Destructive interference in the wavefunction, is this conceptually explainable?

Xilor

Hello, destructive interference seems like an important part of quantum physics, but I'm finding it very hard to grasp it conceptually. For instance in the Elitzur–Vaidman bomb tester, destructive interference in the mirror is used to determine if one of the paths is blocked. What exactly is happening that causes every single photon to choose only one of the paths when the wave comes from both sides? Why is the mirror so significant that it can cause this behavior and how is it decided which of the paths the photons will always take?

Xilor

Bumping as I'm still interested.

Demystifier

The answer is somewhat controversial, because it depends on the interpretation of quantum mechanics, which is not unique. Perhaps the most intuitive interpretation is the Bohmian one, according to which particle is a pointlike object the velocity of which depends on the value of wave function on the position of the particle. Thus, particle takes only one path, depending on the initial position of the particle.

Actually the mirror is not very important, it only plays an auxiliary role by directing the wave beam into a desired direction. A more important peace of apparatus is the beam splitter, which splits the initial wave beam into two beams. In the Bohmian interpretation, the wave is splitted but the particle is not.

Xilor

So the wave goes both ways but the particle only one in Bohmian Mechanics? Is it then possible to combine that 'empty' wave and the particles wave again in a way that affects the particle? That sounds like something that should either be possible in all interpretations, or in none.

And how exactly does the initial position matter? Does it have something to do with the lengths of the paths they take untill hitting the final beamsplitter? (when I said mirror earlier on, I meant the half-silvered one at the point where the beams combine again)

Edit:

Also, how do they even make the beamsplitters precise enough to make exactly 0 photons appear on one side? Wouldn't even the smallest deviation from the 50% reflection on either of the splitters cause different results? Or does the same thing happen if the mirror reflects 25% or 75% of the photons?

Demystifier

Yes.

Yes.

Yes, the two beams can be recombined in all interpretations.

I'm not sure I understand that question, try to reformulate it.

Fine, that's the same as beam splitter.

Well, in reality they have something like 50.1:49.9 beam splitter, which at the end means that the number of photons at one side is close to zero, but not exactly zero.

Xilor

Thank you for your responses! It's starting to make some sense now.

My question was about the following thing which you said:

My question was what exactly the influence of this initial position is, what detail about the position ends up translating to the eventual choice of paths? My own guess in that question was that it might have something to do with the pathlengths towards the final beamsplitter, as that would mean one half of the wave would reach the beamsplitter quicker.

So, wild idea here. What if you direct the empty wave and the particle carying wave towards a double split screen, but you seperate the slits with some wall in between (only on the side where the photons arrive from) so that you know the empty wave always goes through the left, and the particlewave through the right slit. Since you said they can still interact, a interference pattern should still emerge, while you do have which slit information. Am I correct? Because that does sound strange since I've always heard that which slit information destroys the interference pattern.

JamesOrland

I am not a supporter of Bohm so if I'm incorrect please someone correct me, but what you in fact have in this interpretation are two waves, and the knowledge that one of them is real and the other is empty is not available (that is, you don't know which of the waves is empty and which is in fact a particle).

The interference pattern will be destroyed, yes, and you will only find out which wave is the empty one and which is the particle after you detect it beyond the two slits.

Demystifier

The answer is much easier to draw than to tell in words, but let me try. In such a situation, two different possible paths never cross. So, for example, if the initial position of the particle is at at left side of the beam, then after the splitting the particle will end up in the left beam.

Well, if something caries the which-slit information, then it eats up the information about the phase of the wave, which means that the wave decoheres and the interference pattern is destroyed. In some cases that destruction of information is reversible (i.e., information about the phase can be restored), but in most cases it is not.

However, your separation of the slits is something different. While it prevents interference as long as the separation is present, it does not carry any which-slit information. In other words, the mere fact that there is a wall does not enable you to tell which slit the particle passed through. In particular, I don't see how did you conclude that the empty wave will be the left one, and not the right one. Indeed, if the wall is no longer present at some larger distance from the slits, the interference can be seen there.

Xilor

I'm afraid I don't understand.. What do you mean with the left side of the beam? The photons are the beam themselves, so this would suggest to me that there's not only particles on the left side, but also on the right side. Following the logic, some photons would also have to end up travelling the other path, creating a stream of photons on both sides of the final beamsplitter.

Let me draw it out, I might just be misunderstanding what which path information actually is, but it seems to make sense to me.

First the destructive interference setup as I understand it:

http://i.imgur.com/wVTvK.jpg

Now, add the double split setup with a separating wall. Seemingly this altered setup should either not give an interference pattern, or it should retroactively change what way the photons went, or should give an interference pattern while the path is known with a very high degree of certainty (there's the not completely 0 photons thing, and I suppose photons can tunnel through the wall).

http://i.imgur.com/RDuM5.jpg

JamesOrland

In the last situation you drew, there will be an interference pattern, yes. I really don't know how to explain it in Bohmian terms so I will leave it to someone who might, but yes, in that case, the wall doesn't matter at all, and you will observe the interference pattern on the detector.

Xilor

So there's an interference pattern while you can say through which slit the photon went?

What if you go even further and add the delayed choice experiment at the end instead of just a regular detector screen. Am I correct in saying that if the choice were made to detect which slit information (After the photon would've passed through the slits), that it's always going to be the 'right' slit, while if the choice is made to detect interference that there's going to be interference?
I would really love to see the results of an actual experiment like that because it sounds so mind boggling.

JamesOrland

No, not at all. The way you drew it, you cannot say through which slit the photon went until the detector. Even though you drew one of the waves as being photon-containing and the other as being empty, you do not know that. You don't have access to the information of which wave is empty and which wave is not. You didn't put any detectors anywhere that would contain which-wave information. If you had, then yes, the interference pattern would show up, but the way you drew it, you do not have any means of saying 'photon beam' or 'empty wave' before the detector screen.

I will take a raincheck on answering that because I haven't read about the delayed choice experiment yet so I cannot say :P

Xilor

Could you explain why this is the case? Because it does seem like there is information. You know from previous experiments that the particle always went into a certain direction, so why can't you say that you know the particle had gone in that direction?
We didn't actually measure the which path information in any way that would collapse the wavefunction, but to require a wavefunction collapse for a wavefunction collapse seems weird. It would seem to reduce the statements about how knowledge of the path affects things to circular logic like: 'If you collapse the wavefunction by measuring the particles path, then the wavefunction is collapsed'.

Cthugha

This is plain wrong. The setup drawn is a typical Mach-Zehnder interferometer. For such an interferometer the probability distribution at the final output ports is a function of the phase difference along the two paths. If a single coherent beam is used at the entrance of the interferometer, the remaining phase difference is entirely a function of geometry. In this case this is the path length difference between the two paths. As one can measure this difference, one also knows the probability distribution at the exit port. It is not necessary to actually measure it.

JamesOrland

I'm quite very sorry. I was looking at the experiment you drew and thinking of another, different experiment (one that contained another photon-gun). Please disregard my previous explanations!

I am not sure what would happen in the case of the last drawing, but I believe it would be equivalent to the case where there is a single photon-gun shooting at a single-slit wall.

Demystifier

No, by beam I meant the wave beam, and in the Bohmian interpretation the wave is not the photons.

Cthugha

To be honest I also misinterpreted the drawing twice before I understood what it is intended to show. I agree with you. At least if the intentionof the setup is that only one of the two slits is illuminated by the photon beam. If the photon beam illuminates both slits, it gets complicated.