# Are Hilbert spaces still necessary?

1. Jul 23, 2015

### burakumin

I have never been happy with the fact a single quantum state could be encoded by an infinite number of vectors $|\phi\rangle$. Choosing a unit vector limits this overabundance but you have still an infinity of (physically equivalent) possibilities left. I later realized that the projector $|\phi\rangle\langle\phi|$ was unique and had the other advantage of being of the same kind as mixed states. If observable are operators and states can be represented (and in a better way) by operators, then fine, let us speak about operators only! Furthermore if we do not care about what the operators are applied to, let us just forget the whole idea of operators and concentrate on their mutual relations. That's where I discovered C* algebras.

In the end I'm wondering if there are good reasons to keep the idea of Hilbert space or if only its historical primacy keeps its widely used. The only limitation of c* algebra I'm aware of (but I'm really more aware of the mathematical structure than its use in qm) is that they can only represent bounded operators whereas an Hilbert space can be extended to a rigged Hilbert space to deal rigorously with that problem. Are there some known similar structures that extend c*-algebras and address the same issue ? Are they other limitations and situations where Hilbert spaces still remain better tools ?

2. Jul 23, 2015

### Dr. Courtney

There are many QM systems where I think one needs an function space of infinite dimension.

3. Jul 23, 2015

### wle

Well it's still how the mathematical structure of quantum mechanics is defined and I think there are situations where it still allows for the simplest constructions. For example, in a finite-dimensional Hilbert space, you can express any state vector as a linear combination $\lvert \psi \rangle = \sum_{k} c_{k} \lvert \varphi_{k} \rangle$ of a finite number of basis vectors. I don't know of a comparatively simple construction for density operators (which form a convex set with an infinite number of extremal points).

I think concentrating on the density operators also doesn't work so well when considering multipartite quantum systems and entangled states. For example, $\mathcal{S} = \bigl\{\lvert \psi \rangle_{\mathrm{A}} \otimes \lvert \phi \rangle_{\mathrm{B}} \mid \lvert \psi \rangle_{\mathrm{A}} \in \mathcal{H}_{\mathrm{A}} \,, \lvert \phi \rangle_{\mathrm{B}} \in \mathcal{H}_{\mathrm{B}}\bigr\}$ generates the bipartite Hilbert space $\mathcal{H}_{\mathrm{A}} \otimes \mathcal{H}_{\mathrm{B}}$ (i.e., $\mathcal{H}_{\mathrm{A}} \otimes \mathcal{H}_{\mathrm{B}}$ is the smallest Hilbert space that contains $\mathcal{S}$), but $\{\rho_{\mathrm{A}} \otimes \rho_{\mathrm{B}}\}$ doesn't generate the corresponding set of bipartite density operators.

4. Jul 23, 2015

### micromass

Staff Emeritus

The C*-algebra approach and the Hilbert space approach are actually equivalent. Any C*-algebra gives rise to a Hilbert space by the Gelfand-Naimark-Segal theorem. So C*-algebras are just Hilbert spaces in disguise, but rather the C*-algebra focuses on the observables as primary object, while the Hilbert space takes the states as primary object.

There are many ways of dealing with unbounded operators on C*-algebras. But the best approach is just by taking the exponential: if $T$ is a self-adjoint unbounded operator, then $e^{iT}$ is a bounded, unitary operator. So working with unbounded operators (like the position or momentum operator) is equivalent to working with unitary operators in the $C^*$-algebra. The relevant theorems here are the Stone-Von Neumann theorem and the Stone theorem on one-parameter groups.
Another approach studies "unbounded" elements over a C*-algebra, but this is not so popular.

5. Jul 23, 2015

### atyy

As micromass says, the Hilbert space and C* approach are the same.

