photon wave function, double slit, single photon source

Question

There's an old argument by Newton and Wigner, that the photon as a massless particle can't have a position operator and therefore no position space wave function.

How does this tie in with the double slit experiment? If we use a single photon source, can't we simply define the position space wave function based on the interference pattern?

If not, what QM entity does the interference pattern correspond to? I have a vague idea of how an attenuated coherent state doesn't allow us to speak of the wave function of a single photon but I am wondering what happens if we explicitly use a single photon source.

Also, has this experiment been performed? I find only attenuated laser experiments.

Answer 1

There's an old argument by Newton and Wigner, that the photon as a massless particle can't have a position operator and therefore no position space wave function.

The paper you're thinking of is
T. Newton and E. Wigner, "Localized States for Elementary Systems," Rev. Mod. Phys. 21, 400–406 (1949) doi:10.1103/RevModPhys.21.400.

Photons are concepts that arise from second quantization or quantum field theory. This means that field configurations (e.g. $E(\mathbf r)$) become the operators in the theory and $\mathbf r$ is simply a parameter. Note that strictly speaking a massive particle can't be localized either as you can't localize better than the particle's Compton wavelength without creating particle-antiparticle pairs. The difference is at low enough energies $E<mc^2$, you have an effective non-relativistic description (which is the non-relativistic quantum theory that students are first exposed to). However, if $m=0$, such as with photons, there is no non-relativistic limit.

The reason why relativity matters is due to the fact that the generators of motion that transform reference frames is given by Galilean transformations (which commute) in a non-relativistic theory, but Lorentzian transformations (which do not commute) in a relativistic theory. Another equivalent way of stating this non-localizability is the fact that you can localize either the electric or magnetic part of the field but not both due to the transversality condition of EM theory (see e.g. arXiv:0903.3712).

How does this tie in with the double slit experiment?

The interference you see in a double slit experiment is due to interference of the field mode itself (which is why you will also see interference with a classical field). You can roughly think of a classical EM field mode as a single photon wavefunction (see e.g.arXiv:quant-ph/0508202 for a full discussion), in which case two-slit interference can be thought of as simply a "photon only interfering with itself" (to use Dirac's words). So there is little difference between the two-slit with single photons and a classical coherent state (consisting of many photons each prepared in the same state).

Also, has this experiment been performed? I find only attenuated laser experiments.

Yes, single photon experiments with linear optics devices is so common as to be considered routine in quantum optics labs (which may be why you had difficulty finding papers). The best place to find data for these kinds of experiments are either buried in quantum optics papers (because two-slit interference with single photons is so common you couldn't publish with this alone) or educational resources such as here.

Answer 2

Punk Physicist's answer is spot on. But I'd like to add a little to his/her last two paragraphs, in particular, a description of what it is that you see in an interference pattern.

You cannot define a position observable, but you can of course define the state of the second quantized field. Moreover you can describe the probability amplitude for a photon to be absorbed by an ideal detector at a given point in space and time. This absorption probability amplitude is related to a one-photon Fock state $\psi$ of the quantum light field as follows:

$$\begin{array}{lcl}\vec{\phi}_E(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{E}}^+(\vec{r},t)\left|\left.\psi\right>\right.\\ \vec{\phi}_B(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{B}}^+(\vec{r},t)\left|\left.\psi\right>\right. \end{array}\tag{1}$$

where $\psi$ is the (Heisenberg picture) light field quantum state, $\mathbf{\hat{B}}^+,\,\mathbf{\hat{E}}^+$ are the positive frequency parts of the (vector valued) electric and magnetic field observables and, of course, $\left<\left.0\right.\right|$ is the unique ground state of the quantum light field. This relationship is invertible, i.e., given the vector valued $\vec{\phi}_E,\,\vec{\phi}_B$, one can uniquely reconstruct the one-photon light field quantum state, so you can think of it as being a particular representation of the one-photon state. The entities in (1) fulfill Maxwell's equations and thus tie in well with Iwo Bialynicki-Birula's discussion (arXiv:quant-ph/0508202) that Punk Physicist referred you to.

From these vector probability "amplitudes", the probability density to absorb a photon at a given place and time is proportional to the analogue of the classical energy density:

$$p(\vec{r},\,t) = \frac{1}{2}\,\epsilon_0\,|\vec{\phi}_E|^2 + \frac{1}{2\,\mu_0}\,|\vec{\phi}_B|^2\tag(2)$$

This is likely to be a pretty good model, at least qualitatively, of what a photon counting tube, CCD or indeed your eyes "see". Doubtless eyes (photon absorbing atoms) and even photon tubes need a more complicated description than simply a simple lowering ladder operator acting on the quantum field, but there is in principle no problem with an idealized detector along the lines described above, whereas there is a fundamental problem with a position observable as described in the Wigner and Newton paper.

Scully and Zubairy, "Quantum Optics" give a good summary of this in their first and fourth chapters. They also wrote a great summary for article edited by the October 2003 issue of Optics and Photonics News