The path integral in quantum mechanics involves a factor ##e^{iS_{N}/\hbar}##, where(adsbygoogle = window.adsbygoogle || []).push({});

##S_{N}\equiv \sum\limits_{n=1}^{N+1}[p_{n}(x_{n}-x_{n-1})-\epsilon H(p_{n},x_{n},t_{n})].##

In the limit ##N \rightarrow \infty##, ##S_{N}## becomes the usual action ##S## for a given path.

When the Hamiltonian vanishes, the potential energy of the system offsets the kinetic energy of the system and in the limit ##N \rightarrow \infty## the propagator ##\langle x_{b}, t_{b}|x_{a}, t_{a}\rangle## becomes

##S_{N} \equiv \sum\limits_{n=1}^{N+1}\epsilon\bigg[p_{n}\bigg(\frac{x_{n}-x_{n-1}}{\epsilon}\bigg)\bigg]=\int\ dt\ p\dot{x}##

I would like to think of a physical argument to justify this answer. Thoughts?

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# I Vanishing Hamiltonian for quantum path integral

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