# So you think you understand Quantum Mechanics

post by shminux · 2012-12-22T21:16:27.350Z · score: 38 (42 votes) · LW · GW · Legacy · 64 comments

This post is prompted by the multitude of posts and comments here using quantum this and that in an argument (quantum dice, quantum immortality, quantum many worlds...). But how does one know if they understand the concept they use? In school a student would have to write a test and get graded. It strikes me as a reasonable thing to do here, as well: let people test their understanding of the material so that they can calibrate their estimate of their knowledge of the topic. This is an attempt to do just that.

Let's look at one of the very first experiments demonstrating that in the microscopic world things are usually quantized: the Stern-Gerlach experiment, in which measured angular momentum is shown to take discrete values. The gist of the experiment is that in a varying magnetic field the tidal force on a magnet is not perfectly balanced and so the magnet moves toward or away from the denser field, depending on the orientation of its poles. This is intuitively clear to anyone who ever played with magnets: the degree of attraction or repulsion depends on the relative orientation of the magnets (North pole repels North pole etc.). It is less obvious that this effect is due to the spatially varying magnetic field density, but it is nonetheless the case.

In the experiment, one magnet is large (the S-G apparatus itself) and one is small (a silver atom injected into the magnetic field of the large magnet). The experiment shows that an unoriented atom suddenly becomes aligned either along or against the field, but not in any other direction. It's like a compass needle that would only be able to point North and South (and potentially in a few other directions) but not anywhere in between.

If necessary, please read through the more detailed description of the experiment on Wikipedia or in any other source before attempting the following questions (usually called meditations in the idiosyncratic language used on this forum).

Meditation 1. When exactly does the atom align itself? As soon as it enters the field? At some random moment as it travels through the field? The instance it hits the screen behind the field? In other words, in the MWI picture, when does the world split into two, one with the atom aligned and one with the atom anti-aligned? In the Copenhagen picture, does the magnetic field measure the atom spin, and if so, when, or does the screen do it?

Hint. Consider whether/how you would tell these cases apart experimentally.

Meditation 2. Suppose you make two holes in the screen where the atoms used to hit it, then merge the atoms into a single stream again by applying a reverse field. Are the atoms now unaligned again, or 50/50 aligned/anti-aligned or something else?

Hint. What's the difference between these cases?

Meditation 3. Suppose that instead of the reversing field in the above experiment you keep the first screen with two holes in it, and put a second screen (without any holes) somewhere behind the first one. What would you expect to see on the second screen and why? Some possible answers: two equally bright blobs corresponding to aligned and anti-aligned atoms respectively; the interference pattern from each atom passing through both holes at once, like in the double-slit experiment; a narrow single blob in the center of the second screen, as if the atoms did not go through the first part of the apparatus at all; a spread-out blob with a maximum at the center, like you would expect from the classical atoms.

Hint. Consider/reconsider your answer to the first two questions.

Meditation 4. Suppose you want to answer M1 experimentally and use an extremely sensitive accelerometer to see which way each atom is deflecting before it hits the screen by measuring the recoil of the apparatus. What would you expect to observe?

Hint. Consider a similar setup for the double-slit experiment.

This test is open-book and there is no time limit. You can consult any sources you like, including textbooks, research papers, your teachers, professional experimental or theoretical physicists, your fellow LWers, or the immortal soul of Niels Bohr through your local medium. If you have access to the Stern-Gerlach apparatus in your physics lab, feel free to perform any experiments you may find helpful. As they say, if you are not cheating, you are not trying hard enough.

By the way, if anyone wants to supply the pictures to make the setup for each question clearer, I'd be more than happy to include them in this post. If anyone wants to turn the meditations into polls, please do so in the comments.

Footnote: not posting this in Main, because I'm not sure how much interest there is here for QM questions like this.

Comments sorted by top scores.

comment by Benya (Benja) · 2012-12-23T00:41:58.344Z · score: 9 (11 votes) · LW · GW

Thanks for doing this! (I don't think these are properly called "meditations", though: "Research shows that you're much more likely to remember useful info if you try to solve the problem yourself before reading the solution. Succeed or fail, the important thing is to have tried first." I think in this case, the primary point isn't to remember the correct answer once you post it, but to see how far off our own were, to correct our confidence in our understanding of QM.)

Okay, I do not place much confidence in any of the following, and to do it properly I'd probably have to spend far more time on this than I can spare, but I guess it's still useful to find out how wrong I'll be...

My mental image is the following stark simplification: I think of the state at any point in time as two probability distributions in 2D space (2D = the S-G apparatus viewed from the side, with the atom moving left-to-right and being deflected up/down), one probability distribution for the position of the atom if it's aligned along the field, one for if it's aligned against the field. I'm imagining each distribution to be concentrated around a single point at any point in time, but not necessarily the same point for both distributions. By "probability distribution" I do not mean that I'm actually doing measurements that have a certain probability to come out one way or another (though I guess I could), I just mean that I only look at the squared amplitude of the wave function, forgetting about the phase, which I'm guessing is not relevant to the problem (one place where I might be wrong). Before the atom enters the S-G apparatus, the two distributions are the same. I'm guessing that what the magnetic field does is, it makes the center of one distribution move up and the center of the other move down; again, this is a place where I might be going wrong, but my reasoning is that this is what seems to be required to get the right behavior if I put in an atom that is definitely aligned along/against the field.

M1: My intuitive way of thinking about it is that the atom aligns itself when the centers of the two distributions start moving apart, i.e., in the apparatus. But this is not what you're talking about. For the MWI "world" to split, we need entanglement with something big, which happens when the atom hits the screen. Therefore, in the Copenhagen picture, hitting the screen is when the measurement happens.

M2: I assume the probability distributions are very concentrated, and the holes are big enough to let practically all of the "probability mass" through, so that the shape of the distribution looks the same after passing the holes, and continue to move in a straight line (unlike in the double-slit experiment, where after passing the slit, the particle seems to move in all directions from the slit). Then, the reverse magnetic field will make the two blobs come together again, and the atoms are "unaligned" (= aligned in whatever direction each of them was when it entered the apparatus ... no wait, I guess atoms aren't spin-1/2 and I can't think of their state as being given by an alignment in a particular direction -- I think...; but anyway: I think they are again in whatever spin state they were before entering the apparatus).

