# Separability axiom really necessary?

I know other people asked the same question time before, but I read a few posts and I didn't find a satisfactory answer to the question, probably because it is a foundational problem of quantum mechanics.

I'm talking about the Hilbert space Separability Axiom of quantum mechanics. I'd like to understand why it was assumed this condition in the set of postulates of QFT. Is there a physical motivation of this, or was it only a way to simplify computation?

Mathematically speaking such an assumption is understandable. I read the argument about superselection sectors, where, even in presence of a non separable Hilbert space in QFT, every sector can be assumed to be separable and one can work inside this one, agreeing in this way with the said axiom. But the trouble remain unsolved, why this sector has to be separable?

If you know some old post or some book where I can find this answer and I didn't see please notify me.

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Please link to the previous posts. –  Qmechanic Dec 12 '13 at 12:05
Great, probing question. I would have naively said any reasonable theorist would say something like "Fie, begone from mine eyes evil spectre of $\aleph_1$, I shall not behold thee unless I be dragged into thy presence by the whole company of hell!" - like - this is what you'd assume unless forced otherwise wouldn't it be? Could it be something as simple as the convenience of a universal standard like GSM telephones: I mean everyone's working with the same piece of kit with a countable basis Hilbert space given they are all essentially isomorphic. –  WetSavannaAnimal aka Rod Vance Dec 12 '13 at 12:07
I do not understand well your final comment: Even dropping separability all Hilbert bases have the same cardinality and changes of Hilbert bases are isomorphisms. Perhaps I have completely missed your point. –  Valter Moretti Dec 12 '13 at 12:40
Your remark is reasonable. Probably as long as the separability of the state space is not a too low ceiling for physicist's head it remains an adequate postulate, useful for calculation. But was really this the reason that prompted Wightman & co. to assume this postulate? I hope there exist some more deep justification. –  moppio89 Dec 12 '13 at 12:54
–  moppio89 Dec 12 '13 at 12:59

Just a remark from the physical side.

If a physical system is described in a non-separable Hilbert space whatever Hamiltonian operator one chooses, thermal (Gibbs canonical or grand canonical) states cannot be defined as density matrices (mixed states) in the given Hilbert space.

So, if one wants to describe thermodynamics of that system he/she necessarily has only consider a micro canonical description, which means that it is impossible that the system reaches thermal equilibrium if is in contact with a thermal source or another system. Or one has to exploit the algebraic formalism from scratch that makes useless the initial Hilbert space approach.

Proof. If ${\cal H}$ is a generic Hilbert space and $\rho$ is a non-negative self-adjoint trace class operator in it, its non-vanishing eigenvalues, taking their multiplicity into account, must be countable at most, otherwise: $$tr(\rho) = \sum_{\lambda \in \sigma(\rho)} m_\lambda\lambda$$
diverges ($*$). Above $m_\lambda\geq 1$ is the (always finite for $\lambda >0$), multiplicity of $\lambda$.

A mixed state describing a Gibbs canonical or grand canonical ensemble is, by construction a non-negative self-adjoint trace class operator with strictly positive eigenvalues and there is a Hilbertian basis of eigenvectors of $\rho$. If $\cal H$ is non-separable this basis must be uncountable and this, in turn, implies that the associated (non-vanishing) eigenvalues, counted with their multiplicity, form an uncountable set.

footnotes

($*$) if $M= \sup \sigma(\rho) = ||\rho|| <+\infty$, then $$(0, M)=\cup_{n=1}^{+\infty} (M/(n+1), M/n]\:.$$ If each interval $(M/(n+1), M/n]$ contained a finite number of eigenvalues $\lambda >0$ (taking their multiplicity into account), their number would be countable. So, if they are uncountably many, at least some interval, say $(M/(n_0+1), M/n_0]$, has to include and infinite number of them and thus $\sum_{\lambda \in \sigma(\rho)} m_\lambda \lambda$ diverges because $$\sum_{\lambda \in \sigma(\rho)} m_\lambda\lambda \geq \infty M/(n_0+1)\:.$$

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I'd like to add a really elementary level illustration of some of David Bar Moshe's answer when he says:

...They give two major examples of non separable Hilbert spaces: 1) The infinite tensor product of harmonic oscillators, ...

