# Unitary Operator as a complex valued function

A book on Quantum Mechanics states, "A unitary operator can be considered to be a complex valued function of a Hermitian operator."

Please give a hint on how to prove this assertion.

-
I think what the book is saying is this: if $T = T^*$ is a Hermitian operator, then $e^{iT}$ (defined by the usual exponential series) is a unitary operator. So if you have a unitary operator, the question is: can you go backwards, i.e. can you define a logarithm of a unitary operator? Think about the usual series expansion of $\mathrm{log}(1-x)$ and see if that will converge with a unitary operator $U$ in place of $x$. –  MTS Oct 19 '11 at 17:05

Sounds more like a homework/wikipedia problem and not suitable for here but anyways:

First one should maybe mention Stones Theorem which says there is a one-to-one correpsondence between strongly continuous unitary one-parameter group $\lbrace U(t)\rbrace_{t \in \mathbb R}$ and self-adjoint operators $A$ given by $U(t)=\exp(\mathrm i tA)$.

This follows from the more general Borel function calculus from which also follows that for $A$ self-adjoint and $f$ a complex Borel functions with $|f|=1$ follows that $f(A)$ is a unitary. For a hermitian operator this statement is wrong, but in physics literature there is often not made a difference between hermitian and self-adjoint operators and the technichal problems comming with these, I refer to the books of Reed and Simon.

-
What do you mean by "the difference between Hermitian and self-adjoint operators"? Are you referring to boundedness? Because for bounded operators I thought the terms were completely interchangeable. –  MTS Oct 19 '11 at 21:19
I refer to the domain. Ok the problem is that the word "Hermitian operator" is used differently in the literature. For bounded operators there are no problems at all. But normally self-adjoint is stronger then hermitian or symmetric. –  Marcel Bischoff Oct 19 '11 at 21:36

Consider $a=i\cdot\log(u)$ by Borel functional calculus. $a$ is self-adjoint since $a^{*}=-i\log(u^{*})=-i\log(u^{-1})=i\log(u)$ and $\exp(-ia)=u$.

-