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$$ \hat{\Omega}_j{(\tilde{q}_j)}=\Omega_j(\tilde{q}_j-\hat{q}_j) $$

$$ [\hat{q}_j,\hat{q}_l]=ik_{jl} $$

Implies

$$ [\hat{q}_j,\hat{\Omega}_l]= \frac{\partial\Omega_l(\tilde{q}_l-\hat{q}_l)}{\partial\hat{q}_l}.ik_{jl} $$

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    $\begingroup$ Can you clarify what are the symbols in your questions? $\endgroup$ – Andy Bale Nov 20 '13 at 21:03
  • $\begingroup$ The hats refer to arbitrary operators, and the tilde the measured observable $\endgroup$ – user32462 Nov 20 '13 at 21:45
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Hints to the question (v1):

  1. Let $\hat{Z}^I$ be operators that satisfies a Heisenberg algebra $$\tag{1} [\hat{Z}^I,\hat{Z}^J]~=~i\hbar ~\omega^{IJ} ~{\bf 1},$$ where $\omega^{IJ}=-\omega^{JI}$ is an antisymmetric real matrix. The important property will be that the commutators $[\hat{Z}^I,\hat{Z}^J]$ belong to the center of the Heisenberg algebra, i.e. commute with everything.

  2. Let $f(\hat{Z})$ and $(\partial_{J}f)(\hat{Z})$ denote the Weyl-ordered operators corresponding to the functions/symbols $f(z)$ and $$\tag{2} (\partial_{J}f)(z)~:=~\frac{\partial f(z)}{\partial z^J},$$ respectively.

  3. Then using e.g. this formula for the Weyl-ordered operators, it is possible to show$^1$ $$\tag{3} [\hat{Z}^I, f(\hat{Z})]~=~\sum_J [\hat{Z}^I,\hat{Z}^J]~(\partial_{J}f)(\hat{Z}).$$

  4. The corresponding classical formula $$\tag{4} \{z^I, f(z)\}_{PB}~=~\sum_J \{z^I,z^J\}_{PB}~(\partial_{J}f)(z)$$ is a consequence of the Poisson property/Leibniz rule for a Poisson bracket (PB).

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$^1$ Equation (3) actually also holds for many other operator-orderings.

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  • $\begingroup$ Actually after posting this answer, I realized that I have probably misread OP's notation and OP only consider the case where $f(z)$ depends on a single argument, so that no operator ordering issues arise, and the question is much simpler a la this question. Oh, well, here is the generalization to several variables. $\endgroup$ – Qmechanic Nov 20 '13 at 22:56

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