48) ∂ µ j µ over a surface where A µ is not everywhere pure gauge. Only for t = t and t = t does A µ everywhere on the surface become pure gauge and only at thesetimes the winding number is an integer. These initial and final configurations describevacua of the theory in Minkowski spacetime. We can also consider another particular path A II,µ = { (cid:126)A II ( (cid:126)x, t ) , A II, ( (cid:126)x, t ) } , and then we can in the same way create a second classof paths, parametrized again by the function λ ( t ). In this way we generate an infinitecollection of classes of paths.For a given class A ( λ ) µ ( (cid:126)x, t ), we can substitute (cid:126)A ( λ ) and A ( λ )0 into the action, and thenwe obtain, as we shall show, the Lagrangian for a point particle (one dynamical degree offreedom) L = 12 m ( λ ) ˙ λ − V ( λ ) (10.2)where m ( λ ) and V ( λ ) depend on the choice for A µ . We shall then determine for which m ( λ ) and V ( λ ) the tunnelling rate is maximal. The solution of this problem in Minkowskispace involves instantons in Euclidean space. A crucial role is played by the notion of awinding number in Minkowski space, so we first discuss this subject.66ne can define a winding number Q in Minkowski space in the same way as in Euclideanspace because Q does not depend on the metric (in technical terms it is an affine quantity) Q = − π σ (cid:90) σ F aµν F aρσ (cid:15) µνρσ d x = 132 π (cid:90) (tr F µν (cid:15) µνρσ F ρσ ) d x = − π (cid:90) tr (cid:126)E · (cid:126)B d x (10.3)where d x = d x d t and (cid:15) = +1, and we used that A µ = A aµ T a with T a = − i σ a so thattr( T a T b ) = − δ ab and the structure constants are given by [ T a , T b ] = (cid:15) abc T c , so f abc = (cid:15) abc .Furthermore, by definition E j = − F j and B j = (cid:15) jkl F kl . Because we are (and stay all thetime) in Minkowski space, (cid:15) = − (cid:15) = +1 and − F µν F µν = 2 F i − F ij . The integral istaken between two 3-dimensional hypersurfaces σ and σ at t and t .If at t the configuration A µ ( (cid:126)x, t ) describes a vacuum, it has by definition vanishingenergy. Since the energy density is given by H = ( (cid:126)E a ) + ( (cid:126)B a ) , vanishing energymeans F aµν = 0, hence A µ is pure gauge at t = t A µ ( (cid:126)x, t ) = e − α ( (cid:126)x,t ) ∂ µ e α ( (cid:126)x,t ) . (10.4)Similarly, at t we have A µ ( (cid:126)x, t ) = e − β ( (cid:126)x,t ) ∂ µ e β ( (cid:126)x,t ) . We now choose the temporal gauge A ( (cid:126)x, t ) = 0 . (10.5)Having fixed A = 0, there are still residual space-dependent gauge transformations possiblebecause they preserve the gauge A = 0. To check this statement is easy: A (cid:48) ( (cid:126)x, t ) = e − g ( (cid:126)x ) ∂ e g ( (cid:126)x ) = 0 . (10.6)We use these residual gauge transformations to set α ( (cid:126)x, t ) = 0. Then A µ ( (cid:126)x, t ) = 0 forall µ and all (cid:126)r . The gravitational stress tensor is T µν = F aµρ F aρν − η µν F aρσ F a,ρσ and T = ( E a ) + ( B a ) .One can also obtain T µν from canonical methods as follows. Evaluating H = p ˙ q − L with q = A j and p = − E j one finds upon using that ˙ A j = F j + D j A and partially integrating that H = (cid:82) (cid:2) (cid:8) ( E aj ) + ( B aj ) (cid:9) + A a ( D j E aj ) (cid:3) d x plus a boundary term. For solutions of the field equations suchas the vacuum, D j E j = 0. For configurations with finite energy ( E = O r ) the boundary term vanisheswhen A falls off like O ( r ). Moreover in the temporal gauge the last term vanishes. Actually, accordingto the Dirac formalism, the Gausz operator D j E j is a first-class constraint, and should be omitted fromthe Hamiltonian. Thus, H = (cid:82) (cid:2) ( E aj ) + ( B aj ) (cid:3) d x also according to canonical methods. With A j ( (cid:126)x, t ) = e − α ( (cid:126)r,t ) ∂ j e α ( (cid:126)x,t ) we get A (cid:48) j = e − g ( (cid:126)x ) e − α ( (cid:126)x,t ) ∂ j (e α ( (cid:126)x,t ) e g ( (cid:126)x ) ) and clearly A (cid:48) j = 0 ifwe take e g ( (cid:126)x ) to be the inverse of e α ( (cid:126)x,t ) . t = t (such winding at one fixedtime is discussed below (10.13)), one can still gauge it away by a time-independent gaugetransformation, but then the winding at t = t increases by just the same amount. This isas it should be, because the total winding is gauge-invariant.We shall consider paths from σ to σ which at every time t have finite energy (finiteintegral (cid:82) ( E + B )d x ). This means that the energy density for fixed t must tend to zerofor | (cid:126)x | → ∞ (to make the integral (cid:82) ( E + B )d x convergent), hence at large | (cid:126)x | the gaugefields become pure gauge A µ ( (cid:126)x, t ) −−−−→ | (cid:126)x |→∞ e − α ( (cid:126)x,t ) ∂ µ e α ( (cid:126)x,t ) . (10.7)But since A ( (cid:126)x, t ) = 0, we see that α ( (cid:126)x, t ) is independent of t . Because α ( (cid:126)x, t ) = 0 weobtain α ( (cid:126)x, t ) = 0 for all t and | (cid:126)x | → ∞ . This means in particular that at t for large | (cid:126)x | the gauge fields tend to zero A j ( (cid:126)x, t ) −−−−→ | (cid:126)x |→∞ | (cid:126)x | all A j vanish allows us to compactify the 3-dimensionalspacelike hypersurfaces at fixed t into spheres S . The north pole of each sphere correspondsto all points with | (cid:126)x | = ∞ , and at this point on S all A j vanish. Thus, all 3-spaces atfixed t compactify to a sphere S . We summarize the results in a figure (10.9)Everywhere on the boundary of this cylinder the gauge fields vanish, except at the disk at t = t , but there A = 0 and A j are only pure gauge.We now return to Q . First of all, Q can be written as a total derivative, using the samealgebra as in Euclidean space Q = 18 π (cid:15) µνρσ (cid:90) ∂ µ tr [ A ν ∂ ρ A σ + 23 A ν A ρ A σ ]d x . (10.10)68we recall that tr [ A µ A ν A ρ A σ (cid:15) µνρσ ] = 0). Furthermore, since on the boundary F µν = 0, wecan replace ∂ ρ A σ by − A ρ A σ in (10.10). We then find Q = − π (cid:15) µνρσ (cid:90) d σ µ tr [ A ν A ρ A σ ] A ν = e − α ∂ ν e α . (10.11)Since A = 0, there is no contribution from the sides of the cylinder, and since A j = 0 atthe bottom, there is also no contribution from the bottom. Hence in the gauge we havechosen, all contributions to the winding come from the top of the cylinder: Q = − π (cid:15) ijk (cid:90) T r (e − α ∂ i e α )(e − α ∂ j e α )(e − α ∂ k e α )d x . (10.12)At the top of the cylinder the 3-space t = t compactifies to a sphere S (space). Themap from this 3-sphere into the group SU (2) is a map from one S to another S because(i) we can always compactify the R with coordinates (cid:126)x to an S and (ii) the gauge fieldsat | (cid:126)x | = ∞ are equal (and vanish) (10.13)The maps S (space) → S (group) in Minkowski space fall into equivalence classes witha winding number k ∈ Z , just as the maps of instantons in Euclidean space give maps from S (space) → S (group). In the latter case S (space) is the boundary of all of R whilehere it is the compactification of the whole R at t = t . It follows that Q = ± k k ∈ Z . (10.14) The matrix elements of any 2 × g = a µ σ µ with σ µ = { (cid:126)σ, I } and µ = 1 , , ,
0. Unitarity requires that g † = a ∗ µ σ µ equals g − , hence g † g = (cid:80) | a µ | + ( a ∗ j a k i(cid:15) jkl + a ∗ a l + a ∗ l a ) σ l = 1. Hence | a | + | a k | = 1 and the coefficients of σ l must vanish. The determinant yieldsdet g = a − a k , and since also | a | + | a k | = 1, requiring det g = 1 leads to a k = ± i | a k | and a = ±| a | .Then we are left with g = a I + ia k σ k with real a and a k satisfying a + a k = 1 which defines S .
69e now can draw a picture of the energy H = (cid:82) H d x at times t as we move from t = t to t = t . Initially and at the end one has H = 0, but in between we must have H >
H ≥
0) for the following reason. (10.15)There are no paths possible which connect the vacuum at t to the vacuum at t whichare solutions of the field equations because if F µν = 0 on σ (or σ ) and the field equationsare satisfied, one has F µν = 0 everywhere . But, if F µν would vanish everywhere, Q ∼ (cid:82) E · B d x would vanish, hence one could not change the winding number. The conclusionis that paths which go from one vacuum with winding number zero to another vacuumwith nonvanishing winding number necessarily have positive energy at some intermediatetimes.We are now ready to define a subset of paths which depend on one collective coordinate,and to which (we claim) we can restrict our attention. Consider first one given pathcorresponding to a fixed field configuration A j ( (cid:126)x, t ). Instead of this single path, we considerthe set of paths A ( λ ) j ( (cid:126)x, t ), as defined in (10.1). Each path is labeled by a different function λ ( t ), and is defined by A ( λ ) j ( (cid:126)x, t ) = A j ( (cid:126)x, λ ( t )) . (10.16)As we already discussed, for λ ( t ) = t we recover the original path, but for different λ ( t ) weobtain paths which run through the same 3-dimensional configurations (cid:126)A ( (cid:126)x, t ) , (cid:126)A ( (cid:126)x, t ) , (cid:126)A ( (cid:126)x, t ) . . . For the proof, note that if at t one has F µν = 0 and at all t one has D µ F µν = 0, then ∂ µ F µj = ∂ F j = 0 at t . Furthermore, the Bianchi identity D F ij + D i F j + D j F i = 0 yields ∂ F ij = 0. Hence ∂ F µν = 0 at t . Also ∂ j F µν = 0 because F µν = 0 at t = t for all x . Hence ∂ ρ F µν = 0 at t = t for all ρ, µ, ν . We can rewrite this as ( D ρ F µν ) = 0 at t = t . Next we repeat this analysis by notingthat also D µ ( D ρ F µν ) = 0 at t = t , because D µ ( D ρ F µν ) = [ D µ , D ρ ] F µν + D ρ ( D µ F µν ) and D µ F µν = 0everywhere. This shows that ∂ ( D ρ F j ) = 0 at t = t . To also show that ∂ ( D ρ F ij ) = 0 at t = t werewrite ∂ ( D ρ F ij ) = − ∂ D i F jρ − ∂ D j F ρi and then use D ( D i F µν ) = [ D , D i ] F µν + D i ( D F µν ) = 0. Inthis way we get ∂ n F µν = 0 for any n . Hence F µν = 0 at all t .
