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Timeline for $C^*$ algebras and states

Current License: CC BY-SA 3.0

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Sep 4, 2016 at 15:55 comment added Pedro Lauridsen Ribeiro @jjcale but in the case you're describing the corresponding Weyl CCR C*-algebras are different, so one is not really talking about different representations of the same Weyl CCR C*-algebra. Robert Furber's answer is the correct one in the sense that the dimension of the classical configuration space is half the dimension of the classical phase space being quantized by the Weyl CCR C*-algebra. Therefore, in this sense this number is representation-independent.
Sep 4, 2016 at 15:38 comment added jjcale The number of degrees of freedom is representation dependent. For instance all Laplace operators in dimensions $\ge 2$ are unitarily equivalent.
Sep 4, 2016 at 15:24 history edited Pedro Lauridsen Ribeiro CC BY-SA 3.0
Small terminology fix
Sep 4, 2016 at 11:04 comment added Robert Furber The number of degrees of freedom is the dimension of the classical configuration space (in the sense of classical mechanics). Harmonic oscillators are simple examples where the number of degrees of finite (it can even be 1) and the dimension of the quantum Hilbert space is infinite.
Sep 4, 2016 at 8:16 history edited მამუკა ჯიბლაძე CC BY-SA 3.0
primes were missing
Sep 4, 2016 at 5:36 comment added magya_bloom I see. I'll try yet another attempt at degrees of freedom (trying to translate to math terminology): dimensions of the base manifold?
Sep 4, 2016 at 2:14 comment added Pedro Lauridsen Ribeiro Fock space is indeed one possible Hilbert space for QFT (albeit only for free field theories). There are many others, such as those of GNS representations associated to finite-temperature (i.e. KMS) states.
Sep 4, 2016 at 2:01 comment added Pedro Lauridsen Ribeiro Finite degrees of freedom is not the same as a finite-dimensional Hilbert space or a finite-dimensional C*-algebra. Just think of a single, free, non-relativistic quantum particle in one dimension (a system with one degree of freedom) - the Hilbert space of the system is $L^2(\mathbb{R})$, which is definitely infinite-dimensional. In that case, the C*-algebra of observables is generated by the Weyl unitaries $e^{itP}$, $e^{isX}$, where $s,t\in\mathbb{R}$ and $X,P$ are resp. the position and momentum operators. This C*-algebra is also infinite-dimensional.
Sep 4, 2016 at 1:53 comment added magya_bloom Finite degrees of freedom = Finite dimensional C* algebras (or the representation)? QFT translates to infinite tensor products Fock space setting... is how I am interpreting your comment. Is that okay?
Sep 4, 2016 at 1:43 comment added Pedro Lauridsen Ribeiro That isn't really an issue for non-relativistic quantum mechanics due to the Stone-von Neumann uniqueness theorem: all representations of the canonical commutation relations in Weyl form are unitarily equivalent for a finite number of degrees of freedom. Inequivalent representations become important for systems with infinitely many degrees of freedom, such as thermodynamic limits of quantum statistical mechanical systems and quantum field theory. In those cases, the isometric isomorphisms among all separable Hilbert spaces need (and usually do) not intertwine different GNS representations.
Sep 4, 2016 at 1:26 comment added magya_bloom Quite an interesting set of criteria. Thanks! Then this brings up another question: If the representations are not equivalent when states are not unitarily equivalent, then how does one reconcile this to the usual Hilbert space formulation of Quantum Mechanics where there is just one separable Hilbert space as an axiom. Does one have to make a choice of representation depending on the state (prior to the dynamics)? Forgive my ignorance, just don't know enough about this topic. Any suggested readings (for someone with mathematical maturity but no previous background in von Nemann algebras)?
Sep 4, 2016 at 0:56 vote accept magya_bloom
Sep 4, 2016 at 0:55 vote accept magya_bloom
Sep 4, 2016 at 0:56
Sep 3, 2016 at 22:25 comment added Pedro Lauridsen Ribeiro @IgorKhavkine you are correct, of course, but I wanted to emphasize as well that the isomorphism may be chosen to act as the identity on $A/J$. Moreover, your definition of quasi-equivalence relies on Fell's equivalence theorem.
Sep 3, 2016 at 22:22 comment added Igor Khavkine Just a quick remark on how to interpret the definition of quasi-equivalence. For those who might find the definition by non-disjointness of sub-representations a little unintuitive, a quick equivalent definition is as follows: states $\rho$ and $\rho'$ are quasi-equivalent when they generate isomorphic von Neumann algebras, $\pi_\rho(A)'' \cong \pi_{\rho'}(A)''$.
Sep 3, 2016 at 21:53 comment added Pedro Lauridsen Ribeiro Right, good point. Actually that is an easier criterion than the one I gave - I rather had the type III factor case in mind (which is the typical situation for local algebras of observables in relativistic quantum field theory), but since the quasi-equivalence criterion also works in the pure case, I felt I should include it as well.
Sep 3, 2016 at 21:45 comment added Nik Weaver If $\rho$ and $\rho'$ are pure then they induce equivalent representations if and only if they are unitarily conjugate (an easy consequence of Kadison transitivity).
Sep 3, 2016 at 20:43 history answered Pedro Lauridsen Ribeiro CC BY-SA 3.0