In Hamiltonian classical mechanics, we have that the Poisson bracket of position and momentum satisfies $$\{q_i, p_j\} = \delta_{ij}$$

But this implies that momentum and position 'generate' changes in each other. I.e. as we move along the flow of one, the other changes. Since this holds in classical mechanics as well as in quantum mechanics, why is there no uncertainty principle in classical mechanics? Why is it that non-commutativity under the Lie bracket operator gives rise to fundamental uncertainty in one case and not in the other?

Please answer in terms of Lie theory, if possible.


What one requires for the uncertainty principle to arise is that the relevant observables should not commute, i.e., their commutator is non-zero. The Poisson bracket of two observables is not the same as their commutator. Even if the Poisson bracket of the classical observables is non-zero, they do commute in classical mechanics--they are just real numbers. So, no uncertainty principle arises in classical mechanics. What happens is that the commutator of observables in quantum mechanics corresponds to the Poisson bracket of the corresponding classical observables (this is the famous canonical quantization scheme up to some ordering ambiguities which do not really affect the question at hand). And thus, if the Poisson bracket of the classical observables is non-zero, the commutator of the corresponding quantum observables will be non-zero. This gives rise to the uncertainty principle in quantum mechanics. So, in a nutshell, commutators of quantum observables correspond to the Poisson brackets of classical observables, but, the commutator of classical observables (which is always zero) is not the same as their Poisson bracket.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.