6. Jul 23, 2015

### aleazk

It's true that the operators representing observables (if they are unbounded, their exponentials) can be shown to form C*-algebras. It's also true that quantum states are actually positive trace-class operators of trace 1. But to finish the argument that it's reasonable to use only C*-algebras, one also uses the fact that any state $T$ determines a linear functional $\omega:\mathcal{U}\longrightarrow\mathbb{C}$, where $\mathcal{U}$ is the C*-algebra (with unit), via $\omega(a)=tr(Ta)$, for all $a$ in the algebra (seen as operators). This functional satisfies $\omega(a^{*}a)\geq0$ for all $a$ in the algebra (positivity) and $\omega(I)=1$. Furthermore, we know that the physical interpretation of
$\omega(a)=tr(Ta)$ is that it gives the expectation value of observable $a$ in state $T$. Thus, one postulates that a quantum system is characterized by some abstract C*-algebra with unit, that the states are linear positive normalized functionals on this algebra and that the interpretation is that they give expectation values.

As already noted by @micromass, by the GNS construction, one can always get a realization of the abstract C*-algebra as operators in a Hilbert space. Of course, the idea is to work in the abstract C*-algebra as much as you can, in order to obtain very general and powerful results.

But the problem is that the formalism is too abstract. The Hilbert spaces have more clear physical interpretations. For example, consider the Weyl C*-algebra that describes the Klein-Gordon scalar field in flat spacetime. Very (very!) schematically and somewhat sloppy, using the symplectic form associated to the Klein-Gordon equation,

$$\Omega(\varphi,\varphi')=\int_{\mathbb{R}^{3}}\left(\varphi\dot{\varphi}'-\varphi'\dot{\varphi}\right)\mathrm{d}^{3}x$$

(where the phis are solutions to the Klein-Gordon equation) one can define an "inner product", $(\varphi,\varphi')=-i\Omega(\overline{\phi},\varphi')$
on the space of solutions of this equation. But this "inner product" is not positive. By restricting it to certain subspaces, one gets an actual inner product. Cauchy-complete the pair subspace+positive inner product and we get a Hilbert space, $\mathcal{H}_{+}$.

One can build many different Hilbert spaces in this way. Actually, the key element is the projection map $K$ that maps general solutions to the Hilbert space we just built, it "takes" the "positive" part of these solutions. Different constructions have associated different maps, since the chosen subspace changes. Take now the Bosonic Fock space built from $\mathcal{H}_{+}$, $\mathcal{F}_{S}\left(\mathcal{H}_{+}\right)$. Using the creation and annihilation operators there, define the following operator: $\hat{\Omega}(\psi)=A(\overline{K\psi})+A(K\psi)$, for all solutions psi. This operator is self-adjoint in certain domain. By taking the exponentials of these operators, we get an irreducible and faithful representation of the Weyl C*-algebra, i.e., we quantized the field. Usually, the basic Hilbert spaces are isomorphic, but different projection maps $K$ give different, inequivalent represenations of the algebra (the vector space of solutions is infinite dimensional so the Stone-von Neumann theorem does not hold). In certain cases, these constructions have interesting physical interpretations. If the spacetime is flat, one has two interesting notions of time translation: the one given by inertial observers and the one given by observers with uniform acceleration (both are timelike Killing fields, the second case is just the Lorentz boost). If you take the subspace of solutions "which oscillate with positive frequency" with respect to one of these notions of time, then the subspace satisfies the required conditions to make the inner product positive. In the first case, the Hilbert space is interpreted as the one-particle space according to an inertial observer. In the second case, the Hilbert space is interpreted as the one-particle space according to an accelerated observer. The respective Fock spaces represent, of course, their respective quantum field theories, which are not equivalent. So, the Hilbert spaces and the representations give us very concrete quantum systems (more precisely, different realizations of the same and unique quantum system) with very concrete and clear physical interpretations.

So, what's the use of the algebraic formulation in this case? Let's go back to the GNS theorem. In order to build a representation, you need to specify a particular algebraic state. Each states gives a different representation. For the case of the Weyl C*-algebra, there are special states called Gaussian states. The GNS constructions of these states are always over Fock spaces and the chosen states get mapped to the vacuum in the Fock space realization. As you may have guessed now, the Fock space realizations we considered earlier are all examples of this type of GNS construction. The interesting thing to note is that when we take the vacuum states of the inertial and accelerated realizations and view them as algebraic states, they don't coincide. In fact, the state in the accelarated representation that most approximates the vacuum of the inertial realization is a thermal state. This is one way of seeing the Unruh effect. If the field is in the algebraic state represented by the vacuum of the inertial realization, observers in this realization will see no particles, while the ones of the accelerated realization will see a thermal bath of particles. Thus, the algebraic approach allows us to compare the physics of inequivalent representations of the quantum field's C*-algebra.

