19
$\begingroup$

It seems that all the usual physical systems have a separable Hilbert space. Is there any example with a non-separable Hilbert space?

BTW, I am actually always baffled by the fact that a continuous model like the 1d harmonic oscillator defined on $\mathbb{R}$ has a separable Hilbert space. It is well known that $\mathbb{R}$ is uncountable. But on the other hand, the Hermite polynomials are countable. This means while the Hermite polynomials (with some Gaussian prefactor) are a legitimate basis of the Hilbert spade, the more common and sloppy coordinate basis $|x \rangle $ are not.

$\endgroup$
0

3 Answers 3

24
$\begingroup$

The standard formulations of QM and QFT are such that the resulting Hilbert space is always separable, namely there exist a finite or infinite countable Hilbert basis (and thus every Hilbert bases are of the same type correspondingly).

Separability is required as an axiom from scratch or it arises as a consequence of more basic axioms. In particular

  • For elementary non-relativistic systems, all irreducible representations of $X_j$ and $P_k$ CCRs produce separable Hilbert spaces $L^2(\mathbb R^n, d^nx)$ in view of the celebrated Stone-von Neumann theorem. Adding the spin does not alter the result because the space becomes $L^2(\mathbb R^n, d^nx) \otimes \mathbb C^{2s+1}$ which is still separable.

  • If the elementary system is relativistic and therefore supports an irreducible unitary strongly-continuous representation of Poincaré group,separability arises by the classification of the afore-mentioned representations which works on Hilbert spaces of the form $L^2(\mathbb R^n, d^nk) \otimes \mathbb C^{2s+1}$.

  • Finite composite systems are obtained by taking a finite tensor product of elementary systems, so that separability is preserved.

  • In the absence of complicated phenomena as spontaneously broken symmetry (see below), assuming asymptotic completeness, QFT is defined in a Fock space constructed out of separable Hilbert spaces (one particle spaces). These Fock spaces are separable.

  • In curved spacetime, at least in static spacetimes and using the static vacuum as vacuum of the Fock representation, separability is still guaranteed as the one particle space is still a $L^2$ space over a separable (in measure-theory sense) space.

Non separability may arise in presence of continuous superselection rules if picturing them as a direct sum instead of a cumbersome direct integral of sectors. Think of a non relativistic system admitting a mass operator $M$ whose spectrum $\sigma(M)$ is an interval, say $(a,b)$. The Hilbert space results to be the direct orthogonal sum of an infinitely continuous class of eigenspaces $\cal H_m$ of the mass operator $$\cal H = \oplus_{m \in \sigma(M)} \cal H_m$$ so that $\cal H$ cannot be separable as it admits an uncountable sequence of pairwise orthogonal subspaces.

Notice that the spectrum of $M$ is a pure point spectrum made of an interval $\sigma(M) = \sigma_p(M) =(a,b)$ in this picture. This fact is possible in spite of the name "point" spectrum, which is misleading here.

Here, if one admits that the system supports a (projecitve unitary) representation of Galileo group, due to Bargmann's superselection rule of the mass, quantum physics is described in each subspace separately (in the standard way ${\cal H}_m = L^3(\mathbb R, d^3x)$ if the system is a particle with mass $m$ and $m$ appears therein as a fixed parameter) and at most incoherent superpositions of states of different subspaces are permitted.

In each such subspace ${\cal H}_m$ vectors are normalizable and all observables $A$ of the theory admit every ${\cal H}_m$ as invariant subspace, since $A$ commutes with $M$.

The fact that the vectors in each ${\cal H}_m$ are normalizable is the basic difference from the direct integral picture where vectors are instead similar to the kets $|x\rangle$ such that $\langle x| x \rangle$ does not make sense. In this representation $\sigma(M)$ is a continuous spectrum but the theory is quite singular in each coherent sector ${\cal H}_m$ which is not a subspace of the overall Hilbert space. As far as I remember a similar situation arises in loop quantum gravity...

Non-separability arises also when some symmetry spontaneously breaks and you consider all possible Hilbert spaces (continuously parametrized) as orthogonal subspaces of an overall Hilbert space.

Non separable Hilbert spaces have the pathology that quantum statistical mechanics cannot be formulated at least in the standard way, since the trace of usual statistical operators describing equilibrium necessarily diverges. This is because the overall Hamiltonian operator (if assuming to have pure point spectrum) admits an uncountable basis of eigenvectors. Though, in each superselection sector no problem arises.

In presence of non separable Hilbert spaces perhaps the algebraic approach seems more suitable. Thermodynamical equilibrium may be described in terms of KMS condition for an algebraic state over a C*-algebra of observables.

