It's very simple to write down the ground state wave-functional for any free field, and to give it a completely satisfying quasi-rigorous meaning (you will see what I mean by quasi-rigorous). You have to start with the ground state wavefunction for the Harmonic oscillator:
$$\psi(x) = C e^{-\omega {x^2 \over 2}} $$
Where C is a normalizing constant. Then note that each k-mode of a free bosonic field theory (or any mode in curved space) is a Harmonic oscillator, so that it's ground state is an independent Gaussian. The ground state wavefunction is therefore the product over all k of the
$$ \psi(\phi) = C \prod_k e^{-\omega(k) {\phi(k)^2\over 2} \epsilon^3} $$
Which, when you multiply it out (meaning turn the product into a sum in the exponent), becomes the formula in question, Fourier transformed. The $\epsilon^3$ is defining the scale of normalization of the fluctuations of the field mode, it is the size of k-steps, or the volume of space, and it reproduces the integral when you turn the multiplication into a sum.
It is best to say that the integral in the exponent is not really a single integral, because the square root of a differential operator is not a local operator. The ground state wavefunctional does not consist of locally independent fluctuations, but of independent fluctuations at each separate k (or each separate mode).
The quasi-rigorous interpretation associates a constructed measure to this as follows--- to take a statistical sample from a free field theory, simply select a Gaussian random number for each k mode of the appropriate variance, and Fourier transform the random numbers. This produces a sampled distribution $\phi$ in the ground state of the Euclidean continuum field theory. It's a distribution because the variances don't get small very fast at large k.
This is a well-defined randomized computational procedure which produces a mathematical object whose statistics converges in the limit that the random reals you choose are chosen to more precision, and the number of modes simultaneously goes to infinity. There is no problem with either limit, the resulting random picks for $\phi$ converge in distribution to the correct thing.
But because this construction, no matter how concrete, involves picking random real numbers, actually infinitely many, it is not allowed as it stands as a rigorous construction, because modern mathematics does not allow you to speak about picking random numbers freely, because it is not true that every set of real numbers is measurable in standard axiomatizations.
This is why this construction is quasi-rigorous and not rigorous. Usually, when something is not rigorous, this is the fault of the thing. In this case, it is the fault of the axiomatization we use to define rigorous. There is nothing wrong with this definition of free fields, and the headaches in making it formal are simply due to bad foundations for statistics. In a universe where every set of R is measurable, you just say you have described the ground state wavefunction by the above randomized algorithm.
This complete description of the ground state of field theory has been used in several places, most recently in quantum information theoretic analysis of the entanglement content of a quantum field vacuum. There are many references in the last 5 years.
I should also say that you can do the same thing for pure U(1) gauge theory, pure electrodynamics, it is also a free field theory. In this case, the vacuum functional depends on the gauge and you are best off doing it in Dirac gauge, where the independent oscillators are physical.
The interacting analog of the vacuum gauge functional is any convergent stochastic algorithm for sampling distributions from the ground state of the interacting theory, in an infinite limit analogous to "picking more modes" in the free field theory. For example, in pure gauge theory, the algorithm which produces an equilibrated configuration on a lattice, in the limit that the lattice is tiny. This vacuum functional encodes all the information about the correlation functions, so describing it explicitly is out of the question in a field theory that is not exactly integrable.
Q&A (4871)