You don't need rigged Hilbert spaces in order to deal with unbounded operators (if that's what you meant, maybe I misread you?). All of that is well covered by von Neumann's spectral theorem for unbounded self-adjoint operators in ordinary Hilbert spaces. Thanks to this theorem, all of quantum mechanics (including operators with continuous spectrum) can be done without problem in ordinary Hilbert spaces.

The C* structure is not really indispensable. In fact, one can prove an analogous version of the GNS theorem for *-algebras with unit. The representation is in terms of closable operators.

A limitation of the algebraic approach is that it only deals with a very reduced set of observables (in the case of the Weyl algebra, it only gives you a quantum version of the field). In order to obtain more observables, one usually goes to a representation in a Hilbert space, obtains in this way a quantum field operator and then canonically quantizes other observables, like, e.g., the energy (one substitutes the classical field by the quantum operator in the classical expression).

Finally, in the case of the Weyl algebra based on a finite dimensional space of solutions, the Stone-von Neumann theorem holds and then one simply works in the Hilbert space since all realizations of the algebra are equivalent and thus the algebraic approach becomes somewhat superfluous.

7. Jul 23, 2015

### strangerep

If you have a copy of Ballentine, look at section 7.1 where he derives the quantum half-integral spectrum for angular momentum. This stunning result -- using only the properties of the rotation operators represented on an abstract Hilbert space -- cannot be obtained if you outlaw Hilbert space. The proof does not go through.

Until someone shows that this result can be obtained without using a Hilbert space anywhere, this remains compelling evidence that Hilbert space is essential to quantum theory.

8. Jul 24, 2015

### bhobba

That's true. But the realisation of the usual far from rigorous methods physicist's use is via Rigged Hilbert Spaces. Von Neumann presented a perfectly rigorous version of QM - its just not what is usual in the physics literature.

Thanks
Bill

9. Jul 27, 2015

### burakumin

Interesting. However even if they are equivalent I have the impression C*-algebras better satisfies Occam's razor. I'm generally very (over-?) concerned with removing all non physical information from mathematical frameworks in physics. If I'm correct, an infinite number of equivalent representations can be arbitrarily constructed through GNS so C*-algebras seems like an improvement over HS as they exclude this arbitrariness.
Are there some good and standard references that introduce broad aspects of QM using C* algebras as the primary concept (I mean without constructing them over HS)? I would be interested in something that does not assume too much knowledge in QM from the reader but redefine basic notions and formulas with the C* algebras point of view (I know about C* algebras mainly from what I started to read in "introduction to noncommutative spaces and their geometry" which includes GNS construction but I doubt noncommutative geometry is the easiest path to go).
Sorry aleazk, but I think too many notions in this part of your message are beyond my knowledge and understanding.
I believe you, although it's not clear to me how rethinking an unbounded operator as a multiplicative operator on some $L^2(M)$ help you define rigorously its eigenvectors (which cannot belong to the HS). Maybe you don't need them... But I guess this would deserve an independent thread.

10. Jul 27, 2015

### micromass

Staff Emeritus
This might be getting philosophical, but I don't think Occam's razor is very applicable here. All QM wishes to do is to find a purely mathematical model which can make predictions. It does not actually make any claims about what the actual reality is like. Examples of Occam's razor would be that evolution is the best explanation for the diversity of life, and that the earth rotates around the sun in an elliptical orbit. These are factual claims of things that are real. But the Hilbert spaces or C*-algebras cannot be seen as something which actually are real. They are merely a mathematical representation which is used to calculate things. Feel free to disagree though, but that is my interpretation.

Well, C*-algebras are noncommutative geometry (the analytic part of noncommutative geometry that is), and a lot of motivation for noncommutative geometry comes from quantum theory, so there is a definite relation. Anyway, a very nice book would be Strocchi: https://www.amazon.com/Introduction-Mathematical-Structure-Quantum-Mechanics/dp/9812835229 I think this is exactly what you are looking for.

