"Deriving" Poisson bracket from commutator

Question

This Q/A shows that deriving P.B.s from commutators is subtle. Without going into deep deformation quantization stuff, Yaffe manages to show that $$\lim_{\hbar \to 0}\frac{i}{\hbar}[A,B](p,q)=\{a(p,q),b(p,q)\}_{PB} \tag{1}$$ where the capital case letters denote quantum operators and the small case denote classical numbers. Expectation value is implicit on the left hand side i.e. for example $$A(p,q)=\langle p,q | A |p,q \rangle,$$ etc. It is shown in section II of that paper that $$\begin{aligned}(A B)(p, q)=\int & \left(\frac{d p^{\prime} d q^{\prime}}{ 2 \pi \hbar}\right)\left|\left\langle p, q \mid p^{\prime}, q^{\prime}\right\rangle\right|^2 & \times \frac{\left\langle p, q|\hat{A}| p^{\prime}, q^{\prime}\right\rangle}{\left\langle p, q \mid p^{\prime}, q^{\prime}\right\rangle} \frac{\left\langle p^{\prime}, q^{\prime}|\hat{B}| p, q\right\rangle}{\left\langle p^{\prime}, q^{\prime} \mid p, q\right\rangle} .\end{aligned}$$ where $$|\langle p,q |p',q' \rangle|^2=\exp\left(-\frac{1}{2 \hbar}\left[ (p-p')^2+(q-q)'^2\right]\right)$$ is the overlap of Gaussian coherent states. Now, since $\hbar \to 0$ implies that the overlap peaks near $q=q',p=p'$ it is said that expanding the integral about $q=q',p=p'$ we get the result $(1)$ which we are trying to derive. However, I am not sure how to do this expansion of operators. I am unable to understand why the $\frac{i}{\hbar}$ is required on the left hand side of $(1)$ at all. Also how does the cross terms i.e. terms of the form $a \frac{\partial b}{\partial q}$, etc. cancel: do we have to expand the overlap also, how does the integral work out? This might seem a lot of questions, but the issue is really one: I don't understand how to do the calculation. Any help in doing the explicit calculation is appreciated.

Answer 1

Derivation of classical limits relies on comparing apples with apples, and not oranges. That is, one must first devise a faithful invertible map between Hilbert-space operators and phase-space quantities ("symbols"). Larry's coherent-state representation expressions, popular in the large-N community of the early 1980s, have been demonstrated to be less comfortable for elaborate manipulations, and worth translating into the Weyl-Wigner-Moyal-Groenewold expressions of the mainstream, e.g. CTQMPS, fancifully referred to as "deformation quantization".

ℏ enters in the very definition of the expressions A(p,q) involved in your (1), and is thus all over the place. The subleading terms, the terms that trouble you in the semiclassical approximation, are much easier in the deformation-quantization formulation. Such symbols entail elaborate nonlocal averages for which intuition is sparse.

• The PoissonBracket is the easy ℏ→0 limit of the Moyal bracket, the exact translation of the quantum commutator in phase space.

In case you wished to transition from Larry's "apples" to (our ref) CTQMPS, and for the benefit of future readers, I'll just record the translation from his Coherent states' (Husimi) symbols to the standard Weyl symbols, (138) of our linked ref. The thorough transition is found in, e.g., Harriman & Casida (1993), "Husimi representation for stationary states", International journal of quantum chemistry 45 (3), 263-294.

Larry's symbol A(p,q) for the operator $\hat A$ is a convolution of the Weyl symbol a(q,p) of CTQMPS: $$ A(p,q)=\langle p,q |\hat A|p,q \rangle ,\tag{Y 2.5} $$ $$ \langle x |p,q \rangle = (\pi \hbar)^{-1/4} e^{(ipx-(x-q)^2/2)/\hbar} ,\tag{Y 2.5} $$ $$ a(q,p)=\hbar \int \!\! dy ~~e^{-iyp} \langle q+\hbar y/2 |\hat A|q-\hbar y /2 \rangle ,\tag{CTQMPS 138} $$ Below, we'll change variables $x=q'+\hbar y/2$ and $x'= q'-\hbar y/2$.

$$ A(p,q)=\int\!\! dx dx' ~~~\langle p,q|x\rangle \langle x|\hat A|x'\rangle \langle x'|p,q\rangle \\ = {1\over \sqrt {\pi \hbar} } \int\!\! dx dx' ~~ \langle x|\hat A|x'\rangle e^{-ip(x-x')/\hbar} e^{-{(x-q)^2 +(x'-q)^2\over 2\hbar}} \\ = {-\hbar \over \sqrt {\pi \hbar} } \int\!\! dy dq' ~~e^{-iyp} \langle q'+\hbar y/2|\hat A|q'-\hbar y/2\rangle ~~e^{-{(x-q)^2 +(x'-q)^2\over2\hbar }} \\ = { 1 \over \sqrt {\pi \hbar} } \int\!\! dq' ~~ ~~e^{-{(q'-q)^2\over\hbar} +\hbar \partial_p^2/4} ~~a(q',p). $$