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Classical limit of the commutator is a derivative?

  1. Dec 9, 2009 #1
    I just came across the following claim:

    [tex]\lim_{\hbar\rightarrow 0}[\frac{1}{\hbar}(AB-BA)][/tex]

    (which approaches the classical Poincare commutator) is a derivative with respect to [tex]\hbar[/tex]. I know it looks like derivative, but is it really? Please elaborate.
  2. jcsd
  3. Dec 9, 2009 #2
    The commutator can be generally written as a series

    [tex][A,B] = i \hbarK_1 + i\hbar^2K_2 + i \hbar^3 K^3 + \ldots[/tex]

    where [tex]K_i[/tex] are Hermitian operators. This follows simply from the fact that [A,B] is antiHermitian and that it must tend to zero as [tex]\hbar \rightarrow 0 [/tex]. So the limit

    [tex]-\frac{i}{\hbar}\lim_{\hbar\rightarrow 0}[A,B] = K_1[/tex]

    (which is the classical Poisson bracket of A and B) can be regarded as the derivative of [A,B] with respect to [tex]\hbar [/tex] at [tex]\hbar \rightarrow 0 [/tex].

  4. Dec 10, 2009 #3
    Very cool. Thanks, Eugene!
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