On Schwartz’s QFT book, page 232, it’s calculating the cross section for $e^+e^-\rightarrow\mu^+\mu^-$ scattering, assuming spin and polarization are not measured, and it writes

Let us also assume that we do not know the polarization of the initial states. Then, if We do the measurement many times, we will get the average over each polarization. This leads to contractions and traces of the initial state, with a factor of $1/4$ ($1/2$ each for the incoming $e^+$ and incoming $e^-$ ) to average over our ignorance.

I think the polarization vector of electron as those $u_p $, $v_p$ spinors, and my question is

  • What is “the average polarization “ of those particles?
  • What does “contractions and traces of the initial state” mean? (Where does the factor come from?)
  • Why we sum over the $\mu^+$ spins but take the average of polarizations?( They’re both unmeasured quantities.)

1 Answer 1


I am unfamiliar with the details of that text, but the brilliant trick of utilizing spin-energy projection operators (13.59,60), due to Casimir, to convert (13.57) through (13.61) "squared" into 4 times (13.62) is expounded in all QFT texts (Sakurai's Advanced QM Ch 4-3 is the friendliest!). Maybe Schwartz details it in another part of his text, and you could probably just work it out by yourself. (Ignore averages for the time being: a separate issue.)

You start with a pair of contracted spinor index fermion bilinears in half the squared amp; then rearrange the spinors, and end up with a pair of one contraction of the end indices of the composite index bilinears, hence a pair of traces multiplying each other, as in (13.62).

That's for a sum of polarizations s of the incoming electron and the positron, s', statistically equally likely to be + or -. So you literally average polarizations by halving the sum. Your luminosity of unpolarized electrons to be hurling into your interaction vertex (amp) will measure charge (electrons!), likely to be be of polarization half + and half -, so you feed 1/2 of the operator in the s sum; and likewise 1/2 of the s' sum. In all, a factor of 1/4. It's as though you cut down your effective luminosity for each beam by 1/2.

You will then measure a mixture of unpolarized muons leaving the vertex, consisting of both polarizations now, which you might measure, if inclined, or not: but their net number in your detector should be the same, as the input electrons have to go into something.

(Aside: I hope I won't confuse you further. You could veeery alternatively think of the chirality states involved. The photon, being a vector interaction, connects only Left-chiral electrons to Right-chiral positrons, and vice versa. So basically half chiral options (LL and RR) are off the table in this interaction. Likewise for the muons now, so a net reduction by 1/4 over the sum of all logical possibilities.)

Compare to Bhaba.


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