Patterns of flavour symmetry breaking in hadron matrix elements involving u, d and s quarks
J.M. Bickerton, R. Horsley, Y. Nakamura, H. Perlt, D. Pleiter, P.E.L. Rakow, G. Schierholz, H. Stüben, R.D. Young, J.M. Zanotti
AADP-19-19/T1099DESY 19-149Liverpool LTH 1211December 31, 2019
Patterns of flavour symmetry breaking inhadron matrix elements involving u, d and squarks
J. M. Bickerton a , R. Horsley b , Y. Nakamura c ,H. Perlt d , D. Pleiter e , P. E. L. Rakow f , G. Schierholz g ,H. St¨uben h , R. D. Young a and J. M. Zanotti a – QCDSF-UKQCD-CSSM Collaboration – a CSSM, Department of Physics, University of Adelaide,Adelaide SA 5005, Australia b School of Physics and Astronomy, University of Edinburgh,Edinburgh EH9 3FD, UK c RIKEN Center for Computational Science,Kobe, Hyogo 650-0047, Japan d Institut f¨ur Theoretische Physik, Universit¨at Leipzig,04103 Leipzig, Germany e J¨ulich Supercomputer Centre, Forschungszentrum J¨ulich,52425 J¨ulich, GermanyInstitut f¨ur Theoretische Physik, Universit¨at Regensburg,93040 Regensburg, Germany f Theoretical Physics Division, Department of Mathematical Sciences,University of Liverpool, Liverpool L69 3BX, UK g Deutsches Elektronen-Synchrotron DESY,22603 Hamburg, Germany h Universit¨at Hamburg, Regionales Rechenzentrum,20146 Hamburg, Germany a r X i v : . [ h e p - l a t ] D ec bstract By considering a flavour expansion about the SU (3)-flavour symmetricpoint, we investigate how flavour-blindness constrains octet baryon ma-trix elements after SU (3) is broken by the mass difference between quarks.Similarly to hadron masses we find the expansions to be constrained alonga mass trajectory where the singlet quark mass is held constant, which pro-vides invaluable insight into the mechanism of flavour symmetry breakingand proves beneficial for extrapolations to the physical point. Expansionsare given up to third order in the expansion parameters. Consideringhigher orders would give no further constraints on the expansion parame-ters. The relation of the expansion coefficients to the quark-line-connectedand quark-line-disconnected terms in the three-point correlation functionsis also given. As we consider Wilson clover-like fermions, the addition ofimprovement coefficients is also discussed and shown to be included in theformalism developed here. As an example of the method we investigatethis numerically via a lattice calculation of the flavour-conserving matrixelements of the vector first class form factors. ontents f π and f K . . . . . . . . . . . . . 13 SU (2) relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 SU (3) symmetry-breaking expansions . . . . . . . . . . . . . 205.3.1 Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.3.2 Transformations . . . . . . . . . . . . . . . . . . . . . . . . 22 d -fan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318.2 The f -fan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
11 Renormalisation and O ( a ) improvement for the vector current 37 V π µ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3911.1.2 V η R µ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3911.1.3 V η (cid:48) R µ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3911.1.4 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . 3931.2 Determination of ˆ Z V and ˆ b V , ˆ f con V . . . . . . . . . . . . . . . . . . 4011.2.1 V π . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4011.2.2 V η R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4111.2.3 The Ademollo–Gatto theorem . . . . . . . . . . . . . . . . 41
12 Lattice computations of form factors 41
13 Results 45 Z V and ˆ b V , ˆ f con V . . . . . . . . . . . . . . . . . . . . . . 5313.4 Electromagnetic form factor results . . . . . . . . . . . . . . . . . 53
14 Conclusions and outlook 56A Non-zero tensor elements 59B Alternative fan plots 59
B.1 The doubly represented − singly represented fan, the P -fan . . . . 59B.2 The V -fan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 C LO flavour diagonal matrix elements 64D LO disconnected flavour diagonal matrix elements 65 Introduction
Understanding the pattern of flavour symmetry breaking and mixing, and the ori-gin of CP violation, remains one of the outstanding problems in particle physics.The big questions to be answered are (i) What determines the observed patternof quark and lepton mass matrices and (ii) Are there other sources of flavoursymmetry breaking? In [1, 2] we have outlined a programme to systematicallyinvestigate the pattern of flavour symmetry breaking. The program has beensuccessfully applied to meson and baryon masses involving up, down and strangequarks. In this article we will extend the investigation to include matrix elements.The QCD interaction is flavour-blind. Neglecting electromagnetic and weakinteractions, the only difference between flavours comes from the quark massmatrix. We have our best theoretical understanding when all three quark flavourshave the same masses, because we can use the full power of flavour SU (3). Thestrategy is to keep the average bare quark mass ¯ m = ( m u + m d + m s ) / m u = m d = m s . Thus all the quark mass dependence will be expressed as polynomials in δm q = m q − ¯ m , q = u , d , s . It should be mentioned that this is a completelydifferent approach for studying the manifestations of low-energy QCD than chiralperturbation theory. It is a complementary method and based on group theoryrather than effective field theory.The programme has been successfully applied to meson and baryon massesin [1, 2] including an extension to incorporate QED effects [3, 4, 5]. Besides con-straining the quark mass dependence of hadron masses, which helps in extrap-olations to the physical point, it provides valuable information on the physicsof flavour symmetry breaking. For example, the order of the polynomial can beassociated with the order of 1 /N c corrections, [6]. Furthermore, similar to theanalysis of Gell-Mann and Okubo [7, 8], the order of the polynomial classifiesthe order of SU (3) breaking, [1, 2]. As opposed to the conventional method ofkeeping the strange quark mass fixed, our method has the further advantage thatflavour singlet quantities which are difficult to compute can now be disentangledin the extrapolation, and are largely constant on the ¯ m constant line.In this article we shall concentrate on matrix elements for the baryon octetas sketched in the Y – I plane in the left hand panel of Fig. 1. It is easy totranslate the results to octet mesons sketched in the right hand panel of Fig. 1.Furthermore we restrict ourselves to the case of n f = 2 + 1, i.e. the case ofdegenerate u and d quark masses, m u = m d ≡ m l . (Initial results were givenin [9].) However our method is also applicable to isospin breaking effects arisingfrom non-degenerate u and d quark masses. We postpone this analysis to aseparate paper, including electromagnetic effects, [10]. The formalism is general.In our application we consider for definiteness just local currents, but covering5
0− 0 +1−1
Ξ Σ p(uud) Ξ (uds) I Λ (uds) Σ − (dds) n(ddu)(ssd) (ssu) Σ (uus) Y π ++0− π +1−1 I (du) − π K (ds) K (us)K (su) K (sd) Y (ud) η Figure 1:
Left panel: The baryon octet. Right panel: The meson octet. all possible Dirac gamma matrix structure .While of intrinsic interest in itself, an obvious application of this formalismis the determination of semileptonic decay form factors and the associated CKMmatrix element, | V us | . In general disentangling quark mass and momentum de-pendencies is helpful for determining generalised form factors of baryons, as de-scribed for example in the forthcoming Electron Ion Collider (EIC) programme,[11].The structure of this article is as follows. In section 2, we discuss all possiblecurrents (which we call ‘generalised currents’ here) and also their splitting into‘first’ class and ‘second’ class currents. Then in sections 3, 4, 5 we discuss thegroup theory. In section 3 we define our expansion parameter, δm l and thegeneral structure of our expansions. Also discussed there (and at the beginning ofsection 5.1) are simple cases which have previously been determined. In particularthe singlet case will be used later in this article. The next section, section 4 givesour sign conventions (commonly employed in chiral perturbation theory). As wehave mass degenerate u and d quarks then there is an SU (2) isospin symmetry.We then use the Wigner-Eckart theorem to give the reduced matrix elements,contrasting the difference here to the usual conventions. Then in section 5, afterdiscussing the group theory classification of SU (3) tensors, we determine thoserelevant to our study (with complete tables being given in Appendix A), andthen in section 6.1 give the LO expansions. Higher-order terms are given insection 6.2. These sections giving the expansion coefficients form the heart ofthis report. This is followed by section 7 where we briefly restrict ourselves to adiscussion of the amplitudes at the symmetric point.Continuing with the main thread, in section 8 linear combinations of thematrix elements are constructed for the various baryons, leading to functions It can also easily be extended to currents including covariant derivatives. SU (3) flavour symmetric point. Four different‘fan’ plots are constructed, two detailed in section 8 and a further two given inAppendix B.Lattice QCD determinations of matrix elements involve the computation of3-point correlation functions, which fall into two classes – quark-line connecteddiagrams and quark line disconnected diagrams. In section 9, we discuss theimplications of this splitting for the SU (3) symmetry flavour breaking expansionsat LO. In particular for the connected terms, there are further constraints onthe expansion coefficients. In section 10 this is applied to the baryon-diagonalmatrix elements (and as a special case to the electromagnetic current). Thequark-line connected expansions are given there with the general expressionsdescribed in Appendix C while the quark-line disconnected expansions are givenin Appendix D.In section 11 we discuss improvement coefficients for the currents, see e.g. [12],and show that they lead to (small) modifications of the SU (3) flavour symmetricbreaking expansion coefficients. Using the vector current as an example, we showhow we can determine two improvement coefficients (and the renormalisationconstant). Section 12.1 briefly describes how matrix elements (i.e. form factors)are computed from the ratios of 3-point to 2-point correlation functions. Insection 12.2, we describe our n f = 2 + 1 flavour Wilson clover action used andprovide some numerical details. In section 13, specialising to the vector currentagain we give some flavour singlet ‘ X ’-plots, showing their constancy for the F and F form factors. This is followed by some fan plots revealing SU (3)-breakingeffects. The momentum transfer ( Q ) dependence of the expansion coefficients isalso investigated. The numerical values of two improvement coefficients are alsodetermined. Finally in section 14 we give our conclusions. We take here ‘generalised currents’ to be J F ( M ) = qF γ ( M ) q ≡ (cid:88) f ,f =1 F f f q f γ ( M ) q f , (1)where q is a flavour vector, q = ( u, d, s ) T , F is a flavour matrix and γ ( M ) issome Dirac gamma matrix. In particular we have γ ( M ) = γ ( M ) µ , γ ( M ) µ γ ( M )5 , I , iγ ( M )5 and σ ( M ) µν for the vector V ( M ) µ , axial A ( M ) µ , scalar S ( M ) , pseudoscalar P ( M ) and tensor T ( M ) µν generalised currents respectively. The further generalisationto operators including covariant derivatives is straightforward. With our gammamatrix conventions, we obviously have J F( M ) † = ¯ qF T γ ( M ) q , (2)7nd so are Hermitian if the flavour matrix, F , is symmetric and anti-Hermitianif F is antisymmetric.We use Minkowski space , and to emphasise this we use the superscript: ( M ) .The expansion described later will be valid whether we are working in Minkowskior Euclidean space (when we will drop the superscript). We wish to computematrix elements for B → B (cid:48) A ( B → B (cid:48) ) = (cid:104) B (cid:48) , (cid:126)p (cid:48) , (cid:126)s (cid:48) | J F( M ) ( q ) | B, (cid:126)p, (cid:126)s (cid:105) ≡ A ¯B (cid:48) FB , (3)where B and B (cid:48) belong to the baryon octet, the members of which are shown inFig. 1 (the quark content of each baryon is also depicted there). This can thusinclude scattering processes for example Be → Be or semi-leptonic (or β -decays) B → B (cid:48) e ¯ ν e from a parent baryon, B , to a daughter baryon B (cid:48) . For semi-leptonicdecays in the standard model, neutral currents are flavour diagonal, and hencethere is an absence of flavour-changing neutral currents (FCNCs), i.e. s → d transitions. In addition ∆ S = ∆ Q violating modes are not seen. From Fig. 1we see that this means that transitions from right to the left in the picture aresuppressed or absent. For example twelve allowed non-hyperon and hyperon β -decays, are listed in Table 1 of [13]. Of course the present formalism does notincorporate these constraints, but this can motivate our choice of independentmatrix elements, which are transitions from the left to the right in Fig. 1.Momentum transfer p ( M ) − p ( M ) (cid:48) is more natural to take for semi-leptonicdecays, as this is the momentum carried by the lepton and neutrino. Howeverfor scattering processes p ( M ) (cid:48) − p ( M ) is more natural. We wish to adopt a unifiednotation here, so we define the momentum transfer as q ( M ) = p ( M ) (cid:48) − p ( M ) = ( E B (cid:48) ( (cid:126)p (cid:48) ) − E B ( (cid:126)p ) , (cid:126)p (cid:48) − (cid:126)p ) . (4)The decompositions of the matrix elements in eq. (3) are standard, and we write (cid:104) B (cid:48) , (cid:126)p (cid:48) , (cid:126)s (cid:48) | J F( M ) ( q ) | B, (cid:126)p, (cid:126)s (cid:105) = ¯ u B (cid:48) ( (cid:126)p (cid:48) , (cid:126)s (cid:48) ) J ( M ) ( q ) u B ( (cid:126)p, (cid:126)s ) , (5)with for J ( M ) V ( M ) µ = γ ( M ) µ F + iσ ( M ) µν q ( M ) ν F M B + M B (cid:48) + q ( M ) µ F M B + M B (cid:48) , A ( M ) µ = (cid:18) γ ( M ) µ G + iσ ( M ) µν q ( M ) ν G M B + M B (cid:48) + q ( M ) µ G M B + M B (cid:48) (cid:19) γ ( M )5 , S ( M ) = g S , P ( M ) = iγ ( M )5 g P , (6) T ( M ) µν = σ ( M ) µν h + i ( q ( M ) µ γ ( M ) ν − q ( M ) ν γ ( M ) µ ) h M B + M B (cid:48) The conventions used include η µν = diag(1 , − , − , − γ ( M ) µ † = γ ( M )0 γ ( M ) µ γ ( M )0 , γ ( M )5 = iγ ( M )0 γ ( M )1 γ ( M )2 γ ( M )3 and σ ( M ) µν = i/ γ ( M ) µ , γ ( M ) ν ]. i ( q ( M ) µ P ( M ) ν − q ( M ) ν P ( M ) µ ) h ( M B + M B (cid:48) ) + i ( γ ( M ) µ /q ( M ) γ ( M ) ν − γ ( M ) ν /q ( M ) γ ( M ) µ ) h M B + M B (cid:48) , where P ( M ) = p ( M ) + p ( M ) (cid:48) . F i ≡ F ¯B (cid:48) FB i , G i ≡ G ¯B (cid:48) FB i , g S ≡ g ¯B (cid:48) FB S , g P ≡ g ¯B (cid:48) FB P and h i ≡ h ¯B (cid:48) FB i are the form factors and are functions of q ( M ) 2 and the masses ofthe baryons (or alternatively the quark masses). Each combination in eqs. (5, 6)represents a current times a form factor (i.e. the coefficient). For example thefirst term for the vector current reads ¯ u B (cid:48) ( (cid:126)p (cid:48) , (cid:126)s (cid:48) ) γ ( M ) µ u B ( (cid:126)p, (cid:126)s ) × F ¯B (cid:48) FB ( q ( M ) 2 ). Thegoal of this article is to establish ways in which these form factors depend on thetransition taking place and on the quark masses.From eqs. (2, 3) we have A ∗ ¯BFTB (cid:48) = A ¯B (cid:48) FB , (7)and we now apply this to eq. (5) with individual terms defined by eq. (6). Considerfirst the current pieces. For example for the vector currents we find that thefirst and second terms (i.e. currents) are unaltered, (¯ u B γ ( M ) µ u B (cid:48) ) ∗ = ¯ u B (cid:48) γ ( M ) µ u B ,(¯ u B (cid:48) iσ ( M ) µν ( − q ( M ) ν ) u B ) ∗ = ¯ u B iσ ( M ) µν q ( M ) ν u B (cid:48) while the third current changes sign,(¯ u B ( − q ( M ) µ ) u B (cid:48) ) ∗ = − ¯ u B (cid:48) q ( M ) µ u B . Strong interactions are invariant under T -parity and from this it can be shown that the form factors can be chosen to beall real. Hence from eq. (7) we must have F ¯BFTB (cid:48) = F ¯B (cid:48) FB , F ¯BFTB (cid:48) = F ¯B (cid:48) FB , (8)but F ¯BFTB (cid:48) = − F ¯B (cid:48) FB . (9) F and F are called first class form factors while F is called a second class formfactor. This can be applied to all the further currents. These properties of theform factors thus give rise to the notation, [14]first class F , F , G , G , g S , g P , h , h , h second class F , G , h , (10)(with the meaning given by eqs. (8, 9)). Note that when B (cid:48) = B , then thesecond class currents (i.e. form factors) vanish. This occurs, either for a scatteringprocess (i.e. a diagonal current in flavour space, so the matrix F is symmetricand the current is Hermitian) or for semi-leptonic processes at the quark masssymmetric point.We now consider the flavour structures, i.e. the possible flavour matrices ineq. (1). In Table 1 we give the possible octet states, i = 1 , . . . , i = 0. As we are primarily concerned with the9ndex Baryon ( B ) Meson ( F ) Current ( J F )1 n K ¯ dγs p K + ¯ uγs − π − ¯ dγu π √ (cid:0) ¯ uγu − ¯ dγd (cid:1) η √ (cid:0) ¯ uγu + ¯ dγd − sγs (cid:1) + π + ¯ uγd − K − ¯ sγu ¯ K ¯ sγd η (cid:48) √ (cid:0) ¯ uγu + ¯ dγd + ¯ sγs (cid:1) Table 1:
Our numbering and conventions for the generalised currents. For example, B = Σ − , F = π − , J F ≡ J π − . We use the convention that current (i.e. operator)numbered by i has the same effect as absorbing a meson with the index i . γ representsan arbitrary Dirac matrix. flavour structure of bilinear operators, we use the corresponding meson name forthe flavour structure of the bilinear quark currents. So for example the i = 5current is given by the flavour matrix F η = diag(1 , , − / √
6. We shall use theconvention that the current i has the same effect as absorbing a meson with thesame index. In the operator expressions q is the annihilation operator and ¯ q thecreation operator. As an example, we note that absorbing a π + annihilates one d quark and creates a u quark. That is J π + | (cid:105) ∝ | π + (cid:105) , (11)while (cid:104) p | ¯ uγd | n (cid:105) = (cid:104) p | J π + | n (cid:105) represents p = π + n .As an example of this (current) notation the quark electromagnetic currentcan be written by defining an appropriate flavour matrix F or alternatively as J em µ = 23 ¯ uγ µ u −
13 ¯ dγ µ d −
13 ¯ sγ µ s ≡ √ V π µ + 1 √ V ηµ . (12)Furthermore the charged W s currents are a mixture of the charged π and K currents, while the Z current is diagonal and thus a mixture of the π , η and η (cid:48) currents. The K current is a FCNC, so only contributes to beyond the standardmodel (BSM) or higher-order processes.The previous discussion on first and second class currents can now be refor-mulated in terms of these flavour matrices and isospin rotations . The diagonal This discussion follows [15]. i = 4,5 and 0 with F π , F η and F η (cid:48) respectively. As a result F , G , g P , h and h allvanish for these currents. For the off-diagonal currents consider the SU (3)-flavoursymmetric point. As all the quark masses have the same mass, and in particularthe u and d quarks then we first consider isospin, I , invariance. Isospin rota-tions are d - u rotations and relate off-diagonal currents to diagonal currents. (Forexample (cid:104) p | J π + | n (cid:105) is related to (cid:104) p | J π | p (cid:105) , see section 4.2.) Similarly for U -spinrotations s - d , and V -spin rotations s - u . Hence we expect that for transitionswithin a given multiplet (whether I , U or V ) at the SU (3)-flavour symmetricpoint then again F , G , g P , h and h all vanish. Between isospin multipletsthey need not vanish when SU (3) flavour symmetry is broken. We later discussthis in more detail and our coefficient tables, for example Table 6, reflect theseresults. As mentioned already, we follow the strategy used in [2] of holding constant theaverage bare quark mass ¯ m = 13 ( m u + m d + m s ) . (13)This greatly reduces the number of mass polynomials which can occur in Taylorexpansions of physical quantities, and relates the quark-mass dependencies ofhadron masses or matrix elements within an SU (3) multiplet. Since we expandabout the symmetric point where all three quarks have the same mass, it is usefulto introduce the notation δm q ≡ m q − ¯ m , q = u, d, s , (14)to describe the ‘distance’ from the SU (3) flavour symmetry point. Note that itfollows from the definition that we have the identity δm u + δm d + δm s = 0 , (15)so we can always eliminate one of the δm q . In this article we concentrate on the n f = 2 + 1 case, i.e. we keep m u = m d ≡ m l . (16)All our expansion coefficients are functions of ¯ m . The methods developed herecan be generalised to the case of n f = 1 + 1 + 1 non-degenerate quark-massflavours. For this case eq. (15) reduces to2 δm l + δm s = 0 , (17)11hich we use to eliminate δm s . Thus, all mass dependences will be expressed aspolynomials in the single variable δm l . At the physical point m l (cid:28) ¯ m , so δm l isnegative. However on the lattice in principle we are free to choose δm l positive,and look at matrix elements on both sides of the symmetric point. In the following we want to use group theory in flavour space to calculate thepossible quark-mass dependence of baryonic form factors. However for simplicityof notation we shall continue to discuss matrix elements and amplitudes, butit should be noted that for form factors the Lorentz/Dirac structure has beenfactored out. So we shall consider the quark mass expansion for (cid:104) B i | J F j | B k (cid:105) ≡ A ¯ B i F j B k . (18)The indices i and k will run from 1 to 8 for octet hadrons (or 1 to 10 for decuplets).The currents/operators we are interested in are quark bilinears, so the index j willrun from 1 to 8 for non-singlets, or 0 for the singlet. In the following the singletwill be considered separately. When i (cid:54) = k we get transition matrix elements;when i = k within the same multiplet, we get operator expectation values. Thishas already been indicated in Table 1.The allowed quark mass Taylor expansion for a hadronic matrix element mustfollow the schematic pattern (cid:104) B i | J F j | B k (cid:105) = (cid:88) (singlet mass polynomial) × (singlet tensor) ijk + (cid:88) (octet mass polynomial) × (octet tensor) ijk (19)+ (cid:88) (27-plet mass polynomial) × (27-plet tensor) ijk + · · · . The mass polynomials have been determined and given in Table III of [2]. Therelevant part of this table is given in Table 2 where we classify all the polynomialswhich could occur in a Taylor expansion about the symmetric point, δm q = 0, q = u , d , s up to O ( δm q ). The tensors in eq. (19) are 3-dimensional arraysof integers and square-roots of integers; objects somewhat analogous to three-dimensional Gell-Mann matrices. We recover the standard results for unbroken SU (3) by only keeping singlet tensors on the right-hand side of eq. (19). Addinghigher dimensional flavour tensors tells us the allowed mass dependences of matrixelements. The dots in eq. (19) represent terms that are cubic or higher in δm q .We now need to classify the three-index tensors according to their grouptransformations, using the same techniques we used for masses [2]. The newcases to look at will be 8 ⊗ ⊗ ⊗ ⊗
10 for octet and decuplet hadronsrespectively, 10 ⊗ ⊗ SU (3)1 1 δm s δm u − δm d ) 8 δm u + δm d + δm s δm s − ( δm u − δm d ) δm s ( δm d − δm u ) 8 27 δm u δm d δm s δm s ( δm u + δm d + δm s ) 8 27 64( δm u − δm d )( δm u + δm d + δm s ) 8 27 64( δm s − δm u )( δm s − δm d )( δm u − δm d ) 10 10 64 Table 2:
All the quark-mass polynomials up to O ( δm q ), classified by symmetry prop-erties. ⊗ ⊗ f π and f K The vacuum is a singlet, so vacuum to meson, M , matrix elements or decayconstants (cid:104) | J F j | M k (cid:105) , j = 1 , . . . , ⊗ ⊗ ⊗ f π and f K is similar to theallowed dependence of M π and M K , as given in [2]. Results using this approachare given in [16]. For example to LO we have f π = F + 2 Gδm l ,f K = F − Gδm l . (20)The same argument applies in principle to hyperon distribution amplitudes qqq ,and to baryon decays via qqqe Recall from eq. (3) that we have used the notation for the matrix element tran-sition B → B (cid:48) of A ¯ B (cid:48) F B = (cid:104) B (cid:48) | J F | B (cid:105) , (21)where J F is the appropriate operator from Table 1 and F denotes the flavourstructure of the operator. But note that as we are suppressing the Lorentz struc-ture, this includes first and second class form factors as given in eq. (10).13 .1 Sign conventions: Octet operators and octet hadrons In the case of a n f = 2 + 1 simulation we only need to give the amplitudes for oneparticle in each isospin multiplet, and can then use isospin symmetry to calculateall other amplitudes in (or between) the same multiplets. So, for example, wecan calculate the Σ − and Σ matrix elements if we are given all the Σ + matrixelements. Similarly, given the Σ + → p transition amplitude, we can find all theother Σ → N transition amplitudes. All the symmetry factors will be listed insection 4.2.In the next section we will calculate the allowed quark-mass dependencies ofthe amplitudes between the baryons. Within this set there are 7 diagonal matrixelements, and 5 transition amplitudes making 7 + 5 = 12 in total. The 7 diagonalelements are A ¯ NηN , A ¯Σ η Σ , A ¯Λ η Λ , A ¯Ξ η Ξ and A ¯ NπN , A ¯Σ π Σ , A ¯Ξ π Ξ , (22)because there are four I = 0 amplitudes, one for each particle, but only three I = 1 amplitudes, because isospin symmetry rules out an I = 1, Λ ↔ Λ amplitude. There are only 5 transition amplitudes A ¯Σ π Λ and A ¯ NK Σ , A ¯ NK Λ , A ¯Λ K Ξ , A ¯Σ K Ξ , (23)because no octet operator changes strangeness by ±
2, so there is no p ↔ Ξ transition amplitude. See the forthcoming Tables 3 and 4 for the explicit results.To discuss transition matrix elements, we need to specify the hadron statescarefully. If we do not, then the phases and signs of transition matrix elementsbecome ambiguous. (This is not a problem with masses, or diagonal matrixelements such as (cid:104) p | J | p (cid:105) .)We follow a convention commonly used in chiral perturbation theory , e.g.[18, 19] where the mesons transform under SU (3) rotations like the 3 × M = √ π + √ η π + K + π − − √ π + √ η K K − ¯ K − √ η , (24)and octet baryons like the matrix B = √ Σ + √ Λ Σ + p Σ − − √ Σ + √ Λ n Ξ − Ξ − √ Λ , However some papers use different definitions, e.g. in chapter 18 of [17] the meson matrix M is defined the same way as in eq. (24), but in the baryon matrix B the Ξ − appears with aminus sign in comparison to eq. (25). Using the Gasiorowicz convention, [17], would give theopposite sign to all transition matrix elements to or from the Ξ − . B = √ ¯Σ + √ ¯Λ ¯Σ − ¯Ξ − ¯Σ + − √ ¯Σ + √ ¯Λ ¯Ξ ¯ p ¯ n − √ ¯Λ . So for example π + , π , π − are represented by the matrices , √ − √
00 0 0 , , (26)respectively. Under an SU (3) rotation the M , B and ¯ B matrices transform as M → U M U † , B → U BU † , and ¯ B → U ¯ BU † . (27) SU (2) relations As discussed previously we use the convention that operator number i , repre-senting an appropriate flavour matrix, has the same effect on quantum numbersas the absorption of a meson with the index i . So for example, from Table 1operator 6 annihilates a d quark and creates a u , and hence changes a neutroninto a proton, i.e. (cid:104) p | ¯ uγd | n (cid:105) ≡ (cid:104) p | J π + | n (cid:105) ≡ (cid:104) B | J F | B (cid:105) . (28)In Tables 3 and 4 we list the isospin relationships between all of the allowedmatrix elements in the octet, and our standard 7 + 5 = 12 matrix elements.Making the choice given in eqs. (24, 25) which is conventional in chiral pertur-bation theory, the isospin raising and lowering operators do not follow the usualCondon–Shortley sign convention. The Wigner–Eckart theorem applies, but thesigns are not always the ones from the standard Clebsch–Gordan coefficients.To demonstrate this, consider the transformations given in eq. (27) with U =exp( iα i λ i ). Infinitesimal transformations ( α i →
0) correspond to commutators ofthe type [ λ i , B ] or [ λ i , M ]. The isospin operations are constructed from the firstthree λ matrices I = λ ,I + = ( λ + iλ ) , (29) I − = ( λ − iλ ) .I has the expected resultˆ I M = [ λ , M ] = π + 12 K + − π − − K − K − ¯ K , (30)15 (cid:104) n | J η | n (cid:105) A ¯ NηN (cid:104) p | J η | p (cid:105) A ¯ NηN (cid:104) Σ − | J η | Σ − (cid:105) A ¯Σ η Σ (cid:104) Σ | J η | Σ (cid:105) A ¯Σ η Σ (cid:104) Σ + | J η | Σ + (cid:105) A ¯Σ η Σ (cid:104) Λ | J η | Λ (cid:105) A ¯Λ η Λ (cid:104) Ξ − | J η | Ξ − (cid:105) A ¯Ξ η Ξ (cid:104) Ξ | J η | Ξ (cid:105) A ¯Ξ η Ξ I (cid:104) n | J π | n (cid:105) − A ¯ NπN (cid:104) p | J π | p (cid:105) A ¯ NπN (cid:104) n | J π − | p (cid:105) √ A ¯ NπN (cid:104) p | J π + | n (cid:105) √ A ¯ NπN (cid:104) Σ − | J π | Σ − (cid:105) − A ¯Σ π Σ (cid:104) Σ | J π | Σ (cid:105) (cid:104) Σ + | J π | Σ + (cid:105) A ¯Σ π Σ (cid:104) Σ − | J π − | Σ (cid:105) A ¯Σ π Σ (cid:104) Σ | J π − | Σ + (cid:105) − A ¯Σ π Σ (cid:104) Σ | J π + | Σ − (cid:105) A ¯Σ π Σ (cid:104) Σ + | J π + | Σ (cid:105) − A ¯Σ π Σ (cid:104) Λ | J π | Λ (cid:105) (cid:104) Ξ − | J π | Ξ − (cid:105) − A ¯Ξ π Ξ (cid:104) Ξ | J π | Ξ (cid:105) A ¯Ξ π Ξ (cid:104) Ξ − | J π − | Ξ (cid:105) −√ A ¯Ξ π Ξ (cid:104) Ξ | J π + | Ξ − (cid:105) −√ A ¯Ξ π Ξ Table 3:
