C The triangle anomaly. Letter to Jackiw, Jan 20, 1991. C Anomaly 1. Evaluation of the coefficients C(i,j), i=1,2, in terms of C0 and B0. C Anomaly 2. The axial current triangle graphs. Vertex: i*G5*G(al). C Anomaly 3. The axial current triangle graphs. Vertex: i*G5*G(al). An expansion in terms of the external momenta is done (assuming them to be small with respect to the loop mass m). The result shows that the axial current is at least of third order in the momenta. C Anomaly 4. The pseudo scalar graphs. Vertex: 2*m*G5. *end C Anomaly 1. Evaluation of the coefficients C(i,j), i=1,2, in terms of C0 and B0. P ninput BLOCK BFUN{} Id,B22(x~,M~,m~)=(-0.5*Ax(m)+M**2*B0(x,M,m) -0.5*(x+m^2-M**2)*B1(x,M,m))/[1-N] Id,B21(x~,M~,m~)=-((0.5*N-1)*Ax(m) -0.5*N*(x+m^2-M**2)*B1(x,M,m) +M**2*B0(x,M,m) )/x/[1-N] Id,B1(x~,M~,m~)= (0.5*Ax(M)-0.5*Ax(m) -0.5*(x+m^2-M**2)*B0(x,M,m) )/x ENDBLOCK BLOCK CASE6{} C Case: all masses equal. Id,M3=m Al,M2=m Al,M1=m Al,B0(x~,M1~,M2~)=B0(x,m) Id,Ax(M1~)=Ax(m) Al,B0(x~,M1~,M2~)=B0(x,m) ENDBLOCK A Pi,N,N_,m,M,M1,M2,M3,x,y V p,k F Ax D XX(I)= pDp,-pDk,-pDk,kDk X X(I,J)= XX( 2*I-2+J )/Det X f1=M1^2-M2^2-kDk X f2=M2^2-M3^2-pDp-2*pDk *fix D R1(I)=(1/2*f1*c0+1/2*B0(qDq,M1,M3)-1/2*B0(pDp,M2,M3)), (1/2*f2*c0 + B0(kDk,M1,M2)/2 - B0(qDq,M1,M3)/2) Z c11=DS(K,1,2,(X(1,K)*R1(K))) Z c12=DS(K,1,2,(X(2,K)*R1(K))) Id,Multi,pDk^2=kDk*pDp-Det Keep c11,c12 B i,Pi,Det,c0 CASE6{} *next Z c24=i*Pi^2/4-1/2*M1^2*c0+1/4*(B0(pDp,M2,M3)-f1*c11-f2*c12) Id,Multi,pDk^2=kDk*pDp-Det CASE6{} B i,Pi,Det,c0 Keep c11,c12,c24 *next D R3(I)=(1/2*f1*c11+1/2*B1(qDq,M1,M3)+1/2*B0(pDp,M2,M3)-c24), (1/2*f2*c11+1/2*B1(kDk,M1,M2)-1/2*B1(qDq,M1,M3)) D R4(I)=(1/2*f1*c12+1/2*B1(qDq,M1,M3)-1/2*B1(pDp,M2,M3)), (1/2*f2*c12-1/2*B1(qDq,M1,M3)-c24) Z c21=DS(K,1,2,(X(1,K)*R3(K))) Z c23=DS(K,1,2,(X(2,K)*R3(K))) Z C23p=DS(K,1,2,(X(1,K)*R4(K))) Z c22=DS(K,1,2,(X(2,K)*R4(K))) B i,Pi,Det,c0 BFUN{} Id,Multi,pDk^2=kDk*pDp-Det Id,N*[1-N]^-1=-1+[1-N]^-1 Id,Ax(m~)*[1-N]^-1= -1/3*Ax(m) + 2*i*Pi^2*m^2/9 Al,B0(x~,M~,m~)*[1-N]^-1= -1/3*B0(x,M,m) - 2*i*Pi^2/9 CASE6{} Keep c11,c12,c24,c21,c23,C23p,c22 Nprint C23p *next P input C Check: