I have operators satisfying ##[\hat{Z} , \hat{E}] = i \hbar##. The operators ##\hat{Z}## and ##\hat{E}## are taken to be Hermitian. You consider the unitary operator(adsbygoogle = window.adsbygoogle || []).push({});

##U_\lambda = \exp (i \lambda \hat{Z} / \hbar)##.

I have proved that

##U_\lambda \hat{E} U_\lambda^\dagger = \hat{E} - \lambda \quad Eq.1##

I know how to apply Eq.1 to any function of ##\hat{E}##, ##f (\hat{E})##, which may be Taylor expanded. Moreover consider the application of the transformation to some matrix elements

##\int_0^\infty \psi^* (E) f (\hat{E}) \phi (E) dE = \int_0^\infty \psi^* U^\dagger U f (\hat{E}) U^\dagger U \phi dE = \int \psi^* U^\dagger f (\hat{E} - \lambda) U \phi dE##

I'm told that you can deduce from this that (and similarly for ##\psi^*##):

##U \phi (E) = \phi (E - \lambda)##.

I not sure I understand this. What am I missing?

Thanks.

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# Deduction of the action of this unitary on the wavefunction

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