quantum field theory - Chiral anomaly and decay of the pion

I am told that if all classical symmetries were reflected as quantum symmetries, the decay of the neutral pion $$\pi^0 ~\longrightarrow~ \gamma\gamma$$ would not happen. Why would the conservation of the axial current in QED prevent the decay of the pion? What is the non-conserved charge in this decay?

Answer

1) The axial vector current $j^{\mu 5}$ is a pseudovector

$$j^{\mu 5}~:=~\overline{\psi}\gamma^{\mu}\gamma^5\psi~=~j^{\mu}_R-j^{\mu}_L,\qquad j^{\mu}_{R,L}~:=~ \overline{\psi}_{R,L}\gamma^{\mu}\psi_{R,L}, $$ $$\psi_{R,L}~:=~P_{R,L}\psi,\qquad P_{R,L} ~:=~\frac{1\pm\gamma^5}{2} . $$

The $4$-divergence $d_{\mu}j^{\mu 5}$ is a pseudoscalar. That the axial current $j^{\mu 5}$ is conserved classically means that the $4$-divergence $d_{\mu}j^{\mu 5}=0$ vanishes classically, and if one defines the axial charge

$$N^5(t)~:=~N_R(t)-N_L(t),\qquad N_{R,L}(t)~:=~\int {\rm d}^3{x}~j^0_{R,L}(t,\vec{x}),$$

then $N^5(t)$ is conserved over time classically.

2) It follows from the Dirac equation that a spin $1/2$ particle and its antiparticle must have opposite intrinsic parity. Conventionally, for quarks $P(q)=1=-P(\bar{q})$. Thus the parity of a meson is

$$P({\rm meson})~=~P(q)P(\bar{q})(-1)^{L}~=~(-1)^{L+1}.$$

In particular, a pion $\pi^0$ with $J=L=S=0$ is a pseudoscalar, with parity $P(\pi^0)=-1$.

3) A pion is a bound state of a quark and an antiquark, which is difficult to directly relate to the Lagrangian density of the standard model, and ultimately to the two photons $\gamma+\gamma$. In practice, one instead studies how the $\pi^0$ and the two $\gamma's$ couple to the axial vector current $j^{\mu 5}$.

1. 
   

   Quoting Peskin and Schroeder on the bottom of page 669: We can parametrize the matrix element of $j^{\mu5a}$ between the vacuum and an on-shell pion by writing $$ \langle 0 | j^{\mu5a}(x) | \pi^b(p) \rangle~=~ -i p^{\mu} f_{\pi} \delta^{ab} e^{-ip\cdot x}, \qquad (19.88) $$ where $a,b$ are isospin indices and $f_{\pi}$ is a constant [...]. As a consistency check of eq. (19.88), note that lhs = pseudovector $\times$ pseudoscalar=vector=rhs.

   
   


2. 
   
   

   On the other hand, it is e.g. argued in Chapter 76 of Srednecki, QFT, via a LSZ formula and a Ward identity, that the $4$-divergence $$d_{\mu} \langle p,q | j^{\mu5}(x) | 0 \rangle \qquad \qquad \qquad (76.20) $$ vanishes classically, where $\langle p,q |$ is a state with two outgoing photons with $4$-momenta $p$ and $q$.

   
   

So in a nutshell, the pion decay $\pi^0\to \gamma+\gamma$ is classically forbidden because a photon two-state doesn't couple classically to the axial current $j^{\mu5}$.