The isospin relations connecting the set of octet matrix elements with ourstandard subsets A ¯ BF B (each independent set separated by horizontal lines). Lefttable: The I = 0 diagonal relations; right table: the I = 1 transition relations withinthe same isospin multiplet. ˆ I B = [ λ , B ] = + 12 p − Σ − − n − Ξ − Ξ . (31)For example regarding π − as the matrix in eq. (26) givesˆ I π − = − = − π − , (32)(see Fig. 1). Similarly for the baryons, for example ˆ I n = − n , etc... .However ˆ I + and ˆ I − produce results at odds with the Condon-Shortley or CSphase convention, which has positive coefficients for the non-zero matrix elementsof the raising and lowering operators.ˆ I + M = [ λ + iλ , M ] = π − −√ π K − π − − K − . (33)16 (cid:104) Σ − | J π − | Λ (cid:105) A ¯Σ π Λ (cid:104) Σ | J π | Λ (cid:105) A ¯Σ π Λ (cid:104) Σ + | J π + | Λ (cid:105) A ¯Σ π Λ12 (cid:104) n | J K + | Σ − (cid:105) A ¯ NK Σ12 (cid:104) n | J K | Σ (cid:105) − A ¯ NK Σ / √ (cid:104) p | J K + | Σ (cid:105) A ¯ NK Σ / √ (cid:104) p | J K | Σ + (cid:105) A ¯ NK Σ12 (cid:104) n | J K | Λ (cid:105) A ¯ NK Λ12 (cid:104) p | J K + | Λ (cid:105) A ¯ NK Λ12 (cid:104) Λ | J K + | Ξ − (cid:105) A ¯Λ K Ξ12 (cid:104) Λ | J K | Ξ (cid:105) A ¯Λ K Ξ12 (cid:104) Σ − | J K | Ξ − (cid:105) A ¯Σ K Ξ12 (cid:104) Σ | J K + | Ξ − (cid:105) A ¯Σ K Ξ / √ (cid:104) Σ | J K | Ξ (cid:105) − A ¯Σ K Ξ / √ (cid:104) Σ + | J K + | Ξ (cid:105) A ¯Σ K Ξ I (cid:104) Λ | J π + | Σ − (cid:105) A ¯Λ π Σ (cid:104) Λ | J π | Σ (cid:105) A ¯Λ π Σ (cid:104) Λ | J π − | Σ + (cid:105) A ¯Λ π Σ12 (cid:104) Σ − | J K − | n (cid:105) A ¯Σ ¯ KN (cid:104) Σ | J ¯ K | n (cid:105) − A ¯Σ ¯ KN / √ (cid:104) Σ | J K − | p (cid:105) A ¯Σ ¯ KN / √ (cid:104) Σ + | J ¯ K | p (cid:105) A ¯Σ ¯ KN (cid:104) Λ | J ¯ K | n (cid:105) A ¯Λ ¯ KN (cid:104) Λ | J K − | p (cid:105) A ¯Λ ¯ KN (cid:104) Ξ − | J K − | Λ (cid:105) A ¯Ξ ¯ K Λ12 (cid:104) Ξ | J ¯ K | Λ (cid:105) A ¯Ξ ¯ K Λ12 (cid:104) Ξ − | J ¯ K | Σ − (cid:105) A ¯Ξ ¯ K Σ12 (cid:104) Ξ − | J K − | Σ (cid:105) A ¯Ξ ¯ K Σ / √ (cid:104) Ξ | J ¯ K | Σ (cid:105) − A ¯Ξ ¯ K Σ / √ (cid:104) Ξ | J K − | Σ + (cid:105) A ¯Ξ ¯ K Σ Table 4:
The isospin relations connecting the transition set of octet matrix elementswith our standard subsets A ¯ B (cid:48) F B (each independent set separated by horizontal lines).Left table: The ‘forward’ I = 1 and relations; right table: the inverse relations. Again using the π − as an example and comparing this result with eq. (24) we seethat we have ˆ I + π − = − = √ π . (34)Listing all the relations gives ˆ I + π − = √ π , ˆ I + π = −√ π + , (35)ˆ I + K = K + , ˆ I + K − = − ¯ K . Similarly ˆ I + Σ − = √ , ˆ I + Σ = −√ + , (36)ˆ I + n = p , ˆ I + Ξ − = − Ξ . I − is similar. Since these relations are not those usually used to cal-culate the Clebsch-Gordon coefficients, we need to tabulate the isospin relationswithin each multiplet. The signs of the ˆ I + matrix elements follow directly fromthe choice of signs in the chiral perturbation theory representation of the mesonand baryon octets as 3 × − and Ξ are not. Theminus sign tells us that one of the Ξ states must have the opposite phase to theCS convention. Since only relative phases are observable, we could choose the Ξ to have the CS phase, and the Ξ − to have the flipped phase. (Making the otherchoice would not change the final result.) Similarly looking at the Σ baryons wecould choose the Σ + to have the CS phase, and the Σ − and Σ to have flippedphase (or vice versa).One choice of phases that would match eqs. (35, 36) would be to choose the n , p , Σ + and Ξ as standard, and the Σ − , Σ and Ξ − as flipped, and the equivalentchoice for the meson currents (i.e. π − , π , K − flipped). If we look in Tables 3and 4 we see that matrix elements involving an even number of hadrons from theflipped group, the Clebsch-Gordon factor is the same as that in the usual tables,if an odd number of flipped hadrons are involved, the sign is the opposite to thatin the usual tables.As an example of the use of Table 3, we show how the unbroken SU (2) symme-try can be used to find the transition amplitude (cid:104) p | J π + | n (cid:105) from the correspondingdiagonal amplitude (cid:104) p | J π | p (cid:105) . From the table (cid:104) p | J π + | n (cid:105) = √ A ¯ NπN = √ (cid:104) p | J π | p (cid:105) , (37)giving (cid:104) p | ¯ uγd | n (cid:105) = (cid:104) p | (¯ uγu − ¯ dγd ) | p (cid:105) , (38)which is again the simple example showing the relation between off-diagonal anddiagonal currents briefly discussed in section 2. We first consider the simple singlet case (operators with the η (cid:48) flavour structure, i = 0, see Table 1) and then consider the octet states. For matrix elements involving singlet currents, (cid:104) B i | J F | B i (cid:105) ≡ (cid:104) B i | J η (cid:48) | B i (cid:105) , weneed the SU (3) analysis of 8 ⊗ ⊗ ⊗ A ¯ Nη (cid:48) N = a + 3 a δm l ,A ¯Λ η (cid:48) Λ = a + 3 a δm l , (39) A ¯Σ η (cid:48) Σ = a − a δm l ,A ¯Ξ η (cid:48) Ξ = a − a − a ) δm l , with higher orders given in [2]. To find the allowed mass dependence of octet matrix elements of octet hadronswe need the SU (3) decomposition of 8 ⊗ ⊗
8. Using the intermediate result8 ⊗ ⊕ ⊕ ⊕ ⊕ ⊕ , (40)we find8 ⊗ ⊗ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ . With three unequal quark masses, the n f = 1 + 1 + 1 case, I and Y are both‘good’ flavour quantum numbers, so the tensors in eq. (19) will satisfy I = 0, Y = 0, i.e. they will be the central locations (spots) in each multiplet in Fig. 2.Thus in a full n f = 1 + 1 + 1 flavour calculation (three different quark masses)we would see contributions from all the representations in eq. (41).Fortunately in the n f = 2 + 1 case the good flavour quantum numbers are I and Y , giving us the stronger constraint that only tensors with I = 0, Y = 0enter into eq. (19). The 10, 10, 35 and 35 do not contain any I = 0, Y = 0operators, so they no longer contribute in the 2 + 1 case, which means that wecan neglect those representations at present [17, 20]. For example for the Y = 0line for the octet, we have an isospin triplet and singlet of states and similarly forthe 27-plet (isospin 5-plet, triplet and singlet) and 64-plet (isospin 7-plet, 5-plet,triplet and singlet). However for the 10-plet we have just an isospin triplet andfor the 35-plet a 5-plet and triplet. In both cases there is no Y = 0 isospin singlet.We have already seen this phenomenon in [2] for the case of the 10 and 10. Thesimplest quark-mass polynomial with 10, 10 symmetry was ( δm s − δm u )( δm s − δm d )( δm u − δm d ) (see Table 2), which vanishes if any two quark masses are equal.The 10 and 10 only appeared in two quantities we have considered, the violationof the Coleman-Glashow mass relation, and in Σ – Λ mixing [21], both of whichare isospin violating. 19 igure 2: I , Y plots for some of the SU (3) multiplets which appear in the decom-position of 8 ⊗ ⊗
8. The left-hand plot illustrates the octet, 27-plet and 64-plet rep-resentations (clockwise). The right-hand plot shows the 10 and 35-plets (left to right).The number of spots in the central location gives the number of flavour-conservingoperators in each multiplet. SU (3) symmetry-breaking expansions Because 8 × × × × × × × np Σ − Σ Λ Σ + Ξ − Ξ and K K + π − π ηπ + K − ¯ K . (42)The 8 generators of SU (3) are now a set of 8 × λB inthe matrix-vector notation has the same effect as [ λ, B ] in the 3 × λ = √ √ −√ −√ −
10 0 0 0 0 0 − , = i − i i √ − i √ − i √ i √ − i i ,λ = − − − ,λ = 1 √ −√ − −√ −√ − −√ √ √
20 0 0 1 √ √ ,λ = 1 √ i √ i i √ − i √ − i − i − i √ − i √ − i √
20 0 0 i i √ i √ ,λ = 1 √ −√ −√ √ − −√ √ −√ √ − √ , = 1 √ − i i √ i √ − i √ i i − i √ − i √ − i √ i √ − i i √ ,λ = √ − − . (43) These 8 × λ matrices follow similar relations to the familiar 3 × (cid:2) λ i , λ j (cid:3) = 2 if ijk λ k , Tr( λ i λ j ) = 12 δ ij , (44)and I = 12 λ , Y = 1 √ λ , (45)with the difference that the 3 × I and Y for the individualquarks, but the 8 × Under an SU (3) rotation the tensors on the right-hand side of eq. (19) transformaccording to T (cid:48) ijk = U † ia T abc U bj U ck . (46)The change in T under an infinitesimal transformation by the generator λ α isˆ O α T ≡ − λ αia T ajk + T ibk λ αbj + T ijc λ αck . (47)The Casimir operator for the SU (3) representation isˆ CT = (cid:88) α =1 , ˆ O α ˆ O α T , (48)while the Casimir for the SU (2) isospin subgroup isˆ I T = (cid:88) α =1 , ˆ O α ˆ O α T . (49)22he n f = 2 + 1 mass matrix commutes with λ , λ , λ (the generators of isospin)and λ (hypercharge). We are looking for tensors which obey these symmetries,so we require ˆ O α T = 0 , α = 1 , , , . (50)The Casimir operator has the following eigenvalues for the representations occur-ring in 8 ⊗ ⊗
8, see for example chapter 4 of [20] or chapter 7 (exercise 7.12) of[22] representation 1 8 10 10 27 35 35 64Casimir eigenvalue 0 3 6 6 8 12 12 15 . (51)We now want to construct tensors which are eigenstates of the Casimir operator,and which satisfy the conditions in eq. (50). This is analogous to constructingan eigenvector if we know the eigenvalues. We have a large number of simultane-ous linear equations involving the numbers T ijk . The solutions tend to be sparsewith the conditions in eq. (50) forcing many entries to be zero. We calculatethe tensors of a given symmetry with the help of Mathematica, [23]. We beginwith a completely general tensor T ijk with 8 entries, and impose the conditionseq. (50). This forces many entries to be zero, as it eliminates all entries in whichthe flavour quantum numbers of the ‘outgoing’ particle i is not the sum of theflavours of j and k (for example (cid:104) Ξ | J π + | p (cid:105) = 0 because charge and strangenessdo not balance). The conditions eq. (50) are also sufficient to force all the re-lations in Tables 3 and 4 to hold. After imposing eq. (50) we have reduced theinitial general tensor with 8 = 512 entries down to a tensor with only 17 inde-pendent parameters. From the decomposition of 8 ⊗ ⊗ Y = 0 , I = 0 central state, see Fig. 2 and the related discussion. The17 linearly independent tensors remaining after imposing eq. (50) can now befurther classified as eigenstates of the Casimir operator. Finding these tensors isa simple matter of solving simultaneous equations, analogous to determining aneigenvector once the eigenvalue is known.As in the case of degenerate eigenvalues, there is a degree of choice in choos-ing which linear combinations of the eigenstates we choose as our basis. Oftenthere are interchange operations which we can choose to be even or odd. In par-ticular we can choose our tensors to be first class or second class depending onthe symmetry or antisymmetry when the baryons are switched, as discussed insection 2.We can see this by introducing a reflection matrix R which inverts each octet,23eaving the central two states unchanged R = . (52) For the mesons this is the charge conjugation operation. We note that R = I (the unit matrix), so R can only have the eigenvalues ±
1, hence we can classifystates according to whether they are even or odd under operations involving R .Tensors can be divided into first or second class depending on the symmetryfirst class T ijk = + T kai R aj , second class T ijk = − T kai R aj , (53)in which the baryon order is reversed, and R applied to the current (meson)index. Furthermore the definition of first/second class tensors in eq. (53) agreeswith the previous discussion: in eqs. (8, 9) we interchanged B and B (cid:48) and tookthe transpose of the flavour matrix, F . This latter operation is easily seen to beequivalent to the reflection, R in eq. (53).We can further classify tensors by the symmetry when R is applied to all threeindices d − like T ijk = + R ia T abc R bj R ck ,f − like T ijk = − R ia T abc R bj R ck . (54)As can be seen from eq. (41) there must be two singlet eigenstates, eightoctets, six 27-plets and one 64-plet, 17 in total. All tensors, T , are classified bytheir symmetry properties, according to whether first or second class, eq. (53),and whether they are f − like or d − like, eq. (54), and are given by SU (3) T , 1 st class T , 2 nd class d − like f − like d − like f − like1 d f r , r , r s , s t , t u q , q w , w x y z . (55)Furthermore in Appendix A we list all non-zero elements for all 17 tensors, to-gether with their values. For example in eq. (56) we give the non-zero elements24f the tensors T = r and t , T T ijk ijkr t − . (56)The values of the non-zero T ijk elements are given in the second column, whiletheir position is given in the third block. In particular we see that the r tensoronly has 8 non-zero entries, all identical in value, in the positions T i i , where i can take any value from 1 to 8. It can easily be checked, for example, that thetensors r and t with non-zero elements as given in eq. (56) are first and secondclass tensors, respectively.The r i tensors are d -like and can be regarded as responsible for the quark-massdependence of the d coupling (see the d -fan in section 8), while the s i tensors are f -like and act as quark-mass-dependent additions to the f coupling (as seen inthe f -fan – see section 8).We are now finally in a position to present the SU (3) flavour-symmetry-breaking expansions. As we are considering only the isospin limit, eq. (16), thenTable 2 reduces to Table 5. For example, let us consider (cid:104) p | J π + | n (cid:105) ≡ (cid:104) B | J F | B (cid:105) ,Polynomial SU (3)1 1 δm l δm l δm l δm l δm l Table 5:
All the quark-mass polynomials in the isospin limit up to O ( δm l ), classifiedby symmetry properties. eq. (28). From Table 3, this is √ A ¯ NπN . Hence from eq. (19), and using Table 5and Appendix A (for the non-zero 261 component of the appropriate tensor) andusing the same notation for the expansion coefficients as for the tensor gives theLO expansion √ A ¯ NπN = 1 × ( √ f + √ d ) + δm l × ( − √ r + 2 √ s ) . (57)At higher orders, we also need in addition the non-zero elements of the 27- and64-plet. Further examples are given in the next section in eqs. (61, 62).25 Coefficient tables
We use the same notation for the expansion coefficients as for the tensor. Forexample the r tensor (with components T i i ) has expansion coefficient r . The SU (3) singlet and octet coefficients in the mass Taylor expansion of operatoramplitudes are tabulated in Table 6. These coefficients are sufficient for the linear