M3: "Two equally bright blobs corresponding to aligned and anti-aligned atoms respectively", because my two blobs of probability will just move through the holes undisturbed and then hit the second screen just as if the first hadn't been there.

M4: Alright, this makes me question whether the mental model I've been using can be correct, because I've assumed that no entanglement with the apparatus happens, but I guess for the action of the apparatus on the atom there probably has to be an equal and opposite reaction of the atom on the apparatus, of some form... and I'm not sure how to think of that in the context of quantum mechanics. I can't do this one, and wonder whether my answers to the others are wrong because of this.

I hope you'll post a solution set at some point?

comment by shminux · 2012-12-23T03:56:32.576Z · score: 4 (6 votes) · LW · GW

because I've assumed that no entanglement with the apparatus happens, but I guess for the action of the apparatus on the atom there probably has to be an equal and opposite reaction of the atom on the apparatus, of some form... and I'm not sure how to think of that in the context of quantum mechanics.

Yet this entanglement thing is the essence of QM, and a major contention issue still.

comment by shminux · 2012-12-23T05:09:30.429Z · score: 3 (5 votes) · LW · GW

Thank you for posting your attempt! I hope there will be others. I'd award you one shminux point in addition to an upvote, but they aren't worth much.

As for M4, check how well you can understand why in the double-slit experiment trying to spy on the electron ruins the interference pattern.

comment by Wei_Dai · 2012-12-25T00:29:09.277Z · score: 1 (1 votes) · LW · GW

Alright, this makes me question whether the mental model I've been using can be correct, because I've assumed that no entanglement with the apparatus happens, but I guess for the action of the apparatus on the atom there probably has to be an equal and opposite reaction of the atom on the apparatus, of some form...

Here's my guess of why the entanglement between the atom and the apparatus may not cause decoherence. (Although it turns out something else does.) First consider a 10000-dimensional unit ball. If we shift this ball by two units in one of the dimensions, it would no longer intersect at all with the original volume. But if we were to shift it by 1/1000 units in each of the 10000 dimensions, the shifted ball would still mostly overlap with the original ball even though we've shifted it by a total of 10 units (because the distance between the centers of the balls is only sqrt(10000)/1000 = 0.1).

Now consider the amplitude blob of the apparatus and the shift provided to it by an aligned atom as it moves through. The shift is divided among all of the dimensions of the blob (i.e., all the particles of the apparatus), and in each dimension the shift is tiny compared to the spread of the blob, so the shifted blob almost completely overlaps with the original blob. This means the two possible shifted blobs (for aligned and anti-aligned atoms) can interfere with each other just fine.

(This was me trying to answer "how would the math have to work if the entanglement between atom and appartus doesn't cause decoherence?" Maybe someone with better math skills than me can do the actual math and confirm this?)

Now what if we add an accelerometer? I have no understanding of the physics of accelerometers so I don't know if one that can detect such a small acceleration is even theoretically possible, but if we assume that it is, then when an atom passes through the new apparatus (with the accelerometer), its amplitude blob will be shifted a lot in some of the dimensions (namely the dimensions representing the particles that make up the accelerometer output), which would prevent the shifted blobs from interfering with each other.

comment by pengvado · 2013-01-02T11:54:05.253Z · score: 2 (2 votes) · LW · GW

First consider a 10000-dimensional unit ball. If we shift this ball by two units in one of the dimensions, it would no longer intersect at all with the original volume. But if we were to shift it by 1/1000 units in each of the 10000 dimensions, the shifted ball would still mostly overlap with the original ball even though we've shifted it by a total of 10 units (because the distance between the centers of the balls is only sqrt(10000)/1000 = 0.1).

Actually no, it doesn't mostly overlap. If we consider a hypercube of radius 1 (displaced along the diagonal) instead of a ball, for simplicity, then the overlap fraction is 0.9995^10000 = 0.00673. If we hold the manhattan distance (10) constant and let number of dimensions go to infinity, then overlap converges to 0.00674 while euclidean distance goes to 0. If we hold the euclidean distance (0.1) constant instead, then overlap converges to 0 (exponentially fast).

For the ball, I calculate an overlap fraction of 5.6×10^-7, and the same asymptotic behaviors.

(No comment on the physics part of your argument.)

comment by Wei_Dai · 2013-01-02T12:29:22.851Z · score: 1 (1 votes) · LW · GW

For the ball, I calculate an overlap fraction of 5.6×10^-7, and the same asymptotic behaviors.

Hmm, my intuition was that displacing a n-ball diagonally is equivalent to displacing it axially, and similar to displacing a hypercube axially. I could very well be wrong but I'd be interested to see how you calculated this.

comment by pengvado · 2013-01-02T13:28:55.807Z · score: 3 (3 votes) · LW · GW

The intuition: For a high dimensional ball, most of the volume is near the surface, and most of the surface is near the equator (for any given choice of equator). The extremity of "most" and "near" increases with number of dimensions. The intersection of two equal-size balls is a ball minus a slice through the equator, and thus missing most of its volume even if it's a pretty thin slice.

The calculation: Let $v\(n,r\$%20=%202\int%20_{y=0}%5E{r}%20v(n-1,\sqrt{r%5E2-y%5E2})%20dy%20=%20(2%20\pi%5E{n/2}%20r%5En)%20/%20(n%20\Gamma(n/2))) which is the volume of a n-dimensional ball of radius r.
Then the fraction of overlap between two balls displaced by x is $\\tfrac\{2\\int \_\{y=x/2\}^\{r\} v\(n\-1,\\sqrt\{r^2\-y^2\}\$%20dy}{v(n,r)}) (The integrand is a cross-section of the intersection (which is a lower-dimensional ball), and y proceeds along the axis of displacement.) Numeric result.

comment by Wei_Dai · 2013-01-02T22:07:47.831Z · score: 1 (1 votes) · LW · GW

Thanks for both the math and the intuitive explanation. Now I'm really curious what the right answer is to the physics question...

comment by private_messaging · 2013-01-08T20:54:49.622Z · score: -1 (1 votes) · LW · GW

The position of the apparatus has to be uncertain enough for you to be able to measure momentum (i.e. acceleration) precisely enough. It works out just fine to patterns being smeared, an interesting exercise to do mathematically though.

edit: didn't see context, thought you were speaking of the regular double slit experiment. It still applies though.