Last edited by a moderator: May 7, 2017
11. Jul 27, 2015

### aleazk

But it's precisely in the cases in which the GNSs are all equivalent when the algebraic approach is less powerful! For example, for a system in which there are only a finite number of degrees of freedom (the usual non-relativistic particle), all of those (irreducible) GNS constructions (and, in general, any other irreducible representation) are equivalent, a quite non-trivial fact which goes under the name of Stone-von Neumann theorem (this is a heuristic statement of the theorem, the real deal is with Weyl algebras). So, in this case, one simply uses the HS construction since it's unique up to equivalence and it's more manageable/concrete; also, in the HS, one can introduce by canonical quantization other observables that do not belong to the initial algebra (the Weyl algebra used in this case only contains position and linear momentum). You will only score some points in mathematical pedantry if you use the algebraic approach here.

In general, and in particular for quantum field algebras, the GNS representations are usually not equivalent. Therefore, the physics itself in these representations can be very different. In this case, yes, C*-algebras are an improvement exactly for the reasons you mention. In fact, if you do QFT in curved spacetime, then, in the general case, there's no way to choose a prefered GNS construction and that's why the algebraic formulation is used in this theory. The algebraic formulation in this case is indeed superior.

But, even with that, in the example I mentioned before about the Unruh effect, what I was trying to say is that these different GNS constructions often have very clear, but inequivalent, physical interpretations. So, for the most correct and economic mathematical formulation, yes, you use the algebras, but for making physical interpretations it's often necessary to go to the GNSs.

Think about it in the following way: suppose you have a C*-algebra and some state $\omega$. Pick an observable $a$ from the algebra. Thus, we know that the expectation value of $a$ in state $\omega$ is just $\omega(a)$. This is very nice but too abstract. Now, suppose we make some GNS constructions, with associated representations $\pi$ and $\pi'$, and such that there are, in the corresponding HSs, density matrices, $\rho$ and $\rho'$, such that $\omega(a)=tr(\rho\pi(a))=tr(\rho'\pi'(a))$ for all $a$ (the $\omega$ needs to be in what's called the "folium" of the GNSs for this to be possible, but let's leave all the technicalities aside). Since the two representations are not equivalent, the density matrices can represent wildly different physical states in their respective HSs. For example, one could be the vacuum according to some inertial frame (i.e., no particles) and the other a thermal bath of particles according to some accelerated frame! Without the algebraic approach, we wouldn't know how to compare these inequivalent representations; but with it, you can do it by checking to what kind of algebraic states they give rise. I hope that was graspable enough.

On the other hand, in, e.g., Loop Quantum Gravity, they build some abstract C*-algebra and make a GNS construction using a state that is invariant under spatial diffeomorphisms. But they find that this state is unique. So, it seems this GNS construction is the only relevant one and that's why the rest of the theory (e.g., quantization of area/volume observables) is done in this particular HS (I'm not an expert, so my knowledge on this may be outdated).

The book mentioned by @micromass is a very good introduction, I also strongly recommend it. After that, more advanced references (but which cover the most interesting physical applications) can be: Mathematical theory of quantum fields - H.Araki; Quantum field theory in curved spacetime - R.M.Wald.

Yes, you don't need nor use eigenvectors. Everything is done in terms of the projector valued measures associated to the self-adjoint operators via the spectral theorem.

Last edited: Jul 27, 2015
12. Aug 4, 2015

### burakumin

I didn't mean to interprete OR as an argument for or against realism. I view it more as an epistemological principle than an ontological one. To me it is just an invitation to choose when given several theories/explanations/models with equal predictive power, the one that contains the least "number of entities" (which I read "the least information"). And AFAIU C*-algebras assume less than HS. But I guess it's not a big deal if we have different opinions on that point.
Bought. Waiting for it to be delivered.
Pendantry is not what I aim to, understanding is. I suppose what you call "unique up to equivalence" is "unique up to isomorphism" (for an appropriate definition of isomorphism). I have never considered it to be a satisfying notion of uniqueness. I would expect uniqueness to be related to a certain universal property (unique up to a unique isomorphism). This remark is certainly pedantic to you. But this kind of distinctions help me clarifying my ideas and compensating my limited intelligence. Others approaches have never really worked for me.
That was graspable and insightful thanks.