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Question : If $ H $ is hermitian and $ U=e^{iH} $ is unitary, show that any eigenvector of $ H $ with eigenvalue $ h $ is also an eigenvector of $ U $ with eigenvalue $ e^{ih} $.

Let $ \left | h \right \rangle $ as the eigenvector of $ H $.
So $ H \left | h \right \rangle = h \left | h \right \rangle $
To show : $ U \left | h \right \rangle = e^{ih} \left | h \right \rangle $

I tried to prove the above equation using some manipulations but did not succeed.

$ U \left | h \right \rangle = e^{iH} \left | h \right \rangle $
$ U^\dagger \left | h \right \rangle = (e^{iH})^\dagger \left | h \right \rangle = e^{-iH^\dagger} \left | h \right \rangle $

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  • $\begingroup$ This is true for (almost) any function $f$, not just for $e^{ix}$. It is easy if You can use the representation of $f(H)$ as an integral $\int fdE_{\lambda}$. $\endgroup$ – Logic_Problem_42 May 4 '18 at 7:26
  • $\begingroup$ First of all - how is $e^{iH}$ defined? $\endgroup$ – Logic_Problem_42 May 4 '18 at 7:28
  • $\begingroup$ $ e^{iH} $ is defined by its infinite series, $ \sum_{n=0}^\infty \frac{(iH)^n}{n!} $ $\endgroup$ – user1742858 May 6 '18 at 16:08
  • $\begingroup$ Then You can simply do this: $e^{iH}|h\rangle =\sum_{n=0}^{\infty} \frac{(iH)^n}{n!} |h\rangle =\sum_{n=0}^{\infty} \frac{(i)^n}{n!} H^n|h\rangle=\sum_{n=0}^{\infty} \frac{(i)^n}{n!} h^n|h\rangle=e^{ih}|h\rangle$. $\endgroup$ – Logic_Problem_42 May 6 '18 at 16:49
  • $\begingroup$ $H^n|h\rangle=h^n|h\rangle $ follows from $H|h\rangle=h|h\rangle$ by induction. $\endgroup$ – Logic_Problem_42 May 6 '18 at 16:51

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