Managing gamma_5 in Dimensional Regularization and ABJ Anomaly
aa r X i v : . [ h e p - t h ] F e b To Roman Jackiw
Managing γ in Dimensional Regularization and ABJ Anomaly Ruggero Ferrari Center for Theoretical PhysicsLaboratory for Nuclear Scienceand Department of PhysicsMassachusetts Institute of TechnologyCambridge, Massachusetts 02139andINFN, Sezione di Milanovia Celoria 16, I-20133 Milano, Italy(MIT-CTP4517 March 2014)
Abstract
An integral representation is proposed for the trace involv-ing γ in dimensional regularization. Lorentz covariance is pre-served. ABJ anomaly naturally follows. The Local FunctionalEquation associated to x-dependent chiral transformations is ver-ified.PACS: 11.10.Gh, 11.30.Rd, 11.40.Ha This work is supported in part by funds provided by the U.S. Departmentof Energy (D.O.E.) under cooperative research agreement e-mail: [email protected] Introduction
In dimensional regularization (Ref. [1],[2] and [3]) γ has always been a verydifficult object to deal with. Many important contributions to the topicsare present in the literature. We provide an uncommented list of works [4],[1], [5]-[27], which is far from being complete.In this paper an integral representation of the trace involving γ is sug-gested T r ( p . . . p N γ )= i D ( D − K Z d D χ d N ¯ c exp (cid:16) D X µ =1 N X i =1 ¯ c i p iµ χ µ + N X i We consider the trace of a generic product of gamma’s [34]-[38], where theindices are saturated by vectors p jµ ( p j is a standard notation for γ µ p jµ ).The discrete index j runs over the set of integers { , . . . , N } , while the“component index” µ is an element of the set { , . . . , D } (of N ). Our aim isto find an integral representation for the trace with and without γ χ .We prove the integral representation by showing the validity of thegamma’s algebra (Clifford) (with no mention to the dimension D ). Forinteger D we prove the property of cyclicity and evaluate the algebra of γ χ .We use the standard properties of integration on Grassmannian realcoordinates Z d ¯ c = 0 Z d ¯ c ¯ c = 1¯ c ′ ≡ − ¯ c, = ⇒ Z d ¯ c ′ ¯ c ′ = 1 . (3) N is even and no γ χ is present The trace can be written in terms of an integral over a set of Grassmannianvariables ¯ c j T r ( p . . . p N ) = K Z d N ¯ c exp (cid:16) N X i T r ( {6 p , p } 6 p . . . p N ) = K Z d N ¯ c exp (cid:16) N X i T r ( p . . . p N p ) = K Z d ¯ c N . . . d ¯ c exp (cid:16) N − X i =1 ,i T r ( p p . . . p N − γ χ p N )= ( − ) N − i D ( D − K Z d D χ d N ¯ c N − Y j =1 (cid:16) − c j ( p j p N )¯ c N (cid:17) exp (cid:16) D X µ =1 N X i =1 ¯ c i p iµ χ µ + N − X i =1 ,i T r ( p p . . . p N γ χ ) = ( − ) ( D − T r ( p . . . p N p γ χ ) . (41)Let us give a formal proof of the above equation by using the represen-tation (40) T r ( p . . . p N p p N +1 . . . p N + D )= K Z d ¯ c ( N + D ) . . . d ¯ c N . . . d ¯ c exp (cid:16) N − X i =1 ,i We use the algebra for γ χ developed in Section 4 in order to evaluate theABJ anomaly [28][29]. We consider a massless fermion triangle, where onevertex is given by an axial current. Thus we consider the integral ( p is theincoming momentum on the vertex σ and k on ρ ; crossed graph will beconsidered at the end) T µρσ ( k, p ) = − i Z d D q (2 π ) D T r n γ µ γ χ ( q − k ) α γ α γ ρ q β