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I'm trying to do exercise I.2.1. from Zee's QFT in a nutshell but I ran into a problem. The exercise is to derive the QM path integral with a Hamiltonian of the form 1/2 m p^2 + V(q). In the textbook he shows the proof for a free hamiltonian. He gets to a point where he has (I left out the integral for |p><p|)

[itex]e^{-i \delta t (\hat p^2 /2m)} |q> = e^{-i \delta t (\hat p^2 /2m)} |p><p|q> = e^{-i \delta t (p^2 /2m)} |p><p|q>[/itex] ([itex] \hat p [/itex]is an operator) which is obviously true. But in my case I have

[itex]e^{-i \delta t (\hat p^2 /2m + V(\hat q))} |q> \neq e^{-i \delta t( \hat p^2 /2m + V(q))} |q>[/itex]

as the commutator of [itex]\hat p[/itex] and [itex]\hat q[/itex] does not vanish. Thus I have no idea of how to prove this in general. In some QFT lecture notes I found they expand the exponential to first order, substitute[itex] \hat q = q [/itex]and [itex]\hat p = p [/itex]and write it again as an exponential. But I don't like this last step and want to do it more rigorous. Any hints?

Thanks :)

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# Rigorous Feynman pathintegral derivation

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