Non-separability arises from a very abstract viewpoint when considering all non unitarily equivalent representations of a given C*-algebra of observables, e.g., field operators referring to all possible vacua, in a unique Hilbert space made of all the GNS representations of these vacua.

Addendum. As I realized after a discussion with a colleague (at Les Houches school of physics) separability of the Hilbert space arises as soon as the system admits (is) an irreducible strongly-continuous unitary representation of a Lie group of symmetries. (See the sketch of proof in the comments below.) This includes both the case of a non-relativistic and relativistic particle mentioned above in particular.

$\endgroup$
11
  • $\begingroup$ In the argument that a mass operator gives rise to a non-separable Hilbert space, one can replace "mass operator" by "operator with continuous spectrum whose eigenstates we consider physical", right? Or does it really need to be mass? $\endgroup$
    – MBolin
    Commented Oct 5, 2022 at 18:53
  • 1
    $\begingroup$ Yes, you can use any operator whose eigevectors have physical meaning. But the spectrum must be a pure point spectrum made of an uncountable number of eigenvalues, for instance a whole interval $[a,b]$. If the spectrum is continuous the argument does not work. The subtle point is that the point spectrum can be a "continuous" set in spite of its name. The price one pays is that the Hilbert space is not separable. $\endgroup$ Commented Oct 5, 2022 at 20:53
  • $\begingroup$ @ValterMoretti Id like to see the proof of the theorem mentioned in your addednum. Thanks $\endgroup$
    – TheDawg
    Commented Jan 10, 2023 at 13:40
  • $\begingroup$ A Lie group $G$ is separable. So there is a dense countable set of elements $V\subset G$. Now consider the closure $H_0$ of the span of all finite combinations of elements $U_g\psi$ where $g$ varies in $G$ and $\psi$ is a non zero (fixed) vector. This space is invariant under $U$ and closed. Hence it is the whole Hilbert space $H$ due to irreducibility. $\endgroup$ Commented Jan 10, 2023 at 22:05
  • $\begingroup$ Using continuity, we have that every element of $H_0$ is however the limit of a sequence of finite linear combinations with rational (complex) components of elements $U_h\psi$ with $h\in V$. This set of vectors is dense and countable. $\endgroup$ Commented Jan 10, 2023 at 22:08
2
$\begingroup$

I'll address your question about cardinality. (See also the excellent comments by Valter Moretti below.)

Let's say I have a countably infinite orthonormal (and hence linearly independent) set $S$ of vectors in an inner product space $V$, viz. $\left\langle m|n\right\rangle =\delta_{mn}$ with $m,\,n\in\mathbb{N}$. An infinite sequence of vectors whose $n$th element is $u_n:=a_n\left|n\right\rangle$ and $n$th partial sum is $S_n:=\sum_{k\le n}u_k$ satisfies $\left|S_m-S_n\right|^2= \sum_{k=n+1}^{m}\left|a_k\right|^2$ for $m\le n$. If our set is a "basis" of a Hilbert space in quantum mechanics, unitarity requires $\sum_n \left|a_n\right|^2=1$. Thus $\lim_{n\to\infty}\sum_{k=1}^{n}\left|a_k\right|^2=1$ and $\sum_{k=n+1}^{m}\left|a_k\right|^2\le 1-\sum_{k=1}^{n}\left|a_k\right|^2$ can be made arbitrarily small with sufficiently large $m,\,n$. Our sequence of $u_n$ is then a Cauchy sequence.

An inner product space is also a metric space. If each Cauchy sequence in a metric space (inner product space) has a limit therein, we call the space a complete metric space (Hilbert space). We use Hilbert spaces in quantum mechanics to ensure all unitary sums of vectors as above will exist as elements of the state space. Let's see what happens if we don't add the it's-Hilbert assumption.

Since $S$ spans a subspace $W$ of $V$, which may or may not be $V$ (in particular, there may or may not exist $S$ such that $W=V$), $W$ contains all linear combinations of finitely many elements of $S$. In fact, for general inner product spaces we define the span of $S$ as the set of vectors expressible in this form; if it's not Hilbert, we can't in general assume an infinite series has well-defined limit in $V$. The dimension theorem states all $S$ that span a given choice of $W$ have the same cardinality, which is called the dimension of $W$.

In Hilbert spaces, by contrast, we're allowed to use infinitely many members of a "basis" to form a sum; it will exist in the space as long as the coefficients satisfy unitarity. So, strictly speaking, a "basis" in the Hilbert-space sense isn't the type of basis described in the dimension theorem, whose proof (case 1 here) relies on the finitely-many-elements condition.

