Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

146 submissions , 123 unreviewed
3,961 questions , 1,408 unanswered
4,889 answers , 20,762 comments
1,470 users with positive rep
507 active unimported users
More ...

Quantum mechanics formalism and C*-algebras

+ 12 like - 0 dislike
128 views

Many authors (e.g Landsman, Gleason) have stated that in quantum mechanics, the observables of a system can be taken to be the self-adjoint elements of an appropriate C*-algebra. However, many observables in quantum mechanics - such as position, momentum, energy - are in general unbounded operators. Is there any way to reconcile these two apparently contradictory statements?

I have looked at the notion of affiliation for a C*-algebra in the sense of Woronowicz. However, I can't see how you would extend states to elements affiliated with a C*-algebra and so it doesn't seem to solve the problem.

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Naz Miheisi
asked Feb 19, 2011 in Theoretical Physics by Naz Miheisi (60 points) [ no revision ]
retagged Sep 18
If I remember correctly then your question is explained very nicely in Strocchi's lecture notes "An Introduction to the Mathematical Structure of Quantum Mechanics - A Short Course for Mathematicians".

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user user5831
One thing you can do is to consider $U(s) = e^{i s P}$ and $V(t) = e^{i t Q}$ for example, if $P,Q$ are the position and momentum operators and $s,t$ real numbers. $U(s)$ and $V(t)$ are unitaries.You then have the Weyl form of the canonical commutation relations, $U(s) V(t) = e^{ist} V(t) U(s)$.

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Pieter Naaijkens
I don't have anything substantially new to add to Pieter Naaijkens' comment and Tim van Beek's answer. I just want to comment that the use of C*-algebraic techniques in this context should be considered natural and even expected. The real point is that an essentially self adjoint unbounded operator $T$ can be characterized (in a certain sense) by its functional calculus, i.e. by the \textit{bounded} operators $f(T)$ where $f$ is a bounded continuous function on the spectrum of $T$. This is also the reason why C*-algebraic techniques are useful for understanding differential operators.

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Paul Siegel
It's also worth mentioning that von Neumann in fact invented much of what we now consider to be basic C*-algebra and von Neumann algebra theory in order to handle operators coming from quantum mechanics that he knew not to be bounded

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Paul Siegel
If you know the expectation values of all projection operators then you can also calculate the expectation values of the unbounded operators.

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user jjcale

3 Answers

+ 10 like - 0 dislike

In addition to what has already been said I would like to add some more comments. I completely understand your suspicion that the passage from unbounded operators to bounded ones is at least tricky. For the canonical commutation relations of position and momentum operators this can be solved in a reasonable and also physically acceptable way by passing to the Weyl algebra (OK; there are zillions of Weyl algebras around in math, but I'm refering here to the $C^*$-algebra generated by the exponentials of $Q$'s and $P$'s subject to the heuristic commutation relations arising from $[Q, P] = \mathrm{i} \hbar$).

However, there are other situations in physics where this is much more complicated: when one tries to quantize a classical mechanical system which has a more complicated phase space than just $\mathbb{R}^{2n}$ then one can not hope to get some easy commutation relations which allow for a Weyl algebra like construction. To be more specific, for general symplectic of Poisson manifolds the beast quantization scheme one can get in this generality is probably formal deformation quantization. Here the classical observable algebra (a Poisson algebra) is deformed into a noncommutative algebra in a $\hbar$-dependent way such that the new product, the so-called star product, depends on $\hbar$ in such a way that for $\hbar = 0$ one recovers the classical mutliplication and in first order of $\hbar$ one gets the Poisson bracket in the commutator.

Now two difficulties arise: the most severe one is that in this generality one only can hope for formal power series in $\hbar$. Thus one has even changed the underlying ring of scalars from $\mathbb{C}$ to $\mathbb{C}[[\hbar]]$. So there is of course no notion of a $C^\ast$-algebra whatsoever on this ring. Nevertheless, there is a good notion of states in the sense of positive functionals already at this stage. Second, even if one succeeds to find a convergent subalgebra one does usually not end up with a $C^\ast$-algebra on the nose. Worse: in most of the explicit exampes one knows (and there are not really many of them...) the commutation relations one obtains are very complicated. In particular, it is not clear at all how one can affiliate a $C^\ast$-algebra to them. Moreover, it is not clear which of the previous states survive this condition of convergence and yield reasonable representations by the GNS construction.

