Fierz transformations

Transcription

1 Fierz trnsformtions Fierz identities re often useful in quntum field theory clcultions. They re connected to reordering of field opertors in contct four-prticle interction. The bsic tsk is: given four complex fields ψ 1,2,3,4, which crry (spcetime or internl) index, let us consider n interction (ψ 1 Aψ 2 )(ψ 3 Bψ 4 ). (1) The indices of the spinors re suppressed. The sme interction cn be expressed in different wy s (ψ 1 Mψ 4 )(ψ 3 Nψ 2 ). How re the mtrices M, N relted to the mtrices A, B? The obvious nswer is A ij B kl = ±M il N kj, (2) where the plus nd minus signs pply to bosonic nd fermionic field opertors, respectively. However, it is usully inconvenient to use the mtrix elements explicitly nd the mtrices A, B, M, N re often expressed in suitble bsis. Generl Fierz identity Denoting the vector spce of the spinors s R, let s ssume tht we know bsis in the mtrix spce R R; we will cll it Γ. The sclr product of bsis mtrices gives rise to metric, 1 Tr(Γ Γ b ) = g b, which my be used to rise nd lower indices by 2 Γ = b gb Γ b, s usul. Every mtrix M cn be expnded in this bsis s M = M Γ, where M = Tr(MΓ ). This leds immeditely to the completeness reltion, (Γ ) ij (Γ ) kl = δ il δ jk, (3) which is the bsis for the derivtion of ll Fierz identities. The bove introduced mtrices A, B, M, N re ccordingly written s A ij B kl =,b A B b (Γ ) ij (Γ b ) kl, M ij N kl =,b M N b (Γ ) ij (Γ b ) kl. 1 In fct, for generl complex mtrices we should Hermitin conjugte one of the mtrices in the trce to get well-defined sclr product. Therefore, ll our conclusions will hold without further ssumptions in cse the Γ s re Hermitin. Otherwise, we need to suppose t lest tht g b is invertible. This is indeed the cse in ll pplictions. 2 Here nd in the following, ll sums over indices will be indicted explicitly.

2 The tsk to find the reltion (2) between them is therefore equivlent to finding the reltion between (Γ ) ij (Γ b ) kl nd (Γ ) il (Γ b ) kj. This is in generl given by liner combintion (Γ ) ij (Γ b ) kl = c,d C cd b (Γ c ) il (Γ d ) kj. (4) Multiplying by (Γ e ) li (Γ f ) jk, we infer immeditely C bcd = Tr(Γ Γ d Γ b Γ c ). (5) Equtions (4) nd (5) represent the most generl Fierz rerrngement formul. However, in prctice one does not usully need to clculte the coefficients C bcd for ll combintions of indices. Thnks to symmetry there re typiclly just few independent ones. Symmetry constrints The spce of spinors R furnishes n irreducible representtion of symmetry group, under which the interction Lgrngin is required to be invrint. This is most esily ccomplished in two consecutive steps. First the product representtion R R is decomposed into irreducible representtions of the symmetry group using the set of Clebsch Gordn coefficients. For two spinors ψ, χ in the sme representtion, the set of biliners ψγ A χ, = 1,...,dim A, form bsis of the irreducible representtion A tht lies in the decomposition of R R. For ny selected representtion A we cn form unique invrint of the symmetry group, 3 (ψ 1 Γ A ψ 2 )(ψ 3 Γ A ψ 4 ). We therefore need not rerrnge the products of ll pirs of two individul bsis mtrices, but rther the sum, ΓA ΓA, over ll mtrices in given irreducible representtion. The Fierz trnsformtion nlogous to the generl formul (4) will then red (Γ A ) ij(γ A ) kl = C AB (Γ B b ) il(γ Bb ) kj, (6) B where the Fierz coefficients now depend only on the representtions in question. Summing Eq. (4) over ll = b in given representtion nd relizing tht by symmetry considertions, C c d cn only be nonzero for c = d, we find (Γ A ) ij (Γ A ) kl = C b b(γ B b ) il (Γ Bb ) kj, B nd from (5) then C AB = C b b =,b b Tr(Γ A ΓB b ΓA Γ Bb ). (7) 3 The sitution would become slightly more complicted in cse the decomposition of R R contined more equivlent irreducible representtions. For the ske of simplicity, we neglect this possibility here.