must be zero. Z Diff23=c23-C23p *end C Anomaly 2. The axial current triangle graphs. Vertex: i*G5*G(al). C The triangle anomaly. Computing the axial current. The expressions for the C(i,j) have been computed separately, and are contained in the block CFU. P ninput BLOCK CFU{} P ninput Id,C(1,1,m) = + Det^-1*C0(m) * ( 1/2*pDp*pDk + 1/2*pDp*kDk ) - C0(m) + B0(pDp,m)*Det^-1 * ( - 1/2*pDp ) + B0(kDk,m)*Det^-1 * ( - 1/2*pDk ) + B0(qDq,m)*Det^-1 * ( 1/2*pDp + 1/2*pDk ) Al,C(1,2,m) = + Det^-1*C0(m) * ( - 1/2*pDp*kDk - 1/2*pDk*kDk ) + B0(pDp,m)*Det^-1 * ( 1/2*pDk ) + B0(kDk,m)*Det^-1 * ( 1/2*kDk ) + B0(qDq,m)*Det^-1 * ( - 1/2*pDk - 1/2*kDk ) + 0. Al,C(2,4,m) = + 1/4*i*Pi^2 + Det^-1*C0(m) * ( - 1/4*pDp*pDk*kDk - 1/8*pDp*kDk^2 - 1/8*pDp^2*kDk ) + C0(m) * ( - 1/2*m^2 ) + B0(pDp,m)*Det^-1 * ( 1/8*pDp*pDk + 1/8*pDp*kDk ) + B0(kDk,m)*Det^-1 * ( 1/8*pDp*kDk + 1/8*pDk*kDk ) + 1/4*B0(qDq,m) + B0(qDq,m)*Det^-1 * ( - 1/8*pDp*pDk - 1/4*pDp*kDk - 1/8*pDk*kDk ) + 0. Al,C(2,1,m) = + i*Pi^2*Det^-1 * ( - 1/4*pDp ) + Det^-2*C0(m) * ( 3/4*pDp^2*pDk*kDk + 3/8*pDp^2*kDk^2 + 3/8*pDp^3*kDk ) + Det^-1*C0(m) * ( 1/2*m^2*pDp - pDp*pDk - pDp*kDk - 1/4*pDp^2 ) + C0(m) + B0(pDp,m)*Det^-2 * ( - 3/8*pDp^2*pDk - 3/8*pDp^2*kDk ) + B0(pDp,m)*Det^-1 * ( pDp ) + B0(kDk,m)*Det^-2 * ( - 3/8*pDp*pDk*kDk - 3/8*pDp^2*kDk ) + B0(kDk,m)*Det^-1 * ( 1/4*pDp + 3/4*pDk ) + B0(qDq,m)*Det^-2 * ( 3/8*pDp*pDk*kDk + 3/8*pDp^2*pDk + 3/4*pDp^2*kDk ) + B0(qDq,m)*Det^-1 * ( - 5/4*pDp - 3/4*pDk ) Al,C(2,3,m) = + i*Pi^2*Det^-1 * ( 1/4*pDk ) + Det^-2*C0(m) * ( - 3/8*pDp*pDk*kDk^2 - 3/8*pDp^2*pDk*kDk - 3/4*pDp^2*kDk^2 ) + Det^-1*C0(m) * ( - 1/2*m^2*pDk + pDp*kDk + 1/2*pDk*kDk ) + B0(pDp,m)*Det^-2 * ( 3/8*pDp*pDk*kDk + 3/8*pDp^2*kDk ) + B0(pDp,m)*Det^-1 * ( - 1/8*pDp - 1/2*pDk ) + B0(kDk,m)*Det^-2 * ( 3/8*pDp*pDk*kDk + 3/8*pDp*kDk^2 ) + B0(kDk,m)*Det^-1 * ( - 5/8*kDk ) + B0(qDq,m)*Det^-2 * ( - 3/4*pDp*pDk*kDk - 3/8*pDp*kDk^2 - 3/8*pDp^2*kDk ) + B0(qDq,m)*Det^-1 * ( 1/8*pDp + 1/2*pDk + 5/8*kDk ) Al,C(2,2,m) = + i*Pi^2*Det^-1 * ( - 1/4*kDk ) + Det^-2*C0(m) * ( 3/4*pDp*pDk*kDk^2 + 3/8*pDp*kDk^3 + 3/8*pDp^2*kDk^2 ) + Det^-1*C0(m) * ( 1/2*m^2*kDk - 