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

  1. Dec 29, 2016 #1
    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. jcsd
  3. Dec 30, 2016 #2

    vanhees71

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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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