70t different speeds. For example if λ ( t ) is constant for some time interval, the correspond-ing (cid:126)A ( (cid:126)x, t ) do not change, but if λ ( t ) changes rapidly, the sequence of A ( (cid:126)x, t ) is traversedrapidly.Each path A j ( (cid:126)x, λ ( t )) should begin at A j ( (cid:126)x, t ) and end at A j ( (cid:126)x, t ), so we require λ ( t ) = t , and λ ( t ) = t , but between t and t the function λ ( t ) is arbitrary. We shalllater take t = −∞ and t = + ∞ , and then also require that λ ( t ) = −∞ and λ ( t ) = + ∞ .Given a path A ( λ ) j ( (cid:126)x, t ) we can compute the electric and magnetic fields − E j = F j = ∂ A ( λ ) j ( (cid:126)x, t ) = ∂A j ∂λ ( (cid:126)x, λ ( t )) ˙ λ because A ( (cid:126)x, t ) = 0 B i = 12 (cid:15) ijk F jk = 12 (cid:15) ijk ( ∂ j A k ( (cid:126)x, λ ( t )) + A j ( (cid:126)x, λ ( t )) A k ( (cid:126)x, λ ( t )) − j ↔ k ) . (10.17)The Lagrangian L = (cid:82) L d x with L = g tr F µν = − g tr ( (cid:126)E − (cid:126)B ) can then be writtenas L = 12 m ( λ ) ˙ λ − V ( λ ) ,m ( λ ) = − g (cid:90) tr (cid:32) ∂ (cid:126)A∂λ (cid:33) d x ≥ ,V ( λ ) = − g (cid:90) tr (cid:126)B d x ≥ λ ( t ) is p ( λ ) = ∂∂ ˙ λ L = m ( λ ) ˙ λ . Hence H = ( p ( λ )) m ( λ ) + V ( λ ) . (10.19)For a given path A ( λ ) j ( (cid:126)x, t ) one can plot H as a function of t , and one finds then the profilein figure (10.15).We have thus isolated a class of paths A ( λ ) ( (cid:126)x, t ) which depends on one collective coor-dinate λ ( t ). For one given A ( (cid:126)x, t ), this still yields an infinite set of paths, but all thesepaths run through the same set of 3-configurations A j ( (cid:126)x, t ) , A j ( (cid:126)x, t ) , . . . . These are, ofcourse, infinitely many other collective coordinates which describe a general path A j ( x, t ),but the idea is that λ ( t ) is the relevant coordinate to describe tunnelling, while the othercollective coordinates describe variations away from the paths A ( λ ) j ( (cid:126)x, t ) which give onlysmall corrections to the results obtained from λ ( t ). It is, of course, difficult to prove thisassertion; one could begin with two collective coordinates as a start, but even this wouldlead to a complicated analysis.The action for λ ( t ) in (10.18) can be viewed as the action for one point particle. Thisparticle feels the potential barrier V ( λ ), and to go from the vacuum at t = t with V ( λ ) =71 ( λ ) = 0 to the vacuum at t with also V ( λ ) = m ( λ ) = 0, we need tunnelling. Thetunnelling rate R in quantum mechanics is proportional to e − R where R = λ (cid:90) λ d λ (cid:112) m ( λ )( V ( λ ) − E ) , (10.20)with λ ( t ) ≡ λ = t , λ ( t ) ≡ λ = t and V ( λ ( t )) = V ( λ ( t )) = 0 and m ( λ ( t )) = m ( λ ( t )) = 0. We also set E = 0 because we consider tunnelling from one vacuum (with E = 0) to another.There is, of course, an important difference with ordinary quantum mechanics. Thepoint particle λ ( t ) feels a potential V ( λ ), but both are derived from the same object, thefields A j ( x, λ ( t )). In addition the mass is here “position”-dependent, m = m ( λ ). One canshow that in quantum mechanics the formula for R also holds if the mass m ( λ ) dependson the point particle λ ( t )). The crucial step is now to pose the question: for which set ofpaths (cid:126)A ( (cid:126)x, λ ( t )) is the tunnelling rate maximal? The tunnelling rate for the quantummechanical particle λ ( t ) can be described by Minkowski path integrals, so we ask: for which (cid:126)A ( (cid:126)x, t ) is there least destructive interference of the associated paths A ( (cid:126)x, λ ( t )) in the pathintegral? Clearly, V ( λ ) should be as small as possible, but it cannot be too small becauseit must produce winding.The tunnelling rate is e − R where according to (10.20) R = λ (cid:90) λ d λ g (cid:90) tr (cid:32) ∂ (cid:126)A∂λ (cid:33) d x (cid:18) g (cid:90) tr (cid:126)B d x (cid:19) / = 2 g t (cid:90) t d t [(tr (cid:90) (cid:126)E d x )(tr (cid:90) (cid:126)B d x )] / . (10.21)We replaced dλ by dt ˙ λ and brought ˙ λ inside the square root. The fields (cid:126)E and (cid:126)B stilldepend on λ ( t ). Since tr (cid:82) (cid:126)a ( (cid:126)x ) (cid:126)b ( (cid:126)x ) d x is an inner product, while (cid:82) tr (cid:126)E · (cid:126)B is proportionalto the winding number according to (10.3), we have the triangle inequality R ≥ g (cid:12)(cid:12)(cid:12) t (cid:90) t (tr (cid:126)E · (cid:126)B )d x (cid:12)(cid:12)(cid:12) = 8 π g | Q | . (10.22)Hence the tunnelling amplitude is bounded from above bye − R ≤ − e − π g | Q | . (10.23)The inequality is saturated when (cid:126)E is parallel to (cid:126)B at each vector (cid:126)x and at each time t : (cid:126)E ( (cid:126)x, t ) = α ( t ) (cid:126)B ( (cid:126)x, t ). The claim is that among all paths with the same Q , the pathswith the smallest R are the paths with (cid:126)E parallel to (cid:126)B .72et us discuss the meaning of this result. Paths which interpolate between vacua withdifferent winding number must produce electric and magnetic fields (cid:126)E and (cid:126)B in between atfinite (cid:126)x and t which cannot be too small, namely | (cid:82) ( E aj B aj )d x | should be equal to 8 π | Q | .On the other hand, the tunnelling rate is proportional to the length of E a times the lengthof B a , so to make the tunnelling rate as large as possible, the product of these lengthsshould be as small as possible. One could set up a variational problem for R under theconstraint that (cid:82) tr (cid:126)E · (cid:126)B d x be equal to 4 π Q , but we shall not work this out.The bound is reached, namely the tunnelling rate is maximal, when the set of paths A j ( (cid:126)x, λ ( t )) produces parallel electric and magnetic fields (cid:126)E ( (cid:126)x, λ ( t )) = α ( t ) (cid:126)B ( (cid:126)x, λ ( t )) . (10.24)Of course, α ( t ) can also be viewed as a function of λ ( t ) because λ ( t ) is just anotherparametrization of the time interval. Note that this condition does not change if onechanges the parametrization from λ ( t ) to another function λ (cid:48) ( t ), because under suchreparametrizations (cid:126)E scales by a constant factor ∂λ (cid:48) /∂λ , which cancels the Jacobian in(10.20) for this change of integration variables. We use this scaling property to select aparticular λ ( t ) such that (cid:126)E ( (cid:126)x, λ ( t )) = ± (cid:126)B ( (cid:126)x, λ ( t )). The property of (cid:126)E and (cid:126)B beingparallel is also a gauge-invariant property, and L and R are of course gauge-invariant. So,our characterization of paths with maximal tunnelling rate is gauge-invariant, as it shouldbe. Thus the use of temporal gauge did not restrict the generality of the results.We now can establish the connection between tunnelling and instantons. The fieldsfor which (cid:126)E and (cid:126)B in Minkowski space are parallel are closely connected to instantonsin Euclidean space. Namely, among the class of paths (cid:126)A ( (cid:126)x, λ ( t )) parametrized by λ ( t ),there is the path (cid:126)E ( (cid:126)x, λ ( t )) = (cid:126)B ( (cid:126)x, λ ( t )) (and another path with another λ (cid:48) ( t ) suchthat (cid:126)E ( (cid:126)x, λ ( t )) = − (cid:126)B ( (cid:126)x, λ ( t ))). If we then define Euclidean gauge fields A Eµ ( x, t ) by A Ej ( (cid:126)x, t ) = A j ( (cid:126)x, λ ( t )) and A E ( (cid:126)x, t ) = A ( (cid:126)x, λ ( t )) dλdt then this A Eµ ( (cid:126)x, t ) is self dual. Theparameter t is Minkowski time, but in the expressions for A Ej ( (cid:126)x, t ) we should interpret t asthe Euclidean time.Summarizing: the most probable tunnelling paths are given by the set of paths A j ( (cid:126)x, λ ( t ))with parallel (cid:126)E and (cid:126)B fields. A given class of paths with (cid:126)E parallel to (cid:126)B contains one pathwhich, when viewed as a configuration in Euclidean space, is an instanton. Conversely,given an instanton A Eµ ( (cid:126)x, t ) in Euclidean space, one can construct a corresponding set ofpaths A Mµ ( x, λ ( t )) in Minkowski space by setting A M, ( λ ) j ( x, t ) = A Ej ( (cid:126)x, λ ( t )) A ( M, ( λ )0 ( x, t ) = A E ( (cid:126)x, λ ( t )) ˙ λ . (10.25)As an example we take the Q = − A µ =73 σ µν x ν / ( x + ρ ), see (3.41), which yields the following set of paths in Minkowski space A ( λ )0 ( (cid:126)x, t ) = − i(cid:126)x · (cid:126)σ(cid:126)x + λ ( t ) + ρ ˙ λ ( t ) (cid:126)A ( λ ) ( (cid:126)x, t ) = iλ ( t ) (cid:126)σ − i(cid:126)x × (cid:126)σ(cid:126)x + λ ( t ) + ρ λ ( t → −∞ ) = −∞ λ ( t → + ∞ ) = + ∞ . (10.26)We are clearly not in the temporal gauge, but since our results are gauge-invariant, it doesnot matter which gauge we use. We still have A µ → | (cid:126)x | , so that we still havethe notion of winding as a map from S (space) into S (group) at each time.Straightforward calculation yields for the curvatures in Minkowski space F = ∂ A − ∂ A + [ A , A ] = 2 iρ σ ( (cid:126)x + λ + ρ ) ˙ λ ,F = ∂ A − ∂ A + [ A , A ] = 2 iρ σ ( (cid:126)x + λ + ρ ) . (10.27)Hence (cid:126)E = − iρ (cid:126)σ ( (cid:126)x + λ + ρ ) ˙ λ ; (cid:126)B = 2 iρ (cid:126)σ ( (cid:126)x + λ + ρ ) , (10.28)which depend on x = (cid:126)x + λ ( t ) ( not on (cid:126)x − t ). Hence, (cid:126)E is indeed parallel to (cid:126)B (infact, anti-parallel).The winding number Q can be written in two ways Q = − π ∞ (cid:90) −∞ [tr (cid:126)E · (cid:126)B d x ]d t = − π (cid:15) µνρσ (cid:90) ∂ µ tr [ A ν A ρ A σ ]d x . (10.29)In the latter expression Q receives only a contribution from the boundary, but in theformer expression we compute Q by integrating over all space and time. It is then naturalto define a t -dependent function by integrating only up to a time tq ( t ) = − π (cid:90) t −∞ (cid:20)(cid:90) tr (cid:126)E · (cid:126)B d x (cid:21) = − π (cid:90) λ −∞ d λ (cid:90) d x ρ [ (cid:126)x + λ + ρ ] = − (cid:90) λ −∞ ρ d λ ( λ + ρ ) / For example, the contribution to Q from the surface at t = t is proportional to (cid:82) t ( (cid:126)x + t )d x ( t + (cid:126)x + ρ ) which isnonvanishing. On the other hand, the contribution to Q from the sides of the cylinder converges for large | t | . − (cid:90) λ/ρ −∞ d y ( y + 1) / = − (cid:18) t − t (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) x − with x = λ (cid:112) λ + ρ . (10.30)Clearly, q ( t ) is gauge-invariant and has the following form (10.31)It only receives contributions from regions where (cid:126)E and (cid:126)B are nonvanishing, hence where A aµ is not pure gauge.To obtain the action for λ ( t ) in this example we evaluate L = − g tr ( (cid:126)E − (cid:126)B ) = 24 g ρ (cid:34) ˙ λ ( x + ρ ) − x + ρ ) (cid:35) . (10.32)Doing the space integral we obtain L = 12 m ( λ ) ˙ λ − V ( λ ) = 3 π ρ g ( λ + ρ ) / ( ˙ λ − , (10.33)where we used (cid:90) d x ( (cid:126)x + λ + ρ ) = 4 π ( λ + ρ ) / ∞ (cid:90) −∞ y d y ( y + 1) = 4 π
32 1( λ + ρ ) / . (10.34)In this example, we were dealing with a gauge with A (cid:54) = 0. We can map to a gauge inwhich A = 0 by a suitable large gauge transformation A (cid:48) µ = U − ( ∂ µ + A µ ) UU = exp (cid:34) i(cid:126)x · (cid:126)σ (cid:112) (cid:126)x + ρ arctg λ ( t ) (cid:112) (cid:126)x + ρ (cid:35) . (10.35)75ndeed, using the expression for A in (10.26) A = − i(cid:126)x · (cid:126)σ(cid:126)x + λ ( t ) + ρ ˙ λ ( t ) , (10.36)one finds that A (cid:48) = U − ( ∂ t + A ) U = U − ∂ t U + A vanishes A (cid:48) = i(cid:126)x · (cid:126)σ (cid:112) (cid:126)x + ρ
11 + λ ( t ) (cid:126)x + ρ ˙ λ ( t ) (cid:112) (cid:126)x + ρ + A = 0 , (10.37)where we used that A commutes with U . Of course, Q is gauge invariant because it canbe written as a trace over (cid:126)E · (cid:126)B but it is instructive to see what happens if one writes Q as a surface integral and makes a gauge transformation with U . On the boundary ofMinkowski space the A µ = V − ∂ µ V transform into ( V U ) − ∂ µ V U and the winding numberof
V U is the sum of the winding numbers of V and U . However, U is connected to theidentity element: U ≡ exp α [ i(cid:126)x · (cid:126)σ √ (cid:126)x + ρ arctg λ ( t ) √ (cid:126)x + ρ ] traces an orbit as α runs from 0 to 1which begins at the identity element and ends at U . Thus U does not produce any winding,and thus the answer for Q from the total derivative is the same, whether one uses a gaugein which A vanishes or a gauge in which A is nonvanishing. Note, however, that when A (cid:54) = 0 one gets contributions from the timelike part of the boundary of the spacetimecylinder.
