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Expansion 1=Nc for Excited Baryons Dan Pirjol CNLS 97/1500 TECHNION-PH 97-06 Department of Physics, Technion - Israel Institute of Technology, 32000 Haifa, Israel Floyd R. Newman Laboratory of Nuclear Studies, Cornell University, Ithaca, New York 14853 (July 27, 1997) Tung-Mow Yan hep-ph/9707485 27 Jul 1997 Abstract We derive consistency conditions which constrain the possible form of the strong couplings of the excited baryons to the pions. The consistency conditions follow from requiring the pion-excited baryon scattering amplitudes to satisfy the large-Nc Witten counting rules and are analogous to consistency conditions used by Dashen, Jenkins and Manohar and others for s-wave baryons. The consistency conditions are explicitly solved, giving the most general allowed form of the strong vertices for excited baryons in the large-Nc limit. We show that the solutions to the large-Nc consistency conditions coincide with the predictions of the nonrelativistic quark model for these states, extending the results previously obtained for the s-wave baryons. The 1=Nc corrections to these predictions are studied in the quark model with arbitrary number of colors Nc . Typeset using REVTEX 1 I. INTRODUCTION The successes of the nonrelativistic quark model (NRQM) in describing the baryon spectroscopy and couplings 1] have remained for a long time something of a mystery. Recent work by several groups 2, 3, 4, 5, 6, 7, 8, 9], most notably by Dashen, Jenkins and Manohar (see also earlier related work in 10]) helped to shed light on this problem and clarify the relation of the NRQM to QCD. These works showed that the predictions of the NRQM for low-lying s-wave baryons follow from QCD in the large-Nc limit 11] as a consequence of the counting rules of Witten 12, 13, 14] for pion-baryon scattering amplitudes. In this way they have been able to derive consistency conditions which constrain the mass splittings, pion couplings and magnetic moments of ground-state baryons up to order O(1=Nc ) in the 1=Nc expansion. The nonrelativistic quark model has been used to describe also the properties of the orbitally excited baryons. The realization of the fact that these states can be accounted for in the quark model has been one of the rst signi cant arguments in its favor 15]. Later works applied the quark model to explaining the phenomenology of the strong decays of the L = 1 baryons to the ground state baryons. The measured decay widths have been found to be well described by a t to the quark model predictions 16, 17, 18]. When supplemented with dynamic assumptions, the quark model can be also used to make more detailed predictions about the mass spectrum and decay properties of these states 19, 20, 21]. In addition to the quark model, various other approaches have been employed to describe the orbital excitations of baryons. Among them the Skyrme model, which is closely related to the large-Nc approximation, has been used to construct these states as bound states of a soliton and a meson 22, 23, 24, 25, 26]. A bag model description of these states has been given in 27]. The properties of the negative parity baryons have been investigated also with the help of the QCD sum rules in 28, 29, 30]. More recently, in 31] the structure of the mass spectrum of the excited baryons has been studied using an e ective Hamiltonian motivated by large-Nc arguments. Following the recent progress in understanding the predictions of the quark model for ground state baryons, some e ort has been also directed into explaining the analogous predictions for the excited baryons sector. Thus, in 18] the data on the strong decays of these states have been used to test the idea that the large-Nc limit might provide an explanation for the validity of the quark model description. The authors of 18] adopted a Hartree description with the number of quarks in the baryon xed to its physical value Nc = 3. The large-Nc expansion has been implemented at the level of operators mediating the strong decays, which can be classi ed according to their order in 1=Nc . A t to the experimental data on strong decays of the L = 1 baryons in the 70 of SU(6) gave the result that the naive quark model, containing only one-body operators, reproduces the experimental data to a good precision. On the other hand, two-body operators which could contribute to same order in 1=Nc as those kept in the quark model, appear to be suppressed in Nature for reasons seemingly unrelated to the large-Nc expansion. From this, the authors of 18] concluded that there might be more than large-Nc to the success of the quark model relations. In this paper we study the strong pion couplings of the orbitally excited baryons, both light and heavy, in the large-Nc limit using as input constraints on pion-baryon scattering amplitudes following from the counting rules of Witten. This approach is closer in spirit to 2 the one used in 2, 3] by Dashen, Jenkins and Manohar. We derive in this way consistency conditions which constrain the possible form of the strong coupling vertices, which are then solved explicitly. Our nal conclusion is that the model-independent results obtained from solving the consistency conditions are the same as those following from the quark model in the large-Nc limit, thus extending the statements of 2, 3, 4, 5, 6] to the excited baryons' sector. We stress that our results do not con ict with the conclusions of 18]. Rather, the ndings of 18] can be formulated in the light of our results as a puzzle: why does the quark model work better than it should? Our paper is structured as follows. We begin by introducing in Section II the spectrum of the orbital excitations and constructing its generalization to the large-Nc limit. The structure of these states is more complex than for the case of the s-wave baryons. We introduce the concept of P-spin to deal with the mixed symmetry spin- avor states and point out an additional problem connected with the appearance of spurious unphysical states in the Nc > 3 case. Section III contains the derivation of the consistency conditions for strong coupling vertices. These arise from a mismatch between the scaling power with Nc of the meson-baryon vertices and the Witten scaling law for the meson-baryon scattering amplitudes. The consistency conditions are explicitly solved in Section III giving the most general solution for S-, P- and D-wave pion couplings in the large-Nc limit. We show in Section IV that the solutions to the consistency conditions actually coincide with the predictions of the constituent quark model in the large-Nc limit. The orbital excitations are rst explicitly constructed in the quark model with u and d quarks only and arbitrary number of colors Nc. Armed with these wavefunctions, we develop the machinery necessary to compute the strong coupling vertices of these states. A by-product of this quark model calculation is a determination of the large-Nc scaling law of the decay vertices, which exhibits a surprising dependence on the symmetry type of the excited state involved. We conclude in Section V with a summary and outlook on our results. Appendix A contains an exact calculation in the quark model of the strong coupling vertices for an arbitrary value of Nc . These results are used to examine the structure of the 1=Nc corrections to the large-Nc relations obtained in Section III. Appendix B presents the quark model calculation of the strong couplings among excited states transforming under the mixed symmetry representation of SU(4). II. SPIN-FLAVOR STRUCTURE OF THE EXCITED BARYONS In the large-Nc limit, the s-wave baryons containing only u; d quarks form I = J towers of degenerate states. Both possibilities I = J = 1=2; 3=2; and I = J = 0; 1; are of physical signi cance, the former corresponding to the light baryons and the latter to baryons with one heavy quark (in this case J is to be interpreted as the angular momentum of the light degrees of freedom). Baryons with strangeness can be also incorporated in the largeNc limit as separate towers of states, each labeled by a quantum number K related to the number of strange quarks as K = 1 ns . For each K tower, the spin J and isospin I take 2 values restricted by the condition jI Jj K. This picture is precisely the same as the one predicted by the NRQM with SU(4) spinavor symmetry. In NRQM language the s-wave baryons have orbital wavefunctions which are completely symmetrical under permutations of two quarks. This constrains their spinisospin wavefunction to transform also under the completely symmetric representation of 3 SU(4), which contains the I; J values given above. Spin- and avor-independence of the interquark forces in the NRQM is responsible for the degeneracy of all these states. Fig.1 shows the Young diagram of the totally symmetric representation of SU(4) for Nc = 3 and its extension to the case of arbitrary Nc . ! z Nc }| { Nc = 3 and in the large-Nc limit. The spectrum of the p-wave baryons has a more complicated structure. The spin- avor wavefunction of the light baryons has mixed symmetry, transforming for Nc = 3 as a 70 under SU(6) and as a 20 under SU(4). To keep our results as general as possible and to avoid some ambiguities connected with the identi cation of the large-Nc states with physical states, we will not assume SU(3) symmetry. Just as in the case of the s-wave baryons, we will divide the p-wave states into sectors with well-de ned strangeness and assume only isospin symmetry. We extrapolate the mixed symmetry representation from Nc = 3 to the large-Nc case as shown in Fig.2. Fig.1. Young tableaux for the SU(4) representation of the s-wave baryons for ! large-Nc limit. Under the isospin-spin SU(2) SU(2) group this representation splits into (I; S) representations which satisfy jI Sj 1 (except for I = S = Nc =2 which is only contained in the totally symmetric representation). Fig.2. SU(4) representations for light p-wave baryons, for Nc = 3 and in the z Nc 1 }| { = z Nc }| { z Nc 1 }| { of the (I; S) content of the mixed symmetry representation. This can be proven by considering the product of SU(4) representations shown in Fig.3 and its decomposition into irreducible representations of SU(2)spin SU(2)isospin . For definiteness we will take Nc to be odd, although the argument is equally valid also for even values of Nc . The isospin-spin (I; S) content of the product of representations on the l.h.s. can be obtained from the corresponding product 1 (0; 0) ; (1; 1) ; ( Nc 2 1 ; Nc 2 1 ) ( 2 ; 1 ) (2.1) 2 4 Fig.3. Product of SU(4) representations used in the text for the determination 1 and includes all representations of the form (i 1 ; i 2 ) with i = 1; ; (Nc 1)=2. All the 2 representations with I 6= S occur with multiplicity 1. The respresentations with (I; S) = 1 1 1 (i + 2 ; i+ 2 ) ; (i 2 ; i 1 ) appear twice, except for (I; S) = (Nc=2; Nc =2) which appears only 2 once. On the other hand, the symmetric representation on the r.h.s. of Fig.3 contains only the I = S representations described above, but with unit multiplicity. Subtracting them from the (I; S) representations on the l.h.s. of Fig.3 we are left with the representation content mentioned above for the mixed symmetry SU(4) representation. This can be further checked by comparing the dimensionality of the SU(4) representation given by the Young diagram in Fig.2 for arbitrary Nc with the sum of the dimensions of the (I; S) representations described above Nc 1 h i X2 1 dim = 2 (Nc 1)(Nc + 1)(Nc + 2) = n + 2n(n + 2) : (2.2) n=2 ~~~ The total baryon spin J is given by J = S + L with L = 1. The lowest-lying observed p-wave light baryons containing only u; d quarks are listed in Table 1 together with their quantum numbers in the quark model 32]. Note that the states (I; S) = (3=2; 3=2) which would be present in the large-Nc limit are forbidden in the Nc = 3 case for the reason mentioned above. State (I; J P ) L2I;2J (I; S) (SU(3); SU(2)) 1; 1 ) 11 N(1535) ( 2 2 S11 (2; 2) (8; 2) 1 N(1520) ( 2 ; 3 ) D13 2 1; 1 ) 13 N(1650) ( 2 2 S11 (2; 2) (8; 4) 1; 3 ) N(1700) ( 2 2 D13 1 N(1675) ( 2 ; 5 ) D15 2 3; 1 ) 31 (1620) ( 2 2 S31 (2; 2) (10; 2) 3; 3 ) (1700) ( 2 2 D33 Table 1. The p-wave light baryons containing only u; d quarks and their quantum numbers. It is not di cult to introduce also strangeness in this picture. Because the strange quark is now di erent from the other Nc 1 quarks in the baryon, the Pauli principle constrains only the symmetry properties of the wavefunction for the latter. In this case both SU(4) representations shown in Figs.1 and 2 are possible. We show in Table 2 the lowest-lying observed and expected p-wave hyperons with one strange quark together with their quark model quantum numbers. For example, the states with (I; S) = (1; 3=2) in Table 2 are completely symmetric under a permutation of the u; d quarks, whereas the states (I; S) = (0; 3=2) are antisymmetric under the same transformation (for Nc = 3 the mixed symmetry state is in fact antisymmetric). The symmetric representation corresponds to 10 and the antisymmetric one to 6 of SU(4). The other states in Table 2 are mixtures of both representations. To construct the analogs of these states in the large-Nc limit, it is convenient to introduce ~ ~ two vectors K and P , which will be called K-spin and P-spin respectively. The K-spin counts the number of strange quarks as described above and takes the value 1/2 for hyperons with 5 one s quark 5]. The P-spin labels the type of permutational symmetry of the u; d quarks' wavefunction in the baryon and is equal to 0 