# Show expectation value doesn't change with time

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.

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"

• I had a classmate say he used a triple integral from the textbook (different from the one advised by the staff) and was able to avoid using integration by parts, so I think that's just been furthering my confusion. I think I'm being overwhelmed by all the different pieces, I'll try the more systematic approach. Apologies for not tagging correctly, it's my first post. – h_bear Sep 28 at 19:59
• @h_bear don't worry about it, you will get used to it very soon. I like to see his triple integral. But integration by parts is very easy approach to deal with this kind of problems, because after all, we know that wave function become zero at infinity, so you can easily eliminate the first part of integration by parts. I can't explain in more details due to the rules of this forum. – Paradoxy Sep 28 at 20:10
• That's true, it's the method we used in the problem before this. I'll share the triple integral in a follow up answer to this question. – h_bear Sep 28 at 20:13
• I've updated the post with the integral suggested by a classmate. It only further serves to confuse me. – h_bear Sep 28 at 20:31

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 $$\hbar$$.

• So this makes sense to me except how you got from the wave function I provided to your solution, unless you handwaved away some of the complications involved in the function I provided as they are not relevant to the variable we're integrating over for the sake of simplicity? – h_bear Sep 29 at 0:16
• $f(p)=e^{-ip^2 t/2m\hbar}\phi(p)$ and $\int dx e^{-ix(p-p’)/\hbar}\sim \delta(p-p’)$ with some factors like $\sqrt{2\pi \hbar}$. – ZeroTheHero Sep 29 at 0:22
• I see, alright in that case this is a very clear answer. Thank you for the help. – h_bear Sep 29 at 0:49