1, 1 st class 8, 1 st class 8, 2 nd class O (1) O ( δm l ) O ( δm l ) f d d d d f f d d fI A ¯ B (cid:48) F B f d r r r s s t t u N ηN √ − − η Σ 0 2 1 0 2 √ η Λ 0 − η Ξ −√ − N πN √ − π Σ 2 0 0 0 0 − √ π Ξ 1 −√ π Λ 0 2 0 1 −√ π Σ 0 2 0 1 −√ − ¯ N K Σ −√ √ √ √ √ √ ¯ N K Λ −√ − −√ √ − ¯Λ K Ξ √ − √ − − −√ − ¯Σ K Ξ √ √ √ −√ −√ √ ¯Σ ¯ KN −√ √ √ √ −√ −√ ¯Λ ¯ KN −√ − −√ − −√ ¯Ξ ¯ K Λ √ − √ − √ ¯Ξ ¯ K Σ √ √ √ −√ √ −√ Table 6:
Coefficients in the mass Taylor expansion of A ¯ B (cid:48) F B operator amplitudes: SU (3) singlet and octet, for both first-class and second-class currents. The first rowgives whether singlet or octet and first or second class, and the second row gives theorder in δm l . The third row gives whether the associated tensor is f − like or d − likeaccording to the definition given in eq. (54). These coefficients are sufficient for thelinear expansion of hadronic amplitudes. expansion of hadronic amplitudes on the constant ¯ m line. (If ¯ m were not keptconstant there would be two more linear terms.)26he table is to be read: for first-class currents the f and d terms are indepen-dent of the quark mass, while the r , r , r and s , s coefficients are the leadingorder (LO) or δm l terms. For second-class currents, as discussed previously, thereare no leading f and d terms, the expansion starts at O ( δm l ) for the off-diagonalcurrents or completely vanishing for the diagonal currents.Thus for example to first order in δm l (i.e. LO) we can read off from Tables 3,4 and 6 (cid:104) p | J η | p (cid:105) = A ¯ NηN = √ f − d + ( r − s ) δm l , (cid:104) n | J K + | Σ − (cid:105) = A ¯ NK Σ = −√ f + √ d + ( √ r + √ s ) δm l , (cid:104) Σ + | J η | Σ + (cid:105) = A ¯Σ η Σ = 2 d + ( r + 2 √ r ) δm l , (58)for first-class currents (for example for the vector current the form factors F and F from eq. (10)) and (cid:104) n | J K + | Σ − (cid:105) = A ¯ NK Σ = ( √ t + √ u ) δm l , (cid:104) Σ − | J K − | n (cid:105) = A ¯Σ ¯ KN = − ( √ t + √ u ) δm l , (59)for second-class currents (for example for the F vector form factor).A notational comment: we shall usually suppress arguments and indices, buteach coefficient in Table 6 is a function of the (momentum transfer) , Q , aswell as being renormalised or not. Thus for example for the renormalised vectorcurrent, the f coefficient in Table 6 is to be understood as f → f V R ( ¯ m, Q ).Note that the clean separation of amplitudes and form factors into first andsecond class depends on the fact that we have defined our amplitudes in waysthat treat the parent and daughter baryons symmetrically. If we had used anunsymmetric definition, for instance always normalising amplitudes in terms ofthe parent baryon’s mass, we would find t i and u coefficients appearing in theexpansions of quantities which ‘should’ only involve the symmetric terms. For completeness in Table 7 we detail the additional quadratic and cubic coef-ficients in the mass Taylor expansion of the operator amplitudes for the 27 and64-plets.For first-class currents in Table 5 the singlet terms do not contribute at thelinear O ( δm l ) level, but are present at the quadratic O ( δm l ) and cubic O ( δm l )levels. Similarly the octet terms are missing at the O (1) level, but are presentat higher orders. Hence these terms are also present at the higher orders in the SU (3) flavour-breaking expansion. There are 5 + 7 = 12 amplitudes, and atthe O ( δm l ) level 11 free parameters, so there is one constraint. (Alternativelyat the O ( δm l ) level one can have all possibilities which are orthogonal to the64-plet, so there is again just one constraint.) At the O ( δm l ) level one has 1227
7, 1 st class 64, 1 st
27, 2 nd class O ( δm l ) O ( δm l ) O ( δm l ) d d f f d d fI A ¯ B (cid:48) F B q q w w z x y N ηN √ √ η Σ − −
10 0 0 −√ η Λ −
18 18 0 0 − √ η Ξ 9 3 0 − √ √ N πN − √ √ − π Σ 0 0 − π Ξ 5 √ −√ − − π Λ 14 − −√ π Σ 14 − −√ − ¯ N K
Σ 0 2 √ − √ √ √ √ √ ¯ N K Λ − √ √ − √ ¯Λ K Ξ − − √ √ √ ¯Σ K Ξ 0 2 √ √ − √ √ −√ √ ¯Σ ¯ KN √ − √ √ √ −√ −√ ¯Λ ¯ KN − √ √ − √ ¯Ξ ¯ K Λ − − √ √ − − √ ¯Ξ ¯ K Σ 0 2 √ √ − √ √ √ −√ Table 7:
Additional coefficients in the mass Taylor expansion of operator amplitudes: SU (3) 27-plet and 64-plet. These additional terms first appear at the quadratic andcubic levels respectively. The same notation as for Table 6. free parameters for the 12 amplitudes (11 previous and one extra one from the64-plet, the z term). Hence there are now no more constraints available at thisand higher orders in δm l .For second-class currents, there are constraints at the O ( δm l ) order as we have5 amplitudes, but only 3 expansion coefficients. However at the next O ( δm l )level we have additional 2 parameters, so there are no more constraints available.Hence for second-class operators there is no point in going higher than linear inthe quark mass in the SU (3) flavour-breaking expansion.Thus, for example, from Tables 6 and 7 we would have for the first-classcurrent (cid:104) p | J η | p (cid:105) = A ¯ NηN = √ f − d + ( r − s ) δm l (60)+( √ f x − d x + r x − s x + 6 q + 3 q + 3 √ w ) δm l √ f xx − d xx + r xx − s xx + 6 q x + 3 q x + 3 √ w x + 3 √ z ) δm l , where f , d is the leading coefficients, and f x , f xx and d x , d xx are the additionalsubdominant coefficients of the same form as the LO singlet, see Table 5. (Weuse x and xx superscripts to distinguish them.) Similarly for r , s , q , q and w and the octet. For the second-class current (cid:104) n | J K + | Σ − (cid:105) = A ¯ NK Σ (61)= ( √ t + √ u ) δm l + ( √ t x + √ u x + √ x + √ y ) δm l . However as just discussed the O ( δm l ) term for the first-class currents and the O ( δm l ) term for the second-class currents have no constraints between the coef-ficients and hence contain no new information.From eqs. (40, 41) and as previously discussed we see that there is one 64-plet in the decomposition of 8 ⊗ ⊗
8, but none in 8 ⊗ O ( δm l ) as shown in Table 5. In [2] we have seenthat the 64-plet combination of decuplet baryon masses is extremely small andwe should probably expect that the 64-plet combination of amplitudes will alsoremain very small all the way from the symmetric point to the physical point.By using Mathematica we construct the 64-plet flavour tensor, and find that itcorresponds to the combination Q ≡ A ¯ NηN − A ¯Σ η Σ − A ¯Λ η Λ + 2 A ¯Ξ η Ξ + √ ( A ¯ NπN − A ¯Ξ π Ξ ) − ( A ¯Σ π Λ + A ¯Λ π Σ ) + 2 ( A ¯Λ K Ξ + A ¯ NK Λ + A ¯Λ ¯ KN + A ¯Ξ ¯ K Λ )+ (cid:113) ( A ¯ NK Σ + A ¯Σ K Ξ + A ¯Ξ ¯ K Σ + A ¯Σ ¯ KN )= O ( δm l ) , (62)and as expected the linear and quadratic terms in δm l vanish. We also note thatthis quantity should be zero at the 1-loop level in chiral perturbation theory [6].In the remainder of this article we shall not consider these next-to-leading-order (NLO) and next-to-next-leading-order (NNLO) higher orders further. We now further discuss amplitudes at the symmetric point. From eq. (41) thereare two octets and one singlet in the decomposition of 8 ⊗
8, so there will be twosinglets in 8 ⊗ ⊗
8. This means that at the symmetric point there are two waysto couple an octet operator between octet baryons. These correspond to the firsttwo columns of Table 6. These two couplings are traditionally given the letters F and D . The F coupling has a pattern related to the SU (3) structure constant f ijk and the D coupling is related to d ijk . In terms of the 3 × F coupling is proportional to Tr( M [ ¯ B, B ]), the D coupling to Tr( M { ¯ B, B } ).29et us first look at the pattern of amplitudes at the symmetric point (withno breaking of SU (3) flavour symmetry). We can read off the correspondinghadronic matrix elements from Table 6 and can construct many matrix elementcombinations which have to be equal at the symmetric point, for example √ (cid:104) p | J η | p (cid:105) + 12 (cid:104) p | J π | p (cid:105) = (cid:104) Σ + | J π | Σ + (cid:105) = − √ (cid:104) Ξ | J η | Ξ (cid:105) + 12 (cid:104) Ξ | J π | Ξ (cid:105) = 2 f , (63) − (cid:104) p | J η | p (cid:105) + √ (cid:104) p | J π | p (cid:105) = (cid:104) Σ + | J η | Σ + (cid:105) = −(cid:104) Λ | J η | Λ (cid:105) = − (cid:104) Ξ | J η | Ξ (cid:105) − √ (cid:104) Ξ | J π | Ξ (cid:105) = 2 d . These relations become more transparent if we write the operators out in ¯ qγq form, following Table 1 giving1 √ (cid:104) p | (¯ uγu − ¯ sγs ) | p (cid:105) = 1 √ (cid:104) Σ + | (¯ uγu − ¯ dγd ) | Σ + (cid:105) = 1 √ (cid:104) Ξ | (¯ sγs − ¯ dγd ) | Ξ (cid:105) (64)= 2 f , from the first line of eq. (63). Written out in this form, it is clear why these threematrix elements have to be the same at the symmetric point. The u content ofthe proton is the same as the u content of the Σ + or the s content of the Ξ ,because in each case it is the ‘doubly represented’ valence quark. Likewise the s in the proton is the same as the d in the Σ + or the d in the Ξ because ineach case it is the non-valence flavour. So the relations in eq. (64) are simpleconsequences of flavour permutation (the S subgroup of SU (3)). Similarly, thesecond line of eq. (63) implies1 √ (cid:104) p | (¯ uγu + ¯ sγs − dγd ) | p (cid:105) = 1 √ (cid:104) Σ + | (¯ uγu + ¯ dγd − sγs ) | Σ + (cid:105) = 1 √ (cid:104) Ξ | (¯ sγs + ¯ dγd − uγu ) | Ξ (cid:105) (65)= 2 d . All these matrix elements have the same pattern, ‘doubly represented + non-valence − × singly represented’, so again we can understand why they all have30o be the same at the symmetric point. Note that the operator in the d equation,eq. (65), is always orthogonal to the operator in the f equation, eq. (64). Wecould also look at the pattern ‘doubly represented − singly represented’, whichis just a linear combination of eq. (64) and eq. (65). Thus1 √ (cid:104) p | (¯ uγu − ¯ dγd ) | p (cid:105) ≡ √ (cid:104) Σ + | (¯ uγu − ¯ sγs ) | Σ + (cid:105)≡ √ (cid:104) Ξ | (¯ sγs − ¯ uγu ) | Ξ (cid:105) (66)= f + √ d . Of course we can not deduce the full structure at the symmetric point fromflavour permutations alone, identities such as A ¯Σ η Σ = − A ¯Λ η Λ = A ¯Λ π Σ , (67)connecting diagonal matrix elements to transition amplitudes require more gen-eral SU (3) rotations to establish them. If we move away from the symmetric point, keeping ¯ m fixed, non-singlet tensorscan contribute to eq. (19). To first order in δm l we only need consider the octets,so we can then read the mass terms off from Table 6 with an example being givenin eq. (58). We can examine the violation of SU (3) symmetry caused by the m s − m l mass difference by constructing quantities which must all be equal inthe fully symmetric case, but which can differ in the case of n f = 2 + 1 quarkmasses.We now discuss two so–called ‘fan’ plots – the d -fan plot and the f -fan plot. InAppendix B we discuss some further fan plots (called there the doubly represented– singly represented fan plots, namely the P -fan plot and the V -fan plot). d -fan Using Table 6 we can construct seven quantities, D i , which all have the samevalue (2 d ) at the symmetric point, but which can differ once SU (3) is broken D ≡ − ( A ¯ NηN + A ¯Ξ η Ξ ) = 2 d − r δm l ,D ≡ A ¯Σ η Σ = 2 d + ( r + 2 √ r ) δm l ,D ≡ − A ¯Λ η Λ = 2 d − ( r + 2 r ) δm l ,D ≡ √ A ¯ NπN − A ¯Ξ π Ξ ) = 2 d − √ r δm l , (68)31 ≡ A ¯Σ π Λ = 2 d + ( r − √ r ) δm l ,D ≡ √ A ¯ NK Σ + A ¯Σ K Ξ ) = 2 d + 2 √ r δm l ,D ≡ − ( A ¯ NK Λ + A ¯Λ K Ξ ) = 2 d − r δm l . Plotting these quantities gives a fan plot with seven lines, but only three slopeparameters ( r , r and r ), so the splittings between these observables are highlyconstrained. Of course, these seven quantities are not a unique choice, otherlinear combinations of them could be chosen. At the next order (quadratic) in δm l there is one constraint, from eq. (62). In terms of the D i this reads − D − D + 3 D + 2 D − D + 4 D − D = O ( δm l ) . (69)In the d -fan we can thus choose six independent quadratic coefficients, and fixthe seventh from this constraint.A useful ‘average D ’ can be constructed from the diagonal amplitudes X D ≡
16 ( D + 2 D + 3 D ) = 2 d + O ( δm l ) , (70)chosen so that the O ( δm l ) coefficient vanishes. Other average D quantities arepossible if we also incorporate transition matrix elements. These average quan-tities can be useful for helping to set the lattice scale, [24].It is useful to construct from this fan plots of D i /X D . However for our laterexample of the vector current, X D vanishes at Q = 0 and is always small, so weconsider alternatively here ˜ D i ≡ D i /X F . f -fan Again using Table 6 we can construct five quantities F i , which all have the samevalue (2 f ) at the symmetric point, but which can differ once SU (3) is broken. F ≡ √ A ¯ NηN − A ¯Ξ η Ξ ) = 2 f − √ s δm l ,F ≡ ( A ¯ NπN + A ¯Ξ π Ξ ) = 2 f + 4 s δm l ,F ≡ A ¯Σ π Σ = 2 f + ( − s + √ s ) δm l , (71) F ≡ √ A ¯Σ K Ξ − A ¯ NK Σ ) = 2 f − s δm l ,F ≡ √ A ¯Λ K Ξ − A ¯ NK Λ ) = 2 f + 2 √ √ s − s ) δm l . Plotting these quantities gives a fan plot with 5 lines, but only two slope param-eters ( s and s ), so the splittings between these observables are again highlyconstrained. At quadratic and higher level there are no constraints between thecoefficients for the f -fan. 32gain a useful ‘average F ’ can be constructed from the diagonal amplitudes X F ≡
16 (3 F + F + 2 F ) = 2 f + O ( δm l ) , (72)and again we can we can construct fan plots of ˜ F i ≡ F i /X F .The f -fan has the nice property that, to linear order, there is no error fromdropping quark-line-disconnected contributions. This is because r is the onlyparameter with a quark-line-disconnected piece, and none of the r i parametersappear in the f -fan. We shall prove and expand on this point in the followingsections by considering the connected and disconnected expansions separately. In lattice QCD for the three point function and its associated matrix element (seesection 12.1 for some further details) we have two classes of diagrams to compute:quark-line connected (left panel of Fig. 3) and quark-line disconnected (the right τ t B ′ B t B ′ B Figure 3:
The three point quark correlation function for a baryon. The cross rep-resents the current insertion. Left panel: the quark-line-connected piece; right panel:the quark-line-disconnected piece. panel of Fig. 3). We first write (cid:104) B (cid:48) | J F | B (cid:105) = (cid:104) B (cid:48) | J F | B (cid:105) con + (cid:104) B (cid:48) | J F | B (cid:105) dis , (73)corresponding to the left and right panels of Fig. 3 respectively. Note that analternative notation for the quark-line-connected piece is the valence matrix ele-ment (cid:104) B (cid:48) | J F | B (cid:105) con ≡ (cid:104) B (cid:48) | J F | B (cid:105) val . However we shall usually just say connectedmatrix element.The quark-line-disconnected diagrams cannot occur for transition matrix el-ements, B (cid:48) (cid:54) = B , but can for diagonal matrix elements B (cid:48) = B . From Table 133e see that disconnected diagonal matrix elements can only happen for the cur-rents J π , J η and J η (cid:48) (indices 4, 5 and 0 respectively). As we are only consideringmass degenerate u and d quarks then for the J π operators, the u -loop and d -loopquark-line-disconnected pieces always cancel. Thus apart from the singlet oper-ator J η (cid:48) , this leaves only the J η operator to consider. At the symmetric point,the disconnected contribution to J η will cancel. If one moves to m s (cid:54) = m l , thendisconnected η contributions will become non-zero, as twice the strange loop willnot be equal to the u loop + d loop. However, at leading order, this effect isgoing to be the same for all baryons, so it has the pattern only of r in Table 6.Hence r must have a disconnected piece.More explicitly first consider the flavour diagonal amplitudes. In each baryonthe disconnected u and d terms are equal (as m u = m d ), so (cid:104) p | J π | p (cid:105) dis , (cid:104) Σ + | J π | Σ + (cid:105) dis , (cid:104) Ξ | J π | Ξ (cid:105) dis , (74)all vanish. Hence f dis + √ d dis = 0 , f dis = 0 , f dis − √ d dis = 0 (75)and − r dis + s dis = 0 , − s dis + √ s dis = 0 , r dis + s dis = 0 (76)giving f dis , d dis , r dis , s dis , s dis = 0 . (77)This was briefly considered for the axial current in [25] but the results here aremore general than given there.Consider now the transition amplitudes. As stated previously disconnectedterms cannot cause a transition that changes flavour. In particular considering K current transitions they must all be connected, so from Table 6 this again showsthat all the above coefficients in eq. (77) have no disconnected piece, togetherwith the additional result r dis = 0 , (78)which means that indeed only r dis contributes. Thus in future we need only dis-tinguish between connected and disconnected contributions for the r coefficient.Differences between the disconnected pieces in different baryons will thereforefirst contribute at quadratic order in the SU (3) flavour-symmetry-breaking ex-pansion.We shall now develop and make these considerations more explicit in thefollowing section. 34 In the previous sections we have developed SU (3) flavour-breaking expansionsfor (cid:104) B (cid:48) | J F | B (cid:105) , which are sufficient for transition matrix elements. However fordiagonal matrix elements we need the additional expansion (cid:104) B | J η (cid:48) | B (cid:105) as discussedin section 5.1. This will now enable all diagonal matrix elements to be given foreach individual quark flavour.From Table 1 we see that the diagonal flavour states are given by π (index4) and η (index 5), together with the singlet flavour state, η (cid:48) (index 0). Thesecan be inverted to give ¯ uγu , ¯ dγd and ¯ sγs in terms of J η (cid:48) , J π and J η as¯ uγu = 1 √ J η (cid:48) + 1 √ J π + 1 √ J η , ¯ dγd = 1 √ J η (cid:48) − √ J π + 1 √ J η , (79)¯ sγs = 1 √ J η (cid:48) − (cid:114) J η . As discussed previously in section 5.1, the additional expansion for the singletcurrent J η (cid:48) is the same as the mass expansion presented in [2]. We shall onlyconsider LO here (higher orders are also given in [2]). We take the expansion asalready given in eq. (39).Using eq. (79), together with eq. (39) and Tables 3 and 6 allows us to givethe SU (3) flavour-breaking expansion for flavour diagonal matrix elements. InAppendix C we give this expansion to LO for the representative octet baryons p ,Σ + , Λ and Ξ (the others n , Σ − , Σ and Ξ − can be similarly determined).While it appears from eq. (39) that we now have extra coefficients a , a and a that have to be determined, this can be somewhat ameliorated when thequark-line-connected and -disconnected matrix elements are considered. Therewas a general discussion in section 9. We now consider this in more detail byconsidering separate expansions for both the connected and disconnected pieces.So the previous equations are doubled, as given in eq. (73). For example (cid:104) p | ¯ uγu | p (cid:105) = (cid:104) p | ¯ uγu | p (cid:105) con + (cid:104) p | ¯ uγu | p (cid:105) dis , (80)corresponding to the left and right panels of Fig. 3 respectively. There are nowsome additional constraints.For completeness we list the disconnected matrix element results in Ap-pendix D, using a dis , a dis , a dis and eqs. (77, 78). For p ( uud ), Σ + ( uus ) and Ξ ( ssu ) there are no connected pieces for (cid:104) p | ¯ sγs | p (cid:105) , (cid:104) Σ + | ¯ dγd | Σ + (cid:105) and (cid:104) Ξ | ¯ dγd | Ξ (cid:105) . Thus there are now conditions on a con , a con and35 con from the previous expansion parameters. We find a con = √ f − √ d , a con = √ r con − √ s , (81)3 a con = 1 √ r con + √ r + √ s − √ s . (These consistently satisfy all the previous equations.) Using these expressionsfor a con , a con and a con gives for the octet baryons p , Σ + , Λ and Ξ (cid:104) p | ¯ uγu | p (cid:105) con = 2 √ f + (cid:32)(cid:114) r con − √ r + √ s − (cid:114) s (cid:33) δm l , (82) (cid:104) p | ¯ dγd | p (cid:105) con = √ (cid:16) f − √ d (cid:17) + (cid:32)(cid:114) r con + √ r − √ s − (cid:114) s (cid:33) δm l , (cid:104) Σ + | ¯ uγu | Σ + (cid:105) con = 2 √ f + (cid:16) − √ s + √ s (cid:17) δm l , (cid:104) Σ + | ¯ sγs | Σ + (cid:105) con = √ (cid:16) f − √ d (cid:17) (83)+ (cid:32) − (cid:114) r con − √ r − √ s + (cid:114) s (cid:33) δm l , (cid:104) Λ | ¯ uγu | Λ (cid:105) con = (cid:104) Λ | ¯ dγd | Λ (cid:105) con = √ (cid:18) f − √ d (cid:19) + (cid:32)(cid:114) r con + (cid:114) r + √ r + √ s − (cid:114) s (cid:33) δm l , (84) (cid:104) Λ | ¯ sγs | Λ (cid:105) con = √ (cid:18) f + 1 √ d (cid:19) + (cid:32) − √ r con − √ r + √ r + √ s − (cid:114) s (cid:33) δm l , and (cid:104) Ξ | ¯ uγu | Ξ (cid:105) con = √ f − √ d ) + (cid:16) √ r + 2 √ s (cid:17) δm l , (cid:104) Ξ | ¯ sγs | Ξ (cid:105) con = 2 √ f + (cid:32) − (cid:114) r con + √ r + √ s − (cid:114) s (cid:33) δm l . (85)Without Λ there are six equations, together with six parameters, so no con-straint. Adding the Λ gives two more equations and one extra parameter, sothis is now constrained. In addition off-diagonal matrix elements would also givemore constraints. 36 Using the previous results of this section, we can also give the results for theelectromagnetic current, eq. (12). Using this equation we find, for example, thatfor the octet baryons p , Σ + , Λ and Ξ (cid:104) p | J em | p (cid:105) con = √ f + (cid:114) d + (cid:18) √ r con − √ r + √ s − √ s (cid:19) δm l , (cid:104) Σ + | J em | Σ + (cid:105) con = √ f + (cid:114) d + (cid:32) √ r con + √ r − √ s − (cid:114) s (cid:33) δm l , (cid:104) Λ | J em | Λ (cid:105) con = − (cid:114) d + (cid:32) √ r con + (cid:114) r (cid:33) δm l , (86) (cid:104) Ξ | J em | Ξ (cid:105) con = − (cid:114) d + (cid:18) √ r con + √ r + √ s + 1 √ s (cid:19) δm l , for the quark-line-connected terms, and for the quark-line-disconnected terms (cid:104) p | J em | p (cid:105) dis = (cid:104) Λ | J em | Λ (cid:105) dis = (cid:104) Σ + | J em | Σ + (cid:105) dis = (cid:104) Ξ | J em | Ξ (cid:105) dis = 1 √ r dis δm l . (87)Similar expansions hold for the n , Σ , Σ − and Ξ − electromagnetic matrix ele-ments.
11 Renormalisation and O ( a ) improvement forthe vector current The computed matrix elements are bare (or lattice) quantities and must be renor-malised and O ( a ) improved. We would expect that the effect of the O ( a ) improve-ment terms is simply to modify the SU (3) flavour-breaking expansion coefficients.In this section we shall show that this expectation is indeed correct. Again, forillustration, we shall only consider the diagonal sector ( B (cid:48) = B ) of the vectorcurrent here. By using the results and notation in [12] (see also [26]) we have foron-shell improvement V π µ = Z V (cid:2) b V + 3¯ b V ) ¯ m + b V δm l (cid:3) V π µ ,V η R µ = Z V (cid:104)(cid:0) b V + 3¯ b V ) ¯ m − b V δm l (cid:1) V ηµ + √ b V + 3 f V ) δm l V η (cid:48) µ (cid:105) ,V η (cid:48) R µ = Z V r V (cid:104)(cid:0) d V + 3 ¯ d V ) ¯ m (cid:1) V η (cid:48) µ + 2 √ d V δm l V ηµ (cid:105) . (88)37here V for the local vector current denotes V Fµ = V Fµ + ic V ∂ ν T Fµν , (89)with T Fµν = ¯ qF σ µν q and ∂ µ φ ( x ) = [ φ ( x + ˆ µ ) − φ ( x − ˆ µ )] /
2. This additional termonly plays a role in non-forward matrix elements. Note that all the improvementcoefficients b V , d V , ¯ b V , ¯ d V and c V are just functions of the coupling constant, g .Thus we do not have to be precisely at the correct (physical) ¯ m to determine thecoefficients. The r V parameter accounts for the fact that the singlet renormali-sation is different to the non-singlet renormalisation, Z V ( g ). r V also depends onthe chosen scheme and scale. Tree level gives for the relevant coefficients b V ( g ) = 1 + O ( g ) , f V ( g ) = O ( g ) , c V ( g ) = O ( g ) , (90)(together with Z V ( g ) = 1 + O ( g ) and d V ( g ) = O ( g )) where ¯ b V ( g ), ¯ d V ( g ),being connected with the sea contributions are ∼ O ( g ), and are usually takenas negligible. Furthermore we can write V π µ = ˆ Z V (cid:104) b V δm l (cid:105) V π µ ,V η R µ = ˆ Z V (cid:104) (1 − ˆ b V δm l ) V ηµ + √ b V + 3 ˆ f V ) δm l V η (cid:48) µ (cid:105) ,V η (cid:48) R µ = ˆ Z V ˆ r V (cid:104) V η (cid:48) µ + 2 √ d V δm l V ηµ (cid:105) , (91)where for constant ¯ m we have absorbed these ¯ m terms into the renormalisationconstant and improvement coefficients. For example we have ˆ Z V = Z V (1 + ( b V + 3¯ b V ) ¯ m ) , ˆ b V = b V (1 + ( b V + 3¯ b V ) ¯ m ) − , ˆ f V = f V (1 + ( b V + 3¯ b V ) ¯ m ) − . (92)We take eq. (91) as our definition of the improvement coefficients, as the SU (3)flavour-breaking expansion coefficients are already functions of ¯ m . To avoid con-fusion with the previous SU (3) flavour-breaking expansion coefficients we havedenoted them with a caret. Note that in any case we have also numericallythat | ¯ mδm l | (cid:28) m (cid:28) There is a further improvement coefficient, g → ˆ g = g (1 + b g ¯ m ), where b g is a functionof g . Little is known about the value of b g , however perturbatively it is very small, so we shallignore it here. Note that as we always consider ¯ m = const . , then the value of g is only slightlyshifted by a constant. Similarly ˆ r V = r V (1+( d V +3 ¯ d V ) ¯ m )(1+( b V +3¯ b V ) ¯ m ) − and ˆ d V = d V (1+( d V +3 ¯ d V ) ¯ m ) − . V π µ Let us first consider V π µ in eq. (91), together with (for example) (cid:104) p | V π | p (cid:105) R , (cid:104) Σ + | V π | Σ + (cid:105) R , (cid:104) Ξ | V π | Ξ (cid:105) R . From the expansion for F = π given in Table 6for A ¯ NπN , A ¯Σ π Σ and A ¯Ξ π Ξ we see that as expected the effects of the expansioncoefficients simply change their value slightly s → s (cid:48) = s + 12 f ˆ b V ,s → s (cid:48) = s + √ f ˆ b V ,r → r (cid:48) = r − √ d ˆ b V . (93)Furthermore, as a reminder, from eq. (77) the disconnected pieces for f , d , r , r , s , s all vanish, which implies that ˆ b V also has no disconnected piece. Inparticular this means that the results for V π µ remain valid when just consideringthe connected matrix elements. V η R µ We can repeat the process for V η R µ , which gives in addition to the results ofeq. (93), the further results r → r (cid:48) = r + d ˆ b V + √ a (ˆ b V + 3 ˆ f V ) ,r → r (cid:48) = r + d ˆ b V . (94)In addition splitting r into r con and r dis pieces gives upon using a con from eq. (81) r con → r con (cid:48) = r con + 2 √ f (ˆ b V + 3 ˆ f con V ) − d (ˆ b V + 6 ˆ f con V ) ,r dis → r dis (cid:48) = r dis + 3 √ a dis ˆ f dis V . (95) V η (cid:48) R µ Lastly, considering V η (cid:48) R µ , we find a → a (cid:48) = a + 2 (cid:114) (cid:18) f − √ d (cid:19) ˆ d V ,a → a (cid:48) = a − √ d ˆ d V . (96) As expected, all improvement coefficients are terms in the SU (3) symmetryflavour-breaking expansion, and indeed upon inclusion leads to slightly modi-fied expansion coefficients, as given in eqs. (93, 94, 96). We anticipate that theadditional improvement term, ˆ c V , is also of this form.39 ˆ Z V and ˆ b V , ˆ f con V There is an exact global symmetry of the lattice action, q → e − iα q q , valid for eachquark separately. Using Noether’s theorem this leads to an exactly conservedvector current, CVC. Practically the operator counts the number of u quarksand the number of d quarks in the baryon. The local current considered here isnot exactly conserved, so that V CVC = V + O ( a ). We can use this to define therenormalisation constant and several improvement terms. (A similar method wasused for two flavours and quenched QCD in, e.g., [27].) Thus we shall see thatimposing CVC is equivalent to determining some improvement coefficients.Practically here we restrict our considerations to the forward matrix elementsfor V at Q = 0 (no momentum transfer, so there is no additional ˆ c V term). V π First for the CVC, we consider the representative matrix elements (cid:104) p | V π | p (cid:105) R = A R ¯ NπN = 1 √ − , (cid:104) Σ + | V π | Σ + (cid:105) R = A R ¯Σ π Σ = 1 √ − , (cid:104) Ξ | V π | Ξ (cid:105) R = A R ¯Ξ π Ξ = 1 √ − . (97)Using this together with V π in eq. (91) gives f = 1 √ Z V , d = 0 . (98)One possibility is thus to determine f from X F at Q = 0, see eq. (72) asˆ Z V = √ X F . (99)Also from eq. (93) and due to the lack of O ( δm l ) terms in eq. (97) we have s (cid:48) = 0, s (cid:48) = 0 and r (cid:48) = 0 or s = − f ˆ b V , s = −√ f ˆ b V , r = 0 . (100)Using ˜ s i = s i /X F , which to leading order is s i / (2 f ), gives directly the ˆ b V im-provement coefficient. 40 V η R Additionally using the equivalent results from eq. (97) but now for V η R namely (cid:104) p | V η | p (cid:105) R = A R ¯ NηN = 1 √ − , (cid:104) Σ + | V η | Σ + (cid:105) R = A R ¯Σ η Σ = 1 √ − , (cid:104) Ξ | V η | Ξ (cid:105) R = A R ¯Ξ η Ξ = 1 √ − , (101)not only gives consistency with the previous results eqs. (98, 99), but in additionwe have r con (cid:48) = 0, r (cid:48) = 0 or from eqs. (94, 95) r con = − √ f (cid:16) ˆ b V + 3 ˆ f con V (cid:17) , r = 0 . (102)Again using ˜ r con = r con /X F = r con / (2 f ) automatically eliminates f . We observethat once ˆ Z V , ˆ b V (and ˆ f con V ) have been determined then by using eq. (92) andvarying ¯ m , then it is in principle possible to determine ˆ¯ b V . The Ademollo–Gatto theorem [28] (see also [29, 13]) in the context of our flavour-breaking expansions states that the O ( δm l ) terms vanish for the F ¯ B (cid:48) F B formfactor at Q = 0 and B (cid:48) (cid:54) = B . This means that r , r , s , s vanish at Q = 0(or the primed versions if we include the improvement coefficients). This agreeswith the results of this section.
