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Angular momen. conserv

  1. May 18, 2008 #1
    Suppose you have an EPR pair, it's in a state [tex]|\uparrow>|\uparrow> - |\downarrow>|\downarrow>[/tex].
    Then you measure one particle in the [tex]\uparrow[/tex] direction, and the other particle's measured in the [tex]\rightarrow[/tex] direction. So the result is something like [tex]L\uparrow, R\rightarrow[/tex] or [tex]L\uparrow, R\leftarrow[/tex] etc.
    The pair started out with 0 angular moment. But the "measured" particles don't have a net 0 angular moment. How is this consistent with angular moment. conservation?
    Laura
     
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  3. May 18, 2008 #2

    malawi_glenn

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    The operator is not the total angular momentum operator, so you are not to expect to get the same result as the initial state.
     
  4. May 30, 2008 #3
    What happens to the angular momentum? The EPR pair starts out with 0 total spin. After being "measured", the particles in the pair have nonzero total angular momentum, don't they?
    Where does the angular momentum come from? Does it get transferred from the measuring device?
    I don't think that angular momentum is only conserved on average in quantum mech., so what's happened to it?
    Laura
     
  5. May 31, 2008 #4
    Suppose the particles have a magnetic moment that goes along with their spin and you're measuring the spin by seeing how they're diverted in a magnetic field, and the magnetic fields "measuring" the spin of the 2 particles are at right angles to each other.
    I'm not sure if this is a good example, because they'd have to have a charge to have a magnetic moment? and the charges in an EPR pair would be opposite. I'm not sure if you can have a charged EPR pair, is there a problem with creating a magnetic moment out of nothing?
    If this doesn't work for some reason, what other "measuring" device works?
    It's a really basic question that would apply to a lot of different qm situations.
    Laura
     
  6. May 31, 2008 #5

    vanesch

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    Suppose first a simpler question: imagine you have a particle in the |up_x> state. Now you measure it in the z-direction: 50% of the time you find up_z and 50% of the time you find down_z. For each individual measurement, you find a non-zero z-component, nevertheless, the original state didn't have any "z-component". And where is now my original x-component ? Is this a violation of conservation of angular momentum ? No, of course not, because 1) on average, the angular momentum is 0, and 2) case-by-case, a |x-up> state doesn't imply "no angular momentum" in the z-direction as both are incompatible measurements. An x-up state is simply a superposition of a z-up and a z-down state.

    Remark: I suppose there's a typo in your expression, I guess you meant: |up_z>|down_z> - |down_z>|up_z>, didn't you ? I will take that as a hypothesis.

    What we have here, is that for each term, indeed, the total angular momentum along the z-axis, or (simple verification by a base transformation), indeed, any axis is 0, for the two particles. This will come out too on average with your measurement: in 25% of the time, you will find up-up, in 25% of the time, you will find down-up, in 25% of the time you will find up-down and in 25% of the time, you will find down-down.

    But on a case-by-case basis, you might wonder: *if* I had a z-up state, then the second particle must be in a z-down state, but I measure its x-component (of a z-down state), so I will get half of the time an x-up and half of the time an x-down, and I cannot "compensate" for the z-up. But that's the same as with our simpler system, where we had an x-up state, and we went measuring the z-component. We also had "one or the other" outcome along x, and we could evidently not "find a z-component" that way.

    So you must see it in the same way: when you have a z-up and x-down outcome, for instance, this means that the x-down can be seen as a "component of the corresponding z-down" state - or, the z-up was a component of the x-up state: you can choose.

    As the x-measurement is not compatible with the z-measurement, you can think of the system "having the liberty" to have compensated for the z-component, and this with just any result for the x-component.
     
  7. May 31, 2008 #6
    Right :)

    Does all this mean that angular momen. is only conserved on average in quantum mech.? Obviously it *would* be conserved on average, but is it only on average?
    Because this business of having a particle in a definite state of angular momentum along the x-axis, say, I thought is the quantum version of having a top spinning around the x-axis. Something that if magnified a great deal would become the angular momen. we're familiar with.

