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This is something that I've been curious about for some time. A coherent, monochromatic electromagnetic wave is well described by a coherent state $|\alpha\rangle$. The quantum treatment of the interaction between the field and matter then reduces at mean-field level (i.e. neglecting fluctuations) to the usual description of a classical external field acting on quantum matter, so long as $\alpha\gg 1$.

I want to know: does there exist a similar quantum state description for a DC field? For example, the electric field in between two capacitor plates. The expectation values of the field operators in such a state should of course reproduce the classical field strength. If this state (which may not be a pure state) cannot be written down, then I would be curious to know why.

(Feel free to consider, say, a bosonic scalar field rather than vector fields if that makes things simpler.)

Mark Mitchison
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  • Note that I am roughly familiar with the theory of scattering of matter by virtual photons in a DC field (at the level of Zee's book). It seems that here one is able to calculate the interaction energy and some scattering amplitudes for the matter fields. This is not what I am after: I want the amplitudes for the force-carrier field. – Mark Mitchison Jan 22 '15 at 02:25
  • I'm unclear what you are looking for. If you just had a static field for all time and space, that has infinite energy. And it seems like if you didn't have that (or a travelling wave, etc.), then you'd need matter too, so you'll have to have both. – Timaeus Jan 22 '15 at 04:48
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    @Timaeus I am looking for any reasonable quantum state describing an approximately static field. For example the electric field in between two capacitor plates. Obviously you need matter around: you find the state of the field by tracing over the matter degrees of freedom. That is why I do not think the state will necessarily be representable as pure. – Mark Mitchison Jan 22 '15 at 05:11
  • An electromagnetic field with constant amplitude in a finite region of space looks like an electromagnetic cavity – glS Jan 22 '15 at 12:37
  • @glance Could you elaborate on this? I can imagine trying to write the state in terms of the eigenstates of a fictitious cavity. It's still not obvious what the state looks like. We need something with an uncertain number of photons, so that the expectation value of the field operator is non-zero. But this expectation value also needs to be constant in time, which is what I find confusing, since the electric field should not be a constant of the motion for the usual field Hamiltonian. I think that the constant charge density probably needs to be explicitly included to solve this problem. – Mark Mitchison Jan 22 '15 at 12:46
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    Can you not just take a linear superposition of coherent states (plane waves) to get any EM field you like? Just integrate over the different momenta weighted by the Fourier transform. – Holographer Jan 22 '15 at 13:01
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    @Holographer, I think the problem is that the frequency spectrum of the states doesn't include a zero mode, so no matter what superposition you take, you won't get a zero mode. – lionelbrits Jan 22 '15 at 13:07
  • @lionelbrits Yes, although physically one expects that a coherent state corresponding to a very low frequency $\omega \to 0$ should give similar predictions for observables over observation times much less than $1/\omega$. This will be an interesting thing to check. I will try to work out the state as per your suggestion and write up the answer at some point. – Mark Mitchison Jan 22 '15 at 13:16
  • The problem with this is that your charge distribution might not be very symmetrical, so the field might have non-zero spatial frequency while having zero temporal frequency. But for plane waves, you can't achieve that. – lionelbrits Jan 22 '15 at 13:20
  • @lionelbrits Well, near the node of a plane wave (or indeed standing wave) the electric field amplitude should vary approximately linearly in space. Actually this is probably more straightforward if one assumes a very large fictitious cavity as per glance's suggestion, and then looks at the electric field near a node over short times. – Mark Mitchison Jan 22 '15 at 13:22
  • The answer half-works for parallel plates, but how will you approximate the static electric field near a point charge using time-varying plane waves? – lionelbrits Jan 22 '15 at 13:24
  • @lionelbrits Sure, that probably won't work. – Mark Mitchison Jan 22 '15 at 13:26
  • I wouldn't expect a parallel plate capacitor to be 100% stable (charge can go from one plate to the other), so some of those matter states you sum over would be matter states that are about to go. A system with interaction just isn't very static. – Timaeus Jan 24 '15 at 16:50
  • @Timaeus Are you saying that you don't believe that the field inside a parallel plate capacitor is static? Of course I don't believe that it is perfectly static: this is the whole point of finding the quantum state so I can understand the dynamical effects that are usually ignored. Nevertheless, I expect fluctuations to be small, which is why a classical static electric field is a good and useful approximation to the true quantum state. – Mark Mitchison Jan 25 '15 at 17:09
  • @MarkMitchison I didn't think your question was well defined enough. There is a whole spectrum of matter states, some more static than others, and if you some over "all" of them then where you draw that line matters. As for a real capacitor, I'd expect phonons in the metal lattice, conduction electrons with a spectrum of drift velocities each with dynamic fields associated with that motion, even bleeding of electrons. The classical fields I solve for a capacitor would be macroscopically averaged fields, and even then, the electrostatic fields I solve for would just be an equilibrium solution – Timaeus Jan 25 '15 at 18:23
  • @Timaeus Theoretical physics is frequently concerned with idealised situations. I find it obvious that the phonons don't contribute anything interesting so wouldn't even bother putting them in the model in the first place. Unless I am specifically interested in the effect of loss of electrons, I also won't include this. I am interested in the case where the macroscopic matter distribution is sufficiently dense and uniform that it may be treated as classical. This is called a mean-field approximation. The result, as explained in lionelbrits' answer, is a coherent field state. – Mark Mitchison Jan 25 '15 at 22:02
  • This state interacts with matter identically to a classical static field in many ways. The fact that field states exist which interact with quantum matter in the same way as classical static fields is an extremely well established experimental result, think of the Stern-Gerlach experiment for a magnetic-field example. Of course, the kind of model you are talking about would be more accurate, but also probably intractable. I am planning on posting an answer fleshing out the details of what I am interested in, so you can see what I mean. I will tag you when it's finished, if you're interested. – Mark Mitchison Jan 25 '15 at 22:07

