Scattering matrix to traces/dirac spinors

I've been searching online and in my qft books (im an early phd student) and I can't find a clear explanation. If you have one, or can simply direct me to a page that does please do so.

For a generic scattering/decay matrix :

[tex]
\sum _{polarization}
\left|M|^2\right.=\sum _{polarization} \bar{u}_i\left(F^*\left[\gamma ^5\right]\right) u_f \bar{u}_f \left(F\left[\gamma ^5\right]\right)u_i
[/tex]
Where F[gamma5] is just a linear function of gamma5.

You simplify by "knowing" that since they're Dirac spinors you can rewrite to a trace of the sums over the u's.

[tex]\text{Tr}\left[\left(F^*\left[\gamma ^5\right]\right)\sum _{\epsilon } u_f \bar{u}_f \left(F\left[\gamma ^5\right]\right){\sum _{\epsilon } \bar{u}_i}u_i\right]
[/tex]

Then using completeness
[tex]
{\sum _{\epsilon } \bar{u}_f}u_f\right = p+m
[/tex]
[tex]
{\sum _{\epsilon } \bar{u}_i}u_i\right = p-m
[/tex]

or something like that.

My questions is how to get the first step.
How are we rearranging the operators and distributing the sums and using the trace.
Is it that the sum over polarization is a sum over both polarizations:
[tex]
\sum _{\epsilon } =\sum _{\epsilon 1} \sum _{\epsilon 2}
[/tex]

so they only operate on their spinors. So then the trace must come in when we need to get the ui's next to eachother.

Anyone know offhand?

 

Hmm maybe its just the index technique I'm not familiar with.
In Peskin and Schroeder, just below equaiton 5.3 they state :
"Working with the frst half of 5.2 and writing in spinor indices so we can
freely move the v next to the vbar we have"
Then they have :

[tex]
\sum _{s,s'} \bar{v}_a^{s'} \gamma _{\text{ab}}^{\mu } u_b^s \bar{u}_c^s \gamma _{\text{cd}}^{\nu } v_d^{s'} = (p-m)_{\text{da}}\text{ }\gamma _{\text{ab}}^{\mu } (p+m)_{\text{bc}}\text{ }\gamma _{\text{cd}}^{\mu } = \text{trace}\left[(p-m) \gamma ^{\mu } (p+m) \gamma ^{\mu }\right]
[/tex]

 

Right I could see that, but where do we get the trace from anyway? How do we just throw it in? You cant say:
M*M = Sum[ABCD] = Tr[BCAD]

I'm starting to reason out that it has to do with suming over the indices.