With regards to the M1 I don't quite understand the question as the spin is not an arrow that snaps from arbitrary orientation to parallel or anti-parallel. When it interacts with field, after the speed of light lag, there's recoil.

comment by rocurley · 2012-12-24T06:26:19.220Z · score: 6 (4 votes) · LW · GW

Alright, I'll take a crack at this. I haven't read the comments, so likely (I hope?) there's a lot of duplicate information here.

1: When does the atom align itself?

I'm not 100% sure what you mean by this, so let me know if I misinterpreted the question. Consider a single electron. Think of the wavefunction as the product of the spin wavefunction (which is representable as some linear combination of spin up and spin down), and the position-space wavefunction (which is probably a pretty tight gaussian wavepacket). This goes propagating along happily until it reaches the SG aperatus. Up until now, the spin wavefunction didn't affect the time evolution of the position wavefunction at all: this is no longer true.

Quantum mechanics is linear: If you calculate what happens to one half of the wavefunction, and what happens to the other half, and add them together, you get the whole thing. So, you start with:

$\\left | \\text\{packet moving forward\}\\right>\\times \\left\(A\\left|\\uparrow\\right>\+B\\left|\\downarrow\\right>\\right\$=%0AA\left%20%7C%20\text{packet%20moving%20forward}\right%3E%20\left%7C\uparrow\right%3E+%0AB\left%20%7C%20\text{packet%20moving%20forward}\right%3E\left%7C\downarrow\right%3E)

A and B are some complex coefficients whose squares add to one. Now, the first term is pure spin up. As such, it will move up, while the second term will move down. This means your new wavefunction will be:

$A\\left | \\text\{packet moving forward and up\}\\right> \\left|\\uparrow\\right>\+ B\\left | \\text\{packet moving forward and down\}\\right>\\left|\\downarrow\\right>$

Now we get into interpretational differences. I believe in the Copenhagen interpretation, it goes something like this:

The atom hits the screen: this is a position measurement. The wavefunction collapses, and one of the position eigenstates is chosen. It's either going to be an eigenstate of a position in the top section of the screen, or of a position in the bottom section of the screen. Now, there are multiple eigenstates for the same position: spin up or spin down, because positon measurements don't care about spin. However, if you wind up with a top section position eignenstate, it must be spin up, as your pre-measurent wavefunction didn't have any component that was both in the top section and spin down. Likewise, if you measure the position of the atom as in the bottom section, you must be spin down.

Now on to MWI. We need to describe the state of the screen now, so I'm going to add another term to the wavefunction. Before the silver atom hits the screen:

$\\left|\\text\{blank screen\}\\right>\\left\(A\\left | \\text\{packet moving forward and up\}\\right> \\left|\\uparrow\\right>\+ B\\left | \\text\{packet moving forward and down\}\\right>\\left|\\downarrow\\right>\\right\$)

Now, similarly to how the position of the atom got entangled with the spin of the particle, the state of the screen is going to become entangled with both of them, resulting in:

$A\\left | \\text\{packet moving forward and up\}\\right> \\left|\\uparrow\\right> \\left|\\text\{screen with top mark\}\\right>\+ B\\left | \\text\{packet moving forward and down\}\\right>\\left|\\downarrow\\right> \\left|\\text\{screen with bottom mark\}\\right>$

Now here comes the "world splitting" bit. When you look at the screen, or otherwise become causally entangled with the state of the screen in any way (that is to say, when your wavefunction depends on the state of the screen in some way). Before this happens, you have:

$\\left|\\text\{You\}\\right>\\left\(A\\left | \\text\{packet moving forward and up\}\\right> \\left|\\uparrow\\right> \\left|\\text\{screen with top mark\}\\right>\+ B\\left | \\text\{packet moving forward and down\}\\right>\\left|\\downarrow\\right> \\left|\\text\{screen with bottom mark\}\\right>\\right\$)

And afterwards, you have:

$A\\left | \\text\{packet moving forward and up\}\\right> \\left|\\uparrow\\right> \\left|\\text\{screen with top mark\}\\right> \\left|\\text\{You\}\_A\\right>\+ B\\left | \\text\{packet moving forward and down\}\\right>\\left|\\downarrow\\right> \\left|\\text\{screen with bottom mark\}\\right> \\left|\\text\{You\}\_B\\right>$

Now, remember that QM is linear. As such, you can treat each term completely separately. The first term looks like a world where the screen is marked near the top, and the atom is purely spin up. The second term looks like a world where the screen is marked near the top, and the atom is purely spin down. As soon as you become entangled with the state of the screen, the spin no longer seems to be in superposition at all, but is simply up or down, depending on if we're discussing the experiences of You_A or You_B. I would say that the world splits when you interact with the screen.

I may continue, but this level of detail is excruciating and I'm a bit burned out from it atm.

comment by shminux · 2012-12-27T22:39:59.783Z · score: 1 (3 votes) · LW · GW

Your MWI analysis is close to the mark. One thing that is not quite right is multiple states inside each term. Note that once an atom interacts with the screen, it no longer has a definite spin or even position. It becomes a part of the blob on screen, entangled with the atoms around it. Thus the interaction is better described as

blank screen(Awave packet moving forward and upatom spin up + Bwave packet moving forward and downatom spin down) -> Ascreen with top mark + Bscreen with bottom mark.

The atom state is buried somewhere inside the mark on screen.

The observer's interaction with the screen is similarly simplified:

You(Ascreen with top mark + Bscreen with bottom mark) -> AYou-topscreen with top mark+BYou-bottom*screen with bottom mark.