3 Properties of Fierz coefficients [1] By multiplying Eq. (6) with Γ C c,jk nd using the orthogonlity condition in the form Tr(Γ A Γ B b ) = δ AB g b, (8) we derive very useful formul Γ A Γ B b Γ A = C AB Γ B b, (9) which is often more convenient to evlute the coefficients C AB thn the definition (7). [2] Another distinguishing property of the Fierz trnsformtion is tht performing it twice, we get bck to the originl interction, tht is, the Fierz trnsformtion is equl to its inverse. In terms of the mtrix of coefficients C AB, this cn be seen by pplying Eq. (6) to itself, (Γ A ) ij(γ A ) kl = C AB (Γ B b ) il(γ Bb ) kj = C AB C BC (Γ C c ) ij(γ Cc ) kl, B b B C c C AB C BC = δ AC. (10) B [3] In prctice the representtion R often is representtion of direct product of groups, corresponding to different quntum numbers such s spin, flvor, or color. We therefore need to know how to perform the Fierz trnsformtion with respect to severl indices. Let us denote the bsis of mtrices in the product representtion A A s Γ AA Eq. (9), we obtin Γ AA ΓBB bb = (Γ A ΓAA Γ B b Γ A ) (Γ A ΓB b ) = ΓA nd consequently ( ) ( = Γ A Γ B b Γ A ΓA ΓA Γ A ) ΓB b = C ΓA AB C A B ΓB b Γ B b C AA,BB = C ABC A B.. Applying repetedly = C ABC A B ΓBB bb, This is gret simplifiction which tells us tht the Fierz trnsformtion cn be performed on ech index seprtely. [4] Summing Eq. (7) over b, we get result invrint under the exchnge A B, which implies nice reciprocity reltion C AB dim B = C BA dim A. (11) [5] The representtion R R lwys contins the unit representtion, I, nd the corresponding Clebsch Gordn coefficients re conveniently defined by the unit mtrix, Γ I = 11. Assuming

4 tht ll other bsis mtrices re chosen orthogonl to 11, i.e. trceless, we find Γ I = 11/ Tr 11 = 11/ dim R. Substituting A = I in (9) then leds to C IA = 1 dim R, the second reltion following immeditely from (11). C AI = dim A dim R, (12) Exmples [1] su(n) lgebr In this cse, R will be the fundmentl representtion of su(n) with the genertors T normlized by Tr(T T b ) = ξδ b. The representtion R R decomposes into the sum of the djoint representtion T, with Γ T = T, nd the unit representtion I, with Γ I = 11. Note tht the normliztion of 11, given by g II = N, in generl differs from the normliztion of T! With the help of equtions (7) nd (11) we esily deduce C II = 1 N, C IT = 1 N, C T I = N2 1 N. The sme result is n immedite consequence of Eq. (12). The lst missing Fierz coefficient, C T T, follows most esily from (10): C T T = 1/N. We cn now summrize the Fierz trnsformtions for su(n) in the conventionl wy, without using rised indices, (11) ij (11) kl = 1 N (11) il(11) kj + 1 (T ) il (T ) kj, ξ (T ) ij (T ) kl = ξ N2 1 N 2 (11) il (11) kj 1 (T ) il (T ) kj. N From here, or directly from Eq. (9), we then immeditely obtin nother useful identity, T T b T = ξ N T b. Equtions (13) were derived for the fundmentl representtion of su(n). Even for higher representtions our generl formuls cn still yield some (restricted) informtion. Let us therefore consider the su(n) genertors T R in n rbitrry representtion R. From the group-theoretic point of view, they define the Clebsch Gordn coefficients for the djoint representtion T in the decomposition of the direct product R R. Their norm is usully denoted s C(R), tht is, Tr(T RT b R) = C(R)δ b. Eq. (12) gives C T I = (N 2 1)/ dim R. Applying Eq. (9) to B = I then results in the conventionl qudrtic Csimir invrint, expressed s T R T R = C 2(R)11 R, where C 2 (R) = C(R) N2 1 dim R. (13)