1/4*kDk^2 ) + B0(pDp,m)*Det^-2 * ( - 3/8*pDp*pDk*kDk - 3/8*pDp*kDk^2 ) + B0(pDp,m)*Det^-1 * ( - 1/4*pDk + 1/4*kDk ) + B0(kDk,m)*Det^-2 * ( - 3/8*pDp*kDk^2 - 3/8*pDk*kDk^2 ) + B0(qDq,m)*Det^-2 * ( 3/8*pDp*pDk*kDk + 3/4*pDp*kDk^2 + 3/8*pDk*kDk^2 ) + B0(qDq,m)*Det^-1 * ( 1/4*pDk - 1/4*kDk ) + 0. P input ENDBLOCK P input V p,r,q,k I al,mu,nu,L1,L2,L3 A N,N_,Pi,m,Ax,Det F Fx,Cx,C,c,C0,Bx,B0,B1 C Triangle graphs. (k,al) => (p,mu),(q,nu) with all momenta pointing inwards. There are two graphs, differing with respect to each other by reversal of the fermion direction (or by the interchange p <=> q and mu <=> nu). Z A(al,mu,nu) = - i*Cx(m)*G5(1,2)*G(2,3,al)* (i*G(3,4,k) + i*G(3,4,r) + m*Gi(3,4)) *G(4,5,mu)* (i*G(5,6,r) + i*G(5,6,k) + i*G(5,6,p) + m*Gi(5,6)) *G(6,7,nu)* (i*G(7,1,r) + m*Gi(7,1))*Ax - i*Cx(m)*G5(1,2)*G(2,3,al)* (- i*G(3,4,r) + m*Gi(3,4)) *G(4,5,nu)* (- i*G(5,6,r) - i*G(5,6,k) - i*G(5,6,p) + m*Gi(5,6)) *G(6,7,mu)* (- i*G(7,1,r) - i*G(7,1,k) + m*Gi(7,1))*Ax Id,Gammas,"C *yep C Try to work out terms with three r, to avoid the C(3,i). Such terms have more than one r inside the trace. Move them towards each other, to produce rDr Do this by moving them to the right. Id,G(1,"t,"4,G5,al,r,L1~,L2~,L3~,L4~) = - G(1,"t,"4,G5,al,L1,r,L2,L3,L4) + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4) Id,G(1,"t,"4,G5,al,L2~,r,L1~,L3~,L4~) = - G(1,"t,"4,G5,al,L2,L1,r,L3,L4) + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4) Id,G(1,"t,"4,G5,al,L2~,L3~,r,L1~,L4~) = - G(1,"t,"4,G5,al,L2,L3,L1,r,L4) + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4) Id,G(1,"t,"4,G5,al,L2~,L3~,L4~,r,L1~) = - G(1,"t,"4,G5,al,L2,L3,L4,L1,r) + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4) Id,Gammas,"C Id,G(1,"t,"4,G5,al,r,L1~,L2~,L3~,L4~) = - G(1,"t,"4,G5,al,L1,r,L2,L3,L4) + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4) Id,G(1,"t,"4,G5,al,L2~,r,L1~,L3~,L4~) = - G(1,"t,"4,G5,al,L2,L1,r,L3,L4) + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4) Id,G(1,"t,"4,G5,al,L2~,L3~,r,L1~,L4~) = - G(1,"t,"4,G5,al,L2,L3,L1,r,L4) + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4) Id,G(1,"t,"4,G5,al,L2~,L3~,L4~,r,L1~) = - G(1,"t,"4,G5,al,L2,L3,L4,L1,r) + 