for the symmetric representation and to 1 for the mixed symmetric representation. With these de nitions the total quark spin S of a p-wave baryon takes all the values compatible with ~~~~ S =I+K +P : (2.3) In addition to this, an exclusion rule must be imposed for P = 1, forbidding the following symmetric states ~~ jS K j = I = Nc K : (2.4) 2 This exclusion rule is operative only at the top of the large-Nc towers and therefore can be neglected in the large-Nc limit. One should keep however in mind the fact that new unphysical states are introduced in the large-Nc limit which would be otherwise forbidden by this rule. The classi cation of the states into symmetric and mixed representations is even more transparent for the p-wave baryons with one heavy quark. In the heavy mass limit the spin and parity of the light degrees of freedom become good quantum numbers. Furthermore, in the NRQM the total spin of the light quarks S` is also conserved and can be used together with the isospin to identify the permutational symmetry of the state. State (1405) (1520) (1670) (1690) (1620) (1670) (1800) (?) (1830) (1750) (?) (1775) (?) (?) numbers. (I; J P ) (0; 1 ) 2 (0; 3 ) 2 (0; 1 ) 2 (0; 3 ) 2 1 (1; 2 ) 3 (1; 2 ) (0; 1 ) 2 3 (0; 2 ) (0; 5 ) 2 1 (1; 2 ) 3 (1; 2 ) 5 (1; 2 ) 1 (1; 2 ) 3 (1; 2 ) (I; S) (SU(3); SU(2)) (0; 1 ) (1; 2) 2 (0; 1 ) 2 (1; 1 ) 2 (0; 3 ) 2 (1; 3 ) 2 (1; 1 ) 2 (10; 2) (8; 4) (1; 2) Table 2. The p-wave hyperons containing one strange quark and their quantum Thus, in the large-Nc limit the symmetric representation will give rise to an I = S` tower of states, in analogy to the situation for the light s-wave baryons (with the total spin of the light quarks S` taking the place of the total spin J). The mixed symmetry representation 6 will generate also a tower with jI S`j 1, as in the case of the light p-wave baryons. From this the states with I = S` = (Nc 1)=2 will have to be excluded. The total heavy baryon spin J will be given in the general case including also strangeness by ~~~~ ~~~ J = I + S` + SQ + K + P + L (2.5) with SQ = 1=2 the heavy quark spin. We emphasize that the use of quark model quantum numbers such as S; S` ; SQ, etc. does not imply any dynamical assumption on our part and is made with the sole purpose of counting states. All our main results below will be obtained without any assumption of whether these quantities are conserved or not. We use the NRQM just as a convenient language which serves to guide our intuition about the spin and avor structure of the states of interest. In the next Section we will study the strong couplings of the excited baryons in the large-Nc limit. III. CONSISTENCY CONDITIONS FOR EXCITED BARYONS We will obtain constraints on the pion couplings of the excited baryons by studying both elastic pion scattering on these states and inelastic scattering among s-wave and excited states. The results will follow from a set of consistency conditions, derived by requiring the total scattering amplitude to satisfy large-Nc counting rules 12, 13, 14]. We start by reviewing the large-Nc scaling properties of the di erent couplings which will be needed. Pions couple to baryons with a strength proportional to the matrix element of the axial current taken between the corresponding states. In the case of the s-wave baryons this matrix element was parametrized in 2, 3, 5] as 1 hJ 0; m0; 0jq i 5 2 aqjJ; m; i = Ncg(X)hJ 0; m0; 0jX iajJ; m; i (3.1) with X ia an irreducible tensor operator of spin and isospin 1 and g(X) a reduced matrix element of order 1 in the large-Nc limit. X ia has a large-Nc expansion of the form X ia = X0ia + X1ia=Nc + . The matrix element (3.1) grows linearly with Nc because the axial current couples to each of the Nc quarks in the baryon. We will use a similar parametrization for the matrix element of the axial current taken between two excited baryons 1 (3.2) hJ 0; I 0; m0; 0jq i 5 2 aqjJ; I; m; i = Nc g(Z)hJ 0; I 0; m0; 0jZ iajJ; I; m; i where Z ia is again an irreducible tensor operator with J = I = 1. This matrix element grows also with Nc for the same reason as in the preceding case. On the other hand, the axial current matrix elements taken between s-wave and p-wave baryons grow slower than Nc . We parametrize the matrix elements of the time and space components of the axial current as 7 (3.4) with q the momentum of the current. In the quark model the scaling power is equal to 1/2 for p-wave baryons transforming under the completely symmetric representation of SU(4) and 0 for baryons transforming according to the mixed symmetry representation of SU(4). This will be proved in Section IV. Y a is a tensor operator with spin 0 and isospin 1 and Qij;a has spin 2 and isospin 1 (Qij;a = Qji;a, Qii;a = 0). The operators Z; Y; Q have expansions in powers of 1=Nc of the same form as X. The pion coupling to the states appearingp (3.1-3.4) is obtained by dividing these matrix in elements by the pion decay constant f = O( Nc) 2, 3, 5, 10]. The consistency conditions of DJM were obtained by considering pion scattering on s-wave baryons a(q)+B ! b(p)+B 0 2, 3, 5]. The leading contribution to this amplitude arises from two tree graphs with the pions coupling in either order and is given by 22 qipj T = Nc g 2(X) E(~) X0jbyX0ia X0iaX0jby : (3.5) f q This scattering amplitude is of order Nc, in apparent contradiction with the large-Nc counting rules of Witten according to which it should be of order 1. One concludes therefore that one has h jby iai X0 ; X 0 = 0 : (3.6) This is the leading order consistency condition of DJM 2, 3, 5, 10]. Taking as target a p-wave baryon, the above reasoning can be extended immediately to the couplings Z, for which one obtains the analogous condition h jby iai (3.7) Z0 ; Z0 = 0 : The operators Z ia act only on the space of the p-wave states which are degenerate among themselves and have vanishing matrix elements between p-wave states of di erent mass. We would like next to derive consistency conditions involving the couplings Y and Q. In order to do so we consider the scattering amplitude for the process a(q)+ (p-wave) ! b(p)+ (s-wave). The mass splitting between s-wave and p-wave states is of order 1 in the large-Nc limit, so that the initial and nal pions will not have the same energy. Adding together the contributions of the diagrams with intermediate s-wave and p-wave baryons we obtain for this case ( Nc1+ g(Y ) piE(~) g(X)X ia Y by g(Z)Y byZ ia q (3.8) T = f2 E(~) p ) qiE(~) g(X)X ibyY a g(Z)Y aZ iby p + E(~) q 1+ g(Q) ( pi q j q k ia jk;by g(Z)Qjk;byZ ia + Nc f 2 E(~) g(X)X Q p ) qkpi pj g(X)X kbyQij;a g(Z)Qij;aZ kby : + E(~) q 8 hJ 0; m0; 0jq hJ 0; m0; 0jq i 0 5 1 a qjJ; I; m; 2 i = Nc g(Y )hJ 0; m0; 0jY ajJ; I; m; i 1 5 2 aqjJ; I; m; i = Nc g(Q)q j hJ 0; m0 ; 0jQij;a jJ; I; m; i (3.3) This scattering amplitude is apparently of order 0 which again violates the counting rules of Witten, according to which it should be at most of order Nc 1=2 13]. This requires all the independent kinematical structures to vanish to leading order g(X)X0ia Y0by g(Z)Y0byZ0ia = 0 ; g(X)X0ibyY0a g(Z)Y0aZ0iby = 0 g(X)X0ia Qjk;by g(Z)Qjk;byZ0ia = 0 ; g(X)X0kbyQij;a g(Z)Qij;aZ0kby = 0 : 0 0 0 0 (3.9) (3.10) All of our conclusions about the pion couplings of the excited baryons in the large-Nc limit will follow from the set of consistency conditions (3.7,3.9,3.10). In the present paper we restrict ourselves to the leading order in the large-Nc expansion. Therefore, to simplify the notation, we will drop the index 0 on the coupling operators throughout in the following. A. Consistency condition for Z The consistency condition for Z ia (3.7) is completely identical in form to the one for X ia (3.6) which has been studied in detail in 2, 5]. These authors showed that X ia forms, together with the generators of the spin-isospin SU(2) SU(2) group J i; I a a contracted SU(4) algebra. Every possible solution for X ia corresponds to a particular irreducible representation of this algebra. The most general irreducible representation can be labeled by a spin vector ~ , in terms of which the basis states of the representation are constructed as ~~ J =I+~. In principle it would be possible to take over the results of 5] for X ia and write down directly the matrix elements of Z ia. We will prefer however to construct the solution for Z ia by using a NRQM-inspired ansatz. Besides reproducing the result of 5], this approach has the advantage of suggesting a method for obtaining the solution of the consistency conditions (3.9,3.10). In retrospect, this will furnish also a proof of the validity of the NRQM predictions for excited baryons in the large-Nc limit. We begin by parametrizing the matrix elements of Z ia taken between states belonging to - and 0-towers respectively as hJ 0; I 0; m0; 0jZ iajJ; I; m; i = q ( 1)J+I (2I + 1)(2J + 1)Z(J 0; I 0; J; I)hJ 0; m0jJ; 1; m; iihI 0; 0jI; 1; ; ai : (3.11) The notation adopted anticipates a result to be proven below, according to which Z only connects towers with = 0. The reduced matrix element Z(J 0; I 0; J; I) depends on the common value of , although for the sake of simplicity this is not made explicit. The normalization coe cient is chosen such that the reduced matrix element is symmetric under a permutation of the initial and nal indices Z(J 0; I 0; J; I) = Z(J; I; J 0; I 0). The consistency condition (3.7) can be used to obtain constraints on the reduced matrix elements Z(J 0; I 0; J; I). For this it will be sandwiched between two general states hJ 0; I 0; m0; 0j jJ; I; m; i and a complete set of intermediate states is inserted. We obtain X 0 0 0 0 jby hJ ; I ; m ; jZ jJ1; I1; m1; 1ihJ1 ; I1; m1; 1jZ iajJ; I; m; i (3.12) J1 I1 m1 1 (Z jby $ Z ia) = 0 : 9 This equation can be projected, as in 2], onto the channel with total angular momentum H and isospin K by multiplying it with hH 0 ; h0jJ 0; 1; m0; jihH; hjJ; 1; m; iihK 0; 0jI 0; 1; 0; bihK; jI; 1; ; ai (3.13) and summing over m; m0; i; j; ; 0; a; b. The resulting consistency condition takes the form ) )( ( X I 1 K Z(J 0; I 0; J ; I )Z(J ; I ; J; I) J1H (3.14) (2J1 + 1)(2I1 + 1) J 0 1 J 11 11 I 0 1 I1 1 J1 ;I1 = ( )2(J +I )Z(H; K; J 0; I 0)Z(H; K; J; I) : We will try to guess the solution of this consistency condition by using as guidance the nonrelativistic quark model. Once found, the solution will be seen to be unique by using for example numerical solution of the consistency condition (3.14) or the method of the induced representations 5]. Let us consider for simplicity the case of baryons without strange quarks. Also, let us ~ ~~ rst assume that the total baryon spin is given by J = I + L, which is to say that the baryon will be regarded as containing a \core" of u; d quarks transforming under the symmetric representation of SU(4). The \core" spin S is therefore equal to its isospin I. In addition to ~ ~ this, the orbital angular momentum L is added to make up the total spin J. This corresponds to the case of a heavy baryon transforming under the symmetric representation of SU(4), provided that J is interpreted as the angular momentum of the light degrees of freedom. The basis states can be easily constructed and are given by X jI; L; mS ; mL; ihJ; mjI; L; mS ; mLi : (3.15) jI; L; J; m; i = 0 0 mS mL ~ ~ where i acts only on the spin of the u; d quarks S and a acts only on the isospin I . Therefore the matrix element of Z ia between the states (3.15) can be expressed as hI 0; L0;X0; m0; 0jZ iajI; L; J; m; i = J (3.17) 0 ; L0 ; m0 ; m0 ; 0 j i a jI; L; m ; m ; ihJ 0 ; m0jI 0; L0; m0 ; m0 ihJ; mjI; L; m ; m i : hI SL SL SL SL The matrix element in the basis jI; L; mS ; mL; i can be parametrized with the help of the Wigner-Eckart theorem in terms of a new reduced matrix element Z(I 0; I) hI 0; L0; m0S ; m0L; 0j i ajI; L; mS ; mL; i = (3.18) 1 00 0 00 0 + 1 Z(I ; I)hI ; mS jI; 1; mS ; ii LL mL mL hI ; jI; 1; ; ai : 2I 0 0 The current Z ia becomes in the quark model Z ia ! i a (3.16) mS mL mS mL 0 0 With this normalization the reduced matrix element is symmetric Z(I 0; I) = Z(I; I 0). Inserting this expression in (3.17) it is possible to compute the matrix element of Z ia taken between jI; L; J; m; i states. Comparing with the parametrization (3.11) we obtain the following connection between Z(J 0I 0; JI) and Z(I 0; I) 10 q (2J + 1)(2I + 1)Z(J 0I 0; JI) = (3.19) s ) (0 2J 1I Z(I 0; I) LL L ( )2J +J I +1 2I 0 + 1 I L J 0 : J +1 We can nd a consistency condition for Z(I 0; I) by inserting this expression into (3.14). The sum over J1 can be performed explicitly and we nd 8 9 < = X 2I1 ( I 1 I1 ) > I 10 I10 > 0 (2K + 1) ( ) (3.20) 0 1 K > J I > Z(I ; I1 )Z(I1; I) I :J H 1 ; I1 )( ) ( K 1 I 0 K 1 I Z(K; I 0)Z(K; I) : = ( )2I 2K J 0 H JH It is easy to see, by making use of the relation (Eq.(6.4.8) in 33]) ( ) 8 j11 j12 9 ) )( ( > > < = X j11 j12 j31 j32 j33 : (3.21) 2 j21 j22 j23 (2 + 1) j j j > j21 j22 j23 > = ( ) j12 j11 j21 j32 23 33 : 31 j32 j33 ; q that this equation is satis ed by the solution Z(I 0; I) = (2I + 1)(2I 0 + 1) (up to a constant which can be absorbed into g(Z)). We obtain in this way the result ) (0 0; I 0; J; I) = ( ) I+I +1 I 1 I (3.22) Z(J L: J L J0 We consider next the slightly more complicated case of the baryons transforming under the mixed symmetry representation of SU(4). This is relevant for the light baryons containing only u; d quarks. In this case, the total spin of the excited baryon is given by ~~~~ J = I + P + L. There is an important di erence in the application of the quark model to this situation, connected with the fact that i in (3.16) acts on the spins of the u; d quarks. ~~~ The total spin of the u; d quarks S = I + P is not equal to I as before. Therefore the natural set of states for doing the quark model calculation is j(IP)S; L; J; m; i. On the other hand, we would like to classify the states in (3.11) according to the value ~ ~ of the spin vector ~ , such that I + ~ = J. This requires a di erent coupling of the vectors ~; P ; L: jI; (PL) ; J; m; i. The connection between these two sets of states is a well-known ~~ I recoupling problem in the theory of angular momentum and is given by Eq.