It takes quite some effort to first represent the observables by typically very unbounded operators in a reasonable way and then show that they give rise to some self-adjoint operators still obying the relevant commutation relations. This is far from being obvious. To get a flavour for the difficulties it is quite illustrative to take a look at the book of Klimyk and Schmüdgen on Quantum Groups adn their Representations. Note that in these examples one still has a lot of structure around which helps to understand the analysis.

However, physics usually requires still much more general situations. Most important here are systems with gauge degrees of freedoms where one has to pass to a reduced phase space. Even if one starts with a geometrically nice phase space the reduced one can be horribly complicated. This problem is present in any contemporary QFT really relevant to physics :(

For other quantization schemes things are similar, even though I'm not quite the expert to say something more substantial :)

So one may ask the question why one should actually insist on $C^\ast$-algebras and this strong analytic background. There is indeed a physical reason and this is that quantum physics predicts not only expectation values of observables in given states (here the notion of a ${}^\ast$-algebra and a positive functional is sufficient) but also the possible outcomes of a measurement: they are given by particular numbers called the physical spectrum of the observable. To get a good description with predictive power the (as far as I know) only way to achieve this is to say that the physical spectrum is givebn by the mathematical spectrum of a self-adjoint operator in a Hilbert space. If one is here at this point, then the passage from a (unbounded) self-adjoint operator to a $C^\ast$-algebra is comparably easy: one has the spectral projections and takes e.g. the von Neumann algebra generated by them...

So the point I would like to make is that it is very desirable from a physical point of view to have the strong analytic framework for observables as either self-adjoint operators or Hermitian elements in a $C^\ast$-algebra. But the quantum theory of many nontrivial systems requires a long long and nontrivial way before one ends up in this nice heavenly situation.

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Stefan Waldmann
answered Feb 20, 2011 by Stefan Waldmann (440 points) [ no revision ]

I like this suggestion, that one should understand the passage from perturbative formal power series algebras of observables to C-star algebras as accounting for the measurement outcome on top of the probability amplitudes.

It would be nice if one could find a quasi-concrete setup where this actually matters in field-theory experiment; this would be a good further incentive for understanding non-perturbative quantization of field theory.

One problem seems to be that in the usual high energy experiments where the lack of a non-perturbative quantization via C-star algebras is most manifest, all that one sees in the experiments is scattering amplitudes.

@UrsSchreiber: In high energy experiments one sees the discrete spectrum (as masses of the particles) and the continuous spectrum (as scattering cross sections) of the Hamiltonian in the rest frame (global momentum zero). The relation between spectrum (or actually the representation in which the Hamiltonian is diagonal) and scattering is described, for ordinary quantum mechanics, e.g., in Thirring's mathematical Physics, Vol. 3. In all treatments of quantum field theory I know of, this connection is obscured due to the difficulty of defining bound states perturbatively,  but nevertheless it is fully present. 

@ArnoldNeumaier, thanks for the hint. I'll try to look into this...

+ 8 like - 0 dislike

The book mentioned by Bora,

  • F. Strocchi: "An Introduction to the Mathematical Structur of Quantum Mechanics"

is indeed a good reference for an axiomatic approach to quantum mechanics where the observables of a given system are assumed to form a $C^*$-algebra. The physical motivation for this approach is that every detector is representable by an observable (a self-adjoint operator) that is bounded, because every detector has an upper bound of values it can measure. A detector is therefore a bounded function of an unbounded, essentially self-adjoint observable as the impuls operator in, say, the usual (space or impuls) representation of the Heisenberg comutation relations.

This is the physical motivation of the basic assumption of an axiomatic approach to quantum physics via $C^*$- algebras: The assumption is that a quantum system, whose observables are essentially self-adjoint - maybe unbounded - operators, can equivalently be described by an operator algebra containing all bounded functions of the original observables. (The function $e^{i s A}$ for a real number s and an essentially self-adjoint operator A is of course an example, as mentioned in the comment by Pieter).

