Quantum Measurements
post by paulfchristiano · 2010-12-19T01:59:24.944Z · LW · GW · Legacy · 10 commentsRelated to: The Quantum Physics Sequence, particular Decoherence is Pointless.
If you really understand quantum mechanics (or the linked post on decoherence) you shouldn't get anything out of this post, but understanding this really cleared things up for me, so hopefully it will clear things up for someone else too.
Here at lesswrong we are probably all good Solomonoff inductors and so tend to reject collapse. We believe that a measurement destroys interference because it entangles some degrees of freedom from a quantum system with its environment. Of course, this process doesn't occur sharply and discontinuously. It happens gradually, as the degree of entanglement is increased. But what exactly does "gradually" mean here? Lets focus on a particular example.
Suppose I run the classic two-slit experiment, but I measure which slit the electron goes through. Of course, I don't observe an interference pattern. What happens if we "measure it less"? Lets go to the extreme: I change the polarization of a single photon depending on which slit the electron went through (its either polarized vertically or horizontally), and I send that photon off to space (where its polarization will not be coupled to any other degrees of freedom). Do I now see just a little bit of an interference pattern?
A naive guess is that strength of the interference pattern drops off exponentially with the number of degrees of freedom entangled with the measurement of which slit the electron went through. In a certain sense, this is completely correct. But if I were to perform the experiment exactly as I described---in particular, if I were to polarize the photon perfectly horizontally in the one case and perfectly vertically in the other case---then I would observe no interference at all. (This may or may not be possible, depending on the way nature chose to implement electrons, photons, and slits. Whether or not you can measure exactly is really not philosophically important to quantum mechanics, and I feel completely confident saying that science doesn't yet know the answer. So I guess I'm not yet decided on whether decoherence really happens "gradually." )
The important thing is that two paths leading to the same state only interfere if they lead to exactly the same state. The two ways for the electron to get to the center of the screen interfere by default because nature has no record of how the electron got there. If you measure at all, even with one degree of freedom, then the two ways for the electron to get to the same place on the screen don't lead to the exactly same state and so interference doesn't occur.
The exponential dependence on the number of degrees of freedom comes from the error in our measurement devices. If I prepare one photon polarized either horizontally or vertically, and I do it very precisely, then I am very unlikely to mistake one case for the other and I will therefore see very little interference. If I do it somewhat less precisely, then the probability of a measurement error increases and so does the strength of the interference pattern. However, if I create 1000 photons, each polarized approximately correctly, then by taking a majority vote I can almost certainly correctly learn which slit the electron went through, and the interference pattern disappears again. The probability of error drops off exponentially, and so does the interference.
Another issue (which is related in my head, probably just because I finally understood it around the same time) is the possibility of a "quantum eraser" which destroys the polarized photon as it heads into space. If I destroy the photon (or somehow erase the information about its polarization) then it seems like I should see the interference pattern---now the two different paths for the electron led to exactly the same state again. But if I destroy the photon after checking for the interference pattern, how can this be possible?
The answer to this apparent paradox is that erasing the data in the photon is impossible if you have already checked for an interference pattern, by the reversibility of quantum mechanics. In order to erase the data in the photon, you need to measure which slit the electron went through a second time in a way that precisely cancels out your first measurement; there is no way around this. This conveniently prevents any sort of faster than light communication. In order to restore the interference pattern, you need to bring the photon back into physical proximity with the electron.
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comment by lucidfox · 2010-12-20T02:59:12.992Z · LW(p) · GW(p)
Here at lesswrong we are probably all good Solomonoff inductors and so tend to reject collapse.
Isn't that hindsight bias?
I wonder how many Less Wrongers would be preferring Copenhagen by now if Eliezer originally argued for it...
Replies from: paulfchristiano, wedrifid↑ comment by paulfchristiano · 2010-12-23T01:55:54.175Z · LW(p) · GW(p)
I thought collapse was extremely unlikely before Overcoming Bias existed, as I strongly suspect did most less wrongers who had thought seriously about the question.
comment by mwengler · 2010-12-20T21:44:14.388Z · LW(p) · GW(p)
I haven't read through all the interpretation stuff yet, but I have done a lot of work in the past on quantum properties of radio waves, i.e., photons.
I believe you could build an interference slit which would rotate the polarization of the photon (or part of the photon however you want to think about it) going through one of the slits 90degrees compared to the polarization going through the other. I think you merely need to put what is called quarter-wave plate covering one of the slits and not the other.
BUt I am sure that on the other side of the intereference slit, you will see NO two-slit interference. Indeed, consider the more general case that you rotate the polarization of the photon by phi through one slit vs the other. The amplitude on the other side the slit will be proportional to sin(r) oneslitpattern + cos(r)twoslitpattern which means the intensity will be sin^2(r) oneslitpatteern + cos^2(r)twoslitpattern because orthogonal polarizations to do not at all interfere with each other.
I provide this to help insure your gedanken experiments are realistic before you do too much more gedanking.
comment by Manfred · 2010-12-19T04:28:58.550Z · LW(p) · GW(p)
This actually may be a good place to use the machinery of the Copenhagen interpretation - amount of entanglement between the electron and the photon is exactly what you're trying to describe, and should (notably) give you the interference pattern regardless of amount of entanglement as long as you're entangling things but not measuring them yourself. Measuring the partially-entangled photon will then give you a mixed state on the electron that changes the interference pattern in exactly the right way.
Also, you can't create 1000 photons because of the no-cloning theorem.
Replies from: paulfchristiano, JoshuaZ↑ comment by paulfchristiano · 2010-12-19T05:58:50.140Z · LW(p) · GW(p)
I think this is wrong. If you entangle the photons with the electron, even if you don't measure them, you won't see an interference pattern.
You can create as many photons as you want, each entangled with the original one. The no-cloning theorem says that you can't create 1000 unentangled but identical photons. Actually, you can even create 1000 unentangled copies of a starting photon if you know that it is initially in one of two orthogonal states.
Replies from: Manfred, Manfred↑ comment by Manfred · 2010-12-20T00:12:28.308Z · LW(p) · GW(p)
Okay, after some consideration:
Any reversible (by which I mean not leaking heat to outside) process should entangle the photon and the electron, leading to the consequences I described. This condition is a fairly reasonable way to read your post, but apparently not what you intended - you intended a non-reversible process, which has the consequences you describe, and explains how an apparently uninformed observer would see the interference pattern go away: the information leaked out.
↑ comment by Manfred · 2010-12-19T06:27:18.787Z · LW(p) · GW(p)
I think this is wrong. If you entangle the photons with the electron, even if you don't measure them, you won't see an interference pattern.
It's possible I'm wrong. Sadly, I'd have to actually do the math to demonstrate it with entanglement, and I am feeling astoundingly lazy right now. Probably less lazy tomorrow.
You can create as many photons as you want, each entangled with the original one. The no-cloning theorem says that you can't create 1000 unentangled but identical photons.
Hmm, you're right that if it's only along 1 of 2 axes the theorem wouldn't apply, since you could just measure it. I guess this would only apply to an actual electron-photon entanglement process that could output a mixture of horizontal and vertical. Since just having it be discrete requires realio trulio measuring the electron, that would seem to support it destroying the interference pattern, too.