# Dimensions of Hilbert Spaces confusion

1. May 31, 2014

If I understand it, Hilbert spaces can be finite (e.g., for spin of a particle), countably infinite (e.g., for a particle moving in space), or uncountably infinite (i.e., non-separable, e.g., QED). I am wondering about variations on this latter. The easiest uncountable to imagine is the cardinality of the continuum. What examples are there of other uncountable cardinalities in quantum physics (either more or, if assuming the negation of the continuum hypothesis, less)? Thanks.

2. May 31, 2014

### George Jones

Staff Emeritus
Which Hilbert space in QED is non-separable?

3. May 31, 2014

For example
"In QED it is (at least) the continuum of directions in 3-space. One needs this nonseparable space to define Lorentz transformations of charges states, as charged states moving in different directions are in different superselection sectors. Thus the dimension of the Hilbert space of the universe should be at least the cardinality of the continuum." from http://physics.stackexchange.com/qu...rse-have-to-be-infinite-dimensional-to-make-s, or http://en.wiktionary.org/wiki/quantum_electrodynamics

More cryptically
"quantum electrodynamics (uncountable)" in http://en.wiktionary.org/wiki/quantum_electrodynamics

Then, we have field theory:
http://en.wikipedia.org/wiki/Partition_function_(quantum_field_theory)

4. May 31, 2014

### Bill_K

A quantum field is Fourier transformed into normal modes, a set of N simple harmonic oscillators, one for each value of k. Each mode represents a dimension in the Fock space. If N is finite, the Fock space is finite dimensional. If N is infinite but countable (k values are discrete), the Fock space is infinite dimensional and separable. If the allowed k values are continuous, N is infinite and uncountable, and the Fock space is non-separable.

5. May 31, 2014

### George Jones

Staff Emeritus
I guess my question is, when, specifically, does this actually happen, as opposed to, appear to happen?

As a much simpler example, consider the state space for single particle in one-dimensional non-relativistic quantum mechanics. I have seen some folks say, incorrectly, that this state space is non-separable because it has $\left\{ \left| x \right> \right\}$ is an uncountable basis.

For me, the mathematics is very subtle.

6. May 31, 2014

### Bill_K

Yikes, I did get it wrong,. Let me erase everything I said and start over.

Consider first a single particle, with just one degree of freedom and one canonical pair of operators p, q or raising/lowering operators a*, a. Starting with the vacuum state |0> application of the raising operator produces a countably infinite sequence of basis states (these are all linearly independent, and a general state in the space is a linear combination of all of them) and the Hilbert space is infinite dimensional and separable.

Next, for a system with a finite number of degrees of freedom, N. Each of the N raising operators generates a sequence of states. A basis vector in the Hilbert space can be represented by specifying N occupation numbers, e.g. |n1, n2,... nN> This is an infinite basis but still a countably infinite basis, and the Hilbert space is still separable.

However if N → ∞, a system with an infinite number of degrees of freedom, aka a field, the basis becomes uncountably infinite, |n1, n2,... > (aleph-0 to the aleph-0 power is aleph-1) and the Hilbert space is non-separable.

7. May 31, 2014

### dextercioby

As George says, there's a confusion here which stems from incorrectly using the terminology.

The Fock space of a free quantum field is built upon countable products and sums of rigged Hilbert spaces (more precisely, distributional spaces from the 3 spaces which make up a rigged Hilbert space of a single particle).

So I would like to see a Hilbert space used in physics which is uncountable.

8. May 31, 2014

### micromass

But $\aleph_0^{\aleph_0}$ is not $\aleph_1$. In fact, it is unknown which $\aleph_\alpha$ exactly it is. Under the continuum hypothesis, it is $\aleph_1$, but this is not universally accepted.

Also, you described something like a tensor product of Hilbert spaces with infinitely many factors. How exactly is this defined?
I'm not very familiar with Dirac notation, but I think that for finitely many factors, we have something like
$$<n_1n_2|m_1m_2> = <n_1|m_1><n_2|m_2>$$

But how would this be with infinitely many factors?

