How Coulomb's potential is related to the propagator?

Question

Based on this question and in the answers we see how the tree-level amplitude allows ust to relate the bare photon propagator to the Coulomb potential using basically just the Born approximation and the non-relativistic limit.

I wanted to expand that a little bit to do the same thing with the dressed propagator, in order to be able to expand it order by order and get corrections to the Coulomb potential.

For that matter I've considered the t-channel process $e^{-}(p_1)e^{-}(p_2)\to e^{-}(p_3)e^{-}(p_4)$ but instead of the bare propagator I've used the dressed propagator.

By applying Feynman's rules the amplitude I get is:

$$i\mathcal{M}=(-ie)\overline{u}(p_3)\gamma^{\mu}u(p_1)G_{\mu\nu}(p_1-p_3)(-ie)\overline{u}(p_4)\gamma^\nu u(p_4).$$

Now I use the non-relativistic limit. Since we have

$$u(p)=\begin{pmatrix}\sqrt{p\cdot \sigma}\xi \\\sqrt{p\cdot\overline{\sigma}\xi}\end{pmatrix}$$

if $m >> |\mathbf{p}|$ we have $p^0=\sqrt{m^2+|\mathbf{p}|^2}\approx m$ and $p\cdot\sigma=p^0+\mathbf{p}\cdot\vec{\sigma}\approx m$. With this I get

$$u(p)=\sqrt{m}\begin{pmatrix}\xi \\ \xi\end{pmatrix},\quad \overline{u}(p')=\sqrt{m}\begin{pmatrix}\xi \\ \xi \end{pmatrix}^\dagger\gamma^0$$

thus we have

$$i\mathcal{M}=(-ie)^2 m^2\begin{pmatrix}\xi_{s'} \\ \xi_{s'} \end{pmatrix}^\dagger\gamma ^0 \gamma^\mu\begin{pmatrix}\xi_s \\ \xi_s \end{pmatrix}\begin{pmatrix}\xi_{\sigma'} \\ \xi_{\sigma'} \end{pmatrix}^\dagger \gamma^0 \gamma^\nu \begin{pmatrix}\xi_\sigma \\ \xi_\sigma \end{pmatrix}G_{\mu\nu}(p_1-p_3)$$

Now I've read that the relation is with $G_{00}$ so everything else must vanish. I just don't see how that happens. I do see that if I forget everything and leave just $G_00$ the result becomes very simple

$$i\mathcal{M}=-e^2m^2 \delta_{ss'}\delta_{\sigma\sigma'} G_{00}(p_1-p_3)$$

but I'm unsure since I don't know how to justify droping everything. Relating to Born's approximation this is like is done in the other question.

So how is the correct way to get $\mathcal{M}$ in the non-relativistic limit to compare to Born's approximation and relate $V$ to $G$? Have I done anything wrong? Do we need to use the resummed expression in terms of the one-particle irreducible sum?

Answer 1

Let's consider the term $\overline{u}_{s'}(p')\gamma^iu_{s}(p)$, where $i$ is a spatial index. In the chiral representation, we have

$$\gamma^0\gamma^i= \begin{pmatrix} 0 & \textbf{1}\\ \textbf{1} & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^i\\ -\sigma^i & 0 \end{pmatrix}= \begin{pmatrix} -\sigma^i & 0\\ 0 & \sigma^i \end{pmatrix},$$

so that

$$\overline{u}_{s'}(p')\gamma^{\mu}u_{s}(p)= m^2\begin{pmatrix} \xi_{s'}^{\dagger} & \xi_{s'}^{\dagger} \end{pmatrix} \begin{pmatrix} -\sigma^i & 0\\ 0 & \sigma^i \end{pmatrix} \begin{pmatrix} \xi_s\\ \xi_s \end{pmatrix}=0.$$

Thus, we have

$$\overline{u}_{s'}(p')\gamma^{\mu}u_{s}(p)=m^2\delta_{ss'}\delta^{0\mu}.$$

Thus, we obtain the desired result.

In relation to the rest of your question, there is no need to worry about 1PI diagrams or resummation of a perturbation series. The Coulomb potential is a purely tree-level result. There are corrections to the Coulomb potential that are very interesting (I recommend section 16.2 of Schwartz's Quantum Field Theory and the Standard model or chapter 6 of these lecture notes from a class I took at Cambridge by David Skinner for excellent discussions of this).

(Based on the statement of your problem, it seems like you're cool going from this result to the Coulomb potential. If not, leave a comment and I'll explain that bit too.)

I hope this helped!