Can quantum fields be viewed as superpositions of classical fields?

Question

At the end of my undergraduate quantum mechanics class, we looked at phonons. You can let $x_i$ be the position operator of an nth quantum harmonic oscillator, and couple the harmonic oscillators with a potential. The Hamiltonian looks something like: $$H=\sum_i \left(\frac{p_i^2}{2m}+\frac{1}{2} m \omega^2 x_i^2\right)+\sum_{i} \frac{1}{2}m\omega^2 (x_i-x_{i+1})^2$$

You can do the procedure of raising and lowering operators to find:

$$H=\sum_k \hbar \omega_k\left(a_k^\dagger a_k+\frac{1}{2}\right)$$

and even figure out that the ground state in the position basis looks something like the following (ignoring exact frequencies and normalization and all of that):

$$\langle x_1,\cdots,x_N|\psi\rangle=e^{-x_1^2-x_2^2-\cdots-x_N^2}$$ or equivalently,

$$|\psi\rangle=\int \mathrm d^N~ x e^{-x_1^2-x_2^2-\cdots-x_N^2}|x_1,\cdots,x_N\rangle$$

The $a_k^\dagger$ operators are interpreted as creating one phonon in the kth mode.

Taking the continuum limit, we've quantized the scalar wave equation. I want to write something like the following for the ground state: $$|\psi\rangle=\int \mathcal{D}[\phi] e^{-\int d^dy\phi(y)^2}|\phi\rangle$$

I'm studying elementary quantum field theory, and I'm finding it hard to get a straight answer as to whether that's a correct interpretation of a quantum field theory. For electrodynamics, I'd have commuting observables $\bf \vec{E}$ and $\bf \vec{B}$ at every point in space. (Commuting because I want them to form a complete basis of states). I could represent the field as superpositions of eigenvectors of these operators, and get something as above. If $E(y)$ and $B(y)$ are vector fields which are eigenvalues of the above operators at every point in space, that means I could imagine writing any quantum state of my quantized electrodynamics as something like:

$$|\psi\rangle=\int \mathcal{D}[(E,B)] f(E,B)|(E,B)\rangle$$

where the six components of the vector field $(E(y),B(y))$ play the same role as $\phi(y)$ did in the previous example, and as $(x_1,\cdots,x_N)$ did in the example before that. $f(E,B)$ would be a complex number which is a functional of the fields $E$ and $B$.

Sure, I expect ill-definedness and infinities and cut-offs necessary everywhere, but is this intuitively what is going on in defining quantum fields?

Answer 1

As stated in some of the comments, the weighting functional inside the functional integral is sometimes called a wave functional. This type of functional representation can be called the functional Schroedinger representation. Please see the following review by Roman Jackiw.

For a general interacting field theory, the exact solution for this wave functional is of course unknown.But an educated guess of an approximate wave functional with possibly a finite number of free parameters may lead to useful approximate solutions.

This method was suggested by Jackiw in the above review. This variational method has been used to reproduce the perturbation results of QED and QCD. (Plesae see this article by Heinemann, Iancu, Martin, Vautherin).

The variational approach is under active research for an explanation of quark confinement please see a recent work by Vastag, Reinhardt and Campagnari. However in 3+1 dimensions it hard to work without gauge fixing which casts doubts on the non-perturbative conclusions due to the Gribov problem. In most cases the trial wave functional is taken as Gaussian.

It is very plausible that this method can be adapted to all models in 1+1 dimensions which can be solved by a Bogoliubov transformation.

Also, there is a known non-Gaussian solution to the pure Yang-Mills theory known as the Kodama state given by the exponential of the Chern-Simons functional. This solution exactly satisfies the Schroedinger functional equation. This wave functional is considered to be unphysical, however, please see the following article by Witten describing some interesting properties of this state.

A quantum field theory defined by a Lagrangian and boundary conditions may lead to different solutions each describing an inequivalent quantization. Thus, there is the possibility that these different quantizations correspond to different solutions of the vacuum wave functionals, thus the solutions to the functional Schroedinger equations are plausibly non-unique.

Answer 2

I wanted to directly answer your final question.

Yes, you're on the right track here. You should expect for any quantum field theory that there is a set of observables (much like the coordinate operators $X_i$ in point particle quantum mechanics) whose eigenvectors form a basis the Hilbert space. These observables should even be local observables, like the time-zero scalar field value observables $\phi(\vec{x},0)$.

There's two sources of complication.

First, there's analytic fine print needed to deal with the fact that the observables' locations $\vec{x}$ are continuous. This leads to distributions, regularization, and infinities.

Second, most of the interesting QFTs have constrained dynamics. For scalar field theory, you get lucky and you can (modulo analytic fine print) span the Hilbert space with eigenvectors of the field-value observables $\phi(\vec{x}, 0)$. But for most other Lagrangian QFTs (Dirac, Maxwell, Yang-Mills, and their various combinations), the constraints imply that the 'basis' of all simultaneous field-value operators eigenvectors is over-complete or incoherent. (For example, the electric and magnetic field-value basis you suggest has too many degrees of freedom. The electric field is basically a momentum variable conjugate to the vector potential. Using it in addition to the B-value operators is like trying to span a point-particle Hilbert space with momentum and position operators.)