# Insights Mathematical Quantum Field Theory - Fields - Comments

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1. Dec 22, 2017

### samalkhaiat

This is not correct. Notions such as completeness (by a norm) and continuity (i.e., boundedness) of any element of an operator algebra need to be defined with respect to some vector space topology. Hermitian adjoint can only be defined on a vector space with scalar product. Moreover, every (abstract) non-commutative $C^{*}$-algebra can be realized as (i.e., isomorphic to) a norm-closed , *-closed subalgebra of $\mathcal{L}(\mathcal{H})$, the algebra of bounded operators on some Hilbert space $\mathcal{H}$. Precisely speaking, for every abstract $C^{*}$-algebra $\mathcal{A}$, there exists a Hilbert space $\mathcal{H}$ and injective *-homomorphism $\rho : \mathcal{A} \to \mathcal{L}(\mathcal{H})$. That is $\mathcal{A} \cong \rho (\mathcal{A}) \subset \mathcal{L}(\mathcal{H})$, as every *-homomorphism is continuous (i.e., norm-decreasing).

In general, one can say the following about quantization: Given a locally compact group $G$, its (projective) unitary representation on some Hilbert space $\mbox{(p)Urep}_{\mathcal{H}}(G)$ and the group (Banach) *-algebra $\mathcal{A}(G)$, then you have the following bijective correspondence $$\mbox{(p)URep}_{\mathcal{H}}(G) \leftrightarrow \mbox{Rep}_{\mathcal{H}}\left(\mathcal{A}(G)\right) \ , \ \ \ \ (1)$$ where $\mbox{Rep}_{\mathcal{H}}\left(\mathcal{A}(G)\right)$ is the representation of the (Banach) *-algebra $\mathcal{A}(G)$ on the same Hilbert space $\mathcal{H}$, i.e., *-homomorphism from $\mathcal{A}(G)$ into the algebra of bounded operators $\mathcal{L}(\mathcal{H})$ on $\mathcal{H}$. Similar bijective correspondence exists when $\mathcal{A}$ is a C*-algebra. And both ends of the correspondence lead to quantization. When $G = \mathbb{R}^{2n}$ is the Abelian group of translations on the phase-space $S = T^{*}\left(\mathbb{R}^{n}\right) \cong \mathbb{R}^{2n}$ (or its central extension $H^{(2n+1)}$, the Weyl-Heisenberg group) then (a) the left-hand-side of the correspondence leads (via the Stone-von Neumann theorem) to the so-called Schrodinger representation on $\mathcal{H} = L^{2}(\mathbb{R}^{n})$ [Side remark: of course Weyl did all the work, but mathematicians decided (unjustly) to associate Heisenberg’s name with the group $H^{2n+1}$], while (b) the right-hand-side of the correspondence leads to the Weyl quantization which one can interpret as deformation quantization (in effect, Weyl quantization induces a non-commutative product (star product) on the classical observable algebra, thus deforming the commutative associative algebra of functions $C^{\infty}(\mathbb{R}^{2n})$).

2. Dec 22, 2017

### Staff: Mentor

I have said it before and will say it again - I wish Samalkhaiat had the time to post more. His answers cut straight though.

The c*Algebra approach is, as it must be, equivalent to the normal Hilbert-Space approach - but can be formulated in a way where its association with classical mechanics is clearer:
http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Gleason.pdf

Thanks
Bill

3. Jan 17, 2018

### A. Neumaier

Actually it is more general, as the same algebra may have states corresponding to different Hilbert spaces (more precisely, unitarily inequivalent representations).

Thus it is able to account for superselection rules (restrictions of the superposition principle), which have no natural place in a pure Hilbert space approach.

Also it accounts for quantum systems having no pure states (such as those required in interacting relativistic quantum field theory).

Last edited: Jan 19, 2018