(6.1.5) in 33] jI; (PL) ; J; m; i = (3.23) ( ) q X I ( )IPLJ (2S + 1)(2 + 1) L P S j(IP)S; L; J; m; i : J ( )J+I 0 0 0 0 0 0 The matrix element of Z ia taken between the j(IP)S; L; J; mi states can be written as h(I 0P 0)S 0; L0; J 0; m0; 0jZ iaj(IP)S; L; J; m; i = (3.24) X 0S 0 L0 ; m0 ; m0 ; 0 j i a jISL; m ; m ; i hI SL SL mS mL mS mL 0 0 S 0 hJ 0; m0jS 0; L0; m0S ; mLihJ; mjS; L; mS ; mLi : 11 An application of the Wigner-Eckart theorem gives hS 0I 0L0; m0S ; m0L; 0j 1 q (2S 0 + 1)(2I 0 + 1) i a jSIL; m ; m ; i = SL Z(S 0I 0; SI) LL mLmL hS 0; m0S jS; 1; mS ; iihI 0; 0jI; 1; 0 0 (3.25) ; ai ; with Z(S 0I 0; SI) a new reduced matrix element. With this choice for the normalization factor, it transforms under a permutation of the initial and nal indices as Z(S 0I 0; SI) = ( )S+I SI 0 0 Z(SI; S 0I 0) : (3.26) It is easy to compute now the matrix element of Z ia between the jI; (PL) ; J; m; i states by inserting (3.25) into (3.24) and using the expansion (3.23). We obtain 0 hI 0; (PsL0) 0; J 0; m0; 0jZ iajI; (PL) ; J; m; i = ( ) X (2S + 1)(2 + 1)(2 0 + 1) I P S Z(S 0I 0; SI) I0 P 0 S LL L0 J 0 0 0+1 LJ 2I SS X hS 0; m0S jS; 1; mS ; iihJ 0; m0jS 0; L; m0S ; mLihJ; mjS; L; mS ; mLihI 0; 0jI; 1; ; ai : 0 0 ( 0 )( I P L J IPLJ 0 0 0 0 ) (3.27) mS mL mS 0 Let us pause for one moment and compare this expression with (3.11). One can see that the isospin CG coe cient is the same on the r.h.s. of these two relations. We extract Z(J 0I 0; JI) by multiplying both equations with hJ 0; m0jJ; 1; m; ii and summing over (m; i). The resulting sum over 4 CG coe cients can be expressed as a 6j symbol. We obtain nally q ( )J+I (2J + 1)(2I + 1)Z(J 0; I 0; J; I) = (3.28) s 0 ( ) I +J I P P L+1 (2J + 1)(22I 0 + 1)(2 + 1) ) )( ( 0 0 + 1)( X Sq I P S S 0 1 S Z(S 0I 0; SI) : I P S0 ( ) (2S + 1)(2S 0 + 1) L J 0 0 L J J L J0 SS 0 0 0 0 0 We insert the following ansatz for the reduced matrix element Z(S 0I 0; SI) (inspired by (3.22) with the identi cation (ILJ) ! (IPS)) ) ( q 0 I 0; SI) = ( ) S I (2S + 1)(2S 0 + 1)(2I + 1)(2I 0 + 1) I P S (3.29) Z(S S0 1 I 0 0 which has the required symmetry property (3.26). If we assume that P = P 0 we can perform the sums over S and S 0 in (3.28) with the help of the identity (Eq.(C.35e) in 34]) ( )( )( )( )( ) X abx cdx efx = ghj ghj ; ( ) (2x + 1) c d g e f h b a j (3.30) ead f bc x with = a + b + c + d + e + f + g + h + x + j. The nal result for Z(J 0I 0; JI) is 12 Z(J 0; I 0; J; I) = ( )I 2J+I +P 0 ( I0 1 I J J0 ) 0 PP 0 : (3.31) For P = P 0 this reduced matrix element vanishes because the operator (3.16) is totally 6 symmetric and the initial and nal states have di erent permutational symmetry in the spin- avor of the Nc quarks1. One can see that, in spite of their quite di erent detailed structure, both cases considered lead to the same answer (3.22) and (3.31), which also coincides with the result obtained by 5] for the case of X ia. The two most important properties of this solution are now apparent: The excited states can be classi ed in towers of states labeled by a spin vector such that jJ Ij . To enforce the cancellation of the leading Nc-dependence among the di erent intermediate states, expressed by the consistency condition, all the members of a -tower must be degenerate among themselves. Pions do not couple towers of excited states with di erent values of , as the corresponding nondiagonal matrix elements of Z ia vanish. The second property will be useful in the study of the consistency conditions for Y and Q (3.9,3.10), as it allows us to consider the couplings of each -tower at a time. We turn now to the rst of them, the coupling Y responsible for S-wave pion couplings between pand s-wave baryons. B. Consistency condition for 0 Y p hJ 0; I 0; m0; 0jY ajJ; I; m; i = ( )I+I 2I + 1Y (J 0I 0; JI) JJ mm hI 0; 0jI; 1; ; ai ; (3.32) (s-wave) (p-wave) p hJ 0; I 0; m0; 0jY ajJ; I; m; i = ( )2I 2I + 1Y (J 0I 0; JI) JJ mm hI 0; 0jI; 1; ; ai ; (3.33) 0 0 0 0 We parametrize the matrix elements of the Y a operator as (p-wave) (s-wave) : With this choice for the normalization coe cients we have Y (J 0I 0; JI) = Y (JI; J 0I 0). The same de nitions (3.32) and (3.33) will be used for transitions between other orbital excitations. We proceed next in complete analogy to the derivation of the consistency condition for Z ia (3.14). The relation (3.7) is sandwiched between states belonging to 0- and -towers respectively hJ 0; I 0; m0; 0 (s-wave)jrX ia Y by Y byZ iajJ; I; m; (p-wave)i = 0 : (3.34) the model-independent framework of the present work. 1In the quark model this matrix element receives a nonvanishing contribution starting at order (v=c)2 in the nonrelativistic expansion 38]. However, a study of these e ects would take us beyond 13 We denoted here r = g(X)=g(Z). Then a complete set of intermediate states is inserted between each two operators. The necessary matrix elements of X and Z are expressed with the help of the general result (3.22). The resulting equation is nally projected onto the particular channel with total spin-isospin (H; K). We obtain in this way the consistency condition )( ( ) X I 1 K I1 1 I 0 Y (JI ; JI) r (2I1 + 1) I 0 1 I (3.35) 1 H 0J 1 I1 ( ) I 1 K Y (HI 0; HK) : I K+ =( ) HJ 0 In addition to determining the structure of the reduced matrix element Y (J 0I 0; JI), this relation will x also the value of the ratio r. It is straightforward to check that the solution of the consistency condition (3.35) is given by ) (0 I1I ; 0; JI) = ( )I +J+ Y (JI (3.36) J0 0 0 provided that r = 1. After substituting this solution in (3.35) the sum over I1 can be done with the help of the identity (3.30). In particular, for decays into s-wave baryons containing only u; d quarks, we obtain the solution ( ) 2J J 1 I = ( )1 J+I 1 q 1 : (3.37) Y (J; JI) = ( ) J0 3(2J + 1) Let us trace again how the same result arises in the quark model. The quark model counterpart of the operator Y a is X 1X Y a ! h0j1; 1; j; ii j ri a = p ( )1 j j r j a ; (3.38) 3j i;j where the light quark operator j acts only on the spins, ri acts on the orbital degrees of freedom and a acts on the isospins. We consider rst the coupling of a p-wave state transforming under the symmetric representation of SU(4). The matrix element (3.32) is written in the quark model as X hJ; m; 0jY ajJ; I; m; i = hJ; m; 0jY ajI; L; mS ; mL; ihJ; mjI; L; mS ; mLi : (3.39) mS mL The Wigner-Eckart theorem can be used to parametrize the matrix element of the operator on the r.h.s. of (3.38) in terms of a new reduced matrix element T (J; I) hJ; m; 0j j ri ajI; L; mS ; mL; i = (3.40) 1 T (J; I)hJ; mjI; 1; m ; jih0jL; 1; m ; iihJ; 0jI; 1; ; ai : S L 2J + 1 Inserting this expression into (3.39) we obtain the result 14 = 3(2J1+ 1) T (J; I)hJ; 0jI; 1; ; ai L1 ; which can be compared with the de ning matrix element of Y a (3.32). Taking into account the fact that for the quark model states considered = L, we nd the following expression for Y (J; JI) in terms of the quark model reduced matrix element T (J; I) 1p T (J; I) : (3.42) Y (J; JI) = ( ) J I 3(2J + 1) 2I + 1 1 It will be shown in Section IV.D by an explicit calculation in the quark model that the reduced matrix T (J; I) is given in the large-Nc limit, up to a numerical factor, by q T (J; I) = ( )2I+1 3(2J + 1)(2I + 1) : (3.43) This leads to the same expression (3.37) for Y (J; JI) as the model-independent approach based on the consistency conditions. It is possible to generalize this argument by keeping the orbital angular momentum of the quark model state arbitrary L; L0. They are only constrained by the requirement of parity conservation ( )L = 0( )L . The relevant quark model matrix element can be parametrized in this case as 0 hJ; m; 0jY ajJ; I; m; i (3.41) 1 T (J; I) X hJ; mjI; 1; m ; m ihJ; mjI; L; m ; m ihJ; 0 jI; 1; ; ai = SL SL L1 3(2J + 1) mS mL hI 0; L0; m0S ; m0L; 0j j ri ajI; L; mS ; mL; i = (3.44) 1 00 00 00 00 0 + 1)p2L0 + 1 T (I L ; IL)hI ; mS jI; 1; mS ; jihL ; mLjL; 1; mL ; iihI ; jI; 1; ; ai ; (2I with T (I 0L0; IL) another reduced matrix element. We assumed again that the spin- avor wavefunction of the u; d quarks in the baryon is completely symmetric. With the help of this relation it is possible to compute the matrix element of Y a between eigenstates of the total spin p hJ; I 0; m; 0jY ajJ; I; m; i = (2I 0 + 1)1 2L0 + 1 T (I 0L0; IL) (3.45) X h0j1; 1; j; iihJ; mjI 0; L0; m0S ; m0LihJ; mjI; L; mS ; mLi hI 0; m0S jI; 1; mS ; jihL0; m0LjL; 1; mL;(iihI 0; 0jI;)1; ; ai IL T (I 0L0; IL)( )1 I J L L0 I 0 J hI 0; 0jI; 1; ; ai : = q 10 1 3(2I + 1) 0 This has the same structure as the model-independent solution (3.36), which is reproduced provided one takes q T (I 0L0; IL) = ( )1+2I 3(2I + 1)(2I 0 + 1) : (3.46) 0 15 The phase can be equivalently rewritten as 1 + 2I 0 = 1 + 2I which gives an expression identical to (3.43). A similar result is obtained also for the case of the excited baryons whose spin- avor wavefunction transforms according to the mixed representation of SU(4). The relevant matrix element of Y a can be expressed, with the help of the recoupling relation (3.23) in terms of the matrix element 0L hJ 0IX0; m0; 0jY aj(IP)S; L; J; m; i = (3.47) hI 0; L0; m0S ; m0L; 0jY aj(IP)S; L; mS ; mL; ihJ; mjS; L; mS ; mLihJ 0; m0jI 0; L0; m0S ; m0Li : mS mL mS mL 0 0 The matrix element on the r.h.s. can be expressed with the help of (3.38) in terms of a new reduced matrix element T (I 0L0; SIL) de ned by hI 0; L0; m0S ; m0L; 0j j ri aj(IP)S; L; mS ; mL; i (3.48) p = 0 1 0 T (I 0L0; SIL)hI 0; m0S jS; 1; mS ; jihL0 ; m0LjL; 1; mL; iihI 0; 0jI; 1; ; ai : (2I + 1) 2L + 1 Inserting this relation into (3.47) we nd for the matrix element of Y a in the j(IP)S; L; J; m; i basis (3.49) hJ 0I 0L0; m0; 0jY aj(IP)S; L; J;( i = JJ mm hI 0 0jI1; ; ai m; 0 1) L q 10 ( )1+L S J I 0 L J T (I 0L0; SIL) : S 3(2I + 1) Next we transform to the jI; (PL) ; J; m; i basis with the help of the recoupling relation (3.23). We adopt the following ansatz for the quark model matrix element T (I 0L0; SIL) ( ) q 0L0 ; SIL) = ( ) I+I (2I + 1)(2I 0 + 1)(2S + 1) I 1 S I(L0 ; L) (3.50) T (I 1 I0 1 with I(L0; L) an arbitrary function of its arguments2. This will be derived in Section IV.D by explicit calculation in the quark model in the large-Nc limit. Inserting (3.50) into (3.23) we can perform the sum over S with the help of (3.30). The nal result for the matrix element of Y a has the form, with P = 1, ) ( 10 0 )( )I +J+ 0I 0 ; JI) = (3.51) Y (J L JJ c(LL I0 J I with c(LL0 ) a numerical coe cient given by ) ( 1 p2 + 1( ) L 1 1 L0 : c(LL0 ) = p (3.52) 1L1 3 The result (3.51) can be seen to coincide with the general solution of the consistency condition for Y (3.36). 0 0 0 0 0 0 0 0 0 0 2For simplicity we will omit I(L0; L) throughout in the following. 16 C. Consistency condition for Q The operator Qij;a parametrizes pion coupling in a D-wave. It will prove convenient to de ne a modi ed operator Qka with only one index k = 2; 1; 0; 1; 2, by X Qka = h2; kj1; 1; i; jiQij;a : (3.53) ij It is easy to see that Qka satis es the same consistency condition (3.10) as Qij;a. We introduce reduced matrix elements associated with this operator, de ned by ( )J+I+J +I (2J + 1)(2I + 1)Q(J 0I 0; JI)hJ 0; m0jJ; 2; m; kihI 0; 0jI; 1; ; ai ; hJ 0; I 0; m0; 0 (p-wave)jQkajJ; I; m; (s-wave)i = (3.55) q ( )2J+2I (2J + 1)(2I + 1)Q(J 0I 0; JI)hJ 0; m0jJ; 2; m; kihI 0; 0jI; 1; ; ai : 0 0 hJ 0; I 0; m0; 0 (s-wave)jQkajJ; I; m; (p-wave)i = q (3.54) As usual, the choice for the normalization coe cients is made such that Q(J 0I 0; JI) = Q(JI; J 0I 0). The same de nitions (3.54) and (3.55) apply to transitions between other orbital excitations. We derive a consistency condition for Q(JI; J 0I 0) by taking the following matrix element of the relation (3.10) hJ 0; I 0; m0; 0 (s-wave)jrX iaQjby QjbyZ iajJ; I; m; (p-wave)i = 0 : hH 0 ; h0jJ 0; 2; m0; jihH; hjJ; 1; m; iihK 0 ; 0jI 0; 1; 0; bihK; jI; 1; ; ai : 0 (3.56) We insert a complete set of intermediate states between the two operators and project this relation onto the particular channel with total spin-isospin (H; K) by multiplication with (3.57) We obtain nally the set of constraints ) )( )( (0 X I 1 K Q(J I ; JI) (3.58) J 1H I 1 I1 r ( ) J +J1 (2J1 + 1)(2I1 + 1) J 0 J 0 J 0 2 J 11 I 0 1 I1 1 1 J1 I1 ( ) K 1 I Q(J 0I 0; HK) : = ( )2H I K+ +1 J H 0 We quote directly the solution of this consistency condition. We will attempt later to make it plausible using a quark model construction. The most general solution can be written as a sum of 9j symbols of the form 8 0 0 09 > > X < I J= cy > I J > (3.59) Q(J 0I 0; JI) = y=1;2;3 : y 1 2 ; which satis es (3.58) provided that r = 1. In particular, for nal s-wave states containing only u; d quarks one has 0 = 0, J 0 = I 0 and the 9j symbols reduce to 6j symbols 17 ) ) ( ( ( )J+J +1 2 J J 0 ( )J+J 2 J J0 (3.60) 1 + c2 q 2: 1q 3(2J 0 + 1) I 1 1 5(2J 0 + 1) I 1 2 It is not completely straightforward to check that (3.58) is indeed satis ed by (3.59). Therefore it might be useful to sketch the steps of this derivation. First, the 9j symbols on the l.h.s. are written as a sum over 3 6j symbols with the help of Eq.