1
$\begingroup$

When the electromagnetic field is quatized for a single mode, we first take Maxwell equations and proceed to write the electric and magnetic field as a stationary wave, since we consider the electromagnetic field to be in a cavity. (Introductory quantum optics, Gerry and Knight).

After some commutation rules, the electric and magnect field are written as operators.

Classically, the electric and magnetic field are wave functions, since they are solution of the wave equation and Maxwell equations.

So, they start classically as a wave function but when quantized they become observables. How's that possible? I can undestand the position and momenta to be observables when quantized since those are the quantities we want to measure, but when talking about the electromagnetic field we are talking about the entity itself. I don't know if I'm just overthinking or it's really strange that a wave function becomes an observable.

$$\langle \psi| \hat{E}| \psi\rangle $$ is the expected value of the electric field but now there is another object taking the role of wave function, $\psi$. How can we understand this wave function $\psi$ in the classical description?

$\endgroup$
  • 1
    $\begingroup$ In this article you will find how the classical electromagnetic wave emerges from an enormous number of quantum mechanical entities/photons . motls.blogspot.com/2011/11/… $\endgroup$ – anna v Oct 16 '15 at 5:21
3
$\begingroup$

You're confusing the technical term wavefunction (i.e. a function $\psi:M\to\mathbb C$ defined on some configuration space $M$ which obeys a Schrödinger equation and which gives the probability of finding the system in some patch $N\subseteq M$ as $\int_N|\psi(q)|^2\mathrm d\mu(q)$) with the much more loosely-defined "function which obeys some form of linear wave equation". The electric field is the latter but it is definitely not the former. There is nothing crazy about quantizing it.

Quantization means taking our observables, which used to have definite values, and making them operators on some suitable Hilbert space. For the electromagnetic field, our observable was a function of position, so what you get is an operator-valued function of position. Nothing all that mysterious, really. If you find the idea weird, though, then welcome to quantum field theory!

In going forward, it's a lot easier to detach the wavefunction from its pedestal of The Description of a given quantum system. In general, you simply have a quantum state $|\psi⟩$ which lives in some abstract Hilbert space $\mathcal H$, and it doesn't need to represent a function of position. The wavefunction is simply one possible representation of the state, $⟨x|\psi⟩$, when that makes sense. When it doesn't, the state is just the state.

Doing this will slightly blunt the pain and confusion that you'll feel when the time comes for you to see quantum field theorists turn the actual wavefunction (i.e. $\psi(\mathbf r)$) into an operator $\hat \psi(\mathbf r)$ as well.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ My confusion comes from the fact that classically the electric field is the solucion of the Maxwell equations which can be considered as the schrödinger equation for photons. That means that the electric field is a wavefunction in the sense you pointed out before. But there is kind of an inconsistency in considering the electric field as an operator when quantized. Either it is a wave or an operator but not both. $\endgroup$ – E.phy Oct 15 '15 at 22:13
  • 1
    $\begingroup$ That is a misconception. The Maxwell equations are not the Schroedinger equations for a photon - there isn't really such a thing. They do produce, when suitably handled, the Schroedinger equations for the quantized field and is interaction with matter, but that takes place over the (abstract) Hilbert space for the field. Keep on reading, Gerry & Knight do a good job of explaining it if I remember correctly. $\endgroup$ – Emilio Pisanty Oct 15 '15 at 23:22
  • 1
    $\begingroup$ Well if you have the free electromagnetic field, the modes decouple, so you can get an equation for each mode, but it will just say that the number of photons in that mode is constant. If you add some charged matter to your model, the modes don't decouple anymore and you're in trouble. This paper (arxiv.org/abs/1304.6842) contains a calculation from a Schrödinger equation in QED, on one-particle states, and you can see that even then it's quite involved and technical. $\endgroup$ – Robin Ekman Oct 16 '15 at 0:52
  • $\begingroup$ @EmilioPisanty Very nice answer. I have a similar but more specific query of the way this is dealt with in Gerry Knight's book. This is my post. $\endgroup$ – user110903 Jul 7 '17 at 11:03
4
$\begingroup$

