Parity in $\rho^0 \rightarrow \pi^0+\gamma$ decay

Question

I am doing a homework question for $\rho^0 \rightarrow \pi^0+\gamma$ decay.

It is given the $J^{PC}=1^{--}$ for the $\rho^0$ meson and that parity is conserved for this process.

To calculate the parity for $\pi^0\gamma$ system, I tried $$P_{\pi^0}P_{\gamma}(-1)^L=(-1)(-1)(-1)^0=+1$$ where $L$ is the angular momentum. But this is wrong.

I chose $L=0$ because the intrinsic spin of $\pi_0=0$ and intrinsic spin of $\gamma$ photon is 1. By conservation of angular momentum the total angular momentum $J$ of $\pi^0\gamma$ needs to be $1$. So this means orbital angular momentum $L$ for $\pi^0\gamma$ is $0$.

What is wrong with my line of thought?

Edit: Also, if the question did not explicitly state that parity is conserved, can one still deduce the parity of the $\pi^0\gamma$ system?

Answer 1

Related: Why can the pion decay into two photons?

You've shown by parity that you must have $L=1$. And remember quantum mechanically spin $S=1$ and $L=1$ can add to $J=0, 1,2$, depending on the details of the wavefunction. So there is no contradiction with having total $J=1$ as required by conservation.

Answer 2

$\let\g=\gamma \def\ve{\vec\varepsilon} \def\vp{\vec p}$ Decays with photons in the final state require additional care because of photon's zero mass, which forbids 0 helicity. Then you can't simply deal with photons as if they were spin-1 particles.

As an instance, consider a decay very like the one you proposed: $$K^0 \to \pi^0 + \g$$ (I'll neglect $C$ to simplify argument. After all, $K^0$ exists in both $C$-eigenstates.)

Knowing that $K^0$ is $J^P=0^-$ and reasoning as you did for $\rho^0$ we could say that parity conservation requires odd $L$ and it's ok, since by composing $L=1$ and $S=1$ (for photon) $J=0$ can be formed.

Yet that decay doesn't exist. Why?

The simplest way to understand it is to try to build the final wavefunction in momentum representation. Take as initial state 0-momentum for $K^0$. Let's call $\vp$ photon's momentum (pion momentum is $-\vp$). Photon polarization (helicity) can be represented by a polarization vector $\ve$. Then a $J=0$ (i.e. scalar) wavefunction must be $$\ve\cdot\vp\>f(p^2).$$ But e.m. waves are transverse, so real longitudinal photons do not exist: $$\ve\cdot\vp=0$$ qed.

BTW, the same argument for $\rho^0$-decay works well. We have to build a $1^-$ (vector) wavefunction and it's simply $$\ve\>f(p^2).$$ (Note a catch: although $\ve$ represents the so-called "spin" of a photon, it's a polar vector, not an axial one as one could naïvely believe for a spin.)