From the Wikipedia article, the electric $\mathbf{E}$ and magnetic $\mathbf{B}$ fields can be quantised as:

$$ \mathbf{E}(\mathbf{r}) = i \sum_{\mathbf{k},\mu} \sqrt{\frac{\hbar \omega}{2 V \epsilon_0}} \left ( \mathbf{e}^{(\mu)}a^{(\mu)}(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{r}} - \overline{\mathbf{e}}^{(\mu)}a^{\dagger{(\mu)}}(\mathbf{k})e^{-i\mathbf{k}\cdot\mathbf{r}} \right ), $$

$$ \mathbf{B}(\mathbf{r}) = i \sum_{\mathbf{k},\mu} \sqrt{\frac{\hbar }{2\omega V \epsilon_0}} \left ( (\mathbf{k}\times\mathbf{e}^{(\mu)})a^{(\mu)}(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{r}} - (\mathbf{k}\times\overline{\mathbf{e}}^{(\mu)})a^{\dagger{(\mu)}}(\mathbf{k})e^{-i\mathbf{k}\cdot\mathbf{r}} \right ). $$

How do I get a dynamical (in time t) field?
And how would I get a static field, e.g. the Coulomb field, from these equations?

  • $\begingroup$ I think that the static forms of the electric and magnetic field, like Coulomb's Law do not correspond to the quantized fields you are describing. $\endgroup$ – freecharly Jan 31 '18 at 1:00
  • $\begingroup$ Here you'll find a derivation of the Coulomb Law from QED: physics.stackexchange.com/questions/142159/… $\endgroup$ – freecharly Jan 31 '18 at 1:08

How do I get a dynamical (in time $t$) field?

You will have to impose the quantized equations of motion. Their form depends on the field content of your theory. In general this is a hard problem, which can only be solved perturbatively through the use of Feynman diagrams. In a special case of the free Maxwell field (i.e. there's no dynamical charged quantum fields interacting with your electromagnetic field), the problem can be solved exactly though.

Another complication comes from the gauge invariance (which is actually not specific to the quantum case, as the same problem arises in the Hamiltonian formulation of classical electromagnetism). One way of solving this is by imposing a gauge-fixing condition. Let's suppose that you are working in the Lorentz gauge $$ \partial_{\mu} A^{\mu} = 0. $$

In this gauge, the dynamics is encoded in the time-dependence of $a({\bf k})$ and $a^{\dagger}({\bf k})$:

$$ a({\bf k}, t) = a({\bf k}) e^{- i |{\bf k}| \cdot t}, \quad a^{\dagger}({\bf k}, t) = a^{\dagger} ({\bf k}) e^{i |{\bf k}| \cdot t}. $$

But note that in this gauge there's an additional requirement that you have to impose on your Hilbert space. Generally, if you write down the expression for the gauge condition operator, it does not vanish: $$ \partial_{\mu} A^{\mu} \neq 0. $$

This additional constraint has to be imposed weakly (this is called Gupta-Bleuler method). I will not describe it here as it is an interesting topic on its own and not directly related to the question.

And how would I get a static field, e.g. the Coulomb field, from these equations?

If you are quantizing a free electromagnetic field in vacuum, Coulomb potential is not actually a solution of the equations of motions (since it has a singularity at the center). The real solutions are given by superpositions of plane waves.

You have two options here. Either add a static classical charge at position zero (which would break Lorentz invariance) and write down the modified quantum equations of motion for the electromagnetic field, or consider a full interacting quantum theory of matter and light, such as QED.

In second case, the problem of reconstructing a quantum state which corresponds to a given classical solution is a hard one. For the linear field, the answer is actually known (it is given by coherent states), but AFAIK for nonlinear theories this problem is unsolved (please correct me if I am wrong). You can still derive the properties of the Coulomb interaction (including the form of the potential, and the Lamb shift as a next-order correction) using perturbative QED.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.