11 False vacua and phase transitions
In spontaneously broken gauge theories, the potential has a local maximum and an absoluteminimum. These extrema form a metastable and a stable vacuum, respectively. If a systemis in the metastable vacuum at all points in spacetime, it could at some point and at sometime, say (cid:126)x = 0 and t = 0, make a quantum fluctuation to the stable vacuum. Thistransition costs energy, but if the region around x (“the bubble”) is large enough, theenergy needed for creation of a bubble (this energy is located in the boundaries of thebubble) is less than the energy gained by tunnelling to the lower vacuum (this energy isliberated in the volume of the bubble), and then the bubble will rapidly expand. In fact,since the rate of energy production increases the larger the bubble, the bubble will spreadthrough space, with accelerating speed, converting the false vacuum to a true vacuum. Asan application of this process one may consider the universe just after the Big Bang; athigh temperature the universe is in the symmetric vacuum, but as cooling due to expansionsets in the potential develops a lower (true) vacuum, and if for some reason the universeremains stuck in the false vacuum, one can study the decay of the universe towards thetrue (asymmetric) vacuum. We shall consider another example: the perturbed double-wellpotential, with two classically stable minima, but one minimum (the true vacuum) below76he other minimum (the false vacuum). We shall study the decay of the false vacuum inthis model into the true vacuum [79, 80]. We follow [81].As a preliminary to the calculation of the phase transition in field theory, we first revertto quantum mechanics and study the double-well. Let us pretend that we do not knowthat there are big differences between the double-well potential and the following potential.(11.1)We can then repeat the calculation of the nonperturbative corrections to the energy ofthe ground state. Already at this point it is clear that we should not blindly repeat allsteps because previously we were dealing with two perturbatively degenerate vacua, andthe kink-instantons provided the energy shift between both vacua. In the present case,the degeneracy is already broken at the classical level. Proceeding nevertheless we find aclassical solution of the Euclidean equation − ∂ x∂t + ∂V∂x = 0 describing a point particle x ( t )in the inverted potential and use path integral methods. (11.2)77he particle starts at t = −∞ in the point x = a , rolls to the point x = c , “bounces” attime t = X , and ends up at t = + ∞ at the same point x = a . Clearly, X is the collectivecoordinate for this classical solution x cl ( t ). We then get for the “one-bounce solution” T ≡ < x = a | e − h Hτ | x = a > = e h S cl τ (cid:112) − S cl I ,I = N (cid:90) n.z. d q ( τ )e h S (2) E with q ( ± τ / ) = 0 , (11.3)where we used the Faddeev-Popov trick, and “n.z.” indicates that the path integral is overthe solutions of the field equation for the fluctuations about x cl ( t ) in the space orthogonalto the almost-zero mode. Assuming again that I can be written as a factor K times thepath integral of the harmonic oscillator we get I = K (cid:114) ωπ ¯ h e − ωτ ; K = (cid:115) det( − ∂ t + ω )det (cid:48) ( − ∂ t + V (cid:48)(cid:48) ( x cl )) (11.4)Continuing without further thought we would sum over multi-bounces and obtain T = (cid:114) ωπ ¯ h e − ωτ ∞ (cid:88) n =0 ( √− S cl τ K e h S cl ) n n != ωπ ¯ h e − ωτ exp( Kτ e h S cl ) . (11.5)Using the same arguments as used before for the unperturbed double-well potential, wewould conclude that the ground state energy is given by E = 12 ¯ hω − ¯ hK e h S cl . (11.6)However, at this point we note that there are problems with this result(i) first a small problem: the nonperturbative correction is exponentially suppressed, henceit should be neglected compared to the perturbative correction.(ii) a more serious problem (actually a virture, as we shall see) is that K has a negativeeigenvalue. This is easy to prove: ddX x cl ( t − X ) is the zero mode fluctuation. It has amode because x cl bounces: unlike the kink, x cl ( τ ) moves first forward and then backwards,78ielding a kind of kink-antikink solution. (11.7)Hence there exists one mode for the fluctuations with lower eigenvalue and without a node,and since ∂∂X x cl ( x − X ) has zero eigenvalue, there exists an eigenfunction for the fluctuationwith negative eigenvalue. Thus the nonperturbative correction is imaginary, reflecting thefact that the perturbative ground state near x = 0 is nonperturbatively unstable ImE = ¯ h | K | e h S cl ≡ Γ / . (11.8)So, instantons (or rather bounces, still solutions of the classical field equations with finiteaction) yield in this case the width Γ of the unstable state.Having seen that in quantum mechanics the path integral approach to nonperturba-tive corrections to the vacuum energy leads to the correct result that the ground state isunstable, we now return to the problem of phase transitions.As a toy model for studying such decays we need a system with at least one spacecoordinate because bubbles have a finite extension in space. The simplest choice is a 1 + 1dimensional field theory. We choose the double-well potential with an extra term to destroythe degenacy between both minima. Since the double-well potential is symmetric under ϕ → − ϕ , the extra term should be antisymmetric, and if it is to be a small perturbationcompared to the leading λϕ term, we need either a term linear in ϕ or cubic in ϕ , or both.It simplifies the mathematics if we keep the local minima of the perturbed potential at thesame place as the minima of the unperturbated potential, namely at ϕ = ± µ/ √ λ . We arethen led to the following model L = 12 ˙ ϕ −
12 ( ϕ (cid:48) ) − λ (cid:18) ϕ − µ λ (cid:19) − B (cid:18) ϕ − µ λ ϕ (cid:19) + 23 B (cid:18) µ √ λ (cid:19) , (11.9)79here we take B small and positive. For constant ϕ , the solutions of the classical fieldequations occur at ∂V∂ϕ = λϕ (cid:18) ϕ − µ λ (cid:19) + B (cid:18) ϕ − µ λ (cid:19) = 0 , (11.10)and from this result it is clear that the values ϕ = ± µ/ √ λ are indeed extrema. Thepotential has the following form (11.11)It vanishes at ϕ = − µ/ √ λ because we added the constant B ( µ/ √ λ ) , but at ϕ = µ/ √ λ it is negative. Thus ϕ = − µ/ √ λ is the unstable vacuum and ϕ = µ/ √ λ is the stablevacuum. The value of the potential at the stable minimum is V (cid:16) ϕ = µ/ √ λ (cid:17) = − (cid:15) = − B (cid:16) µ/ √ λ (cid:17) . (11.12)There is a relative maximum a bit below the maximum of the symmetric potential V ( B =0 , ϕ ) at ϕ = 0; for small B it occurs at ϕ (cid:39) − B/λ and its value is µ /λ − (cid:15) + O ( B ).These results are intuitively clear: if one pulls ϕ down at µ/ √ λ by an amount (cid:15) , then themaximum at ϕ = 0 is pulled down half as much, and moves of course a bit to the left.In addition to the three solutions of the classical field equations with constant ϕ ( ϕ = − µ/ √ λ, ϕ = µ/ √ λ , and ϕ ∼ − B/λ ), there is an exact kink-antikink solution. This is clear80y inspection of the inverted potential (11.13)A ball at rest at ϕ = − µ/ √ λ at x = −∞ starts rolling down to the hill and up the otherhill; it reaches the point where V ( ϕ ) = 0 at x = 0 and then returns and comes to rest at ϕ = − µ/ √ λ at x = + ∞ . The classical solution ϕ cl ( x ) is thus a soliton of the followingform (11.14)We approximate ϕ cl ( x ) by the following expression ϕ cl ( x ) = µ √ λ (cid:104) tanh (cid:16) m x + X c ) (cid:17) − tanh (cid:16) m x − X c ) (cid:17) − (cid:105) . (11.15)This is a static soliton in 1 + 1 dimensions, which can also be viewed as an instanton in x -space. (In the quantum mechanical models we considered previously, we dealth withinstantons in Euclidean time). Near x = − X c the antikink is exponentially suppressed81nd the mass of the kink is M . Between the kink and antikink ϕ is equal to µ/ √ λ (upto exponentially suppressed corrections), and near x = X c we have an antikink with mass M . For large x we find the correct asymptotic value ϕ cl ( x → ±∞ ) = − µ/ √ λ . We fix thevalue of X c such that the total energy of ϕ cl ( x ) (which is the energy of the ball rolling upand down the hills in (11.13)) vanishes E = 2 M − (cid:15)X c = 0 , (11.16)where M = m λ is the classical mass of a single kink. Hence, the separation between thekink and antikink is given by 2 X with X = M/(cid:15) .The exact solution begins at V = 0, climbs the hill, and comes down on the other sidewhere it reaches the value V = 0, and then it returns, climbing the hill once more, andending at V = 0. The approximate solution comes down to V = − (cid:15) after climbing the hill,but it has more energy in the kink (and antikink) region, such that in both cases the totalenergy is zero.We now compute the transition amplitude from the unstable vacuum ϕ = − µ/ √ λ tothe kink-antikink solution (the bubble). Once a bubble has formed, it will rapidly grow(the kink and antikink move increasingly fast away from each other, i.e., X exponentiallyincrease).This is a tunnelling process because classically it is forbidden but quantum mechanicallyallowed. If the field ϕ at x = 0 starts making a transition from the metastable vacuumto the stable vacuum, it must first climb the potential barrier, but when it comes downin the true vacuum energy density − (cid:15) is gained. However, as we already mentioned, ittakes energy to distort the field in order to go from one vacuum to another; this is just theenergy (mass) of a kink and of an antikink. These energies are located at the boundary ofthe bubble (around the centers of the kink and the antikink). Once in a while there occursa quantum mechanical transition to a bubble which is large enough that (cid:15) X is larger than2 M ; in that case the bubble does not collapse but grows increasingly rapidly.Note that we do not tunnel from the state ϕ ( x ) = − µ/ √ λ to the state ϕ ( x ) = µ/ √ λ because the energy difference of these states is infinite (namely (cid:15) times the volume of x -space, so 2 L(cid:15) with L → ∞ ). When we discussed the unperturbed kink, the vacua ϕ = ± µ/ √ x were exactly degenerate, and in such cases the true vacuum is a linear combinationof these vacua which can be determined by tunnelling from one vacuum to another.The intermediate configuration with the kink and antikink moving away from eachother can be described by Lorentz boosting the kink to a velocity − v and the antikink toa velocity + vϕ cl ( x, t ) = µ √ λ (cid:20) tanh m (cid:18) x + X c + vt √ − v (cid:19) − tanh m (cid:18) x − X − vt √ − v (cid:19) − (cid:21) . (11.17)82or constant ˙ X the boost of the kink is again a solution because the field equation userelativistically invariant. However, since ˙ X itself is expected to change with time, wedenote X + ˙ Xt by λ ( t ) and obtain then ϕ cl ( x, t ) = µ √ λ (cid:34) tanh m (cid:32) x + λ ( t ) (cid:112) − ˙ λ (cid:33) − tanh m (cid:32) x − λ ( t ) (cid:112) − ˙ λ (cid:33) − (cid:35) . (11.18)The distance between the kink and antikink is now 2 λ ( t ). The Lagrangian for thisapproximate solution is obtained by substituting ϕ cl into the action. The calculation ofthe first two terms is straightforward. Taking twice the result for a single kink yields ∞ (cid:90) −∞ (cid:34)
12 ˙ ϕ − (cid:18) ∂ϕ∂x (cid:19) (cid:35) d x = 2 ∞ (cid:90) −∞ d x
