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Showing That $\frac{d}{d_a} F_a(\hat{X}) \cdot \psi = F'(x) \psi$ at a=0
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[QUOTE="John Greger, post: 6074752, member: 648951"] [h2]Homework Statement [/h2] 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}? [h2]The Attempt at a Solution[/h2] 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! [/QUOTE]
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Showing That $\frac{d}{d_a} F_a(\hat{X}) \cdot \psi = F'(x) \psi$ at a=0
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