Note that you still have to apply the Born rule to calculate the prior probabilities of You becoming You-top and You-bottom.

* any potential double entendre is completely accidental.

comment by Luke_A_Somers · 2013-01-16T14:19:24.206Z · score: 1 (1 votes) · LW · GW

Note that once an atom interacts with the screen, it no longer has a definite spin or even position.

I don't follow, on position. The atom had a distribution over position which was a smooth wave while it was in transit, and then it has a distribution over position which is some complicated function of position along the screen. It hasn't lost any definiteness in position - and indeed it is much less spread out in space than before (went from 3 extended dimensions to 2). It has only lost simplicity of representation.

comment by shminux · 2013-01-16T17:06:54.344Z · score: 1 (3 votes) · LW · GW

How do you know which atom inside the blob is the one that hit it a moment ago? After all, they are indistinguishable. All you have left is the Hamiltonian of the blob of interacting silver atoms (and the screen, and the field) and its eigenstates, which might not even correspond to single atoms, but instead to some lattice states.

comment by Luke_A_Somers · 2013-01-16T17:34:43.621Z · score: 1 (1 votes) · LW · GW

What? The note I quoted was in the approximation of 'atoms have identity'.

If you're going that far from the first, then my point about reduction of dimensions still applies.

Plus, these energy eigenstates are each for a well-defined number of atoms.

comment by shminux · 2013-01-16T17:44:25.747Z · score: 1 (3 votes) · LW · GW

Sorry, I don't follow...

comment by Luke_A_Somers · 2013-01-17T15:06:07.283Z · score: 1 (1 votes) · LW · GW

You said, Once an atom interacts with the screen, it no longer has a definite spin or even position.

In no event did any atom become LESS constrained in position by hitting the screen, and this applies whether or not you take individual identities or not. That's the first two lines.

The last line points out that you seem to think that the energy eigenstates of the screen and field might correspond to single atoms - but the eigenstates for an extended object will be multiparticle states of extreme complexity - and, at least within the energy regime we're talking about, a fixed number of particles.

comment by ShardPhoenix · 2012-12-23T00:17:33.764Z · score: 4 (8 votes) · LW · GW

Some of these "quantum" concepts that you mention don't require actual quantum physics in the first place. "Quantum dice" is just a short-hand for "truly random decision making method", and a quantum multiverse is for decision-theoretical purposes equivalent to having an infinite ensemble of classical universes with small differences between them, plus self-location uncertainty.

That's my impression anyway.

comment by shminux · 2012-12-23T05:17:15.134Z · score: 4 (10 votes) · LW · GW

"Quantum dice" is just a short-hand for "truly random decision making method", and a quantum multiverse is for decision-theoretical purposes equivalent to [...]

Unfortunately, some people take it overly seriously, just look at the recent posts about infinities and death in many worlds, and about whether a copy of you preserves your identity. Some are filled with quantum angst.

comment by Wei_Dai · 2012-12-23T13:04:08.263Z · score: 3 (3 votes) · LW · GW

If, as ShardPhoenix says, "quantum multiverse is for decision-theoretical purposes equivalent to having an infinite ensemble of classical universes with small differences between them, plus self-location uncertainty", (BTW I would use "similar" instead of "equivalent"), "posts about infinities and death in many worlds, and about whether a copy of you preserves your identity" seem perfectly justified, so I find your comment puzzling as a response to his comment.

comment by Baughn · 2012-12-23T17:55:03.163Z · score: 0 (2 votes) · LW · GW

Is there reason not to take it seriously, from a quantum-mechanical perspective?

I may not agree with their conclusions, but I'm fairly sure the MWI worlds actually exist.

comment by aaronde · 2012-12-23T16:43:39.325Z · score: 0 (0 votes) · LW · GW

Agreed. The multiverse idea is older than, and independent of quantum theory. Actually, a single infinitely large classical universe will do, since statistically, every possibility should play out. Nietzsche even had a version of immortality based on an infinitely old universe. Though it's not clear whether he ever meant it literally, he very well could have, because it was consistent with the scientific understanding of the time.

That said, I like the idea of sminux's post. I try to steer clear of quantum language myself, and think others should too, if all they mean by "quantum" is "random".

comment by shminux · 2012-12-30T07:19:52.616Z · score: 3 (5 votes) · LW · GW

For those interested in the proper QM calculation of what is observed in the classic S-G experiment, I warmly recommend this old paper (PDF), especially Appendix 1. It shows why it is possible to treat the wave function of a spin-half atom as a superposition of spin-up and spin-down trajectories, why the spin precession in magnetic field can be ignored, and how to account for the velocity variation and beam width in calculating the deflection distribution.

comment by kpreid · 2012-12-24T16:32:08.220Z · score: 3 (7 votes) · LW · GW

I object to “the following questions (usually called meditations in the idiosyncratic language used on this forum).” The term “meditation” was introduced relatively recently, not as an alternative to “question”, but a question serving a certain role, which these questions do not. I find this both a misuse of jargon and unnecessarily bizarre in this context. I request that you use what you think is ordinary language, instead.

comment by Luke_A_Somers · 2012-12-24T05:01:00.399Z · score: 3 (3 votes) · LW · GW

1: Whatever the original alignment of the atom was, the N and S components experience a different lateral force, so the amplitude blobs corresponding to those two cases will be shoved apart sideways. I guess you could say they begin aligning the moment the blobs become distinguishable - and that'll depend on a variety of other factors like how you're measuring the deflection. This is the case whether you are working in MWI or Copenhagen.

2: back to unaligned, if you can completely reverse the forces in all cases and equalize the path-lengths - though the way the S-G apparatus really works, that's not going to happen. So you're going to end up with a complicated mishmash of altered phases, and thus reorientations.

3: The atoms heading through the hole in the first screen would end up with two single-slit interference patterns - though since atoms are really heavy I wouldn't count on this being a very noticeable interference pattern. It's two single-slit patterns because the left and right contributions are not feeding into the same quantum state (different spin), so they can't interfere.