12 1cosh (cid:18) m x + λ √ − ˙ λ (cid:19) µ λ m (cid:32) ˙ λ (cid:112) − ˙ λ + ˙ λ ¨ λ ( x + λ )(1 − ˙ λ ) / (cid:33) − µ λ m − ˙ λ . (11.19)The calculation of the contribution from the nonderivative terms splits into two parts:from the region between the kink and antikink we obtain a term (cid:15) λ , while from each ofthe two walls we find a term M (cid:112) − ˙ λ as we now explain. Around x = − λ and x = + λ ,the integral (cid:82) V ( ϕ )d x with ϕ = µ √ λ tanh m x + λ √ − ˙ λ can be evaluated as follows. The integral (cid:82) ∞−∞ U ( ϕ )d x with ϕ = µ √ λ tanh m ( x ∗ + λ ∗ ) with x ∗ = x √ − ˙ λ and λ ∗ = λ √ − ˙ λ is equalto ( (cid:82) U ( ϕ ( x ∗ ))d x ∗ ) (cid:112) − ˙ λ . From equipartion of energy for a static kink we know thatthe integral (cid:82) U ( ϕ ( y ))d y equals M . Thus (cid:82) around kink ( U )( ϕ cl )d x = M (cid:112) − ˙ λ .Hence, neglecting term with ¨ λ , we find L = − m λ m (cid:112) − ˙ λ ∞ (cid:90) −∞ d y cosh y + (cid:15) λ − M (cid:112) − ˙ λ = − M (cid:112) − ˙ λ + (cid:15) λ . (11.20)The Hamiltonian follows from p = ∂L∂ ˙ λ = M ˙ λ √ − ˙ λ and reads H = 2 M (cid:112) − ˙ λ − (cid:15) λ = (cid:112) p + 4 M − (cid:15) λ . (11.21)We can split H into a kinetic term K and a potential term VK = (cid:112) p + 4 M − M = 12 p /M + O ( p ) , ( λ ) = 2 M − (cid:15) λ . (11.22)This formula for V ( λ ) is valid when the bubble is reasonably large: when λ is larger thanthe kink size (when the bubble is larger than the thickness if its walls). For smaller x weexpect that V ( λ ) rises from 0 till a maximum value when the bubble is formed, and thendecreases as the bubble gets larger (11.23)The value X c corresponds to the classical solution, with energy E = 0 and constant X ,corresponding to the ball rolling in the inverted potential. For this case, p = 0. Quantumfluctuations with X < X c produce only bubbles which collapse since their potential energyis positive, but bubbles with X = X c are metastable (they have constant X = X c so p = 0),while for X > X c the bubble expands.We now treat H as the Hamiltonian of a point particle which sees the potential V ( λ )and has energy zero. We find with the WKB approximation for the tunneling amplitude A = exp (cid:104) − X c (cid:90) | p | d λ (cid:105) = exp (cid:104) − X c (cid:90) (cid:112) M − ( (cid:15) λ ) d λ (cid:105) , (11.24)where we used that H = 0 = (cid:112) p + 4 M − (cid:15) λ . Since X c = M(cid:15) , we have A = exp (cid:104) − M (cid:15) (cid:90) (cid:114) − (cid:15) M λ d (cid:16) (cid:15)M λ (cid:17) (cid:105) = exp (cid:104) − M (cid:15) (cid:90) (cid:112) − y d y (cid:105) = exp( − πM (cid:15) ) . (11.25)84ence, the rate of the transition to the true vacuum is exp − πM (cid:15) per second and per unitvolume. (To evaluate the integral we set y = cos ϕ ).We end this section with a few comments. 1. The decay of the false vacuum per unittime and per unit volume is of the form Γ /V = A e − B/ ¯ h (1 + O (¯ h )). We computed B . For A see [80, 76].2. We used energy conservation to determine how fast a bubble expands. However, weneglected radiation of mesons. In general, when the false vacuum collapses to the truevacuum, mesons will be created, and thus the bubble will expand less rapidly.3. Above we considered the critical bubble: a static solution of the classical field equationswhich describes a bubble which has just the correct form and size that it is metastable.For larger sizes there is no static solution, but one can consider the creation at t = 0 of alarge bubble which then expands. This is an initial value problem: ϕ ( x ) is given and also ∂ϕ∂t = 0 at t = 0. One can define the size of a bubble for example as Q = (cid:82) ∞−∞ ( ϕ + µ √ λ ) d x .Far away, ϕ = − µ √ λ , so Q is finite for bubbles. A problem we now want to solve is: giventhe size Q of a bubble, for which shape is its action minimal. (Minimal action in Euclideanspace means maximal tunnelling rate). This will yield a one-parameter parametrization ofbubbles; the parameter is a collective coordinate λ ( t ), and having found the solution, wecan then compare our ansatz in (11.15) and see how good the ansatz was. Mathematically,we can formulate this problem as a variational problem with a constraint. Introducing aconstant Lagrange multiplier α we consider the action for the variational problem L = −
12 ( ∂ x ϕ ) − λ (cid:18) ϕ − µ λ (cid:19) − B (cid:18) ϕ − µ λ ϕ (cid:19) + 23 B (cid:18) µ √ λ (cid:19) + 12 α (cid:18) ϕ + µ √ λ (cid:19) . (11.26)The equation of motion ∂∂x ∂ L ∂ϕ x − ∂ L ∂ϕ = 0 , (11.27)has a first integral due to equipartition of energy12 (cid:18) dϕdx (cid:19) = λ (cid:18) ϕ − µ λ (cid:19) + B (cid:18) ϕ − µ λ ϕ (cid:19) − B (cid:18) µ √ λ (cid:19) − α (cid:18) ϕ + µ √ λ (cid:19) . (11.28)The integration constant vanishes for bubbles. Introducing a field ˜ ϕ which vanishes forlarge x ˜ ϕ = ϕ + µ √ λ , ϕ = ˜ ϕ − µ √ λ , (11.29)85e obtain d ˜ ϕdx = (cid:114) λ ϕ (cid:115)(cid:18) ˜ ϕ − µ √ λ (cid:19) + 4 Bλ (cid:18)
13 ˜ ϕ − µ √ λ (cid:19) − αλ . (11.30)For α = B = 0 the solution is the kink, but for α (cid:54) = 0 we get bubbles. One can actuallysolve this equation exactly by using (see Gradhstein and Resznik, page 84, 2.266) (cid:90) d yy (cid:112) a + by + cy = 1 √ a arc cosh 2 a + byy √− ac , (11.31)which holds if a > b − ac >
0. This corresponds to 0 < α < µ . By writing thedifferential equation as (cid:90) d (cid:32)(cid:114) λ x (cid:33) = (cid:90) d ˜ ϕ ˜ ϕ (cid:114)(cid:16) µ λ − αλ − Bµλ √ λ (cid:17) + (cid:16) − µ √ λ + B λ (cid:17) ˜ ϕ + ˜ ϕ , (11.32)we obtain for the bubble with fixed size and minimum actioncosh (cid:32) √ a (cid:114) λ x (cid:33) = 2 a + b ˜ ϕ ˜ ϕ √ ac − b ˜ ϕ = 2 a √ ac − b cosh (cid:16) √ a (cid:113) λ x − x (cid:17) − b (cid:39) √ λ (2 µ − α ) √ α cosh( √ µ ( x − x )) + 2 µ . (11.33)The constant α lies in the domain 0 < α < µ . For α = 0 we find ˜ ϕ = µ √ λ (cid:16) or ϕ = µ √ λ (cid:17) while for α = 2 µ we find ˜ ϕ = 0 (cid:16) or ϕ = − µ √ λ (cid:17) . In between, we have bubbles of finiteextent; for small α the function ˜ ϕ remains constant for a long time (the bubble) and thenit falls rapidly off to zero (due to the cosh). This is the same behaviour as displayed byour ansatz in (11.15).
12 The strong CP problem
The vacua | n > of Yang-Mills theory in Minkowski space with winding number n all havethe same energy, namely zero (because they are vacua). We recall that at fixed time spacewas compactified to an S which was mapped to the S of the group manifold of SU (2).Since there is tunnelling as we have discussed, the physical vacuum is a linear combinationof all of them. Since they all appear on equal footing, we expect that the generator T forlarge gauge transformations which change the winding number, defined by T | n > = | n +1 > ,86ommutes with the Hamiltonian. Hence T maps the physical vacuum into itself. It followsthat T | vac > = e iϕ | vac > with ϕ some phase. The solution of this equation is | vac > ≡| θ > = (cid:88) n e inθ | n > . (12.1)Indeed T | θ > = (cid:80) e inθ | n + 1 > = e − iθ | θ > .Instead of using the infinite set of states in (12.1), we can work with the ordinary vacuum | > if at the same time we add a term L θ = − θ QCD g π tr F µν ∗ F µν (12.2)to the action. This term yields a factor e inθ in the action e i ¯ h S if one is in the vacuum withwinding number n . We shall set ¯ h = 1. Note that we are in Minkowski space and that L θ is hermitian.Strictly speaking, we should first make a Wick rotation to Euclidean space because wecan only discuss instantons in Euclidean space, but L θ has the same form in Euclideanspace: one gets a factor i from d x and another factor i from F i . Together with thefactor i ¯ h in e i ¯ h S one gets the same factor e inθ in Euclidean space. The θ -term is a totalderivative, and usually one discards total derivatives in Lagrangians because fields vanishat infinity, but for instanton backgrounds one finds of course a nonvanishing contributiondue to winding.The θ -term clearly violates parity P. It conserves charge conjugation symmetry C , henceit violates CP. The strong interactions described by QCD are not supposed to violate P orPC, hence θ QCD should be very small. However, the observed θ parameter contains morethan only θ QCD . There is a second origin for a θ -angle coming from the electroweak sector:the manipulations leading to the CKM matrix. Recall that the mass terms of the quarksin the Standard Model come from Yukawa couplings L = − (cid:88) m,n g ( qu ) mn (cid:32) ¯ q L,m ¯ q (cid:48) L,m (cid:33) T (cid:32) ( h ) ∗ − ( h + ) ∗ (cid:33) q R,n + g (cid:48) ( qu ) mn (cid:32) ¯ q L,m ¯ q (cid:48) L,m (cid:33) T (cid:32) h + h (cid:33) q (cid:48) R,n + h.c. , (12.3)where g ( qu ) are the Yukawa couplings to quarks, and h + , h are the two components of thecomplex SU (2) Higgs doublet. Furthermore m = 1 , , q denotesthe up quark while q (cid:48) denotes the down quark. When h gets a vacuum expectation value < h > = √ v , one obtains mass matrices M for the ( u, c, t ) quarks and M (cid:48) for the ( d, s, b )87uarks, where M mn = v √ g mn , M (cid:48) mn = v √ g (cid:48) mn . (12.4)These matrices are in general arbitrary complex 3 × × U L M U − R = diag( m u , m c , m t ) ≡ D ,U (cid:48) L M (cid:48) U (cid:48)− R = diag( m d , m s , m b ) ≡ D (cid:48) . (12.5)The mass matrix for the quarks becomes then diagonal with real masses¯ q L,m M mn q Rn = ( U L q L ) D ( U R q R ) (12.6)and similarly for ¯ q (cid:48) L,m M (cid:48) mn q (cid:48) Rn . So, the physical quarks are Q L = U L q L and Q R = U R q R ,and similarly for Q (cid:48) L and Q (cid:48) R .If one rescales q L to Q L , and q R to Q R , three things happen(i) the quark mass terms are diagonalized as we have discussed, yielding real physical quarkmasses(ii) a phase δ appears in the CKM matrix which describes electroweak CP violation(iii) a new term is produced in the action by the Jacobian for these chiral rescalings. Thisnew term is again proportional to (cid:82) F µν ∗ F µν d x with a coefficient which we call − θ EW .Hence, now the action contains the sum θ = θ QCD + θ EW L θ = − ( θ QCD + θ EW ) g π (cid:90) ( F aµν ∗ F aµν )d x . (12.7)There is no reason that θ strong = θ QCD + θ EW vanishes, yet, as we now discuss, this seemsto be the case.We can make a final chiral rescaling of the 3 light quarks (the u, d and s quarks) suchthat the θ -term is entirely removed. Rescaling the left-handed quarks by U (1) factorse iϕ u , e iϕ d and e iϕ s , the Jacobians for these rescalings yield a term − ( ϕ u + ϕ d + ϕ s ) g s π tr F µν ∗ F µν , (12.8)which cancels the θ -term if ϕ u + ϕ d + ϕ s = θ strong . Because the action is invariant exceptfor the mass terms, only the transformation of the mass terms yields a new term in theaction. In the diagonal mass term m u ¯ uu + m d ¯ dd + m s ¯ ss , (12.9) A complex matrix M can always be written as V H where V is unitary and H hermitian. This is thegeneralization to matrices of the decomposition z = e iϕ ρ of complex numbers. Then H can be diagonalizedby a unitary matrix, H = U R DU − R , and U L is given by V U R . ϕ u , ϕ d , ϕ s , a new term in the action L CP violation = iϕ u m u ¯ uγ u + iϕ d m d ¯ dγ d + iϕ s m s ¯ sγ s . (12.10)The ϕ ’s are only constrained by ϕ u + ϕ d + ϕ s = θ strong , so we can still choose them suchthat this new term is SU (3) V invariant. Namely if ϕ u = θm d m s m u m d + m u m s + m d m s , and cyclicallyfor ϕ d and ϕ s , then L CP violation = iθ strong m u m d m s m u m d + m u m s + m d m s (¯ uγ u + ¯ dγ d + ¯ sγ s ) . (12.11)This term is hermitian and SU (3) V invariant, but it violates P, and since it conserves C,it also violates CP. The original θ -term in the action in (12.2) has been transformed intothe masslike terms in (12.11). No longer does one have to deal with total derivatives, butan ordinary extra masslike term has appeared in the QCD action. There is no reason that θ strong should be small, but one can compute the electric dipole moment of the neutronwhich is nonzero if θ strong is nonzero, and since experimentally the electric dipole momenthas a very small upper bound, one finds that θ strong is incredibly small θ strong < − . (12.12)The problem why θ strong is so small is called the strong CP problem. Note that it hasnothing to do with the CP violation due to the phase δ in the CKM matrix, which is anelectroweak effect. Also the electroweak CP violation is very small; it can be parametrizedby the area of the unitarity triangles (each of the 6 unitarity triangles has the same area2 J in the Standard Model) J = (3 . ± . − . (12.13)
13 The U (1) problem In this section we discuss an application of instantons in QCD.In the 1960’s, in the absence of a renormalizable theory of the strong interactions,current algebra was developed as a method to derive information about matrix elementsof currents, mostly the vector and axial-vector Noether currents which correspond to the(approximate) rigid flavor symmetry of the up, down and strange quarks. In terms ofmodern QCD, the action for the strong interactions reads L = −
