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

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?

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.

• Thanks. This makes sense. Actually the original homework question was to determine whether the process can happen by the electromagnetic interaction without it saying that parity is conserved or not. So in that case I have to determine whether parity was conserved given $J^{PC}=1$ for $\rho^0$. Now how can I determine the value of $L$? – TaeNyFan Apr 5 at 20:58
• @TaeNyFan, Well, you know if parity is conserved $L$ must be odd. Can $L\geq 3$ plus $S=1$ ever produce $J=1$? The electromagnetic interaction conserves parity, just to be clear. – octonion Apr 5 at 21:00

$$\let\g=\gamma \def\ve{\vec\varepsilon} \def\vp{\vec p}$$ 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.)