I have a neutral pion of mass $m_{\pi}$, and it decays into two photons. In it's reference frame the decay is isotropic. One of the photons has a helicity of $+\hbar$ and the other $-\hbar$. In another frame the pion has relatavistic momentum $\vec p=p \hat z$, and the probability to find the + helicity photon at angular position $(\theta,\phi)$ is:

$$dP_+=f(\theta,\phi)\sin\theta d\theta d\phi$$

and then we have (because it's a probability)

$$P=\int_0^{\pi}\sin\theta d\theta \int_0^{2\pi} d\phi \ f(\theta,\phi)=1$$

The goal is to find $f(\theta,\phi)$. I'm looking at this and it reminds me a lot of spherical harmonics. Mostly because spherical harmonics are denoted by


Which are functions of the angles $\theta$ and $\phi$. And $f(\theta,\phi)$ is also a function of the same angles. And another connection I see is that the probability to find a particle in a certain state, within the solid angle $d\Omega$ at the angles $\theta$ and $\phi$ is $$\big|Y_{l,m}(\theta,\phi)\big|^2d\Omega$$ and

$$P=\int_0^{\pi}\sin\theta d\theta \int_0^{2\pi} d\phi \ \big|Y_{l,m}(\theta,\phi)\big|^2=1$$

These last two things are from my quantum mechanics textbook, and they look very similar to the first equations I wrote because $\sin\theta d\theta d\phi$ is an infinitesimal solid angle, just like $d\Omega$. Is $f(\theta,\phi)$ the magnitude of a spherical harmonic? Or is it not because we are dealing with a different reference frame than the decay and we need to take relativity into account?


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