# Requirement of Separability of Hilbert Space

1. Oct 13, 2014

### Kushagra Nigam

I have started reading formal definitions of Hilbert Spaces. I don't understand the requirement of separability postulate. I have proved that it leads to count ability of basis but again why is that required at first place.

2. Oct 13, 2014

### bhobba

Hilbert spaces aren't really the actual space used by physicists - its what is known as a Rigged Hilbert Space:
http://en.wikipedia.org/wiki/Rigged_Hilbert_space

Physically the states that actually exist are finite dimensional. But for mathematical convenience you consider the space of all sequences of finite length (they are the physically realizable states) and extend it by considering the linear functionals defined on the space. They are the limit of all sequences in its weak topology - but that is just by-the-by.

By this means you end up with a Gelfland Triple - and a separable Hilbert space always sits in the middle of the triple.

Thanks
Bill

3. Oct 13, 2014

### dextercioby

I remember an argument by DarMM here: if the Hilbert space is not separable, then Stone's theorem won't hold. I hope my memory won't betray me :D

Indeed, the Hilbert space 'in the middle of a RHS' is always separable by construction.

4. Oct 13, 2014

### DarMM

Yeah, if the Hilbert space is non-separable, then it is possible to have a set of symmetry transformations represented on the Hilbert space by unitary operators whose generator does not have a self-adjoint representation, a failure of Stone's theorem.

Basically you might have a symmetry without a corresponding conserved quantity.

5. Oct 13, 2014

### rubi

Are we talking about "Every strongly continuous one-parameter group of unitary operators is of the form $U_t = e^{\mathrm i t H}$, where $H$ is self-adjoint"? I'm pretty sure that also holds in non-separable Hilbert spaces. At least the proof in Reed&Simon doesn't require separability. Maybe you are thinking of the Stone-von-Neumann theorem instead?

Also, it's not true that the rigged Hilbert space construction requires a separable Hilbert space.

6. Oct 13, 2014

### DarMM

Well, it's a failure of the strong version of Stone's theorem (or Von Neumann's extension of the theorem), where the group is required to be weakly measurable only. If it is strongly continuous, then yes, even on a non-separable Hilbert space the generator exists. However for the Weyl-Algebra this is not the case on a non-separable Hilbert space (i.e. it is not strongly continuous), so for example the fields themselves (being the generators of the algebra) might not have a self-adjoint representation. I know this is a major issue in the Hilbert space version of Loop Quantum Gravity.

The Stone-VonNeumann theorem more relates to the dimension of the Weyl-Algebra than the separability of the Hilbert space.

7. Oct 13, 2014

### dextercioby

Can you expand on that comment? Thanks.

8. Oct 13, 2014

### rubi

Since unitary representation theory of groups is usually studied in the context of strongly continuous representations, the failure of the stronger version of Stone's theorem in the non-separable case is usually not problematic. The Weyl algebra (finitely many degrees of freedom) can't be realized on a non-separable Hilbert space because of the Stone-von-Neumann theorem (there is a unique (up to unitary equivalence) weakly continuous representation and its Hilbert space is separable). However, in the field theory case, there is no uniqueness theorem for the representations of the analogue of the Weyl algebra, so there can also be strongly continuous representations on non-separable Hilbert spaces, which then have generators due to Stone's theorem.

It's true that there are non-strongly continuous representations in LQG, but that's not problematic either. The diffeomorphism group isn't represented strongly continuous, but the diffeomorphism invariant states can be found nevertheless by a group averaging method. Also, the gauge potential isn't realized as an operator, but only in terms of holonomies. That's also not a problem, because it's not an observable anyway and all observables can be regularized in terms of the holonomies (and fluxes).

A rigged Hilbert space is the collection $(\Phi,\mathcal H,\Phi^\prime)$, where $\mathcal H$ is any Hilbert space, $\Phi$ is a so called countably Hilbert nuclear space and $\Phi^\prime$ is it's topological dual space. Maybe you think that the word countably refers to some separability requirement. That's not the case. It rather refers to the fact that $\Phi$ is equipped with a countable family $\left<\cdot,\cdot\right>_n$ of inner products (with some additional compatibility requirements). But neither $\Phi$ nor $\mathcal H$ need to be separable. Of course, in the standard example $(\mathcal S, L^2(\mathbb R), \mathcal S^\prime)$, the Hilbert space is separable, but it's just a special case. (Strictly speaking, that particular example is pretty overrated, since everything can be (and usually is) done using standard Fourier theory and no fancy math is needed).

