I'm reading the following article:

Feynman's derivation of the Schrödinger equation

In this article, the autor claims that Feynman derivation of the Schrödinger equation was a key aspect of the development of the Path integral approach to Quantum mechanics. However there is a step in the derivation that I don't understand, it is the argument of this step:

(Page 883): Putting this expression into Dirac's integral equation gives: $$\psi(x,t+\epsilon) \approx \int \exp\left(\frac{mi(x-y)^2}{2\hbar \epsilon}\right) \times \left[ 1-\frac{i}{\hbar}\epsilon U\left( \frac{x+y}{2} \right)\right]\psi(y,t)dy \tag{4.7}$$ Although the integration extends from $-\infty$ to $\infty$, Feynman felt that the rapid oscillation of the exponential factor (due to the small sizes of Planck's constant and the time interval) would cause the integrand to be small except where $x-y$ was likewise small. So he decided to rewrite the integral in terms of the difference $x-y=\xi$; $$\psi(x,t+\epsilon) \approx \int \exp\left(\frac{mi\xi^2}{2\hbar \epsilon}\right) \times \left[ 1-\frac{i}{\hbar}\epsilon U\left( x-\frac{1}{2}\xi \right)\right]\psi(x-\xi,t)d\xi \tag{4.8}$$ Since the integral is large only when $\xi$ is small, it made sense to Feynman to expand $\psi(x-\xi)$ in a Taylor series...

I don't understand why can we make such an approximation, My intuition in complex integration is small as well. Some references are welcome. But I've read that in perturbation theory one get things like $e^{i\phi}$. If $\phi$ is rapidly oscillating then we can neglect the term, but I've never really understood why. Furthermore in an integral I don't understand why when we make $\xi$ big, the integral goes to zero.


1 Answer 1


The exponential factor is a phase factor, where a change in the exponent represents a rotation in the complex plane. Because $\xi^2$ multiplies a very large number (both $\epsilon$ and $\hbar$ being very small), when $\xi$ is not small any $d\xi$ makes the rotation "fast" compared to the corresponding change in the other factor (the phase is proportional to the quadratic $\xi^2$ while $U$ and $\psi$ are functions of a linear $\xi$).

Over a $\Delta\xi$ corresponding to a full-circle rotation (a phase period), if the variation of the non-exponential part is negligible ("rapid" rotation) then each contribution in the overall expression is cancelled by a contribution of opposite phase.

So when $\xi$ is not small each $\Delta\xi$ period counts as zero. Overall the integral takes its value from small values of $\xi$, which justifies the Taylor expansion.

  • $\begingroup$ It seems like you know what you're talking about, but I find your answer hard to follow. Perhaps you could clarify/reword? $\endgroup$
    – anon01
    Jun 23, 2016 at 3:03
  • $\begingroup$ @ConfusinglyCuriousTheThird. I rewrote the answer. $\endgroup$ Jun 23, 2016 at 8:29

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