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.)

  • $\begingroup$ 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. $\endgroup$ – Mark Mitchison Jan 22 '15 at 2:25
  • $\begingroup$ 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. $\endgroup$ – Timaeus Jan 22 '15 at 4:48
  • 1
    $\begingroup$ @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. $\endgroup$ – Mark Mitchison Jan 22 '15 at 5:11
  • 1
    $\begingroup$ 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. $\endgroup$ – Holographer Jan 22 '15 at 13:01
  • 1
    $\begingroup$ @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. $\endgroup$ – lionelbrits Jan 22 '15 at 13:07

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.


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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Mark Mitchison Jan 22 '15 at 12:42
  • $\begingroup$ I have edited to clarify $\endgroup$ – lionelbrits Jan 22 '15 at 13:05
  • $\begingroup$ Thanks, I also just came to the same conclusion about adding classical sources. I think this is the answer I am looking for. $\endgroup$ – Mark Mitchison Jan 22 '15 at 13:09
  • $\begingroup$ @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. $\endgroup$ – Mark Mitchison Jan 22 '15 at 13:29
  • $\begingroup$ 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. $\endgroup$ – Dragonsheep Mar 7 at 21:16

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.