Help Proving the Momentum Shift Operator

[Danielk010]

TL;DR

I need help proving given

I am stuck on proving:
$\langle p_x \rangle \rightarrow \langle p_x \rangle + p_0$ given $\langle x \mid \psi \rangle \;\rightarrow\; e^{i p_0 x / \hbar}\,\langle x \mid \psi \rangle$
I was able to prove given the change in wave function mentioned above: $\langle x \rangle \rightarrow\; \langle x \rangle$ by using
$$ \langle x \rangle = \int dx \, \langle \psi | x \rangle x \langle x | \psi \rangle $$ and plugging in the modified wave function and the complex conjugate of the modified wave function.

The textbook, A Modern Approach to Quantum Mechanics, 2nd edition by John S. Townsend, gives me these equations:

$$
\langle p_x \rangle = \langle \psi | \hat{p}_x | \psi \rangle = \int dx' \, \langle \psi | x' \rangle \frac{\hbar}{i} \frac{\partial}{\partial x'} \langle x' | \psi \rangle

= \int dx' \, \psi^*(x') \frac{\hbar}{i} \frac{\partial}{\partial x'} \psi(x') = \int dx \, \psi^*(x) \frac{\hbar}{i} \frac{\partial}{\partial x} \psi(x)
$$
I tried taking the deriative of $e^{i p_0 x / \hbar}$ to get the $p_0$ term, but that changes the $\langle p_x \rangle$ equation. Am I looking at the right equations or was I going in the right direction? Is there any hint you could give for this problem? Thank you any help on this problem.

 

[renormalize]

Is there any hint you could give for this problem? Thank you any help on this problem.

Under the transformation $\psi\left(x\right)\rightarrow e^{ip_{0}x/\hbar}\,\psi\left(x\right)$ we have:$$\left\langle p_{x}\right\rangle \equiv\int dx\,\psi^{*}\left(x\right)\frac{\hbar}{i}\frac{d}{dx}\psi\left(x\right)\rightarrow\int dx\,\left[e^{-ip_{0}x/\hbar}\,\psi^{*}\left(x\right)\right]\frac{\hbar}{i}\frac{d}{dx}\left[e^{ip_{0}x/\hbar}\,\psi\left(x\right)\right]=\,?$$Can you "turn the crank" from here to finish the derivation?

 

[Danielk010]

Under the transformation $\psi\left(x\right)\rightarrow e^{ip_{0}x/\hbar}\,\psi\left(x\right)$ we have:$$\left\langle p_{x}\right\rangle \equiv\int dx\,\psi^{*}\left(x\right)\frac{\hbar}{i}\frac{d}{dx}\psi\left(x\right)\rightarrow\int dx\,\left[e^{-ip_{0}x/\hbar}\,\psi^{*}\left(x\right)\right]\frac{\hbar}{i}\frac{d}{dx}\left[e^{ip_{0}x/\hbar}\,\psi\left(x\right)\right]=\,?$$Can you "turn the crank" from here to finish the derivation?

From the rightmost part of the equation,
$$\int dx \, \psi^*(x) \frac{\hbar}{i} \frac{\partial}{\partial x} \psi(x) (e^{{-i p_0 x / \hbar} + {i p_0 x / \hbar}})$$
which should make $\langle p \rangle$ = $\langle p \rangle$, not $\langle p \rangle$ = $\langle p \rangle + p_0$. Did I make a mistake in my calculation? Is this what you meant by "turn the crank"? Am I missing some mathematic trick? There is probably something wrong with my calculation given the problem statement. Thank you for the help.

 

[renormalize]

Did I make a mistake in my calculation?

Yes you did. You should have:$$\frac{\hbar}{i}\frac{d}{dx}\left[e^{ip_{0}x/\hbar}\,\psi\left(x\right)\right]=e^{ip_{0}x/\hbar}\left[p_{0}\psi\left(x\right)+\frac{\hbar}{i}\frac{d}{dx}\psi\left(x\right)\right]$$Do you understand how to apply the product rule for differentiation and how to take the derivative of an exponential function?

 

[Danielk010]

I do understand how to apply the product rule for differentiation and take the derivative of the exponential function. Thank you so much for the help. I got the answer.