The Hilbert space that had you confused, viz. "is its dimension $\aleph_0$ or $c$?", is a good example of this subtlety. The countably infinite set of vectors that "span" it do so only with the machinery that makes Hilbert spaces special. In dimension-theorem terms, any "basis" of that space has cardinality $c$ because the countably infinite "Hilbert-basis" only "spans" (as opposed to "Hilbert-spans") a countably infinite subspace: namely, the set of vectors expressible using finitely many of its elements. (You can verify that set satisfies the definition of a vector space.) For example, this subspace contains $\sum_{k=1}^{10}\frac{1}{\sqrt{10}}\left| k\right\rangle$ but not $\sum_{k=1}^{\infty}\frac{\sqrt{6}}{k\pi}\left| k\right\rangle$, whereas the full Hilbert space contains both.

Note: none of the phrases I've placed in scare quotes are technical terms; they're just a way to emphasize here the difference between concepts with the same name.

$\endgroup$
5
  • $\begingroup$ Sorry I am not sure to understand what you wrote. In a Hilbert space $H$ there are two notion (at least) of basis: Hamel bases and Hilbert bases their existence follows from Zorn's lemma. The first class exist for every vector space nomatter topological issues. A set of vectors $S \subset H$ is a Hamel basis if every vector $x\in H$ is a finite linear combination of elements of $S$ and the elements of $S$ are linearly independent. All Hamel basis have the same cardinality. $\endgroup$ Commented Jan 7, 2017 at 10:13
  • $\begingroup$ A Hilbert basis $T\subset H$ is a set of pairwise orthonormal vectors such that the finite span of these vectors is dense in $H$. This is equivalent to say that every $x\in H$ can be (uniquely) expanded as an orthogonal sum of elements of $T$ with coefficients whose at most a countable subset is nonvanishing, and the convergence is in the topology of the natural norm of the Hilbert space. Also Hilbert bases have the same cardinality, finite, countable or not. $\endgroup$ Commented Jan 7, 2017 at 10:16
  • $\begingroup$ If the dimension of the space is not finite Hamel bases and Hilbert bases are disjoint sets. The formal bases like this $\{|x\rangle\}_{x \in \mathbb R}$ are not Hilbert nor Hamel bases. They are not bases at all, strictly speaking, because they are made of objects which do not belong to the Hilbert space. If you want to interpret them as bases in some sense you have to replace the sum with a direct integral. A way to do it is thinking of $L^2(\mathbb R)$ as a rigged Hilbert space. $\endgroup$ Commented Jan 7, 2017 at 10:21
  • $\begingroup$ (sorry I am writing from my phone, I cannot correct the various typos I see in my previous comments) $\endgroup$ Commented Jan 7, 2017 at 10:23
  • $\begingroup$ I forgot to say: for a Hilbert space, separability as a metric space (there exist a dense countable subset) is equivalent to the existence of a finite or infinite countable Hilbert basis. $\endgroup$ Commented Jan 7, 2017 at 10:26
2
$\begingroup$

Separability is a very serious problem. In the limit as $N\to\infty$, the Hilbert space for a quantum mechanical system with $N$ degrees of freedoms is not separable. The solution in this case is to select a vacuum vector and build out a separable sector around the vacuum by applying a countable number of operators to it. The operators correspond to the countable family of all local observables.

If the vacuum vector is invariant under relativistic transformations, the Galileo or the Poincare group will be defined in the sector and the sector will be decomposable in irreducible representations of the relevant relativistic group. If instead the vacuum vector changes as a function of time or is not invariant under the relativistic group, we have to apply the relativistic transformations also to the vacuum itself.

Non-separable Hilbert spaces arise in quantum mechanics in the thermodynamic limit of infinite volume. Also a quantum field confined to a finite volume and with no ultraviolet cutoffs is defined on a non-separable Hilbert space. In both cases, it becomes necessary to specify a vacuum vector as part of the system definition.

The deeper question is whether the mathematical inconsistencies of non-separable spaces have a physical meaning also for finite but macroscopic system with a number of particles of the order of an Avogadro number or even of meso-scopic scale. If there was physical meaning, this would imply that Quantum Mechanics cannot alone give a complete description of a macroscopic system as, in addition to giving the Hamilton operator, one would also have to define the vacuum vector and describe its dynamics. Of course the vacuum vector would be equivalent to the state of a classical system and follow classical Physics laws. We know that there are Laws of Classical Physics which are fundamentally inconsistent with Quantum Mechanics, namely the Second Law of Thermodynamics and General Relativity. This inconsistency could simply mean that Quantum Mechanics is an incomplete description of the world and the nature and dynamics of the vacuum vector is not a problem that can be decided within the context of Quantum Theory alone.

$\endgroup$
1
  • $\begingroup$ Welcome to the site! $\endgroup$ Commented Sep 14, 2023 at 9:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.