12 Lattice computations of form factors
We now need to determine the matrix elements from a lattice simulation whichcomputes two- and three-point correlation functions. For completeness as well asform factors with B = B (cid:48) , we are developing a formalism for semileptonic decays, B (cid:54) = B (cid:48) so we first consider the general method here.The baryon two-point correlation function is given by C B Γ ( t ; (cid:126)p ) = (cid:88) αβ Γ βα (cid:10) B α ( t ; (cid:126)p ) ¯ B β (0; (cid:126)p ) (cid:11) , (103)while the three-point correlation function generalises this and is given by C B (cid:48) B Γ ( t, τ ; (cid:126)p, (cid:126)p (cid:48) ; J ) = (cid:88) αβ Γ βα (cid:10) B (cid:48) α ( t ; (cid:126)p (cid:48) ) J ( τ ; (cid:126)q ) ¯ B β (0; (cid:126)p ) (cid:11) , (104)41ith J at time τ either the vector, axial or tensor current, and where the sourceis at time 0, the sink operator is at time t andΓ ≡ Γ unpol = (1 + γ ) , or Γ ≡ Γ pol = (1 + γ ) iγ (cid:126)γ · (cid:126)n , (105)where (cid:126)n is the polarisation axis.To eliminate overlaps of the source and sink operators with the vacuum, webuild ratios of 3-point to 2-point correlation functions. More explicitly let us set R Γ ( t, τ ; (cid:126)p, (cid:126)p (cid:48) ; J )= C B (cid:48) B Γ ( t, τ ; (cid:126)p, (cid:126)p (cid:48) ; J ) C B (cid:48) Γ unpol ( t ; (cid:126)p (cid:48) ) (cid:115) C B (cid:48) Γ unpol ( τ ; (cid:126)p (cid:48) ) C B (cid:48) Γ unpol ( t ; p (cid:48) ) C B Γ unpol ( t − τ ; (cid:126)p ) C B Γ unpol ( τ ; (cid:126)p ) C B Γ unpol ( t ; (cid:126)p ) C B (cid:48) Γ unpol ( t − τ ; (cid:126)p (cid:48) ) . (106)This is designed so that any smearing for the source and sink operators is cancelledin the ratios, e.g. [30, 31]; of course smearing the baryon operators improves theoverlap with the lowest-lying state, so the relevant overlaps for the two- andthree-point correlation functions must match.Inserting complete sets of unit-normalised states in eq. (106) and for 0 (cid:28) τ (cid:28) t (cid:28) T gives R Γ ( t, τ ; (cid:126)p, (cid:126)p (cid:48) ; J ) = (cid:115) E B ( (cid:126)p ) E B (cid:48) ( (cid:126)p (cid:48) )( E B ( (cid:126)p ) + M B ) ( E B (cid:48) ( (cid:126)p (cid:48) ) + M B (cid:48) ) F (Γ , J ) , (107)with F (Γ , J ) = 14 tr Γ (cid:18) γ − i (cid:126)p (cid:48) · (cid:126)γE B (cid:48) ( (cid:126)p (cid:48) ) + M B (cid:48) E B (cid:48) ( (cid:126)p (cid:48) ) (cid:19) J (cid:18) γ − i (cid:126)p · (cid:126)γE B ( (cid:126)p ) + M B E B ( (cid:126)p ) (cid:19) (108)(with J being given from the Euclideanised version of eq. (5)). The transferred(Euclidean) momentum from the initial, B , to final, B (cid:48) state is given by Q =( i ( E B (cid:48) ( (cid:126)p (cid:48) ) − E B ( (cid:126)p )) , (cid:126)p (cid:48) − (cid:126)p ) so Q = − ( M B (cid:48) − M B ) + 2 ( E B (cid:48) ( (cid:126)p (cid:48) ) E B ( (cid:126)p ) − M B (cid:48) M B − (cid:126)p · (cid:126)p (cid:48) ) . (109)To illustrate the previous SU (3) flavour symmetry-breaking results, we shall nowconsider here only the vector current. Furthermore in general for arbitrary mo-menta geometry, the kinematic factors can be complicated; in this article we shallonly be considering the simpler case (cid:126)p (cid:48) = (cid:126)
0. The technical reason is that in thelattice evaluation, it requires less numerical inversions and is hence computation-ally cheaper. (Physically, of course it is more natural to start with a stationarybaryon, but computationally of course it does not matter.) Evaluating Q in thisframe, eq. (109), shows that for flavour diagonal matrix elements form factors Q is always positive, while for semileptonic decays for small momentum it can also42e negative. For the vector current with (cid:126)p (cid:48) = (cid:126) . R Γ unpol ( t, τ ; (cid:126)p, V ) = (cid:115) E B(cid:126)p + M B E B(cid:126)p (cid:34) F ¯ B (cid:48) F B − E B(cid:126)p − M B M B + M B (cid:48) F ¯ B (cid:48) F B − E B(cid:126)p − M B (cid:48) M B + M B (cid:48) F ¯ B (cid:48) F B (cid:35) ,R Γ unpol ( t, τ ; (cid:126)p, V i ) = − ip i (cid:113) E B(cid:126)p ( E B(cid:126)p + M B ) (cid:34) F ¯ B (cid:48) F B − E B(cid:126)p − M B (cid:48) M B + M B (cid:48) F ¯ B (cid:48) F B − E B(cid:126)p + M B M B + M B (cid:48) F ¯ B (cid:48) F B (cid:35) , (110) R Γ pol ( t, τ ; (cid:126)p, V i ) = ( (cid:126)p × (cid:126)n ) i (cid:113) E B(cid:126)p ( E B(cid:126)p + M B ) (cid:104) F ¯ B (cid:48) F B + F ¯ B (cid:48) F B (cid:105) ,R Γ pol ( t, τ ; (cid:126)p, V ) = 0 . In particular for (cid:126)p = 0 then the only non-zero ratio is R Γ unpol ( t, τ ; 0 , V ) = F ¯ B (cid:48) F B − M B − M B (cid:48) M B + M B (cid:48) F ¯ B (cid:48) F B , (111)so we see that in this case for B (cid:48) (cid:54) = B then we cannot disentangle F ¯ B (cid:48) F B from F ¯ B (cid:48) F B . However to LO (i.e. O ( δm l ) effects in the matrix elements) and as M B − M B (cid:48) ∝ δm l then from eq. (111) we can write R Γ unpol ( t, τ ; 0 , V ) = F ¯ B (cid:48) F B + O ( δm l ) , (112)for all B and B (cid:48) , where the O ( δm l ) term is not present when B (cid:48) = B . As a demonstration of the method we apply the formalism outlined in the previoussections to the form factors published in [33, 34]. Further details of the numericalsimulations can be found there. The simulations have been performed using n f = 2+1, O ( a ) improved clover fermions [35] at β ≡ /g of 5 .
50 and on 32 × ∼ O (1500)configurations).As discussed previously and particularly in section 3.1 our strategy is to keepthe bare quark-mass constant. Thus once the SU (3) flavour degenerate sea quarkmass, m , is chosen, subsequent sea quark-mass points m l , m s are then arranged We use the Euclideanisation conventions given in [32]. In particular V = V ( M )0 , V i = − iV ( M )i with γ = γ ( M )0 , γ i = − iγ ( M )i , γ = − γ ( M )5 , σ µν = i/ γ µ , γ ν ].
43n the various simulations to keep ¯ m (= m ) constant. This then ensures thatall the expansion coefficients given previously do not change. In [2], masses wereinvestigated and it was seen that a linear fit provides a good description of thenumerical data on the unitary line over the relatively short distance from the SU (3) flavour symmetric point down to the physical pion mass. This proveduseful in helping us in choosing the initial point on the SU (3) flavour symmetricline to give a path that reaches (or is very close to) the physical point.The bare unitary quark masses in lattice units are given by m q = 12 (cid:18) κ q − κ c (cid:19) with q = l, s , (113)and where vanishing of the quark mass along the SU (3) flavour symmetric linedetermines κ c . We denote the SU (3) flavour symmetric kappa value, κ , asbeing the initial point on the path that leads to the physical point. m is givenin eq. (113) by replacing κ q by κ . Keeping ¯ m = constant = m then gives δm q = 12 (cid:18) κ q − κ (cid:19) . (114)We see that κ c has dropped out of eq. (114), so we do not need its explicit valuehere. Along the unitary line the quark masses are restricted and we have κ s = 1 κ − κ l . (115)So a given κ l determines κ s here. This approach is much cleaner than the moreconventional approach of keeping (the renormalised) strange quark mass constant,as this necessitates numerically determining the bare strange quark mass. Inaddition the O ( a ) improvement of the coupling constant is much simpler, in ourapproach as it only depends on ¯ m , [2]. Thus here, the coupling constant remainsconstant and hence the lattice spacing does not change as the quark mass ischanged. In the more conventional approach this can be problematical as youmust in principle monitor the changing of the coupling constant as the quarkmasses vary.An appropriate SU (3) flavour symmetric κ value chosen here for this actionwas found to be κ = 0 . X π to anestimate for the pion mass of ∼
465 MeV at our chosen SU (3) flavour symmetricpoint and from X N an estimation of the lattice spacing of a N ( κ = 0 . .
074 fm.Specifically as indicated in Table 8 we have generated configurations, [33, 34],at the ( κ l , κ s ) values listed, all with κ = 0 . R , the appropriate formfactor. As described in [33, 34], we bin Q to directly compare each configuration44 l κ s M π MeV0 . . . . . . Table 8:
Outline of the ensembles used here on the 32 ×
64 lattices together with thecorresponding pion masses. and using the bootstrapped lattice configurations, we set up a weighted leastsquares to extract the linear fit parameters and weighted errors at each Q value.The lattice momenta used here in this study in units of 2 π/
32 are given by a(cid:126)q =(0 , , , , , , , , , , , , , , , ,
0) togetherwith all permutations (where different) and all possible ± values.
13 Results
We now illustrate some of the features that we have described in previous sections,using our lattice calculations and the ensembles in Table 8.
We first consider the lattice quantities X F D , X F F and X F D , X F F . As discussedpreviously we only consider diagonal form factors to construct the X s, i.e. theequations: D con , D con and D in eq. (68) and F , F and F in eq. (71) . Usingthe method of section 12.2 allows us to create the appropriate D con , D con and D defined in eq. (68) and hence X F D , X F D in eq. (70) or F , F and F ineq. (71) and thus again X F F , X F F in eq. (72). In Fig. 4 we consider X F D and X F F for the F form factor for Q = 0 and 0 .
49 GeV . First, as we expect theyare constant and show little sign of O ( δm l ) or curvature effects. Although notso relevant on this plot, as an indication of how far we must extrapolate in thequark mass from the symmetric point to the physical point, we also give this,using the previous determination, [21], of δm ∗ l = − . Q = 0, X F D vanishes as d = 0, which we also see on the plot.This constancy of X does not depend on the form factor used. In Fig. 5we show similar plots, but now for the F form factors: X F D and X F F , for Q = 0 . and 0 .
