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?


2 Answers 2


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.

  • $\begingroup$ Very good answer! $\endgroup$
    – Michael
    Jan 6, 2013 at 9:24
  • $\begingroup$ Thanks! Nice point about Higgs. A follow up question: is the vacuum expectation of a real Klein Gordon field zero? $\endgroup$
    – hwlin
    Jan 10, 2013 at 6:05
  • $\begingroup$ @hwlin: No, not necessarily. $\endgroup$
    – user1504
    Jan 10, 2013 at 14:58
  • 1
    $\begingroup$ @user1504: Can you gimme a reference where the eigenvector and eigenvalues of a quantum field operator is discussed? $\endgroup$
    – rainman
    Apr 12, 2013 at 8:59
  • $\begingroup$ @Ome Glimm & Jaffe. $\endgroup$
    – user1504
    Apr 12, 2013 at 13:48

The eigenvectors of a quantum field are states with a definite value of the field:

$$ \hat{\phi}(x) \left| \Phi \right> = \phi(x) \left| \Phi \right> $$

These are not states which have a definite number of particles, i.e., they are superpositions of Fock space states with different numbers of particles. The easiest way to see this is to write the field operator in terms of creation and annihilation operators (schematically):

$$ \hat{\phi}(x) \sim \sum_k a_k + a^\dagger_k $$

and note that this does not commute with the particle number operator $ \hat{N} = \sum_k a^\dagger_k a_k$:

$$ \left[ \hat{N}, \hat{\phi}(x) \right] \neq 0 $$

So you can't have a state which is simultaneously a state with a definite value of the field and a state with a definite number of particles.

  • 4
    $\begingroup$ Nicely put. Maybe worth mentioning that, in the particle picture, the eigenvectors of the field operators are a kind of coherent state, being eigenvectors of a creation/annihilation operator. $\endgroup$
    – user1504
    Jan 5, 2013 at 5:36
  • $\begingroup$ $\hat{\phi}(x) \equiv \hat{\phi}(x,t)$ ? $\endgroup$
    – rainman
    Apr 11, 2013 at 15:28
  • $\begingroup$ What does the eigenstate $|\Phi\rangle $ represent? $\endgroup$
    – rainman
    Apr 20, 2013 at 16:15
  • $\begingroup$ @Ome Same as always in QM: a state with a definite value for the observable. Note that in general the field operators at different times do not commute. You have to take the fields on a spacelike hypersurface to get a complete set of commuting observables. $\endgroup$
    – Michael
    Apr 21, 2013 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.