4: By measuring the deflection, you've provided an earlier mechanism for distinguishing the blobs, so it would be at that earlier time.

comment by Wei_Dai · 2012-12-23T12:16:28.849Z · score: 3 (3 votes) · LW · GW

M2: Apparently it's not possible to recover the original spin, so I guess you just end up with a random spin.

Abstract: In this report, we investigate the spin dynamics of a neutron beam in a Stern-Gerlach experiment. In contrast to the simple constant gradient magnetic field assumed in most literatures which violates Maxwell Equations, we work with a model of magnetic fi eld which satis es Maxwell Equations. The spin dynamics is investigated by solving the Schrodinger equation using Fast Fourier Transform method. The spin coherence of the neutron beam is found to exhibit the humpty-dumpty behaviour[1, 2, 3] even though there is no fluctuation to the magnetic field nor any environmental noise. The main cause of the spin decoherence is identi fied as the inhomogeneity of the magnetic field in the x and y components. In addition, the nonlinear terms in the z component of the magnetic fi eld, zn, n > 1 also contribute to the loss of the spin coherence. Although only magnetic field model considered is very speci fic, the cause of the loss of spin coherence identi fied using this model is a common features of any realistic magnetic to be used in a SGA experiment. Therefore the humpty-dumpty nature of the spin coherence exist even without any fluctuation to the magnetic field and is inherent to the SGA experiment itself.

And see also this earlier paper, Is spin coherence like Humpty-Dumpty? I. Simplified treatment (free full text), which gives a more understandable qualitative argument:

Why should it be so difficult to maintain spin coherence? The essential features of the following qualitative argument are due to Heisenberg.

I won't try to quote the actual argument, which is on page 3 of the PDF, but it seems that the main problem isn't entanglement between the particle and the apparatus.

comment by [deleted] · 2012-12-27T18:30:07.678Z · score: 2 (2 votes) · LW · GW

shminux,

I think this test demonstrates how little I understand quantum mechanics, but the test also serves as an impetus for me to learn physics at a more rigorous level. So thank you for that!

Do you have any textbook recommendations for teaching myself quantum mechanics? Ideally, I need an introductory textbook suitable for someone already studying mathematics at an undergraduate level.

(I also welcome others to answer my question and debate each others' recommendations. Answers in this style get extra Paragon points.)

comment by shminux · 2012-12-27T22:53:13.530Z · score: 1 (3 votes) · LW · GW

I learned the basics of QM, as well as a few interpretational bits from Griffiths. I went through three or four other undergrad and grad QM texts, as well, but Griffiths is still my favorite by far, even though it does not address some of the standard grad curriculum topics, like partial waves. YMMV.

Bear in mind that none of the standard QM texts address the interpretations in any depth, mostly embracing the instrumental "shut up and calculate" approach. The most you will get is the Bell theorem and the discussion of hidden variables. I am not sure which texts address this issue head on, but my guess is that it would be in some of the Quantum Information ones. Maybe someone else can chime in.

comment by crap · 2012-12-23T04:54:01.093Z · score: 2 (10 votes) · LW · GW

You're doing a world of good with this. People take their uneducated guesses far, far too seriously (QM and morality etc).

comment by shminux · 2012-12-23T05:10:48.256Z · score: 3 (5 votes) · LW · GW

Thank you!

comment by jsteinhardt · 2012-12-23T02:23:09.445Z · score: 2 (4 votes) · LW · GW

Thanks! Great post.

comment by shminux · 2012-12-23T03:45:07.783Z · score: 4 (6 votes) · LW · GW

You are welcome. One of my goals was to show how non-trivial Quantum Mechanics is, especially where "measurement" is concerned.

comment by [deleted] · 2012-12-23T00:24:00.267Z · score: 2 (4 votes) · LW · GW

I was expecting more math-looking stuff in physics problems. Is it implicit?

comment by jsteinhardt · 2012-12-23T02:22:40.061Z · score: 8 (8 votes) · LW · GW

Yes. Part of the problem is to understand what mathematical formalisms are relevant to the question in the first place.

comment by Wei_Dai · 2012-12-26T18:45:12.303Z · score: 1 (1 votes) · LW · GW

When will the solutions be posted? If you post any more of these tests, could you mention that in the post itself, so that we know how much time we have to finish them, and also when to remind you to post the answers in case you forget?

comment by shminux · 2012-12-26T21:54:31.773Z · score: 2 (4 votes) · LW · GW

This basically nailed it, I'll try to write it up in more detail this week.

comment by [deleted] · 2012-12-23T13:55:40.922Z · score: 1 (1 votes) · LW · GW

In other words, in the MWI picture, when does the world split into two, one with the atom aligned and one with the atom anti-aligned?

This one I'm going to do un-rot13ed because this misconception bothers me: The world was split all along according to the modern understanding of the MWI.

comment by Manfred · 2012-12-23T16:28:04.825Z · score: 2 (2 votes) · LW · GW

Well, once you want to solve real problems, talking about "worlds" becomes less helpful than talking about entanglement. Quantum mechanics is continuous along a dimension that thinking using the word "worlds" tends to be discrete. In fact, the name "many-worlds" was proposed about 13 years after Everett's seminal paper.

comment by EHeller · 2012-12-23T16:43:19.396Z · score: 1 (1 votes) · LW · GW

i think you still have to talk about worlds to make actual predictions. You need to specify "a world where observer X sees Y" and a probability weight. of some sort for that world (or that observer, however you want to phrase it).

comment by Manfred · 2012-12-23T18:46:14.561Z · score: 1 (1 votes) · LW · GW

Nay. In fact, being able to make predictions just from a quantum state, no labels on "worlds," is an important part of making a quantum-mechanics-only interpretation a reductionist success.

comment by EHeller · 2012-12-23T21:11:54.090Z · score: 1 (1 votes) · LW · GW

How can you predict 'observer A sees the electron aligned with the magnetic field' without being able to point to the section of the wavefunction that represents 'observer A sees the electron aligned with the magnetic field'?