5 [2] Dirc lgebr Proceeding in the sme mnner one cn derive the Fierz identities for n rbitrry mtrix lgebr. Here we quote some results for the lgebr of Dirc mtrices, without detiled derivtion, which is strightforwrd, though bit tedious. The stndrd Lorentz-covrint bsis of 4 4 mtrices is creted from the sclr, vector, tensor, xil-vector, nd pseudosclr combintions of Dirc γ-mtrices, {11, γ µ, σ µν, γ µ γ 5, iγ 5 }, where we use the conventions σ µν = i 2 [γµ, γ ν ] nd γ 5 = iγ 0 γ 1 γ 2 γ 3. In mny physicl pplictions, the Lorentz symmetry is nevertheless broken by the presence of dense medium down to mere rottion symmetry. One therefore needs to work with the bsis of rottion-covrint mtrices, tht is, {11, γ 0, γ, σ 0, σ b, γ 0 γ 5, γ γ 5, iγ 5 }. The indices, b now run from one to three. The resulting Fierz identities for rottion-invrint biliners re summrized in Fig. 1. The first mtrix gives the Fierz coefficients defined by (6), while the second mtrix correspond to the Fierz trnsformtion to the prticle prticle chnnel, discussed below. Fierz trnsformtion in the prticle prticle chnnel Of the four fields in the expression (1), two trnsform in the representtion R nd two in its complex conjugte R. In the bove two-step construction we creted invrints of given symmetry from the product representtions (R R) nd (R R). The two possible wys to compose R R gve rise to the Fierz rerrngement identity (6). However, we cn lso construct n invrint by first composing (R R) nd (R R), nd then combining those. We will for simplicity refer to this rerrngement s the prticle prticle one, for obvious resons. The product representtion R R nturlly hs different bsis thn Γ A ; we will denote the corresponding mtrices s Ξ A. The new Fierz trnsformtion will therefore be defined nlogously to (6) s (Γ A ) ij(γ A ) kl = D AB (Ξ B b ) ik(ξ Bb ) lj. (14) B The mtrices Ξ, forming the bsis of R R, re most generlly defined by (ψξχ T ) = χ T Ξψ. Note tht we will not derive the Fierz trnsformtion to the prticle prticle chnnel in the most generl cse, nlogously to Eq. (4). The generliztion is obvious, but of little prcticl utility. b The bsis mtrices Ξ A will be ssumed to be normlized similrly to (8), Tr(Ξ A ΞB b ) = δab h b, nd the metric h b will gin be used to rise nd lower indices. Anlogously to Eqs. (7) nd (9) we derive the identities D AB = Tr [ Γ A Ξ B b (Γ A ) T Ξ Bb ], (15)

6 Γ A ΞB b (ΓA ) T = D AB Ξ B b. (16) The coefficients C AB nd D AB rise from rerrngement of the sme mtrices. It is therefore not surprising tht they re relted to ech other. To see this, let us pply the identity (6) to the left-hnd side of Eq. (16), [ Γ A Ξ B b (Γ A ) T] = (Γ A ij ) ik (Γ A ) jl (Ξ B b ) kl = C AC (Γ C c) il (Γ Cc ) jk (Ξ B b ) kl =,k,l C c,k,l = [ C AC Γ C c (Ξ B b )T (Γ Cc ) T]. ij C c Being bsis of the representtion R R, the mtrices Ξ B b re lwys either symmetric or ntisymmetric, tht is, (Ξ B b )T = η B Ξ B b, where the sign η B = ± depends only on the irreducible representtion B. Applying Eq. (16) to the first nd lst expression bove, we obtin the reltion D AB = η B C AC D CB. (17) Setting A to the unit representtion I, we obtin useful specil cse D IA = 1 dim R, C 1 = η B D AB. (18) The second identity follows from (12). Although the reltions (17) nd (18) only constrin the coefficients D AB, in some specil cses they my ctully be sufficient to determine D AB completely. A su(n) lgebr The product R R of two fundmentl representtions of su(n) decomposes into two irreducible representtions, the symmetric nd ntisymmetric tensors, S nd A. We will denote the corresponding bsis mtrices respectively s S nd A. Note tht A re simply the ntisymmetric T mtrices, while S re the symmetric T mtrices together with the unit mtrix, normlized to hve the sme norm ξ, tht is, 11 ξ/n. Now the first identity in (18) immeditely yields D IS = D IA = 1/N. Given tht η S = 1 nd η A = 1, the second identity then implies D T S = N 1 N, D T A = N + 1 N. We cn therefore summrize the Fierz identities nlogous to (13), (11) ij (11) kl = 1 (S ) ik (S ) lj + 1 (A ) ik (A ) lj, ξ ξ (T ) ij (T ) kl = N 1 (S ) ik (S ) lj N + 1 (A ) ik (A ) lj. N N