    And if it's only conserved on average it brings up the question of whether it would be possible to have a lot of particles all clumping together in a given angular moment. state, and weird macroscopic behavior violating angular moment. conservation. Like if you could somehow make a sextillion EPR photon pairs and the photons going off in a given direction all have the same polarization ... You measure the sextillion pairs, and your lab starts rotating? And someone else's lab measuring the other sextillion pairs at right angles to your lab, starts rotating at right angles to your lab?

    If you measure just one particle to be in an up-x state, it would transfer a tiny bit of angular moment. in the x-direction to the detector, right?

    I suppose this weirdness is particular to angular moment., there would not be this confusion of directions with linear moment. or energy.

    Laura
     
    Last edited: May 31, 2008
  8. Jun 1, 2008 #7

    vanesch

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    The point is that due to the complementarity of Sx and Sz, if we are in a pure state of Sx, that the z-angular momentum is undefined. There are different ways of expressing that, and in the end, it will depend on an interpretational picture - which doesn't matter, as long as we remain within the same picture.
    We can see an |x-up> state as a superposition of z-up and z-down. As such, an x-up state does not have a well-defined momentum along z, and you can say that if you've measured an x-up state, that you (or nature, or whatever) is "free" to consider this as coming from a z-down or a z-up state.
    So a measured x-up state can be seen as "having compensated" any z-up or z-down state. If conservation of angular momentum required us to have a z-down state, then consider that your x-up "came from" a z-down so that the balance was right etc...

    In other words, if the outcome is z-up and x-down, say, you can say that angular momentum has (potentially) been conserved as well along the z-axis (as the x-down could have "come from" a z-down state for the second particle) as well as along the x-axis (as the first z-up could have "come from" an x-up state).

    You mean, improbable but large statistical fluctuations ? Actually, they are possible but they don't mean a violation of conservation of angular momentum, for the reason I tried to illustrate. If you measure a z-component and an x-component, then the measure of the x-component "allows quantum uncertainty" in its z-component and vice versa, which is enough to compensate for the balance of angular momentum. Now in my preferential interpretational scheme, which is MWI, this comes down to having my lab in different states, of which the most probable are of course those where I don't have a macroscopic angular momentum, but of which there are small states which rotate indeed - however, they are the (improbable) result all these uncertainties not averaging out, and it will be combined with a similar improbable state of the emitter of the pairs rotating in the opposite sense.

    Again, this is inherently quantum-mechanical, and you don't need to go to entangled pairs in order to see this. Let us consider a single 2-state system, which is emitted to be:

    |psi> = |a> + |b>, and let us say that there is "conservation of wobble", with the wobble quantum number of a being +1 and of b to be -1.

    Clearly, |psi> is a state of 0 wobble. And if we send out a lot of |psi> it means that upon measurement, we will get on average about as many |a> (wobble = 1) as |b> (wobble = -1).

    But of course, each individual measurement "violates wobble" if we take it that the emission was 0 wobble: we will find on one single measurement a +1 or a -1 for wobble. If we are "statistically unlucky", we might find a billion times in a row the state |a>, and hence have accumulated a wobble = + a billion. Is this a violation of the law of conserved wobble ?

    No, it isn't, and here it depends upon the interpretational scheme. In MWI, this is very simple: in each "world", there is a compensation between the "emitted branch" and the "observed branch": for the world in which we observed a billion times |a>, we had also that the emitter has emitted a billion times |a>, and this in parallel with another branch (or world) in which the emitter has emitted a billion times |b>, and where also a billion times |b> was observed ; and both these branches are very small (not very probable) to be observed.

    Indeed, the emitted overall multiverse state was (|a1> + |b1>)(|a2>+|b2>)(|a3> + |b3>)... (|an>+|bn>) for the n emitted particles. One branch of the emitter is:
    |a1,a2,a3,...,an> . This is an emitter state of wobble = +n. This state will evolve on the observer side into an observed state of, well, (a,a,a,a,...a). So in this improbable branch, there was actually a state of wobble = +n emitted, and also observed. Most branches will have a combination of the kind (|a1,b2,b3,a4,a5,b6,a7....>) which averages out to about 0 wobble, and this will also be observed.