The only case I know, where a rigged Hilbert space construction for a non-separable Hilbert space is (almost) involved, is again in LQG. It is used for the construction of the diffeomorphism invariant Hilbert space. ("Almost", because one uses the algebraic dual, rather than the topological dual.)

9. Oct 13, 2014

### dextercioby

Well, actually if the embedding operator of $$\Phi$$ into $$\mathcal{H}$$ is nuclear (as in the original Gelfand-Kostyuchenko construction), then H is necessarily separable (countable orthonormal base), see the argument of Gelfand-Vilenkin, Vol.4, Page 107-108. To my knowledge, every treatment of compact, H-S and nuclear/trace-class operators takes place in a separable Hilbert space.

So it's not really a RHS, because the (anti)dual is not taken with respect to the topology, but a much larger space, the alegebraic one. Nonethesess, can you make an exact book reference?

Thanks!

10. Oct 13, 2014

### rubi

If you refer to the fact that there is a countable sum on that page: That's only because trace-class operators are usually defined with respect to a separable Hilbert space. It can be generalized to non-separable Hilbert spaces, though (see Reed&Simon). In the general case, there would in principle be a sum over an uncountable index set, but only countably many terms can be non-zero, so you would still have a countable sum there. The arguments in the book actually don't depend on the cardinality of the index set because of this fact.

One could also use the topological dual. The reason for choosing the algebraic dual is that there is no physical choice for the topology and the algebraic dual guarantees that no solutions are discarded apriori. One could say, it is the topological dual with respect to the discrete topology. The LQG method is described in Thiemann's book, section 9.2 and section 30.1.

11. Oct 14, 2014

### DarMM

Is there examples of this. All the known representations of the field theory Weyl-Algebra (infinite-dimensional Weyl Algebra) have representations on separable Hilbert spaces. Such as the $\phi^{4}_{3}$ Hilbert space constructed by Glimm and any other actually constructed Hilbert space. Do you have an example of a representation of the Weyl-Algebra on a non-separable subspace which is known to be strongly continuous (and hence the generators exist).

In quantum field theory, the Wightman reconstruction theorem for n-point functions obeying the Wightman axioms, ensures the Hilbert space will be separable*. So only theories which break the reconstruction theorem (i.e. the theory cannot be recovered from its n-point functions) can take place on non-separable Hilbert spaces.

Of course, nobody would say that quantum systems with non-separable Hilbert spaces do not exist, as examples can be found. However they are very much edge cases in both quantum mechanics and quantum field theory and in many cases separability can be proven from other assumptions which are unlikely to fail. And there are difficulties with defining a Hamiltonian as you can no longer be assured a unitary group has representations for its generators (you would have to prove strong continuity, which is known to fail in certain examples, or restrict to a space with strong continuity, which in all examples studied so far turns out to be a separable sub-space.)

*See:
Maurin, K. Mathematical Structure of Wightman Formulation of Quantum Field Theory, Bull. Acad. Polon. Sci. 9 (1963) 115-119
(Proof that if the n-point functions are continuous as linear functionals, then the Hilbert space is separable)
Wyss, W. The Field algebra and its positive linear functionals. Comm. Math. Phys. 27 (1972) p. 223-234
(Removes assumption of continuity)

12. Oct 14, 2014

### dextercioby

Thanks for the input, as far I saw R-S also don't remove the countability of the H space basis.

13. Oct 14, 2014

### rubi

I don't know any example. I'm just pointing out that it isn't forbidden apriori. However, the fact that Wightman QFT's have a separable Hilbert space isn't surprising, since it is an axiom in the Hilbert space version of the axioms and the n-point function and CQFT versions are equivalent to the Hilbert space Wightman axioms. :) There don't seem to be any technical obstructions to dropping the separability axiom and indeed, the motivation for having that axiom (according to Streater&Wightman) isn't of technical nature. It rather seems to be convenience. As we know, finding actual interacting QFT's that satisfy the Wightman axioms is a really hard problem and no non-trivial examples are known in the 3+1 dimensional case, so it seems reasonable to weaken the axioms to make progress. (Of course, nobody wants to do that, since then there is no prize money to be won anymore :) )

I'm not saying that we should be using non-separable Hilbert spaces. I'm just saying that they can't be discarded for technical reasons. And you can also have unitary representations of groups that fail to have generators in the separable case (if you violate weak measurability), so you have to check the assumptions in any case. But as I said, the theory of strongly continuous unitary representations is well developed.