All decoherence can tell me is that my density matrices evolve towards being diagonal. I still need a framework to extract actual meaning from it.

I've never seen a paper make a prediction without some implicit definition of world (or sometimes equivalently of 'observer')- if you have a reference, please point me to one.

comment by Manfred · 2012-12-23T22:08:23.490Z · score: 0 (0 votes) · LW · GW

Yes, definitely, one needs to "point to the section of the wavefunction that represents 'observer A sees the electron aligned with the magnetic field'." So to the extent doing that is an implicit definition of "world," then I agree on that part too.

The reason why I replied, er, emphatically, to "you need to specify [...]" is because any theory treating worlds as basic should not be the simplest way to understand what's going on - that should be plain ol' quantum mechanics if the relative state interpretation is to be believed, even though that interpretation is sometimes called "many-worlds." So you shouldn't have to talk about worlds.

And sometimes it's a good idea not to talk about worlds, like when shminux asks you "when does the atom get polarized when it's in a magnetic field?" One definitely has to use math for that one.

comment by EHeller · 2012-12-24T04:49:35.884Z · score: 1 (1 votes) · LW · GW

The only way I know to answer shminux's question rigorously (with the decoherence time scale) in MWI is to make the assumption that a diagonal density of states of represents a classical ensemble of worlds (with weights corresponding to probabilities)- which means explicitly talking about worlds.

Then you can define a decoherence time-scale of some sort. To the best of my knowledge, without some rule-of-thumb for what a world is, all you can say is that "the system evolves to a a density matrix that has 1/sqrt(2) Aligned on one diagonal element, and 1/sqrt(2) anti-aligned on the other."

comment by Manfred · 2012-12-24T16:01:10.563Z · score: 0 (0 votes) · LW · GW

The only way I know to answer shminux's question rigorously (with the decoherence time scale) in MWI is to make the assumption that a diagonal density of states of represents a classical ensemble of worlds (with weights corresponding to probabilities)- which means explicitly talking about worlds.

Yeah, that's a cool realization: "Hey, this all works if we assume that a density matrix is what an entangled state looks like from inside!" And this happens to be pretty true, though if it's your definition of "world" then there will be different "worlds" from different perspectives.

But anyhow, to answer shminux I wouldn't use worlds, I'd just give the decay rate of the top state coupled to a photon mode, to the bottom state plus a photon. The assumption there is not about worlds, but simply that the amplitude-squared measure should be used to describe the properties of the atom.

comment by EHeller · 2012-12-24T17:44:05.013Z · score: 1 (1 votes) · LW · GW

And this happens to be pretty true, though if it's your definition of "world" then there will be different "worlds" from different perspectives.

I can't parse what you mean by perspectives here- do you mean different non-relativistic observers (no relativity, we are using Schroedinger quantum), or do you mean putting a different basis on the Hilbert space?

But anyhow, to answer shminux I wouldn't use worlds, I'd just give the decay rate of the top state coupled to a photon mode, to the bottom state plus a photon.

Shminux's question involves an unpolarized beam entering a magnetic field and aligning, not polarized atoms flipping state. These are very different problems.

In the shminux question, you have an entirely off-diagonal density matrix that evolves toward diagonal very quickly when the system becomes entangled with the screen. To extract information, you implicitly or explicitly assume that a diagonal density of states represents an ensemble of classical worlds.

In the question you are answering, you start with only one diagonal element in the density matrix non-zero, and over time the amplitude of that element shrinks while the element of the other diagonal element grows. This is a totally different problem. You still are implicitly thinking about worlds, you've just created a different problem where there is no entanglement so it dodges the messy question.

comment by Manfred · 2012-12-24T18:41:26.216Z · score: 2 (2 votes) · LW · GW

I can't parse what you mean by perspectives here- do you mean different non-relativistic observers (no relativity, we are using Schroedinger quantum), or do you mean putting a different basis on the Hilbert space?

Hm. So what I mean is that if you have several particles entangled with each other, and you want to know what that "looks like" to one subsystem (or, experimentally, if you're going to produce a beam of these particles and do a bunch of single-particle measurements), then you have to trace over the other particles in the entangled state. This gives you a reduced density matrix, which is then interpreted as a mixed state. A mixed state between what? Well, "worlds," of course.

This the definition of "world" I meant when I said "if that's your definition of world..."

But anyhow, why did I say that different observers will see different lists of worlds? Well, because when you take the partial trace, what you trace over depends on which perspective you want (experimentally, what partial measurement you're making). If you're an electron, your worlds are boring - we traced all the complicated externals away and now your perspective just looks like a distribution over single-electron states. If you're a person, your perspective is much more interesting, your density matrix is much bigger. Or to put it another way, you have more "worlds" to have a distribution over.

You still are implicitly thinking about worlds,

And you're implicitly thinking about waterfalls, because every cognitive algorithm is isomorphic to a thoght about a waterfall :P

comment by EHeller · 2012-12-23T16:23:13.260Z · score: 1 (1 votes) · LW · GW

Whose modern understanding of MWI? Give me a mathematical formalism for "world" that allows world count to be conserved as interactions go forward (if the worlds 'were split all along', then something like 'world count' is constant).

Your understanding also seems to have implications for ("the modern understanding of") MWI and Bell's theorem- since you are applying if I know enough about the degrees of freedom "I don't care about" I can track which world I am in. This should reduce to some sort of hidden variable approach, I would think.

comment by [deleted] · 2012-12-23T18:29:33.407Z · score: 0 (0 votes) · LW · GW

Give me a mathematical formalism for "world" that allows world count to be conserved as interactions go forward (if the worlds 'were split all along', then something like 'world count' is constant).

As Manfred says, thinking of worlds as discrete isn't helpful. Anyway, the time evolution dictated by Schrödinger's equation is unitary, so if it always applies (i.e. there's no such thing as objective collapse), the measure stays constant.

I know enough about the degrees of freedom "I don't care about" I can track which world I am in.

