# I Shankars derivation of equivalence between path integral and Schroedinger's eqn

1. Dec 29, 2016

### 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 $\eta$?

2. Dec 30, 2016

### 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.