    In a more Copenhagen kind of view, you can say that the emitted particle HAD AN UNCERTAINTY of emitted wobble of +/- 1, and hence the total uncertainty of a billion emitted particles is +/- 1 billion for wobble, although its probability was very small.

    So given that we had an uncertainty in emission, the value of wobble was not fixed with certainty to be 0 at emission, and hence it is not a violation if it is observed not to be 0 at reception.
     
  9. Jun 1, 2008 #8
    If you have an EPR pair though you don't have an uncertainty in emission, the total angular momentum is definitely 0, and as you say, there could be a sextillion measurements of the particle going off in one direction, where it comes out in the up direction by pure chance, and a sextillion measurements of the particle going off in the other direction, where it's measured in the right direction, by pure chance. That can't be attributed to the source.
    So, I'll try to sort all this out and try to say what the situation seems to be:
    Apparently, at the quantum level, what conservation of angular momentum turns into would be 3 separate conservation laws, about each axis; and the separate laws can't be applied at the same time, because you can't measure angular moment. around the different axes at the same time.
    Macroscopically this would also be true, but in everyday life you would not notice this and angular moment. becomes something that appears to be measurable in any direction at any time. Why? Because almost all the time, the uncertainties of the angular momentum would statistically average out to something sensible. Or something like that.
    There are coherent quantum states with angular moment. Bose Einstein condensates that rotate, things like that. Funny things must happen when they are suddenly forced to go from a coherent state with angular moment. in one direction to a coherent state with angular moment. in another direction!
    What would happen, by the way, if the state got slowly changed from a state of definite angular moment. in the x direction - rotated by increments to a state that has definite angular moment. in the y direction? In the same way that light polarization can be changed incrementally. I guess then you are getting angular moment. transferred to the particle.
    Laura
     
  10. Jun 2, 2008 #9

    vanesch

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    Sure, but that's not the problem. Even in the improbable case of observing a sextillion times |z-up> |z-down> when the initial state was |z-up> |z-down> - |z-down> |z-up>, there is conservation of angular momentum along the z-axis, which is what you are alluding to.

    However, we are not measuring Z and Z, we are measuring Z and X. So what happens now is that, after the measurement of a sextillion times |z-up> of the first particle, we can consider that the second particle has been a sextillion times |z-down>, which is to be written as |x-down> + |x-up>. So we can consider this case as just a beam of |x-down> + |x-up> for the second particle, as if it were "living on its own".

    So now consider a beam of |z-down>, measured along X. Normally we would get half of the time |x-up> and half of the time |x-down>. That's not in contradiction with the large overall z-down angular momentum transfer that should happen from the source to the target, because each X-measurement has an uncertainty in the Z-component, which can accomodate conservation of z-angular momentum. If we are doubly unlucky, then we measure |x-up> all the time too. So we have received a definite amount of x-angular momentum, but that's not in contradiction with conservation of momentum too, because our certainty of zero z-momentum at emission implied for an uncertainty (see superposition!) of x-angular momentum, which can account for the total received x-angular momentum.

    Or, you can say: the 3 laws apply, given that we cannot check them simultaneously, so there's enough liberty in the uncertainties for them to "apply behind the scene".

    In other words, you cannot contradict them with measurements, because to contradict them would mean: we have a certainty of its value at the start, we have a certainty of its value at reception, and we don't find the same value.

    But I see your question of course: imagine a spaceship, emitting |z-down> particles to a second spaceship. In doing so, it will acquire of course an angular momentum along Z.

    Now, imagine that the second spaceship receives all these particles, and measures along X. Is it going to rotate in the opposite direction along Z now ? The answer should be yes, and where does it come from ? I should think a bit about it, but I guess it comes from the interaction of the z-state and and the apparatus...
     
