I have read this question:

EM wave function & photon wavefunction

Wave function of a photon?

where Arnold Neumaier says:

photons do not have a spatial probability density

He specifically says that in the case a laser is shone on a surface, the beam does have a spatial probability density, for the surface where the beam hits it. That is understandable, but the double slit experiment is about single photons being shot one at a time, and they have to pass through slits, not shone directly on the screen.

Though, this paper talks about spatial probability density and photons.


Now the double slit experiment works well with both electrons and photons.

Though, electrons do have a spatial probability density.


To summarize, the probability distribution of the outcome is the normalized square of the norm of the superposition, over all paths from the point of origin to the final point, of waves propagating proportionally to the action along each path. The differences in the cumulative action along the different paths (and thus the relative phases of the contributions) produces the interference pattern observed by the double-slit experiment.


As I understand the main difference is that photons do not have a position operator.

Why doesn't there exist a wave function for a photon whereas it exists for an electron?

where Chiral Anomaly says:

Saying that a photon doesn't have a wavefunction can be misleading. A more accurate way to say it is that a photon doesn't have a strict position observable.

photon wave function, double slit, single photon source

where Punk_Physicist says:

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 interference you see in a double slit experiment is due to interference of the field mode itself


Without strict photon position observable (spatial probability density), how is the double slit experiment possible?

  • $\begingroup$ Arnold also says in the very answer you link that they do have a probability of crossing a spatial surface. Can you explain why you think this is insufficient to account for the double slit (hint: the detection screen is a surface)? $\endgroup$ – ACuriousMind Apr 7 '20 at 20:22
  • $\begingroup$ @ACuriousMind good point but "probability density of hitting any given surface crossing the beam at a particular point of the surface." He talks about a laser beam hitting a surface. That is clear. But in the double slit, the photons, shot one at a time, have to pass through the slits, and not as he describes. I thought he was talking about a direct laser beam being shone on a surface, and then the probability density is OK. But I will edit to make that clear. $\endgroup$ – Árpád Szendrei Apr 7 '20 at 20:33
  • $\begingroup$ @ACuriousMind can you please reopen? $\endgroup$ – Árpád Szendrei Apr 7 '20 at 22:47
  • $\begingroup$ I won't vote to leave closed to give other people a chance to voice their opinion, but this honestly doesn't meet my standard for clarity. You simply seem to be misreading Arnold's post. The phrase "photon (in a laser beam, say)" means "photons, for example in a laser beam". It is not clear why you think that giving a laser beam as an example somehow makes the statements in it invalid for situations that are not a laser beam. I would expect an explicit argument that explains what you specifically think is lacking in order to account for the double slit. $\endgroup$ – ACuriousMind Apr 8 '20 at 7:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.