0
$\begingroup$

I'm trying to understand the nature of the wavefunction within an interferometer as described below, so I can make accurate predictions about it's behavior.

The experiment goes like this. A laser light source shines through two beam splitters in series, and then onto a detector. The first splitter sends a portion of the beam on a long detour, greater than one meter, before joining the direct path at the second beam splitter. The optical elements are angled in just such a way to cause interference between the direct and lagged beams across the width of the detector. The last part of the apparatus is a series of neutral density filters placed directly in-front of the laser output that attenuates the beam so that there is only one photon in the apparatus at a time. Even with this condition, and the fact that the two light paths differ in length by a very large amount, an interference pattern still occurs at the detector.

schematic of the basic setup

And the results of interference with the attenuators in place, averaged over many photons, with and without one of the paths blocked

Now, I'd like to make one modification to this experiment.

Instead of a continuous source of light, that is attenuated to deliver single-photon levels of light, I'd use a pulsed light source, with a pulse duration so that each light pulse was only like a centimeter long. Then I'd still pass this pulsed beam through the ND filter to reduce it to single-photon levels. My prediction is that with and without the ND filter, there would no longer be interference.

In the first experiment, I imagine that the wave function is distributed evenly throughout both arms of the setup and for a very long time, so that even though the two paths differ by a significant distance, interference still occurs.

In the second experiment, I imagine that the wave function does not overlap in time, so that the amplitude at the screen would not exhibit interference bands.

Are these predictions accurate, or incorrect? Why?

edit:

I thought to add a little more context. This issue came up after I saw the first of my two described experiments performed by Huygans optics on youtube, see the video here https://www.youtube.com/watch?v=SDtAh9IwG-I

$\endgroup$

2 Answers 2

0
$\begingroup$

The term "interference' is very misleading especially as taught in high school/1st year university, basically 1930s material. 2 photons never cancel each other, that would be a violation of energy conservation. Famous physicists Dirac (1940s) and Feynman(1960s) discussed the problem in terms of the double slit experiment ... as is known today each photon determines and takes its own path! The EM field controls everything ... an excited electron, before it gives up a photon, is already interacting with the field.

In your "interferometer" experiment it is possible to set up optical paths where the EM field does not allow transmission (max interference) or does allow transmission (no interference). IT does NOT matter if you pulse or use one photon at a time ... the EM field guides all photons.

$\endgroup$
10
  • $\begingroup$ So, you would say that there is no difference between using a pulsed laser source or a continuous source? if I angle one of the return beams just so, I get interference fringes across the CCD sensor in the usual continuous beam without attenuation. Would you say that these fringes should be present also in the pulsed case? $\endgroup$ Commented Jun 12, 2023 at 15:32
  • $\begingroup$ Fringes on the CCD ... are you looking at the CCD visually or is this the recorded image you are talking about? 2 issues 1) just shining a beam on a ccd makes "interference" as the beam spreads the distance from source to ccd varies in circles by multiples of 1/2 lambda. 2) the surface of the ccd is a regular array of pixel geometries and this can cause interference (like looking at a CD under light). $\endgroup$ Commented Jun 12, 2023 at 15:39
  • $\begingroup$ Yes you will observe in both cases .... pulsed or continuous. $\endgroup$ Commented Jun 12, 2023 at 15:42
  • $\begingroup$ I added some pictures to try and make the setup more clear. $\endgroup$ Commented Jun 12, 2023 at 15:56
  • $\begingroup$ I don't know what Feynman and Dirac came up with, but the solution is trivial and already completely contained in Copenhagen. There simply are no quanta in the free fields. Quanta are the results of irreversible energy exchanges (i.e. they only "exist" in the theory AFTER the Born rule is applied). Hence the question "which path the electron or photon took" is entirely unphysical. Energy doesn't have a position and it doesn't have a path. That is simply an objectification fallacy that historically goes all the way back to Einstein's 1905 paper on the photoelectric effect. $\endgroup$ Commented Jun 12, 2023 at 16:17
0
$\begingroup$

Your logic is correct. You will not in general see fringes with a pulsed laser, unless pulses overlap temporally on the screen/camera matrix. This is typically achieved by equating the length of interferometer arms.

As matter of fact, under certain circumstances you will not see fringes even in the setup you draw. Namely, if the difference between path-lengths of the two interferometer arms is more than the so-called coherence length of the source laser. For a typical off-shelf HeNe laser from e.g. Thorlabs that would be something like 10-15cm.

$\endgroup$
1
  • $\begingroup$ Perhaps you can just mention that there are ways to make different pulses mutually coherent. The result is called a frequency comb. In such a case one can get interference from different pulses. $\endgroup$ Commented Jun 15, 2023 at 13:05

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.