Im working on a problem from my textbook about self-adjoint operators. I am asked the following: Let T be a compact self-adjoint operator in the hilbert space H and calculate:
$A = e^{iT}$
with help of the spectral theorem.
I found out that the eigenvalues of T are related to the eigenvalues of A So if $\lambda$ is an eigenvalue of T, then $e^{i\lambda}$ is an eigenvalue of A. When I try to find help online i often find the following solution:
A = $\sum e^{i\lambda} P_\lambda$
But in order to now "represent" A with the spectral theorem, A needs to be compact. I found out that this is only the case for finite dimensional Hilbert spaces. Why am I still allowed to do this?
I also tried to calculate it by representing T with the spectral theorem, but this didnt work out either.
Can anyone guide me in the right direction?
Since $T$ is compact and self-adjoint, you can represent it as a sum $\sum \lambda P_\lambda$ where $P_\lambda$ is the orthogonal projection onto the eigenspace of $T$ with eigenvalue $\lambda$. It follows that $T^n=\sum\lambda^nP_\lambda$ for all $n$ (since $P_\lambda^2=\lambda$ and $P_\lambda P_{\lambda'}=0$ if $\lambda\neq\lambda'$), and so $e^{iT}=\sum e^{i\lambda}P_\lambda$. In other words, you get the representation for $A$ not by applying the spectral theorem to $A$ directly, but by applying it to $T$ and then using that to compute $A$.
(Note that there are some issues making sense of the sum $\sum e^{i\lambda}P_\lambda$ (and the sum $\sum \lambda^n P_\lambda$ when $n=0$), since the terms all have norm $1$ and so the sum does not converge in norm. You should instead interpret this sum in some weaker topology such as the strong operator topology, since it does converge if you apply it to any fixed vector.)