2
$\begingroup$

I have a problem with understanding what is the role of coherence in such experiment. Taking the Dirac's statement that photon interferes only with itself, it's fairly understandable, that single photon virtually splits to pass trough different slits and interferes with itself, therefore the optical path difference (if I can even talk about it in this example) is pretty much 0, because it can't actually split. It might expand or something, but definitely not split. But my interpretation must be wrong, because in such situation coherence wouldn't matter. Interference would always occur, because every photon would interfere with itself, not with the photons from delayed beam, and OPD would always be 0. I think I missed something really simple...?

$\endgroup$
1
  • $\begingroup$ Could you clarify why a photon could not split in your opinion? And what do you mean by an expanding photon? $\endgroup$
    – fffred
    Commented May 29, 2013 at 22:04

1 Answer 1

6
$\begingroup$

You are correct that every photon will interfere with itself for sure. But for the whole interference pattern to be observed, you need a large number of photons independently interfering with themselves. And, these large number of photons should be identical in every respect so that they can be represented by the same single photon wavefunction. This makes sure that all interference pattern formed on the screen is same for all the photon wavefunctions.

In other words, good temporal coherence means that the independent photons emitted have a very narrow bandwidth. So, if we assume that the photon source has perfect temporal coherence, then it means that all the photons emitted have the same wavelength. Since the fringe width after interference depends on wavelength, therefore all the independent photons obey the same probability distribution of fringe pattern for that particular wavelength. If, the temporal coherence of source goes down, then it means that the bandwidth increases. So, the photons of different wavelengths follow different interference fringe pattern probability distribution. Therefore, the overall effect will be blurring of the interference pattern. Therefore, to observe the interference "pattern", you need a coherent source, even though each individual photon interferes with itself.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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