In the last decade there were several papers claiming that they've measured a "transverse quantum state" / "quantum wave-function" / "spatial Wigner function" of a single photon:

Most of them refer to a Iwo Bialynicki-Birula's paper "Photon wave function" [Prog. Opt. 36, 245 (1996), arXiv:quant-ph/0508202] when describe the measured object (or don't refer to anything). Having read these papers and some other literature discussion, as well as this forum (see links below), I still cannot really understand of what exactly was measured by the authors of the experimental papers and weather it makes sense to call that a "photon wavefunction", so I assume I am missing something important.

I wonder if

  • they've measured the electric field / amplitude of Maxwell mode with a single-photon in it (then what's so quantum about that, other than you have to do long counting)?

  • they've measured the spatial momentum quantum state, but by the means of Fourier optics transformed it into distribution over spatial points (they why to call it the way they do)

  • there is a reason to introduce a real quantum spatial wave-function of a single photon (than how it lines up with the absence of position operator for a photon and other problems discussed in the topics below?)

and I would greatly appreciate if someone can help.


2 Answers 2


The questions seems to boil down to "what is so quantum about a single photon?" In the absence of interactions, a single photon is described by the same mathematics as a classical electromagnetic (EM) field. (Helmholtz equation for the electric field for instance when one considers the propagating field without charges and currents.) The only difference is the interpretation of the quantity in the equations. (The modulus square of the wave functions gives a probability density while the modulus square of the electric field gives the intensity.) So, if one can have a classical EM field (bright laser light) that has a Laguerre-Gaussian mode, then one can also have a single photon with that mode describing its spatial wave function.

Those experiments that measure single photon wave functions are therefore almost the same as those that would measure the classical EM fields, with the only differences being the source that would produce (effectively) single photons and the physical detection process. For the single photon detection, one would typically use avalanche photo diodes that are sensitive to single photons.

The issue with the lack of a position operator for the photon does not imply that one cannot represent its wave function in the spacetime domain. All one needs for the latter is a coordinate basis. This basis does not have to be eigenstates of some position operator. In fact, it would not make sense for such a coordinate basis to be eigenstates of a position operator, because the coordinate basis needs to be four-dimensional whereas the set of eigenstates would only be three-dimensional.

  • $\begingroup$ a single photon is described by the same mathematics as a classical electromagnetic (EM) field [...] So, if one can have a classical EM field [...], then one can also have a single photon with that mode describing its spatial wave function. This doesn't quite make sense to me. A wavefunction is a complex scalar, which you can square to get probabilities. A solution to Maxwell's equations is a real tensor, and squaring a tensor won't give you a probability measure. (It will be of even rank ≥2, but the relativistic density of a scalar needs to be the timelike component of a rank-1 tensor.) $\endgroup$
    – user4552
    Commented Sep 8, 2019 at 15:55
  • $\begingroup$ BTW, I wasn't the downvoter. I think there's some very helpful material in this answer, but I don't think it's 100% right. $\endgroup$
    – user4552
    Commented Sep 8, 2019 at 15:57
  • $\begingroup$ @BenCrowell: Is there any reason why a wave function cannot carry internal degrees of freedom? Is there any reason why you cannot why you cannot compute the modulus square of a vector field (or tensor field)? In the latter case you'll get the intensity as opposed to the probability distribution, but I have state in the answer that the interpretation of the quantities are different. $\endgroup$ Commented Sep 9, 2019 at 4:10
  • $\begingroup$ @BenCrowell: I added some clarifications $\endgroup$ Commented Sep 9, 2019 at 4:15

Here is a paper giving the maxwell wavefunction of a photon.


(I am aware that a similar derivation using the A potential exists, but do not remember where I saw it.)

So it is a wavefunction, which is complex.

It has a chapter on measuring the wave function.

If a single-photon state of the field is created, then to know its quantum state means to know its electric and magnetic field distributions in space and time. Such a state is a single-photon wave-packet state, and its generation is an important goal in quantum-information researc

Recently, a technique has been developed to measure the transverse spatial quantum state of an ensemble of identically prepared photons . The single-photon light beam is sent into an all-reflecting, out-of-plane Sagnac interferometer, which performs a relative rotation of 180° and a mirror inversion on the wave fronts of the counter-propagating beams. The Sagnac performs a two-dimensional parity operation on one of the beams relative to the other. The fields are recombined at the output beam splitter and are interfered on a photon-counting photomultiplier tube (PMT), allowing the emerging beams to be detected at the single-photon level. The mean photo-count rate is directly proportional to the transverse spatial Wigner function at a phase-space point that is set by the tilt and translation of a mirror external to the interferometer.

So as all probability distributions, it needs an ensemble at the end.(mean photo count rate)

  • $\begingroup$ This doesn't answer the question. $\endgroup$ Commented May 13, 2019 at 7:38
  • $\begingroup$ @EmilioPisanty "what exactly was measured " is answered in the quote "The mean photo-count rate" , i.e. a distribution mathematically connected to the formula of the wavefunction., not for each individual photon an electric or magnetic field..( which is in a complex number state) $\endgroup$
    – anna v
    Commented May 13, 2019 at 8:06
  • $\begingroup$ @annav, thank's for the reply! I saw this paper and some others (including Iwo's), that interpret Riemann–Silberstein vector as photon wave function. My confusion is that it's essentially a classical quantity,and I cannot see any difference (or additional challenge) between the performed measurements and waveform measurement of classical light pulses (wo phase stabilization) . Basically, the only difference seems to be that one has to count photons or just integrate with much longer exposure, as nothing specific about the photon statistics is observed. $\endgroup$
    – Basil
    Commented May 13, 2019 at 14:29
  • $\begingroup$ @basil, the amazing thing is that when maxwell;s differentials are used as qm operators the photon wave function can be modeled consistently, so that the normal electric and magnetic fields appear from the QED formulation of photons with creation and annihilation operators acting on the photon wavefunction. see this blog motls.blogspot.com/2011/11/… . For consistent physics, all classical should emerge from the quantum underlying fields.. The existence of single photons is enough to make the problem quantum mechanical at root. $\endgroup$
    – anna v
    Commented May 13, 2019 at 14:37
  • $\begingroup$ see sps.ch/en/articles/progresses/… $\endgroup$
    – anna v
    Commented May 13, 2019 at 14:40

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