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As it pointed out on page 133 of Peskin and Schroeder, any QED amplitude involving external fermions, when squared and summed or averaged over spins, can be converted to traces of products of Dirac matrices.

Does the Trace Technology in QED for calculating cross section have any physical interpretation? Or it's just a mathematical technique?

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closed as unclear what you're asking by ZeroTheHero, Kyle Kanos, Jon Custer, Aaron Stevens, Martin Mar 8 at 12:34

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    $\begingroup$ It might be useful to quote the context of the Peskin for those that don't have a copy. $\endgroup$ – Kyle Kanos Feb 15 at 11:04
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My answer is kind of, there is a physical reason why Peskin mentions this.

Let $g_1(s_1,t_1,u_1), g_2(s_2,t_2,u_2), \ldots$ be functions of mandalstam variables $s_i, t_i, u_i$, and let $\Gamma_1, \Gamma_2$ be dirac bilinear. Then the quantity

$$ \overline{|\mathcal{M}|^2} = \mathrm{tr}(\not g_1(s,t,u)\cdot \Gamma_1 \cdots) \times \mathrm{tr}(\not g_2(s,t,u)\cdot \Gamma_2 \cdots)\times \cdots$$

is a Lorentz invariant quantity. So at this point it appears that there indeed is a physical significance to mentioning this.

However, the reason my answer is "kind of" is because the matrix element $\mathcal{M}$ was already Lorentz invariant to begin with (i.e. before squaring and summing over polarizations etc.), so it would seem to be a rather moot point. This leads me to conclude that he's just trying to let us know that squaring and averaging etc. will just always result in a scalar in dirac space (which doesn't have any physical significance as far as I know).

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