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Canonically conjugate operators

  1. Dec 17, 2012 #1

    ShayanJ

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    I've searched for this but found nothing,so I ask it here.

    What are canonically conjugate operators?
    Is [itex] [A,B]=cI [/itex] a definition for A and B being canonically conjugate?

    Thanks
     
  2. jcsd
  3. Dec 17, 2012 #2

    tom.stoer

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    Yes

    Canonically conjugate operators A, B follow from canonically conjugate variables A, B in classical mechanics; their Poisson bracket is {A,B} = 1; they span the phase space of the system, can be used to formulate the Hamilton function H(A,B) and therefore their Hamilton e.o.m. fully define the dynamics of the theory.

    In QM (canonical quantization) the variables on phase space are replaced by operators acting on Hilbert space; the commutators are defined as

    [tex]\{A,B\}_\text{Poisson} = c \;\to\; [\hat{A},\hat{B}] = ic;\;c = \text{const.}[/tex]
     
    Last edited: Dec 17, 2012
  4. Dec 17, 2012 #3

    ShayanJ

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    http://en.wikipedia.org/wiki/Canonical_commutation_relation
    You see,looks like its related to the fourier transform too.
    In the things I've read,such relationship exists in classical mechanics too.
    Just extending it to QM is a little hard for me.

    Thanks tom
     
  5. Dec 18, 2012 #4

    tom.stoer

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    I would not start with the Fourier transform.

    It's correct, that iff one uses a representation like a wave function in x, then p acts as a derivative and this is related to the Fourier transform; and it's correct that the relation between x- and p-space wave functions is just the Fourier transform. But the defining operator equation [x,p] = i is more general than a specific representation and therefore does not require a Fourier transform in its definition.
     
  6. Dec 18, 2012 #5
    Does the prescription of turning canonically conjugate variables into operators whose commutator is "i" hold for any pair of classically conjugate variables? I vaguely recall that this is not true for all pairs of classically conjugate variables (it's certainly true for Cartesian variables), but I don't remember exactly what the issue was.
     
  7. Dec 18, 2012 #6

    tom.stoer

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    That is the definition of canonical quantization. The problem with non-cartesian coordinates (like polar coordinates) is that already classically their Poisson bracket is not = 1, so it's the wrong starting point. But yes, you are right, it is not allowed to replace classical Poisson bracket by the commutator in general; it works for "elementary" canonical pairs like x and p but fails for general functions f(x) and g(p). The reason is that first calculating the Poisson bracket {f,g} and then replacing this with the commutator is not the same as replacing {x,p} with the commuattor and then calculating [f,g].

    Not knowing (globally) cartesian coordinates is a difficult starting point. On a manifold (with non-trivial metric) one may define "covariant derivatives" instead of standard ones; usually this results in a reasonable quantum theory. If you start with polar coordinates on the sphere using ∂Ω does not makes sense, but when using covariant derivatives one e.g. arives at the generalized Laplace-Beltrami operator Δg (g is the metric on the manifold) which is equivalent to the standard 3-dim. Laplacian expressed in polar coordinates plus ∂r set to zero (fixed radius).

    Have a look at http://en.wikipedia.org/wiki/Canonical_quantization as a starting point
     
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