2*D(L1,r)*G(1,"t,"4,G5,al,L2,L3,L4) Id,Gammas,"C *yep C There may still be some traces with two r's left. Id,G(1,"t,"4,G5,al,r,L1~,L2~) = - G(1,"t,"4,G5,al,L1,r,L2) Id,Gammas,"C *yep C Work out the rDr terms, writing rDr = rDr+m^2 - m^2. In the resulting two point function one has 1/((r+k)^2+m^2)*((r+p+k)^2+m^2). Shift r => r-k. IF Cx(m)*rDr=Bx(m)*Shift - Cx(m)*m^2 IF Shift=1 Id,r(mu~)=r(mu)-k(mu) Al,Func,r(mu~)=r(mu)-k(mu) Al,Dotpr,r(mu~)=r(mu)-k(mu) ENDIF ENDIF Id,Gammas,"C *yep C Now do the integration over the loop momentum r. Id,All,r,N,Fx Id,Adiso,Cx(m)*Fx(L1~,L2~)= + k(L1)*k(L2)*C(2,1,m) + p(L1)*p(L2)*C(2,2,m) + k(L1)*p(L2)*C(2,3,m) + p(L1)*k(L2)*C(2,3,m) + D(L1,L2)*C(2,4,m) Id,Adiso,Cx(m)*Fx(L1~) = k(L1)*C(1,1,m) + p(L1)*C(1,2,m) Al,Adiso,Bx(m)*Fx(L1~) = p(L1)*B1(pDp,m) Id,Cx(m)= C0(m) Al,Bx(m)= B0(pDp,m) Id,Gammas *yep C Work out the C(i,j). CFU{} Id,B1(x~,m) = -0.5*B0(x,m) Id,p(al)*Epf(mu,nu,la~,ka~)= p(mu)*Epf(al,nu,la,ka) + p(nu)*Epf(mu,al,la,ka) + p(la)*Epf(mu,nu,al,ka) + p(ka)*Epf(mu,nu,la,al) Al,k(al)*Epf(mu,nu,la~,ka~)= k(mu)*Epf(al,nu,la,ka) + k(nu)*Epf(mu,al,la,ka) + k(la)*Epf(mu,nu,al,ka) + k(ka)*Epf(mu,nu,la,al) Id,Multi,pDk^2=kDk*pDp-Det Id,Det=kDk*pDp-pDk^2 *yep C Write the result, i.e. the axial current, in the more standard form. Id,k(mu~)=-q(mu)-p(mu) Id,Func,k(mu~)=-q(mu)-p(mu) Id,Dotpr,k(mu~)=-q(mu)-p(mu) Id,Multi,pDq^2=qDq*pDp-Det P output *yep C Take any one of the following three options. Id,Ax=k(al) ! Taking the divergence of the axial current. C Id,Ax=p(mu) ! Must be zero: gauge invariance. C Id,Ax=q(nu) ! Must be zero: gauge invariance. Id,k(mu~)=-q(mu)-p(mu) Id,Func,k(mu~)=-q(mu)-p(mu) Id,Dotpr,k(mu~)=-q(mu)-p(mu) Id,Multi,pDq^2=qDq*pDp-Det *end C Anomaly 3. The axial current triangle graphs. Vertex: i*G5*G(al). An expansion in terms of the external momenta is done (assuming them to be small with respect to the loop mass m). The result shows that the axial current is at least of third order in the momenta. C The triangle anomaly. Computing the axial current to order 3 in the external