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Angular momentum conservation in pion decay?

Physics Asked by amc on July 19, 2021

I have seen the charged pion decay $$pi^{-}~to~ bar{nu}_{ell} +ell^{-}$$ represented with diagrams containing a $W^-$ in the $s$-channel. The $pi^-$ and $W^-$ have angular momentum $0$ and $1$ respectively, though. How does this process conserve angular momentum?

I see that this question has been asked before, but I haven’t found an answers.

4 Answers

The W is massive so can be in the spin 0 state (or $s=-1, 0, 1$ in general). The photon is massless so does not have this "longitudinal" polarization. For the massive vector boson, the relevant symmetry group is the little group $SO(3)$, and for the photon it is $SO(2,1)$.

Answered by lionelbrits on July 19, 2021

Here is the diagram you are discussing:

charged pion decay

It seems you are worried by the angular momentum carried by the W+. The W+ is a virtual particle in this reaction.

In virtual paths the particle is off mass shell and its mass is unphysical, and angular momentum as a part of a four vector will be a complicated function also having unphysical measure, so conservation is moot.

One imposes conservation laws to the ingoing and out going particles . Angular momentum is conserved once the Jz of the muon added to the Jz of neutrino is zero. In the center of mass system shown in the middle picture it means that the J of the two particles must also be oriented equal and opposite so as to match the J=0 of the pion. This constraint will be included in the integral for calculating the decay rate, which the Feynman diagram represents.

Answered by anna v on July 19, 2021

The $W^-$ has to have zero angular momentum, since the $pi^-$ does in its rest frame. Since the $W^-$ has spin $1$, it must be in states of orbital angular momentum $1$ as well. In particular, the spin-orbital states have to combine to give zero total angular momentum. This specific combination is:

$$|+-rangle+|-+rangle-|00rangle$$

where $+$ is $L_z$ or $S_z=+1$, $-$ is $L_z$ or $S_z=-1$, and $0$ of course is $L_z$ or $S_z=0$.

As a side effect, this proves that the pion can't be a point particle, something already known, because the $W^-$ in its p-state would have no chance of having been at the origin if the pion were a point particle.

Answered by John Morrison on July 19, 2021

Anna V gave the correct explanation. Only when weak bosons are created on mass shell, e.g. at collisions @ Ecm = M, can you apply total angular momentum conservation at a single vertex (production and decay). On the other hand, even for off mass shell bosons chirality (read helicity for ultra relativistic particles) imposses costraints at each vertex.

Answered by Toni on July 19, 2021

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