Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
4  
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". –  user5831 Feb 19 '11 at 16:39
1  
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, $U(s) V(t) = e^{ist} V(t) U(s)$. –  Pieter Naaijkens Feb 19 '11 at 16:50
2  
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. –  Paul Siegel Feb 19 '11 at 19:08
1  
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 –  Paul Siegel Feb 19 '11 at 19:12
    
If you know the expectation values of all projection operators then you can also calculate the expectation values of the unbounded operators. –  jjcale Jun 30 '13 at 8:42

3 Answers 3

up vote 15 down vote accepted

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.

share|improve this answer

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.

share|improve this answer
    
The reference should be: F. Strocchi, "An introduction to the mathematical structure of quantum mechanics". –  Matthew Daws Feb 21 '11 at 9:01

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 Frechet). 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

P.S.: How do you make the accent in Frechet on mathoverflow?

share|improve this answer
    
My (rash?) assumption was that the OP was referring to work of Andrew Gleason (any relation?), cf. en.wikipedia.org/wiki/Gleason%27s_theorem –  Yemon Choi Aug 13 '11 at 20:57
1  
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 :). –  Jonathan Gleason Aug 13 '11 at 21:29
4  
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. –  Tom LaGatta Aug 13 '11 at 22:18
    
Thanks Tom! Glad you enjoyed the article :) –  Jonathan Gleason Aug 14 '11 at 20:21
    
the command for the "accent aigu" you look for is \'{...} –  Adrien Hardy Nov 5 '11 at 13:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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