The example below is clearly well below the level the OP is looking for, but hopefully it shows to a wider audience what a most excellent and surprisingly practical question the OP's really is. It shows that in this particular case, that the separability axiom has a clear and practical physical meaning here - "we must only put a finite number of particles into a finite quantisation volume".

Just take a Fermionic Fock space for a finite quantisation volume, so that there are countably infinitely many plane wave as the "modes". Then an arbitrary basis member is of the form:

$$\left|\left.0,1,1,0,1,1,1,0,1,0,\cdots\right>\right.$$

that is, a countably infinitely long string of 0s and 1s showing which modes are filled. This set is mapped bijectively to the interval $[0,1]$ - it's simply a binary expansion of a number in $[0,1]$. So boom! The old $\aleph_1$-in-the-Fock trick: we've got ourselves a Hilbert space with $\aleph_1$ basis states just like that! (call it $\beth_1 = 2^{\aleph_0}$ if you want to forswear the continuum hypothesis!)

The problem "only gets worse" for Bosonic Fock spaces of course (although we're still "only" dealing with $\aleph_1$).

One can tame this space by saying we consider the space of states with a finite number of particles in them. Then our basis Fermionic Fock states are equivalent to finite binary expansions of numbers in $[0,1]$, i.e. to a proper subset (those with nonrecurring finite binary expansions) of the rationals $\mathbb{Q}\cap[0,1]$ in that interval. So now we have our friendly separable Hilbert space back again.

How necessary is the the separability axiom? Clearly it makes life easier and is mathematically reasonable. This example shows that there are readily conceivable quantum spaces (although one might argue how "physical" an infinite number of particles in a finite quantisation volume is) that have distinctly different basis cardinality. So whether or not you choose to take it on does make a difference to your physics. The example also shows how in this case you can make a physically reasonable argument that our real quantum space is spanned by a meagre subset of the uncountable basis states. It's worth saying again to emphasise that that separability axiom means real, practical physics here: - "only a finite number of particles for a finite quantisation volume".

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From the viewpoint of the Wightman axioms, the separability assumption on the Hilbert space can be actually derived from a few of the other axioms if you adopt a formulation using $n$-point functions.

The reasoning goes as follows. A (say, scalar) quantum field theory on $\mathbb{R}^d$ can be thought of as being specified by a sequence of $n$-point distributions $\omega_n\in\mathscr{D}'(\mathbb{R}^{nd})$, $n=1,2,3,\ldots$ satisfying a condition of positivity which will be specified shortly (the other Wightman axioms are not relevant for the purpose of establishing separability of the "vacuum" Hilbert space).

The point is that this sequence of distributions can be thought of as a linear functional $\omega$ on the algebra $\mathfrak{F}=\mathbb{C}\oplus\left(\bigoplus^{\infty}_{n=1}\mathscr{D}(\mathbb{R}^{nd})\right)$ - an element $f=(f_0,f_1,f_2,\ldots)\in\mathfrak{F}$ is a sequence such that $f_0\in\mathbb{C}$, $f_n\in\mathscr{D}(\mathbb{R}^{nd})$ are zero for all but finitely many $n\in\{0,1,2,\ldots\}$. The algebraic operations on $\mathfrak{F}$ are defined in the following way: if $f,g\in\mathfrak{F}$ and $\alpha\in\mathbb{C}$, then we write:

• Sum: $f+g=(f_0+g_0,f_1+g_1,f_2+g_2,\ldots)$;
• Scalar multiplication: $\alpha f=(\alpha f_0,\alpha f_1,\alpha f_2,\ldots)$;
• Product: $fg=((fg)_0,(fg)_1,(fg)_2,\ldots)$, where $(fg)_0=f_0g_0$ and $(fg)_n(x_1,\ldots,x_n)=\sum_{i+j=n}f_i(x_1,\ldots,x_i)g_j(x_{i+1},\ldots,x_n)$;
• Involution: $f^*=(f_0^*,f_1^*,f_2^*,\ldots)$, where $f_0^*=\overline{f_0}$ and $f^*_n(x_1,\ldots,x_n)=\overline{f_n(x_n,\ldots,x_1)}$.

