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Gedankenexperiment on quantum entanglement

  1. Feb 22, 2009 #1
    Hello,

    could someone please tell what mistake I am doing by the following gedankenexeperiment.

    Imagine two Stern-Gerlach-like devices that measure spin projections of some spin-1/2 particles. Right between them in the middle there is some trap (I don't care about technical details) which contains an entangled state of two identical spin-1/2 particles with a total spin 0. At some point the entangled particles fly apart and go through the detectors, which simultaneously measure the spin projections (yeah, I know, simultaneity is relative, let's say that the measurements happen exactly simultaneously in the center mass system.

    But here's the important point. Let's say it all happens on the z-axis (trajectories of the particles, orientation of devices). And let now the first device measure the x-projection of the spin and the second the y-projection!

    If there were only one device, measuring, say, the x-projection of the spin of the first particle, then we would know that if we measure, say, "spin down", then we know that the other one is "spin up". Ok.

    And now we measure both particles - different projections that do not commute. When we measure the first particle, say, "spin down" in the x-direction then the other one should be "spin up" in x-direction, but we also measure its y-projection of the spin (which gives us the y-projection of the spin of the first particle)!!! This means, that we know BOTH the x- and the y-projection of the spin of both particles. How can it be? The corresponding operators do NOT compute!

    What's the point that I am missing? Where is my mistake?

    Thanks
     
  2. jcsd
  3. Feb 22, 2009 #2

    Doc Al

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    In the standard interpretation of quantum mechanics, by measuring the spin of the particle along the y-axis you destroy any information about its spin along the x-axis.
     
  4. Feb 22, 2009 #3
    Exactly my point. But if we perform the two measurements simultaneously, what happens then?
     
  5. Feb 22, 2009 #4

    Vanadium 50

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    Well, you yourself pointed out the problem with "simulaneous".

    But consider the following situation: you have a setup where one leg measures Sx and the other Sy. You are handed a list of measurements. Is there any way to tell from that list which one was measured first?
     
  6. Feb 22, 2009 #5
    I cannot tell it from the list. But I can arrange the measurement in such a way that in some frame of references the two measurement do happen simultaneously. In any other frame of reference one measurement would happen after the another and, of course, "destroy" the results of the first measurement. But what happens in the frame of reference where the measurements are indeed simultaneous?

    By the way, this brings me to another question. I can find a frame of reference where the x-measurement happens first and I can find one where the y-measurement happens first. In the first case the x-measurement result is "destroyed", in the second the y-measurement result. But the "physics" should somehow be invariant of the frame of the reference, shouldn't it? I mean, if we, for example, calculate the "twin paradox" in any arbitrary frame of reference, we shall still get the absolute sign of the age difference. But in this case, sometimes the particles will have a definite spin x-projection and sometimes y-projection. What is physical about it? What is invariant?
     
  7. Feb 22, 2009 #6

    DrChinese

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    Welcome to PhysicsForums!

    The "mistake" is simple. No real issue with the setup as you describe it, nor how you define simultaneous. The issue is: how would you know if you *really* had learned more about a particle than the Heisenberg Uncertainty Principle (HUP) allows?

    Let's say Alice is measured in the x direction as UP. Bob must be DOWN in x, right? And you measure Bob in the y direction and get UP. Alice must be DOWN in y, right?

    So next you measure Alice in y and Bob in x - AFTER making the first measurements per above. Do you think you then get the predicted values per above? Definitely not, you get the predicted answer only 50% of the time - the same as by pure chance! That is because you really learned nothing "extra" from the other measurement. And the HUP is magically left intact. So the answer is you didn't beat the HUP after all. A certain measurement on one observable means complete uncertainty in a non-commuting observable. And the particle will respond accordingly. (And the order of the measurements will not matter in this example either.)
     
  8. Feb 22, 2009 #7
    Thanks, DrChinese, that was a great answer! Now I see, where I am wrong. Thanks again!
     
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