Square root of a positive operator

Remarks

Example: on $L^2[0, 1]$ let $A$ be multiplication by $x$, so $(Af)(x) = x f(x)$. Multiplication by $\sqrt{x}$ is a positive operator $B$, since $\ip{Bf}{f} = \int_0^1 \sqrt{x} \abs{f(x)}^2 \, dx \ge 0$, and $B^2 f = x f$; so $A^{1/2} = B$.

Non-example: $-I$ satisfies $(-I)^2 = I$, but $\ip{-x}{x} = -\norm{x}^2 < 0$ for $x \ne 0$, so $-I$ is a square root of $I$ but not the positive square root.

The square root is given by the continuous functional calculus, $A^{1/2} = f(A)$ with $f(t) = \sqrt{t}$ on the spectrum of $A$. It is also the norm limit of polynomials in $A$, so it commutes with every bounded operator that commutes with $A$.