The spin of neutral pion

1. Nov 4, 2012

karlzr

In Perkins's Introduction to High Energy Physics, the author obtained the spin of neutral pions from the decay $$\pi^0 → 2\gamma$$ He argued that the z-component of the total photon spin in the above decay can have the value $S_z=0$ or $2$. If $s_\pi=1$, then only $S_z=0$ is possible, and the two-photon amplitude must behave under rotation like $P_1^m(\cos \theta)$ with $m=0$.
Why does the orbit angular momentum $l=1$ in his reasoning? Acually I do not understand why only $S_z=0$ is possible. Since if there is a $l_z$, then $S_z$ can be both 0 and 2.

2. Nov 4, 2012

Staff: Mentor

How do you get an lz with two photons flying back to back in z-direction? I can imagine how you get lx and ly.

3. Nov 4, 2012

karlzr

What about the total angular momentum $l=1$? This is the real issue that bothers me.

4. Nov 4, 2012

Staff: Mentor

I think the idea is that you get the total spin as combination of the photon spin ($S_z=0$) and the angular momentum (in a different direction, therefore $s_\pi \geq S_z$). But that is a bit speculative.

5. Nov 5, 2012

andrien

photon's mass is zero.the spin component along z-axis can be only +1 or -1.Since it still behaves like boson(integral spin),you should use the same rotation rotation matrix for m=0.If s=1,for pion then of course,0 is the only possible value.