  11. Jun 2, 2008 #10
    I don't think it can, because then the apparatus is essentially measuring the angular moment. in both the z direction and in the x direction, which it isn't supposed to be able to.

    If we are just thinking of angular moment. in the z-direction, the second spaceship is then a superposition of x-up and x-down, because that's what z-down is. So you'd have a superposition of a spaceship rotating in an x-up direction and in an x-down direction. What happens if it becomes macroscopic?
    Maybe even at the macroscopic level we can't think of angular moment. conservation as applying along more than one axis at once.
    In reality I suppose it's impossible to emit a stream of just z-down particles. They would be a mixture of z-down and z-up. I don't know that this helps, because if the particles are charged you can sort them by their magnetic moments, right? and you could have a beam of z-down particles going out a port of the spaceship. And you can sort particles by charge by passing them through an electric field.
    Laura
     
    Last edited: Jun 2, 2008
  12. Jun 3, 2008 #11

    vanesch

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    On an individual basis it is not possible of course, but after having send out a billion z- states, obviously there is a macroscopic z-angular momentum transferred. It is no violation of the uncertainty principle, as the *increment* in z-angular momentum per individual measurement is undetectable.

    Well, that's the point. If we have a lot of particles, then the number of x-up should on average compensate the number of x-minus, and the expectation value of the x-angular momentum should be about 0. However, there should definitely be a macroscopic transfer of z-angular momentum, and I have to say I'm not clear myself on how this is realised on the reception part. It must be due to the interaction of the z-minus particles and the X-measurement apparatus, but I fail to see how.

    No, that's clearly not the case, or we wouldn't obtain classical mechanics in the end.

    No, it is for uncharged particles but with magnetic moment that you can select them using a Stern-Gerlach apparatus. And it is definitely possible to emit a beam of z-down particles only, simply by blocking the z-up stream, and sending them back off to the source. So we CAN have a spaceship emitting a pure beam of z-up particles.
     
  13. Jun 3, 2008 #12
    That seems fishy, because it only relies on the angular moment. of the particles being really tiny. What if you *did* have individual particles with a lot of angular moment., say? What the quantum mechanics is saying is that the z-momentum is a mixture of x-up and x-down, when you measure in the x-direction. If the particles were transferring z-momentum as well as getting "measured" in the x-direction, you are essentially measuring them in 2 directions at once.
    And what if you did do the measurements a quadrillion quadrillion times? You would have a long string of measurements along the x-axis, and, you believe, you would have transferred obvious z-momentum to the spaceship. So you would know what the z-momentum of the particles had to have been, at least for the great majority of them. So you would have measured the angular moment. along 2 axes at once.
    No, it seems to me you can't have the stream of z-down states transferring z-angular moment, on an individual basis, unless the system remains in a superposition as far as the x-momentum measurements go. So you would have to consider it as not measured in the x-direction, when thinking about z-momentum.

    What happens to this situation when it gets macroscopic, I don't know. How do you go from a quantum state where x-momentum definite [itex]\leftrightarrow[/itex] z-momentum indefinite, to a macroscopic state where both momentums are definite? Maybe this "paradox" with the spaceship would be one of the rare examples where you could have a macroscopic quantum superposition, of x-up and x-down states.

    Yes, that's what I was saying, I don't see a way to get out of it by saying that the experiment couldn't be done. The QM demon seems to have it easy, compared to the Maxwell demon.
    I asked about this elsewhere and somebody told me they'd asked the same question I did, of their professor and the answer was "It depends on what you mean by 'observe'".
    Laura
     
    Last edited: Jun 3, 2008
  14. Jun 3, 2008 #13
    And it seems to me, if when you measure a z-down particle in the x direction, you are transferring z-angular moment. *from* the particle, by the same token, you are transferring x-momentum *to* the particle. So you think this measurement process is an exchange of angular moment.? Maybe it is - I hadn't heard it described that way.
    Laura
     
  15. Jun 3, 2008 #14

    reilly

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    There is no evidence that angular momentum is not conserved -- no torques allowed.