You are right, I missed the remark in the beginning of the chapter that restricts the analysis to separable spaces. However, "Locally convex spaces" by Jarchow deals with the general case and the german book "Funktionalanalysis" by Werner remarks that the proofs can be adapted to the non-separable case with minor modifications. (Also, the Wikipedia definition of nuclear operators doesn't require separability.)

14. Oct 15, 2014

### DarMM

You are confusing the Wightman axioms with the Gårding-Wightman axioms.

The Wightman axioms are axioms for a collection of distributions meant to be the n-point functions.

The Gårding-Wightman axioms are axioms for operator-valued distributions, with the operators they map to being ones on a Hilbert space carrying a representation of the Poincaré group.

If you read PCT, Spin and Statistics and all that, you will see the Gårding-Wightman axioms on pages 97-102 and the Wightman axioms on pages 117-118 listed as conditions for theorem 3-7.
In the pages between they show that a field theory obeying the Gårding-Wightman axioms has n-point functions obeying the Wightman axioms.
Theorem 3-7 is the converse, with a set of distributions obeying the Wightman axioms you can construct a field theory obeying the Gårding-Wightman axioms.

What the two papers I cited above prove is that separability of the Hilbert space in the field theory approach actually follows from a weaker set of assumptions on the n-point functions than the Wightman axioms.

Specifically, theorem 3-7, commonly known as "The Reconstruction Theorem", builds the Hilbert space by viewing the n-point functions as a linear functional on the extended test function space:

$\mathcal{G} = \mathbb{C} \oplus \left(\oplus_{n = 1}^{\infty} \mathcal{D}\left(\mathbb{R}^{nd}\right)\right)$

$\mathcal{D}\left(\mathbb{R}^{nd}\right)$ being the space of test-functions over $\mathbb{R}^{nd}$.

$d$ being the dimension of spacetime. For instance the 2-point function acts on the space $\mathcal{D}\left(\mathbb{R}^{2d}\right)$.

We can use the n-point functions to define an inner-product on this space.
If $f,g \in \mathcal{G}$ and $f_{i}$ denotes the component of $f$ from the space $\mathcal{D}\left(\mathbb{R}^{id}\right)$ and $W^{j}$ the j-point function, then the inner product is:
$(f,g) = \sum_{j,k} \int{dx_{1}\cdots dx_{n} dy_{1}\cdots dy_{n} f^{*}_{j}(x_{1} \cdots x_{n}) W^{j + k}(x_{1} \cdots x_{n}, y_{1} \cdots y_{n}) g_{k}(y_{1} \cdots y_{n})}$

This also defines the n-point functions as a linear functional, $\omega$, on the space $\mathcal{G}$.

If the difference between two functions $u = f - g$ has norm zero under this inner product, $(u,u) = 0$, then we identify them. The resultant space after the zero-vectors have been quotient out is the Hilbert space.
If the n-point functions obey the list of conditions given in theorem 3-7, then the Hilbert space has all the expected properties of carrying a rep of the Poincaré group e.t.c.

What the first paper above shows is that if you assume (W denoting Wightman):
(Wa) The space $\mathcal{G}$ is nuclear.
(Wb) $\omega$ is positive semi-definite, i.e. $(f,g) \geq 0$
(Wc) $\omega$ is $C^{0}$ as a functional on $\mathcal{G}$.

Then the Hilbert space is separable. The second paper goes one step further and shows that if $\mathcal{G}$ is constructed from the space of Schwartz functions, $\mathcal{S}$, rather than the test functions $\mathcal{D}$, i.e. that the n-point functions are tempered distributions. Then assumption (Wc) can be dropped as it follows from (Wb).

I would find it impossible for (Wb) to be dropped, as it guarantees that the Hilbert space has a positive norm. The only question is with (Wa) and (Wc). Again (Wa) can be directly proven. $\mathcal{G}$ is nuclear. So that leaves only (Wc).

Since (Wc) follows directly from (Wb) if the n-point functions are tempered, this can only be dropped by saying that the n-point functions are not tempered distributions, only distributions. This will automatically cause the fields not to possess Fourier transforms, so there will be no momentum space picture of the field theory.

So really, unless the field theory has no Fourier space representation, then separability is forced to be true. It follows from only (in field theory language, hence GW = Gårding Wightman):
(GWa) Fields are tempered distributions.
(GWb) Hilbert space has positive semi-definite norm.

