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I have designed an experiment. Without going into detail it resolves around the double slit quantum eraser experiments. If we can infer the location of a particle without actually measuring it, does it remain in a quantum state? After all, we have not measured or detected that particle in any way. If not, it can not be the act of measuring it, but merely the fact we know about it that effects its quantum state. This does not fit any model I know of. If it is true that you do not need to measure or detect a particle to affect its state, what implications does that have for quantum mechanics?

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How will you check that you correctly inferred the location of the particle? –  Raskolnikov Aug 25 '12 at 5:58

3 Answers 3

A particle is always in a quantum state. The only thing that causes that quantum state to change in the manner known as a "measurement" (or wavefunction collapse) is an actual interaction with another particle. The mere fact that you know or don't know something about the quantum state is irrelevant.

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Thanks for the responses guys. I guess what I am getting at is: In the classic double slit experiment you get the interference pattern. But if you put a detector, lets say at slot B, which will detect the passage of the photon but not block it, you get what I call the particle stripes pattern. Now, we only measured the photon at slot B. If we get a hit on the screen, but don't see the photon go through slot B, we know that the photon must have gone through slot A. So, we can infer with confidence the route of the photon without having measured it. Does it's wavefunction collapse? If so why? –  Matt Aug 25 '12 at 9:28
    
@Matt: It collapses because the detector entangles the photon wavefunction with the detector wavefunction, so that you can't get interference, because interference is only between paths which end up in exactly the same state, and the detector is in a different state in this case in the two paths. –  Ron Maimon Aug 25 '12 at 11:35
    
@Matt: but the detector does measure the photon. That's the only way you can tell it went through one slit or the other. –  David Z Aug 25 '12 at 23:31
    
@DavidZaslavsky: The point is that the detector doesn't detect anything when the electron only went through the slit where the detector isn't present, and yet the wavefunction still collapses. –  Ron Maimon Aug 26 '12 at 4:40
    
The absence of a "hit" on the detector is still a detection, though. –  David Z Aug 26 '12 at 4:57

This is a famous intuitive paradox in quantum mechanics, the null-result measurement. There are many examples:

  • If you take a particle in a spherical wave, and have a hemispherical detector, and it fails to detect the particle, you know the particle is going in a certain direction. This means that the failure to detect the particle leads the particle to collapse to carry a certain net momentum. I think this example is discussed by Peierls.
  • If you have a detector in a two-slit experiment, and it fails to detect the particle, the interference still goes away. This is your example.
  • If you shine a laser to continuously detect if an atom is in it's ground state, the first excited state never decays, because the (failing) measurement keeps decohering the ground state.

All of these examples have the property that the quantum measurement fails, nothing happens, and yet the wavefunction collapses becuase the null result reveals to you that the quantum wavefunction is not in the state you are checking.

The ultimate version of this is the Elitzur-Vaidman bomb-tester, which is a device which confirms whether a mirror is doing a measurement, without the mirror ever successfully getting a measurement outcome. The way it is usually stated, you have a bunch of bombs set off by a very sensitive mirror, that detects the recoil of a single photon. You also have duds, which are just ordinary mirrors that don't do a measurement on the photon. Can you tell them apart without exploding the bombs? The answer is yes, and this is an extreme example of doing a counterfactual measurement in QM--- you can tell if the mirror would measure the photon, were you to shine it, without having any appreciable probability of the mirror actually detecting anything.

This property is not paradoxical in many-worlds derived intepretations, because all that is going on is that the detector is entangling with the particle, and the branch where the detector doesn't go off is special, in that it only has a particle relative state which is collapsed. This is the standard way of understanding these effects in modern interpretations, and these only differ in philosophy regarding what causes the collapse, or if it happens at all.

In other interpretations, you can consider this as an indication that the quantum state is a measure of information about the particle, since getting information about the particle's position affects the wavefunction, or else not. This is all philosophy, so I don't think it is so important.

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Thanks so much Ron, that really helps. I will be back with a follow up question, once I clear my head. Did you read about the theory that the wave function may be something which actually exists, rather than just a handy mathematical construct? I feel a crossover here but can't see it yet, but it is there I am sure. –  Matt Aug 26 '12 at 2:46
    
@Matt: On those days that I think QM is exact, I interpret it many worlds. In this case, collapse is just reduction of the wavefunction relative to the observer's state, and it is somewhat mystical, since the observer's state is classical, while the quantum state is quantum, and the two don't mesh exactly. Objective collapse theories are generally no good, there is no real mechanism for making collapse happen, you can't make the SE nonlinear. The only alternative is Bohm's theory, but this is also not great, because it is too big, it has wavefunction plus position variables. –  Ron Maimon Aug 26 '12 at 4:44
    
.... unfortunately, last week I figured out this new way to embed quantum mechanics in a normal probability space in a natural way, so I sometimes believe that too. That's described in the answer to 'tHooft's most upvoted question. Still, I haven't recovered full QM from this, only some toy systems, and I am not confident in this formulation. If this alternate view works 100% correctly, then wavefunction collapse would be nothing more than Baysian probability collapse, and quantum mechanics is only an approximate description, and wouldn't work for quantum computation. –  Ron Maimon Aug 26 '12 at 4:46

So basically we can view this as a particle being carried on a wave. The wave is an actual physical thing, in the same way that a magnetic field is real, or a gravitational field is real, albeit nothing we can actually see - we can just infer it is there because we see how it affects other things. So I see this wave spreading out in every direction from the source, like when you throw a pebble into a pool, but the particle it carries is only going in one direction, ie, through one or other of the slits. Of course we don't know this until later when certain measurements are not made/made. Until then it is everywhere on that wave front. And the act of measuring, but failing to detect a particle as the wave carrying that particle passes through the detector causes the whole wave to collapse, leaving in it's wake a Newtonian particle which just continues on it's way all the way to the detector.

So we can - again - have an effect on a particle remotely. Sort of the opposite of twinning. You collapse the wave by failing to detect a particle here, and the particle which is over there loses uncertainty. So the question is: how do we measure a particle to find out if it is in a quantum state or not?! LOL IS this a philosophical quantum paradox? ;O)

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Without a wave to carry it the particle can not move. failing to measure a particle you are looking for as the wave carrying that particle passes collapses the wave. Immediately a new wave appears to the carry the particle to it's destination. this wave originated at the slit the particle passed through, so there is no wave to pass through the other slit, and this is why we do not see any interference. But there is always a wave. I can't express this mathematically, but I'll bet an arm and a leg that it is correct. someone cleverer than me should do the maths. –  Matt Sep 28 '12 at 16:23

protected by Qmechanic Jun 23 '13 at 1:10

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