Operator relations

  • #1

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!
 

Answers and Replies

  • #2
Orodruin
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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?
 
  • #3
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 dosen't ring a bell.
 
  • #4
Orodruin
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If the operators are hermitian I know they commute, but other than that it dosen'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## ...
 
  • #5
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.
 
  • #6
Orodruin
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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##.
 
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