How Does Shankar's PQM Derive Schrödinger's Equation?

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SUMMARY

The discussion centers on Shankar's derivation of Schrödinger's equation in his 1994 edition of "Principles of Quantum Mechanics" (PQM). The key equation presented involves a path integral formulation, specifically the integral of the wave function $\psi (x,\epsilon)$, which incorporates oscillatory behavior due to the small parameter $\epsilon \hbar$. A critical point of contention arises when Shankar restricts the range of integration for the variable $\eta = (x-x')$, yet later integrates over the entire range of $\eta$, raising questions about the validity of this step. The method of steepest descent is identified as the technique used for evaluating the integral, despite the apparent contradiction in the treatment of the variable's limits.

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hideelo
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On page 230 in Shankar's PQM (1994 edition) he is trying to show that the path integral formulation reduces to Schroedinger's eqn. The equation he comes up against is the following

$$\psi (x,\epsilon) = \sqrt{\frac{m}{2 \pi i \epsilon \hbar}}\int_{-\infty}^{\infty}dx' \psi (x',0) \exp\left[\frac{im(x-x')^2}{2\epsilon \hbar} \right] \exp\left[\frac{-i\epsilon}{ \hbar} V\left( \frac{x+x'}{2},0\right) \right]$$

He makes the argument that the first exponential is going to oscillate like hell because ##\epsilon \hbar## is so small. He says that in order to keep this under control we need to restrict the range of x' so that

$$\frac{m \eta^2}{2\epsilon \hbar} \leq \pi$$
where
$$\eta = (x-x')$$

which I follow. He then changes the variable of integration from x' to ##\eta##, no big deal. He expands everything inside the integral to second order in ##\eta## because that corresponds to first order in ##\epsilon##. I'm still on board. He then integrates over ##\eta## from ##-\infty## to ##\infty## and this is where he loses me. What happened to ##\frac{m \eta^2}{2\epsilon \hbar} \leq \pi## ? The way he expressed everything inside the integral assumed that ##\eta## is small. So why is he integrating over the whole range of ##\eta##?
 
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It doesn't matter, because ##\epsilon \hbar## is so small. That's the usual way to evaluate an integral approximately with the method of steepest descent. You usually get an asymptotic series with that technique.
 

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