If I understand decoherence well enough (probably I don't), the answer to that is “in you could, yes, but you can't, because thermodynamics”. The differential equations of quantum field theory are more-or-less time-symmetric (i.e. reversing time is equivalent to boring stuff such as conjugating complex phases and swapping particles with anti-particles and left with right), so the reason stuff appears to be irreversible is the boundary condition that the past had very little entropy. (And I think I've seen the argument that given that T-symmetry is equivalent to CP-symmetry, the fact that the past had that little entropy may (or may not) have something to do with the fact that there are so many more particles than antiparticles.)

comment by EHeller · 2012-12-23T18:59:34.440Z · score: 2 (2 votes) · LW · GW

You have to have a mechanism to separate world as discrete or you have a theory that can't make predictions. If you want to talk about the aligned/anti-aligned beam in the Stern-Gerlach experiment you have to be able to point and say "this represents the world where observers measure aligned, and this bit over here represents the world where observers measure anti-aligned." If you can't do that, you have no theory.

If I understand decoherence well enough (probably I don't), the answer to that is “in you could, yes, but you can't, because thermodynamics”.

This has to be wrong, otherwise MWI would predict violations of Bell inequalities.

I think your 'the world is already split' interpretation is actually the fundamental misunderstanding- I can't make any sense of it other than as a hidden variable theory of the type already experimentally ruled out by Aspect-like experiments.

Edit: Unrelated, but to clarify- you can show that (assuming energy is bounded below) a Lorentz invariant Hamiltonian has a combined CPT symmetry, which can mean a lot of things, depending on dimension. T has to be related to CP, but not necessarily the-same-as, unless you have a state where CP^2 = 1.

comment by [deleted] · 2012-12-23T19:13:00.063Z · score: 0 (0 votes) · LW · GW

unless you have a state where CP^2 = 1

How can it be anything else? Even then, T would equal (CP)^-1.

comment by EHeller · 2012-12-23T21:07:05.807Z · score: 2 (2 votes) · LW · GW

Generally, you pick up a phase factor after CP^2. The story is exactly like parity (P) if you can embed P^2 in a continuous symmetry, you can define away the phase factor, but if you can't you are just stuck with it.

comment by [deleted] · 2012-12-24T10:33:05.191Z · score: -2 (2 votes) · LW · GW

That's still in the reference class I called “boring stuff”, though.

comment by satt · 2012-12-23T12:56:59.243Z · score: 1 (1 votes) · LW · GW

Wasn't sure if posting my not-very-considered answers was in the spirit of the exercise, but since you're hoping for more answers, here are mine. (And despite taking a QM class that discussed the Stern-Gerlach experiment, I'd somehow forgotten that S-G used an inhomogeneous magnetic field. So I might as well solicit some schoolin' from LW on this topic.)

Q1. If "When exactly does the atom align itself?" means "When does the atom start changing trajectory in response to the magnetic field?", I'd say it does that as soon as it enters the field. If it means "When can we say the atom has a single, unambiguous spin direction?", I'd say that's not until the atom hits the screen — before then it's in a superposition of spin up and spin down states.

Q2. As the atoms all travel unimpeded through the first S-G field, through the holes, and then through the reverse field, I guess they'd remain in a 50/50 superposition the whole way along. So they're just as "unaligned" or "aligned" as they were when they started. Presumably most of us would say the silver atoms were originally "unaligned" when they came flying out of the furnace, in which case we'd have to say they're still "unaligned".

Q3. I'd see two equally bright blobs, like those I'd have seen on the original screen, only further apart (because the two groups of atoms move further apart as they propagate, since the spin down particles have downward velocity relative to the spin up particles).

Q4. I'd observe either an atom continuously accelerating upwards or an atom continuously accelerating downwards. Once the atom enters the S-G apparatus and I see the accelerometer needle start moving, it's no longer in a superposition; it's either spin up or spin down, and I observe it accelerate accordingly.

comment by Vaniver · 2012-12-22T22:47:29.070Z · score: 1 (3 votes) · LW · GW

It's not clear to me what you're envisioning for M4. I'm able to translate M2 and M3 into questions that I would have seen in my quantum classes, but M1 (and by extension M4) remind me more of electrodynamics. Is that intentional?

Similarly, the compass needle only pointing up or down doesn't fit with the standard description what underlies Stern-Gerlach, and I can't tell if discovering that subtlety is an intended effect of the meditations or not.

comment by shminux · 2012-12-23T01:06:53.589Z · score: 3 (5 votes) · LW · GW

It's not clear to me what you're envisioning for M4.

The question is straightforward, predict what an accelerometer signal would show. For example, classically one expects to see the signal show near-constant recoil level during the time the small magnet travels through this inhomogeneous field, assuming it's aligned. The situation is not necessarily the same in QM. If you expect the measurement to happen at the screen, you'd predict no signal until the spike at the moment of collision. Note that M4 is not necessarily the same experiment as M1.

the compass needle only pointing up or down doesn't fit with the standard description what underlies Stern-Gerlach

Right, the setup is not exactly the same, but it determines essentially the same physical property,

comment by Vaniver · 2012-12-23T01:28:53.549Z · score: 0 (0 votes) · LW · GW

The question is straightforward, predict what an accelerometer signal would show.

Reading Benja's answers, I think my issue was that I thought "apparatus" referred to the accelerometer, not the large magnet used in S-G. Replacing that word with "large magnet" makes the intended question clear to me, and I see how to answer it with just QM.

comment by Oscar_Cunningham · 2012-12-26T21:59:18.872Z · score: 0 (0 votes) · LW · GW

M1: The worlds split over a short period of time starting as soon as the atom enters the field. They don't split very far. In the Copenhagen interpretation no measurement occurs (EDIT: until they hit the screen) (you can tell because later on we'll "unmeasure" this alleged measurement, which would be impossible if the wavefunction had collapsed.). EDIT: The worlds don't split very far while the atoms are in the field (they only split in one direction). When they hit the screen they suddenly split a lot further. In Copenhagen this would constitute a measurement.