7 Prticle prticle chnnel for (pseudo)rel representtions Often, the representtion R is (pseudo)rel, i.e., it is equivlent to its complex conjugte. Then the mtrices Ξ A cn be directly relted to ΓA nd the coefficients D AB to C AB. Let us ssume tht the equivlence is provided by the unitry mtrix Q, tht is, ψ trnsforms in the sme wy s Qψ T. Then the nturl choice for the mtrices Ξ A is ΞA = ΓA Q, nd ccordingly Ξ A = QΓA. (We tcitly ssume ΓA = ΓA, which cn lwys be ensured by suitble definition.) (Anti)symmetry of the mtrices Ξ A, encoded in the sign η A, then implies (Γ A ) T = η A (Q T ) 1 Γ A Q. According to the generl theory of Lie lgebr representtions, the mtrix Q itself is either symmetric or ntisymmetric. In fct, it is symmetric if the representtion R is rel, nd ntisymmetric if R is pseudorel. We shll therefore write generlly Q T = η Q Q. Then (Γ A )T = η Q η A Q 1 Γ A Q. The coefficients D AB now follow from Eq. (15), D AB = Tr [ Γ A Ξ B b (Γ A ) T Ξ Bb ] = ηa η Q Tr [ Γ A Γ B b QQ 1 Γ A QQΓ Bb]. This lredy looks lmost like the expression (7) for C AB, were it not for the mtrix product QQ. However, this must ctully be proportionl to the unit mtrix. To see this, note tht since ψ = Qψ T trnsforms in the sme wy s ψ, ψ ψ must be group invrint. An esy mnipultion shows tht ψ ψ = ψ T QQψ T = ±ψqqψ, the ± sign referring to bosons or fermions. The invrince of the lst expression requires tht QQ commutes with ll mtrices of the representtion R, hence by Schur s lemm must be proportionl to the unit mtrix s long s the representtion R is irreducible. Let us write QQ = η11, where η is complex unit since Q is unitry. We then immeditely obtin the simple finl result D AB = η A η Q ηc AB. (19) Specificlly for the lgebr of Dirc γ-mtrices, Q is the chrge conjugtion mtrix. In the stndrd Dirc representtion we find η Q = η = 1, the coefficients D AB re thus relted to C AB simply by the sign, defining the (nti)symmetry of the representtion A. The vlues of ll Fierz coefficients re summrized in Fig. 1.

8 δ ij δ kl (γ 0 ) ij (γ 0 ) kl (γ ) ij (γ ) kl (σ 0 ) ij (σ 0 ) kl 1 2 ( σ b ) ij (σ b) kl (γ 0 γ 5 ) ij (γ 0 γ 5 ) kl (γ γ 5 ) ij (γ γ 5 ) kl (iγ 5 ) ij (iγ 5 ) kl = 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 3/4 3/4 1/4 1/4 1/4 3/4 1/4 3/4 3/4 3/4 1/4 1/4 1/4 3/4 1/4 3/4 3/4 3/4 1/4 1/4 1/4 3/4 1/4 3/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 3/4 3/4 1/4 1/4 1/4 3/4 1/4 3/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 δ il δ kj (γ 0 ) il (γ 0 ) kj (γ ) il (γ ) kj (σ 0 ) il (σ 0 ) kj 1 2 ( σ b ) il (σ b) kj (γ 0 γ 5 ) il (γ 0 γ 5 ) kj (γ γ 5 ) il (γ γ 5 ) kj (iγ 5 ) il (iγ 5 ) kj δ ij δ kl (γ 0 ) ij (γ 0 ) kl (γ ) ij (γ ) kl (σ 0 ) ij (σ 0 ) kl 1 2 ( σ b ) ij (σ b) kl (γ 0 γ 5 ) ij (γ 0 γ 5 ) kl (γ γ 5 ) ij (γ γ 5 ) kl (iγ 5 ) ij (iγ 5 ) kl = 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 3/4 3/4 1/4 1/4 1/4 3/4 1/4 3/4 3/4 3/4 1/4 1/4 1/4 3/4 1/4 3/4 3/4 3/4 1/4 1/4 1/4 3/4 1/4 3/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 3/4 3/4 1/4 1/4 1/4 3/4 1/4 3/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 C ik C lj (γ 0 C) ik (Cγ 0 ) lj (γ C) ik (Cγ ) lj (σ 0 C) ik (Cσ 0 ) lj 1 2 ( σ b C ) ik (Cσ b) lj (γ 0 γ 5 C) ik (Cγ 0 γ 5 ) lj (γ γ 5 C) ik (Cγ γ 5 ) lj (iγ 5 C) ik (icγ 5 ) lj Figure 1: Fierz trnsformtion of rottion-invrint Dirc biliners into the prticle ntiprticle nd prticle prticle chnnels.