Adar Kahiri

Problem 1.7

May 2020

Problem: Show that \(\frac{d\langle p \rangle}{dt} = \left\langle - \frac{\partial V}{\partial x}\right\rangle .\)

Solution: In the book, Griffiths derives momentum to be

\[\langle p \rangle = m \frac{d\langle x \rangle}{dt} = -i\hbar \int_{-\infty}^{\infty} \left(\Psi^* \frac{\partial \Psi}{\partial x}\right)dx\]

And so we have

\[\frac{d \langle p \rangle}{dt} = m \frac{d^2\langle x \rangle}{dt^2} = -i\hbar \int_{-\infty}^{\infty} \frac{\partial}{\partial t}\left(\Psi^* \frac{\partial \Psi}{\partial x}\right)dx\]

\[= -i\hbar \int_{-\infty}^{\infty} \left( \frac{\partial \Psi^*}{\partial t} \frac{\partial \Psi}{\partial x} + \Psi^* \frac{\partial}{\partial t} \left( \frac{\partial \Psi}{\partial x}\right)\right)dx\]

\[= -i\hbar \int_{-\infty}^{\infty} \left( \frac{\partial \Psi^*}{\partial t} \frac{\partial \Psi}{\partial x} + \Psi^* \frac{\partial}{\partial x} \left( \frac{\partial \Psi}{\partial t}\right)\right)dx\]

Now, we can substitute in the Schrödinger equation and its complex conjugate to get (I'm temporarily ignoring the integral sign for clarity)

\[\frac{\partial \Psi^*}{\partial t} \frac{\partial \Psi}{\partial x} + \Psi^* \frac{\partial}{\partial x} \left( \frac{\partial \Psi}{\partial t}\right) = \frac{\partial \Psi}{\partial x} \left(-i \frac{\hbar}{2m} \frac{\partial^2 \Psi^*}{\partial x^2} + \frac{i}{\hbar} V \Psi^*\right) + \Psi^* \frac{\partial}{\partial x} \left(i \frac{\hbar}{2m} \frac{\partial^2 \Psi}{\partial x^2} - \frac{i}{\hbar} V \Psi \right)\]

\[= -i \frac{\hbar}{2m}\frac{\partial \Psi}{\partial x} \frac{\partial^2 \Psi^*}{\partial x^2} + \frac{i}{\hbar} \frac{\partial \Psi}{\partial x} V \Psi^* + i \frac{\hbar}{2m} \Psi^* \frac{\partial}{\partial x} \left(\frac{\partial^2 \Psi}{\partial x^2}\right) - \frac{i}{\hbar} \Psi^* \frac{\partial}{\partial x} \left(V \Psi\right)\]

\[= -i \frac{\hbar}{2m}\frac{\partial \Psi}{\partial x} \frac{\partial^2 \Psi^*}{\partial x^2} + \frac{i}{\hbar} \frac{\partial \Psi}{\partial x} V \Psi^* + i \frac{\hbar}{2m} \Psi^* \frac{\partial^3 \Psi}{\partial x^3} - \frac{i}{\hbar} \Psi^* \Psi \frac{\partial V}{\partial x} - \frac{i}{\hbar} \Psi^* V \frac{\partial \Psi}{\partial x}\]

\[= -i \frac{\hbar}{2m}\frac{\partial \Psi}{\partial x} \frac{\partial^2 \Psi^*}{\partial x^2} + i \frac{\hbar}{2m} \Psi^* \frac{\partial^3 \Psi}{\partial x^3} - \frac{i}{\hbar} \Psi^* \Psi \frac{\partial V}{\partial x}\]

\[\implies \frac{d \langle p \rangle}{dt} = i \hbar \int_{\infty}^{\infty} -i \frac{\hbar}{2m}\frac{\partial \Psi}{\partial x} \frac{\partial^2 \Psi^*}{\partial x^2} + i \frac{\hbar}{2m} \Psi^* \frac{\partial^3 \Psi}{\partial x^3} - \frac{i}{\hbar} \Psi^* \Psi \frac{\partial V}{\partial x}dx\]

\[= \int_{-\infty}^{\infty} -\frac{\hbar^2}{2m}\frac{\partial \Psi}{\partial x} \frac{\partial^2 \Psi^*}{\partial x^2} + \frac{\hbar^2}{2m} \Psi^* \frac{\partial^3 \Psi}{\partial x^3} - \Psi^* \Psi \frac{\partial V}{\partial x}dx\]

Now, let's split the terms of the expression into 3 different integrals

\[= - \frac{\hbar^2}{2m} \int_{-\infty}^{\infty} \frac{\partial \Psi}{\partial x} \frac{\partial^2 \Psi^*}{\partial x^2} dx + \frac{\hbar^2}{2m} \int_{-\infty}^{\infty} \Psi^* \frac{\partial^3 \Psi}{\partial x^3}dx - \int_{-\infty}^{\infty} \Psi^* \Psi \frac{\partial V}{\partial x}dx\]

Let's evaluate the first (leftmost) integral.

\[\int_{-\infty}^{\infty} \frac{\partial \Psi}{\partial x} \frac{\partial^2 \Psi^*}{\partial x^2} dx = \left. \frac{\partial \Psi}{\partial x} \frac{\partial \Psi^*}{\partial x}\right|_{-\infty}^{\infty} - \int \frac{\partial \Psi^*}{\partial x} \frac{\partial^2 \Psi}{\partial x} dx = \left[ \frac{\partial \Psi}{\partial x} \frac{\partial \Psi^*}{\partial x} - \frac{\partial \Psi}{\partial x} \frac{\partial \Psi^*}{\partial x}\right]_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \frac{\partial \Psi}{\partial x} \frac{\partial^2 \Psi^*}{\partial x^2} dx\]

Basically, after integrating by parts twice, we are left with two terms and the same integral we started with. Besides the fact that these two terms cancel each other out, evaluating either one of them from \(-\infty\) to \(\infty\) yields zero (remember that virtually any normalizable wave function must approach zero at \(\pm \infty\), and by implication, so must its derivative), and so the entire integral evaluates to zero.

Now, let's evaluate the second (middle integral).

\[\int_{-\infty}^{\infty} \Psi^* \frac{\partial^3 \Psi}{\partial x^3} dx = \left. \Psi^* \frac{\partial^2 \Psi}{\partial x^2}\right|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{\partial \Psi^*}{\partial x} \frac{\partial^2 \Psi}{\partial x^2}dx = \left[ \Psi^*\frac{\partial^2 \Psi}{\partial x^2} - \Psi^*\frac{\partial^2 \Psi}{\partial x^2} \right]_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \Psi^* \frac{\partial^3 \Psi}{\partial x^3} dx\]

By the exact same reasoning as with the previous integral, this one evaluates to zero as well. And so, we are left with

\[\frac{d \langle p \rangle}{dt} = - \int_{-\infty}^{\infty} \Psi^* \Psi \frac{\partial V}{\partial x}dx \]

This is exactly the negative of the expectation value of the derivative of the potential energy function with respect to position, or

\[\left\langle - \frac{\partial V}{\partial x}\right\rangle\]