Transpose of an operator If $\hat O$ is an operator and $c\in \mathbb C$
Then can I safely say that $[\exp(c\hat O)]^\dagger=\exp(c^*\hat O^\dagger)$?
My reasoning is by taking the transpose term by term in the power series definition of $\exp$ but I don't know if that is valid or not!
Thanks!
George
 A: So I suppose that $\hat O\colon X \to X$ is a continous operator on some complex Hilbert space $X$. Then one usually defines 
\[
  \exp(\hat O) := \sum_{k=0}^\infty \frac 1{k!} \hat O^k = \lim_{K \to \infty} 
  \sum_{k=0}^K \frac 1{k!} \hat O^k 
\]
where the limit is meant with respect to the operator norm 
\[
  \|T\| = \sup_{\|x\| = 1} \|Tx\| 
\]
The above series converges in this norm as we have 
\[
   \left\|\sum_{k=L+1}^K \frac 1{k!}\hat O^k\right\| \le \sum_{k=L+1}^K \frac 1{k!}\|\hat O\|^k \to 0, \quad K,L \to \infty\]
So $\exp(\hat O)$ is a well defined operator $X \to X$. 
Concernig it's transpose, if we try to compute it formally we have 
\begin{align*}
\exp(\hat O)^\dagger
  &= \left(\sum_{k=0}^\infty \frac 1{k!} \hat O^k \right)^\dagger
  \\&= \left(\lim_{K\to \infty} \sum_{k=0}^K \frac 1{k!} \hat O^k \right)^\dagger
  \\&\stackrel?= \lim_{K\to \infty} \left(\sum_{k=0}^K \frac 1{k!} \hat O^k \right)^\dagger
  \\&= \lim_{K\to \infty} \sum_{k=0}^K \frac 1{k!} {\hat O^\dagger}^k 
  \\&= \exp(\hat O^\dagger)
\end{align*}
The only step which doesn't follow from the linearity and antimultiplicativity of $(-)^\dagger$ is marked with $\stackrel?=$, we want to know if $(-)^\dagger$ exchanges with limits with respect to the operator topology, that is, if $(-)^\dagger\colon L(X) \to L(X)$ is continuous (where $L(X)$ denotes the set of linear continous functions $X \to X$). As taking the transpose is linear, it suffices to prove it is bounded. We have - denoting the scalar product of $X$ by $(-,-)$ - for $T \in L(X)$:
\begin{align*}
 \|T^\dagger\| &= \sup_{\|x\| = 1} \|T^\dagger x\|
  \\&= \sup_{\|x\| = 1} \sup_{\|y\| = 1} \left|(T^\dagger x, y)\right|
  \\&= \sup_{\|x\| = 1} \sup_{\|y\| = 1} \left|(x, Ty)\right|
  \\&= \sup_{\|y\| = 1} \sup_{\|x\| = 1} \left|(Ty, x)\right|
  \\&= \sup_{\|y\| = 1} \|Ty\|
  \\&= \|T\|,
\end{align*}
so $T \mapsto T^\dagger$ is bounded (even an isometry) and hence, $\stackrel?=$ is verified, we do indeed have $\exp(\hat O)^\dagger = \exp(\hat O^\dagger)$. 
The continuity of $(-)^\dagger$ implies of course, that for every convergent series of operators we can apply the transpose termwise.
