Vertex function, quantum action

[The black vegetable]

I am looking at Srednicki ch 64 , how does equation 64.1 follow from 64.3 as stated.

Explicitly in QED how does
$u_{s'}(p')V^{u}(p',p)u_{s}(p)=e\bar{u'}(F_{1}(q^{2})\gamma ^{u}-\frac{i}{m}F_{2}(q^{2})S^{uv}q_{v})u$

follow from the quantum action
$\Gamma =\int d^{4}x(eF_{1}\bar{\varphi }\not{A}\varphi+\frac{e}{2m}F_{2}(0)F_{uv}\bar{\varphi }S^{uv}\varphi + ...$
Where the… represent more derivatives

Is it from the derivative expansion of the quantum action, (chapter 21 equation 21.19)

Many thanks

 

[vanhees71]

The expression for $\Gamma$ indeed follows from writing down the quantum action considering the symmetries of QED (gauge symmetry, P invariance, C invariance, T invariance). Then going to the 1st equation is done as usual by mode expansion of the Dirac field to get the corresponding vertex (photon-electron-positron vertex). The $F_1$ and $F_2$ are form factors. In general they are function of $q^2$ as indicated in the first formula, but here obviously Srednicky considers only the on-shell limit of the photon. The most important thing is that you can extract the anomalous magnetic moment from these form factors. Since QED is renormalizable, only $F_1$ needs renormalization while $F_2$ is finite at any order of perturbation theory.

 

[The black vegetable]

ok thanks , i tried to look up mode expansion of dirac field, couldn't find how it gives the vertex function,