The product is just the tensor product of test functions, whereas the involution (a sort of noncommutative analogue of complex conjugation) makes $\mathfrak{F}$ into a unital $*$-algebra. This *-algebra inherits a (locally convex) topology from the test function spaces $\mathscr{D}(\mathbb{R}^{nd})$ which makes it a so-called nuclear *-algebra. The sequence $\omega=(\omega_0,\omega_1,\omega_2,\ldots)$, where $\omega_0=1$, becomes a (continuous) linear functional on $\mathfrak{F}$ if we set $\omega(f)=f_0+\sum_{n=1}^\infty\omega_n(f_n)$ (the sum is always finite by our definition of $\mathfrak{F}$ above). The *-algebra $\mathfrak{F}$ is sometimes called a Borchers-Uhlmann algebra.

Now we can state our condition of positivity on $\omega$: for all $f\in\mathfrak{F}$, we must have $\omega(f^*f)\geq 0$. In other words, $\omega$ is a(n algebraic) state on the *-algebra $\mathfrak{F}$.

At this point, we can invoke a result due to K. Maurin ("Mathematical Structure of Wightman Formulation of Quantum Field Theory", Bull. Acad. Polon. Sci. 9 (1963) 115-119), which essentially tells us that the nuclearity of $\mathfrak{F}$ and the continuity of $\omega$ entail that the Hilbert space obtained by the Wightman(-GNS) reconstruction theorem is separable. Notice that the actual construction of the Hilbert space and the "vacuum" vector need only positivity to work. The other axioms (covariance, causality) are needed to obtain the unitary representation of the Poincaré group, spectrum condition, and so on. Therefore, Maurin's argument is stronger (and simpler) than the one found in Streater-Wightman's book.

The argument can be extended to fields of any spin, provided we define the tensor product of test sections of vector bundles in the appropriate way. I don't know of an analogue of this argument for fields where the requirement of positivity is not satisfied (e.g. electromagnetic fields in a covariant gauge). However, for free fields the Wightman reconstruction theorem is just the construction of the vacuum Fock space from the one-particle space, which always yields a separable Hilbert space. In general, one may think of such Hilbert spaces as "sectors", as argued in David Bar Moshe's answer.

Finally, one must recall that there are other Hilbert spaces which are interesting for quantum field theory and are non-separable. Of course, these cannot be obtained by the Wightman reconstruction theorem alone. One such example is provided by all coherent states of a free field. One can use Wightman reconstruction with a single coherent state, but the resulting Hilbert space cannot contain all coherent states. Such spaces also appear indirectly in the Bloch-Nordsieck approximation in QED, used to deal with infrared problems. They are a particular case of the continuous tensor product of Hilbert spaces mentioned in David Bar Moshe's answer, which also discusses other examples of interest.

