Vanishing Hamiltonian for quantum path integral

[spaghetti3451]

The path integral in quantum mechanics involves a factor $e^{iS_{N}/\hbar}$, where

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

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?

 

[secur]

Well, for an intuitive physical argument you could consider the classical formulation. Leaving aside various caveats, L = 2T - H (T = kinetic energy; 2T is what Leibniz called "vis viva"). When H = 0 then L = 2T which (integrated through time) is precisely your final answer, representing "action". I.e., (ignoring factor of 2) the integral of energy through time, or integral of momentum through distance.

So the answer makes perfect sense classically. I don't know if that's what you're looking for? Perhaps you already knew that and want a physical argument for the principle of least action? There are a few of those.

 

[Fightfish]

What you have there seems to be the abbreviated action $S_{0} \equiv \int \mathbf{p} \cdot d\mathbf{q}$, which is linked to Maupertuis' principle of least action - i.e. "shortest" path.