And nothing else, not any of the other axioms. As I said before (GWb) is impossible to remove I think, we wouldn't have a quantum theory. (GWa), well I think a theory with no notion of momentum space would be an odd one. Again, they can exist, this is what theories with non-separable Hilbert spaces break, but they are a bit strange to say the least.

By the way theories with non-separable Hilbert spaces actually tend to violate (Wa) above, i.e. the function space they use is not even $\mathcal{D}$. So the fields are operator-valued hyperfunctions or operator-valued ultradistributions. The fact that these are not local objects is related to the difficulty of defining time evolution.

Last edited: Oct 15, 2014
15. Oct 15, 2014

### rubi

I'm not confusing them. I just referred to the Gårding-Wightman axioms as "Hilbert space version" and to the Wightman axioms as "n-point function version". The book explains how to go from one version to the other (either way). What I'm saying is that if this is supposed to work, then the Wightman axioms ("n-point version") must of course imply separability, since the Gårding-Wightman axioms ("Hilbert space version") require it.

You are probably right that in the Wightman framework, the separability follows mainly from the choice of Schwarz space test functions. However, i can also imagine another loophole: Maybe one can weaken some other axiom to make the uniqueness in the reconstruction theorem fail, so we don't need to use $\mathcal G / \sim$ in the first place?

On a different note: Do you know any similar results on separability in the Haag-Kastler framework?

16. Oct 15, 2014

### DarMM

True, although as discussed above, far less than the full list of axioms is needed in the n-point function version to get separability in the Hilbert space version.

Well the only axioms used are that (summing up the previous post):
(1) The Hilbert space can be reconstructed from the n-point functions, i.e. the theory is completely specified by the n-point functions.
(2) The resulting Hilbert space should have a positive semi-definite norm.
(3) The fields should have a Fourier transform.

This is really far weaker than the Wightman framework. As I said I think (2) cannot be altered, it's not a quantum theory if it is. So really you only have (1) and (3). I discussed (3) above. (1) is also an interesting case.

For (3) we get theories with fields which are ultradistributions or hyperfunctions (note, these are nonlocal objects). Attempts I've seen though, such as those of Jaffe, to work with these fields always find out that a we only have a sensible theory if additional locality constraints are required, and it turns out with these assumptions (at least in examples so far, I'm not aware of a general proof) that the Hilbert space is separable.

For violations of (1) we would need to look at thermal states, or any kind of quantum field state where the number of particles per unit volume is infinite. This links in with your second question:

Yes, in the Haag-Kastler framework we have the GNS reconstructions theorem, where one can construct the Hilbert space with a quantum field from the Algebra of observables. The algebra of observables can come in several types relevant to physics, which actually correspond to the classification of Von-Neumann algebras mathematically. They are:

Type $I_{n}$: A quantum theory with only discrete quantities, like spin.

Type $I_{\infty}$: A typical non-relativistic quantum theory with continuous quantities like position.

Type $II_{n}$: The thermodynamic limit of a gas of particles of type $I_{n}$.

Type $II_{\infty}$: The thermodynamic limit of a gas of particles of type $I_{\infty}$

Type $III$: The algebra of observables of a quantum field theory.

If these algebras are abelian, you get a classical Kolmogorov probability theory. For example the abelian Type $I_{2}$ is a coin flip, the nonabelian Type $I_{2}$ is a spin-1/2 quantum particle (without position and momentum, just spin). So all probability theories are included in this, with flipping a coin at the abelian Type $I_{2}$ end and relativistic quantum field theory at the non-abelian Type $III$ end. (Type $I_{1}$ is just the complex numbers.)

So for Types $II$ and $III$ we can have states whose GNS reconstruction gives a non-seperable Hilbert space, either due to a failure of (1) or (3) above. Although as I said before failure of (3) requires extra conditions which produce separability anyway. So let's only look at states which violate (1). It turns out that these are thermal states with non-finite particle number per unit volume.

So perhaps the way to think about separability is as (ultimately) the requirement that particle number be finite per unit volume.

17. Oct 15, 2014

### A. Neumaier

What are the Kolmogorov probability theories corresponding to types II and III? Is there a convenient reference?

18. Oct 15, 2014

### dextercioby

The book by Jarchow is highly unreadable for me. :D, but I saw the remark by Werner on page 293 (7. edition), so I assume indeed you can drop the countability of the basis. But, as one knows, the standard RHS are built with separable Hilbert spaces.