Prove that $e^{iT}$ is unitary operator if $T$ is adjoint My exercise says: Prove that $e^{iT}$ is unitary operator if $T$ is self adjoint on a complex Hilbert space $H$.
My attempt was: By  definition we have that
$$e^{iT}=\sum_{k=0}^{\infty}\frac{(iT)^k}{k!}$$
then i want to prove that for any $x,y\in H$ then $\langle e^{iT}x,e^{iT}y\rangle= \langle x,y\rangle$. In effect,
\begin{equation}
\begin{split}
\langle e^{iT}x,e^{iT}y\rangle&= \langle \sum_{k=0}^{\infty}\frac{(iT)^k}{k!}x,\sum_{k=0}^{\infty}\frac{(iT)^j}{j!}y\rangle\\
\text{by continuity of the inner product}& = \sum_{k=0}^{\infty}\frac{(iT)^k}{k!}\overline{\sum_{j=0}^{\infty}\frac{(iT)^j}{j!}}\langle x,y\rangle \\
\text{continuity of conjugate operator}&=\sum_{k=0}^{\infty}\frac{(iT)^k}{k!}\sum_{j=0}^{\infty}\overline{\frac{(iT)^j}{j!}}\langle x,y\rangle.
\end{split}
\end{equation}
So, we reduce the proof to show that the product of the last series is the identity right?
Then, i know that
for a product of series $\sum a_n \sum b_n= \sum c_n$ where $c_n= \sum_{i=0}^n a_{n-k}b_k$.
In my case
$c_n= \sum_{k=0}^{n}\frac{(iT)^{n-k}}{(n-k)!}\overline{\frac{(iT)^{k}}{(k)!}}= \frac{1}{n!}\sum_{k=0}^n \binom{n}{k}i^n T^{n-k}\overline{T^k}$. I am stuck here because i do not know how to use that $\langle Tx,y\rangle= \langle x,Ty\rangle$ or $T=T^{*}$. Please, i would like to know how to continued or hints. Thank you
 A: Although, your idea is the right one, your second equality sign makes no sense. $T$ is acting on vectors and not on scalars. The whole matter works a little bit different: First observe
\begin{align*}
\langle \mathrm e^{\mathrm iT} x, y \rangle &= \bigg\langle \sum_{k = 0}^{\infty} \frac{(\mathrm i T)^k}{k!}x, y \bigg\rangle = \sum_{k = 0}^{\infty} \frac{\mathrm i^k}{k!} \langle T^k x, y \rangle \\
&= \sum_{k = 0}^{\infty} \frac{\overline{-\mathrm i}^k}{k!} \langle x , T^k y \rangle = \sum_{k = 0}^{\infty} \bigg\langle x , \frac{(-\mathrm i T)^k}{k!}y \bigg\rangle = \langle x, \mathrm e^{-\mathrm iT} y \rangle
\end{align*}
for all $x, y \in H$. Thus, $(\mathrm e^{\mathrm iT})^\ast = \mathrm e^{-\mathrm iT}$.
Using the Cauchy product and the binomial theorem, we see
\begin{align*}
\mathrm e^{\mathrm iT} \mathrm e^{-\mathrm iT} x &= \sum_{k = 0}^{\infty} \frac{(\mathrm i T)^k}{k!} \sum_{j = 0}^{\infty} \frac{(-\mathrm i T)^j}{j!}x = \sum_{k = 0}^{\infty} \sum_{j = 0}^{k} \frac{(\mathrm i T)^j}{j!} \frac{(-\mathrm i T)^{k - j}}{(k - j)!} x \\
&= \sum_{k = 0}^{\infty} \frac{1}{k!} \sum_{j = 0}^{k} \frac{k!}{(k - j)! j!} (\mathrm i T)^j (-\mathrm i T)^{k - j} x \\
&= \sum_{k = 0}^{\infty} \frac{1}{k!} (iT - iT)^k x = \frac{1}{0!} x = x 
\end{align*}
for all $x \in H$ since $(\mathrm iT - \mathrm iT)^0 = I$. Note here that we can apply the bonomial theorem only since $\mathrm i T$ and $-\mathrm i T$ commute as operators. Thus, $\mathrm e^{\mathrm iT} \mathrm e^{-\mathrm iT} = I$ and changing the roles of the two operators you get similarly that $\mathrm e^{-\mathrm iT}\mathrm e^{\mathrm iT} = I$. As we already observed $(\mathrm e^{\mathrm iT})^\ast = \mathrm e^{-\mathrm iT}$, it follows that $\mathrm e^{\mathrm iT}$ is unitary.
