Showing That $\frac{d}{d_a} F_a(\hat{X}) \cdot \psi = F'(x) \psi$ at a=0

[John Greger]

Homework Statement

Consider the operator $F_a(\hat{X}) =e^{ia \hat{p} / \hbar} \cdot F(\hat{X}) e^{-ia \hat{p} / \hbar}$ where a is real.

Show that $\frac{d}{d_a} F_a(\hat{X}) \cdot \psi = F'(x) \psi$ evaluated at a=0.

And what is the interpretation of the operator e^{i \hat{p_a} / \hbar}?

The Attempt at a Solution

By just starting taking the derivative I find that,
$\frac{d}{d_a} F_a(\hat{X}) = (\frac{i \hat{p}}{ \hbar}) e^{ia \hat{p} / \hbar} \cdot F(\hat{X}) e^{-ia \hat{p} / \hbar} - e^{ia \hat{p} / \hbar} \cdot F(\hat{X}) (\frac{i \hat{p}}{\hbar}) e^{-ia \hat{p} / \hbar}$

Plugging a=0 gives,

$\frac{i}{h}[ \hat{p}, \hat{F}]$. But how do I take it from here to get the required form?And finally, what is the interpretation of $e^{i \hat{p_a} / \hbar}$?
Thanks in advance!

 

[Orodruin]

But how do I take it from here to get the required form?

Are you perhaps familiar with some commutation relation involving the argument of $\hat F$?

And finally, what is the interpretation of ei^pa/ℏeipa^/ℏe^{i \hat{p_a} / \hbar}?

What are your thoughts?

 

[John Greger]

Are you perhaps familiar with some commutation relation involving the argument of ^FF^\hat F?

If the operators are hermitian I know they commute, but other than that it doesn't ring a bell.

 

[Orodruin]

If the operators are hermitian I know they commute, but other than that it doesn't ring a bell.

This is not correct. Hermitian operators need not commute. Just look at, just off the top of my head for no apparent reason whatsoever, say $\hat x$ and $\hat p$ ...

 

[John Greger]

This is not correct. Hermitian operators need not commute. Just look at, just off the top of my head for no apparent reason whatsoever, say $\hat x$ and $\hat p$ ...

Okey. Anyway I'm pretty stuck at the problem. I only get as far as in my attempt of solution.

 

[Orodruin]

Insertion gives
$$
\frac{i}{\hbar}[\hat p, F(\hat x)]\psi,
$$
which means that you now need to show that $(i/\hbar)[\hat p, F(\hat x)] = F'(\hat x)$. The absolutely easiest way to do that is via the commutation relation between $\hat p$ and $\hat x$.