M2: The atoms become unaligned again. EDIT: You can't necessarily tell this apart from 50/50 aligned unaligned. If I could be bothered I'd calculate the density matrices. The situations are distinguishable iff the density matrices are different. My intuition says they're the same.

M3: Two blobs. No interference. EDIT: Because the spins are different.

M4: The detector answers soon after the atom enters the field. It will of course end up agreeing with the light formed by the atom hitting the screen. EDIT: In particular the acceleration would vary with time exactly as if the atom were classical.

comment by shminux · 2012-12-27T22:57:26.092Z · score: 0 (2 votes) · LW · GW

The worlds split over a short period of time starting as soon as the atom enters the field. They don't spilt (sic.) very far

Neither statement is clear to me. How short a time? How do you define the distance between worlds?

Re M2: Assume you are dealing with a single atom, so it's a pure state, no need for density matrices. What would be your answer?

comment by Oscar_Cunningham · 2012-12-28T15:54:28.597Z · score: 0 (0 votes) · LW · GW

M1: The atom has a wavefunction, a function from {up,down} x R^3 to C. We can view this as two spacial wavefunctions (i.e. from R^3 to C) one for spin up, one for spin down. Before entering the field the up and down wave functions are the same, and localised in space (i.e. all the amplitude is in a small lump around a point). This lump of amplitude moves along until it reaches the field. At this point the two wavefunctions cease to be the same (the spin of the particle becomes entangled with its position). The one associated with spin up translates upward, its velocity increasing just like a particle in a classical field. Similarly the spin down wavefunction moves downward. The time that the worlds take to split is the time until the two lumps (corresponding to up and down) cease to overlap spatially. I don't view the "worlds" as ontologically fundamental, only the wavefunction, so the previous sentence is close to tautology. If we allow the wavefunction to have thin tails off to infinity then the lumps never truly split, but they do still mostly separate.

Since the two lumps haven't separated very far (say at most a few cms, or however big the S-G apparatus is) and the atom hasn't entangled with anything else, it will be easy to remove the entanglement between spin and position by reversing the field. This is what I mean by "the worlds aren't very far apart". To formalise the notion of distance I suppose one could take the root of the sum of the squares of the displacements of every particle in the universe between the two worlds. So in this case the distance between worlds would be the literal distance between the two lumps of amplitude.

Once they hit the screen we suddenly have that lots of particle's positions differ between the two worlds, and so the distance between them becomes very great. This is decoherence.

M2: In the experiment given, any single particle is essentially returned to exactly the same superposition of states it was in before it entered the fields. Exactly like neither of the fields was ever there. (Also, I hold that I can still use density matrices to deal with my subjective uncertainty, even about single particles).

comment by shminux · 2012-12-28T19:07:36.236Z · score: 0 (2 votes) · LW · GW

At this point the two wavefunctions cease to be the same (the spin of the particle becomes entangled with its position).

So it seems that you define the degree of separation of worlds as the spatial overlap of the spin-up and spin-down components of the wave function, probably the inner product of the two normalized terms. I did not follow your musings on "displacements of every particle in the universe between the two worlds", however.

M2: In the experiment given, any single particle is essentially returned to exactly the same superposition of states it was in before it entered the fields.

Are you saying that the screen with the two holes has no effect whatsoever? Just wondering.

(Also, I hold that I can still use density matrices to deal with my subjective uncertainty, even about single particles).

You sure can, but it seems unnecessary for a pure state. Or maybe I misunderstand what you mean by "subjective uncertainty".

comment by [deleted] · 2012-12-23T18:43:32.626Z · score: 0 (0 votes) · LW · GW

D1: Gur ngbz vzzrqvngryl ragref va n fhcrecbfvgvba bs fcva-hc naq fcva-qbja, naq rnpu grez bs gur fhcrecbfvgvba trgf qvfcynprq, ohg (cneqba gur unaqjnil Pbcrauntral ynathntr) Angher qbrfa'g qrpvqr juvpu grez bs gur fhcrecbfvgvba lbh bofreir (va ZJV + qrpburerapr, gur fhcrecbfvgvba qbrfa'g qrpburer) hagvy vg uvgf gur fperra.

D2: Gurer'f ab jnl gb gryy. Gur qrafvgl zngevk (|hc><qverpgvba|), fb gurer'f ab jnl gb gryy na rafrzoyr jurer unys gur ngbzf unir fcva hc naq unys unir fcva qbja sebz bar jurer nyy gur fcvaf ner qvfgevohgrq ng enaqbz vfbgebcvpnyyl. Guvf obguref zr orpnhfr gur oryvrsf “gur ngbz vf rvgure fcva-hc be fcva-qbja jvgu cebonovyvgl 50% rnpu” naq “gur ngbz vf rvgure fcva-rnfg be fcva-jrfg jvgu cebonovyvgl 50% rnpu” srry vapbzcngvoyr ohg gurl yrnq gb gur rknpg fnzr nagvpvcngrq rkcrevraprf.

D3: Ubj jvqr ner gur ubyrf?

D4: Gur gjb ornzf qba'g vagresrer fvtavsvpnagyl ba gur fperra, fb vg qbrfa'g znggre gung lbh qb gung. BGBU vs lbh qvq gung jvgu n gjb-fyvg rkcrevzrag, lbh'q qrfgebl gur vagresrerapr cnggrea.

comment by [deleted] · 2012-12-23T00:05:41.700Z · score: 0 (0 votes) · LW · GW

Wouldn't a meaningful test include some sort of math?

comment by Decius · 2012-12-22T22:56:32.272Z · score: 0 (2 votes) · LW · GW

This experiment doesn't seem to require any non-classical behavior beyond quantized basic properties and a better understanding of electromagnetism.

comment by Vaniver · 2012-12-22T23:59:17.074Z · score: 0 (2 votes) · LW · GW

Right; S-G gets interesting when you chain it.

comment by Decius · 2012-12-23T04:41:38.452Z · score: 0 (0 votes) · LW · GW

I think that indicates interesting nonclassical features of the electromagnetic effect. There's a handful of interesting experiments I'd like to see the results of before going any further.