The question about commutator $[\hat{x},\hat{p}]=i\hbar$ at $\hbar\rightarrow 0$ seemingly can't match with Poisson bracket $\{x,\,p\}=1$

Question

At the limit $\hbar\rightarrow 0$, all "quantum" should tend to "classical", but why is the quantum commutator $[\hat{x},\hat{p}]=i\hbar$ at $\hbar\rightarrow 0$ equal to $0$, but the classical Poisson bracket is $\{x,p\}=1$? Why does it seem that $[\hat{x},\hat{p}]=i\hbar$ does not match with $\{x,p\}=1$ in the classical limit?

Answer 1

There is a systematic invertible change of language (Weyl correspondence) between Hilbert space operators and phase-space q-number variables, $$ \hat A \leftrightarrow A, \qquad \hat B \leftrightarrow B,\\ {1\over i\hbar} [\hat A, \hat B] \leftrightarrow \{\{A,B\}\}=\{ A,B\}+ O(\hbar^2) $$ where {{•,•}} is the Moyal bracket.

In your case, $$ {1\over i\hbar} [\hat x, \hat p] \leftrightarrow \{\{x,p\}\}=\{ x,p\}=1, $$ where no limit has been taken. The subleading terms in ℏ happen to vanish identically.

Answer 2

Well... $[x,p]=0$ for classical variables so there's no formal problem there. The simplest quantization procedure is due to Dirac in his classic

P.A.M.Dirac, The Principles of Quantum Mechanics

wherein he shows that, at least for the simplest observables, one should have $$ [\hat A,\hat B]_q=i\hbar \widehat{\{A,B\}_c} $$ i.e. the quantum commutator should be $i\hbar$ times the classical Poisson bracket of the corresponding observables. Dirac's proof leverages the similar properties of Poisson brackets and commutators, such as antisymmetry in the arguments and the so-called derivative property.

Of course the Groenewold van Hove theorem shows Dirac's procedure will eventually fail.