Going from scattering amplitudes to classical potentials

Question

1. In Maggiore's textbook, he defines the amplitude $$\langle k_1,..,k_n|S-1|p_1,..,p_m\rangle=(2\pi)^4\delta^{(4)}\left(\Sigma_i k_i-\Sigma_ip_j\right)\ i\mathcal{M},\tag{5.115}$$ where $S-1=iT$. When he then tries to calculate the classical potential for a $2\rightarrow 2$ scattering, he writes (eq. 6.68) $$i\mathcal{M}=\left( \langle k_1, k_2|iT|p_1,p_2 \right)^{(R)}\tag{6.68}$$ where the superscript $R$ stands for relativistic. Where did the factors of $(2\pi)^4\delta^{(4)}\left(\Sigma_i k_i-\Sigma_ip_j\right)$ go?

We can then use $|k_1, k_2\rangle^{(R)}=\sqrt{2E_{k_1}}\sqrt{2E_{k_2 }}|k_1, k_2\rangle^{(NR)}$, where $NR$ stands for non-relativistic. One can then use the Born approximation to show that the classical potential $V(x)$ is minus the Fourier transform of $\mathcal{M}$ (eq. 6.71, p.169).

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2. The second question is in getting the corrected Coulomb potential in QED from the following process:

Schwartz (from QFT and the Standard model, chapter on vacuum polarization) first defines the second diagram to be $i\Pi^{\mu\nu}_2$ and then writes it in the form $$i\Pi^{\mu\nu}_2=i(-p^2g^{\mu\nu}+p^\mu p^\nu)e^2\Pi_2(p^2)\tag{16.48}$$ One can now compute the full diagram and find (eq. 16.50, p.309) $$-i\dfrac{\left[1-e^2\Pi_2(p^2)\right]g^{\mu\nu}}{p^2} +(\text{terms $\propto p^\mu p^\nu$ that have to do with the gauge})\tag{16.50}$$

Now, as I understand it, if we also include the contribution from the external legs, which would be in the form of the polarization vectors, $\epsilon_1^\mu, \epsilon_2^{\nu*}$, the $\mu,\nu$ indices in (16.48) would contract and we get something reminiscent of $i\mathcal{M}$ and we can follow the process from the first part of the question to find the classical potential.

Instead, Schwartz just takes the part of (16.50) that's proportional to $g^{\mu\nu}$, slaps an extra $e^2$ in front and defines $$\tilde{V}(p)=e^2\dfrac{\left[1-e^2\Pi_2(p^2)\right]}{p^2}\tag{16.51}$$ as being the Fourier transform of the classical potential.

Can somebody explain this step? I don't see why we haven't contracted the extra indices and where the $e^2$ comes from. I mean, I can kind of make sense of it (maybe attach vertices at the end which would give rise to the $e^2$), but I would like to have a clear picture of how to handle these situations to get classical potentials so I can handle more difficult cases myself.

Answer 1

Eq. 6.68 as you quote it seems wrong. Here's how to argue better that the delta functions should "vanish": The Born approximation says that the scattering amplitude for a momentum $\vec p$ into a momentum $\vec p'$ is $$ A_\text{QM}(p,p') = (2\pi)\delta(E_p - E_{p'})(-\mathrm{i})\langle \vec p\vert V\vert \vec p'\rangle,$$ and the corresponding expression from QFT is the non-relativistic limit of $$ A_\text{QFT}(p,p') = \int \langle p',k'\vert \mathrm{i}T\vert p,k\rangle\mathrm{e}^{\mathrm{i}\vec k\cdot\vec r}\frac{\mathrm{d}^3k}{(2\pi)^3},$$ where the integration over $k$ is because the scattering needs to treat one of the electrons as the "fixed" source of the potential at some location $\vec r$ to be comparable to the Born approximation scattering off a potential $V$ that doesn't consider the source of the potential to incur a recoil, either. So with the definition of $\mathrm{i}\mathcal{M}$ you give first, this is $$ A_\text{QFT}(p,p') = \int(2\pi)^4\delta^{(4)}(p' + k' - p + k)\mathrm{i}\mathcal{M}\mathrm{e}^{\mathrm{i}\vec k\cdot\vec r}\frac{\mathrm{d}^3k}{(2\pi)^3}$$ and now the integral over $\vec k$ kills off the spatial part of the $\delta^{(4)}$ and just imposes $\vec k' = \vec p - \vec p'$, leaving a $\delta(E' - E)$ for the final and initial energies. Now you see that $$ \langle \vec p\vert V\vert \vec p'\rangle = -\mathcal{M} = -f^\mu e \Pi_{\mu\nu} e f^\nu,$$ by comparing the QM and QFT expressions, where $f^\mu = \bar u_{s'}(p') \gamma^\mu u_s(p)$ are the fermionic contributions (however you want to write them) and $\Pi^{\mu\nu} = \Pi g^{\mu\nu}$ is the photon propagator. In the non-relativistic limit we get $f^\mu \approx 2m \delta_{s's}\delta^{\mu0}$, and so after discarding the resulting $(2m)^2$ because the normalization between non-relativistic and relativistic momentum states differs we finally get $$ \langle \vec p\vert V\vert \vec p'\rangle = -e^2\Pi^{00} = e^2\Pi. \tag{*}$$ Spelling it out this way directly answers your second question - when computing the 1-loop correction, nothing in this logic changes except for the expression for $\Pi$, so we can directly jump to using eq. $({}^*)$.

(No guarantee is taken in this answer for missing signs or spurious factors of $\mathrm{i}$, because they do not matter for the overall logic)