Axiomatic Quantum Theory Is there any practical advantage/disadvantage between using the axioms for the Hilbert Space formulation as opposed to the C* Algebra formulation of Quantum Theory?
If not, then is it possible to establish the equivalence between the two, at least in principle?
 A: The algebra of (bounded) operators acting on the Hilbert space $\mathcal{B}(\mathcal{H})$ is a $C^{*}$-algebra. Therefore, its subalgebra generated by the observables of a particular quantum mechanical model is itself a $C^{*}$-algebra.
Given a $C^{*}$-algebra and an algebraic state $\omega$, one can reconstruct a $*$-representation of the algebra on the Hilbert space via GNS reconstruction. This gives the Hilbert space formulation for a $C^{*}$-algebraic model, as long as there is an algebraic state. For example, Wightman reconstruction in QFT uses the vacuum expectation value as the algebraic state on the $C^{*}$-algebra of polynomials in the quantum field, and its GNS $*$-representation is the Hilbert space of Quantum Field Theory.
Hence, Hilbert space formulations and algebraic formulations are very much related, though not completely equivalent. You need an algebraic state to pass from one to another (and different choices of states can lead to different results). Also, algebraic formulations deal with "nets" of $C^{*}$-algebras as e.g. in Haag-Kastler axiomatics (where there's a separate algebra for each open region of spacetime), not with a single $C^{*}$-algebra, so the algebraic axiomatics is more general from a mathematical point of view.
In quantum mechanics (where there's a finite number of degrees of freedom), the Stone-von-Neumann theorem guarantees that there is only one representation of the Weyl $C^{*}$-algebra, which makes the two formalisms equivalent. The same is not true for QFT.
