If $AT=TA$ with $A\geq0$. Why $A^{1/2}T=TA^{1/2}$? Let $\mathcal{H}$ be a complex Hilbert space. Let $T\in \mathcal{B}(\mathcal{H})$ and let $A\in \mathcal{B}(\mathcal{H})^+$ (i.e. $A^*=A$ and $\langle Ax\;| \;x\rangle \geq0,\;\forall x\in \mathcal{H}$).

Assume that $AT=TA$. Why $A^{1/2}T=TA^{1/2}$?

Thank you

You can check that $p(A)T=Tp(A)$ for any polynomial $p$.
For any continuous function $f$ on $\sigma(A)$, there is a sequence $\{p_n\}$ of polynomials such that $p_n$  is convergent to $f$ uniformly. Hence $p_n(A)$ is convergent to $f(A)$ and thus $f(A)T=Tf(A)$. 
Specially, $f(x)=\sqrt{x}, x\in \sigma(A)\subset [0,\infty)$ is your case.
