# Eigenvectors of Hermitian and unitary operators

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.

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

• 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 $\int fdE_{\lambda}$. – Logic_Problem_42 May 4 '18 at 7:26
• First of all - how is $e^{iH}$ defined? – Logic_Problem_42 May 4 '18 at 7:28
• $e^{iH}$ is defined by its infinite series, $\sum_{n=0}^\infty \frac{(iH)^n}{n!}$ – user1742858 May 6 '18 at 16:08
• Then You can simply do this: $e^{iH}|h\rangle =\sum_{n=0}^{\infty} \frac{(iH)^n}{n!} |h\rangle =\sum_{n=0}^{\infty} \frac{(i)^n}{n!} H^n|h\rangle=\sum_{n=0}^{\infty} \frac{(i)^n}{n!} h^n|h\rangle=e^{ih}|h\rangle$. – Logic_Problem_42 May 6 '18 at 16:49
• $H^n|h\rangle=h^n|h\rangle$ follows from $H|h\rangle=h|h\rangle$ by induction. – Logic_Problem_42 May 6 '18 at 16:51