Consider the QED. When we write the covariant derivative of the theory and couple vector field of gauge boson and the spinor of fermion we think of electric charge of electron $e$ as a coupling constant. Using the Feynman diagram we can compute scattering amplitude for a photon from an electron.

Therefore $e$ is related to the probability that light be detected if a freely propagating electron hits a surface.

Is this understanding correct? If so what happens to the dimension of $e$ and can we reconcile here between classical electrostatic and QED in our understanding of the electric charge because in the latter case it is simply a probability related constant of interaction between two real valued and complex valued vectors?


Indeed, in quantum field theory, $e$ has the interpretation of a coupling constant, as well as electric charge. What I believe you are describing is Compton scattering, $e^- \gamma \to e^- \gamma$. We have that,

$$\frac14 \sum_{\mathrm{spins}} |\mathcal M|^2 = -2e^4 \left(\frac{u}{s} + \frac{s}{u} \right)$$

after simplification, where $s,u$ are the usual Mandelstam variables. Now, as an example, the cross section takes the form,

$$\frac{\mathrm d\sigma}{\mathrm d \cos\theta} = -\frac{e^4}{16\pi s} \left( \frac{u}{s} + \frac{s}{u} \right).$$

We know the differential cross section goes as the amplitude squared, and thus is proportional to the probability. Thus, knowing the kinematic variables, we can deduce $e$ in an experiment, in theory from the outcome of such a scattering process.

It should be noted as well that $e$ has a non-zero beta function, which implies it flows to different values as you change the energy scale of the process. Thus, there is no fixed value of $e$. However, these changes are small, and in non high energy physics, one can stick with the classical value.

As for units, there is no disagreement. $[m\bar\psi \psi] = 4$, in terms of mass dimension. Obviously $[m] = 1$ so $[\psi] = \frac32$. We have $[A_\mu] = 1$. Thus in natural units, in four dimensions, $[e] = 0$.

The Planck charge goes as $\sqrt{\hbar c/k_e}$, so in units wherein $\hbar=c=1$ it is easy to see that the coupling constant is dimensionless. If you introduce the factors of $\hbar, c$ in any QFT formula, you will find $e$ becomes a charge again.

  • $\begingroup$ What about the dimension of e? Do they match in classical electrostatic and QED? $\endgroup$ – user56963 Jul 15 '18 at 11:20
  • $\begingroup$ @VictorVahidiMotti Of course. $e$ has dimensions of charge, and if we did not switch to natural units in QED, it would remain so. $\endgroup$ – JamalS Jul 15 '18 at 11:23

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