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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?

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up vote 7 down vote accepted

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 $j^{\mu5}$.

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This is only approximately true in QCD, because the axial current is conserved. The violation of the Ward identity is not surprising because it is nonzero, but because it is an order of magnitude bigger than other chiral symmetry non-conserving things, which are suppressed by the smallness of the quark mass (this one isn't). – Ron Maimon Nov 26 '11 at 4:27

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