The electric field is really a whole bunch of observables, one for each point in space.1 The usual thing to do in QFT is to work with the 4-potential $A_\mu$, so let's do that. By taking the Fourier transform and some technicalities, we find the representation $$A_\mu(x^\nu) = \sum {e_\mu^i}^*(\mathbf k) \exp(-ik^\nu x_\nu)a^\dagger _i(\mathbf k) + e_\mu^i(\mathbf k) e^{ik^\nu x_\nu} a_i(\mathbf k) $$ where the sum is over allowed $k^\nu$ and polarizations $e_\mu^i(\mathbf k)$. The operators have the commutation relation $$[a^\dagger_i(\mathbf k), a_j(\mathbf k') ] = \delta_{\mathbf k \mathbf k'}\delta_{ij}. \tag{1}$$ The familiar argument from the harmonic oscillator then shows that the state space of this system has a basis, the number representation, where the basis vectors are labeled by the series of integer eigenvalues of $a^\dagger_i(\mathbf k) a_i(\mathbf k)$. More concretely: by the number of photons in each mode.

The Hamiltonian is something like $$H = \sum \hbar\omega(\mathbf k) a^\dagger(\mathbf k) a(\mathbf k)$$ so there is indeed a Schrödinger equation $i\hbar \partial_t |\psi\rangle = H|\psi\rangle$. Since the algebra (1) is a direct sum of CCR algebras $[x, p] = i\hbar$, you can at least formally write the Schrödinger equation for the electromagnetic field as a differential equation. But it will be a PDE on an infinite-dimensional space. I'm not sure if that makes sense mathematically and I don't think anyone really wants to do that.

This should not really be surprising. Because you know that the wave function is a function on configuration space, the dimension of which is the number of degrees of freedom. For $N$ particles in three spatial dimension, $3N$. But the electromagnetic field has an infinite number of degrees of freedom -- an infinite number of modes -- so its wavefunction must be a function on an infinite-dimensional space.

But then how do we get to the classical electromagnetic field, i.e., the function on physical space, $\mathbf E(\mathbf x,t)$? (Or the 4-potential ) Well, consider the following state, $$|\tilde{A}_\mu(\mathbf{k})\rangle = \exp \big[ \sum_\mathbf k \tilde{A}^\mu(\mathbf k)\epsilon(\mathbf k)^i a_i^\dagger(\mathbf k) \big ]| 0 \rangle$$ (note: all the terms in the exponential commute). This is an eigenstate of each $a_i(\mathbf k)$ with eigenvalue $\tilde{A}_\mu e^\mu_i(\mathbf k)$. We call it a coherent state.

Now if you calculate the expectation value of $A_\mu$ in the state, it will be $$ \langle\tilde{A}_\mu(\mathbf{k})| A_\mu(x^\nu) |\tilde{A}_\mu(\mathbf{k})\rangle = \sum \tilde{A}^\rho(\mathbf k) e^i_\rho(\mathbf k) e^i_\mu(\mathbf k) \exp(i k^\nu x_\nu) + \text{c.c.} $$ where $\text{c.c}$ is complex conjugate. But this is (up to constants) a real function $\tilde{A}_\mu(x^\nu)$ expressed in terms of its Fourier transform $A_\mu(\mathbf k)$.

This is really entirely analogous to the harmonic oscillator where the coherent states $\exp(\alpha a^\dagger)|0\rangle$ have an expectation value $\langle x\rangle \propto \alpha + \alpha^*$. Also, completely analogously, a coherent state remains a coherent state under time evolution, but expectation values evolve according to classical dynamics. This is the proper way to pass to the classical limit.


1 Well, technically, we should talk about distributions that don't necessarily have point values, but this is not really relevant for the present discussion.

| cite | improve this answer | |
$\endgroup$

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.