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

[hideelo]

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 $\eta$?

 

[vanhees71]

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.