How we define $M^s$ for $s\in (0,1)$ and $M$ a positive operator Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$).


*

*The square root of $M$ is defined to be the unique operator $N$ such that
$$N^2=M.$$

*In this case we write $N=M^{1/2}$

Let $M\in \mathcal{B}(F)^+$ and $s\in (0,1)$ be a real number. How we define $M^s$ ?

 A: Since $M$ is normal, you can use the Continuous Functional Calculus: the C$^*$-algebra $C^*(M)$, generated by $M$ is isomorphic to $C(\sigma(M))$. As $\sigma(M)\subset[0,\infty)$, the function $f(t)=t^s$ is continuous on $\sigma(M)$, and so by pulling back $f$ with the isomorphism you get $f(M)=M^s$. 
If you know the Spectral Theorem, since $M$ is selfadjoint you can define
$$
M^s=\int_{\sigma(M)}\lambda^s\,dE(\lambda).
$$
This last approach is nice in that it works on a Banach algebra, so it is irrelevant if $M$ is normal and/or positive. If $M$ is invertible and its spectrum does not surround $0$, then $g(z)=z^s$ (defined as $g(z)=e^{s\log z}$ is analytic. Then you can use the Riesz Functional Calculus to get an explicit expression: 
$$
M^s=\frac1{2\pi i}\int_\gamma z^s\,(zI-M)^{-1}\,dz,
$$
where $\gamma$ is any curve that contains $\sigma(M)$ and does not contain $0$. 
The important fact is that all three definitions agree wherever they make sense, so there is no way to say that one is better than the others. 
