Peskin & Schroder QFT book page 187 On Peskin & Schroder QFT book page 187, at eq.(6.34):
$$
i \mathcal{M}=-i e F_{1}(0) \tilde{\phi}(\mathbf{q}) \cdot 2 m \xi^{\prime \dagger} \xi \tag{6.34}
$$
which gives the amplitude for electron scattering from an electric field with the approximation $\bf{q}\rightarrow 0$.
What really puzzle me is that the book says the Born approximation for scattering from a potential in following equation:
$$
V(\mathbf{x})=e F_{1}(0) \phi(\mathbf{x})
$$
So why do we specify this term as the potential in Born approximation?
The book says $F_{1}(0)$ is the electric charge of the electron, in unit $e$. I can understand this by the electric potential definition, however, in the leading order, $F_{1}(0)=1$, (maybe not in unit of $e$), how to understand this?
If you have any comment to this, I am really appreciate it.
 A: 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 $\langle p'|M|p \rangle$ 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.