14 ( F aµν ) − (cid:88) ¯ ψ i (cid:54) ¯ Dψ i , (13.1)where i = 1 , ..., N f labels the flavors. One can consider either two very light quarks ( u and d ), or three rather light quarks ( u , d and s ). Decomposing the massless quarks into89eft-handed and right-handed parts, their action becomes L (quarks) = ¯ ψ i,L (cid:54) ¯ Dψ iL − ¯ ψ i,R (cid:54) ¯ Dψ iR . (13.2)It has clearly a rigid U L ( N f ) × U R ( N f ) symmetry group, where U L acts only on ψ iL and U R only on ψ iR . Instead of U L and U R we consider the vector part U V ( N f ) and the axial vectorpart U A ( N f ). The vector part transforms ψ iL and ψ iR the same way, while they transformoppositely under U A . The total number of symmetries and group parameters has notchanged, but physically U V and U A are very different. The SU (2) V part of the symmetryis realized in Nature, and yields the SU (2) classification scheme for quark hadroscopy.The U (1) V corresponds to baryon-number conservation which is also (very well) satisfied.The SU (2) A symmetry is spontaneously broken, and the corresponding Goldstone bosonsform an SU (2) multiplet of pseudoscalars (the pions and the η meson). One might beinclined to apply the same reasoning to the U (1) A symmetry, and argue that it, too, mustbe spontaneously broken because there is no doubling of multiplets with opposite parityobserved in nature. However, the U (1) A symmetry is explicitly violated by the presence ofinstantons in QCD, leading to the instanton-induced six-fermion interaction in the effectiveaction. This solves “the U (1) problem” that no isoscalar Goldstone boson exists in Nature[82]. There is a pseudoscalar meson, the η with a mass of 478 M eV . It cannot be theGoldstone boson because from current algebra S. Weinberg has shown that the mass ofsuch a Goldstone boson has to be smaller than √ m π , far below the mass of the η meson[83]. (The η meson can still be made of a quark and an antiquark, so the usual SU (2)scheme is still applicable - only this η meson is not a Goldstone boson) .The axial-vector isoscalar current associated with the U A (1) symmetry is j µ = (cid:80) i ¯ ψ i γ γ µ ψ i .It has an Adler-Bell-Jackiw chiral anomaly ∂ µ j (5) µ = N f g π (cid:15) µνρσ F aµν F aρσ , (13.3)where F aµν denotes the field strengths of the gluons, and g is the QCD coupling constant.In a QCD instanton background, integration over spacetime yields (cid:90) ∂∂t Q d t = N f k , (13.4)where k is the winding number. To make sense of this equation one should first integratein Euclidean space to obtain a non-vanishing expression for the right -hand side in terms ofthe winding number k of the QCD instanton, and then Wick-rotate so that the left-handside can be written as (cid:82) d x∂ µ j µ ∝ (cid:82) ∂∂t Q d t . The conclusion is that Q is not conservedbecause k can be different from zero. For further discussion of the U (1) problem we referto [82], [84, 85], or to the lecture notes by Coleman in [2]. One can extend this discussion to U L (3) × U R (3) with pions, kaons and η now 8 Goldstone bosons,and the η (cid:48) with mass 958 M eV taking the place of η . This η (cid:48) is an SU (3) singlet. In this section we present an application of instantons to the gauge fields of the electroweaksector of the Standard Model.In an instanton background with winding number k , massless (or approximately mass-less) fermions in the fundamental representation of SU ( N ) have | k | zero modes, see (4.37).In the electroweak SU (2) w × U (1) theory (the subscript w stands for weak), quarks andleptons are in the fundamental representation (doublets) of SU (2) w . In Euclidean spacethe integration over zero modes of these quarks and leptons has dynamical consequenceswhich we shall derive, but of course real quarks and leptons live in Minkowski space and notin Euclidean space. We assume that the Green functions in Minkowski spacetime can beobtained from those in Euclidean space by analytic continuation. Ideally we should provethat the Euclidean results give the main contribution to processes in Minkowski space inthe same way as this was shown for tunnelling, but as far as we know this has not been done.Since processes involving electroweak instantons are suppressed by a factor exp (cid:16) − h π g | k | (cid:17) with g the electroweak SU (2) coupling constant, we only consider instantons with | k | = 1made from W + , W − and W bosons. Then the left-handed quark doublets (cid:0) ud (cid:48) (cid:1) and (cid:0) cs (cid:48) (cid:1) each have 3 zero modes because there are 3 colors, while the lepton doublets (cid:0) ν e e − (cid:1) L and (cid:0) ν µ µ − (cid:1) L each have one zero mode. The primes on d (cid:48) and s (cid:48) denote Cabibbo-rotated quarks d (cid:48) = d cos θ c + s sin θ c s (cid:48) = s cos θ c − d sin θ c (14.1)with θ c = 13 the Cabibbo angle. As we shall explain, this Cabibbo rotation makes itpossible for a neutron and a proton (six quarks together) to decay into two antileptons [5] p + n → e + + ¯ ν µ (or µ + + ¯ ν e ) . (14.2)In these instanton- induced processes, the electron number E , muon number M , up plusdown number, and charm plus strangeness number change as follows∆ E = ∆ M = 1 , ∆ u + ∆ d (cid:48) = 3 , ∆ c + ∆ s (cid:48) = 3 . (14.3)The decay of a proton with ( u, u, d ) and neutron with ( u, d, d ) quarks into e + and ¯ ν µ , orinto µ + and ¯ ν e , can be described by a local vertex operator with 3 up-quark fields withdifferent colors from (cid:0) ud (cid:48) (cid:1) doublets, and 3 down-quark fields also with different colors from (cid:0) cs (cid:48) (cid:1) doublets, and further one field from each of the two lepton doublets. This operator is ofcourse nonrenormalizable, but it can be used in effective field theories for phenomenologicalpurposes. Although this efective operator is derived from field theory in the sector with an91nstanton, once it is obtained one can add it to the effective action and then forget aboutthe existence of instantons. We now derive these results.The U (1) A symmetry has at the perturbative level an anomaly. There are trianglegraphs with an anomaly: one vertex of the triangle graph is given by j (5) µ = (cid:80) s ¯ ψ s γ γ µ ψ s (where s = 1 , . . . , N f and N f is the number of flavors, 3 in our case if we restrict ourattention to the lightest quarks u, d and s ). The one-loop perturbative chiral anomaly isthen given by ∂ µ j (5) µ = iN f g π G aµν ∗ G aµν , (14.4)where G aµν is the W -boson field strength and g the coupling constant of the SU (2) weakinteractions. (This is thus the abelian flavor U (1) A anomaly. The nonabelian anomaly forthe rigid flavor group vanishes because it is proportional to the trace of T a of the flavorgroup, which vanishes).If one integrates over space and time, the anomaly equation becomes (cid:90) dd t Q (5) ≡ (cid:90) dt dd t (cid:90) d x ( ij ) = 2 N f k . (14.5)The instanton number k counts the number of left-handed fermions minus the numberof right-handed fermions, and in ordinary perturbation theory (with k = 0) for masslessquarks, this difference is thus conserved. However, in an instanton background ( k (cid:54) = 0), thechiral charge of the vacuum at t = −∞ changes to a different chiral charge of the vacuumat t = + ∞ : ∆ Q (5) = 2 N f k . The conclusion is that the perturbative anomaly, and theviolation of the axial charge which occurs when one tunnels from one vacuum to another,are related! Both are different aspects of the same chiral anomaly. The perturbativeanomaly occurs when fields are small, so the winding vanishes and one is in the k = 0sector. The nonperturbative anomaly is due to the same axial-vector current but nowin the background of instantons which cannot be viewed as small and tending to zero atinfinity, since they must produce winding.One may at this point wonder whether the Higgs effect which gives the W -bosons amass, in such a way that they vanish exponentially at large distances, does at the sametime destroy the concept of winding. There is no contradiction. When we discussed thelarge instanton problem, we chose the regular gauge for the instanton to simplify thecalculations. However, exponential fall-off only occurs in the singular gauge. In that case,the winding takes place at the origin, as we discussed in the introduction. In this sectionwe use the regular gauge and then there is winding at infinity even in Higgs models.To saturate the integrations over the Grassmann collective coordinates, one needs 6 chi-ral quark fields in a correlator (one for each zero mode). Each field has a mode expansion92nto a zero mode and all nonzero modes, but the integration measure d K over the Grass-mann variable K in the mode expansion picks out only the zero mode. Then the integrationover collective coordinates gives as result the product of the 6 zero mode functions. If weput one SU (2) w doublet with one up quark and one down quark at a point x , a secondpair a x , and a third pair at x , and the instanton is at x , we find from (5.35) for largeseparations ( x >> ρ ) the factor
48 3 (cid:89) i =1 x i − x ) . (14.6)So if one computes some correlator in a theory with instantons, six quark fields from thecorrelator are needed to saturate the Grassmann integrals, and the remaining fields are thentreated as in ordinary field theory (with propagators and vertices). Thus instantons inducea term proportional to (cid:81) i =1 1( x i − x ) in the effective action which describes the annihilationof 6 quarks. Further there are σ matrices and other constants which are also due to thezero mode function.One can now construct an effective local 6-quark vertex V at a point x which yieldsthe same results in a theory without instantons as one obtains in a theory with instantonsif one integrates over the fermionic collective coordinates of the quarks. This vertex mustcontain 6 quark fields which contain the 6 different collective coordinates, hence it has theform V = u α, L u β, L u γ, L d δ, L d (cid:15), L d ζ, L T αβγδ(cid:15)ζ where T is a numerical tensor. Contraction of 6“probe-quarks” at positions x , x , x with V at x using ordinary flat space propagators x i − x ) for massless quarks in a trivial vacuum precisely reproduces the result for thecorrelation function of the 6 probe-quarks in an instanton background centered around x ,provided the form of T is correctly chosen.These new vertices lead to anomalies in the baryon currents and lepton currents. Inparticular, the rigid U (1) A symmetry is explicitly broken by the presence of the interaction V in the action, and as we discussed in the previous section, this solves the U (1) A broken.As we already mentioned, a proton and a neutron (two baryons equal six quarks) mayannihilate to form two antileptons (an e + or a µ + , and an anti-neutrino). However, due to A massless complex Dirac spinor contains two Weyl spinors which are decoupled from each other¯ ψ D / Dψ D = ¯ ψ L / Dψ L + ¯ ψ K / Dψ R . Each has a zero mode. However, since only left-handed quarks coupleto the W gauge fields, only left-handed quarks feel the presence of instantons, and so we neglect theright-handed quarks in this discussion. The down quark is contained in the s (cid:48) of the doublet ( c, s (cid:48) ). This is an SU (2) w doublet, and theinstanton is an SU (2) w instanton. Although the c quark is heavier than the s quark, one can still viewthem as massless compared to the scale 250 GeV of electroweak interactions. Massive spinors in an instantonbackground have no zero modes, as one may show by adding a mass term to (4.17) and (4.18). We assumethat for such a broken SU (2) w doublet there still exists approximately a zero mode. (cid:16) − π g | k | (cid:17) , where g is the SU (2) weak coupling constant,these processes are not observable.
15 Discussion
In this chapter we have reviewed the general properties of single Yang-Mills instantons,and have given tools to compute non-perturbative effects in (non-) supersymmetric gaugetheories. However, we have not discussed several other important or interesting topics: • Perturbation theory around the instanton: the methods described here enable us tocompute non-perturbative effects in the semi-classical approximation where the couplingconstant is small. It is in many cases important to go beyond this limit, and to study sub-leading corrections that arise from higher order perturbation theory around the instanton[58, 57]. Apart from a brief discussion about the one-loop determinants in section 7, wehave not really addressed these issues. • Multi-instantons: we have completely omitted a discussion of multi-instantons. Thesecan be constructed using the ADHM formalism [31]. The main difficulty lies in the explicitconstruction of the collective coordinates in an instanton solution and of the measure ofcollective coordinates beyond instanton number k = 2. However, it was demonstrated thatcertain simplifications occur in the large N limit of N = 4 SYM theories [25], where one canactually sum over all multi-instantons to get exact results for certain correlation functions.For reviews on the ADHM construction in super Yang-Mills theories, see e.g. [32, 25, 86].The same techniques were later applied for N = 2 , N non-supersymmetrictheories. For a review on instantons in QCD, see for instance [89]. A Winding number
