# Proving that $\sqrt{A}^2=A$

Let $A \in \mathfrak{U}$ where $\mathfrak{U}$ is a $C^*$ algebra and let $A=A^*$ with $\sigma(A)\subseteq \mathbb{R}^+$. Let $f := \sqrt{\cdot}\in \mathcal{C}(\sigma(A))$ where $\mathcal{C}(X)$ is the set of continuous functions from $X$ to $\mathbb{R}$

1.Prove that $f(A)$ is well-defined using the continuous functional calculus

1. Show that $\sqrt{A}^2 = A$

These are my questions, I'm studying elements of mathematical theory between quantum mechanics and classical mechanics. I don't quite understand how we can apply the continuous functional calculus usually I'm having trouble to show it.

These are my goals: I really think that this is not something difficult. But still I could not use the correct line of reason here. Using the continuous functional calculus we know that for an $f$ like the one in the question there exists an Gelfand homomorphism $\phi: \mathcal{C}(\sigma(A)) \to \mathfrak{U}$ such that, by the Weierstrass theorem, $\phi(f)$ is the limit of the operator norms of $\phi(p)$ with $p$ the polynomials that approximate to $f$ in the sup norm. So, we define $\phi$ in this way.

Then, we use for the second part of the problem that $\sqrt{A}^2 = \phi(f)\phi(f) = \phi(f\cdot f) = \phi(f^2) = A$

Can someone help me in showing some technicalities behind these types of questions? Thanks in advance.

Another try: 1. Let's use the continuous functional calculus to say that it is properly defined. So, what we have is that, for any function $f \in \mathcal{C}(\sigma(A))$ there is a $\phi: \mathcal{C}(\sigma(A))\to \mathfrak{U}$ such that $\phi(f)$ is defined as the limit of $\phi(p)$ for the $p$ polynomials in Weierstrass approximation theorem. This makes $\phi(f)$ well-defined. Now, for $f := \sqrt{\cdot}$ we define $\phi(f):=\sqrt{A}$. This concludes the first part.

2.With this an the properties of $\phi$ as a $\star$-homomorphism we have that, (see that $f^2 = \text{id}_{\mathcal{C}(\sigma(A))}$) $$\tag{1}\sqrt{A}^2 = \phi(f^2) = \phi(\text{id}_{\mathcal{C}(\sigma(A))}) = \text{id}_{\mathfrak{U}}(A) = A$$ because we have that $\mathfrak{U}$ is a $C^*$ unital algebra.

The $(a)$ line I think that there is a problem in here. Because how can I say the last equality? This seems something that I'm cheeting to get the correct answer.

## 2 Answers

You arguments are correct. When you consider a selfadjoint (normal is enough) $A$ as you do, the Gelfand transform maps the identity function to $A$. So you don't need to write $\text{id}_{\mathfrak{U}}(A)$, it is enough to write $$\tag{1}\sqrt{A}^2 = \phi(f^2) = \phi(\text{id}_{\mathcal{C}(\sigma(A))})= A.$$ Note, though, that there is no "cheating" in $\text{id}_{\mathfrak{U}}(A)$: you are just applying the polynomial $p(t)=t$ to $A$, so $p(A)=A$.

This may help. A commutative operator algebra of observables is equivalent to the set of continuous functions C(X) on a compact Hausdorff space. This equivalence arises through the Gelfand transform of a quantum operator A, an observable in a set of commuting observables. Since the Gelfand transform is an algebra isomorphism, the spectrum (the set of measurement eigenvalues) of the operator A, forms the set X See introduction part of

Moffat J, Oniga T, Wang CHT (2017) Unitary Representations of the Translational Group Acting as Local Diffeomorphisms of Space-Time. J Phys Math 8: 233. doi: 10.4172/2090-0902.1000233

for more on this. More generally we take as general context the modern algebraic theory of relativistic quantum fields in which we denote the set of all local observables O(D) as representing physical operations performable within the space-time constraints D. In particular, if O(D) is weakly closed (i.e., a von Neumann algebra) and A is an observable in O(D) then we assume that A is both bounded and self-adjoint, thus A generates an abelian subalgebra.