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Joint distribution of position and momentum

  1. Mar 7, 2014 #1


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    In quantum mechanics, there doesn't seem to be a joint distribution of position and momentum.

    But in Bohmian mechanics there is.

    But Bohmian mechanics is quantum mechanics, so what is the error in my reasoning?
  2. jcsd
  3. Mar 7, 2014 #2


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    Isn't this what they call the Wigner distribution function?
  4. Mar 7, 2014 #3


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    No, it isn't. The Wigner distribution is the closest thing, but in general it has negative bits, so it isn't a true probability distribution. In special cases like the free Gaussian wave packet, the Wigner distribution is positive, which explains why in those cases one can think of classical trajectories.
  5. Mar 7, 2014 #4


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    Heh, well, quantum probability is not the same thing as classical probability anyway. :biggrin:

    The "joint" distribution stuff is one of the places where differences arise: Cox's 4th axiom is not applicable as-is, since the notion of ##A \& B## is problematic in QM. So one must use a time-ordering instead.

    Not sure about Bohmian.
  6. Mar 8, 2014 #5


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    Not quite - its an interpretation of QM.

    It has hidden variables not in standard QM and they have properties different to the usual quantum properties eg the particle has both a definite momentum and position. But because of that pesky pilot wave that guides it you can't determine what it is - that's why its hidden.

    For more details check out:

  7. Mar 10, 2014 #6


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    There seem to be several ways to define an "actual" momentum in dBB but it doesn't seem possible to match this actual momentum with the outcome of momentum measurements. From http://arxiv.org/abs/quant-ph/0408113 (p13): "Insisting on the belief that Newtonian momentum (energy, angular momentum) measurements reveal the momentum (energy, angular momentum) leads to the orthodox view of quantum mechanics."

    I don't know any details. I just googled it because I've asked myself similar questions before.
  8. Mar 10, 2014 #7


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    I think that must be the answer.
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