Original illustration by me. Use as you please

In short: the question is, does the length of the path affect the outcome of detecting a photon?

Consider the single photon beam splitter experiment. Does the probability of detecting the photon change if the distance between the detectors is unequal? Because light has a fixed velocity c. If the photon is detected at a place that is nearer (detection means absorption unless some special means are used), then it can no longer be anywhere else.

This might be the case in Wheeler's delayed choice experiment: http://www.sciencenews.org/pictures/112010/essay_delayed_zoom.gif As the particle detection path is shorter when that path is chosen, the photon will always be detected there?

If so, could temperature gradient on the detection plate cause unequal expansion of the plate on quantum scales affecting the outcome of detecting photon because the place where they reach on the plate first is different?

  • 2
    $\begingroup$ Where is the interference happening in this experiment? I don't see the beams being recombined. $\endgroup$
    – Dan Piponi
    Mar 22, 2013 at 23:12
  • $\begingroup$ The beams recombine as shown in this experiment: sciencenews.org/pictures/112010/essay_delayed_zoom.gif $\endgroup$ Mar 23, 2013 at 3:38
  • $\begingroup$ But even without the interference, the main question is, will the probabilities of detecting the photon change if the path difference to detectors is unequal. Logically, they should still be 50%, but I would ideally like to know experimental findings if possible. $\endgroup$ Mar 23, 2013 at 3:53
  • $\begingroup$ Because if the probabilities experimentally turn out to be different, it would explain a lot of things! $\endgroup$ Mar 23, 2013 at 4:13

3 Answers 3


There is the basic confusion here, in my opinion, of the wave/particle identity.

When we speak of photons we are in the quantum mechanical regime, it is an elementary "particle". The quotation marks are necessary because it is not a particle like a billiard ball, and it is not a wave like an acoustic wave, or even a classical electromagnetic wave. It is a mathematically described "entity" which, depending on the experiment will act either like a billiard ball, i.e. a point in four dimensional space but with specific quantum numbers, (in the case of the photon spin , polarization and zero mass) or like a probability wave .

Probability wave in bold to emphasize that the energy of the entity when appearing as a wave is not distributed, as the energy in sound waves, in space. The entity will appear always with a specific (x,y,z,t) ( within the heisenberg uncertainty principle) but the probability of finding it there will display the properties of waves, interference patterns..

The experiment you are setting up does not have the ability to detect the wave nature of the probability distribution of the photon. If it is done in vacuum the distance will play no role to the efficiency of detection. It will take a little longer due to the velocity of light but the material of the detector will not make a difference: the photon will either be there or not.

In the delayed choice experiment the description makes the same mistake. The individual photon does not take both paths. The interference pattern appears because of the statistical accumulation of many photons which then display the probability wave aspect of the photon wave function. Each individual photon takes a specific path, but the probability of taking it is affected by the interference setup.

  • $\begingroup$ Thanks for taking the time to answer this. My question really stemmed from the setup of the experiment as shown in the link: sciencenews.org/pictures/112010/essay_delayed_zoom.gif When, the particle path was chosen, then only the particle was observed and no interference. My question arose becuase of the unequal path lengths between the particle path and the wave interference path. If the particle was observed (and absorbed) at the shorter path, how can we observe the interference on the other path. $\endgroup$ Mar 27, 2013 at 18:01
  • 1
    $\begingroup$ That is what I am explaining. A "particle" is not split or spread out or whatever in space. It always will appear at a definite (x,y,z,t) ( within the HUP) , whole. It is the probability distribution that displays wave properties, not the particle. Take a dice and throw it 100 times , you get a probability distribution from 1 to 6. The dice is not split between 1 and 6. The interference appears because the wave function carries the information for the probability from the boundary conditions at the origin of the beams, and the interference will appear statistically. $\endgroup$
    – anna v
    Mar 27, 2013 at 19:38
  • $\begingroup$ have a look at the wiki link I gave in the answer of the interference build up of individual electrons. The same is true for photons $\endgroup$
    – anna v
    Mar 27, 2013 at 19:56
  • $\begingroup$ youtube.com/… The Quantum Eraser video at the youtube link also helped a great deal to further my understanding. Thanks for the help. $\endgroup$ Mar 27, 2013 at 23:45
  • 2
    $\begingroup$ this last video also carries the misundarstanding that if you know the slit it went through then the interference pattern disappears. Read the wiki link in my answer about the experiments that do know which slit the particle/photon went through and the interference does not disappear. It is the change in the boundary conditions of the experiment that change the interference pattern: too destructive. There are new experiments with non destructive detection that keep the pattern. $\endgroup$
    – anna v
    Mar 28, 2013 at 4:20

This is not an interference experiment (the paths are not combined). It is a setup to measure the two-photon correlation function (a Hanbury Brown-Twiss experiment). It measures $g^{(2)}(t=L/c,t'=3L/c)$ which is zero if the input state is a single-photon state.

What happens is that the "single-photon wave function" (WF) propagates to the beam splitter and gets split into to components there, one is transmitted, the other one reflected. At some time $L/c$, the transmitted wave function arrives at the close by detector and the wave function is projected onto the device. Then, with a particular probability (which is 1/2 for a balanced beam splitter), the detector will click. In this case, the reflected wave component disappears (instantaneously). If the detector does not click, the "whole probability" moves into the reflected wave component and a little later, at time $3L/c$, the other detector will click. So, the first detector measurement changes the wave function (in a similar vein as in the Einstein-Podolsky-Rosen experiment, the difference is that the wave function there is a two-particle wave function).


Classically, if you had a source that sent light in all directions, in double the distance the light would spread out to four times the area. So the intensity would be 1/4 as much.

For a single photon, the path it follows fits a probability wave. There is some chance that it arrives at your detector, and some chance that it misses. The bigger the distance, the more chance it misses the detector.

If both detectors are large enough that the photon will not miss, then I think you should get an equal probability that the photon is detected at either detector, regardless of the distance.

Remember that the beam-splitter has been calibrated. If it did not give an equal probability with the same distance, then it was adjusted until it did.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.