Quantum Mechanics – What is the Quantum State of a Static Electric Field?

electromagnetismfield-theoryquantum mechanicsquantum-field-theory

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.)

Best Answer

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.

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