2

According to a post on a mailing list here: https://groups.google.com/d/msg/sci.physics.research/n7Ao7N9aeq0/wqgj20-jBQAJ

In general, quatum states |Psi> of the electromagnetic field, no matter if free radiation field or interacting field, can be represented as wave functionals

Psi[A^mu(x),t]

on the space of classical field configurations A^mu(x).

What does a one-photon state look like when expressed as such a wave functional? I have tried to develop the theory myself by canonically quantizing the field, but I'm stuck because I can't figure out how to mathematically work with such wave functionals---they belong to some confusing uncountably-dimensional vector space.

Brian Bi
  • 6,491
  • 27
  • 62

1 Answers1

4

That's a naïve point of view; however there is a rigorous construction of "wavefunctionals". It is a point of view initiated by Segal and then continued by Nelson and called the (free) Markoff field. It is rigorously understood for free fields, and in some special case also for interacting ones.

The idea is related to the fact that it is possible to link functorially the Fock space (representation of free fields) with a suitable $L^2$ space of "functionals" in a way such that the free field is the multiplication by the coordinate of the space, and the corresponding momentum is (functional) derivation, plus a correction term that is proportional to the coordinate as well.

The key is to exploit the bilinear nature of free field Hamiltonians to obtain gaussian measures. Without entering too much into details, the free classical scalar field $\phi: \mathscr{S}(\mathbb{R}^d)\to \mathbb{R}$ is a distribution that obeys some dynamical equation. Such equation translates (I am omitting a lot of details here) into a quadratic form $Q$ acting on $\mathscr{S}(\mathbb{R}^d)$. The exponential $e^{-\frac{1}{2}Q(f)}$ of such quadratic form defines a gaussian probability measure $\nu$ on the space of distributions $\mathscr{S}'(\mathbb{R}^d)$.The wavefunctionals' space is taken to be $L^2(\mathscr{S}',d\nu)$, and it is proved to be equivalent to the Fock representation. An easy example of such functionals is the constant functional $1(\phi)=1$. It is square-integrable, since the gaussian measure is a probability measure, i.e. $\int_{\mathscr{S}'}d\nu=1$, and it corresponds to the (Fock) vacuum state of the free scalar field. Coherent states will be of the form $C_f(\phi)=e^{i\phi(f)}$, where $\phi\in\mathscr{S}'$ is the coordinate and $f\in\mathscr{S}$ is fixed.

yuggib
  • 11,987