momenta. V p,r,q,k I al,mu,nu,L1,L2,L3,L4 A N,N_,Pi,m,Ax,ax,Det,Den F Ln,Fx,Cx,C,c,C0,Bx,B0,B1 X DY(L1,L2,L3,L4)=D(L1,L2)*D(L3,L4)+D(L1,L3)*D(L2,L4)+D(L1,L4)*D(L2,L3) X DZ(L1,L2,L3,L4,L5,L6)= D(L1,L2)*DY(L3,L4,L5,L6) + D(L1,L3)*DY(L2,L4,L5,L6) + D(L1,L4)*DY(L3,L2,L5,L6) + D(L1,L5)*DY(L3,L4,L2,L6) + D(L1,L6)*DY(L3,L4,L5,L2) C Triangle graphs. (k,al) => (p,mu),(q,nu) with all momenta pointing inwards. There are two graphs, differing with respect to each other by reversal of the fermion direction (or by the interchange p <=> q and mu <=> nu). Z A(al,mu,nu) = - i*Cx(m)*G5(1,2)*G(2,3,al)* (i*G(3,4,k) + i*G(3,4,r) + m*Gi(3,4)) *G(4,5,mu)* (i*G(5,6,r) + i*G(5,6,k) + i*G(5,6,p) + m*Gi(5,6)) *G(6,7,nu)* (i*G(7,1,r) + m*Gi(7,1))*Ax - i*Cx(m)*G5(1,2)*G(2,3,al)* (- i*G(3,4,r) + m*Gi(3,4)) *G(4,5,nu)* (- i*G(5,6,r) - i*G(5,6,k) - i*G(5,6,p) + m*Gi(5,6)) *G(6,7,mu)* (- i*G(7,1,r) - i*G(7,1,k) + m*Gi(7,1))*Ax Id,Gammas,"C C Cx(m) is the three point function: 1 / (r^2+m^2)*((r+k)^2+m^2)*((r+k+p)^2+m^2) Expand the second and third denominator, use k+p = -q. Below Den stands for 1/(r^2+m^2). Id,Cx(m)=Den^3*Dev2*Dev3 Id,Dev2= 1 - (2*kDr+kDk)*Den + (2*kDr+kDk)^2*Den^2 - (2*kDr+kDk)^3*Den^3 Al,Dev3= 1 - (-2*qDr+qDq)*Den + (-2*qDr+qDq)^2*Den^2 - (-2*qDr+qDq)^3*Den^3 Id,Multi,Den^7=0 *yep Id,All,r,N,Fx C Keep up to and including third order. Id,Count,ax,"F,G,"F,Fx,k,1,q,1,p,1 Id,Multi,ax^4=0 Id,ax=1 C Doing the momentum integrals. Id,Fx(L1~,L2~,L3~,L4~,L5~)=0 Al,Fx(L1~,L2~,L3~)=0 Al,Fx(L1~)=0 Id,Fx(L1~,L2~,L3~,L4~,L5~,L6~)*Den^6 = DZ(L1,L2,L3,L4,L5,L6)*i*Pi^2/960/m^2 Al,Fx(L1~,L2~,L3~,L4~,L5~,L6~)*Den^5 = DZ(L1,L2,L3,L4,L5,L6)*(i*DEL/192-i*Pi^2/192*Ln(m)) Al,Fx(L1~,L2~,L3~,L4~,L5~,L6~)*Den^4 = DZ(L1,L2,L3,L4,L5,L6)*m^2*(-i*DEL/48+i*Pi^2/48*(-1+Ln(m))) Id,Fx(L1~,L2~,L3~,L4~)*Den^6 = i*Pi^2/480/m^4*DY(L1,L2,L3,L4) Al,Fx(L1~,L2~,L3~,L4~)*Den^5 = i*Pi^2/96/m^2*DY(L1,L2,L3,L4) Al,Fx(L1~,L2~,L3~,L4~)*Den^4 = DY(L1,L2,L3,L4)*(i*DEL/24 - i*Pi^2/24*Ln(m)) Al,Fx(L1~,L2~,L3~,L4~)*Den^3 = DY(L1,L2,L3,L4)*m^2*(-i*DEL/8 + i*Pi^2/8*(-1 + Ln(m))) Id,Fx(L1~,L2~)*Den^5 = i*Pi^2/48/m^4*D(L1,L2) Al,Fx(L1~,L2~)*Den^4 = i*Pi^2/12/m^2*D(L1,L2) Al,Fx(L1~,L2~)*Den^3 = D(L1,L2)*(i*DEL/4 - i*Pi^2/4*Ln(m)) Id,Den^5=i*Pi^2/12/m^6 Al,Den^4=i*Pi^2/6/m^4 Al,Den^3=i*Pi^2/2/m^2 *yep C Take the trace. B DEL,i,Pi Id,Gammas *yep Id,Multi,m^-4=0 Id,N=4+N_ Id,N_*DEL=-2*Pi^2 Id,N_=0 Id,k(mu~)=-q(mu)-p(mu) Al,Func,k(mu~)=-q(mu)-p(mu) Al,Dotpr,k(mu~)=-q(mu)-p(mu) *yep Id,p(al)*Epf(mu,nu,la~,ka~)= p(mu)*Epf(al,nu,la,ka) + p(nu)*Epf(mu,al,la,ka) + p(la)*Epf(mu,nu,al,ka) + p(ka)*Epf(mu,nu,la,al) Al,q(al)*Epf(mu,nu,la~,ka~)= q(mu)*Epf(al,nu,la,ka) + q(nu)*Epf(mu,al,la,ka) + q(la)*Epf(mu,nu,al,ka) + q(ka)*Epf(mu,nu,la,al) P output *yep C Take any one of the following three options. C Id,Ax=k(al) ! Taking the divergence of the axial current. C Id,Ax=p(mu) ! Must be zero: gauge invariance. Id,Ax=q(nu) ! Must be zero: gauge invariance. Id,k(mu~)=-q(mu)-p(mu) Id,Func,k(mu~)=-q(mu)-p(mu) Id,Dotpr,k(mu~)=-q(mu)-p(mu) *end C Anomaly 4. The pseudo scalar graphs. Vertex: 2*m*G5. C The triangle anomaly. Computing the pseudo scalar triangle diagrams. V p,r,q,k I al,mu,nu,L1,L2,L3 A N,N_,Pi,m,Ax,Det F Fx,Cx,C,c,C0,Bx,B0,B1 C Triangle graphs. (k) => (p,mu),(q,nu) with all momenta pointing inwards. There are two graphs, differing with respect to each other by reversal of the fermion direction (or by the interchange p <=> q and mu <=> nu). Z A(al,mu,nu) = - Cx(m)*G5(1,3)*2*m* (i*G(3,4,k) + i*G(3,4,r) + m*Gi(3,4)) *G(4,5,mu)* (i*G(5,6,r) + i*G(5,6,k) + i*G(5,6,p) + m*Gi(5,6)) *G(6,7,nu)* (i*G(7,1,r) + m*Gi(7,1))*Ax - Cx(m)*G5(1,3)*2*m* (- i*G(3,4,r) + m*Gi(3,4)) *G(4,5,nu)* (- i*G(5,6,r) - i*G(5,6,k) - i*G(5,6,p) + m*Gi(5,6)) *G(6,7,mu)* (- i*G(7,1,r) - i*G(7,1,k) + m*Gi(7,1))*Ax Id,Gammas,"C *yep C Now do the integration over the loop momentum r. Id,All,r,N,Fx Id,Adiso,Cx(m)*Fx(L1~,L2~)= + k(L1)*k(L2)*C(2,1,m) + p(L1)*p(L2)*C(2,2,m) + k(L1)*p(L2)*C(2,3,m) + p(L1)*k(L2)*C(2,3,m) + D(L1,L2)*C(2,4,m) Id,Adiso,Cx(m)*Fx(L1~) = k(L1)*C(1,1,m) + p(L1)*C(1,2,m) Id,Cx(m)= C0(m) Id,Gammas C Write the result in the more standard form. Id,k(mu~)=-q(mu)-p(mu) Id,Func,k(mu~)=-q(mu)-p(mu) Id,Dotpr,k(mu~)=-q(mu)-p(mu) Id,Multi,pDq^2=qDq*pDp-Det *end