49 GeV . Again these are all constant, within our statistics. We note that care needs to be taken to distinguish the F i corresponding to a form factorand the F i defined in eq. (71). This corresponds to a lattice momentum of a(cid:126)q = (2 π ) /
32 (1 , , This corresponds to a lattice momentum of a(cid:126)q = (2 π ) /
32 (1 , , .012 0.010 0.008 0.006 0.004 0.002 0.000 δm l X F D a n d X F F Q = 0 . GeV X F F X F D δm l X F D a n d X F F Q = 0 . GeV X F F X F D Figure 4: X F D and X F F for F at Q = 0, top panel and for Q = 0 .
49 GeV , lowerpanel. The lower filled circles in each plot are X F D , the upper filled triangles are X F F . The dashed lines are constant fits and the stars represent the physical point. .012 0.010 0.008 0.006 0.004 0.002 0.000 δm l X F D a n d X F F Q = 0 . GeV X F F X F D δm l X F D a n d X F F Q = 0 . GeV X F F X F D Figure 5: X F D and X F F for F at Q = 0 .
25 GeV , top panel and for Q =0 .
49 GeV , lower panel. The same notation as for Fig. 4. X F D at Q = 0 via an extrapolation, so we show Q = 0 .
25 GeV instead.)Finally we can plot the dependence of X on Q . In Fig. 6 we show X F D and X F F and similarly for X F versus Q (using the previously determined fittedvalues). This gives the Q dependence of d and f respectively. For X F F , d isinitially zero and remains small for larger Q , while f drops monotonically. Weexpect d and f to drop like ∼ /Q for large Q for all the form factors. We now turn to ‘fan’ plots, as defined by eqs. (68) and (71). Note that again weonly consider lattice quantities, the improved operator would have small changesto the SU (3) flavour-breaking expansion, as discussed in section 11.1. Againwe only consider diagonal form factors in these equations: D con , D con and D in eq. (68) and F , F and F in eq. (71). We construct the system of linearequations in eq. (68) with parameters r con , r and d for the d -fan and eq. (71)with parameters s , s and f for the f -fan. In Fig. 7 we show ˜ D F i = D F i /X F for i = 1, 2 and 4 and ˜ F F i = F F i /X F for i = 1, 2 and 3. Note that as d vanishesfor the F form factor at Q = 0, and even away from Q = 0 it remains small,see the lower panel of Fig. 4, then dividing by X F D is not possible or very noisy,so we use X F F . Although for X F D this is not the case (as seen in Fig. 5) howeverfor consistency we still use X F F . The only change in these cases is that the valueat the symmetric point is no longer one.The lines shown in Fig. 8 correspond to linear fits to the D F i using eq. (68)(upper plot) and F F i using eq. (71) (lower plot). The fits to D F i determine r con , r using three fits and are hence constrained. Furthermore determining thesetwo parameters also allows us to plot the off-diagonal hyperon decays for i = 6,which is also shown. Similarly for F F i , we first determine the constrained fitparameters ˜ s = s /X F , ˜ s = s /X F and then plot the off-diagonal hyperondecays for i = 4, 5.Similarly in Fig. 8 we show the equivalent results for F . As previously wehave normalised the parameters, ˜ r con = r con /X F , ˜ r = r /X F and ˜ s = s /X F ,˜ s = s /X F . Again we have some constraints. In addition off-diagonal hyperondecays for i = 6, d -fan plot and i = 4, 5, f -fan plot are also shown.From these fan plots at various Q we can determine the dependence of theexpansion coefficients as a function of Q . In Fig. 9 we show the expansioncoefficients r con , r , s , s for the F con and F form factors as function of Q . Asdiscussed previously in section 11.1, at Q = 0 the expansion coefficients for F con vanish, which determines the improvement coefficients b V , f con V . Thus in the toppanel of Fig. 9 the negative values of the r con , s , s are a clear indication of thenature of the improvement coefficients. For rather small Q , these all change signrather quickly and also their order inverts. We have (approximately) | r | , | s | ≈ | r con | is a factor of 2–4 larger than | s | . For F the expansion coefficients tend48 .0 0.5 1.0 1.5 Q ( GeV ) X F i X F F X F D Q ( GeV ) X F i X F F X F D Figure 6:
Top panel: X F F (filled circles) and X F D (filled triangles) versus Q . Lowerpanel: Similarly for F . .012 0.010 0.008 0.006 0.004 0.002 0.000 0.002 0.004 0.006 δm l D F i / X F F Q = 0 . GeV D /X F D /X F D /X F D /X F D /X F D /X F D /X F δm l F F i / X F Q = 0 . GeV F /X F F /X F F /X F F /X F F /X F F /X F F /X F F /X F Figure 7:
Top panel: ˜ D F i ≡ D F i /X F F for i = 1 (filled circles), 2 (filled squares) and4 (filled triangles) for Q = 0 .
49 GeV . The three fits are from eq. (68), the line for i = 6 is also shown. The vertical dotted line represents the physical point. Lower panel:˜ F F i ≡ F F i /X F F again at Q = 0 .
49 GeV for i = 1 (filled circles), 2 (filled squares)and 3 (filled triangles), together with fits from eq. (71) normalised by X F F . The linefor i = 5 is also shown. .012 0.010 0.008 0.006 0.004 0.002 0.000 0.002 0.004 0.006 δm l D F i / X F F Q = 0 . GeV ) D /X F D /X F D /X F D /X F D /X F D /X F D /X F δm l F F i / X F F Q = 0 . GeV ) F /X F F /X F F /X F F /X F F /X F F /X F F /X F F /X F Figure 8:
Top panel: ˜ D F i for i = 1 (filled circles), 2 (filled squares) and 4 (filledtriangles) for Q = 0 .
49 GeV . The three fits are from eq. (68) normalised by X F D , alsoshown is the i = 6 line. The vertical dotted line represents the physical point. Lowerpanel: ˜ F F i for i = 1 (filled circles), 2 (filled squares) and 3 (filled triangles), also for Q = 0 .
49 GeV , together with fits from eq. (71) normalised by X F F . Also shown arethe lines i = 4 (upper line), 5 (lower line). .0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 Q ( GeV ) r c o n , r , s , s r con r s s Q ( GeV ) r c o n , r , s , s r con r s s Figure 9:
Top panel: r con (filled circles), r (filled triangles), s (filled squares) and s (filled diamonds) expansion coefficients for the vector F con form factor as a functionof Q . Lower panel: Similarly for the F form factor.
52o be flatter. Also s ≈
0, indicated in Fig. (8) by the small difference between˜ F F and ˜ F F . ˆ Z V and ˆ b V , ˆ f con V X F F at Q = 0 determines the renormalisation constant ˆ Z V via eq. (99). Theconstant fit described in eq. (72) and shown in Fig. 4, see also Fig. 6, leads to f = 0 . Z V = 0 . . (116)Our previous non-perturbative estimates of Z V at β = 5 .
50 are given in [36, 37]of 0 . . Z V in eq. (116). Notethat the different determinations can have O ( a ) differences. Also ˆ Z V has beenmeasured rather than Z V . The difference is ∼ b V ¯ m . Here we have b V ∼ O (1)and ¯ m ∼ .
01 (using the κ c found in [2]), so there a further possible difference(and reduction from the ˆ Z V value) of ∼ Q = 0 value for r is 0 . Q = 0 values for s , s are s = − . s = − . s /s = 3 .
42, which is in goodagreement with the theoretical value for the ratio from eq. (100) of 2 √ ∼ . b V = 1 . , (117)which is about a 15% increase from the tree-level value. Although a strict com-parison with other determinations of this improvement coefficient is not possible,it is interesting to note that compared to other computations , e.g. [26] and for n f = 0, 2, [27] the value determined here is much closer to its tree-level valueeq. (90). This suggests that improvement coefficients are small, including possiblyˆ c V . Using the value of ˆ b V from s , s and using eq. (102) together with r con = − . f con V = 0 . . (118)As expected this is quite small. With a knowledge of f , d and r con , r , d , s , s we can find the electromagneticDirac form factor F con ( Q ) and Pauli form factor F con ( Q ) using the electromag-netic current J con em µ (see section 10.2) and results of eq. (86). Also we shall useˆ Z V , ˆ b V and ˆ f con V (i.e. equivalent to CVC) from section 13.3.53 .0 0.5 1.0 1.5 Q ( GeV ) ˜ d ( Q ) = X D ( Q )2 X F ( Q ) X F ( Q ) X F (0) X F ( Q ) X F (0) ˜ d ( Q ) = X D ( Q )2 X F ( Q ) Figure 10: X F ( Q ) /X F (0) (filled circles) and ˜ d ( Q ) (filled triangles) for F con against Q . The interpolation formulae used are given in eq. (121). It is interesting to determine the various contributions to the form factorsfrom the expansion coefficients. For illustrative purposes, we shall just consider F con here and for p and Ξ . From eq. (86) we can write (cid:104) p | J em | p (cid:105) con R = X F ( Q , ¯ m ) X F (0 , ¯ m ) (cid:20) √ d ( Q , ¯ m ) + ˜ (cid:15) (cid:48) p ( Q , ¯ m ) δm l (cid:21) , (cid:104) Ξ | J em | Ξ (cid:105) con R = − X F ( Q , ¯ m ) X F (0 , ¯ m ) (cid:20) √ d ( Q , ¯ m ) − ˜ (cid:15) (cid:48) Ξ ( Q , ¯ m ) δm l ) (cid:21) , (119)with ˜ (cid:15) (cid:48) p = 1 √ r con (cid:48) − ˜ s (cid:48) ) + 2(˜ s (cid:48) − ˜ r (cid:48) ) , ˜ (cid:15) (cid:48) Ξ = 1 √ r con (cid:48) + ˜ s (cid:48) ) + 2(˜ s (cid:48) + ˜ r (cid:48) ) , (120)where, for example, ˜ r con (cid:48) = r con (cid:48) ( Q , ¯ m ) /X F ( Q , ¯ m ) and similarly for the otherexpansion coefficients. The prime includes the improvement terms, see eqs. (93,94). In this form, we can investigate the contributions to the form factors. InFig. 10 we show the results for the terms of eq. (119): X F ( Q ) /X F (0) and ˜ d . InFig. 11 we show ˜ r con (cid:48) , ˜ s (cid:48) , ˜ r (cid:48) and ˜ s (cid:48) . All the interpolation formulae (fits) are of54 .0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 Q ( GeV ) ˜ r , ˜ r , ˜ s , ˜ s ˜ r ˜ r ˜ s ˜ s ˜ r ˜ r ˜ s ˜ s Figure 11: ˜ r con (cid:48) (filled circles), ˜ s (cid:48) (filled diamonds), ˜ s (cid:48) (filled squares) and ˜ r (cid:48) (filledtriangles) against Q together with interpolation formulae also given by eq. (121). the form AQ BQ + C ( Q ) . (121)From Fig. 10 and the leading term in eq. (119) for the proton form factor, thedominant contribution comes from X F ( Q ) /X F (0) – the f term, while there isa small contribution from the d term (as ˜ d ). Furthermore from Fig. 11 we seethat for the ˜ (cid:15) coefficients, ˜ r (cid:48) and ˜ s (cid:48) are essentially negligible and most of thecontribution comes from ˜ r con (cid:48) and ˜ s (cid:48) .We illustrate this for the F form factor for the p and Ξ . In Fig. 12 we show F con R for these baryons at the physical point δm ∗ l = − . X F ( Q ) /X F (0), The dashed-dotted lines arethe complete leading terms: X F ( Q , ¯ m ) /X F (0 , ¯ m )(1 + 2 / √ d ( Q , ¯ m )) for p and X F ( Q , ¯ m ) /X F (0 , ¯ m ) × / √ d ( Q , ¯ m )) for the Ξ , while the full lines are thecomplete expressions in eq. (119).We see that for the proton the f term (represented by X F ( Q , ¯ m ) /X F (0 , ¯ m ))gives a result very close to the numerical result; the addition of the ˜ d term pullsit slightly away in the +ve direction. The inclusion of the O ( δm l ) term, being − ve pushes it back. However the additional terms to the f term contributesvery little (only a few percent) to the final result. For the Ξ the O ( δm l ) termimproves the agreement. 55 .0 0.5 1.0 1.5 2.0 Q ( GeV ) F c o n P h y s X F ( Q ) X F (0) X F ( Q ) X F (0) [1 + p ˜ d ] − X F ( Q ) X F (0) [ p ˜ d ] › Ξ | J em | Ξ fi conR › p | J em | p fi conR Figure 12: F con R for the proton (filled circles) and Ξ (filled triangles) at the physicalpoint. The dashed line is X F ( Q ) /X F (0). The dashed-dotted lines are the completeleading terms, for the proton: X F ( Q , ¯ m ) /X F (0 , ¯ m )(1 + 2 / √ d ( Q , ¯ m )) and for Ξ : X F ( Q , ¯ m ) /X F (0 , ¯ m ) × / √ d ( Q , ¯ m )), while the full lines are the complete expressionsin eq. (119).
