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

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?

• Source for the first statement? Commented Feb 23 at 12:50
• Note that [$\hat q,\hat p] = i\hbar \{q,p\}$ Commented Feb 23 at 13:28
• I've fixed your dollar placements. Note we put them around full expressions, not individual symbols. In particular, $y=z$ shouldn't have dollar signs around the $=$.
– J.G.
Commented Feb 23 at 13:36
• i.e. the commutator $[x, p]=iℏ$, because $\{x,p\}=1$. Commented Feb 23 at 13:36
• Possible duplicates: What is the connection between Poisson brackets and commutators? and links therein. Commented Feb 23 at 16:07

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.
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
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.