[Physics] Do protons exchange photons with electrons

Question

I'm sorry for this question but, I just don't get it. According to the electromagnetic field theory, electrons repel each other by exchanging photons. How do protons attract electrons, by photon exchange?

Answer 1

ELECTRON-PROTON ATTRACTION: a simple, semi-classical analysis to avoid full scale QFT.

The exchange of the photon between the proton and the electron leads to attraction, only because the total energy of the electron is negative.

Let us consider the hydrogen atom for simplicity, and imagine the funnel-like shape of the electron energy. The total energy of the electron in the hydrogen atom at distance $r_0$ from the proton is

$E(r_0)=-\frac{e^2}{8\pi\epsilon_0 r_0}$.

If the distance $r_0$ is sufficiently short, then the electron will emit a photon which will be absorbed by the proton, and the amount of energy of the exchanged photon will be dictated by the uncertainty principle:

$\Delta E\Delta t=\hbar.$

But $\Delta t=\frac{r_0}{c}$ so that

$\Delta E r_0=\hbar c\rightarrow \Delta E= \frac{\hbar c}{r_0}$

So the new energy of the electron will be

$E_1=-\frac{e^2}{8\pi\epsilon_0 r_0}-\frac{\hbar c}{r_0}=-\frac{e^2+8\pi\epsilon_0\hbar c}{8\pi\epsilon_0r_0}$

or the equivalent amount of energy corresponding to some new position $r_1$

$-\frac{e^2}{8\pi\epsilon_0 r_1}=-\frac{e^2+8\pi\epsilon_0\hbar c}{8\pi\epsilon_0 r_0}$

from which we get $r_1$ in terms of $r_0$

$r_1=r_0\frac{e^2}{e^2+8\pi\epsilon_0\hbar c}<r_0$

Therefore the electron moves closer to the proton rather than farther from it (an attractive force.)