    You start, assuming that you did make a typo in the initial state, with two fermions with Sz = 0. Whatever total S is, the final state must have Sz=0. We need to figure out what total Z is.For two fermions, S can be 0 or 1; the S=0 state is antisymmetric, while the S=1 state is symmetric. So, the easiest states to construct are J=L+S =0 with L=0; S=0, or, L=1, S=1 -recall that an S-wave(L=0) is symmetric, and a P-wave(L=1). Let's look at the L=0, S=0 case. Rotate one of the states by 90 degrees to get the Sz eigenstates in terms of "Sx" states. The state of interest is not an eigenstate of Sx, so it has a probability to be in both Sx states. That's all you need to assert that angular momentum is conserved unless the data show otherwise. (Do you really need L?)
    Regards,
    Reilly Atkinson
     
  16. Jun 3, 2008 #15
    vanesch-

    Re: lark's last two posts; measuring a Z-down particle in the X direction. Say, that the measuring device were configured, by some some means, so that all free electrons in the device were Z-down. Would it cease to sort electrons, but pass them through unchanged?
     
  17. Jun 3, 2008 #16

    vanesch

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    Ah, but then you cannot say anymore that the Z and X component are totally incompatible. They still have an incompressible uncertainty, like momentum and position, but it is not because you know position to some extend that momentum is entirely uncertain. In other words, it is not forbidden to know the Z-component with limited accuracy if we measure X when the angular momentum is large. It is only for the case 1/2 hbar that the uncertainty is total.

    Again, in the classical limit, there's obviously no problem in observing both the X and the Z component of angular momentum (with tiny uncertainties). So (correspondence) there's no doubt that for large overall L, the possibility exists quantum mechanically to measure Lx and Lz simultaneously within some limits of precision.

    This is also manifest with the Heisenberg relationship:
    [Lz,Lx] = i hbar Ly from which follows that delta Lz * delta Lx > 1/2 hbar E [ Ly ]

    which doesn't help of course in the specific case because we have 0 on both sides (delta Lz is 0 and E [Ly] is 0 for a z-minus state), but which indicates that for a random, large L, we have something of the kind of (delta)^2 on the left, and hbar L on the right, meaning that the error is going to be the geometrical mean of hbar and L, so that the relative error is going to be like 1/sqrt(N), where N is the number of "hbars" we have in the total angular momentum.
     
    Last edited: Jun 3, 2008
  18. Jun 4, 2008 #17

    reilly

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    1 As I pointed out in my previous post, angular momentum conservation imposes a precise probability structure to the Sz-Sx measurements.(If I'm not mistaken this notion is important for Bell's Thrm.) This is about as basic a QM problem as you can find. See any reputable QM textbook, and satisfy for yourself that angular momentum is conserved in the problem you describe. No "ands, ifs or buts" about it.

    2. Check out the Euler Equations for rigid body motion. They are not so easy due to the fact that the 3D rotations form a non-commutative group. They are closely related to the Quantum Theory of the rigid rotator, used in the study of molecules among other things. In fact, this is a difficult problem, solved for a symmetric top for example, by wave functions that equal the spatial representations of finite rotations. This is very old stuff.

    3. For a spinless particle, the rotation group allows only L*L, and Lz to be conserved in QM( L is orbital angular momentum). Look in Goldstein or any advanced Classical Mechanics text to see what happens in the classical case.

    4. Old as it is, Edmonds' Angular Momentum in Quantum Mechanics gives a superb account of angular momentum theory from the very basics,to representations of finite rotations, and further to arcane things like fractional parentage coefficients and 9-j symbols, used in nuclear physics -- these are elegant versions of Clebsch-Gordon coefficients. He also gives a very nice account of the classical limit of quantum orbital angular momentum states.

    No mysteries here. Just plain old QM, including the 3D rotation group.
    Regards,
    Reilly Atkinson
     
    Last edited: Jun 4, 2008
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