Deriving the Coulomb force equation from the idea of virtual photon exchange?

Question

Since Newton's law of gravitation can be gotten out of Einstein's field equations as an approximation, I was wondering whether the same applies for the electromagnetic force being the exchange of virtual photons. Is there an equation governing the force from the exchange of virtual photons? Are there any links which would show how the Coulomb force comes out of the equations for virtual photon exchange? I know that my question is somewhat similar to the one posted here The exchange of photons gives rise to the electromagnetic force, but it doesn't really have an answer to my question specifically.

Answer 1

The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles.

Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is

$$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p - E_{p'})(-\mathrm{i})\int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r$$

This is to be compared to the amplitude obtained from the Feynman diagram:

$$ \int \mathrm{e}^{\mathrm{i}k r_0}\langle p',k \rvert S \lvert p,k \rangle \frac{\mathrm{d}^3k}{(2\pi)^3}$$

where we look at the (connected) S-matrix entry for two electrons scattering off each other, treating one with "fixed" momentum as the source of the potential, and the other scattering off that potential. Using the Feynman rules to compute the S-matrix element, we obtain in the non-relativistic limit with $m_0 \gg \lvert \vec p \rvert$

$$ \langle p',k \rvert S \lvert p,k \rangle \rvert_{conn} = -\mathrm{i}\frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}(2m)^2\delta(E_{p,k} - E_{p',k})(2\pi)^4\delta(\vec p - \vec p')$$

Comparing with the QM scattering, we have to discard the $(2m)^2$ as they arise due to differing normalizations of momentum eigenstate in QFT compared to QM and obtain:

$$ \int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r = \frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}$$

where Fourier transforming both sides, solving the integral and taking $\epsilon \to 0$ at the end will yield

$$ V(r) = \frac{e^2}{4\pi r}$$

as the Coulomb potential.

Answer 2

There is the (non-genetal) relation between the free energy of interacting of two currents $J^{a}, J^{b}$ and the propagator: $$ U = -\frac{1}{2} \int d^{4}xd^{4}y J^{a}(x) D_{ab}(x - y)J^{b}(y). $$ It's not general, but it realizes the simple example which can help you to understand how to get the expression for force.

The structure of field which causes structure of propagator helps us to get the expression for the force. For example, for intercation via scalar field (by setting $D_{ab}(x - y) = \frac{1}{p^{2} - m^{2}}$) after simple transformations for "point-like" currents $J(x) = \delta (\mathbf x - \mathbf x_{0})$ we can get $$ U = -\frac{1}{4 \pi |\mathbf r|}e^{-mr}. $$ For case $m = 0$ we get the Coulomb-like law of interaction.

Absolutely the same thing you may do with the case of EM field (in Feynman gauge $D_{\mu \nu} = -\frac{g_{\mu \nu}}{p^{2}}$).

If you need some explicit derivation I'll give it later.

Answer 3

I've gone through the same thought process as you, so I suspect the question you're really trying to answer is:

"Can the formula for the electric Coulomb potential be generated as a natural consequence of quantum wave functions alone?"

If that's what you're wondering, then the answer is "no".

After Schrodinger had developed the wave function of the electron, he had hoped that it would somehow represent the electric charge and field. Alas, it does not. It is nothing more than a probability. Particle wave function doesn't give us the field potential.

Force potentials have to be added to the calculation of the interaction. The formulas that define the potential are experimentally derived. Physicists have come up with formulas that match observed behavior. Different forces have different formulas to define the potential. And that formula gets baked into the scattering calculation.

For a massless particle exchange, as in the photon, the potential simplifies to 1/r, as in classical physics. But this is really just because the potential was deliberately written to behave that way, in order to match experimental observation.

This leaves us no better or worse off than we were before with classical physics. The force potential is arbitrary, but this was also the case with the Coulomb potential. There's no particular reason the Coulomb potential should be 1/r in classical physics. It just is.

By contrast, the potential of the gravitational field potential naturally arises from solutions to the Einstein field equations. This is very aesthetically pleasing. Nothing comparable to this exists in quantum physics to derive the behavior of the other forces.

So if you were hoping that the potentials would naturally jump out of the particle wave functions, as Schrodinger had - sorry, no luck.