Every normal operator on a separable Hilbert space has a square root that commutes with it Show that every normal operator on a separable Hilbert space has a square root that commutes with it. Uniqueness?


My attempt: Let $T$ be a normal operator. By polar decomposition $T=U|T|$  where $U$  is a partial isometry and $|T|$  is positive. Now using functional calculus $\phi: C(\sigma(|T|))\to C^∗ (|T|,1)$ , there is a sequence $\{f_n \}$  of continuous functions in $C(\sigma(|T|))$  such that $x^{ 1/2} =\lim f_n (x)$  so $|T|^ {1/2}$   is the unique square root of $ |T|$ .

But what about square root of $T$ ? Is it $|T|^{ 1/2}? 
I do not use of separablity of Hilbert space $H$ in my attempt. Also I think the  exercise is always true, not just for separable Hilbert space. And $|T|^{1/2}$ is always unique. Where is my mistake? 
Also if I have mistaken, Please give me an example of a normal operator on a non-separable Hilbert space that its square root is not unique. Thanks.
 A: A square root for an operator is not unique because square roots of complex numbers are not unique. If $A$ is a diagonal matrix on $\mathbb{C}^{n}$, then there are $2^{n}$ possible square roots for $A$ that you can spot right away. It's worse for a general Hilbert space.
If $N$ is bounded and normal on a Hilbert space, then the Spectral Theorem for $N$ gives a Borel spectral measure $E$ for which
$$
                        N = \int \lambda dE(\lambda).
$$
If $\sqrt{\lambda}$ is some branch of the square root, then you can define $\sqrt{N}$ by
$$
                      \sqrt{N} = \int \sqrt{\lambda} dE(\lambda).
$$
By the Borel functional calculus, $\sqrt{N}^{2}=\int\lambda dE(\lambda) = N$. By the way, you can't get this type of thing using the $C^{\star}$ algebra continuous functional calculus because $\sqrt{\lambda}$ cannot be assumed to be continuous on the spectrum of $N$. It's not that the result fails; it's that the technique of continuous functional calculus fails.
Check what assumptions you have for your version of the spectral theorem, especially concerning separability.
A: By Theorem 12.35(b) of Walter Rudin's Functional Analysis, if $T\in B(H)$ is normal, then it
has a polar decomposition $T=UP$, where $U$ and $P$ commute with each other and with $T$ and $U$ is unitary and $P≥0$. The spectrum of $P$ is a bounded subset of $[0,\infty)$, hence $\sqrt{z}$ can defined as a nonnegative square root of a nonnegative number. In this way, via the cont. functional calculus for $P$, we have unique positive $\sqrt{P}$. The spectrum of $U$ is a subset of the unit circle $C=\{ e^{it}; -\pi<t\leq \pi\}$. Define square root on $C$ by $\sqrt{e^{it}}=e^{it/2}$. This is well defined Borel function on $\sigma(U)$. Hence there
is $\sqrt{U}$ in the von Neumann algebra generated by $U, U^*$ and $I$. By Fuglede theorem $\sqrt{U}$ and $\sqrt{P}$ commute. Hence $(\sqrt{U}\sqrt{P})^2=UP=T$. 
Note that square roots are not unique. Already the $2\times 2$ identity matrix has a lot of
square roots (for each idempotent $P$ one has $(2P-I)^2=I$.
