Exponent of an operator - Existence/Uniqueness? I have the following questions: When I can define an Expression $A^p$ with an Operator $A$ and a fractional Exponent $p$?
Is the root (or fractional or even real exponent) existing for arbitrary Operators $A$? 
Is it possible to obtain a unique $B$ with $B:=A^p$?
 A: The answer is definitely yes if the function $x \mapsto x^p$ is a bounded and measurable function on the spectrum $\sigma(A)$ of the operator $A$ and the operator $A$ is bounded and self-adjoint. (It might also be true if $A$ is unbounded, but I haven't yet studied unbounded operators, so I can't say anything about that case). This is because of the spectral theorem that gives you 
$$ A = \int_{\sigma(A)} \lambda\; dE$$
Where $E$ is a spectral measure. The operator you are looking for can then be written as 
$$ A^p = \int_{\sigma(A)} \lambda^p \;dE $$
A: Preliminary Remark
There's no need of bounded operators nor bounded functions!
Banach Spaces
Here the natural operators are the closed ones.
One then has the holomorphic functional calculus.
Example
(Does someone have an interesting example therefor?)
Hilbert Spaces
Here the natural operators are the normal ones.
One then has the measurable functional calculus.
Example
Consider a Hamiltonian:
$$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$
Then one can regard its density:
$$\rho:=\frac{1}{1+e^{\beta H}}:\mathcal{D}(\frac{1}{1+e^{\beta H}})\to\mathcal{H}$$
(One has to be careful about its domain, however.)