9. May 31, 2014

### Bill_K

Granted, I should have just said that in any case, $\aleph_0^{\aleph_0}$ is clearly greater than $\aleph_0$, and therefore uncountable.

Not so hard. Write the basis vectors using occupation numbers, |n1, n2,... >.
Then <m1, m2 ...|n1, n2 ...> = 1 if and only if m1 = n1 and so on.

10. May 31, 2014

### rubi

The Hilbert space of standard QFT is constructed this way: Fix a separable one-particle Hilbert space $\mathcal H$, define $\mathcal H^{\otimes 0} = \mathbb C$ and $\mathcal H^{\otimes n} := P \bigotimes_{k=1}^n \mathcal H$, where $P$ is a projector onto the symmetric or totally antisymmetric subspace. The Hilbert space of QFT is then $\mathcal F = \bigoplus_{n=0}^\infty \mathcal H^{\otimes n}$ (the definition of the inner product should be clear now). It's also clear that $\mathcal F$ is separable since $\mathcal H^{\otimes n}$ is separable and the direct sum of countably many separable spaces is separable as well. The difference between field theory and many-particle QM is that the observables become operator-valued distributions.

11. May 31, 2014

### Bill_K

No, it isn't the direct sum, it's the direct product.

12. May 31, 2014

### micromass

Reed and Simon's Methods of Mathematical physics, page 53 says that it's the direct sum.

Could you give me a reference to where the direct product of Hilbert spaces is actually defined?

13. May 31, 2014

### rubi

It's the direct sum of ((anti-)symetrized) direct products (as i wrote). The direct products are all finite. Only the direct sum is over infinitely many spaces. It's also written here: https://en.wikipedia.org/wiki/Fock_space

14. May 31, 2014

### Fredrik

Staff Emeritus
I have no idea, but I was pretty sure that A. Neumaier had mentioned something about it, and probably included a reference. I did a quick search and didn't find any references, but I found a post where he made a comment about it. https://www.physicsforums.com/showthread.php?p=3988874#post3988874

15. May 31, 2014

### George Jones

Staff Emeritus
In post #3, nomadreid also referenced Neumaier, but did not give a text reference. QED has been around along time, so, if one exists, I would like to see an example from a text of a standard use of non-separable Hilbert spaces in QED.

As has already been mentioned, a Fock space based on a separable Hilbert space is itself separable.

16. May 31, 2014

### Bill_K

Again, this is simply not the case. Here's what you're missing: the nonseparability only arises when there can be an infinite number of nonzero occupation numbers. Such states cannot be reached from the vacuum state in a finite number of steps. They come into play when you're dealing with coherent states, which contain an infinite number of photons.

17. May 31, 2014

### dextercioby

Bill, I don't understand. Non-separability has to do with whether the vector space is endowed with a countable basis or not. What is the vector space and what is the basis ?

18. May 31, 2014

### rubi

Neumaier is both right and wrong in the SE posting. He is right about the fact that the Hilbert space of QED can be written as a direct integral (it arises if you apply Mackey's theory of unitary representations to the Poincare group). It's not true however, that this implies that it is non-separable. For example, the space $L^2(\mathbb R) = \int_{\mathbb R}^\oplus \mathbb C \,\mathrm d \lambda(x)$ can be written as a direct integral over an uncountable domain as well (where $\lambda$ is the Lebesgue measure), yet we know that it is separable.

It's just wrong what you're saying. The Fock space is a direct sum of spaces $P\mathcal H^{\otimes n}$ and each of them has a countable basis $(\psi^n_k)_k$. The collection $(\psi^n_k)_{nk}$ is then a basis for the direct sum and it is countable since the index is in $\mathbb N^2$, which is a countable set.

A coherent state is a superposition of states with a finite number of particles. It is given by $\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} \left|n\right>$, where each $\left|n\right> \in P\mathcal H^{\otimes n}$. So a coherent state doesn't contain an infinite number of photons, but instead it's an infinite superposition of states that contain finitely many photons.

19. May 31, 2014

### Bill_K

Because the basis states with a finite number of nonzero occupation numbers span a subspace which IS separable. It has a countable basis. Like the rationals as a subspace of the reals.