For an explanation of how and why this works in quantum mechanics, see the book by Strocci. (The short answer is that the $C^*$ - algebra of a massive particle described by position and momentum as observables is the Weyl algebra which is generated by the bounded operators occuring in the Weyl communtation relations mentioned by Pieter.)

In axiomatic quantum field theory, the difference of bounded versus unbounded observables manifests itself in two different sets of axioms:

  • the Wightman axioms use unbounded operators,

  • the Haag-Kastler axioms use bounded operators.

In this setting it is not completely clear that both sets of axioms are equivalent, although the relations Wightman => Haag-Kastler and vice versa have been proven with certain additional "technical" assumptions, see for example:

  • H.J. Borchers, Jakob Yngvason: “From quantum fields to local von Neumann algebras”, Rev.Math.Phys. Special issue, 1992, p.15-47.

("Quantum fields" refer to the Wightman axioms and "local von Neumann algebras" to the Haag-Kastler axioms.)

When you compare the situation in quantum field theory to the situation in quantum mechancis, you'll see that the situation is more complicated because in quantum field theory the von Neumann uniqueness theorem cannot be applied.

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Tim van Beek
answered Feb 19, 2011 by Tim van Beek (745 points) [ no revision ]
The reference should be: F. Strocchi, "An introduction to the mathematical structure of quantum mechanics".

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Matthew Daws
+ 3 like - 0 dislike

I wrote a paper about this at an REU I attended two years ago. At the time, I had the goal of making the transition from classical mechanics to quantum mechanics as natural as possible, motivating the axioms of quantum mechanics from those of classical mechanics, which are a lot more intuitive. Unfortunately, given the time constraints of the REU (just a couple of months) and being relatively new to the subject, I found myself mostly following other sources (in particular, Strocchi), which made use of the $C^*$-algebraic formalism. I have to admit, at the time, I did feel a bit uncomfortable throwing away all the unbounded operators, because, having studied "physicist's" quantum mechanics before, these are some things I naturally wanted to include.

Come two years later at the same REU, I tried to tackle the same problem, but this time I wanted to develop the axioms of quantum mechanics so to as allow unbounded operators in the theory. This led me to develop the notion of what I call an $F^*$-algebra ($F$ for Fréchet). I am actually still working on the paper at the moment, but, after reading your post, I decided to upload a preliminary copy to my academia.edu account. You should read it, check it out, and see what you think. I'd be very happy to hear any comments you have.

Be warned though, I am still in the process of editing and revising it, so dare I say, there may be some errors. Read it with a skeptical eye and let me know if you catch anything.

Cheers! Jonny Gleason

Here's the link: http://chicago.academia.edu/JonathanGleason/Papers/857632/From_Classical_to_Quantum_The_F_-algebraic_Approach

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Jonathan Gleason
answered Aug 13, 2011 by Jonathan Gleason (255 points) [ no revision ]
Most voted comments show all comments
My (rash?) assumption was that the OP was referring to work of Andrew Gleason (any relation?), cf. en.wikipedia.org/wiki/Gleason%27s_theorem

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Yemon Choi
He definitely could be. If that is the case, I would be incredibly embarassed for being so presumptuous =P. In any case, I did spend the summer two years ago and this summer working to solve the very question the OP asked, and my best (current) solution that I have found I am currently writing up and would be very interested in hearing what the OP (or anybdy else for that matter) has to say about it :). Also, if you know anything Andrew has written up on the $C^*$-algebraic foundations of QM I would love to see it. P.S.: As far as I know, no, there is no relation :).

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Jonathan Gleason
Hi Jonny, don't fret: we all make silly mistakes from time to time. A consequence of your mistake is that your very nice survey article is now disseminated to a wide audience. I enjoyed reading it. May all your mistakes be of this caliber.

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Tom LaGatta
Thanks Tom! Glad you enjoyed the article :)

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Jonathan Gleason

Could you please post your paper on the arXiv or another place where one does not have to register to download it?

Most recent comments show all comments
I bet, $F^*$-algebras will not save quantum mechanics, but stereotype ones will do: arxiv.org/abs/1303.2424 :)

This post imported from StackExchange MathOverflow at 2017-09-18 17:14 (UTC), posted by SE-user Sergei Akbarov

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverf$\varnothing$ow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...