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 $