I have read many answers on the correspondence between classical Poisson brackets and quantum commutators, but I was wondering if it was able to be spelled out in simple words.
In this answer (What is the physical interpretation of the Poisson bracket) they say 'you can consider the Poisson bracket {π,π} as expressing the rate of change of π as a consequence of a flow induced by π...When π=π and π=π, since the momenta are the generators of translations, the flow generated by π can be interpreted as translations, so that the canonical bracket {π,π}=1 implies a variation of πΏπ={π,π}π=π'.
So would I be at all correct in saying the reason why this implies that the quantum commutation relation $[\hat x,\hat p]=i\hbar$ (and ultimately that position and momentum are not simultaneously measurable) is because QM uses operators and therefore the above Poisson bracket shows that $\hat x \hat p$ generates a translation 'flow' before measuring position whereas $\hat p \hat x$ generates a translation 'flow' after measuring position (and this is not the case where the Poisson bracket is zero)?
What exactly then is the distinction here between classical and quantum mechanics. If $p$ generates a flow in $x$ in both theories, but in QM this causes $p$ and $x$ to not commute, but in CM they can still commute?