γ β γ σ ( q + p ) ι γ ι o ( q − k ) q ( q + p ) (64)Now we use Feynman parametrization and get − iT r (cid:16) γ µ γ χ γ α γ ρ γ β γ σ γ ι (cid:17) Z dx Z x dy Z d D q (2 π ) D ( q + r − k ) α ( q + r ) β ( q + r + p ) ι h q + k y + p x − p y − ( ky − px + py ) i − , (65) r ν ≡ ( yk − xp + yp ) ν . (66)We use the simplified case k = p = 0 . (67)After symmetric integration over q we can split the integral into a divergent − i D Z dx Z x dy (cid:16) δ αβ ( r + p ) ι + δ αι r β + δ βι ( r − k ) α (cid:17)(cid:16) − i (4 π ) (cid:17)h D − γ + 2 − ln 4 π + ln 2 pky ( y − x ) i (68)and finite part − i Z dx Z x dy ( r − k ) α r β ( r + p ) ι Z d D q (2 π ) D q − kpy ( y − x )) = − π ) pk Z dx Z x dy ( r − k ) α r β ( r + p ) ι y ( y − x ) . (69)In front of the two amplitudes (68) and (69) the gamma’s trace must beexpanded in powers of ( D − 4) as required by eq. (64). For the finite partin eq. (69) we can use the D = 4 expression, but for the divergent partone needs also the linear part. Let us use the representation of γ χ provided16y eq. (18) in order to tackle the problem. According to the discussionof the Section 4 we know that the algebra of γ χ for non-integer D is rathercomplicated as shown in eqs. (24) and (25). Then we use instead the algebraof the gamma’s in eq. (11) which has been proved valid for generic D ; i.e.we do not change the relative position of γ χ with respect to the remainingfactors in the trace. Thus in eq. (64), according to eq. (68) we evaluate T r (cid:16) γ µ γ χ γ α γ ρ γ β γ σ γ ι (cid:17) δ αβ = [ − − ( D − T r (cid:16) γ µ γ χ γ ρ γ σ γ ι (cid:17) T r (cid:16) γ µ γ χ γ α γ ρ γ β γ σ γ ι (cid:17) δ βι = [ − − ( D − T r (cid:16) γ µ γ χ γ α γ ρ γ σ (cid:17) T r (cid:16) γ µ γ χ γ α γ ρ γ β γ σ γ ι (cid:17) δ αι = T r (cid:16) γ µ γ χ h (2 − ( D − γ ρ γ β γ σ − δ ρβ γ σ − δ ρσ γ β + δ σβ γ ρ ) i(cid:17) . (70)We collect the non-zero part associated to the amplitude (68). Start with eq. (68) − i D Z dx Z x dyT r (cid:16) γ µ γ χ γ α γ ρ γ β γ σ γ ι (cid:17)(cid:16) δ αβ ( r + p ) ι + δ αι r β + δ βι ( r − k ) α (cid:17)(cid:16) − i (4 π ) (cid:17)h D − γ + 2 − ln 4 π + ln 2 pky ( y − x ) i = 1(4 π ) D T r (cid:16) γ µ γ χ γ ρ γ σ γ ι (cid:17) Z dx Z x dy (cid:16) r + p − k ) + ( D − r + p − k ) (cid:17) ι h D − γ + 2 − ln 4 π + ln 2 pky ( y − x ) i . (71)By using Z dx Z x dy h yk − xp + yp ) + p − k i = 0 Z dx Z x dy ( yk − xp + yp + p − k ) = 13 ( p − k ) (72)one gets − i i (4 π ) D T r (cid:16) γ µ γ χ γ ρ γ σ γ ι (cid:17) Z dx Z x dy (cid:16) r + p − k ) + ( D − r + p − k ) (cid:17) ι D − γ + 2 − ln 4 π + ln 2 pky ( x − y ) i = 1(4 π ) T r (cid:16) γ µ γ χ γ ρ γ σ γ ι (cid:17) Z dx Z x dy r + p − k ) ι ln y ( x − y ) + 23 ( p − k ) ι ! . (73)Finally we use Z dx Z x dy ln[ y ( x − y )] = − Z dx Z x dy y ln[ y ( x − y )] = − Z dx x Z x dy ln[ y ( x − y )] = − 89 (74)and get − i i (4 π ) T r (cid:16) γ µ γ χ γ ρ γ σ γ ι (cid:17) Z dx Z x dy r + p − k ) ι ln y ( x − y ) + 23 ( p − k ) ι ! = 1(4 π ) T r (cid:16) γ µ γ χ γ ρ γ σ γ ι (cid:17) − 13 ( p − k ) ι + 23 ( p − k ) ι ! = 1(4 π ) T r (cid:16) γ µ γ χ γ ρ γ σ γ ι (cid:17) 13 ( p − k ) ι . (75)The divergence of the axial current is obtained by multiplying with ( p + k ) µ − i ( k + p ) µ i (4 π ) T r (cid:16) γ µ γ χ γ ρ γ σ γ ι (cid:17) 13 ( p − k ) ι = − π ) T r (cid:16) γ χ γ ρ pγ σ k (cid:17) . (76)By adding the crossed graph( k + p ) µ (cid:16) T DIV µρσ ( k, p ) + T DIV µσρ ( p, k ) (cid:17) = − 23 1(4 π ) T r (cid:16) γ χ γ ρ pγ σ k (cid:17) . (77) Further finite terms can be evaluated directly at D = 4. From eq. (65)the finite integral contribution to the triangular graph is (always in the case18 = p = 0) T FIN µρσ ( k, p )= − iT r n γ µ γ γ α γ ρ γ β γ σ γ ι o Z dx Z x dy ( k ( y − 1) + p ( y − x )) α ( ky + p ( y − x )) β ( ky + p ( y − x + 1)) ι ( − i (4 π ) ) 14 pky ( y − x ) (78)By repeated use of the identity kγ ρ k = − k γ ρ + 2 k ρ k. (79)one shows that only two forms T r n γ µ γ γ α γ ρ γ β γ σ γ ι o p α k β p ι = − p µ T r n γ γ ρ k γ σ p o T r n γ µ γ γ α γ ρ γ β γ σ γ ι o k α p β k ι = − k µ T r n γ γ ρ p γ σ k o (80)give non-zero contribution to the divergence of the current( k + p ) µ T FIN µρσ ( k, p )= − iT r n ( k + p ) µ γ µ γ γ α γ ρ γ β γ σ γ ι o Z dx Z x dy ( k ( y − 1) + p ( y − x )) α ( ky + p ( y − x )) β ( ky + p ( y − x + 1)) ι ( − i (4 π ) ) 14 pky ( y − x )= 1(4 π ) T r n γ γ ρ p γ σ k o Z dx Z x dy (( y − − ( y − x + 1))= − π ) T r n γ γ ρ p γ σ k o . (81)The contribution of the finite integral to the divergence is then( k + p ) µ ( T FIN µρσ ( k, p ) + T FIN µσρ ( p, k )) = − π ) T r n γ γ ρ p γ σ k o 43 (82)Finally the sum of the contributions in eq. (77) and (82) is( k + p ) µ ( T µρσ ( k, p ) + T µσρ ( p, k )) = − π ) T r n γ γ ρ p γ σ k o , (83)which agrees with the ABJ anomaly.19 Local Functional Equation Once we have discovered that γ χ has a complicated behavior in D dimension,we must test our formalism in the path integral. The LFE has been discussedat length in Ref. [30]. Here we give the essential steps. The functional is Z [ A ] = Z Y x Y µ d ¯ ψ µ ( x ) Y µ ′ dψ µ ′ ( x ) e i S [ A ] (84)where the action ( e = 1) is function of the external vector field A µ ( x ) S = Z d D x ¯ ψ ( i ∂ − 6 A ) ψ. (85)The path integral measure is Lorentz invariant. Moreover it is invariantunder the U (1) local chiral transformations ψ → e iα ( x ) γ χ ψψ † → ψ † e − iα ( x ) γ χ (86)since the Jacobian of the transformation is equal one. In fact Y µ dψ µ → det( e iα ( x ) γ χ ) Y µ dψ µ = e iα ( x ) T r ( γ χ ) Y µ dψ µ . (87)Thus if we perform a substitution in the path integral variables accordingto eq. (86) the functional Z does not change. For infinitesimal parameter α one gets D(cid:16) − ¯ ψγ γ χ γ ( i ∂ − 6 A ) ψ + ¯ ψ ( i ∂ − 6 A ) γ χ ψ − i∂ µ ( ¯ ψγ µ γ χ ψ ) (cid:17)E = 0 , (88)where the brackets h· · ·i denote the mean value with the path integral mea-sure of eq. (84).If one uses the naive commutation relations of γ χ (i.e. { γ χ , γ µ } = 0) thefirst two first terms from the left in eq. (88) cancel out.In Section 2 we have found that γ χ has complicated behavior. Then onemust evaluate at one loop the expressions in eq. (88) according the rules ofeqs. (24) and (25). The results will be compared with eq. (83).A single interaction insertion gives zero since it depends only on k or p ; never on both. Consequently no completely antisymmetric tensor canemerge. We need two insertions: the triangular graph. We consider onlythe one that can provide some non-zero contributions T ρσ ( k, p ) = Z