(6.4.3) in 33] 9 80 ) (0 )( )( > I1 J1 > X = < y I1 I 1 J1 J 2 : 2x (3.61) > y I J > = x ( ) (2x + 1) 1 2 x xJ x 0 I1 : 1 2; Q(J 0; JI) = c 0 0 This allows the sum over J1 to be performed with the help of (3.30) ) )( )( (0 X J +J1 J 1 J1 I1 0 J1 2 J J1 (3.62) () (2J1 + 1) I 0 I 0 1 J0 H 2Jx 1 J1 ) )( 0 (0 I xH I xH = ( ) 1 2 J0 0 J 1 I 1 0 J x H. Next, the sum over I1 can be done, also with 1 = 2J 0 + 1 + I1 + I 0 with the help of (3.30). As a result, the l.h.s. of (3.58) takes the form ) )( (0 )( )( 0 X 2(x+ ) KH y I0 K 1 JH2 2 r( ) () (2x + 1) 1 2 x J I 1 : (3.63) xH x 0 I0 x 0 K I + 1. We added a phase factor identically equal to 1 under with 2 = 2H the summation sign, which allows the x-sum to be performed with the help of a relation analogous to (3.61). The result is a 9j symbol identical to the one on the r.h.s. of (3.58). It is easy to check that also the total phase factor and the remaining 6j symbol are the same as the ones on the r.h.s. of (3.58), which proves the validity of (3.59). The solution (3.58) satis es the consistency condition for Q regardless of the value of y, which can take therefore all values compatible with the nonvanishing of the 9j symbol in which it appears. We will try now to make the result (3.59) plausible, by examining the structure of this coupling in the quark model. The operator Qka is given in the quark model by X Qka ! h2; kj1; 1; j; ii j ri a (3.64) 0 0 ij with the same j ; ri; a as in (3.38). Let us consider rst baryons containing only u; d quarks and whose avor-spin wavefunction transforms under the symmetric representation of SU(4). We will keep the orbital angular momenta of the initial and nal states L; L0 completely general, subject only to the requirement of parity conservation ( )L = 0( )L . The matrix element of the quark model operators on the r.h.s. of (3.64) between eigen~ ~ states of S and L has been already parametrized in (3.44) in terms of the reduced matrix ~~~ element T (I 0L0; IL). The matrix element of Qka between eigenstates of J = I + L can be easily obtained as X h2; kj1; 1; j; iihJ 0 ; m0jI 0; L0; m0S ; m0Li (3.65) hJ 0; I 0; m0; 0jQkajJ; I; m; i = 0 ijmS mL mS mL hJ; mjI; L; mS ; mLihI 0; L0; m0S ; m0L; 0j j ri 0 0 ajI; L; m ; m ; SL i: 18 Comparing with (3.54) we see that it is possible to extract Q(J 0I 0; JI) by multiplying the r.h.s. with hJ 0; m0jJ; 2; m; ki and summing over m; k. The resulting sum over 6 CG coe cients can be written in terms of a 9j symbol by using Eq.(6.4.4) in 33]. We obtain nally q ( )J+I+J +I Q(J 0I 0; JI) (2J + 1)(2I + 1) (3.66) 8 0 0 09 s >L I J > = < I J+I +J +L+L 5 2J + 1 T (I 0L0 ; IL) L I J : =( ) >1 1 2> 2I 0 + 1 ; : For the quark model states considered one has = L, so that the 9j symbol corresponds to y = 1 in (3.59). Requiring equality with the model-independent solution (3.59) of the consistency condition for Q gives for the quark model reduced matrix element T (I 0L0; IL) the expression 1q 0L0 ; IL) = ( )L+L p (2I 0 + 1)(2I + 1) : T (I (3.67) 5 This agrees, up to an unimportant overall coe cient, with the result (3.46) obtained from considering the matrix element of the s-wave operator Y a. The 9j-symbol with y = 2 in (3.59) arises when considering initial states transforming under the mixed symmetry representation of SU(4). The calculation for this case proceeds in close analogy to the one for the Y a operator. First we compute the matrix element of Qka in the j(IP)S; L; J; m; i basis with the help of the relation 0 0 0 0 0 0 0L hJ 0IX0; m0; 0jQkaj(IP)S; L; J; m; i = (3.68) 0; L0 ; m0 ; m0 ; 0 jQka j(IP)S; L; m ; m ; ihJ; mjS; L; m ; m ihJ 0 ; m0jI 0; L0 ; m0 ; m0 i hI SL SL SL SL mS mL mS mL 0 0 followed by the application of (3.64,3.48). We obtain L; hJ 0I 0L0; m0; 0jQkaj(IP)S;9 J; m; i = hJ 0m0jJ2; mkihI 0 0jI1; ai 8 s 0> <S I 2J + 1 > L 1 L0 = T (I 0L0; SIL) : 5 2I 0 + 1 > 1 > : J 2 J0 ; (3.69) We are eventually interested in the matrix elements of Qka in the basis jI; (PL) ; J; m; i. Using the recoupling relation (3.23) we get hJ 0I 0L0qm0; 0jQkajI; (PL) ; J; m; i = hJ 0m0jJ2; mkihI 0 0jI1; ai ; 5(2 + 1)(2I + 1)(2J + 1)( ) L J+2J I +1 ( )( ) 8 S 1 I0 9 < = X 2S I1 1 1 1 > L 1 L0 > ( ) (2S + 1) L J S I I0 S > J 2 J0 > : ; S 0 0 (3.70) where we used the ansatz (3.50) for T (I 0L0; SIL). To do the sum over S we rst combine the two 6j symbols with the help of (3.30) such that S appears only in one 6j symbol 19 ( I1S 1 I0 1 )( ) ( )( )( ) I 1 S = X( ) (2x + 1) I 0 J x 1x 1Lx LJ 1I 1 L 1 J I0 S x (3.71) with = I 0 + J + + L + I + S + 1 + x. Now the sum over S can be performed using (3.21) 9 8 )( 0 ) ( ( 0 ) > 1 I0 S > <0= X 1 1I S 2x 2 J 0 J (3.72) (2S + 1) J L x > 2 J J > = ( ) I 0 x L0 x L L : 12 : 1 L0 L ; S We obtain for the matrix element of Qka in the jI; (PL) ; J; m; i basis the following expression containing a sum over 4 6j symbols hJ 0I 0L0qm0; 0jQkajI; (PL) ; J; m; i = hJ 0m0jJ2; mkihI 0 0jI1; ai ; 0 0 0 5(2 + 1)(2I + 1)(2J + 1)( ) I J I + +J+L +L )( 0 ) (0 )( )( Xx 2Lx 1x 2 L0 x I Jx ( ) (2x + 1) 1I 1 L 1 I0 J J0 L 1 1 : x (3.73) This can be put in a form resembling (3.59) by rst combining the second and fourth 6j symbols with (3.30) ( )( )( ) ( )( ) 11y L0 y 1 Lx 1 L x = X ( ) (2y + 1) 1 2 y (3.74) 111 L0 L 21x 1 1 L0 2 1 y=1;2 0 with 0 = + L L0 x + y. The sum over x can be now done in terms of a 9j symbol similar to those in (3.59) ) 8 L0 I 0 J 0 9 (0 )( )( 0 = < X 2x I Jx 1 x L 2 x => I J >: (3.75) ( ) (2x + 1) >y 1 2> 1I 2 L0 y J I0 J0 ; : x When inserted into (3.73) this gives a result for the reduced matrix element Q(J 0I 0; JI) of the same form as (3.59) 8 0 0 09 >L I J > = < X 0) 0I 0; JI) = (3.76) cy (LL > I J > Q(J :y 1 2; y=1;2 with coe cients cy given by ( )( ) q 11y : 0 ) = 5(2 + 1)( ) 2J +y (2y + 1) 1 2 y cy (LL 111 L0 L 0 (3.77) Finally, the most general solution for Q(J 0I 0; JI) containing also 9j symbols with y = 3 is obtained if one considers transitions among two states with mixed symmetry. This situation is not very relevant from a phenomenological point of view so that its discussion is relegated to Appendix B. 20 IV. QUARK MODEL MATRIX ELEMENTS A. Symmetric states In this Section we compute the reduced matrix elements of the operator i a on quark model states with arbitrary number of colors. It will be seen that in the limit Nc ! 1 these reduced matrix elements coincide with those required by the consistency conditions discussed in Section III. We start by computing the reduced matrix element Z(I 0; I) de ned by hI 0; L0; m0S ; m0L; 0j 2I 0 + 1 1 a jI; L; m ; m ; i = SL Z(I 0; I)hI 0; m0S jI; 1; mS ; ii LL mLmL hI 0; 0jI; 1; i 0 0 (4.1) ; ai : The states on the l.h.s. transform under the completely symmetric representation of SU(4). For simplicity we will take them to contain only u- and d-type quarks, although additional quark avors can be included in a straightforward way. In the quark model with Nc colors they are given by X jI; m; i = hI; mj Nu ; Nd ; i; m iiSj Nu ; iiuj Nd ; m iid (4.2) 22 2 2 i X N N N N = hI; mj Nu ; Nd ; i; m iiS(u ") 2u +i (u #) 2u i(d ") 2d +m i (d #) 2d m+i ; 22 i with the numbers of u and d quarks, respectively, in the baryon state. The symbol S means complete symmetrization under permutation of all quarks. The explicit form of the wavefunction (4.2) has been given without proof in 35] and a particular case was previously considered in 36]. For a simple method of computing matrix elements in the quark model with Nc colors see 37]. In the following, for completeness of the presentation we give a detailed derivation of (4.2). Proof. Any completely symmetric state of Nc quarks can be constructed as a linear combination of symmetrized products of one-quark states (4.4) S(n ; n ; n ; n ) = p1 ((u ")n1 (u #)n2 (d ")n3 (d #)n4 + permutations) 1234 Nu = Nc + ; 2 Nd = Nc 2 (4.3) N with 3 N = (n1 + n2 + n!n+! n4)! : n !n !n 1234 (4.5) It is easy to see that the action of spin and isospin operators on these states is given by 21 X i q + n4(n3 + 1)S(n1; n2; n3 + 1; n4 1) q Xi S(n1; n2; n3; n4) = n1(n2 + 1)S(n1 1; n2 + 1; n3; n4) i q + n3(n4 + 1)S(n1; n2; n3 1; n4 + 1) Xi z S(n1; n2 ; n3 ; n4) = (n1 n2 + n3 n4 )S(n1 ; n2; n3 ; n4 ) i q Xi + S(n1; n2 ; n3 ; n4) = n3 (n1 + 1)S(n1 + 1; n2 ; n3 1; n4 ) i q + n4(n2 + 1)S(n1; n2 + 1; n3; n4 1) q Xi S(n1; n2; n3; n4) = n1(n3 + 1)S(n1 1; n2; n3 + 1; n4) i q + n2(n4 + 1)S(n1; n2 1; n3; n4 + 1) Xi z S(n1; n2 ; n3 ; n4) = (n1 + n2 n3 n4 )S(n1 ; n2; n3 ; n4 ) : i + S(n1; n2 ; n3 ; n4) = n2 (n1 + 1)S(n1 + 1; n2 1; n3 ; n4 ) i q (4.6) (4.7) (4.8) (4.9) (4.10) (4.11) It will prove more convenient to express the arguments of the symmetrized products of one-quark states in terms of four angular momentum-like variables de ned as n1 = j1 + m1 (4.12) n2 = j1 m1 (4.13) n3 = j2 + m2 (4.14) n4 = j2 m2 : (4.15) In terms of these variables, the action of the spin and isospin operators can be expressed as q Xi (4.16) + S(j1 ; j2; m1 ; m2) = (j1 m1)(j1 + m1 + 1)S(j1; j2 ; m1 + 1; m2) i q + (j2 m2)(j2 + m2 + 1)S(j1; j2; m1; m2 + 1) q Xi S(j1; j2; m1; m2) = (j1 + m1)(j1 m1 + 1)S(j1; j2; m1 1; m2) (4.17) i q + (j2 + m2)(j2 m2 + 1)S(j1; j2; m1; m2 1) Xi (4.18) z S(j1 ; j2; m1 ; m2) = 2(m1 + m2 )S(j1; j2 ; m1; m2) i q Xi 1 1 1 1 + S(j1 ; j2; m1 ; m2) = (j2 + m2)(j1 + m1 + 1)S(j1 + 2 ; j2 2 ; m1 + 2 ; m2 2 ) (4.19) i q 1 1 + (j2 m2)(j1 m1 + 1)S(j1 + 2 ; j2 2 ; m1 1 ; m2 + 1 ) 2 2 q Xi 1 ; j + 1 ; m 1 ; m + 1 ) (4.20) S(j1; j2; m1; m2) = (j1 + m1)(j2 + m2 + 1)S(j1 2 2 2 1 2 2 2 i q 1 1 + (j1 m1)(j2 m2 + 1)S(j1 1 ; j2 + 2 ; m1 + 1 ; m2 2 ) 2 2 22 X i i z S(j1 ; j2; m1 ; m2) = 2(j1 j2)S(j1; j2; m1; m2) : (4.21) A state of well-de ned spin is constructed by taking appropriate linear combinations of symmetrized products of one-particle states X jI; m; i = c(m1; m2; m; j1; j2)S(j1; j2; m1; m2) (4.22) m1 ;m2 with m1 + m2 = m. The quantum numbers of the state x j1 and j2 through the conditions = j1 j2 (4.23) Nc = 2(j1 + j2) (4.24) which give (4.3) with j1 = Nu=2 and j2 = Nd=2. The coe cients c in (4.22) can be determined by requiring the states jI; m; i to satisfy the relations q J jI; m; i = (I m)(I m + 1)jI; m 1; i : (4.25) Inserting the expansion (4.22) one nds, with the help of (4.16-4.21), the following recursion relations among the coe cients c q (I m)(I + m + 1)c(m1; m2; m + 1; j1; j2) (4.26) q = (j1 + m1)(j1 m1 + 1)c(m1 1; m2; m; j1; j2) q + (j2 + m2)(j2 m2 + 1)c(m1; m2 1; m; j1; j2) q (I + m)(I m + 1)c(m1; m2; m 1; j1; j2) (4.27) q = (j1 m1)(j1 + m1 + 1)c(m1 + 1; m2; m; j1; j2) q + (j2 m2)(j2 + m2 + 1)c(m1; m2 + 1; m; j1; j2) : These relations can be seen to coincide with the familiar recursion relations for the ClebschGordan coe cients, with the identi cation c(m1; m2; m; j1; j2) = hI; mjj1; j2; m1; m2i : (4.28) It is known that these recursion relations x uniquely the CG coe cients up to an overall phase. To complete our proof of (4.2) we still have to show that this state is also an eigenstate ~ ~ ~ of I 2, with the same eigenvalue as J 2. This can be done by comparing the action of J 2 on 2 . We obtain ~ the state (4.2) with that of I ~ (4.29) J 2jI; m; i = 1 J+J + 1 J J+ + Jz2 jI; m; i 2 2 X = c(m1; m2; m; j1; j2) f(j1(j1 + 1) + j2(j2 + 1) + 2m1m2) S(j1; j2; m1; m2) m1 ;m2 q + (j1 + m1)(j1 m1 + 1)(j2 m2)(j2 + m2 + 1)S(j1; j2; m1 1; m2 + 1) q + (j1 m1)(j1 + m1 + 1)(j2 + m2)(j2 m2 + 1)S(j1; j2; m1 + 1; m2 1) 23 ~ which also coincides with the result of applying I 2 on the same state. The knowledge of the states (4.2) can be used to calculate the matrix element (4.1). We will choose for this calculation the spherical component (i; a) = (0; 0) of the current in (4.1). The corresponding quark model operator can be written as a sum over Nc one-quark operators Nc 0 0 =X i i: (4.30) 33 to calculate: Z(I; I) and Z(I; I 1). We obtain for them the results i=1 Because of the symmetry property of Z(I 0; I), there are only two independent quantities Z(I; I) = (2I + 1)(Nc + 2) q q Z(I; I 1) = (2I 1)(2I + 1) (Nc + 2 + 2I)(Nc + 2 2I) : (4.31) (4.32) In order to obtain Z(I; I) we consider the following matrix element of the type (4.1) Nc Xi I (4.33) hI; I; Ij 3 3ijI; I; Ii = (I + 1)(2I + 1) Z(I; I) : i=1 The quark model matrix element on the l.h.s. can be computed with the help of the wavefunction (4.2) with the result Nc X Xii hI; I; Ij 3 3jI; I; Ii = jhI; Ij Nu ; Nd ; m; I mij2(4m 2I) (4.34) 22 m i=1 2) = Nu(Nu +2(I +Nd(Nd + 2) : 1) Nu ; Nd are given by (4.3). Inserting this expression into (4.33) one obtains the result (4.31) for Z(I; I). For Z(I; I 1) we consider the matrix element Nc Xi hI; I 1; I 1j 3 3ijI 1; I 1; I 1i = I(2I1+ 1) Z(I; I 1) : (4.35) i=1 The quark model matrix element can be computed with the result Nc Xi (4.36) hI; I 1; I 1j 3 3ijI 1; I 1; I 1i i=1 X = 4 mhI; I 1j Nu ; Nd ; m; I 1 mihI 1; I 1j Nu ; Nd ; m; I 1 mi 22 22 m s 1 = 2 2I + 1 Nc + 1 + I Nc + 1 I : I 2I 2 2 Comparing with (4.35) gives immediately the result (4.32). The results (4.31), (4.32) can be put into a common form 24 q q Z(I 0; I) = (2I 0 + 1)(2I + 1) (Nc + 2)2 (I 0 I)2(I 0 + I + 1)2 (4.37) q = (Nc + 2) (2I 0 + 1)(2I + 1) + O(1=Nc ) : We have made here apparent the fact that the corrections to the lowest-order result come only at sub-subleading order in 1=Nc . This is an illustration, on the example of the quark model, of a model-independent result obtained by Dashen and Manohar 2] using the counting rules for pion-baryon scattering. B. Mixed symmetry states In this Section we construct quark model states whose spin- avor wavefunctions transform under the mixed symmetry representation of SU(4) shown in Fig.2. They can be built using the procedure described in Sect.II, by adding one extra quark to a symmetric state of Nc 1 quarks. We write the state obtained by adding the j th quark to a symmetric