12
$\begingroup$

The algebraic formulation of quantum mechanics (and related stuff, like quantum thermodynamics & dynamical systems etc.) via C*-algebras provides a viewpoint based mostly on abstract functional analysis. However, I haven't really seen a working application of this approach, i.e. an example of a (preferably physical) problem which is difficult to solve or even to formulate in the standard formalism, while considerably easier to tackle with all this algebraic stuff. Any ideas? Without such concrete examples the whole field seems to be interesting mathematically, perhaps, but lacking any physical substance (even if it's sometimes masqueraded as having connections with physics).

edit: by "standard formalism" I mean "observables, i.e. operators, acting on a Hilbert space of physical states", either in Schroedinger picture (time dependence of states) or Heisenberg (time dependence of operators). C*-algebraic approach starts from a C*-algebra and defines states as positive functionals on the elements of algebra, time evolution as a *-automorphism etc. (for an introduction to this formalism, see http://hal.archives-ouvertes.fr/docs/00/12/88/67/PDF/qds.pdf)

This might be of less interest to non-physicists than a typical MO post, but I hope it's still relevant.

$\endgroup$
4
  • $\begingroup$ Could you please clarify, for those of us with limited physics background, what the "standard formalism" is and how it differs from "this algebraic stuff"? Are you talking about AQFT or something at lower levels? $\endgroup$
    – Yemon Choi
    Feb 3, 2010 at 21:24
  • $\begingroup$ OK, thanks, that helps. Not sure I can offer any useful answer, but at least I understand the question better now. $\endgroup$
    – Yemon Choi
    Feb 3, 2010 at 22:56
  • $\begingroup$ I think it shouldn't be called algebraic because it is actually analytic. C* algebras are not part of algebra. $\endgroup$
    – MBN
    Feb 5, 2010 at 14:13
  • $\begingroup$ I think the "standard formalism" is probably to deal with finite-dimensional Hilbert spaces and certain examples of infinite-dimensional Hilbert spaces, but not actually prove anything rigorous about infinite-dimensional Hilbert spaces in general. $\endgroup$
    – Peter Shor
    Oct 17, 2010 at 2:42

3 Answers 3

6
$\begingroup$

If you want to consider quantum statistical physics properly this approach is necessary. The KMS condition gives a generalization of the quantum Gibbs postulate that allows for the treatment of phase transitions, coexistence of multiple phases, etc.

The functional-analytic approach is also important for other reasons (besides the rigged Hilbert space that is necessary even to understand a free particle): Fock space depends on it, as does axiomatic quantum field theory, and pages 12-13 of Bratteli and Robinson v.1 give some quick background to the KMS approach (both volumes are worth looking at). A book by Sewell called Quantum mechanics and its emergent macrophysics also gives quite a bit of relevant physical background.

BTW, the KMS condition is not as obscure as it might at first seem (the Wikipedia article is one of the few places I recall seeing that demystifies it). In the Heisenberg picture observables evolve under the time evolution map

$\tau_t : A \mapsto e^{iHt/ \hbar}Ae^{-iHt/ \hbar}.$

The appropriate generalization of the classical Gibbs rule is

$\langle A \rangle = Z^{-1}\mbox{Tr}(e^{-\beta H} A), \quad Z := \mbox{Tr}(e^{-\beta H}).$

To see this, consider the projection observable $\Pi_k := \lvert k \rangle \langle k \rvert$. We have that

$\langle \Pi_k \rangle = Z^{-1}\mbox{Tr}(e^{-\beta E_k} \lvert k \rangle \langle k \rvert) = Z^{-1}e^{-\beta E_k}$

in accordance with classical statistical physics. Now for generic observables $A$ and $C$, we have that

$\left \langle \tau_t(A)C \right \rangle = Z^{-1}\mbox{Tr}(e^{-\beta H} e^{iHt/\hbar}Ae^{-iHt/\hbar}C)$

$= Z^{-1}\mbox{Tr}(Ce^{iH(t+i\hbar\beta)/\hbar}Ae^{-iHt/\hbar})$

$= Z^{-1}\mbox{Tr}(Ce^{iH(t+i\hbar\beta)/\hbar}Ae^{-iH(t+i\hbar\beta)/\hbar}e^{-\beta H})$

$= \left \langle C\tau_{t+i\hbar\beta}(A) \right \rangle$

which gives the KMS condition:

$\left \langle \tau_t(A)C \right \rangle = \left \langle C\tau_{t+i\hbar\beta}(A) \right \rangle.$

$\endgroup$
3
  • $\begingroup$ I am quoting from Sewell here: "The recasting of statistical mechanics within the framework of algebraic quantum mechanics offers radical advantages in that it yields both an enriched picture of thermodynamic phase structure and some dynamical basis for the Gibbsian assumptions, even for finite systems." $\endgroup$ Feb 3, 2010 at 21:50
  • $\begingroup$ Browsing Sewell, I find applications to superfluidity, crystalline phases, superconductivity, and lasers. I think this is a pretty good set of physically relevant phenomena. $\endgroup$ Feb 3, 2010 at 21:57
  • $\begingroup$ While I am not as familiar with algebraic QM, I am familiar with categorical QM (some of my research) as well as the standard formalism (I teach it). The standard formalism leaves a lot to be desired even with the "improvements" that have been made to it in the context of information theory. But it boils down to a personal preference in the end since there's no single consistent conceptual layer to QM. There are lots of "interpretations" and approaches but until one predicts something physical that the others can't, it's really just a question of preference and maybe efficiency. $\endgroup$
    – Ian Durham
    Feb 4, 2010 at 2:25
5
$\begingroup$

The algebraic approach subsumes the standard formulation of QM (everything that can be done in the latter can be done in the former).

One important feature of the algebraic framework is that it allows you to handle inequivalent representations of the algebra of observables (which invariably come up in field theory, or any theory with an infinite number of degrees of freedom).

An example of a problem tackled only in the algebraic framework is the perturbative construction of interacting quantum field theory on arbitrary globally hyperbolic spacetimes. The construction is perturbative because every quantity of interest is taken to be a formal power series in the interaction strength. AFAIK, this construction has only been done using algebraic methods.

$\endgroup$
2
$\begingroup$

I think one answer ist to gain conceptual insight. Since Dirac it was usual to quantize via
Classical theory in a Hamiltonian framework => Quantum theory
Since there are a lot of problems with this approach (to QFT), it would be nice not to have to take the route over the classical theory. Hence the algebraic theory was born.

Well, I attended a workshop on algebraic QFT last year - it is not my research area, though. They said that they are still looking very hard for a working model in 4d (they have good models for dimension 2 and 3). That's also the reason why you did not encounter any concrete computable examples.
That's a critical issue - if they don't find a good model in the future, the field might be as good as dead (as a physics discipline).

$\endgroup$
1
  • $\begingroup$ The question wasn't about QFT but about quantum mechanics, which is on much solider ground mathematically. $\endgroup$
    – Peter Shor
    Oct 17, 2010 at 2:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.