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? $$[\Lambda,\Omega]=i\hbar \{\lambda,\omega\}_{x,p}$$
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Qmechanic
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Young Kindaichi
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Is this in reference to canonical quantisation? The classical algebra with the Poisson bracket becomes the quantum algebra with the commutator. – Charlie Mar 23 '21 at 15:43
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$\lambda(x,p)\rightarrow \hat{\Lambda}(\hat{X},\hat{P})$. Is that makes it clear? – Young Kindaichi Mar 23 '21 at 15:44
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3Does this answer your question? What is the connection between Poisson brackets and commutators? – Mar 23 '21 at 15:46
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1It 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. – mike stone Mar 23 '21 at 16:51