For a gauge field configuration with finite classical gauge action the field strength musttend to zero faster than x − at large x . For vanishing F µν , the potential A µ becomes thenpure gauge, A µ x →∞ −→ U − ∂ µ U . All configurations of A µ which become pure gauge at infinityfall into equivalence classes, where each class has a definite winding number. As we nowshow, this winding number is given by k = − π (cid:90) d x tr ∗ F µν F µν , (A.1)where ∗ F µν = (cid:15) µνρσ F ρσ and T a are the generators in the fundamental representation of SU ( N ), antihermitean N × N matrices satisfying tr T a T b = − δ ab . This is the normalization94e adopt for the fundamental representation. The key observation is that ∗ F µν F µν is atotal derivative of a gauge variant current tr ∗ F µν F µν = 2 ∂ µ tr (cid:15) µνρσ (cid:8) A ν ∂ ρ A σ + A ν A ρ A σ (cid:9) . (A.2)According to Stokes’ theorem, the four-dimensional space integral becomes an integral overthe three-dimensional boundary at infinity if one uses the regular gauge in which there areno singularities at the orgin. Since F µν vanishes at large x , one may replace ∂ ρ A σ by − A ρ A σ , and since A µ becomes a pure gauge at large x , one obtains k = 124 π (cid:73) S (space) d Ω µ (cid:15) µνρσ tr (cid:8)(cid:0) U − ∂ ν U (cid:1) (cid:0) U − ∂ ρ U (cid:1) (cid:0) U − ∂ σ U (cid:1)(cid:9) , (A.3)where the integration is over a large three-sphere, S (space), in four-dimensional Euclideanspace. To each point x µ on this large three-sphere in space corresponds a group element U in the gauge group G . If G = SU (2), the group manifold is also a three-sphere S (group).Then U ( x ) maps S (space) into S (group), and as we now show, k is an integer whichcounts how many times S (space) is wrapped around S (group). Choose a parametrizationof the group elements of SU (2) in terms of group parameters ξ i ( x ) ( i = 1 , , ξ i ( x ) map x into SU (2). Consider a small surface element of S (space).According to the chain ruletr (cid:8)(cid:0) U − ∂ ν U (cid:1) (cid:0) U − ∂ ρ U (cid:1) (cid:0) U − ∂ σ U (cid:1)(cid:9) = ∂ξ i ∂x ν ∂ξ j ∂x ρ ∂ξ k ∂x σ tr (cid:8)(cid:0) U − ∂ i U (cid:1) (cid:0) U − ∂ j U (cid:1) (cid:0) U − ∂ k U (cid:1)(cid:9) , (A.4)and using ∆Ω µ = (cid:15) µαβγ ∆x α ∆x β ∆x γ , (A.5) Note that ∗ F µν F µν is equal to 2 (cid:15) µνρσ { ∂ µ A ν ∂ ρ A σ + 2 ∂ µ A ν A ρ A σ + A µ A ν A ρ A σ } but the last termvanishes in the trace due to the cyclicity of the trace. The elements of SU (2) can be written in the fundamental representation as U = a i (cid:80) k a k τ k where τ k are the Pauli matrices and a and a k are real coefficients satisfying the condition a + (cid:80) k a k = 1. Thisdefines a sphere S (group). (If the a ’s are not real but carry a common phase, one obtains the elementsof U (2)). There is actually a complication. Far away A µ = U − ∂ µ U but in order that U be only a function on S (space) it should only depend on the 3 polar angles but not on the radius. Hence A r = U − ∂ r U shouldvanish. We can make a gauge transformation with a group element V such that A (cid:48) r = V − ( ∂ r + A r ) V vanishes. The V which achieves this is the path ordered integral along the radius from the origin, V = P exp − (cid:82) r A r dr . Note that U is only defined for large r , but V must be defined everywhere, and V (cid:54) = U .In fact, V does not have winding since it can be continuously deformed to the unit group element. Thewinding number is computed in the text for U V , but since k in (A.1) is gauge invariant, k is also thewinding number of the original gauge field A µ . For example, Euler angles, or Lie parameters U = a
1l + i (cid:80) k a k τ k with a = (cid:112) − (cid:80) k a k . For example, if the surface element points in the x -direction we have ∆Ω = ∆y∆z∆τ if (cid:15) = 1. (cid:15) µνρσ (cid:15) µαβγ = δ νρσ [ αβγ ] and ∆ ξ [ i ∆ ξ j ∆ ξ k ] = (cid:15) ijk ∆ ξ , we obtain for the contribution ∆k ofthe small surface element to k∆k = 124 π (cid:15) ijk tr (cid:8)(cid:0) U − ∂ i U (cid:1) (cid:0) U − ∂ j U (cid:1) (cid:0) U − ∂ k U (cid:1)(cid:9) ∆ ξ , (A.6)where k = (cid:72) S (space) ∆k . The elements ( U − ( ξ ) ∂ i U ( ξ )) lie in the Lie algebra, and define thegroup vielbein e ai ( ξ ) by (cid:0) U − ∂ i U (cid:1) = e ai ( ξ ) T a . (A.7)With (cid:15) ijk e ai e bj e ck = (det e ) (cid:15) abc , we obtain for the contribution to k from a surface element ∆Ω µ ∆k = 124 π (det e ) tr (cid:0) (cid:15) abc T a T b T c (cid:1) ∆ ξ = − π (det e ) ∆ ξ . (A.8)We used that for SU (2) we have [ T a , T b ] = (cid:15) abc T c . As we have demonstrated, the originalintegral over the physical space is reduced to one over the group with measure (det e ) d ξ .The volume of a surface element of S (group) with coordinates d ξ i is proportional to(det e ) d ξ (called the Haar measure). Since this expression is a scalar in general relativity, we know that the value of the volume does not depend on which coordinates one uses exceptfor an overall normalization. We fix this overall normalization of the group volume suchthat near ξ = 0 the volume is ∆ ξ . Since e ai = δ ai near ξ = 0, we have there the usualEuclidean measure d ξ . Each small patch on S (space) corresponds to a small patch on S (group), ∆ k ∼ Vol(∆ ξ ). Since the U ’s fall into homotopy classes, integrating once over S (space) we cover S (group) an integer number of times. To check the proportionalityfactor in ∆k ∼ Vol (∆ ξ ), we consider the fundamental map U ( x ) = ix µ σ µ / √ x , U − ( x ) = − ix µ ¯ σ µ / √ x . (A.9)where σ µ denotes the 2 × (cid:126)σ, i ) with (cid:126)σ the Pauli matrices, and ¯ σ µ = ( (cid:126)σ, − i ).This is clearly a one-to-one map from S (space) to S (group) and should therefore yield | k | = 1. Direct calculation gives U − ∂ µ U = − x µ x + x ν ¯ σ ν x σ µ = − σ µν x ν /x , (A.10)where σ µν is defined in (B.8). Substitution into (A.3) leads to k = − π (cid:72) d Ω µ x µ /x = − To obtain k = 1 one has to make the change σ ↔ ¯ σ or x ↔ − x inEq. (A.9). Under a change of coordinates ξ = ξ ( ξ (cid:48) ) at the point ξ , the vielbein transforms as e ai ( ξ ) = ∂ξ (cid:48) j ∂ξ i e (cid:48) aj ( ξ ( ξ (cid:48) )),hence det e ( ξ ) = (cid:16) det ∂ξ (cid:48) ∂ξ (cid:17) det( e (cid:48) ( ξ (cid:48) )), while d ξ is equal to | det ∂ξ∂ξ (cid:48) | d ξ (cid:48) . For small coordinate transfor-mations det ∂ξ/∂ξ (cid:48) is positive, hence det e d ξ is invariant. Only the commutator of the first two matrices in tr ( σ να σ ρβ σ σγ ) x α x β x γ contributes because the anti-commutator is proportional to the unit matrix. In the result only the anticommutator gives a nonvanishingresult, because the commutator yields term proportional to σ αβ whose trace vanishes. A sing µ vanishes fast at infinity, but becomes pure gauge near x = 0. Inthe region between a small sphere in the vicinity of x = 0 and a large sphere at x = ∞ we have an expression for k in terms of a total derivative, but now for A sing µ the onlycontribution to the topological charge comes from the boundary near x = 0: k = − π (cid:73) S x → (space) d Ω µ (cid:15) µνρσ tr (cid:8)(cid:0) U − ∂ ν U (cid:1) (cid:0) U − ∂ ρ U (cid:1) (cid:0) U − ∂ σ U (cid:1)(cid:9) . (A.11)The extra minus sign is due to the fact that the normal to the S (space) at x = 0 pointsinward. Furthermore, A sing µ ∼ U − ∂ µ U = − ¯ σ µν x ν /x near x = 0, while A reg µ ∼ U ∂ µ U − = − σ µν x ν /x for x ∼ ∞ . There is a second extra minus sign in the evaluation of k fromthe trace of Lorentz generators. As a result k sing = k reg , as it should be since k is a gaugeinvariant object. The gauge transformation which maps A reg µ to A sing µ transfers the windingfrom a large to a small S (space). B ’t Hooft symbols and Euclidean spinors
In this appendix we give a list of conventions and formulae useful for instanton calculus.Let us first discuss the structure of Lorentz algebra so (3 ,
1) in Minkowski space-time. Thegenerators can be represented by L µν = ( x µ ∂ ν − x ν ∂ µ ) and form the algebra [ L µν , L ρσ ] = − η µρ L νσ − η νσ L µρ + η µσ L νρ + η νρ L µσ , with the signature η µν = diag( − , + , + , +). The spatialrotations J i ≡ (cid:15) ijk L jk and boosts K i ≡ L i satisfy the algebra [ J i , J j ] = − (cid:15) ijk J k , [ J i , K j ] =[ K i , J j ] = − (cid:15) ijk K k and [ K i , K j ] = (cid:15) ijk J k .There exist two 2-component spinor representations, which we denote by λ α and ¯ χ ˙ α ( α = 1 , α = 1 , σ µν and ¯ σ µν ,where σ µν ≡ ( σ µ ¯ σ ν − σ ν ¯ σ µ ) , ¯ σ µν = (¯ σ µ σ ν − ¯ σ ν σ µ ), with σ α ˙ βµ = ( (cid:126)τ , I ) , ¯ σ µ ˙ αβ = ( (cid:126)τ , − I ) , µ =1 , , ,
0, and I denotes the identity matrix. The matrices τ i with i = 1 , , σ ij = i(cid:15) ijk τ k and σ i = τ i for λ α , and ¯ σ ij = i(cid:15) ijk τ k and¯ σ i = − τ i for ¯ χ ˙ α . The rotation generators σ ij are clearly antihermitian, but the boostgenerators are hermitian.Under a rotation or boost, both spinors simultaneously transform. Most importantly, thetwo spinor representations are complex. In fact, they are each other’s complex conjugateup to a similarity transformation: ( σ µν ) ∗ = σ ¯ σ µν σ . The matrices iτ k and τ k form the2 × Sl (2 , C ), which is the covering group of SO (3 , δ µν = diag(+ , + , + , +)) with SO (4) in-stead of the Lorentz group SO (3 , L µν , L ρσ ] = δ νρ L µσ + 3 terms, and [ J i , J j ] = − (cid:15) ijk J k , [ J i , K j ] = − (cid:15) ijk K k but [ K i , K j ] = − (cid:15) ijk J k where obviously J i ≡ (cid:15) ijk L jk and97oosts K i ≡ L i . The linear combinations of ( ij ) and (4 , i )-plane rotations M i ≡
12 ( J i + K i ) , N i ≡
12 ( J i − K i ) , (B.1)give the algebras of commuting SU (2) subgroups of SO (4) = SU (2) × SU (2) in view of theanti-hermiticity M † i = − M i , N † i = − N i . We now denote the two spinor representationsby λ α and ¯ χ α (cid:48) . Because M and N are represented by generators i(cid:126)σ M and i(cid:126)σ N whichact in different spaces, one can transform λ α while ¯ χ α (cid:48) stays fixed, or vice versa. Thetwo spinor representations in Euclidean space are each pseudoreal: as we shall discuss( σ µν ) ∗ = σ σ µν σ and (¯ σ µν ) ∗ = σ ¯ σ µν σ .It is an easy exercise to check that we can represent the operators M and N by M i = ¯ η iµν , and N i = η iµν , (B.2)where we introduced ’t Hooft symbols [4] η aµν ≡ (cid:15) aµν + δ aµ δ ν − δ aν δ µ , or η aij = (cid:15) aij , η aj = δ aj ¯ η aµν ≡ (cid:15) aµν − δ aµ δ ν + δ aν δ µ , or ¯ η aij = (cid:15) aij , ¯ η aj = − δ aj (B.3)and ¯ η aµν = ( − δ µ + δ ν η aµν . They form a basis of anti-symmetric 4 by 4 matrices and are(anti-)selfdual in vector indices ( (cid:15) = 1) η aµν = (cid:15) µνρσ η aρσ , ¯ η aµν = − (cid:15) µνρσ ¯ η aρσ . (B.4)The η -symbols obey the following relations (cid:15) abc η bµν η cρσ = δ µρ η aνσ + δ νσ η aµρ − δ µσ η aνρ − δ νρ η aµσ ,η aµν η aρσ = δ µρ δ νσ − δ µσ δ νρ + (cid:15) µνρσ ,η aµρ η bµσ = δ ab δ ρσ + (cid:15) abc η cρσ ,(cid:15) µνρτ η aστ = δ σµ η aνρ + δ σρ η aµν − δ σν η aµρ ,η aµν η aµν = 12 , η aµν η bµν = 4 δ ab , η aµρ η aµσ = 3 δ ρσ . (B.5)The same holds for ¯ η except for the terms with (cid:15) µνρσ ,¯ η aµν ¯ η aρσ = δ µρ δ νσ − δ µσ δ νρ − (cid:15) µνρσ ,(cid:15) µνρσ ¯ η aστ = − δ σµ ¯ η aνρ − δ σρ ¯ η aµν + δ σν ¯ η aµρ . (B.6)Obviously η aµν ¯ η bµν = 0 due to different duality properties. In matrix notation, we have[ η a , η b ] = − (cid:15) abc η c , [¯ η a , ¯ η b ] = − (cid:15) abc ¯ η c , { η a , η b } = − δ ab , { ¯ η a , ¯ η b } = − δ ab , (B.7)98nd the two sets of matrices commute, i.e. [ η a , ¯ η b ] = 0 (this is equivalent to the statementthat the generators M and N commute).The two inequivalent spinor representations of the Euclidean Lorentz algebra are givenby σ µν ≡ [ σ µ ¯ σ ν − σ ν ¯ σ µ ] , ¯ σ µν = [¯ σ µ σ ν − ¯ σ ν σ µ ] , (B.8)in terms of Euclidean matrices σ αβ (cid:48) µ = ( τ a , i ) , ¯ σ µ α (cid:48) β = ( τ a , − i ) , µ = 1 , , , , (B.9)obeying the Clifford algebra σ µ ¯ σ ν + σ ν ¯ σ µ = 2 δ µν . Since σ µν contains σ ij = (cid:15) ijk iτ k and σ i = − iτ i , while ¯ σ µν contains ¯ σ ij = (cid:15) ijk iτ k and ¯ σ i = iτ i , they are not each others complexconjugate, contrary to the Minkowski case. Rather, they are pseudo-real, meaning thattheir complex-conjugates are related to themselves by a similarily transformation σ ∗ µν = σ σ µν σ ; (¯ σ µν ) ∗ = σ ¯ σ µν σ . (B.10)To prove these, and other, spinor relations, one needs some formulas which we nowpresent. As in Minkowski space, also in Euclidean space σ µ and ¯ σ µ are related by trans-position σ µαα (cid:48) = ¯ σ µα (cid:48) α (B.11)where ¯ σ µα (cid:48) α is obtained from ¯ σ µβ (cid:48) β by raising indices¯ σ µα (cid:48) α ≡ (cid:15) α (cid:48) β (cid:48) (cid:15) αβ ¯ σ µβ (cid:48) β (B.12)We use everywhere the north-west convention for raising and lowering the spinor indices (cid:15) αβ ξ β = ξ α , ¯ ξ β (cid:48) (cid:15) β (cid:48) α (cid:48) = ¯ ξ α (cid:48) , (B.13)with (cid:15) αβ = − (cid:15) α (cid:48) β (cid:48) , (cid:15) αβ = (cid:15) αβ , and (cid:15) α (cid:48) β (cid:48) = (cid:15) α (cid:48) β (cid:48) . However, the relation between σ µ and¯ σ µ under complex conjugation is different (as expected because σ = I but σ = iI ). InMinkowski space we have ( σ α ˙ βµ ) ∗ = ¯ σ ˙ βαµ , while in Euclidean space ( σ αβ (cid:48) µ ) ∗ = ¯ σ µ,β (cid:48) α = σ µ,αβ (cid:48) and (¯ σ µ,α (cid:48) β ) ∗ = σ βα (cid:48) µ = ¯ σ α (cid:48) βµ .Let us now apply these formulas to give another proof that σ µν and ¯ σ µν are pseudorealin Euclidean space (( σ µν ) αβ ) ∗ = 12 ( σ µαβ (cid:48) ) ∗ (¯ σ ν,β (cid:48) β ) ∗ − µ ↔ ν = 12 σ µ,αβ (cid:48) ¯ σ β (cid:48) βν − µ ↔ ν = − σ µ,αβ (cid:48) ¯ σ ν,β (cid:48) β − µ ↔ ν = − (cid:15) γα ( σ µν ) γδ (cid:15) βδ = ( − iσ )( − σ µν )( − iσ ) = σ σ µν σ (B.14)99nd idem for ¯ σ µν .The two spinor and vector representations of the su (2) algebra are all given in terms ofanti-hermitian 2x2 matrices σ µν , ¯ σ µν and iτ a and they are related by the ’t Hooft symbols,¯ σ µν = iη aµν τ a , σ µν = i ¯ η aµν τ a . (B.15)Furthermore, ¯ σ µν is selfdual whereas σ µν is anti-selfdual. Some frequently used identitiesare ¯ σ µ σ νρ = δ µν ¯ σ ρ − δ µρ ¯ σ ν − (cid:15) µνρσ ¯ σ σ , σ µ ¯ σ νρ = δ µν σ ρ − δ µρ σ ν + (cid:15) µνρσ σ σ ,σ µν σ ρ = δ νρ σ µ − δ µρ σ ν + (cid:15) µνρσ σ σ , ¯ σ µν ¯ σ ρ = δ νρ ¯ σ µ − δ µρ ¯ σ ν − (cid:15) µνρσ ¯ σ σ . (B.16)The Lorentz generators are antisymmetric in vector and symmetric in spinor indices σ µν αβ = − σ νµ αβ , σ µν αβ = σ µν βα , (B.17)and obey the algebra[ σ µν , σ ρσ ] = − { δ µρ σ νσ + δ νσ σ µρ − δ µσ σ νρ − δ νρ σ µσ } , { σ µν , σ ρσ } = − { δ µρ δ νσ − δ µσ δ νρ − (cid:15) µνρσ } . (B.18)The same relations hold for ¯ σ µν but with + (cid:15) µνρσ . In spinor algebra the following contrac-tions are frequently used σ αα (cid:48) µ ¯ σ µ β (cid:48) β = 2 δ βα δ β (cid:48) α (cid:48) , σ αρσ β σ γρσ δ = 4 (cid:110) δ αβ δ γδ − δ αδ δ γβ (cid:111) . (B.19)so that ξ α (1) ξ (2) α = ξ α (2) ξ (1) α . For hermitean conjugation we define (cid:16) ξ α (1) ξ (2) α (cid:17) † = ( ξ (2) α ) † ( ξ α (1) ) † (cid:16) σ αβ (cid:48) µ (cid:17) ∗ = σ µ αβ (cid:48) , (¯ σ µ α (cid:48) β ) ∗ = ¯ σ α (cid:48) βµ . (B.20)Throughout the paper we frequently use the following integral formula (cid:90) d x ( x ) n ( x + ρ ) m = π (cid:0) ρ (cid:1) n − m +2 Γ ( n + 2) Γ ( m − n − Γ ( m ) , (B.21)which converges for m − n > C The volume of the gauge orientation moduli space
The purpose of this appendix is to prove equation (6.15). Let us consider an instantonin SU ( N ) gauge theory. Deformations of this configuration which are still self-dual and We thank R. Roiban for help in writing this appendix. A µ → U − A µ U preserve self-duality and transversality but not all constant SU ( N ) matrices U change A µ . Those U which keep A µ fixed form the stability subgroup H of the instanton, hence we want to determine the volume of the coset space SU ( N ) /H . Ifthe instanton is embedded in the lower-right 2 × N × N SU ( N ) matrix,then H contains the SU ( N −
2) subgroup in the left-upper part, and a U (1) subgroup withelements exp ( θA ) where A is the diagonal matrix A = i (cid:114) N − N diag (cid:18) − N , . . . , − N , , (cid:19) . (C.1)All generators of SU ( N ) (and also all generators of SO ( N ) discussed below) are normalizedaccording to tr T a T b = − δ ab , as in the main text.At first sight one might expect the range of θ to be such that the exponents of all entriescover the range 2 π an integer number of times. However, this is incorrect: only for thelast two entries of exp ( θA ) we must require periodicity, because whatever happens in theother N − SU ( N −
2) part of the stabilitysubgroup. Thus all elements h in H are of the form [52] h = e θA g, with g ∈ SU ( N −
2) and 0 ≤ θ ≤ θ max = 4 π (cid:114) NN − . (C.2)For N = 3 the range of θ is larger than required by periodicity of the first N − N = 4 it corresponds to periodicity of all entries, but for N ≥ θ is less thanrequired for periodicity of the first N − Thus H (cid:54) = SU ( N ) × U (1) for N ≥ N − kθ max A ) with integer k are given by exp (cid:0) − ik πN − (cid:1) and lietherefore in the center Z N of SU ( N − SU ( N ) group elements h = exp ( θA ) g with 0 ≤ θ ≤ θ max and g in SU ( N −
2) form a subgroup H . We shall denote H by SU ( N − × “ U (1)” where “ U (1)” denotes the part of the U (1) generated by A which liesin H . We now use three theorems to evaluate the volume of SU ( N ) /H :(I) Vol SU ( N ) SU ( N − × “ U (1)” = Vol ( SU ( N ) /SU ( N − U (1)” , (II) Vol SU ( N ) SU ( N −
2) = Vol SU ( N ) SU ( N −
1) Vol SU ( N − SU ( N − , (C.3) For example, consider SU (5) with exp[ iθ (cid:113) diag ( − , − , − , , θ runs from 0 to (cid:113) π ,last two entries repeat, but the first three entries only reach exp( − πi/ SU ( N −
2) = SU (3), namely they yield an element z of the center Z . So when θ rangesbeyond (cid:113) π , these SU (5) elements can be written as a product of z and exp iθA with θ smaller than (cid:113) π . So, the range of θ is bounded by (cid:113) π . SU ( N ) SU ( N −
1) = Vol SU ( N )Vol SU ( N − . It is, in fact, easiest to first compute Vol ( SU ( N ) /SU ( N − SU ( N ) /H (with Vol SU ( N ) as a bonus).In general the volume of a coset manifold G/H is given by V = (cid:82) (cid:81) µ dx µ det e mµ ( x )where x µ are the coordinates on the coset manifold and e mµ ( x ) are the coset vielbeins.One begins with “coset representatives” L ( x ) which are group elements g ∈ G such thatevery group element can be decomposed as g = L ( x ) h with h ∈ H . We denote the cosetgenerators by K m and the subgroup generators by H i . Then L − ( x ) ∂ µ L ( x ) = e mµ ( x ) K m + ω iµ ( x ) H i . We shall take the generators K m and H i in the fundamental representationof SU ( N ): antihermitian N × N matrices. Under a general coordinate transformationfrom x µ to y µ ( x ), the vielbein transforms as a covariant vector with index µ but alsoas a contravariant vector with index m at x = 0. Hence V does (only) depend on thechoice of the coordinates at the origin . At the origin, L − ∂ µ L = e mµ (0) K m , and we fix thenormalization of K m by tr K m = − for K m in the N × N matrix representation of SU ( N ).To find the volume of SU ( N ) /SU ( N −
1) we note that the group elements of SU ( N )have a natural action on the space C N and map a point (cid:0) z , . . . , z N (cid:1) ∈ C N on the com-plex hypersphere (cid:80) Ni =1 | z i | = 1 into another point on the complex hypersphere. The“south-pole” (0 , . . . , ,
1) is kept invariant by the subgroup SU ( N − L ( z )of SU ( N ) /SU ( N − SU ( N ) the generators for SU ( N −
1) inthe upper-left block, and further the following coset generators: N − T k and T k +1 each of them containing only two non-zero elements . . . ·· ... i/
2. . . ...0 i/ . . . , . . . ·· ... 1 /
2. . . ...0 − / . . . , (C.4)and further one diagonal generator T N − = i (cid:115) N ( N −
1) diag ( − , . . . , − , N − . (C.5)(For instance, for SU (3) there are two pairs, proportional to the usual λ and λ and λ and λ , and the diagonal hypercharge generator λ .) The idea now is to establish a naturalone-to-one correspondence between points in C N and points in R N , namely we write allpoints ( x , . . . , x N ) in R N as points in C N as follows: ( ix + x , . . . , ix N − + x N ). In102articular the south pole (0 , , ..., ,
1) in R N corresponds to the south pole (0 , , ..., , C N and the sphere (cid:80) Ni =1 ( x i ) = 1 in R N corresponds to the hypersphere (cid:80) Ni =1 | z i | = 1in C N . Points on the sphere S N − in R N correspond one-to-one to coset elements of SO (2 N ) /SO (2 N − SO (2 N ) /SO (2 N −
1) are antisymmetric2 N × N matrices A I ( I = 1 , . . . , N −
1) with the entry +1 / − / δg = 1 + d t I A I maps the south pole s =(0 , . . . , ,
1) in R N to a point s + δs in R N where δs = 1 / t , . . . , d t N − , C N correspond to points in R N , so we can ask which coset element in SU ( N ) /SU ( N −
1) maps the south-pole in C N to the point in C N which corresponds to s + δs . In C N the corresponding point is s + δs with δs = 1 / i d t + d t , . . . , i d t N − ).The coset generators of SU ( N ) /SU ( N −
1) act in C N as follows: g = 1 + d x µ K µ maps thesouth-pole s to s + δs where now δs = 1 / i d x + d x , . . . , i (cid:113) N − N d x N − ). We can cover SO (2 N ) /SO (2 N −
1) = S N − with small patches. Similarly we cover SU ( N ) /SU ( N − S N − can be brought by the action of a suitablecoset element to the south-pole, and then we can use the inverse of this group elementto map this patch back into the manifold SU ( N ) /SU ( N − S N − and SU ( N ) /SU ( N −
1) are covered by patches which are in a one-to-one correspondence.Each pair of patches has the same ratio of volumes since both patches can be broughtto the south pole by the same group element and at the south pole the ratio of theirvolumes is the same. To find the ratio of the volumes of S N − and SU ( N ) /SU ( N − S N − with coordinates (d t , . . . , d t N − (cid:1) andvolume d t . . . d t N − . The same patch at the south pole in C N has coordinates d x µ where (cid:0) i d t + d t , . . . , i d t N − (cid:1) = (cid:18) i d x + d x , . . . , i (cid:113) N − N d x N − (cid:19) . The volume of a patch in SU ( N ) /SU ( N −
1) with coordinates d x , . . . , d x N − is d x . . . d x N − . It follows that thevolume of SU ( N ) /SU ( N −
1) equals the volume of S N − times (cid:113) N N −
1) 58 ,Vol SU ( N ) SU ( N −
1) = (cid:115) N N −
1) Vol S N − . (C.6)From here the evaluation of Vol SU ( N ) /H is straightforward. UsingVol S N − = 2 π N ( N − l , (C.7) This result yields the same answer for (6.15) as [52], but it yields π N / ( N N !) for the volume of thecomplex projective space CP ( N ) = SU ( N + 1) / ( SU ( N ) × U (1)) which differs from the result Vol[ U ( N +1) / ( U ( N ) × U (1))] = Vol S N given in [90]. l = 1 if one uses the normalization tr K m = −
2, but l = 2 N − with our normalizationof tr K m = − , we obtain Vol SU ( N ) = √ N N (cid:89) k =2 √ π k ( k − k − . (C.8)We assumed that Vol SU (1) = 1 which seems a natural value but must be, and will be,justified below. ThenVol H = Vol SU ( N − U (1)” , Vol “ U (1)” = 4 π (cid:114) NN − , Vol SU ( N ) /H = π N − ( N − N − N − N − . (C.9)This then produces formula (6.15).As an application and check of this analysis let us derive a few relations between thevolumes of different groups. From now on till the end of this appendix we adopt thenormalization tr( T a T b ) = − δ ab for the generators of all groups involved. Let us check thatVol SU (2) = 2Vol SO (3), Vol SU (4) = 2Vol SO (6) and Vol SO (4) = (Vol SU (2)) (thelatter will follow from SO (4) = SU (2) × SU (2) /Z ). We begin with the usual formula forthe surface of a sphere with unit radius (given already above for odd N )Vol S N = 2 π ( N +1) / Γ (cid:0) N +12 (cid:1) . (C.10)In particular Vol S = 2 π andVol S = 4 π , Vol S = 2 π , Vol S = π , Vol S = π , Vol S = π , Vol S = π . (C.11)Furthermore Vol SO (2) = 2 π since the SO (2) generator with tr T = − T = (cid:0) − (cid:1) andexp( θT ) is an ordinary rotation (cid:0) cos θ sin θ − sin θ cos θ (cid:1) for which 0 ≤ θ ≤ π . The vielbein is unityfor an abelian group, and thus the Haar measure is simply d θ .With Vol SO ( N ) = Vol S N − Vol SO ( N −
