Classical limit of Moyal bracket in integral representation

Question

It is well-known that the Poisson bracket can be recovered out of the Moyal bracket under the limit when $\hbar$ goes to zero $$\lim_{\hbar\rightarrow 0} \lbrace f,g\rbrace_M=\lbrace f,g \rbrace_P.$$ This is easy to verify in the differential form of the brackets, but the Moyal bracket admit an integral representation as $$\lbrace f,g\rbrace_M\\ =\frac{2}{\hbar^3\pi^2}\int dp' dp'' dx' dx'' f(x+x',p+p')g(x+x'',p+p'')\sin\left (\frac{2}{\hbar}(x'p''-x''p')\right ).$$ How to recover an integral form of the Poisson bracket performing the limit here?

Answer 1

There are no compact integral forms of the PBs, of course, unless you consider conversions of derivatives into powers of the Fourier conjugate variable, as you might be insinuating in the comments. The genius of G Baker's 1958 construction is his appreciation of the $\star$-product as the basis of phase-space translations, which results in the MB integral expression you have (corrected by me).

One may reduce the classical limit of that expression to the standard differential rep of the PBs (only), provided for the curious student of the future, in case you yourself were not interested anymore, as commented.

You must first non-dimensionalize the primed and double primed dummy variables of the integral (only), so, for $$ (x',x'',p',p'')\mapsto \sqrt{\hbar}~(x',x'',p',p''), $$ you have, for the Baker MB, $$ =\frac{2}{\hbar\pi^2}\!\!\int\!\! dp' dp'' dx' dx'' f(x+\sqrt{\hbar}x',p+\sqrt{\hbar}p')g(x+\sqrt{\hbar}x'',p+\sqrt{\hbar}p'')\sin\left ({2}(x'p''-x''p')\right )\\ = \frac{1}{i\hbar\pi^2}\!\!\int\!\!\! dp' dp'' dx' dx'' \left ( e^{{2i}(x'p''-x''p')}-e^{{-2i}(x'p''-x''p')}\right )e^{\sqrt{\hbar}(x'~_f\partial_x + p'~_f\partial_p+ x''~_g\partial_x + p''~_g\partial_p)} f(x,p)g(x,p) . $$ The last line reflects the Baker translation operator, where the left subscript of the partial derivatives signifies the respective functions f or g on which these partial derivatives act! note all four terms of the last exponential commute among themselves, so there are no ambiguities.

Now performing the x' and p' integrals nets two evident δ-functions, which then collapse the x'' and p'' integrals $$ = \frac{1}{i\hbar}\!\!\int\!\! dp'' dx'' \left ( \delta(p''-i\sqrt\hbar ~_f\partial_x /2)\delta(x''+i \sqrt\hbar ~_f\partial_p /2)-\hbox{ c.c.}\right ) e^{\sqrt{\hbar}( x''~_g\partial_x + p''~_g\partial_p)} f(x,p)g(x,p)\\ = \frac{2i}{i\hbar}\sin \left({\hbar \over 2}(_f\partial_x~_g\partial_p -_g\partial_x ~_f\partial_p)\right )~~fg. $$

It is evident that this expression collapses to the the PB $\{f,g\}$, for small $\hbar$, and it should be evident why this differential expression is 37000 times simpler than any indirect integral expression, in sharp contrast to the MB...