state of Nc 1 quarks with spin and isospin i, as X 1 jSI; m; ij = hS; mji; 1 ; m1; m2ihI; 1 ji; ; 1; 2iji; m1; 1i j 2 ; m2; 2ij : (4.38) 2 2 m1 m2 1 2 The states of mixed symmetry under SU(4) must be antisymmetric under permutations of the two quarks corresponding to the rst column of the Young diagram. There are Nc(Nc 1)=2 ways to choose such a pair of quarks, but not all the states obtained in this way will be linearly independent. In fact there are only Nc 1 independent states with mixed symmetry, and we will choose them such that they are antisymmetric under a permutation of the rst quark with any of the remaining Nc 1 quarks in the baryon. The corresponding spin- avor wavefunction will be denoted as 1 jSI; m; i j;1] = p2 (jSI; m; ij jSI; m; i1) ; j = 2; 3; : : : ; Nc : (4.39) The space part of the wavefunction must transform also under the mixed symmetry representation of the permutation group, corresponding to the same Young diagram as in Fig.2. There are again Nc 1 linearly independent wavefunctions, which can be chosen to be antisymmetric under a permutation of the j th and 1st quarks. Their generic form is 1 r r jL; mLi j;1] = p2 ( (rj )YLmL (^j ) (r1) (r1)YLmL (^1) (rj )) S (r2; ; rj 1; rj+1; ; rNc ) (4.40) with S (r2; ; rj 1; rj+1 ; ; rNc ) a symmetric function of its arguments. In (4.40), we have assumed that the orbital angular momentum is carried by a single quark. This is strictly true only for the lowest orbital excitations. It is easy to combine now the spatial and the spin- avor parts into a completely symmetric wavefunction of well-de ned spin and isospin. Our nal result for such a quark model state is X jJI; m; i = hJ; mjS; L; mS ; mLijSIL; mS ; mL; i (4.41) mS ;mL 25 with jSIL; mS ; mL; i = ( ) (SIi) p Nc 1 X jSI; m ; i S j;1] jL; mL i j;1] : Nc 1 j=2 (4.42) The phase of these states (SIi) will be chosen later for convenience. These states have a peculiar normalization, due to the fact that the spatial wavefunctions (4.40) with j 6= j 0 are not orthogonal. They satisfy instead 1 0 (4.43) j ;1] hL; mL jL; mLi j;1] = ( jj + 1) mL mL I 2 with I an overlap integral. Using this expression we obtain the following exact result for the norm of the states (4.42) hS 0I 01; m0S ; m0L; 0jSI1; mS ; mL; i = SS mS mS mLmL II Nc 4+ 2 I (4.44) 8 ( ( )2 )2 < 2(2i + 1) S I 0 SI1 1 :3(2i + 1) 1 1 i Nc 1 2 1 i 22 2 ) 5 + 2i(i + 1) S(S + 1) I(I + 1) : 1 + 2(N 1) 2 c The derivation of this relation will be presented in some detail, as it illustrates a few techniques useful in dealing with the mixed symmetry states. We start by computing the scalar product of two direct product states X 0 001 0 0 0 001 0 0 1 00 0 0 j hS I ; m ; jSI; m; ij = hS ; m ji ; ; m1; m2ihI ; ji ; ; 1 ; 2 ihS; mji; ; m1; m2i (4.45) 2 2 2 1 ; ; i h 1 ; m0 ; 0 j hi0; m0 ; 0 ji; m ; i j 1 ; m ; i : hI; ji; 2 1 2 j 2 2 2 11 11 2 2 2j 0 0 0 0 0 0 0 0 0 0 1 (4.47) Pjj = 4 (1 + ~ j ~ j ) (1 + ~j ~j ) is an operator which exchanges the spins and isospins of the j; j 0 quarks. We obtain in this way 100 1 000 (4.48) j h ; m2; 2 j hi ; m1 ; 1ji; m1; 1 i j ; m2; 2 ij 2 2 1 = 4 j h 1 ; m02; 02j 1 ; m2; 2ij hi0; m01; 01ji; m1; 1i 2 2 0 0 0 0 where The matrix element on the r.h.s. can be written as 1 100 000 j h ; m2; 2 j hi ; m1 ; 1ji; m1; 1i j ; m2; 2 ij 2 2 1 ; m0 ; 0 j hi0; m0 ; 0 jP ji; m ; i j 1 ; m ; i = jh2 2 2 11 1 1 jj 2 2 2j 0 0 (4.46) 26 1 1X + 4 ( )k j h 1 ; m02; 02j j k j 2 ; m2; 2ij hi0; m01; 01j jk ji; m1; 1i 2 k 1X 0 + 4 ( )b j h 1 ; m02; 02j j bj 1 ; m2; 2ij hi0; m1; 01j jb ji; m1; 1i 2 2 b 1 X( )k+b h 1 ; m0 ; 0 j k bj 1 ; m ; i hi0; m0 ; 0 j k b ji; m ; i : + 11 j 1 1jj 4 k;b 2 2 2 j j 2 2 2j 0 0 0 0 The matrix elements on the one-quark states are computed easily with the results 1 0 01 j h ; m2; 2j ; m2; 2ij = m2 m2 2 2 2 2 1 ; m0 ; 0 j k j 1 ; m ; i = p3h 1 ; m0 j 1 ; 1; m ; ki 2 jh 22 2 2 2 j 2 2 2j 2 22 p 1 01 1 0 0 b1 j h ; m2; 2j j j ; m2 ; 2ij = 3h ; 2 j ; 1; 2 ; bi m2 m2 2 2 22 1 ; m0 ; 0 j k bj 1 ; m ; i = 3h 1 ; m0 j 1 ; 1; m ; kih 1 ; 0 j 1 ; 1; ; bi : jh 2 2 2 2 j j 2 2 2j 2 22 2 22 2 0 0 0 0 (4.49) (4.50) (4.51) (4.52) The matrix elements of the one-quark operators taken on symmetric states containing Nc 1 quarks can be obtained with the help of the wavefunction (4.2) of these states. For example, the matrix element of jk is parametrized as hi0; m0; 0j jk ji; m; i = F(i) ii hi; m0ji; 1; m; ki : The state ji; m; i has the explicit form X ji; m; i = hi; mjj1; j2; k; m kiSNc 1(j1 + k; j1 k; j2 + m k; j2 m + k) 0 0 (4.53) (4.54) k with j1 = Nu =2, j2 = Nd=2 and Nu;d = (Nc 1)=2 using the relation . Next, we single out the quark j by SNc (n1; n2; n3; n4) = (4.55) s s n1 (u ") S (n 1; n ; n ; n ) + n2 (u #) S (n ; n 1; n ; n ) j Nc 1 1 234 j Nc 1 1 2 34 N N sc sc n n + N3 (d ")j SNc 1(n1; n2; n3 1; n4) + N4 (d #)j SNc 1(n1; n2; n3; n4 1) : c c The reduced matrix element F(i) can be computed by taking the spherical component k = 0 in (4.53). The matrix element on the l.h.s. of this relation can be written with the help of (4.55) as X hi; m; j j0ji; m; i = jhi; mjj1; j2; k; m kij2 (4.56) k ) ( j1 + k j1 k + j2 + m k j2 m + k = 2m : N1N1 N1 N1 N1 c c c c c Comparing with (4.53) we obtain 27 q F(i) = N 2 1 i(i + 1) : c In a completely analogous way we write the other needed matrix elements as (4.57) (4.58) (4.59) hi0; m0; 0j hi0; m0; 0j jaji; m; i = G(i) ii s ka j j ji; m; 0 2i i = 2i0 + 1 H(i0; i)hi0; m0ji; 1; m; kihi0; 0ji; 1; ; ai : +1 mm hi; 0 0 ji; 1; ; ai The corresponding reduced matrix elements can be computed with the results q G(i) = N 2 1 i(i + 1) (4.60) c N H(i; i) = Nc + 1 (4.61) c1 s 2 (4.62) H(i; i 1) = N 1 Nc 2 1 + i + 1 Nc 2 1 i + 1 c 2 H(i0; i) = 1 + N + O(1=Nc2 ) : (4.63) c We note from these results that only the unit operator 1 and jk ja give leading contributions to (4.45) in the large-Nc limit. Inserting the individual expressions for the matrix elements into (4.45) we obtain 0I 0 00 j hS " i; m jSIi; m ij 0 (4.64) = SS mm( II ) (1 q 1 i S) q 1 1 1 + 6(2i + 1)F (i)( )i+ 2 +S 2 i+ 1 +I 2 i I 1 1 + 6(2i + 1)G(i)( ) 2 1 i i21 4 ( 2 1 )( )# i1 2 +6( )2i+1+S+I (2i + 1)H(i; i) S i 2 I 1 i : 1i 1 1 0 0 0 0 2 2 The product of two 6j-symbols can be transformed with the help of the identity (3.30) into the form 2( )( )( )3 ( 1) S I 1 25 : S i 1 I i 2 = 1 ( )S+I+1+2i 4 S I 0 2 2 (4.65) 11 11i 11i 11i 2 2 2 22i 22 Furthermore, the second 6j-symbol on the r.h.s. can be eliminated by using the relation ( ) X I S x 2= 1 : (2x + 1) 1 1 i (4.66) 2i + 1 22 x=0;1 We obtain nally for the scalar product of tensor product states (4.64) the simple result 00 0 0 j hS I ; m ; jSI; m; 8 0 < (2i + 1) Nc + 1 S I 0 1 : Nc 1 2 1 i 2 ( ij = SS mm II )2 0 (4.67) 9 1 5 + 2i(i + 1) S(S + 1) I(I + 1) = : ; 2(Nc 1) 2 0 0 0 28 This result only holds if the two external quarks are di erent j 6= j 0. If they are identical, only the rst term in (4.48) contributes (without the factor 1/4). This gives : (4.68) We can use (4.68) and (4.67) to compute the norm of the states jSIL; mS ; mL; i. With the help of the de nition (4.42), it can be written as hS 0I 01; m0S ; m0L; 0jSI1; mS ; mL; i = (4.69) Nc X 00 0 0 0 =N1 1 j ;1]hS I ; mS ; jSI; mS ; i j;1] j ;1]h1; mL j1; mL i j;1] c j;j =2 = Nc + 2 mLmL I (j hS 0I 0; m0; 0jSI; m; ij j hS 0I 0; m0; 0jSI; m; ij ) ; 4 where we used (4.39) and (4.43). To bring this into the nal form (4.44) we only need to insert the expressions (4.68) and (4.67) for the scalar products on the r.h.s. and simplify the resulting expression with the help of (4.66). SS mm II 0 0 0 0 0 0 0 0 0 00 0 0 j hS I ; m ; jSI; m; ij = C. Matrix elements of Z ka on mixed symmetry states In this Section we will compute the matrix element (3.25) of Z ka taken between quark model states with mixed symmetry. It will be shown that the ansatz for Z(S 0I 0; SI) introduced in Sect.III.A can in fact be obtained by an explicit calculation in the quark model. We parametrize the matrix element of Z ka between the quark model states (4.42) as Nc Xk (4.70) hS 0I 01; m0S ; m0L; 0j n najSI1; mS ; mL; i = n=1 q010 Z(S 0I 0; SI) mLmL hS 0; m0S jS; 1; mS ; kihI 0; 0jI; 1; ; ai : (2S + 1)(2I + 1) 0 We obtain for the reduced matrix element the following result q q Z(S 0I 0; SI) = 3 Nc(Nc + 2) (2i + 1)(2i0 + 1) (2S + 1)(2S 0 + 1)(2I + 1)(2I 0 + 1) (4.71) 4 ) ( 0 )( 1 )( i S 2 i0 S 0 1 I : i i+S +I + (SIi)+ (S I i ) 1 S S 2 () 1 1 I0 1 I I0 1 1 I 2 2 This has to be divided with the square roots of the norms of the initial and nal states (4.44). To leading order in Nc the result takes exactly the form (3.29) provided the phase (SIi) of the quark model states (4.42) is chosen as (4.72) (SIi) = i + I + 1 : 2 The derivation of (4.71) proceeds in close analogy to the computation of the norm of the mixed symmetry states. First, we express the matrix element (4.70) of Z ka in terms of matrix elements on direct product states as 0 0 0 0 00 29 hS 0I 01; m0S ; m0L; 0j 0 00 Nc X n=1 ka n n jSI1; mS ; mL; 0 i ja n n jSI; mS ; (4.73) where we denoted the diagonal and nondiagonal matrix elements of Z ia on direct product states by Nc X ka (4.74) Z1 = j hS 0I 0; m0; 0j n n jSI; m; ij Nc Nc X 00 0 0X = ( ) (SIi)+ (S I i ) N 1 1 j ;1] hS I ; mS ; j c n=1 j;j =2 = ( ) (SIi)+ (S I i ) Nc 4+ 2 mLmL I (Z1 Z2) ; 0 0 00 0 i j;1] 0 j ;1]h1; mL j1; mL i j;1] 0 Z2 = j hS 0I 0; m0; 0j 0 n=1 Nc X The nondiagonal matrix element on direct product states (j 0 6= j) can be transformed into a diagonal one with the help of the exchange operator (4.47) Nc 100 0 0 0 X k aji; m ; i j 1 ; m ; i (4.76) j h ; m2 ; 2j hi ; m1; 1j 11 nn 2 2 2 2j n=1 Nc X ka 1 =j h 1 ; m02; 02j hi0; m01; 01jPjj n n ji; m1; 1 i j 2 ; m2; 2 ij 2 n=1 This expression can be computed by expanding the Pjj operator and inserting a complete set of intermediate states. Nc 1 ; m0 ; 0 j hi0; m0 ; 0 j X k aji; m ; i j 1 ; m ; i = (4.77) jh 11 nn 11 2 22 2 2 2j n=1 Nc 1 h 1 ; m0 ; 0 j 1 ; m ; i hi0; m0 ; 0 j X k aji; m ; i 11 nn 11 4j 2 2 2 2 2 2 j n=1 1X 1 + 4 ( )l j h 1 ; m02; 02j j lj 2 ; m2; 2ij 2 0 0 0 0 n=1 ka n n jSI; m; ij : (4.75) l X m1 00 hi0; m01; 01j jl ji0; m00; 01ihi0; m00; 01j 1 1 0 Nc X 1 +4 +1 4 X b 1 ( )b j h 2 ; m02; 02j j bj 1 ; m2; 2ij 2 Nc X 0 0 0 b 0 0 00 0 0 00 X hi ; m1; 1j j ji ; m1; 1 ihi ; m1; 1 j 0 n=1 ka n n ji; m1; 1 i X l+b 1 0 0 ( ) j h 2 ; m2; 2j l;b 1 00 Nc 0 ; m0 ; 0 j l b ji00; m00; 00ihi00 ; m00; 00 j X k a ji; m1; 1 i : hi 1 1 j j nn 11 11 n=1 i ;m1 ; 1 0 0 00 00 00 X l b1 j j j 2 ; m2; 2 ij n=1 ka n n ji; m1; 1i 30 Only completely symmetric states of Nc 1 quarks contribute to the sum over intermediate states since both the operator and the initial state in the last matrix elements of each term are symmetric under permutations of any quarks. One notes that keeping only the rst and the last term in this relation is su cient to obtain the large-Nc limit of this matrix element3. Furthermore, in the sum over quarks in Z ka one can omit the term acting on the j 0th quark, as this will only change the result by an amount nonleading in Nc. This allows us to compute these matrix elements by using the results of Sec.IV.A. Putting all pieces together one obtains for the matrix element of Z ka between quark-model states with well-de ned spin and isospin (S; I) the following result j 0 hS 0I 0; m0S ; 0jZ kajSI; mSq ij = ; q 0 Nc (2i + 1)(2i0 + 1) (2S + 1)(2I + 1)hS 0; m0S jS; 1; mS ; kihI 0; 0jI; 1; ; ai " ( )( ) 1 ( ) 2i +I+S 1 1 S S 0 1 I I 0 10 10 4 2 i i( 2 i i ( ) )( )( )# 3 ( )I+I +S+S X(2i00 + 1) i00 1 S 0 S 1 S 0 i00 1 I 0 I 1 I0 : +2 1 20 1 20 i00 1 i i00 1 i 2i 1 2 2i1 2 i 0 0 00 (4.78) Each of the last two lines corresponds to the contributions of the rst and fourth terms in (4.77), respectively. They can be transformed into the following form by a repeated application of (3.30) )( ) ( 1 S S 0 1 I I 0 = ( )i+i +I+S+I +S (4.79) 10 1 i0 i 2 "( 2 i )i( ) (0 )( )( )# 1 )( S 0 S 1 S i 2 S 0 1 i0 S S 1 S i 1 S 0 1 i0 20 2 20 3 I I0 1 10 11I 1I 1 I I0 0 2( I 2) ( 0 2I1 )( )2 ( 00 1 0 ) X 00 S 1 S 0 i00 1 I 0 I 1 I 0 = ( )1+i i+S+I (4.80) (2i + 1) i1 i20 S 1 20 00 1 1 i00 1 i 2 2)( 2 i )1 ( i 2 i ) ( i ) ( 1 )# "( )( 1 1 1 S0 S 1 S i 2 S 0 2 i0 + S 0 S 1 S i 1 S 0 2 i0 : 1 I0 0 1 I0 1 10I 12 I I0 1 2 I I0 0 2 2 21I 2 0 0 0 0 00 Inserting these expressions into (4.78) we obtain the following result for the nondiagonal matrix element of Z ka between direct product states q q 0 + 1) (2S + 1)(2I + 1)hS 0 ; m0S jS; 1; mS ; kihI 0 ; 0 jI; 1; ; ai (4.81) Z2 = Nc (2i + 1)(2i (0 )( 1 ) ( S 0 1 i0 ) S1 i 2 ( )1+i i+S +I S I 0 0 S 0 I 12 1 I0 0 : I 2 2 0 0 0 For the diagonal case, only the rst term in (4.77) survives (without the factor of 1/4). Using (4.79) we can write for this case 3The exact result for arbitrary Nc is presented in the Appendix A. 31 Z1 = ( )1 i +i+S +I (4.82) q q 0 N (2i + 1)(2i0 + 1) (2S + 1)(2I + 1)hS 0; mS jS; 1; mS ; kihI 0; 0jI; 1; ; ai )( 0 1 0 ) ( 0 )( )( )# "( 0 c ) ( 1 S S 1 S i 1 S 0 1 i0 : S S1 Si2 S 2i 2 2 3 I I0 1 10I 11I 10 1 I0 1 I I0 0 2 2I0 2 2 0 0 0 Inserting (4.81) and (4.82) into (4.73) and using the de nition of Z(S 0I 0; SI) (4.70) gives the nal result for the matrix element of Z (4.71). We will present in the following an alternative method of calculating the matrix element of a current between states with mixed symmetry. Besides reproducing the result (4.71), this method has the advantage of simplifying very much the computation of transition matrix elements between excited and ground state baryons, to be discussed in the next Section. We start by writing the matrix element of the current Z ka taken between two states (4.42) as (4.83) hS 0I 0jZ kajSIi = ( ) (SIi)+ (S I i ) 2(N 1 1) c 9 8N Nc c = <X 0 I 0jZ ka jSIi + X hS 0 I 0jZ ka jSIi j;1] j;1]; I : j ;1]hS j;1] : 0 0 00 jj =2 0 0 j=2 We consider the two terms of this relation in turn. The rst sum can be written as Nc Nc Nc X 0 0 ka 0 0 X A j 0; 1]Z ka X A j; 1]jSIi ; (4.84) 1 j ;1]hS I jZ jSIi j;1] = 1hS I j jj =2 0 0 j =2 0 j=2 1 A j; 1] = p (1 Pj1 ) (4.85) 2 is the antisymmetrization operator for quarks j; 1] and Pj1 has been de ned in (4.47). The spin states de ned in (4.39) can be written in terms of it as jSIi i;1] = A i; 1]jSIi1. An important relation we will use extensively in the following expresses the result of symmetrizing a direct product state jSIi1 under a permutation of any two quarks where tion constant B(Ii) can be computed by taking the norm of the both sides of this relation. We obtain Nc 2(Ii) = X n hSI; m jSI; m in (4.87) SI B = Nc + Nc(Nc 1)n hSI; m jSI; m in : 0 jSI; m i1 = (1 + P12 + + P1Nc ) jSI; m i1 = SI B(Ii)jI; m; i (4.86) with jI; m; i the completely symmetric state constructed in Section IV.A. The normaliza- nn =1 0 0 (n = n0) 6 The nondiagonal matrix element appearing on the r.h.s. has been calculated previously and is given by (4.67). We obtain nally 32 ( 1 1 )2 220 (4.88) ISi Nc 1 + 2i(i + 1) 2I(I + 1) : 22 It will be shown below that the phase of B(Ii) can be chosen such that the leading term in Nc is positive. The sums over the antisymmetrization operators in (4.84) can be written in terms of the complete symmetrization operator as Nc X 1 A j; 1] = p (Nc ) : (4.89) 2 j=2 B 2(Ii) = Nc (Nc + 1)(2i + 1) We will need also the following matrix element 2 Nc 1hS 0I 0j Z kajSIi1 = 1hS 0I 0j Z ka 1 + P12 + = 1hS 0I 0j Z ka jSIi1 : 2 + P1Nc ]jSIi1 (4.90) 2 We used in the rst line the property of the permutation operator Pij = 1. The second ka P 2 = P1j Z ka P1j . When acting to the left on 1hS 0 I 0j, equality is obtained by writing Z 1j this gives 1 hS 0I 0 j P1j = 1hS 0I 0j (4.91) since 1hS 0I 0j is completely symmetric under any permutation of the Nc quarks. This completes the proof of (4.90). The relations (4.89) and (4.90) allow us to express the sum of matrix elements (4.84) as 1 N 2 hS 0I 0jZ kajSIi 1 hS 0I 0j Z ka jSIi (4.92) 1 21 1 2 c1 jj =2 1 = 2 Nc2 1hS 0I 0jZ kajSIi1 1 SI S I B(Ii)B(I 0i0)hS 0 = I 0jZ kajS = Ii : 2 The second term in (4.83) can be computed in an analogous way. We note for this the following useful properties of the antisymmetrization operator A j; 1]. 0 Nc X 0 0 ka j ;1] hS I jZ jSIi j;1] = 0 0 0 1. A j; 1] commutes with Z ka A j; 1] ; Z ka] = 0 : (4.93) This follows from the fact that Z ka is completely symmetric under a permutation of two quarks and commutes therefore with the P operator (4.47). 2. The square of A j; 1] is given by A j; 1]2 = 2A j; 1] : 33 p (4.94) With the help of these relations and (4.89) we can write Nc Nc X X hS 0I 0jZ kajSIi j;1] = 1hS 0I 0jA j; 1]Z kaA j; 1]jSIi1 j;1] j=2 j=2 (4.95) = = Nc 0 I 0jZ ka X A j; 1]2jSIi 1 hS 1 j=2 ) jSIi1 : 1 hS 0 I 0jZ ka (Nc Using a relation similar to (4.90) for the second term, this equation can be put into the form Nc X 0 0 ka (4.96) j;1]hS I jZ jSIi j;1] = 1 Nc 1hS 0I 0jZ kajSIi1 N SI S I B(Ii)B(I 0i0)hS 0 = I 0jZ kajS = Ii : c Combining the two results (4.92) and (4.96) gives the following general expression for the matrix element of the current Z ka taken between two mixed symmetry states hS 0I 0jZ kajSIi = ( )((SIi)+ (S I i ) (4.97) ) Nc(Nc + 2) I hS 0I 0jZ kajSIi 1 1 1 N 2 SI S I B(Ii)B(I 0i0 )hS 0 = I 0jZ ka jS = Ii : 4(N 1) 0 0 0 0 00 j=2 c c 0 0 We are now in a position to compute the phase of the normalization constant B(Ii). This can be done by comparing the two expressions (4.73) and (4.97) for the matrix element hZ iai. We obtain in this way the following exact relation 1 1 00 0 0 ka (4.98) 2 SI S I B(Ii)B(I i )hS = I jZ jS = Ii = Z2 + Nc (Z1 Z2 ) : Nc Using (4.81) for Z2 one nds to leading order in Nc s q q (Nc + 2) 2I0 + 1 B(Ii)B(I 0i0) = Nc3 (2i + 1)(2i0 + 1) (2S + 1)(2I + 1) (4.99) 2I + 1 q (0 )( 1) 1 )( S S 1 S i 2 S 0 2 i0 = N 3 (2i + 1)(2i0 + 1) ( )1+i i+S +I I I 0 0 IS I S : 10I 10 c 2(2I 0 + 1) 2 2I0 From this follows that B(Ii) can be chosen to be positive for all values of its arguments. It is easy to see now with the help of (4.82) and (4.99) that (4.97) gives, to leading order in Nc, the same result for hS 0I 0jZ kajSIi as (4.71). 0 0 0 0 0 0 0 D. Matrix elements of Y a and Q ka in the quark model As already mentioned in Section III, the matrix elements P the operators Y a and Qka of ija in the quark model can be reduced to those of the operator Nc rn n n . Here rn; n; n n=1 are vector operators acting on the orbital, spin and isospin degrees of freedom of the nth 34 quark respectively. In this Section we prove that the quark model reproduces, in the largeNc limit, the results (3.43) and (3.50) expected from the model-independent treatment of Sections III.B and III.C. We consider rst the transitions from an excited baryon state transforming under the symmetric representation of SU(4) to another symmetric baryon state. For generality we leave the orbital momenta of the initial and nal states completely arbitrary L; L0. The dependence on the spin-isospin quantum numbers is contained in the reduced matrix element T (I 0; I) de ned by Nc Xij p hI 0L0; m0S m0L 0j rn n najIL; mS mL i = (2I 0 + 1)1 2L0 + 1 T (I 0; I)I(L0; L) (4.100) n=1 hI 0; m0S jI; 1; mS ; jihI 0; 0jI; 1; ; aihL0; m0LjL; 1; mL; ii : We will restrict our considerations to baryon states for which all the orbital angular momentum is carried by one quark at a time. This is strictly true only for the lowest orbital excitations. In Hartree language the spatial part of the wavefunction for these states has the form Nc p1 X (r1 ) (r2 ) L;mL (~p ) (rNc ) r (4.101) jL; mLi = N c p=1 where (r ) is a s-wave one-particle wavefunction and L;mL (~ ) carries angular momentum r (L; mL). The spatial part of the matrix element (4.100) can be written in terms of the matrix element 1 i 0 hL0; m0Ljrn jL; mLi = N p2L10 + 1 I(L0; L)hL0 ; mLjL; 1; mL; ii ; (4.102) c with I(L0; L) an overlap integral of order Nc0. The case L0 = 0 of a s-wave baryon in the nal state is special, as the scaling law with Nc is di erent i (4.103) h0jrn jL; mLi = p1 I L1h0jL; 1; mL ; ii ; Nc i For both these cases the matrix element of rn is independent of n due to the symmetry of the wavefunction under any permutation of two quarks. Therefore the spin-isospin part of the matrix element (4.100) decouples completely from the spatial part and is given exactly by the formula (4.37) for the ground state baryons. We obtain in this way for the reduced matrix element T (I 0; I) 8 Nc+2 q < Nc (2I + 1)(2I 0 + 1) ; L0 6= 0 0; I) = T (I (4.104) : NcN q(2I + 1)(2I 0 + 1) ; L0 = 0 : p +2 c which can be seen to coincide, up to an unimportant phase and numerical factor, with the result (3.43) anticipated in Section III. We consider next the case of an excited baryon transforming under the mixed symmetry representation of SU(4) in the initial state. The nal state corresponds to the completely symmetric representation. We write the matrix element relevant for this case as 35 hI 0L0; m0m0L 0j Nc X n=1 i rn 1 0 0 ja n n jSIL; mS mL i = (2I 0 + 1)p2L0 + 1 T (I ; SI)I(L ; L) hI 0; m0jS; 1; mS ; jihI 0; 0jI; 1; ; aihL0; m0LjL; 1; mL; ii : (4.105) The scaling law with Nc of the spatial part of this matrix element is again di erent, depending on whether L0 6= 0 or L0 = 0. Both cases can be considered together by writing it as Nc (SIi) X N i hL0; m0LjrnjSIL; mS mL i = q c ( ) ( k n nk k n n1) (4.106) 2Nc (Nc 1) k;k =1 p 10 I(L0; L)hL0; m0LjL1; mL; iijSI; mS i k;1] 2L + 1 ) ( Nc X Nc ( ) (SIi) p 1 I(L0; L)hL0 ; m0 jL1; m ; ii jSI; m i =q jSI; mS i k;1] : L S n;1] n1 L 2Nc (Nc 1) 2L0 + 1 k=2 0 0 0 Here = 1=2 for L0 = 0 and = 0 for L0 6= 0. Adding the spin-isospin part of the operator and summing over the Nc quarks in the baryon gives Nc Xij (4.107) hI 0L0; m0m0L 0j rn n najSIL; mS mL i = N q c( ) p 10 I(L0; L)hL0; m0LjL1; mL; ii 2Nc (Nc 1) 2L + 1 (X ) Nc Nc 0 ; m0 0 j j ajSI; m i 0; m0 0 j j a X jSI; m i hI S n;1] hI S k;1] : 11 nn n=1 k=2 (SIi) n=1 ! Nc Nc 1 X hI 0; m0 0j j ajSI; m i hI 0; m0 0j X j ajSI; m i = p S1 Sn nn nn 2 n=1 n=1 ! Nc 1 N hI 0; m0 0j j ajSI; m i 1 B(Ii)hI 0; m0 0j X j ajI; m i : pc S1 S 11 nn Nc SI 2 n=1 In the second line we used the identity Nc Nc Xj Xja 2 2 jSI; mS i1 = hI 0; m0 0j n na(1 + P12 + + P1Nc )jSI; mS i1 (4.109) NchI 0; m0 0j nn = Nc X n=1 k=1 The rst term in the braces can be written as Nc X 0 00 ja hI ; m j n n jSI; mS i n;1] = n=1 (4.108) hI 0; m0 0jP1k Nc X n=1 n=1 ja n n P1k jSI; mS i1 = hI 0; m0 0j Nc X n=1 ja nn jSI; mS i1 followed by the application of the relation (4.86). In (4.109) we have de ned P11 = 1. The second term in (4.107) can be put into the following form through an application of (4.89) and (4.86) 36 hI 0; m0 0j 1 p j i Nc hI 0; m0 0j 1 1ajSI; mS i1 : 2 Combining (4.108) and (4.110) together we obtain the following general formula for the matrix element (4.105) Nc j a X jSI; m i S k;1] = 11 k=2 0 00 ja SI B(Ii)hI ; m j 1 1 jI; mS ; (4.110) hI 0L0; m0m0L 0j Nc X n=1 i rn ja n n jSIL; mS mL i= ) (4.111) (SIi) N qc ( ) p 10 I(L0; L)hL0; m0LjL1; mL; ii Nc (Nc 1) 2L + 1 ( Nc 1 B(Ii)hI 0; m0 0j X 0; m0 0 j j a jSI; m i Nc hI S1 11 N SI c n=1 ja n n jI; mS ; i: The second term in (4.111) is already known from our analysis of the symmetric states in Section IV.A. The rst matrix element is new. In the following we present the details of its calculation. Using (4.86) one can write hI 0; m0 0j 11 j ajSI; m S 1 i1 = B(I 0i0) 1hI 0I 0; m0 0j Nc X k=1 j P1k 1 1ajSI; m i1 : (4.112) A typical term of the sum over k has the form ja 1hI 0 I 0; m0 0 jP1k 1 1 jSI; m i1 = where we used the de nition of the P operator (4.47) and the fact that F(i) and G(i) computed in Section IV.B are nonleading in 1=Nc . The rst matrix element is easily calculated with the result (4.113) 1 hI 0I 0; m0 0j j ajSI; m i + 1 hI 0I 0; m0 0j(~ ~ )(~ ~ ) j ajSI; m i + O(1=N ) 1 k 1 k 11 1 41 1 c 11 41 ; ai (4.114) q 0 1 I I0 : 6 ii (2S + 1)(2I + 1) 1 S I1 1 i11 i22 22 The second matrix element can be reduced to quantities already known by simplifying the products of two spin and isospin Pauli matrices with the help of the identity p pX 3h0j11; abi 1 2 h1cj11; abi c : (4.115) a b= 0 ja 0 1 hI 0I 0; m0 0 j 1 1 jSI; m i1 = hI 0 ; m0jS; 1; m; jihI 0; )jI; 1; )( ( c We obtain in this way 37 1 hI 0I 0; m0 0 j(~ 1 j ~ k )(~1 ~k ) 1 1ajSI; m i1 = (4.116) q hI 0; m0jS; 1; m; jihI 0; 0jI; 1; ; ai (2i + 1)(2i0 + 1)(2S + 1)(2I + 1) 8 1 1 93 2 ( > > 0) 6 ) i+ 12 +S S 1 I + 6( ) S I < 2 1 i20 =7 i1 4( > S 1 I 0 >5 i0 1 i : ; 2 8 1 1 93 2 ( > > 0) 6 ) i+ 12 +I I 1 I + 6( ) I I < 2 1 i20 =7 i1 4( 1 > I 1 I 0 >5 i0 2 i ; : q = 36hI 0; m0jS; 1; m; jihI 0; 0jI; 1; ; ai (2i + 1)(2i0 + 1)(2S + 1)(2I + 1) ( 0 )2 ( 0 ) ( 0 ) SI 1 II 1 : ii 1 11 11i 1 1 I0 22 22i 22 0 0 We used in the second equality the identity 81 1 9 ( )( ) ( ) > 2 2 1> < 0= i i0 1 S I 0 1 + 1 ( )i +S+ 32 i i0 1 ii1 11i 1 > S I0 1 > = 1 2 I0 I0 S 1 6 ; : 2 22 2 0 (4.117) and a similar one with S ! I, which can be obtained from (3.21) by taking = 1=2 and the j's are the same as in the 9j symbol on the l.h.s. Combining (4.114) and (4.116) we nd the following result for the matrix element (4.113) ja 0; 1 hI 0 Iqm0 0 jP1k 1 1 jSI; m i1 = 3 2I 01+ 1 S 1 1 1 22i (2i + 1)(2i0 + 1)(2I + 1)(2S + 1)hI 0; m0jS; 1; m; jihI 0; 0jI; 1; ( I0 0 )( I 11 22 I0 1; i ) ; ai (4.118) where we have rewritten the ii 0 symbol in (4.114) as ( 0 )2 q i0 = 2 (2i + 1)(2i0 + 1) 1 i1 I 0 22 ii (4.119) and added the two terms with the help of the identity ( 0 )2 X ix (2x + 1) 1 i1 I 0 = 2I 01+ 1 : 22 x=0;1 Next we use (3.30) to write the product of 2 6j-symbols in (4.118) as ( 1 0 )( 1 0 ) SI 2I i 2I i S 1 1 I 1 1 = 6q(2S + 1)(2I + 1) 2 2 )( ( ) 1 +S+I+I +i SI1 : 111 +( ) 2 11i S I I0 22 0 (4.120) (4.121) 38 The sum over k in (4.112) is dominated by the terms with k 6= 1, of which there are Nc 1. Neglecting the contribution of the k = 1 term, we obtain for the matrix element (4.112) to leading order in Nc s q 0 ; m0 0 j j ajSI; m i = 3 2(2i + 1) (2S + 1)(2I + 1) hI (4.122) s1 11 2I 0 + 1 hI 0; m0jS; 1; m; jihI 0; 0jI; 1; ; ai 8 )9 )( ( < 1 +S+I+I +i S I 1 =: 111 SI q +( ) 2 11i ; : 6 (2S + 1)(2I + 1) S I I0 22 0 When inserted into (4.111), the term proportional to SI will be cancelled exactly by the second term in (4.111). As a result, we obtain for the reduced matrix element T (I 0; SI) taken between the unnormalized quark model states q q T (I 0; SI) = Nc 3 2(2i + 1) ) ( + 1)(2I + 1)(2I 0 + 1)( ) 12 +S+I+I +i( ) (SIi) (4.123) (2S ( ) SI1 : 111 11i S I I0 22 0 The physical value of this reduced matrix element is obtained by dividing with the square root of the norm of the initial state (4.44). This gives pq T (I 0; SI)]norm = Nc 1=2( 6 (2S)+ 1)(2I + 1)(2I 0 + 1)( ) 12 +S+I+I +i( ) (SIi)+ (SIi) (4.124) 2 111 : S I I0 0 Finally we insert here (SIi) = 1 + I + S the phase of the 6j-symbol appearing in the formula for the norm (4.44) and (SIi) = i + I + 1 the phase of the mixed symmetry 2 state, which gives for the total phase ( ) I+I . Thus (4.124) can be seen to coincide exactly, up to a numerical factor, with the expression (3.50) expected from the model-independent treatment of Section III.B. An important by-product of this calculation is the large-Nc scaling law of the transition matrix elements into nal states in an s-wave. We obtain that the matrix elements of Y a and Qka from an initial state with mixed symmetry scales like Nc0. On the other hand, the 1 same matrix elements with a symmetric excited state in the initial state scale as Nc2 . This dependence of the scaling law on the symmetry type of the excited state is a new feature, unnoticed previously. As discussed in Section III, for both cases the scaling law for the total scattering amplitude is su ciently restrictive to allow the derivation of useful consistency conditions. In spite of their di erent Nc scaling, the solutions for these matrix elements have the same dependence on spin and avor quantum numbers. The quark model computations of Appendix A illustrate another important asymmetry between the symmetric and the mixed symmetry states. The 1=Nc corrections to the large-Nc results for coupling ratios vanish for the former 2] but not for the latter. Such dependence on the symmetry properties of these states raises the question of how to distinguish states with di erent permutational symmetry beyond the framework of the quark model. 