# Chrial gauge theories and anomalies(Srednicki ch75-76)

1. Oct 25, 2011

### LAHLH

Hi,

Don't know if anyone can help me but have a bit of confusion with Srednicki ch75 p466 just above (75.55). I understand why in non-Abelian gauge theory we get extra factors $Tr(T^aT^bT^c)$ and so on, but I don't understand why the $P_{L}\to1/2$ diagrams then end up with the extra factor $1/2Tr([T^a,T^b],T^c)$, does anyone know?

Also then in ch77, Srednicki says the triangle diagrams analyzed now come with the extra factor $Tr(T^a T^b)$, why not $Tr(T^a T^b T^c)$?, after all they are the same diagrams he talks about of p466 (except for some changes he notes on p470 that dont seem to make a difference to this argument)

Thanks, would be really grateful if anyone is familiar with this...

2. Oct 26, 2011

### Avodyne

In QED, the two diagrams just cancel. In nonabelian theory, one diagram gets an extra factor of $Tr(T^a T^b T^c)$, and the other gets an extra factor of $Tr(T^b T^a T^c)$. These two factors do not cancel, and their difference gives the expression you wrote (up to factors of 2, which I'm not trying to get right here).

Because now one of the three external lines corresponds to the current of the global symmetry, and this does not couple to the nonabelian charge represented by a T^a matrix.

3. Oct 27, 2011

### LAHLH

Calling the first term in (75.16) : $D^{\mu\nu\rho}(p,q,r)$ for brevity ,and also taking the $P_L \to 1/2$ part of it, then in nonabelian guage theory I believe the PL 1/2 term of (75.16) generalizes to:

$$iV^{\mu\nu\rho}(p,q,r)=1/2\text{Tr}\left(T^aT^bT^c\right)D^{\mu\nu\rho}(p,q,r)+1/2\text{Tr}\left(T^bT^aT^c\right)D^{\nu \mu \rho}(q,p,r)+\mathcal{O}(g^5)$$

Now if $D^{\mu\nu\rho}(p,q,r)=-D^{\nu \mu \rho}(q,p,r)$ you could take this D out as a common factor, and be left with $1/2 r\left(T^aT^bT^c\right)-1/2r\left(T^bT^aT^c\right)=1/2Tr([T^a,T^b],T^c)$. But is this relation between the D's (integrals over the loops) true? It's not completely obvious to me if so.....

4. Nov 4, 2011

### Avodyne

Yes. Without the group matrices, and without a $\gamma_5$, the two diagrams are just the ones of QED, and they cancel, as discussed in the paragraph above eq.(75.16).

5. Nov 7, 2011

### LAHLH

Yes, I can see your logic must be correct, as if these D's didn't behave this way the cancellation wouldn't happen in QED, as you say. Seems hard to show directly just from the integrals themselves that they cancel though...

thanks again for the help

6. Nov 8, 2011

### LAHLH

Last edited by a moderator: May 5, 2017
7. Nov 8, 2011

### Avodyne

In QED, you can use charge-conjugation symmetry to show that there can't be a 3-photon amplitude. So inserting $C^{-1}C$ between every pair of gamma matrices in one expression might help you to turn it into minus the other one. ( I have not tried this myself.)