# Photon wavefunction

1. Feb 25, 2006

### jauram

Hi all,

we know that the state of material particles (like electron) can be described by a wavefunction and that the wavefunction has a probabilistic interpretation. I have studied at very introductory level the quantum theory of electromagnetic radiation and it seems that it is built up in a way that does not require the notion of photon wavefunction. So the question is that is it meaningful (at least for pedagogical purposes) to consider the photon wavefunction (giving the probability distribution of finding the photon in space)? And how far can we say that the classical field (the solution of Maxwell's equations) represents that wavefunction? Thanks.

2. Feb 25, 2006

### Perturbation

If one were to look at it from the view of quantum field theory, the wavefunctions of quantum mechanics are classical fields because they aren't quantised.

The "wavefunction" describing a photon would be the electromagnetic potential vector $A_{\mu}$, as it is in QED, satisfying

$$\partial _{\mu}F^{\mu\nu}=j^{\nu}$$
$$F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$$
$$\nabla A_0=\mathbf{E}$$
$$\nabla \times \mathbf{A}=\mathbf{B}$$

Or, equivalently E and B themselves.

If you're considering a wavefunction as a scalar function giving the spatial or momentum distributions and the probabilistic properties of the particle under consideration, then the electromagnetic field vector does exactly this, except it defines the distribution of the electric and magnetic fields (though it's a four-vector, not a scalar, but this doesn't matter). To treat the electromagnetic field quantum mechanically we pass from the notion of a field as the passive mediator of force to a dynamical body in its own right. For this we make the leap from the classical field of Maxwellian Electrodynamics to fields as operators responsible for the creation of state vectors themselves, whose Hamiltonian we then quantise. The state vectors describing the state of the electromagnetic field are eigenstates of this Hamiltonian, having energy $H_m=\hbar \omega (m+\tfrac{1}{2})$, where m is the mode of the field and $\omega$ its frequency.

Last edited: Feb 25, 2006
3. Feb 26, 2006

### Ratzinger

Vanesch wrote two months ago a nice post on this. I take the freedom to quote this here.