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Pedro (Hi I am Valter!) I did not understand if in your general argument you explicitly assume that $\omega$ is continuous (Wightman does for the vacuum state). For $C^*$-algebras positivity implies continuity, but here you only have a topological $^*$-algebra. Does the result hold anyway? –  Valter Moretti Dec 12 '13 at 16:01
Sorry, you assumed it! I did not see. However my question remains. –  Valter Moretti Dec 12 '13 at 16:03
Ciao Valter! No, in the case of topological $*$-algebras continuity does not follow from positivity, unfortunately. The C$*$-condition on the norm is crucial to obtain continuity from positivity. Maurin assumes continuity as a separate condition. Perhaps a similar result can be obtained if you can find a separating family of C$*$-seminorms on the Borchers-Uhlmann algebra. I don't know if this has ever been done... –  Pedro Lauridsen Ribeiro Dec 12 '13 at 16:11
Thanks. An interesting open problem maybe. It could be very useful to find a definite answer. Ciao, V. –  Valter Moretti Dec 12 '13 at 16:16
I really didn't expect this, thank you! –  moppio89 Dec 12 '13 at 19:32

First, let me state that the notion of a Hilbert space is not fundamental in quantum theory. One can realize the same quantum physical system using different Hilbert spaces. This is because quantum states (which are really the objects which physically matter) are only weakly connected to vectors on a Hilbert space. It is true that pure states correspond to rays of vectors in a Hilbert states, but when one considers mixed states, the correspondence becomes more indirect.

A classical argument for the irrelevance of non separable Hilbert spaces in quantum field theory is given by Streater and Wightman in their book: PCT, Spin and statistics and all that (pages 85-87).

First they prove that Hilbert space corresponding to a free field is separable (The Fock space). Secondly, they argue that in an interacting field theory, at least in the case when we are interested only in the scattering states, then these are basically free, thus in the case of a finite number of particle species, they correspond to a finite tensor product of Fock spaces, thus a separable Hilbert space.

They give two major examples of non separable Hilbert spaces: 1) The infinite tensor product of harmonic oscillators, they argue that the observables Bose field can be defined on a tiny subspace of this space mainly the Fock space, 2) Hilbert spaces corresponding to infinite volume finite density quantum statistical systems. Here, the basis vectors of the Hilbert space will be parameterized by the density, therefore the Hilbert space is genuinely non separable. There are known examples of this case.

One can add two types of non separable Hilbert spaces: 3) The infinite spin lattice Hilbert space, where each spin flip corresponds to a different basis vector. 4) Continuous tensor products of Hilbert spaces. The theory of continuous tensor products was initiated by von-Neumann and it has mathematical applications.

In the first case 3), one obtains a separable Hilber space if one divides by diffeomorphisms, and this happen in some physical applications, where the non separable Hilbert space becomes unnecessary. On the other hand the GNS construction can give rise to non separable Hilbert spaces.

Quite recently renewed interest in non separable Hilbert spaces has emerged in connection in quantum gravity (especially loop quantum gravity), where Hilbert spaces of the types 3) and 4) are being used. This approach is advanced by T. Thiemann, see for example the following article. Also, this theory uses Hilbert spaces on graphs which generalize the type 3) case.

There is another application is which systems with nonassociative algebras of operators which appear within string theories are transformed to associative theories on a non separable Hilbert space, please see for example the following article by Sämann and Szabo.

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If I understood well, in Wightman QFT the separability is assumed mainly to ensure the consistency of a scattering theory for standard quantum particle? In other theories indeed a nonseparable Hilbert space could be required. Then, when you say: "the notion of Hilbert space is not fundamental in quantum theory. One can realize the same quantum physical system using different Hilbert spaces" what do you mean exactly? Are you referring to the algebraic theory and the GNS construction? –  moppio89 Dec 12 '13 at 19:07
@moppio89 Yes, I think that their main assumption that what is physically important can be deduced from a scattering process. In addition, the usual notion of vacuum restricts to a subspace of a nonseparable space. Also, may be the mathematical complications constituted a factor due to the need for more advanced measure theoretic analysis on the underlying space than in the case of a countable set. –  David Bar Moshe Dec 15 '13 at 11:36
@moppio89 cont. In quantum mechanics we are interested in computing probabilities. The Hilbert space is an auxiliary tool. The example that I had in mind is the case of bosonization where observable of the same physical system have representations on the Bose and Fermi Hilbert spaces. –  David Bar Moshe Dec 15 '13 at 11:37