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When calculating the matrix element for (let's say) $e^+e^- \to \mu^+\mu^-$ we have to average over initial spins and sum over final spins. I understand the motivation of this, but when the calculation is done, the sum is done for 2 cases: spin up and spin down, so you have to add 8 terms (2 for each particle) and divide by 4 (the 2 spins of the 2 initial incoming particles).

Why is this summation enough? Shouldn't one integrate over all possible values of spin? It is not like the particle will come with either spin up or spin down on a given axis, they can be a linear combination of these.

I also noticed that in the massless limit they do the same approach using $e_R$ and $e_L$ separately and then divide by 4? Why is this enough? You get the same result doing this as you would get by integration or am I missing something?

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  • $\begingroup$ Are you utilizing the vector nature of the pairs' coupling to the photon? That is that the R electron only couples to the L positron, and vice versa? $\endgroup$
    – Cosmas Zachos
    Commented Feb 28, 2018 at 15:28
  • $\begingroup$ You can do that, or use the projection on the R and L component. My confusion is why do they use only the 2 orthonormal states of the spin (R and L) and not all the possible linear combinations when averaging over all initial posibilities $\endgroup$
    – Silviu
    Commented Feb 28, 2018 at 20:09
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    $\begingroup$ They are a complete basis of the nontrivial channels. Combinations that do not contribute to the process are omitted. $\endgroup$
    – Cosmas Zachos
    Commented Feb 28, 2018 at 20:17
  • $\begingroup$ @CosmasZachos I am not sure I understand. We can choose the direction of the incoming particles to be z for simplicity. But we can't choose the spin directions (hence why we do the average). But these initial spin directions don't need to be up or down along the z axis (i.e. (1,0) or (0,1)). They can be any linear combination of these. So why consider only these 2 when doing the average? $\endgroup$
    – Silviu
    Commented Mar 1, 2018 at 8:37
  • $\begingroup$ I can't stop you from pointlessly integrating over all directions. Refer all spins to the z basis and sum incoherently. If you do this rights you should find the same 50-50 answer, as you learned from your orthogonal rotation matrices, e.g. Feynman volume III. A uniform distribution over the globe amounts to 50% of the stuff ending up in the northern hemisphere. $\endgroup$
    – Cosmas Zachos
    Commented Mar 1, 2018 at 12:05

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The reason that you do not need to average over all possible orientations of the z-axis (or the whole xyz-coordinate system) is that any-old orientation will get you the same result, when you sum over the four allowed polarization cases. For one electron: $$\left| +z \right\rangle \left\langle +z \right|+\left| -z \right\rangle \left\langle -z \right|=\left| +z' \right\rangle \left\langle +z' \right|+\left| -z' \right\rangle \left\langle -z' \right|$$ with different choices of z-axis orientation. Likewise for two electrons and four cases.

This is not the same as saying that summing over two polarizations is always the same as summing over all possible classical orientations of a vector. Consider the stat-mech problem of a magnetic dipole (m) in a magnetic field (H). If it has just two states, $\pm m$, the average polarization will be $m\tanh (mH/kT)$, but if it has continuously orientable polarization, the answer will be the Langevin function.

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