If you look on the web for other mentions of nonseparable Hilbert spaces in QM, you'll find them in the context of coherent photon states, as I pointed out. For example this one:

20. May 31, 2014

### rubi

I agree that infinite tensor-product spaces are non-separable. However, QED isn't formulated on such a space. Instead it is formulated on a Fock space, which is separable.

21. May 31, 2014

### Bill_K

Many of the discussions this topic that I'm finding on the web, even though they say basically the same thing, aren't peer reviewed, (e.g. stackexchange threads) and so I can't claim them as an authority. However this one from functionspace.org echoes what I've been saying, and perhaps offers further clarification, so just consider it as words I wish I'd said:

.

22. May 31, 2014

### George Jones

Staff Emeritus
We seem to be going back and forth on this.

I had a vague recollection of years ago reading something relevant, so I went looking.

On page 33 of his book "Quantum Field Theory on Curved Spacetime and Black Hole Thermodynamics", Wald writes (for quantum fields in flat Minkowski spacetime)

23. May 31, 2014

### strangerep

If I understand Kibble's construction, he was trying to introduce an improved QED formulation over the standard Fock-space formulation of QED. In the latter, one makes sense of IR divergences only at the level of cross-sections, not the S-matrix. However, in Kibble's framework he banishes IR divergences from the S-matrix to all orders of perturbation theory, albeit at the cost of having asymptotic fields which are less convenient to work with.

Certainly, Kibble's space is unitarily inequivalent to the usual QED Fock Space, but does he not settle for a particular space among an uncountable set of alternatives?

Separately, (iirc) at least one existence proof for certain low-dimensional scalar field theories relies on progressively moving to a different unitarily-inequivalent space at each order of perturbation theory and then showing that this process converges in some sense.

Sure wish DarMM would stop by...

---------------------------------
For sufficiently masochistic readers, here are references to the other Kibble papers:

T.W.B. Kibble,
Coherent Soft-Photon States & Infrared Divergences. I. Classical Currents,
J. Math. Phys., vol 9, no. 2, (1968), p. 315.

T.W.B. Kibble,
Coherent Soft-Photon States & Infrared Divergences.
II. Mass-Shell Singularities of Green's Functions,
Phys. Rev., vol 173, no. 5, (1968), p. 1527.

T.W.B. Kibble,
Coherent Soft-Photon States & Infrared Divergences.
III. Asymptotic States and Reduction Formulas.,
Phys. Rev., vol 174, no. 5, (1968), p. 1882.

T.W.B. Kibble,
Coherent Soft-Photon States & Infrared Divergences.
IV. The Scattering Operator.,
Phys. Rev., vol 175, no. 5, (1968), p. 1624.

Last edited: May 31, 2014
24. Jun 1, 2014

### Bill_K

I think we've pretty much got it nailed down, although I see two points that deserve further comment.

What's being referred to here is that for basis states that contain an infinite number of particles, the eigenvalues of the observables may well be infinite. That's why the infinite tensor product is physically realizable only on coherent states, in which the infinite number of soft photons still has finite total energy.

This is both important and interesting! The usual representation of the canonical commutation relations contains the vacuum state, and all other states belonging to it can be obtained from the vacuum by application of the raising/lowering operators. This is conventionally what we mean by the Fock space. But states with an infinite number of particles cannot be obtained in this way, and therefore belong to another irreducible representation. The representation over the entire Hilbert space is reducible. In fact, any two states whose occupation numbers differ at an infinite number of places belong to separate representations.

25. Jun 3, 2014

### DarMM

Quantum field theories at zero-density have separable Hilbert spaces. It can be proven that the separability axiom is equivalent to the statement that in the theory all states have a finite particle number expectation value. At non-zero densities this fails and introduces a whole host of difficulties, especially with the mathematical treatment of thermal states.

Also, for non-separable Hilbert spaces, just because a group has a representation on the Hilbert space does not imply that its Lie-Algebra has a representation (failure of Stone's theorem). Physically this would mean that symmetry would not imply the existence of a conserved quantity (failure of Noether's theorem in finite density field theory)