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Physics

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    $\begingroup$ It is not true in general that the quantum-mechanical commutator $[F,G]$ is always just the classical Poisson bracket $\{F, G\}$ times $\lambda=i\hbar$. There may also be further corrections of $\mathcal{O}(\hbar^{2})$, because of operator ordering ambiguities. Note, for example, that $\{x,xp\}=\{x,px\}$, but $[x,xp]\neq[x,px]$. $\endgroup$ Commented 16 hours ago
  • $\begingroup$ In quantum physics the commutator of two operators is defined simply in terms of their product as $[\hat A,\hat B]:=\hat A\hat B-\hat B\hat A$. Thus the statement $[\hat A, \hat B] \neq 0$ is equivalent to $\hat A\hat B \neq \hat B\hat A$. In classical physics the Poisson bracket of two functions is defined by the expression given in the question, but this is not directly related to "multiplication"; even if $\{F,G\}\neq0$, where $F=F(q_i,p_i)$, $G=G(q_i,p_i)$, it still obviously holds $FG=GF$, as both are c-number valued functions. $\endgroup$ Commented 8 hours ago