# Photon wavefunction

## Main Question or Discussion Point

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.

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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.

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

The relationship between the classical EM description and the quantum mechanical description is rather subtle, and depending on what aspect you want to look at, things can be formulated differently and sometimes, at first sight, even contradictory. But when you understand the relationship well enough, you can often see the veracity of the different apparently contradictory statements.

Here are a few.
The classical EM description is of course a 4-vector field over spacetime:
A(x,y,z,t), from which the more well-known description in terms of E and B fields can be derived: E(x,y,z,t) and B(x,y,z,t).

The quantum description of the free EM field consists of Fock space, which has the basis:
|0> the vacuum
|k1,e1>, |k2,e2>, ... the 1-photon states
|k1,e1,k2,e2>,|k3,e3,k4,e4> ... the 2-photon states
....

Each (k,e) pair corresponds to a momentum vector (3-dim) and a polarization vector e which can take on 2 different values.

The above basis is the basis of a hilbert space called "Fock space" and all possible linear combinations of the above states are the possible quantum states of the free EM field.
But only very special combinations correspond to classical EM waves ; these are called "coherent states". Coherent states are special superpositions of the vacuum, 1-photon, 2 - photon ... 28621-photon... states which are eigenvectors of the so-called destruction operator. There is a coherent state that corresponds to each possible state of the classical EM field, but in Fock space there are many, many more possible states.

However, the 1-photon state |k,e> behaves *in certain respects* as the classical EM wave that corresponds to a plane wave with wave vector k and polarization e. It is not the equivalent (that would have been the coherent state, which CONTAINS |k,e> as a term, but has many others in them too) of the classical EM wave, but for certain aspects, one can *pretend* it to be like the classical EM wave. For instance, the probability of detection on a screen of the one-photon state is proportional to the intensity of the classical EM wave with wavevector k and polarization e. So IF YOU KNOW WHAT YOU'RE DOING, you can jump back and forth sometimes between a 1-photon state and a classical EM wave, because things come out the same for certain quantities.
And then people get sometimes a bit sloppy, and say that "the 1-photon state interferes with itself like the classical EM wave" and things like that. In the proper context, this is an operationally correct statement.
And then others say that 1-photon states are totally different from a classical EM wave which needs many photons to be correctly described. This is ALSO correct (the coherent state contains n-photon states).