d D q (2 π ) D T r h ✘✘✘✘ ( q − k ) γ χ ✘✘✘✘ ( q − k ) γ ρ ✁ q γ σ ✘✘✘✘ ( q + p ) i ( q − k ) q ( q + p ) . (89)20he gamma’s algebra gives T r h ✘✘✘✘ ( q − k ) γ χ ✘✘✘✘ ( q − k ) γ ρ ✁ q γ σ ✘✘✘✘ ( q + p ) i = − T r h γ χ ✘✘✘✘ ( q − k ) ✘✘✘✘ ( q − k ) γ ρ ✁ q γ σ ✘✘✘✘ ( q + p ) i + T r h γ χ n ✘✘✘✘ ( q − k ) , ✘✘✘✘ ( q − k ) γ ρ ✁ q γ σ ✘✘✘✘ ( q + p ) oi (90)In the first term of the RHS the dependence on k disappears, thus it can beneglected. We consider the remaining terms T r h γ χ n ✘✘✘✘ ( q − k ) , ✘✘✘✘ ( q − k ) γ ρ ✁ q γ σ ✘✘✘✘ ( q + p ) oi = 2( q − k ) T r h γ χ γ ρ ✁ q γ σ ✘✘✘✘ ( q + p ) i − q − k ) ρ T r h γ χ ✘✘✘✘ ( q − k ) ✁ q γ σ ✘✘✘✘ ( q + p ) i +2( q − k ) qT r h γ χ ✘✘✘✘ ( q − k ) γ ρ γ σ ✘✘✘✘ ( q + p ) i − q − k ) σ T r h γ χ ✘✘✘✘ ( q − k ) γ ρ ✁ q ✘✘✘✘ ( q + p ) i +2( q − k )( q + p ) T r h γ χ ✘✘✘✘ ( q − k ) γ ρ ✁ qγ σ i (91)All terms containing ( q ) or ( ✘✘✘✘ ( q − k ) ) or ( ✘✘✘✘ ( q + p ) ) should be neglectedsince no ǫ term can emerge. Thus T r h γ χ n ✘✘✘✘ ( q − k ) , ✘✘✘✘ ( q − k ) γ ρ ✁ q γ σ ✘✘✘✘ ( q + p ) oi = 2( q − k ) T r h γ χ γ ρ ✁ q γ σ ✁ p i +2( q − k ) ρ T r h γ χ k ✁ q γ σ p i − q − k ) qT r h γ χ ✘✘✘✘ ( p + k ) γ ρ γ σ ✘✘✘✘ ( q + p ) i +2( q − k ) σ T r h γ χ k γ ρ q p i − q − k )( q + p ) T r h γ χ k γ ρ ✁ qγ σ i (92)Now we shift q q −→ q + r r ≡ yk − xp + yp (93)and we drop all terms that are zero as a result of the symmetric integration T r h γ χ n ✘✘✘✘ ( q − k ) , ✘✘✘✘ ( q − k ) γ ρ ✁ q γ σ ✘✘✘✘ ( q + p ) oi = q D T r h γ χ γ ρ ✘✘✘✘✘ ( r − k ) γ σ ✁ p i + DT r h γ χ γ ρ ✁ r γ σ ✁ p i T r h γ χ ✓ k γ ρ γ σ ✁ p i − T r h γ χ ✘✘✘✘ ( p + k ) γ ρ γ σ ✘✘✘✘✘ (2 r − k ) i − D (1 − x ) T r h γ χ ✓ k γ ρ γ σ ✁ p i + T r h γ χ ✓ k γ ρ γ σ ✁ p i − T r h γ χ ✓ k γ ρ ✭✭✭✭✭✭ (2 r + p − k ) γ σ i − DT r h γ χ ✓ k γ ρ ✁ rγ σ i! = q D y − T r h γ χ γ ρ ✓ k γ σ ✁ p i + DyT r h γ χ γ ρ ✓ k γ σ ✁ p i + T r h γ χ ✓ k γ ρ γ σ ✁ p i − (2 x − T r h γ χ ✁ p γ ρ γ σ ✓ k i − D (1 − x ) T r h γ χ ✓ k γ ρ γ σ ✁ p i + T r h γ χ ✓ k γ ρ γ σ ✁ p i − [2( y − x ) + 1] T r h γ χ ✓ k γ ρ ✁ pγ σ i + D ( x − y ) T r h γ χ ✓ k γ ρ ✁ pγ σ i! ≃ q D T r h γ χ γ ρ ✓ k γ σ ✁ p i( y − 1) + yD − − (2 x − − x ) D − − y − x ) − x − y ) D o = q D T r h γ χ γ ρ ✓ k γ σ ✁ p in − D o . (94)The factor D − γ χ anti-commuteswith all γ µ and therefore T ρσ ( k, p ) is zero from start in eq. (89).The integration over x, y gives= q D T r h γ χ γ ρ ✓ k γ σ ✁ p i D − q D − D T r h γ χ γ ρ ✓ k γ σ ✁ p i (95)Now we multiply by 2 (the Feynman parameter), 2 (the crossed graph), − i (4 π ) D − from q integration. We get2 i (4 π ) T r h γ χ γ ρ p γ σ k i . (96)Thus the results in eqs. (83) and (96) do satisfy the LFE identity in eq.(88). Acknowledgements I gratefully acknowledge the warm hospitality of the Department of Physicsof the University of Pisa and of the INFN, Sezione di Pisa.22 eferences [1] G. ’t Hooft and M. J. G. Veltman, “Regularization And Renormalization OfGauge Fields,” Nucl. Phys. B , 189 (1972).[2] C. G. Bollini and J. J. Giambiagi, “Dimensional Renormalization: The Num-ber Of Dimensions As A Regularizing Nuovo Cim. B , 20 (1972).[3] G. M. Cicuta and E. Montaldi, “Analytic Renormalization Via ContinuousSpace Dimension,” Lett. Nuovo Cim. , 329 (1972).[4] L. 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