Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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: 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?

share|cite|improve this question
Where is the interference happening in this experiment? I don't see the beams being recombined. – Dan Piponi Mar 22 '13 at 23:12
The beams recombine as shown in this experiment: – Physics n00b Mar 23 '13 at 3:38
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. – Physics n00b Mar 23 '13 at 3:53
Because if the probabilities experimentally turn out to be different, it would explain a lot of things! – Physics n00b Mar 23 '13 at 4:13
up vote 1 down vote accepted

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.

share|cite|improve this answer
Thanks for taking the time to answer this. My question really stemmed from the setup of the experiment as shown in the link: 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. – Physics n00b Mar 27 '13 at 18:01
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. – anna v Mar 27 '13 at 19:38
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 – anna v Mar 27 '13 at 19:56… The Quantum Eraser video at the youtube link also helped a great deal to further my understanding. Thanks for the help. – Physics n00b Mar 27 '13 at 23:45
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. – anna v Mar 28 '13 at 4:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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