1 Answers1

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At first glance, what you are describing sounds a lot like squeezed coherent state. However, the more I think about it, what you need is to act the displacement operator $D(\alpha)$ on the coherent state and pick (the real part of) $\alpha$ such that the field fluctuates around some value $E_0$ rather than 0. The displacement operator is $$D(\alpha) = e^{\alpha a^\dagger - \alpha^* a},$$ which you can re-write in terms of the real and imaginary parts of $\alpha$ as $$D(\alpha) = e^{ \sqrt{2} \operatorname{Im} \alpha \, i Q + \sqrt{2} \operatorname{Re} \alpha\, i\frac{d}{dQ}}.$$ or $$D(\alpha) = e^{ -\frac{i q_0}{\hbar} P + \frac{i p_0}{\hbar} Q}.$$ As you know, $P$ is the generator of translations in $Q$ space, so the displacement operator translates the state in $PQ$ space (i.e., phase space)

The simplest way to see this is to consider the wave-function of a SHM in its ground state: $\psi(x) = A e^{- \frac{m\omega}{2\hbar}(x-x_0)^2}$. The value $x_0$ represents the point about which the oscillator oscillates, and is conventionally taken to be zero, because we pick our coordinate origin to coincide with the minimum of the potential $V(x)$. However, nothing in principle stops us from writing down such a state for which $x_0$ isn't zero. The state will just not be an energy eigenstate.

Edit:

The problem with the visual you are having of that coherent state fluctuating around zero is that you are using the free-field Hamiltonian. However, if you have capacitor plates with charges on them, then from the point of view of the EM field, you will have to places sources $A_0 J_0$ in your Lagrangian which will change your Hamiltonian. In that case, the minimum for the potential for the fields will no longer lie at zero, but at some other value.

lionelbrits
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  • But a displacement operator acting on a coherent state just gives you another coherent state. The problem with this is that the expectation value of the electric field amplitude of a coherent state oscillates with time. – Mark Mitchison Jan 22 '15 at 12:42
  • I have edited to clarify – lionelbrits Jan 22 '15 at 13:05
  • Thanks, I also just came to the same conclusion about adding classical sources. I think this is the answer I am looking for. – Mark Mitchison Jan 22 '15 at 13:09
  • @Sofia I believe this contains the seed of an answer to my question. When I get time I will actually do the calculation and post it here. – Mark Mitchison Jan 22 '15 at 13:29
  • Thank you for your answer. You describe how both classically electrostatic and electrodynamic fields can be constructed as coherent states. From what I've read about QED, people tend to think of the "electrostatic" part of the EM field as virtual, with virtual photons that in a number of ways aren't real. This is consistent with classical expectations, in that electrostatic fields don't carry energy density. I've previously only thought of coherent states as existing for far-field waves, i.e., real photons. – Dragonsheep Mar 07 '20 at 21:16
  • By constructing coherent states for electrostatic fields, it seems to ascribe to them several properties, e.g., energy, that aren't associated with the virtual photon picture of electrostatics. My question is, if both electrostatic and electrodynamic fields have associated with them coherent states, how are the coherent states different so as to preserve the differences in observables between real/virtual photons (or equivalently, near/far field radiation). Or, in other words, how can electrostatics be transmitted both by "virtual" photons and have associated with them real coherent states? – Dragonsheep Mar 07 '20 at 21:24