14 Conclusions and outlook
In this article we have outlined a programme for investigating the quark-massbehaviour of matrix elements, for n f = 2 + 1 quark flavours starting from a pointon the SU (3) flavour symmetric line when the u , d and s quarks have the samemass and then following a path keeping the singlet quark-mass constant. This isan extension of our original programme for masses, [1, 2], using a generalisationof the techniques developed there.When flavour SU (3) is unbroken all baryon matrix elements of a given oper-ator octet can be expressed in terms of just two couplings ( f and d ), as is wellknown. We find that when SU (3) flavour symmetry is broken, at LO and NLO,the expansions are constrained (but not at further higher orders). By this wemean that there are a large number of relations between the expansion coeffi-cients. Our main results for the expansions are contained in sections 6.1 and 6.2.Although we concentrated on the n f = 2 + 1 case, in which symmetry breakingis due to mass differences between the strange and light quarks our methods arealso applicable to isospin-breaking effects coming from a non-zero m d − m u alongthe lines of [38, 21].The results here parallel those for the mass case. Firstly, for example we have56onstructed ‘singlet-like’ matrix elements – collectively called X here – wherethe LO term vanishes. As noted in [2] these can be extrapolated to the physicalpoint, using a one-parameter constant fit. In this article we constructed severalof these X functions, and indeed can isolate the constant as either the f or d coupling. Secondly again in analogy to the mass expansions we constructed ‘fan’plots, each element of which is a linear combination of matrix elements, where atthe SU (3) flavour symmetric point all the elements have a common value, andthen radiate away from this point as the quark masses change. This is slightlymore complicated than for the mass case as we now have two couplings, f and d .Indeed the ‘fan’ plot expansions can be constructed involving either f or d aloneat the SU (3) flavour symmetric point (more generally we have some combinationof them).Technically important for lattice determinations of matrix elements is thedifference between quark-line-connected and quark-line-disconnected terms in thecalculation of the three-point correlation functions. (The quark-line-disconnectedterms are small, but difficult to compute using lattice methods, due to large gluonfluctuations.) Applying the SU (3) flavour breaking expansion to these casesseparately, we have identified which expansion coefficient(s) have contributionscoming from the quark-line-disconnected terms. We found that at LO there isjust one expansion coefficient which has a quark-line-disconnected piece.As numerically we are using Wilson clover improved fermions, then for O ( a )continuum expansions, improvement coefficients need to be determined. Thegeneral structure for n f = 2 + 1 flavours of fermions has been determined, seee.g. [12]. We showed here these coefficients are equivalent to modifications tothe expansion parameters. Using the subsidiary condition that the relation be-tween the local and conserved vector current is O ( a ) allowed us to determine twoimprovement terms (together with the renormalisation constant).To demonstrate how the expansions work, we discussed numerical resultsusing the vector current and diagonal matrix elements. However these can beextended to include transition hyperon decays (a phenomenological review isgiven in [13]). These would allow an alternative method to the standard K (cid:96) decays of determining | V us | , e.g. [39, 13, 40]. Earlier quenched and n f = 2results for Σ − → n(cid:96)ν and Ξ → Σ + (cid:96)ν can be found in [41, 42], and n f = 2 + 1results have been obtained in [43, 44]. The latter reference also investigates thepossibility of non-linear effects in the quark-mass, which in the SU (3) symmetryflavour-breaking expansion means including terms from Table 7.Future theoretical developments include extending the formalism to partiallyquenched quark masses, when the valence quark mass, δµ q , does not have tobe the same as the sea or unitary quark mass. Then eq. (14) is replaced by δµ q = µ q − ¯ m . In this case the generalisation of eq. (17) does not hold. Thisallows the determination of the expansion coefficients over a larger quark massrange than is possible using the unitary quark masses (and allows, for example,the charm quark to be included, [45]). Furthermore expansions for ‘fake’ hadrons57ould be useful. Possible are a ‘nucleon’ with three mass degenerate strangequarks and a ‘Lambda’ with two mass degenerate strange quarks. Although theyare not physical states, they can be measured on the lattice, and do not introduceany more SU (3) mass flavour-breaking expansion coefficients, so simply add moreconstraints to the coefficient determination. An example of this for the baryonoctet masses is given in [21].Another extension of the SU (3) mass flavour breaking method is to the baryondecuplet with 10 ⊗ ⊗
10 tensors, and also to the meson octet. While the latterextension is straightforward, there are some extra constraints, as due to chargeconjugation, the particles in the meson octet are related to each other.Furthermore generalised currents can be evaluated between quark states. Thisleads to a SU (3) mass flavour-breaking expansion involving 3 ⊗ ⊗ RI (cid:48) − M OM scheme which de-fines the renormalisation constants (and improvement constants) by consideringthe generalised currents between quark states. Useful would also be to considerthe axial current improvement coefficients using a partially conserved axial-vectorcurrent (PCAC) along the lines of [12].Finally, a more distant prospect is to include QED corrections to the matrix el-ements, [10], along the lines of our previous studies of the SU (3) flavour-breakingexpansion for masses, [3, 4, 5]. Acknowledgements
The numerical configuration generation (using the BQCD lattice QCD program[46])) and data analysis (using the Chroma software library [47]) was carried outon the IBM BlueGene/Q and HP Tesseract using DIRAC 2 resources (EPCC,Edinburgh, UK), the IBM BlueGene/Q (NIC, J¨ulich, Germany) and the CrayXC40 at HLRN (The North-German Supercomputer Alliance), the NCI NationalFacility in Canberra, Australia (supported by the Australian CommonwealthGovernment) and Phoenix (University of Adelaide). We would like to thankAshley Cooke for useful discussions at an early stage in this project. RH wassupported by STFC through grant ST/P000630/1. HP was supported by DFGGrant No. PE 2792/2-1. PELR was supported in part by the STFC undercontract ST/G00062X/1. GS was supported by DFG Grant No. SCHI 179/8-1. RDY and JMZ were supported by the Australian Research Council GrantsFT120100821, FT100100005, DP140103067 and DP190100297. We thank allfunding agencies. 58 ppendixA Non-zero tensor elements
The non-zero elements of the tensors T ijk are listed in Tables 9 – 13. tensor value position2 334 463 646 − √ f −√ √ −√ − √ d − √ −√ − Table 9:
Flavour-singlet first-class non-zero elements of the f and d tensors. B Alternative fan plots
B.1 The doubly represented − singly represented fan, the P -fan The traditional way of expressing the two ways of coupling octet operators tooctet hadrons are the f and d couplings. In terms of hadron structure, thischoice is perhaps more natural for octet mesons than it is for octet baryons.Consider the eqs. (64, 65). In the K + , with quark content u ¯ s the f combination (cid:104) K + | (¯ uγu − ¯ sγs ) | K + (cid:105) is very natural (the difference between the two valencequarks), and the d combination (cid:104) K + | (¯ uγu + ¯ sγs − dγd ) | K + (cid:105) is also a natural-looking symmetric combination. For the Λ, the d combination is also the naturalnon-singlet operator to consider, d ∝ (cid:104) Λ | (2¯ sγs − ¯ uγu − ¯ dγd ) | Λ (cid:105) because the u and d in the Λ have the same structure functions, while the s structure is different(even before breaking SU (3)). 59 ensor value position r r √ − √ r − −√ √ − √ − √ − s √ −√ √ −√ − √ s −√ − Table 10:
First-class octet non-zero elements of the r , r , r and s , s tensors. But in the proton, it might be a bit more natural to choose the combinations(¯ uγu − ¯ dγd ) and (¯ uγu + ¯ dγd − sγs ) instead. The first combination is the non-singlet combination normally considered in discussions of proton structure, thesecond is almost (but not exactly) a measure of the total valence contribution,because the quark-line-disconnected (sea) contribution to (¯ uγu + ¯ dγd − sγs ) iszero at the symmetric point, and will probably stay small if the nucleon’s sea isapproximately SU (3) symmetric.We can therefore construct a fan plot for the doubly represented − singlyrepresented quark. P = √ A ¯ NπN = (cid:16) √ f + √ d (cid:17) − √ r − s ) δm l , ensor value position t − √ −√ t √ −√ − √ −√ u √ −√ − Table 11:
Second-class octet non-zero elements of the t , t and u tensors. P = 1 √ A ¯Σ π Σ + √ A ¯Σ η Σ ) = (cid:16) √ f + √ d (cid:17) + 1 √ (cid:16) √ r + 6 r − s + √ s (cid:17) δm l ,P = − √ A ¯Ξ π Ξ + √ A ¯Ξ η Ξ ) = (cid:16) √ f + √ d (cid:17) (122) − √ (cid:16) √ r + 2 r + 2 s + √ s (cid:17) δm l ,P = A ¯Σ K Ξ = (cid:16) √ f + √ d (cid:17) + √ r − s ) δm l . We have based this fan plot on the doubly − singly represented structure, soseveral of the observables have very simple quark structures. P = (cid:104) p | (¯ uγu − ¯ dγd ) | p (cid:105) ,P = (cid:104) Σ + | (¯ uγu − ¯ sγs ) | Σ + (cid:105) , (123) P = (cid:104) Ξ | (¯ sγs − ¯ uγu ) | Ξ (cid:105) ,P = (cid:104) Σ + | ¯ uγs | Ξ (cid:105) . This P fan only includes the ‘outer’ octet baryons. The natural plot for the Λstructure is the d -fan. There are two linear constraints on the P -fan,13 ( P + P + P ) = ( √ f + √ d ) + O ( δm l ) ,
13 ( P + 2 P ) = ( √ f + √ d ) + O ( δm l ) . (124)61 ensor value position −
18 55514 335 445 536 544 563 665 − √ q √ − √ − −
10 353 454 656 − √ q √ − √ √ √ −√ √ − √ √ − √ w √ − √ − − √ − √ √ − √ w − √ −√ − Table 12:
First-class 27-plet non-zero elements of the q , q and w , w tensors. ensor value position − √ √ z −√ √ − − x − √ −√ √ −√ √ − √ y √ −√ − Table 13:
First-class 64-plet and second-class 27-plet non-zero elements of the z and x , y tensors. A fan with just the four lines from eq. (123), P , P , P , P , is a four-line plot withjust two independent slope parameters, ( r − s ) and ( √ r + 4 r + √ s ).The advantage of this fan plot is that some of the quantities are of immediatephysical interest, for example in the weak decay case P gives the neutron decayconstant, while P gives the semileptonic decays Ξ → Σ + l − ¯ ν l , Ξ − → Σ l − ¯ ν l .The disadvantages are that there are fewer constraints than the d -fan. Also, the d -fan and f -fan are independent – they involve different parameters, and thereare no constraints that mix F i and D i quantities. A first attempt to show thisfan plot for the fraction of the baryon’s momentum carried by a quark, i.e. (cid:104) x (cid:105) , isgiven in [48].Finally it is again often useful to note from eq. (124) that for example X P = 13 ( P + P + P ) = ( √ f + √ d ) + O ( δm l ) , (125)and to consider the quantities P i /X P . 63 .2 The V -fan The other natural non-singlet to look at in the proton is (cid:104) p | (¯ uγu + ¯ dγd − sγs ) | p (cid:105) .This is approximately the total valence distribution, the quark-line-disconnected(sea) contribution to (¯ uγu + ¯ dγd − sγs ) is zero at the symmetric point, and willprobably stay small if the nucleon’s sea is approximately SU (3) symmetric. V = √ A ¯ NηN = √ √ f − d ) + √ r − s ) δm l ,V = 3 √ A ¯Σ π Σ − (cid:114) A ¯Σ η Σ , = √ √ f − d ) − √ (cid:16) √ r + 6 r + 6 s − √ s (cid:17) δm l ,V = 3 √ A ¯Ξ π Ξ − (cid:114) A ¯Ξ η Ξ = √ √ f − d ) , (126) − √ (cid:16) √ r − r − s + √ s (cid:17) δm l ,V = √ A ¯ NπN + 2 A ¯Ξ π Ξ ) = √ √ f − d ) + 2 √ r + 3 s ) δm l ,V = ( A ¯Σ K Ξ − A ¯ NK Σ ) = √ √ f − d ) − √ r + 3 s ) δm l . We have the two constraints13 ( V + V + V ) = √ √ f − d ) + O ( δm l ) ,
13 ( V + 2 V ) = √ √ f − d ) + O ( δm l ) , (127)and can again construct an X V from either combination, for example set X V = 13 ( V + V + V ) , (128)and again consider ratios such as V i /X V . C LO flavour diagonal matrix elements
To leading order we have for the representative octet baryons p , Σ + , Λ and Ξ (cid:104) p | ¯ uγu | p (cid:105) = 1 √ (cid:16) a + √ f + √ d (cid:17) + 1 √ (cid:18) a + 1 √ r − √ r + √ s − √ s (cid:19) δm l , (cid:104) p | ¯ dγd | p (cid:105) (129)= 1 √ (cid:16) a − √ d (cid:17) + 1 √ (cid:18) a + 1 √ r + √ r − √ s − √ s (cid:19) δm l , (cid:104) p | ¯ sγs | p (cid:105) = 1 √ (cid:16) a − √ f + √ d (cid:17) + 1 √ (cid:16) a − √ r + √ s (cid:17) δm l , Σ + | ¯ uγu | Σ + (cid:105) = 1 √ (cid:16) a + √ f + √ d (cid:17) + 1 √ (cid:18) − a + 1 √ r + √ r − √ s + 3 √ s (cid:19) δm l , (cid:104) Σ + | ¯ dγd | Σ + (cid:105) (130)= 1 √ (cid:16) a − √ f + √ d (cid:17) + 1 √ (cid:18) − a + 1 √ r + √ r + √ s − √ s (cid:19) δm l , (cid:104) Σ + | ¯ sγs | Σ + (cid:105) = 1 √ (cid:16) a − √ d (cid:17) + 1 √ (cid:16) − a − √ r − √ r (cid:17) δm l , (cid:104) Λ | ¯ uγu | Λ (cid:105) = (cid:104) Λ | ¯ dγd | Λ (cid:105) , = 1 √ (cid:16) a − √ d (cid:17) + 1 √ (cid:18) a + 1 √ r + √ r (cid:19) δm l , (cid:104) Λ | ¯ sγs | Λ (cid:105) = 1 √ (cid:16) a + 2 √ d (cid:17) + 1 √ (cid:16) a − √ r − √ r (cid:17) δm l , (131)and (cid:104) Ξ | ¯ uγu | Ξ (cid:105) = 1 √ (cid:16) a − √ d (cid:17) + 1 √ (cid:18) − a − a ) + 1 √ r + √ r + √ s + 1 √ s (cid:19) δm l , (cid:104) Ξ | ¯ dγd | Ξ (cid:105) = 1 √ (cid:16) a − √ f + √ d (cid:17) (132)+ 1 √ (cid:18) − a − a ) + 1 √ r − √ r − √ s + 1 √ s (cid:19) δm l , (cid:104) Ξ | ¯ sγs | Ξ (cid:105) = 1 √ (cid:16) a + √ f + √ d (cid:17) + 1 √ (cid:16) − a − a ) − √ r − √ s (cid:17) δm l . D LO disconnected flavour diagonal matrix el-ements
From eqs. (77, 78) we have f dis , d dis , r dis , r dis , s dis and s dis all vanishing at LO andonly r dis contributing. Thus we have (cid:104) N | ¯ uγu | N (cid:105) dis = (cid:104) N | ¯ dγd | N (cid:105) dis = 1 √ a dis + (cid:18) √ a dis + 1 √ r dis (cid:19) δm l , (133) (cid:104) N | ¯ sγs | N (cid:105) dis = 1 √ a dis + (cid:32) √ a dis − (cid:114) r dis (cid:33) δm l , n , p ), (cid:104) Σ | ¯ uγu | Σ (cid:105) dis = (cid:104) Σ | ¯ dγd | Σ (cid:105) dis = 1 √ a dis + (cid:18) −√ a dis + 1 √ r dis (cid:19) δm l , (134) (cid:104) Σ | ¯ sγs | Σ (cid:105) dis = 1 √ a dis + (cid:32) −√ a dis − (cid:114) r dis (cid:33) δm l , (for Σ + , Σ , Σ − ), (cid:104) Λ | ¯ uγu | Λ (cid:105) dis = (cid:104) Λ | ¯ dγd | Λ (cid:105) dis = 1 √ a dis + (cid:18) √ a dis + 1 √ r dis (cid:19) δm l , (135) (cid:104) Λ | ¯ sγs | Λ (cid:105) dis = 1 √ a dis + (cid:32) √ a dis − (cid:114) r dis (cid:33) δm l , (for Λ ) and (cid:104) Ξ | ¯ uγu | Ξ (cid:105) dis = (cid:104) Ξ | ¯ dγd | Ξ (cid:105) dis = 1 √ a dis + (cid:18) −√ a dis − a dis ) + 1 √ r dis (cid:19) δm l , (136) (cid:104) Ξ | ¯ sγs | Ξ (cid:105) dis = 1 √ a dis + (cid:32) −√ a dis − a dis ) − (cid:114) r dis (cid:33) δm l , (for Ξ , Ξ − ). References [1] W. Bietenholz, V. Bornyakov, N. Cundy, M. G¨ockeler, R. Horsley, A. D.Kennedy, W. G. Lockhart, Y. Nakamura, H. Perlt, D. Pleiter, P. E. L.Rakow, A. Sch¨afer, G. Schierholz, A. Schiller, H. St¨uben and J. M.Zanotti [QCDSF–UKQCD Collaboration],
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