I am self-studying QFT, I may not get the right answer, I just wonder if I get it correctly.
We consider some potential $V$ as a perturbation, in QM $H=H_o +V$, in QFT we consider $V$ as $H_{int}$.
In QM, first term in the first Born approximation tells us
$\langle p'|M|p \rangle \approx \langle p'|V|p \rangle =\int d^3rV(r)e^{-iqr}={V}(q)$
If we consider $\langle p'|M|p \rangle$ in QM acts like $\langle p'|S|p \rangle$$\langle p'|M|p \rangle$ in QFT, we can write
$\langle p'|iT|p \rangle= -iV(q)(2\pi)^4\delta^4(p-p'-q)$,
and here if we consider classical potential $V(x)=e\phi(x), V(q)=e\phi(q)$, we have
from Born approximation $\langle p'|iT|p\rangle=-ie\phi(q) (2\pi)^4\delta^4(p-p'-q)$
from QFT $\langle p'|iT|p\rangle=iM (2\pi)^4\delta^4(p-p'-q) =-ieF_1(0)\phi(q)(2\pi)^4\delta^4(p-p'-q) 2m \xi'^\dagger \xi$
When we compare terms, $2m$ is dropped due to different normalization in QFT and QM, $\xi'^\dagger \xi$ is irrelavant as it just tells us polarization is conserved. So, we have $F_1(0)=1$
I really doubt I am correct, please comment.