# Feynman's derivation of Schrödinger equation. Potential spatial dependence

I am working on the book "Quantum Mechanics and Path Integrals" from Feynman and Hibbs. When finding the correspondence with Schrödinger equation he takes

\eqalign{&\psi(x,t+\epsilon) = {}\cr &\int_{-\infty}^{\infty} \!\!\exp\left\{\frac{i\,\epsilon}{\hbar} L\left(\frac{x + y}{2},\frac{x - y} {\epsilon}\right) \right\}\, \psi(y,t)\,\frac{\mathrm{d}y}{A(\epsilon)}\cr}

Making the Lagrangian explicitly as $$L = m\dot{x}^2/2 - V(x,t)$$, and making the substitution $$y = x + \eta$$ he gives

\eqalign{ &\psi(x,t+\epsilon) = \int_{-\infty}^{\infty} \exp\left\{ \frac{i m \eta^2}{2\hbar\epsilon} \right\} \cr &\qquad\exp\left\{ -\frac{i\, \epsilon}{\hbar} V\left( x+ \frac{\eta}{2}, t \right) \right\} \psi(x +\eta, t) \, \frac{\mathrm{d}\eta}{A(\epsilon)}\cr}

Now the first exponential varies very rapidly and he says that most of the integral will be contributed by $$\eta$$ in the order of 0 to $$\sqrt{2\hbar \epsilon/m}$$. For a small $$\eta$$ he can now expand the second exponential, as well as $$\psi$$

\eqalign{ &\psi(x,t) + \epsilon\, \frac{\partial \psi}{\partial t} = {}\cr &\quad\int_{-\infty}^{\infty} \exp\left\{ \frac{i m \eta^2}{2\hbar\epsilon} \right\} \left[1 -\frac{i\, \epsilon}{\hbar} V \left( x, t \right)\right] \cr &\qquad\left[\psi(x,t) + \eta \frac{\partial \psi}{\partial x} +\frac{\eta^2}{2} \frac{\partial^2 \psi}{\partial x^2} \right] \, \frac{\mathrm{d}\eta}{A(\epsilon)}\cr}

Here, he replaces $$\epsilon V(x +\frac{\eta}{2},t)$$ for $$\epsilon V(x,t)$$ saying that the error is of higher order than $$\epsilon$$.

My problem is that the expansion of $$V(x +\frac{\eta}{2},t)$$ would have a term of order $$\eta$$, which when multiplied by $$\eta \frac{\partial \psi}{\partial x}$$ would give a term of order $$\eta^2$$ and it's integration would be non-zero. The terms of order $$\eta^2$$ are not neglected, since that going with the second derivative of $$\psi$$ is preserved. The problematic term is then

$$\int_{-\infty}^{\infty} \exp\left\{ \frac{i m \eta^2}{2\hbar\epsilon} \right\} \frac{i\, \epsilon \, \eta^2}{\hbar} \left. \frac{\partial V}{\partial (x + \eta/2)} \right|_{(x,t)} \frac{\partial \psi}{\partial x} \, \frac{\mathrm{d}\eta}{A(\epsilon)}$$

I think that the problem might be I am not working properly the Taylor series.

Thank you for your help.

## 1 Answer

Okay, the problem actually is not there. Both statements are correct, the mistake I've made was in the comparison between orders of the development.

We take just the first order in $$\epsilon$$ in the left hand side $$\psi + \epsilon \partial_t\psi,$$ and the last integral, $$\int_{-\infty}^{\infty} \exp\left\{ \frac{i m \eta^2}{2\hbar\epsilon} \right\} \frac{i\, \epsilon \, \eta^2}{\hbar} \left. \frac{\partial V}{\partial (x + \eta/2)} \right|_{(x,t)} \frac{\partial \psi}{\partial x} \, \frac{\mathrm{d}\eta}{A(\epsilon)},$$ will give something of the order of $$\epsilon^2$$, since we have $$\int_{-\infty}^{\infty} \exp\left\{ \frac{i m \eta^2}{2\hbar\epsilon} \right\} \eta^2 \, \frac{\mathrm{d}\eta}{A(\epsilon)} = \frac{i\hbar \epsilon}{m},$$ where the condition for $$A$$ is found through the correspondence of the terms of zeroth order, and nothing else depends on $$\eta$$, and there is one factor $$\epsilon$$ already present.

The term with the second derivative does have an $$\eta^2$$, but only its product with the 1 in the expansion of the potential is preserved.

The identification of the first order terms on $$\epsilon$$ gives the expression of the Schrödinger equation.