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Poisson Bracket to Commutator, What Does it REALLY Mean?

  1. Dec 28, 2011 #1
    Let me just head off the first waves of posts this thread will likely get. I am very fluent in quantum mechanics. I am completely aware of the behaviour of a commutator structure: simultaneous eigenbasis, etc. I understand how commutators model the structure that quantum mechanics has. My question is, when I have my classical equation

    [tex]\frac{d A}{dt} = \{ A , H \}+ \frac{\partial A}{\partial t}[/tex]

    and I say [tex] \{ A, H \} \rightarrow [A,H] [/tex] what single fundamental ASSUMPTION am I changing about my reality? I've read about symplectic manifolds and moyal brackets and such but what intuitively/physically is occurring in this transition?

    I would like to be able to make a statement like:

    In classical mechanics we assumed our universe has property BLAH, however, in reality it has property BLARG. In our equation we had our classical poisson bracket which represented THIS about our system, however, in light of our new assumptions we can CALCULATE that it must be THIS, which is called the commutator.

    With all the capitalized stuff filled in.

    Thanks for the help
     
  2. jcsd
  3. Dec 28, 2011 #2
    Quantum mechanics is exact theory, while classical mechanics is just an approximation. So, it makes more sense to ask how classical Poisson brackets arise as an approximation to quantum commutators? This question is easier to answer than your original question.

    For a single particle any quantum observable can be represented as a function of position and momentum observables [itex]F = f(x,p) [/itex]. Here I use 1-dimensional case for simplicity. Then one can prove that commutator of two Hermitian observables can be represented as a series in powers of [itex] \hbar [/itex]

    [tex] [F, G] = i \hbar K_1 + i \hbar^2 K_2 + \ldots [/tex]

    Finally, it is not difficult to prove that the most significant term [itex] K_1 [/itex] on the right hand side has exactly the form of Poisson bracket

    [tex] [f(x,p), g(x,p)] = i \hbar \left(\frac{\partial f}{\partial x}\frac{\partial g}{\partial p} - \frac{\partial f}{\partial p}\frac{\partial g}{\partial x} \right) + \ldots [/tex]

    Eugene.
     
  4. Dec 28, 2011 #3
    Hmm... That's very interesting. But whenever we construct a quantum field theory or theory we start with a classical lagrangian which we generally motivate by some assumptions of locality and simplicity and such and then we quantize it. Is it possible to do the reverse? Can one motivate a quantum theory/field theory with a built in commutator structure and then approximate it as above to receive the classical theory?
     
  5. Dec 29, 2011 #4

    tom.stoer

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    The peoblem is to 'guess' the correct quantum theory. In principle this is possible, but in practice we often start with a classical theory we know and then try to quantize it; guessing a classical theoriy is much easier.

    This is like building a house based on a drawing. w/o having ever seen a house the drawing alone will not help; but w/o the dawing even an expert will not be able to build the house correctly. So even if the starting point is only a sketch and even if the construction itself requires some assumptions, the whole procedure seems to make sense.
     
  6. Dec 29, 2011 #5
    But that just gets back to the initial problem. The physical interpretation and justification of the poisson bracket -> commutator transition. Plus I don't much like this notion of "guessing" without appealing to some symmetry or locality requirement, ec.
     
  7. Dec 30, 2011 #6

    strangerep

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    In both the classical and quantum cases, one usually has a set of observable quantities. In CM, different observables can be measured in any order (for a system in a given state), and the state space is essentially a classical phase space. One represents the symmetries, observables, etc, as a commuting algebra of functions on that phase space.

    In QM we start with that same algebra of observables, and demand that there exists a linear mapping from any observable to a number (for the system in a given state), and the state space is a (rigged) Hilbert space. Inner products on the Hilbert can be interpreted as a (nondeterministic) transition amplitude between states. IOW, we're combining the algebra of observables with a statistical interpretation. The latter was absent in the ordinary CM case, hence we were able to use a simpler state space (phase space).

    In CM, the algebra of observables is encoded in the form of Poisson brackets between the functions. In QM it's the commutation relations of operators on Hilbert space. In both cases, we're trying to construct a representation of the same algebra of observables, but on different state spaces.

    For higher order products in the algebra of observables, the transition to operators on Hilbert space might be ambiguous. Sometimes, simple symmetrization of the operators in a product works ok for resolving the ambiguity satisfactorily.

    HTH.
     
  8. Dec 30, 2011 #7
    What is this general algebra of observables?
     
  9. Dec 30, 2011 #8

    strangerep

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    In general, the algebra of observables is specific to the particular class of system being studied. It's not one-algebra-fits-all.

    E.g., for nonrelativistic systems, the generators of Galilean transformations form a (sub)algebra for the overall dynamical algebra. (The latter contains additional stuff related to the potential.)

    But, hmmm, that probably doesn't help, right? I'm not sure at what level I should try to answer this. You'll have to give me a bit more to work with in your question.
     
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