1
$\begingroup$

I am currently working through the problems in Griffiths Intro to Quantum mechanics, with the goal of hopefully understanding how magnets work.

At the moment I am a little stuck on one of the early problems in the first chapter. And I would like a Hint about how to proceed.

The problem is to solve the time derivative of the expectation value of momentum of a particle. ie: $$ \frac{d\left<p\right>}{dt}. $$

My working so far

Just prior to that question the text gives an expression for momentum, $$\int\limits_{-\infty}^{+\infty}{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx, $$ (Where $\Psi$ is a function satisfying Schrödinger's equation, $i$ is the square root of negative 1, and $\hbar$ is Planck's constant divided by $2\pi$.)

So my initial approach was to try and differentiate this with respect to $t$. $$ \frac{d}{dt}\int\limits_{-\infty}^{+\infty}{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx. $$ However I didn't know how to do that, so I asked around in the maths chatroom, and got a lot of help. Specifically this strategy from @RobJohn: $$ \begin{align} \int vv'\,\mathrm{d}x &=vv-\int vv'\,\mathrm{d}x\\ &=\frac12vv+C \end{align} $$

So I suppose it would follow that, $$ \begin{align} \int\limits_{-\infty}^{+\infty}{ \left( \Psi^* \left[ \frac{\partial}{\partial x} \right] \Psi \right) }\,\mathrm{d}x &= \left[\Psi^* \cdot \Psi\right]_{-\infty}^{+\infty} - \int\limits_{-\infty}^{+\infty} {\Psi\frac{\partial \Psi^*}{\partial x} }\,\mathrm{d}x\\ &= \frac{1}{2}|\Psi|^2 \Big|_{-\infty}^{+\infty}, \end{align} $$ And then, $$ \begin{align} &\frac{d}{dt}\int\limits_{-\infty}^{+\infty}{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx \\&=-i\hbar \frac{d}{dt}\left( \frac{1}{2}|\Psi|^2 \Big|_{-\infty}^{+\infty} \right). \end{align} $$ But this feels wrong. I'm not sure exactly why. (Maybe it's just because it is the limits of my maths)


For reference, regarding what I understand, my maths is pretty low-level, I took a course in undergrad calculus over 10 years ago and only just passed that at the time, and my physics knowledge is not much further than that of an enthusiastic lay person.

$\endgroup$
1
  • 1
    $\begingroup$ That was a comment to someone who had also misread the image. Two lines later, I said, "oops, I misread the fuzzy image". Note that there is only $v$ there, not $u$ and $v$ as in the image. $\endgroup$
    – robjohn
    Apr 17 at 5:17
0
$\begingroup$

The strategy you got from @RobJohn is not correct - specifically,

$\begin{align} \int\limits_{-\infty}^{+\infty}{ \left( \Psi^* \left[ \frac{\partial}{\partial x} \right] \Psi \right) }\,\mathrm{d}x \neq \frac{1}{2}|\Psi|^2 \Big|_{-\infty}^{+\infty}, \end{align}$

Since you just asked for hints, here are a couple useful things to know to solve this problem:

(1) Since the variables $t$ and $x$ are independent, you can bring the time derivative into the integral. You can also use the derivative product rule to get two terms in the integral.

(2) Do you know how to evaluate the time derivative of the wavefunction? Hint: Shrodinger equation

The problem involves a bit of algebra, but that should be a start.

$\endgroup$
2
  • $\begingroup$ So I'll spare you the details(though am happy to share through chat if you are interested) but I got this far: $$ i\hbar \left( \int\limits_{-\infty}^{+\infty}{ \frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} + \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x} }dx + \int\limits_{-\infty}^{+\infty}{ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx \right) $$ Any chance of a 3rd hint to get me over the line? $\endgroup$ Apr 17 at 7:02
  • $\begingroup$ In the third term, partial t and partial x can be swapped. Now again you can use hint (2) to get rid of the time derivative. $\endgroup$ Apr 17 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.