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Poisson brackets commutator vs. quantum commtation relation

  1. Feb 2, 2016 #1
    If we have Poisson bracket for two dynamical variables u and v, we can write as it is known ...

    This is for classical mechanics. If we write commutation relation, for instance, for location and momentum, we obtain Heisenberg uncertainty relation.

    But, what is a pedagogical transfer from Poisson bracket to quantum mechanics. Because formulae are very different, Classical one has partial derivations, quantum one has only multiplication of matrices.

    What is transfer to quantum mechanics, or in the opposite direction?

    One example of this question:
    But; i think, that it can be answered more clearly.
  2. jcsd
  3. Feb 2, 2016 #2


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  4. Feb 2, 2016 #3


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    There's no pedagogical transfer. It's only the axiomatic Dirac quantization (PB/DB --> 1/ihbar x commutator) scheme (shown by Groenewold and van Hove to be severely limited) which is standard textbook material then the further contributions of Weyl, Wigner and Moyal which are only presented in advanced texts (they form the basis for the so-called deformation quantization). The link you gave contains more detailed answers. In the Schrödinger picture of QM, commutators also contain differential operators, quite similarly to the classical PB.
  5. Feb 2, 2016 #4


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    One can go a surprisingly long way by simply taking the (classical) dynamical Lie algebra (expressed in terms of Poisson brackets), and re-expressing it by quantum commutators (possibly inserting factors of ##\hbar## to make the dimensions correct). The ambiguities Dex mentioned above, where quadratic and higher products of operators are involved, can often be resolved in practical cases by to simply symmetrizing them. (Indeed this is necessary to obtain a satisfactory Hermitian quantum operator corresponding to the LRL vector in the H-atom problem.) In other cases, one must deform the Lie algebra to obtain a satisfactory quantum version.
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