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Quantum optics all discuss the quantization of free electromagnetic radiation. The result is well established.

But what about an arbitrary electromagnetic field? For example, the simplest case, electric field generated by a point charge. How is that field quantized as an operator?

Siyuan Ren
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  • You might try looking at B. Huttner and S. M. Barnett, “Quantization of the electromagnetic field in dielectrics,” Phys. Rev. A 46, 4306–4322 (1992). I may write an answer summarizing this paper's content in the next few days. – Selene Routley Jul 12 '13 at 13:08

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There is only one electromagnetic field in the Universe – it's the function that assigns each point in ${\mathbb R}^4$ with two vectors $\vec E,\vec B$. When we say that we quantize the electromagnetic field, it doesn't mean that we quantize a particular configuration of the electric and magnetic vectors. It means that we quantize the whole function, namely we declare that the vectors $\vec E,\vec B$ assigned to each point in the spacetime are operators (more precisely: operator distributions – because their commutators may involve the Dirac delta distribution and its derivatives).

This quantized electromagnetic field – the set of observables contained in the theory called Quantum Electrodynamics (QED) – is able to describe all configurations of electromagnetic field we observe in Nature, including those from electric charges, permanent magnets, or electromagnetic waves coming from time-dependent charge configurations. The latter have energy quantized in photons, $E=hf$, if the frequency is fixed. It's one of the simplest physical statements one may make about the quantized electromagnetic field. But the electromagnetic field is quantized – it is a field of operators – at all times.

So I am afraid that the question "What about" is too vague and the last question, "how is that field quantized", has the answer that it is the same field so it is quantized in the same way it always is. This question is exactly analogous to the question "how is the location of an electron quantized if an electron is 2 microns away from a proton". In other words, I am afraid that you are misunderstanding something very basic here but I don't know what it exactly is.

Luboš Motl
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  • So can you give me an generalized expression of $\mathbf{E,B}$ operators? What I get on the book is something like $\mathbf{E}(\mathbf{r})= i\sum_{\mathbf{k},\mu} \sqrt{\frac{\hbar\omega}{2 V\epsilon_0}} \left(\mathbf{e}^{(\mu)} a^{(\mu)}(\mathbf{k}) e^{i\mathbf{k}\cdot\mathbf{r}} - \bar{\mathbf{e}}^{(\mu)} {a^\dagger}^{(\mu)}(\mathbf{k}) e^{-i\mathbf{k}\cdot\mathbf{r}} \right)$, which is derived under the assumption of free field in a box. – Siyuan Ren Oct 03 '12 at 13:01
  • Dear Karsus, the best formula depends on the gauge and we usually write the formula for $\vec A$, not $\vec E$. However, your task to write down $\vec E$ actually simplifies the task. You just add the $c$-number for the classical electric field. So it is $\mathbf{E}(\mathbf{r})= i\sum_{\mathbf{k},\mu} \sqrt{\frac{\hbar\omega}{2 V\epsilon_0}} \left(\mathbf{e}^{(\mu)} a^{(\mu)}(\mathbf{k}) e^{i\mathbf{k}\cdot\mathbf{r}} - \bar{\mathbf{e}}^{(\mu)} {a^\dagger}^{(\mu)}(\mathbf{k}) e^{-i\mathbf{k}\cdot\mathbf{r}} \right)+Q/(4\pi\epsilon r)$ – Luboš Motl Oct 04 '12 at 10:08
  • Note that adding $c$-number to operators doesn't modify their commutators, so the commutators for $E,B$ and the commutators for $a,a^\dagger$ are still the same. I just effectively shifted the vacuum. Here, it was assumed that the source of the electric field was classical. However, one may similarly write down $\vec A$ or $\vec E$ in general fields with charges that are quantized objects, particles, and the impact is just the addition of a Coulomb term to the Hamiltonian. – Luboš Motl Oct 04 '12 at 10:10
  • You should understand that in the normal perturbative treatment of QED, however, it is correct to expand the electromagnetic field to the free modes only - to assume that there are no sources - because the sources are added as perturbations. Still, this perturbative QED is able to describe the full dynamics of the EM field and the charges so there is no loss of generality! – Luboš Motl Oct 04 '12 at 10:11
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    "the best formula depends on the gauge and we usually write the formula for $\mathbf{A}$ , not $\mathbf{E}$". So how is $\phi$ expressed as an operator? Also, is there any reference for simply adding a c-number to $\mathbf{E}$? It seems too easy to be true. – Siyuan Ren Oct 04 '12 at 12:48
  • Dear Karsus, $\phi$ isn't enough to describe the whole electromagnetic field, just the electric one, so one often uses the gauge $\phi=0$. But even if she is not, it doesn't matter, $\phi$ is an operator (distribution) just like any other field. Also, I didn't make any controversial addition of a c-number to $E$. I just wrote down a Fourier-like formula to express it in terms of some new variables, $a$ and $a^\dagger$. You may view may formula as the definition of $a,a^\dagger$. You may also show that they will have the usual commutators. There's nothing to defend here. – Luboš Motl Oct 05 '12 at 08:50
  • In quantum optics, the gauge is usually fixed as $\nabla\cdot\mathbf{A}=0$, so you cannot set $\phi=0$ simultaneously. And I don't know how to show the newly defined $a$ satisfies $[a,a^\dagger]=1$. – Siyuan Ren Oct 05 '12 at 09:06
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The problem of (non-perurbative) quantization of interacting fields is formidable. Not only it contains an infinite number of degrees of freedom, it is nonlinear due to the interaction, thus its classical solutions do not consist of a Hilbert space, in addition, it suffers from the ultraviolet and infrared problems encountered already in perturbation theory.

In principle, to perform a quantization, one needs a space of "modes”: classical solutions of the equations of motion to expand the field with respect to which. In the free case, the modes can be chosen as wave packets. It is true that in many cases in the interacting field theory, there is a one to one correspondence of the interacting solutions to free asymptotic solutions, and this is the principle according to which QED is quantized. But in QED we know these correspondences only in perturbation theory, where the interaction is assumed to be small. In the case of strong interactions one needs to know these maps explicitly to come up with quantitative answers.

Thus in practice, one needs to resort to approximations in treating interacting electromagnetic field quantizations in the presence of charges or matter in general. There are several known approximation schemes such as:

1) Quantization of the electromagnetic field in macroscopically modeled materials such as dielectrics. For linear dielectrics, here, we still have mode expansions but not wave packets in general. In addition the gauge fixing conditions will allow for longitudinal modes in addition to the transversal ones.

2) Quantization of the electromagnetic field in the presence of charges, such as interaction with atoms in theories of radiation emission. Here the electromagnetic field is expanded in a wave packet basis, but non-perturbative approximations of the nonlinear interacting equations are applied such as the rotating wave approximation

David Bar Moshe
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I am no expert here, and I imagine I misunderstand many basic things, but I share your curiosity about quantizing general em fields. The only related item I have found was in Feynman's Theory of Fundamental Processes, where he connects the coulomb interaction to the longitudinal and time-polarizations of the photon field. I tried to present his analysis here: properties-of-the-photon-electric-and-magnetic-field-components.

I guess that this analysis, since it is working with propagators, is actually assuming the results of quantization, and so not answering your question, but it seems relevant. Maybe someone can connect the dots?

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Art Brown
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