# Spin of a neutral pion

I was going through the book High Energy Physics by DH Perkins. In the decay $\pi^{0}\rightarrow 2\gamma \ (photon)$, the possible spin components of the system of photons $S_z$ can be $0$ or $2$.

A statement in the book says, if the spin of of the pion is $1$, then $S_z$ has to be zero. Why is this true?

Thanks in advance.

## 1 Answer

Since it decays to two photons, in the center of mass the photons will be back to back from conservation of momentum, i.e. zero rotational angular momentum. From the two allowed additions of the spin orientation of the collinear opposite momentum photons, one has to chose 0.

I have the 1982 edition, and it goes on to exclude an angular momentum between the two photons ( which I consider self evident by collinearity and the point particle definition of the standard model).

In page 85:

The argument explores the possibility that there exists an angular momentum in addition to the spin of the two photons and excludes it, thus the spin 1 of pi0 has to be 0 or or equal/larger than 2. i.e it excludes spin 1.

• Then in the frame of reference of the pion, I think total angular momentum J=L+S is not conserved. J=1 for the pion and for the photons J=L+S=0+0. So I think we cannot have S_z=0 as well. Where am I going wrong? – Tejas P Feb 6 '18 at 13:22
• well yes; That is why we say the pi0 spin has to be zero. I say there is no L for the gammas in the center of mass of the pi0,. Perkins explores if there could be an L , and still the spin of the pi0 cannot be 1 is his conclusion. Think of the rho0, also with quark antiquark annihilating in its decay. The rho has spin 1, but it does not go to two gamma, because the angular momentum cannot be conserved in the center of mass of the rho0 for two gammas – anna v Feb 6 '18 at 13:29