0 39 The exact large-Nc scaling law for matrix elements of Y and Q following from the calculations of this Section is strictly correct only for the case of the baryons made of heavy quarks, for which the constituent quark picture is known to be exactly valid. Our results following from the consistency conditions discussed in Sect.III rest on the assumption that no important changes occur as the quarks become light and that the modi ed scaling law corresponding to this situation still allows the derivation of consistency conditions. While this assumption seems plausible and is similar to smoothness arguments commonly used in other large-Nc studies 12, 9], it is important to keep it in mind as one of the vulnerable points of an analysis of this type. V. CONCLUSIONS AND OUTLOOK We have studied the strong couplings of the excited baryons in the large-Nc limit with the help of consistency conditions on pion-baryon scattering amplitudes. This method is similar to the one used by Dashen, Jenkins and Manohar 2, 3, 5, 6] in their analysis of the strong couplings of the s-wave baryons. In extending their analysis to the excited baryons' sector one has to deal with additional complications, related to the more complex structure of the spectrum of these states. The consistency conditions are very e ective in constraining the large-Nc spin-isospin dependence of the strong vertices of these states, especially for the S-wave pion coupling, which is completely determined in terms of just one unknown constant. The allowed form of the strong vertices turns out to be exactly the same as the one following from the constituent quark model. In addition to constraining the structure of the strong vertex, the consistency conditions predict also the equality of the pion couplings to excited and to s-wave baryons respectively. This is again what is expected from the constituent quark model. Our ndings extend therefore the results obtained in 2, 3, 5, 6] for the strong couplings of the s-wave baryons and give a natural explanation for the successes of the quark model when applied to strong decays of the excited baryons 1, 16, 17, 18] in terms of the large-Nc expansion. For example, this lends additional support to some predictions made recently for strong decays of excited heavy baryons 38, 39] with the help of the quark model. However, as discussed in Appendix A, the quark model predictions for ratios of strong couplings for these states cannot be expected to hold to the same accuracy as in the s-wave sector, as these ratios are not in general protected against 1=Nc corrections. The exact results in Appendix A provide a speci c framework to study quantitatively how good the large-Nc approximation is by examining their complete Nc dependence as Nc varies from the physical value Nc = 3 to in nity. The results of the present paper can be expanded in a number of directions. We recall that our analysis has only assumed isospin symmetry. Thus, one can attempt to incorporate SU(3) with some amount of symmetry breaking, by studying consistency conditions following from large-Nc counting rules for kaon-baryon scattering amplitudes 5, 6]. In this way one should be able to relate the strong couplings of di erent towers of states with di erent strangeness quantum numbers, which in our present analysis are left completely unrelated. Second, we have only discussed excited states transforming under the symmetric and mixed symmetric representations of SU(4). It is known that excited states exist which transform also under the antisymmetric representation. Extending our analysis to this case should 40 be completely straightforward. Finally, a similar analysis could be performed for the electromagnetic couplings of the excited baryons, with the help of consistency conditions for photon-baryon scattering amplitudes. For the s-wave baryons such constraints on the magnetic moments have been worked out in 4]. We plan to return to some of these problems in a future publication. ACKNOWLEDGMENTS The research of D.P. was supported by the Ministry of Science and the Arts of Israel. The work of T.M.Y. was supported in part by the National Science Foundation. 41 APPENDIX A: QUARK MODEL MATRIX ELEMENTS FOR ARBITRARY N C In this Appendix we give, for completeness, the full expressions for reduced matrix elements in the quark model with arbitrary number of colors Nc. The results presented in the main text are obtained from these expressions by keeping only the leading terms in Nc. We take advantage of our ability to derive exact relations for the quark model matrix elements to study the 1=Nc corrections to the large-Nc predictions. By examining a few simple particular cases we conclude that the results obtained in Section III in the large-Nc limit will receive, in general, 1=Nc corrections. 1. (S 0 Z I ; SI 0 ) We begin by giving the result for the matrix element Z(S 0I 0; SI) de ned by Nc 0 I 0L0 ; m0 ; m0 ; 0 j X k a jSIL; m ; m ; i = hS SL nn SL n=1 q010 Z(S 0I 0; SI) LL mLmL hS 0; m0S jS; 1; mS ; kihI 0; 0jI; 1; (2S + 1)(2I + 1) 0 0 (A1) ; ai : According to (4.73) this matrix element is completely determined in terms of the diagonal and the nondiagonal matrix elements of the current on direct product states. These will be characterized by two quantities z1; z2 de ned by Nc X ka (A2) Z1 = j hS 0I 0; m0; 0j n n jSI; m; ij n=1 q = (2S + 1)(2I + 1)z1hS 0; m0jS; 1; m; kihI 0; 0jI; 1; ; ai Nc X ka (A3) Z2 = j hS 0I 0; m0; 0j n n jSI; m; ij n=1 q = (2S + 1)(2I + 1)z2hS 0; m0jS; 1; m; kihI 0; 0jI; 1; ; ai : 0 The reduced matrix element Z(S 0I 0; SI) (taken between unnormalized quark model states) is expressed in terms of z1 and z2 as q 0 I 0; SI) = ( ) (S I i )+ (SIi) (2S + 1)(2I + 1)(2S 0 + 1)(2I 0 + 1) Nc + 2 (z z ) : (A4) Z(S 12 4 We obtain for the diagonal matrix element the simple result ( q 0 ) ( 1 I I0 ) 0 + 1)z(i0 ; i)( ) 1+2i +S+I 1 S S (A5) z1 = (2i + 1)(2i 10 1 i0 i 2 2i i )( ) ( 1 I I0 ; 1 2i S I 1 S S 0 + 6 ii ( ) 11 1 i22 i21 2 0 00 0 0 0 0 with 42 q z(i0; i) = (Nc + 1)2 (i0 i)2(i0 + i + 1)2 = Nc + 1 + O(1=Nc ) : (A6) The nondiagonal matrix element z2 can be written as a sum over the four terms into which it can be decomposed with the help of (4.77) 1 (A7) z2 = 4 (T1 + T2 + T3 + T4) : We nd T1 = q z1 1 T2 = 6(2i + 1)(2i0 + 1)2F(i0)z(i0; i)( ) 2 +2i+i +I+S+S )( ) ( 0 1 0 )( 1 S S0 1 I I 0 i 2S 10 1 i0 i 10 2i 1 2 2 (i i )( )( ) p p0 1iI 1 S S0 1 I I 0 i i0 1 6 6F (i) ii 2i + 1( ) 2 1 1 1 S0 i12 i11 22 2 22 q 1 0 + 1)2 F(i0)z(i0; i)( ) 2 +2i+i +I+S+I T3 = 6(2i + 1)(2i ( 0 1 0 )( )( ) i 2I 1 S S0 1 I I 0 10 1 i0 i 1 i0 2i 1 2 2( i )( )( ) p p0 1iS i i0 1 1 S S 0 1 I I 0 6 6F (i) ii 2i + 1( ) 2 1 1 I0 1 i12 i11 22 2 22 q S+I+S +I T4 = 6 (2i + 1)(2i0 + 1)( ) ( 00 1 0 ) ( )( )( X z(i0; i00)z(i00; i) 00 i 2S S 1 S 0 i00 1 I 0 I 2 1 1 i0 1 1 i00 2 i i00 Nc 1 (2i + 1) 2 i0 1 2 i )( 0 )( 0 )( 0 ( z(i0; i) q(2i + 1)(2i0 + 1) i i0 1 ii 1 SS 1 II + 36 N 1 1 1 I0 11i 11 1 1 S0 c 22 22 22 22 0 0 0 0 0 0 0 0 0 0 00 (A8) (A9) (A10) 1 1: i 1 2 )i ) I0 (A11) The expression for z2 greatly simpli es if only terms of order 1 are kept, in addition to the leading ones of order Nc , due to the fact that in this approximation the z(i0; i) factors are constants. This allows the sum over i00 to be performed with the help of (4.80). We obtain z1 q (A12) = (Nc + 1)( ) 0 f6j0g3 3f6j1g3] 0 + 1) (2i + 1)(2i )( ) ( 0 )( 0 )( i i 0 1 S S 0 1 I I 0 + O(N 1 ) ii 0 +12 1 1 S 0 1 11 0 1 c i21 i12 22 22I 2 2 z2 q = (Nc + 2)( ) 0 f6j0 g3 + 3( ) 0 f6j1g3 (A13) 0 + 1) (2i + 1)(2i ( )( ) 1 S S0 1 I I 0 SI +3 2I 0 + 1 i 1 1 1 i12 22 2 1 ( ) 0 1 + 2i0(i0 + 1) S 0(S 0 + 1) I 0(I 0 + 1)] f6j g3 3f6j g3] + O(N 1 ) 0 1 c 2 2 0 0 43 with 0 = 1 + i i0 + I 0 + S 0. We denoted here the products of 6j symbols encountered in Section III ( )( )( ) S i 1 S 0 1 i0 : 3 = S0 S 1 2 2 f6j0(1)g (A14) 1 I 0 0(1) 1 I I 0 0(1) 2 0(1) I 2 The di erence of z1 and z2 can be nally written as q z1 z2 = 3(Nc + 2)( ) 0 f6j1g3 ( ) 0 f6j0g3 (A15) (2i + 1)(2i0 + 1) 1 + 1 ( ) 0 2 + 2i0(i0 + 1) S 0(S 0 + 1) I 0(I 0 + 1)] f6j0g3 3f6j1 g3] 2 ( 0 0 )( 0 0 )( )( ) S I 1 1 S S 0 1 I I 0 + O(N 1) : i+i +I +S S I 1 18( ) 11i 1 1 1 i0 c i12 i11 22 22 2 22 The alternative method presented in Section IV.C can be also used to give an exact expression for the reduced matrix element of Z ia. We nd from (4.97) the following result q 2) (A16) Z(S 0I 0; SI) = ( ) (S I i )+ (SIi) (2S + 1)(2I + 1)(2S 0 + 1)(2I 0 + 1) Nc(Nc +1) 4(N !c 0; I) 1 Z(I z1 N 2 B(Ii)B(I 0i0) (2I + 1)(2I 0 + 1) SI S I ; 0 0 0 0 00 c 0 0 where Z(I 0; I) has been de ned in (4.37) and B(Ii) is given by (4.88). We have checked explicitly that both methods lead to the same answer for Z(S 0I 0; SI) up to the next-to-leading order in 1=Nc . We notice that (A16) does not involve any summation over intermediate state quantum numbers. The leading order term in (A15) is written as proportional to Nc + 2, which was seen to give the correct result to two orders in the 1=Nc expansion for the case of the symmetric baryons. It is natural to ask whether a similar result holds also for the reduced matrix element Z(S 0I 0; SI). In the following we will argue that no result of comparable simplicity can be obtained for the 1=Nc corrections to this quantity. Strictly speaking this still does not prove that there are nonvanishing 1=Nc corrections to Z(J 0I 0; JI) (which is the true physical coupling with a meaning beyond the quark model) which is related to Z(S 0I 0; SI) by (3.28). It is still conceivable that the 1=Nc corrections to Z(S 0I 0; SI) add up to zero when inserted into (3.28), although we have not been able to prove it. We will consider for simplicity the case when the quantum numbers of the initial and nal states satisfy S 6= I ; S 0 6= I 0 (A17) and examine the structure of the 1=Nc corrections in the following two particular cases: a) S = S 0 ; I = I 0 and b) S = I 0 ; I = S 0. This constrains i; i0 to be equal: i = i0. The norm of a state satisfying (A17) can be obtained from (4.44) and is given exactly by hSIjSIi = Nc 4+ 2 N Nc 1 : (A18) c 44 This will have to be divided out from the quantity on the r.h.s. of (A15). We obtain Nc + 2 (z z ) = q(2i + 1)(2i0 + 1) n 3(N + 2)( ) 0 f6j g3 (A19) 1 q 12 c 1 hSIijSIiihS 0I 0i0jS 0I 0i0i 4 ( 0 0 )( 0 0 )( )( ) ) S I 1 1 S S 0 1 I I 0 + 3( ) 0 f6j g3 : i+i +I +S S I 1 18( ) 1 11i 1 1 i0 1 i11 i12 22 22 22 2 Next, we note that in the limit (A17) the product of 4 6j symbols on the r.h.s. can be written as ( 0 0 )( 0 0 )( )( ) S I 1 1 S S0 1 I I 0 i+i +I +S S I 1 () (A20) 11 1 1 i0 1 i11 i12 28 2 i 22 22 2 > 1 r (2i 1)(2i+3) case a) < = > 36(2i+1)2 1 i(i+1) : case b) 9(2i+1)2 0 0 0 0 0 0 On the other hand, the leading order term is proportional to f6j1g3. From (4.79) we obtain in the limit (A17) 8 r ( < 0 ) ( 1 I I 0 ) > ( )2i 1 2 (2i 1)(2i+3) case a) 1 3(2i+1) 2i(2i+2) (A21) => f6j1g3 = ( )1+2i 3 1 S0 S 10 1i i : ( )1+2i 6(2i+1)12i(i+1) case b) 2 2i i One can see that for case b) the terms of order 1 in (A19) do not have the same structure as the leading term of order Nc and therefore cannot be generally absorbed into a rescaling of the latter. 