Show expectation value doesn't change with time

Question

Given a wave packet:

$$\psi({\it x,t}) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} \exp\left[\frac{i}{\hbar}\left(px - \frac{p^{2}}{2m}t\right)\right]\phi(p){\it dp} $$

we are asked to find that $\langle p \rangle$ does not change with time. So it's basically asking us to show $\frac{d}{dt}\langle p\rangle=0$. I know that:

$$\frac{d}{dt}\langle p\rangle= -\left< \frac{\partial V}{\partial x}\right>$$

But I'm unsure how to solve for V in this case. Additionally, I've been advised to use the formula:

$$\langle p_{x}\rangle = \int \psi^{*}({\bf r},t)\left(-i\hbar \frac{\partial}{\partial x}\right)\psi ({\bf r},t) d{\bf r}$$

and that I will be using a delta function. I'm just kind of lost and trying to figure out which relevant equations to use for this problem and would appreciate some advice on how proceed.

Edit: The triple integral from the textbook my classmate suggested was:

$$\langle p_{x}\rangle = (2\pi\hbar)^{-3}\int d{\bf p} \int d{\bf r} \int d{\bf r'} e^{i{\bf p}\cdot{\bf r'}/\hbar}\Psi ^{*}({\bf r'},t)\left(i\hbar\frac{\partial}{\partial x}e^{-i{\bf p}\cdot{\bf r}/\hbar}\right)\Psi({\bf r},t) $$

I believe he then used the delta to achieve an exponent value of 1.

Answer 1

That's a very standard question. To solve it, you should use Schrödinger equation. Use the advised formula, and take its derivative with respect to the time. Then use Schrödinger equation to replace time derivation with position derivation. You might need to use integration by part. Also tag these kind of questions with "homework"

Answer 2

Rewrite this as $$ \int dp’ \int dp \int dx e^{-ip’x}f(p’) (\hbar p)e^{ipx}f(p) $$ and use the definition of $\delta(p-p’)$ as an integral of $x$ to remove the $\int dx$ then using the $\delta$ integrate over the $dp’$. You are left with an odd function of $p$ between symmetric boundaries.

Note that you should have $e^{i p x/\hbar}$ in the exponentials or, if you have $\hbar=1$ then $-i\partial_x$ without the $\hbar$.