# C*-algebraic representation of observables vs self-adjoint operators one

I am trying to reconcile the "physicist" definition of an observable: self-adjoint operator on a Hilbert space, and the operational one as given by Strocchi in "An introduction to the mathematical structure of quantum mechanics" : an observable is an element of a C*-algebra.

Clearly one cannot represent the CCR as operators on finite dimensional Hilbert spaces, or as bounded operators on a Hilbert space (even infinite dimensional). The only left option is to represent at least one of the two variables as an unbounded operator on a Hilbert space. This is incompatible with the operational definition of an observable as this unbounded element can't belong to a C*-algebra.

A possible solution to this is to consider bounded functions of $Q,P$ which is operationnally sensible, and people say that you get a C*-algebra, but I know no reference for this.

1) Do you know any? Do we still have self-adjoint operators? Does the commutation relation change?

Another solution is to take Weyl operators $e^{\imath t Q}$ and $e^{\imath s P}$ which are bounded and the Weyl form of CCR: these generate a C*-algebra. So one sees that by starting from an abstract Weyl C*-algebra as an observables algebra you can construct (through Stone theorem) self-adjoint densely defined operators on $\mathcal{H}$ under a small representation regularity assumption (so to verify Stone theorem strong continuity condition) [cf.p.63 Strocchi], with a 1-1 correspondance.

But now the self-adjoint operators that we can identify with an observable thanks to this 1-1 correspondance are defined only on a dense subset of $\mathcal{H}$, whereas no hypothesis was made on $Q,P$ domain in CCR. Hence Weyl operators are also defined on only part of the Hilbert space or you can't associate a self-adjoint generator to them on part of the Hilbert space, in either case you have a problem.

2) Was the starting Hilbert space too big? How to reduce its size so to have a generator for the Weyl operator defined on the whole Hilbert space?

So it seems to me that either there must be positive answers to my questions or one should abandon the identification between observables and self-adjoint operators: the operationally/mathematically correct association is between an observable and a Weyl operator, and you get under a representation regularity assumption a self-adjoint operator representation of the observable only on a dense part of the Hilbert space.

3) But then why bother with this kind of representation at all? Can't quantization be done directly in Weyl form? If not in what sense is Weyl-CCR harder to deduce during quantization than Heisenberg CCR?

I think the best way to answer this question is to direct you to an introductory textbook on the mathematics of quantum mechanics. One book I really like is Hilbert Space Operators in Quantum Physics by Blank, Exner, and Havlicek.

It sounds like the issue you are dealing with is that the canonical commutation relations can't be satisfied by bounded operators. You need to go to unbounded self-adjoint operators, which means they are only defined on a dense subspace of the Hilbert space. That's just the nature of the beast. In the C*-algebra approach we focus attention on C*-algebras of bounded operators to which the unbounded observables are affiliated.

Maybe part of your question is that if $P$ and $Q$ are only densely defined then $e^{itQ}$ and $e^{isP}$ can only be densely defined too? No, when you exponentiate an unbounded self-adjoint operator you get a bounded, and everywhere-defined, unitary operator. An easy way to do this is to use spectral theory to turn your self-adjoint operator into a multiplication operator, say multiplication by a possibly unbounded real-valued function $f$ on $L^2(X,\mu)$, and then define the exponential of this multiplication operator to be multiplication by the bounded function $e^{itf}$. Again, you need to read a good introductory textbook to learn about this.

• Thanks for your answer Nik. Since it is a negative answer to the two first questions, my third question becomes important: why do we bother with the associated self-adjoint operators instead of the Weyl ones? Why is quantization harder to achieve if we want to deal only with Weyl operators? – Issam Ibnouhsein Aug 14 '13 at 10:23
• Here is a partial answer, comment by Stefan Waldmann: mathoverflow.net/questions/55988/…, tough the references he gives are hard as a starting point to understand these issues. – Issam Ibnouhsein Aug 14 '13 at 10:41
• @Issam: because the associated self-adjoint operators are things like position and momentum which have basic physical meaning. – Nik Weaver Aug 14 '13 at 14:22
• After I've asked another question:mathoverflow.net/questions/139540/c-algebras-and-quantum-fields and found this paper: sot2012.pns.aegean.gr/files/summerschools/78/…, I am not so sure anymore self-adjoint operators are the most "natural" (mathematically speaking) path to quantization..any comment is most welcome! – Issam Ibnouhsein Aug 18 '13 at 14:00
• I mean there is certainly a correspondance between the two ways of quantizing as there is one between self-adjoint operators and Weyl operators, but maybe some of the issues one faces when doing "usual" quantization (non-existence or non-unicity of pre-quantization etc) are dual to domain problems etc of unbounded self-adjoint operators, and quantization as done in the paper might have less problems. That's just a guess and I need to study all this in more details! – Issam Ibnouhsein Aug 18 '13 at 14:23

I am basically repeating what Nick already said, but it was to long for a comment:

There is a one-to-one correspondence between self-adjoint operators and strongly continuous one-parameter groups of unitaries (Stone von Neumann Theorem). The one-parameter group for an self-adjoint operator is given by $\{U_A(t):=e^{itA}\}_{t\in\mathbb R}$ and vice versa given $\{U(t)\}_t$ one obtains a self-adjoint (in generally only densely defined) operator $A$ called the generator of $U(t)$ by differentiating $U(t)$ wrt to $t$ and take the value at $t=0$, which can be made precise and can be found in any functional analysis book.

In the usual Fock representation of CCR, $t\to W(t f)$, with $W$ the Weyl operator is strongly continuous, and the generator of this one-parameter group is your self-adjoint operator. In your example $W(f)=e^{i (f_1 P+f_2 Q)}$ with $f=(f_1,f_2)\in\mathbb{R}^2$ is the Weyl operator and the self-adjoint operators $P$ and $Q$ are obtained as above. So in this (unique for finite degrees of freedom) representation of CCR, you see the two quantizations are exactly the same, after you checked that the Weyl relations "infinitesimal" give back the Heisenberg CCR relations. In particular they act on the same Hilbert space.