Chiral anomaly and decay of the pion

Question

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

I) 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.

II) 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 $P(\pi^0)=-1$.

III) 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 $$ \langle 0 | j^{\mu5a}(x) | \pi^b(p) \rangle~=~ -i p^{\mu} f_{\pi} \delta^{ab} e^{-ip\cdot x}, \tag{19.88}\label{eq:19.88} $$ where $a,b$ are isospin indices and $f_{\pi}$ is a constant [...]. As a consistency check of eq. $\eqref{eq:19.88}$, 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 $j^{\mu5}$.

Answer 2

Neutral Pion would not decay (in Your discussed case) only if the constituent quarks forming the neutral pion were massless. The professional answer to Your question can be found here: http://www.scholarpedia.org/article/Axial_anomaly . It clearly states (between Eq. 14 - 15 and Eq. 24 - 25), that zero mass of Fermi-Dirac field is required for the axial current conservation at classical level. If that is the case, quantum effects still allow the pseudoscalar bound state (particles) to decay into two photons due to the quantum anomaly.

However, quarks are not massless, and neutral pion can decay even at classical level into two gammas, in the same way, as para-positronium (e+e-) quantum state annihilates = decays into two gammas. Nobody speaks about anomalies in the para-positronium case, since mass of electron is clearly 0.5 MeV. In this sense, axial anomaly role in the decay of neutral pion may seem to be overestimated. R.Jackiw says (physical pion's mass can be accurately described as "approximately" vanishing) in his explanation (http://www.scholarpedia.org/article/Axial_anomaly). Neutral Pi0 mass is 135MeV, which is 135x more massive than the Positronium is. If one finds 135 MeV to be approximately vanishing, than Pi0 decay can be related to the quantum anomaly, I think.