In R. Shankar's book, He has written $$[X_i,P_j]=i\hbar\{x_i,p_j\}=i\hbar$$ Is there any specific reason to use the Poisson bracket? Is there any general relation which looks like? $$[\Lambda,\Omega]=i\hbar \{\lambda,\omega\}_{x,p}$$

  • $\begingroup$ Is this in reference to canonical quantisation? The classical algebra with the Poisson bracket becomes the quantum algebra with the commutator. $\endgroup$
    – Charlie
    Mar 23, 2021 at 15:43
  • $\begingroup$ $\lambda(x,p)\rightarrow \hat{\Lambda}(\hat{X},\hat{P})$. Is that makes it clear? $\endgroup$ Mar 23, 2021 at 15:44
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    $\begingroup$ Does this answer your question? What is the connection between Poisson brackets and commutators? $\endgroup$
    – user87745
    Mar 23, 2021 at 15:46
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    $\begingroup$ It only works exactly for operators that are at most quadratic in $x$ and $p$. Beyond that there are $O(\hbar^2)$ corrections. This is the Groenewold-van Hove no-go theorem. $\endgroup$
    – mike stone
    Mar 23, 2021 at 16:51