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 1


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.


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