1) we obtain Vol SO (1) = 1 andVol SO (2) = 2 π , Vol SO (3) = 8 π , Vol SO (4) = 16 π , Vol SO (5) = π , Vol SO (6) = π . (C.12)Now consider SU (2). In the normalization T = − iτ , T = − iτ and T = − iτ (so thattr T a T b = − δ ab ) we find by direct evaluation using Euler angles Vol SU (2) = 2 π . This One clearly must specify the normalization of the generators T a ; for example by choosing T a = (cid:18) −
12 0 (cid:19) , the range of θ becomes 0 ≤ θ ≤ π , but the Haar measure is still d θ . Parametrize g = e αT e βT e γT , determine the range of α, β, γ and compute the group vielbeins. N = 2, justifying our assumption that Vol SU (1) = 1.For higher N we getVol SU (2) = 2 π , Vol SU (3) = √ π , Vol SU (4) = √ π . (C.13)The group elements of SU (2) can also be written as g = x + i(cid:126)τ · (cid:126)x with ( x ) +( (cid:126)x ) = 1 which defines a sphere S . Since near the unit element g ≈ i(cid:126)τ · δ(cid:126)x , thenormalization of the generators is as before, and hence for this parametrization Vol SU (2) =2 π . This is indeed equal to Vol S . In the mathematical literature one finds the statementthat Vol SU (2) is twice Vol SO (3) because SU (2) is the double covering group of SO (3).However, we have just found that Vol SU (2) = Vol SO (3). The reason is that in orderto compare properties of different groups we should normalize the generators such thatthe structure constants are the same (the Lie algebras are the same, although the groupvolumes are not). In other words, we should use the normalization that the adjoint representations have the same tr T a T b . For SU (2) the generators which lead to the samecommutators as the usual SO (3) rotation generators (with entries +1 and −
1) are T a = (cid:8) − i τ , − i τ , − i τ (cid:9) . Then tr T a T b = − δ ab . In this normalization, the range of eachgroup coordinate is multiplied by 2, leading to Vol SU (2) = 2 · π = 16 π . Now indeedVol SU (2) = 2Vol SO (3).For SU (4) the generators with the same Lie algebra as SO (6) are the 15 antihermitean4 × ( γ m γ n − γ n γ m ), iγ m / γ m γ / iγ /
2, where γ m and γ are the five4 × γ M obeying the Clifford algebra { γ M , γ N } = 2 δ MN . Now, tr T a T b = − δ ab (for example, tr (cid:8)(cid:0) γ γ (cid:1) (cid:0) γ γ (cid:1)(cid:9) = − T a T b = − δ ab . We must thus multiply the range of each coordinate bya factor √
2, and hence we must multiply our original result for Vol SU (4) by a factor (cid:0) √ (cid:1) . We find then indeed that the relation Vol SU (4) = 2 Vol SO (6) is fulfilled.Finally, we consider the relation SO (4) = SU (2) × SU (2) /Z . (The vector representationof SO (4) corresponds to the representation (cid:0) , (cid:1) of SU (2) × SU (2), but representationslike (cid:0) , (cid:1) and (cid:0) , (cid:1) are not representations of SO (4) and hence we must divide by Z .The reasoning is the same as for SU (2) and SO (3), or SU (4) and SO (6).) We choose thegenerators of SO (4) as follows T (+)1 = 1 √ L + L ) , T (+)2 = 1 √ L + L ) , T (+)3 = 1 √ L + L ) , (C.14)and the same but with minus sign denoted by T ( − ) i . Here L mn equals +1 in the m th columnand n th row, and is antisymmetric. Clearly tr T a T b = − δ ab . The structure constants followfrom (cid:20) √ L + L ) , √ L + L ) , (cid:21) = − ( L + L ) , (C.15) As Dirac matrices in six dimensions we take γ m ⊗ τ , γ ⊗ τ and I × τ . (cid:104) T (+) i , T (+) j (cid:105) = −√ (cid:15) ijk T (+) k , (cid:104) T ( − ) i , T ( − ) j (cid:105) = −√ (cid:15) ijk T ( − ) k , (cid:104) T (+) i , T ( − ) j (cid:105) = 0 . (C.16)We choose for the generators of SU (2) × SU (2) the representation T (+) i = iτ i √ ⊗ , T ( − ) i = 1l ⊗ iτ i √ . (C.17)Then we get the same commutation relations as for SO (4) generators (C.16); however, thegenerators are normalized differently, namely tr T a T b = − δ ab for SO (4) but tr T a T b = − δ ab for SU (2). With the normalization tr T a T b = − δ ab we found Vol SU (2) = 2 π . Inthe present normalization we find Vol SU (2) = 2 π (cid:0) √ (cid:1) . The relation Vol SO (4) = (Vol SU (2)) is now indeed satisfiedVol SO (4) = 16 π = (Vol SU (2)) = 12 (cid:18) π (cid:16) √ (cid:17) (cid:19) . (C.18) D Zero modes and conformal symmetries
The bosonic collective coordinates obtained for gauge group SU (2) and the one-instantonsolution could all be identified with rigid symmetries of the action: a µ with translations, ρ with scale transformations and θ a with rigid gauge symmetries. Similarly, the fermioniccollective coordinates for SU (2)( ξ α and ¯ η ˙ α with α, ˙ α = 1 ,
2) could be identified with ordi-nary supersymmetry and conformal supersymmetry. However, the full conformal algebrain 4 Euclidean dimensions is SO (5 , P µ , K µ , D, M µν , so one mightexpect that the conformal boost transformations K µ and the Lorentz rotations M µν pro-duce further collective coordinates. As we now show, the transformations due to thesesymmetries can be undone by suitably chosen gauge transformations with constant gaugeparameters [29]. So there are no further bosonic collective coordinates, as we already knowfrom the index theorem discussed in the main text.Consider first rigid Lorentz transformations. Here one should not forget that in additionto a spin part which acts on the indices of a field they also contain an orbital part thatacts on the coordinates: M µν = Σ µν + L µν . For example, for a spinor one has δ ( λ mn ) ψ = λ mn γ mn ψ +( λ mn x m ∂ n ) ψ . One may check that only with this orbital part present the Diracaction is Lorentz invariant. In fact, starting with only the spin part or the orbital part,one can find the other part by requiring invariance of the action. We begin by consideringthe field strength F µν = 2¯ σ µν ρ / ( x + ρ ) for an instanton with k = 1 in the regulargauge. Under a Lorentz transformation with parameter λ µν = − λ νµ one has δ M A µ = λ µν A ν + λ mn x m ∂ n A ν . (Note that coordinates transform opposite to fields: δx m = − λ mn x n .One may check this transformation rule by showing that the Maxwell action is Lorentz106nvariant (use the Bianchi identities ), or just by writing down the transformation law fora covariant vector in general relativity. The Lagrangian transforms into ∂ µ ( ξ µ L ), where ξ µ = λ ρµ x ρ .). The field strength of the instanton transforms as follows δ M F µν = λ µρ F ρν + λ νρ F µρ (D.1)There is no contribution from the orbital part because x is Lorentz invariant. On theother hand, under a gauge transformation with parameter Λ ρσ we obtain δ gauge F µν = [¯ σ µν , Λ ρσ ¯ σ ρσ ](2 ρ / ( x + ρ ) ) = Λ νσ F µσ − Λ µσ F νσ (D.2)Thus F µν is invariant under combined Lorentz and gauge transformations with oppositeparameters, Λ ρσ = − λ ρσ . Using ¯ σ ρσ = iη aρσ τ a , it is clear that the SU (2) gauge parameterΛ a is proportional to η aρσ λ ρσ . Only the selfdual part of λ ρσ contributes. For an anti-instanton we would have needed the anti-selfdual part of λ ρσ . So we have only proven that F µν is invariant under combined Lorentz and gauge transformation if the Lorentz parameteris self dual. However, the anti-self dual part of λ ρσ leaves F µν separately invariant, withoutthe need to add compensating gauge transformations. One can prove this directly, usingthat λ µρ F ρν = − ( ∗ λ µρ )( ∗ F ρν ) and then working out the product of two (cid:15) -tensors and finallyantisymmetrizing in µν , but it is already clear from the index structure: F µν is proportionalto (¯ σ µν ) α (cid:48) β (cid:48) while an anti-selfdual λ ρσ has in spinor notation only undotted indices.Let us now repeat this exercise for the gauge field A µ . One finds for the combinedLorentz and gauge transformation δA µ = λ µν A ν + λ ρσ x ρ ∂ σ A µ + [ A µ , Λ ρσ ¯ σ ρσ ] (D.3)The instanton field A µ for k = 1 in the regular gauge is given by A µ = ( − ¯ σ µν x ν ) / ( x + ρ ). The orbital part with λ ρσ now contributes, but there is no term ∂ µ Λ a in the gaugetransformation of A µ since Λ a is constant. One obtains δA µ = λ µν A ν + ¯ σ µν ( λ νρ x ρ ) x + ρ − (Λ νσ ¯ σ µσ − Λ µσ ¯ σ νσ ) x ν x + ρ (D.4)For Λ µν = − λ µν all terms again cancel. Hence, Lorentz symmetry does not yield furtherzero modes.In spinor notation these results are almost obvious. In general the selfdual part of acurvature reads in spinor notation ( F µν ) uv (¯ σ µν ) α (cid:48) β (cid:48) (D.5) One has δ M F µν = F µν ∂ µ ( λ νρ A ρ ) + F µν ∂ µ ( λ mn x m ∂ n A ν ) = F µν λ mn x m ∂ µ ∂ n A ν . Replacing ∂ µ ∂ n A ν by − ∂ n ∂ ν A µ − ∂ ν ∂ µ A n yields ∂ µ ( ξ µ L ). The usual form of an SU (2) gauge transformation is δF µν = [ F µν , Λ a ( x ) τ a i ], but using η aµν η bµν = 4 δ ab and ¯ σ ρσ = iη aρσ τ a , this can be rewritten as δF µν = [ F µν , Λ ρσ ¯ σ ρσ ] where Λ ρσ = − η aρσ Λ a . u, v are the indices of ( τ a ) uv , and α (cid:48) , β (cid:48) are the spinor indices. If we raise/lowerindices by (cid:15) tensors, we get for the instanton solution( F µν ) uv (¯ σ µν ) α (cid:48) β (cid:48) ≡ F uvα (cid:48) β (cid:48) ∼ δ uα (cid:48) δ vβ (cid:48) + δ vα (cid:48) δ uβ (cid:48) (D.6)It is then clear that F µν is invariant under diagonal transformations of SU (2) R and SU (2) gauge ,and separately invariant under SU (2) L . For an anti-instanton, the roles of SU (2) L and SU (2) R are interchanged.We come now to the more complicated problem of conformal transformations. A con-formal transformation of a field ϕ with constant parameter a m is given by δ ( a m K m ) ϕ = (2 a · x x m − a m x ) ∂ m ϕ + δ (2 a · xD (spin) ) ϕ + δ (2 a m x n M (spin) mn ) ϕ (D.7)where D spin and M spin mn act only on ϕ (0) and δ ( a m K m ) ϕ ( x ) is by definition [ ϕ ( x ) , a m K m ]. Asthe notation indicates, only the spin parts of the dilatational generator D and the Lorentzgenerators contribute. For example δ ( D spin ) A µ = [ A µ , D (spin) ] = A µ , δ ( λ mn M (spin) mn ) A µ = λ µν A ν . (D.8)Consider first F µν . We obtain δ ( a m K m ) F µν = (2 a · xx m − a m x ) ∂ m F µν + 4 a · xF µν +4 δ (cid:0) a m x n M (spin) mn (cid:1) F µν with F µν = 2¯ σ µν ρ ( x + ρ ) . (D.9)We already know that the last term can be canceled by a suitable gauge transformation(there are no contributions from M (orb) mn because x is Lorentz invariant). The first termgives − a · xx x + ρ F µν . The first and second term together produce then a · xρ x + ρ F µν . But this isthe opposite of a translation with parameter a m ρ , namely δ ( a m ρ P m ) F µν = − a · xρ x + ρ F µν ; δ ( P m ) ϕ = ∂ m ϕ . (D.10) This formula follows from δ ( a m K m ) ϕ ( x ) = [ ϕ ( x ) , a m K m ], and ϕ ( x ) = e − P · x ϕ (0) e P · x with [ ϕ (0) , P µ ] = ∂ µ ϕ (0). One may then use e P · x K m = ( e P · x K m e − P · x ) e P · x and [ K m , P n ] = − δ mn D − M mn ; [ P m , D ] = P m ; [ P m , M rs ] = δ mr P s − δ ms P r and this yields (D.7). In the same way one may derive the Lorentztransformation rule for a spinor ψ ( x ), with both spin and orbital parts, by using that the spin part isgiven by [ ψ (0) , λ mn M mn ] = λ mn γ mn ψ (0). One finds then the correct result: δ ( λ mn M mn ) ψ ( x ) = λ mn γ mn ψ ( x ) + λ mn x m ∂ n ψ ( x ). Given the spin part of the transformation rule of the field at the origin,one derives in this way the orbital part. In this way one finds that the generators of the conformalalgebra act as follows on the coordinates: δ ( P m ) x n = δ mn , δ ( D ) x n = x n , δ ( M st ) x m = x s δ tm − x t δ sm and δ ( K m ) x n = 2 x m x n − x δ mn . Note that coordinates transform contragradiently to fields. For example,whereas [ δ ( K m ) , δ ( P n )] ϕ = − δ ([ K m , P n ]) ϕ (by definition), one finds [ δ ( K m ) , δ ( P n )] x s = δ ([ K m , P n ]) x s . F µν in variant a m K m + ρ a m P m + δ gauge (Λ mn = − a m x n + 2 x m a n ) (D.11)Let us now check that also A µ itself is invariant under this combination of symmetries.We find by direct evaluation, using A µ = ( − ¯ σ µν x ν ) / ( x + ρ ) and (D.9) and (D.2) δA µ = (cid:18) − a · xx x + ρ A µ − ¯ σ µν (2 a · xx ν − a ν x ) x + ρ (cid:19) + 2 a · xA µ + (2 a µ x ν A ν − x µ a ν A ν ) + (cid:18) − ρ a · xx + ρ A µ − ¯ σ µν a ν ρ x + ρ (cid:19) + ∂ µ ( − a ρ x σ ¯ σ ρσ ) + [ A µ , − a ρ x σ ¯ σ ρσ ] . (D.12)As in the case of F µν , the sum of the first, third and sixth term cancels. This takes care ofthe dilatation term and the denominator of A µ . 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