2. T (I 0 by Nc X n=1 ; SI ) Next we present the exact calculation of the reduced matrix element T (I 0; SI) de ned hI 0L0; m0m0L 0j i rn 1 ja 0 0 n n jSIL; mS mL i = 0 + 1)p2L0 + 1 T (I ; SI)I(L ; L) (2I hI 0; m0jS; 1; mS ; jihI 0; 0jI; 1; ; aihL0; m0LjL; 1; mL; ii ; (A22) relevant for the transitions from orbital excitations with mixed symmetry to symmetric states. We calculate this matrix element starting from the general formula (4.111) and proceeding along the same steps as in Section IV.D. The rst term in (4.111) can be expressed with the help of (4.112) in terms of the two matrix elements (k 6= 1) q j hI 0I 0; m0 0j 1 1ajSI; m i1 = (2S + 1)(2I + 1)t1hI 0; m0jS1; m; jihI 0; 0jI1; ; ai (A23) 1 q ja 1 hI 0I 0; m0 0 jP1k 1 1 jSI; m i1 = (2S + 1)(2I + 1)t2hI 0 ; m0jS1; m; jihI 0; 0 jI1; ; ai : (A24) Expanding the permutation operator P1k and evaluating the resulting matrix elements with the help of the results of Section IV.B we obtain the following exact expressions for the coe cients t1; t2 45 ) )( I0 1 1 I I0 1 1 i12 2i 2 ) )( I0 1 1 I I0 1i 11 i22 2 )( ) 1 I I0 1 : 11i 22 22i 22 The result for t2 simpli es considerably when only the terms of order 1 and 1=Nc are kept t2 q = (A27) (2i + 1)(2i0 + 1) 8 ( 0 )29 <3 1 65 i0= + N 2I 06+ 1 N 2 + 2i0(i0 + 1) 2I 0(I 0 + 1) 1 i1 I 0 ; : 2I 0 + 1 c c 22 ( 0 )( 0 ) S I 1 I I 1 + O(N 2 ) : 11i 11i c 22 22 0 0 0 0 t1 = 6 ii t2 = 3 ii 2 s ( )( 1 3 F(i) p2i0 + 1( )I 2 +i i i0 1 S 3 ii 1 1 I0 1 22 s2 ( 0 )( 2 p 1 i1 3 3 G(i) ii 2i0 + 1( )I 2 +i 1 i1 I 0 S 1 2 22 ( 0 )2 ( 20 q SI i1 + 9H(i0; i) (2i + 1)(2i0 + 1) 1 i1 I 0 11 0 0 ( )( ) 1 S I0 1 I I0 11 i11 2 2 )(i 2 2 ) ( 1 S I0 1 I I0 1 1 i12 i12 2 2 (A25) (A26) The matrix element on the r.h.s. of (4.112) is proportional to the combination q t1 + (Nc 1)t2 = (2i + 1)(2i0 + 1) 2 ( )3 3 N+ 3 1 + 2i0(i0 + 1) 2I 0(I 0 + 1) i i0 0 25 40 1 1 I0 2I + 1 c 2I 0 + 1 6 2 22 ( 0 )( 0 ) S I 1 I I 1 + O(N 1 ) : 11i 11 c 22 22i (A28) After dividing this expression with B(I 0i0) we nd for the rst matrix element in (4.111) the following result hI 0; m0 0j (A29) s q p 2i (2S + 1)(2I + 1)hI 0; m0jS1; m; jihI 0; 0jI1; ; ai 3 2 2I 0+ 1 +1 8 0 ( 0 )219 0+1 1 < = 1 i0 1 + 2N + 2I + 2i0(i0 + 1) 2I 0(I 0 + 1) @ 2(2i03+ 1) 6 1 i1 I 0 A; : 3Nc 2 c 22 ( 0 )( 0 ) S I 1 I I 1 + O(N 1 ) : 11i 11i c 22 22 a 1 1 jSI; mS i1 = j The second matrix element in (4.111) is given by 46 q p s 2i + 1 00 00 2 SI Nc (2S + 1)(2I + 1)hI ; m jS1; m; jihI ; jI1; ; ai 3 2 0 2I )+ 1 ( 5 1 1 1 + 2N 2N2I ++ 1) 1 + 2i(i + 1) 2I(I + 1) : 6(2I + 1) 2 c c (2i The result for the reduced matrix element T (I 0; SI) valid to next-to-leading order in 1=Nc is obtained by inserting (A29) and (A30) into (4.111) and making use of (4.121) for the product of 6j symbols in (A29). Let us examine closer the structure of the 1=Nc corrections to the leading order result for T (I 0; SI) on the simple particular case when S 6= I. After dividing with the norm of the initial state (A18) we nd for this case ) ( q 0; SI) = Nc 1 2p6 (2S + 1)(2I + 1)(2I 0 + 1)( ) I+I 1 1 1 2 (A31) T (I S I I0 8 0 ( 0 ) 2 19 < = 1 + 2I 0 + 1 1 + 2i0(i0 + 1) 2I 0(I 0 + 1) @ 3 i0 1 6 1 i1 I 0 A; : : 2Nc 3Nc 2 2(2i0 + 1) 22 0 Nc 0 0 0 X j a jI; m ; S SI B(Ii)hI ; m j nn n=1 i= (A30) The last term in the braces has the explicit expression 0 ( 0 )21 2I 0 + 1 1 + 2i0(i0 + 1) 2I 0(I 0 + 1) @ 3 ii 0 A (A32) 1 3Nc 2 2(2i0 + 1) 6 1 2 I 0 2 0+ 1 = 2I2N 1 ( ) 2 i +I (1 2 ii ) c which shows that it cannot be absorbed into a rescaling of the leading order term. This quark model calculation suggests therefore that the ratios of the Y and Q couplings of the mixed symmetry states predicted by the consistency conditions in the large-Nc limit will receive nontrivial 1=Nc corrections. 0 0 0 47 APPENDIX B: TRANSITION MATRIX ELEMENTS BETWEEN STATES WITH MIXED SYMMETRY We present in this Appendix the computation in the quark model of the matrix elements of Y a and Qka between excited baryon states with mixed symmetry. This quantity is phenomenologically relevant for strong decays of positive-parity excited baryons into negativeparity states in the 70. We include this calculation here merely for the sake of completeness and because the result provides an explicit realization for the most general solution of the consistency condition for Q(J 0I 0; JI) (3.59). We start by computing the quark model reduced matrix element T (S 0I 0; SI) de ned by Nc Xij 1 hS 0I 0L0; m0S m0L 0j rn n najSIL; mS mL i = q 0 (B1) (2S + 1)(2I 0 + 1)(2L0 + 1) n=1 T (S 0I 0; SI)I(L0; L)hS 0; m0S jS1; mS ; jihL0 ; m0LjL1; mL; iihI 0; 0jI1; ; ai : We proceed in close analogy to the calculation of T (I 0; SI) in Section IV.D. First we take the matrix element of the spatial part of the operator, which is parametrized by the overlap integral I(L0; L) ij hS 0I 0L0; m0S m0L 0jrn n najSIL; mS mL i (B2) Nc (S I i )+ (SIi) X 00 0 0 ja = ( )2(N 1) k ;1]hS I ; mS j n n jSI; mS i k;1] c k;k =2 p 10 hL0; m0LjL1; mL; iiI(L0; L) kn kk + n1] : 2L + 1 After summing over the contributions of the Nc quarks to the transition operator we obtain the following general expression for the reduced matrix element T (S 0I 0; SI) (S I i )+ (SIi) 1 q T (S 0I 0; SI)hS 0; m0S jS1; mS ; jihI 0; 0jI1; ; ai = ( )2(N 1) (B3) c (2S 0 + 1)(2I 0 + 1) (X ) Nc Nc Nc X 0 I 0; m0 0 j j ajSI; mS i 0 I 0; m0 0 j j a X jSI; mS i n;1] hS n;1] + k ;1]hS k;1] : 11 S nn S 0 00 0 0 0 0 00 The rst term in the braces is of order Nc and is therefore suppressed relative to the second one, which is of order Nc2. In this section we work only to leading order in Nc so we keep only the contribution of the second term. It can be computed by expressing the sums over k; k0 with the help of (4.89) Nc Nc X jX (B4) hS 0I 0; m0S 0j 1 1a jSI; mS i k;1] = k ;1] k=2 k =2 1n ja 00 0 0 0 j a 2 00 0 0 2 SI S I B(Ii)B(I i )hI ; mS ; j 1 1 jI; mS ; i Nc 1hS I ; mS ; jP1k 1 1 jSI; mS ; i1 o j j Nc2 1hS 0I 0; m0S ; 0j 1 1aP1k jSI; mS ; i1 + Nc2 1hS 0I 0; m0S ; 0j 1 1ajSI; mS ; i1 + O(Nc ) : Each of the terms on the r.h.s. can be evaluated using the methods of Sec.IV.D. We obtain for the reduced matrix element T (S 0I 0; SI) 0 0 0 0 n=1 k =2 0 0 k=2 48 q010 T (S 0I 0; SI) = (2S + 1)(2I + 1) 2q (S I i )+ (SIi) Nc 0+ 1)(2I () 4(Nc 1) (2i + 1)(2i ) (1)(2S +) + 1) (0 ( 0 I SI S I 3 2I 0S+ 1 S S 1 I I1 1 11i 1 i 2(2I + 1)(2I 0 + 1) ( 2 20 ) ( 2 20 ) 1 1 3( ) I S I S 2I SI 1 S S i0 I I1 i0 11 1 ( 0 + ) ( 2 20 ) ( 2 2 0 ) ( 0 )) ii 0 SS 1 II 1 : 0 +12( ) I S I S i i1 I 1 1 i0 11S 1 1 i0 1 22 22 22 22 0 00 0 0 0 0 0 0 0 0 (B5) The rst two terms can be combined together by using (4.121) for the product of two 6jsymbols in the second term. The last two terms can be also written together such that we obtain for the total sum of the four terms in the braces ) ( ( )" 1 +S+I+I +i 111 SI1 SI (B6) f g= 1 1 i 3( ) 2 2I 0 + 1 S)I I 0 22 )( 0 ( )# ( S S 1 I I0 1 : 1+i+i I S S I 1 +18( ) 110 11 0 1 1 i0 22 22i 22i 0 0 0 0 0 0 The product of three 6j-symbols on the r.h.s. can be transformed into the following form by repeated application of (4.121) ( )( )( ) S I 1 S S0 1 I I 0 1 = (B7) 110 110 11 0 2(i 2 2) 2 i 22i ( )( )# " 111 S0 I 0 1 SI 2S+S +I 1 1 1 1 1 i0 6(2I 0 + 1) + ( ) S I 0 I I 0 S0 S 22 ( ) 1 +S+S +I+i 111 : SI 2 + 6(2I 0 + 1) ( ) S I0 I 0 0 0 0 0 0 When inserted into (B6), the last term in this relation exactly cancels the rst term in (B6). We obtain in this way the following expression for the reduced matrix element T (S 0I 0; SI) taken between unnormalized mixed symmetry states q T (S 0I 0; SI) = 9 Nc ( ) (S I i )+ (SIi) (2i + 1)(2i0 + 1)(2S + 1)(2I + 1)(2S 0 + 1)(2I 0 + 1) (B8) 2 )( 0 0 ) ( SI1 1+i+i I S S I 1 () 110 11i 22 2 2 i)( " ( )# 111 : SI 2S+S +I 1 1 1 S I 0 I I 0 S0 S 6(2I 0 + 1) + ( ) 0 00 0 0 0 0 0 The physical value of T (S 0I 0; SI) is obtained after dividing this expression with the squared roots of the norms for the initial and nal states (4.44). Inserting the appropriate phases of the mixed symmetry states (SIi) = 1 + I + i we obtain our nal result 2 49 111 111 + ( )2S+S +I S I 0 I I 0 S 0 S : This will be used in the following to compute the matrix elements of Y a and Qka between states with mixed symmetry. SI 6(2I 0 + 1) 0 0 T (S 0I 0; SI)]norm = 6( )S+2I+I ((2S + 1)(2I + 1)(2S 0 )#1)(2I 0 + 1) + )( " 0 q (B9) 1. Matrix elements of Y a The matrix element of Y a takes its simplest form in the j(IP)S; L; J; m; i basis, where it is directly proportional to T (S 0I 0; SI) h(I 0P)S 0; L0; J 0; m0; 0jY aj(IP)S; L; J; m; i = (B10) 8 s 09 >S 1 S > < =0 2J + 1 T (S 0I 0; SI) 0 JJ mm > L 1 L > hI ; jI1; ; ai : 2I 0 + 1 : J 0 J0 ; The 9j-symbol with one value of 0 can be reduced to a 6j-symbol. We decided to write it in this form, as it allows us to read o the results from the corresponding expressions for Qka given below by making the replacement 2 ! 0 in the Wigner symbols. We are interested nally in the matrix elements of Y a in the jI; (PL) ; J; m; i basis, which is reached through the recoupling relation (3.23). With P = P 0 = 1, we have a hI 0; (P 0L0) 0; J 0; m0; 0jYsjI; (PL) ; J; m; i = (B11) 2J ( ) I L J I L J 2I 0 + 1 JJ mm hI 0; jI1; ; ai +1 8 09 ( )( 0 q < = 0 )> S 1 S > X I1 I 1S L 1 L0 > T (S 0I 0; SI) : (2S + 1)(2S 0 + 1)(2 + 1)(2 0 + 1) L J S > L0 J 0 0 : J 0 J 0 ; SS All what is left to do is insert here the result of the quark model calculation of T (S 0I 0; SI) (B9) and perform the summations over S; S 0. We write the total result for the reduced matrix element Y (JI 0; JI) as Y (JI 0; JI) = Y (JI 0; JI)]1 + Y (JI 0; JI)]2 (B12) where Y (JI 0; JI)]1;2 stand for the contributions of the two terms in T (S 0I 0; SI) (B9). We nd ) ( ( 0 0) I 1 I0 J+I + 0; JI)] = ( )1+2J p2 + 1 L L (B13) () Y (JI 1 0J 1 1 L0 )( ) ( q 111 0; JI)] = 2p3( )2J+ + 0 + 1) 1 1 0 10 Y (JI (B14) (2 + 1)(2 2 0 L LL ( ) I 1 I0 : J+I + () 0J Their sum can be seen to have the same form as the model-independent solution of the corresponding consistency condition (3.36). 0 0 0 0 0 0 0 0 0 0 0 0 0 0 50 2. Matrix elements of Q ka The matrix element of Qka is given, in the j(IP)S; L; J; m; i basis, by an expression similar to (B10) J 2 J0 This can be transformed to the jI; (PL) ; J; m; i basis with the help of the recoupling relation (3.23). Again with P = P 0 = 1, we have hI 0; (P 0L0) 0; J 0; m0; 0jQkajI; (PL) ; J; m; i = (B16) s 2J ( ) I L J I L J 5 2I 0 + 1 hJ 0; m0jJ2; m; kihI 0; jI1; ; ai +1 8 09 ( )( 0 < = 00 0 )> S 1 S > Xq I1 I 1S 0 (2S + 1)(2S 0 + 1)(2 + 1)(2 0 + 1) L J S 0 J 0 0 > L 1 L > T (S I ; SI) : L : J 2 J0 ; SS In the following we consider the contributions of the two terms in T (S 0I 0; SI) (B9) to this relation in turn. For the rst term the summation over S is trivial and amounts to the substitution S ! I 0. The remaining sum over S 0 can be readily done by using (3.21), which gives for the contribution of this term to Q(J 0I 0; JI) ) )( )( ( q J J0 2 : I 1 I0 L 0 2 0I 0; JI)] = ( )1+2 Q(J 5(2 + 1)(2 0 + 1) L J 1 0 L I0 1 1 L0 (B17) This can be put into a form similar to (3.59) by expressing the product of the rst and last 6j-symbols with the help of (3.21) 8 ( ) > 0 I0 J0 9 ( > = < 0 ) ( J J0 2 ) X 12y I1I I J >: (B18) (2y + 1) 0 L > 0 L I0 = LJ :y 1 2; y=1;2;3 0 0 0 0 0 h(I 0P)S 0; L0; J 0; m0; 0jQkaj(IP)S; L; J; m; i = 8 9 s > S 1 S0 > < = 2J 5 2I 0 + 1 T (S 0I 0; SI) > L 1 L0 > hJ 0; m0jJ2; m; kihI 0; jI1; ; ai : +1 : ; (B15) The contribution of the second term is proportional to the double sum over S; S 0 ( )( 0 ) X S+S +I+I I 1 S0 0 + 1) I 1 S (2S + 1)(2S (B19) ISS = ( ) LJ L0 J 0 0 SS ( )( ) 8 S 1 S0 9 < = 111 1 1 1 > L 1 L0 > : S I 0 I I 0 S0 S > J 2 J 0 > : ; 0 0 0 0 The summations over S and S 0 are analogous to the sum over S encountered in Sec.III.C in Eq.(3.70) and can be performed along similar lines. A slight generalization of the sum over S in (3.70) gives the identity 51 9 ) 8 S 1 I0 > > < = X 2S I1 111 0 ( ) (2S + 1) L J S >J 2 J ; I I 0 S : L y L0 > = S 8 ( )( ) > L0 I 0 J 0 9 > < = Xz y1z y1z I+J +L+1 ( ) (2z + 1) 1 2 1 () I J> L0 L > z 1 2 ; : z=1;2;3 ( )( 0 (B20) where y can take the values y = 1; 2; 3. Applying (B20) twice we obtain ) ( )( Xy Xz 11y 11y ( ) (2z + 1) ( ) (2y + 1) 1 2 1 ISS = 0 L L0 y=1;2 z=1;2;3 ( )( ) 8 0 I0 J0 9 = < y1z y 1 z > I J>: 0L > 121 ; :z 1 2> 0 0 (B21) The contribution of the second term in (B9) to the reduced matrix element Q(J 0I 0; JI) is given, in terms of ISS , by q (B22) Q(J 0I 0; JI)]2 = 6( )L+L 5(2 + 1)(2 0 + 1)ISS : 0 0 The total expression for Q(J 0I 0; JI) is given by Q(J 0I 0; JI) = Q(J 0I 0; JI)]1 + Q(J 0I 0; JI)]2 which can be seen to have the form of the general solution (3.59). (B23) 52 REFERENCES 1] For an overview see F.E. Close, An Introduction to Quarks and Partons, Academic Press, 1979. 2] R. Dashen and A.V. Manohar, Phys.Lett. B315 425, 438 (1993). 3] E. Jenkins, Phys.Lett. B315 431, 441, 447 (1993) 4] E. Jenkins and A.V. Manohar, Phys.Lett. B335 452 (1994). 5] R. Dashen, E. Jenkins and A.V. Manohar, Phys.Rev. D49 4713 (1994); D51 3697 (1995). 6] J. Dai, R. Dashen, E. Jenkins and A.V. Manohar, Phys.Rev. D53 273 (1996). 7] M.A. Luty and J. March-Russell, Nucl.Phys. B426 71 (1994). 8] M.A. Luty, J. March-Russell and M. White, Phys.Rev. D51 2332 (1995). 9] C. Carone, H. Georgi and S. Osofsky, Phys.Lett. B322 227 (1994). 10] T. Cook, C.J. Goebel and B. Sakita, J.Math.Phys. 8 708 (1967). J.L. Gervais and B. Sakita, Phys.Rev.Lett. 52 527 (1984); Phys.Rev. D30 1795 (1984). K. Bardacki, Nucl.Phys. B243 197 (1984). M.P. Mattis and M. Mukherjee, Phys.Rev.Lett. 61 1344 (1988). M.P. Mattis, Phys.Rev. D39 994 (1989). M.P. Mattis and E. Braaten, Phys.Rev. D39 2737 (1989). 11] G. t'Hooft, Nucl.Phys. B72 461 (1974); B75 461 (1974). 12] E. Witten, Nucl.Phys. B160 57 (1979). 13] S. Coleman, 1/N in Aspects of Symmetry: Selected Erice Lectures, Cambridge University Press, Cambridge 1985. 14] C.S. Lam and K.F. Liu, MCGILL-97-03, hep-ph/9704235. 15] O.W. Greenberg, Phys.Rev.Lett. 13 598 (1964). 16] D. Faiman and D. Plane, Nucl.Phys. B50 379 (1972). 17] A. Hey, P. Litch eld and R. Cashmore, Nucl.Phys. B95 516 (1975). 18] C.D. Carone, H. Georgi, L. Kaplan and D. Morin, Phys.Rev. D50 5793 (1994). 19] N. Isgur and G. Karl, Phys.Lett. B72 109 (1977). 20] N. Isgur and G. Karl, Phys.Rev. D18 4187 (1978). 21] N. Isgur and R. Koniuk, Phys.Rev. D21 1868 (1980); (E) D23 818 (1981). 22] K. Dannbom, E.M. Nyman and D.O. Riska, Phys.Lett. B227 291 (1989). 23] J. Schechter and A. Subbaraman, Phys.Rev. D51 2311 (1995). 24] Y.-S. Oh and B.-Y. Park, Phys.Rev. D53 1605 (1996). 25] C.K. Chow and M.B. Wise, Phys.Rev. D50 2135 (1994). 26] C.K. Chow, Phys.Rev. D54 3374 (1996). 27] F. Myhrer and J. Wroldsen, Z.Phys.C25 281 (1984) 28] Y. Chung, H.G. Dosch, M. Kremer and D. Schall, Nucl.Phys.B197 55 (1982). 29] D. Jido, N. Kodama and M. Oka, Phys.Rev.D54 4532 (1996). 30] S.H. Lee and H. Kim, Nucl.Phys.A612 418 (1997). 31] J.L. Goity, JLAB-THY-96-15, hep-ph/9612252. 32] Particle Data Group, R.M. Barnett et al., Phys.Rev. D54 1 (1996). 33] A.R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957. 34] A. Messiah, Quantum Mechanics, North-Holland Publishing Company, 1965. 35] M. Soldate, Int.J.Mod.Phys. E1 301 (1992). 36] A. Kakuto and F. Toyoda, Prog.Theor.Phys. 66 2307 (1981) 53 37] G. Karl and J.E. Paton, Phys.Rev. D30 238 (1984). 38] D. Pirjol and T.M. Yan, CNLS-97-1457, hep-ph/9701291. 39] F. Hussain, J.G. Korner and S. Taw q, IC/96/35, MZ-TH/96-10. 54

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