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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.

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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.

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  • $\begingroup$ Since pion is a bound state, so could it be that the pion first goes to a state of angular momentum 1 before going to the final state via W+ channel? $\endgroup$ – user10001 Dec 10 '13 at 12:39
  • $\begingroup$ The pion is a bound state of an up and antidown with spin 0, i.e. the spins of the quark is opposite to the spin of the antiquark and there is no angular momentu ( l=0, S state). Due to the existence of the weak interaction the 1/2 spin quark(antiquark) line can go to the lower ground state of an antilepton( mu+) and a neutrino , conserving lepton number. In the virtual path as I said energy momentum are not conserved ( off mass shell) and the same happens with angular momentum, it is not a conserved quantity within a virtual line. One should look for conservation btetween input output lines $\endgroup$ – anna v Dec 10 '13 at 14:11
  • $\begingroup$ The pion is pound by the strong interaction. The decay is due to the weak interaction. $\endgroup$ – anna v Dec 10 '13 at 14:20
  • $\begingroup$ Hmm.. but does the state l=1, S=0 exist for a bound state of up and anti-downquark ? $\endgroup$ – user10001 Dec 10 '13 at 14:23
  • $\begingroup$ The pion has spin 0 so the probability of an l=1 state should be 0, though the QCD potential will not be of a simple type. It is bound by the exchange of very many off mass shell gluons, but this is the collective angular momentum obtained. l=1 is a higher energy/mass, the rho meson . $\endgroup$ – anna v Dec 10 '13 at 14:28
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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 $SO(2,1)$.

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  • $\begingroup$ Yes, I understand that, but that only deals with the quantum number $J_z$. I am asking about how $J$ is conserved. $\endgroup$ – amc Dec 10 '13 at 1:00
  • $\begingroup$ I suspect that what's happening is that the W boson couples to some current $J^\mu$ that is a composite of the fields that make up the $\pi$, and it is this that that has spin 1... $\endgroup$ – lionelbrits Dec 10 '13 at 20:28
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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:

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

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.

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  • $\begingroup$ I see a problem with the momentum of the W- being zero. Not sure how to deal with it right now. Off to the pencil and paper... I'm going to use the muon-neutrino end rather than the pion end to keep things as simple as possible. $\endgroup$ – John Morrison Mar 25 '16 at 17:58
  • $\begingroup$ You said that the W must be in a state of orbital angular momentum 1. W is not a composite particle. How can it have an orbital angular momentum? Orbital angular momentum between what and what? $\endgroup$ – Paganini Mar 25 '16 at 18:12
  • $\begingroup$ In the simplest way of dealing with the hydrogen atom, the electron is the only particle there. (The nucleus is treated as purely the source of potential energy.) The electron can be in an l=1 state. (Sorry, this seems not to let me insert newlines. So continuing in another comment.) $\endgroup$ – John Morrison Mar 25 '16 at 19:55
  • $\begingroup$ Other than the issue of zero momentum, there is no problem of putting the W- in l=1 states, in the rest frame of the pion, for the same reason. I suspect that the zero momentum issue is related to your questions. I hope that applying the Feynman rules at the muon-W vertex should resolve the problem. (We have to put the muon and neutrino into a state of zero total momentum and zero relative angular momentum.) $\endgroup$ – John Morrison Mar 25 '16 at 20:03
  • $\begingroup$ @JohnMorrison: your analogy with the hydrogen doesn't hold: the orbital angular momentum is between the electron and the proton. But for the W, contrary to the electron there is apparently nothing else with which it can have an orbital angular momentum (nothing equivalent to the proton). $\endgroup$ – Paganini Mar 25 '16 at 20:40
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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.

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