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.
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.
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.
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.)
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.
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
%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
%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.
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.
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?
Yet this entanglement thing is the essence of QM, and a major contention issue still.
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.
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.
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.)
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.
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
Then the fraction of overlap between two balls displaced by x is
Thanks for both the math and the intuitive explanation. Now I’m really curious what the right answer is to the physics question...
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.