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.