How does non-commutativity emerge from path integral?

[kof9595995]

It is not obvious to see the non-commutative nature of QM in path integral formulation. I've read something on Wiki:
http://en.wikipedia.org/wiki/Path_integral_formulation#Canonical_commutation_relations
But I can't work out the math fully, can someone guide me a bit?

 

[genneth]

Consider constructing the path integral from the Hamiltonian. You need the Hamiltonian to be in ordered a particular way --- if you're using coherent states, then normal ordered (or anti-), or if you're using q and p, then again some definite ordering of polynomial terms. You sandwich exp(-Ht) between the initial and final states, then break this up into N small pieces, inserting resolution of the identity with the states |x> and |p> or whatever else you need, remembering that you want the correct states to hit the right things in the Hamiltonian. This then bring out the term which looks like $p \partial_t q$ in the final expression, and we identify the whole shebang along with -H as the lagrangian.

 

[kof9595995]

I see, when trying to order the operators the commutation relation is naturally invoked. But I don't see how to use the argument in the link I cited, to show the non-commutativity.