# Going from scattering amplitudes to classical potentials

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).

1. 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.

## 1 Answer

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)

• Thanks for the answer! Just a question. Based on the scattered particles and their charge, we should consider the appropriate contributions from each of the s-, u-, t-channels. For example for $e^- e^-\rightarrow e^- e^-$, we consider the t and u (Moller scattering). For $e^- e^+\rightarrow e^- e^+$, we consider the s and t channels (Bhabha scattering). Peskin & Schroeder (p.125) only considers the s-channel and Schwartz (p.308) considers a single channel as well. Shouldn't we add up the contributions of the two channels? Wouldn't that give us $2$ times the Coulomb potential? – TheQuantumMan Jan 15 at 10:12
• @TheQuantumMan: The electrons here are not really indistinguishable if we want to compare to the Born approximation scattering - one of them is a fixed potential source, the other is scattering off it, and really the fixed one is usually not a single electron but some far heavier and complicated thing that contains an electron. So the u-channel does not contribute because of distinguishability. There is no s-channel for electron-electron scattering, you only have that for electron-positron scattering. – ACuriousMind Jan 15 at 10:16
• I see. Thanks a lot for your comment! – TheQuantumMan Jan 15 at 10:17