Eigenvalues of a quantum field?

Fields in classical mechanics are observables. For example, I can measure the value of the electric field at some (x,t).

In quantum field theory, the classical field is promoted to an operator-valued function of space-time. But no one talks about eigenvectors of the quantum field! If I try to measure the field at one point in spacetime, I should get a real value which should be an eigenvalue of the quantum field, right? I guess the eigenvectors of the quantum field also live in Fock space?

Answer

If I try to measure the field at one point in spacetime, I should get a real value which should be an eigenvalue of the quantum field, right? I guess the eigenvectors of the quantum field also live in Fock space?

Yes, that's basically correct. If the value of the field at a point is observable, the eigenvalues of the operator representing it are the values the field can attain at that point. And the eigenvectors live in the Hilbert space of states, which you can think of (at least conceptually) as $L^2(\{\mbox{initial boundary conditions}\})$. This Hilbert space is a Fock space in free field theories.

There's a couple of subtleties worth mentioning:

The value of the field at a point might not be a physical observable. In electrodynamics, for example, you can't actually measure the value $A_\mu(x)$ of a component of the connection 1-form; instead, you can measure gauge invariant quantities like the curvature $F_A(x)$ and the holonomy $Hol_L(A)$ along a loop $L$. Likewise, in nonlinear sigma models, where the classical fields are maps $\phi: \Sigma \to X$ to some curved manifold, you can't measure the value $\phi(x)$. Eigenvalues are complex numbers, not points on a manifold. But you do get a real observable $\mathcal{O}_f(x)$ for every function $f: X \to \mathbb{R}$; measure the value of $f(\phi(x))$.

It's also not strictly correct to say that quantum fields are operator-valued functions on spacetime. The physical problem is that if you measure the value of the field at one point, you'll disturb the field near that point, affecting the values at other nearby points. The closer you look to the place where you made the measurement, the bigger the disturbance; even in free scalar field theory, the 2-point correlation function $\langle \phi(x) \phi(y) \rangle$ blows up as $x \to y$. This tells you that the fields aren't quite functions, because you can't multiply the 'value at a point' observables when they live at the exact point.

The mathematically correct thing to do is to think of the field (and more generally local observables constructed from fields) as an operator-valued distribution. Distributions are mild generalization of functions; they are objects which don't have values at a point, but which do have average values in an arbitrarily small (but finite) region. Basically, for any test function $f$ on your spacetime, you get an operator $\phi(f)$ which you can think of as measuring the value "$\int f(x) \phi(x) dx$" of $\phi$ sampled by a probe with resolution $f$. Distributions can only be multiplied when their singularities don't coincide; they exhibit the same obnoxious behavior that quantum field operators do.

Probably you don't have to worry about this too much. For one thing, even if you can't (strictly speaking) define an operator $\phi(x)$, you can still safely talk about the correlation function $\langle \phi(x)\phi(y)\rangle$. (It's the kernel function of the multilinear map $(f,g)\mapsto \langle \phi(f)\phi(g) \rangle$.)

Physicists don't spend a lot of time worrying about solving the eigenvalue problem for the field operators. Usually the spectrum is all of $\mathbb{R}$, and finding the eigenvectors isn't worth the trouble. There is one important exception though: In the Standard Model, it's pretty important that the vacuum vector is an eigenvector of the Higgs field operators, with non-zero eigenvalue.