In the text I'm using for QM, one of the properties listed for Hilbert space that is a mystery to me is the property that it is separable. Quoted from text (N. Zettili: Quantum Mechanics: Concepts and Applications, p. 81):

There exists a Cauchy Sequence $\psi_{n} \ \epsilon \ H (n = 1, 2, ...)$ such that for every $\psi$ of $H$ and $\varepsilon > 0$, there exists at least one $\psi_{n}$ of the sequence for which $$ || \psi - \psi _{n} || < \varepsilon.$$

I'm having a very hard time deciphering what this exactly means. From my initial research, this is basically demonstrating that Hilbert space admits countable orthonormal bases.

  1. How does this fact follow from the above?

  2. And what exactly is the importance of having a countable orthonormal basis to the formalism of QM?

  3. What would be the implications if Hilbert space did not admit a countable orthonormal basis, for example?


2 Answers 2


I usually see it in the reverse way, but it is a matter of taste. Hilbert spaces, in general, can have bases of arbitrarily high cardinality. The specific one used on QM is, by construction, isomorphic to the space L2, the space of square-integrable functions. From there you can show that this particular Hilbert space is separable, because it is a theorem that a Hilbert space is separable if and only if it has a countable orthonormal basis, and L2 has one.

  • $\begingroup$ I see, thanks. A remaining question is, how is this fundamental to the underlying formalism of QM? I just want to see where is the physics exactly in working with separable Hilbert spaces. $\endgroup$
    – Zack
    Nov 25, 2014 at 4:07
  • $\begingroup$ The physics is not about separable physics spaces, that came later. The physics is about square-integrable functions. But after that, mathematicians formalize the theory by choosing their preferred axioms. So, one thing is how the theory arised historically, another is how you like to formalize it in the "simplest" or more "aesthetic" way after you have it. $\endgroup$
    – user65081
    Nov 25, 2014 at 4:14
  • $\begingroup$ I am assuming you know that any formal system, or theory T is equivalent to anothet theory T', in which axioms and theorems are exchanged. Both theories are equivalent from a formal or practical point of view. I have seen formalizations of QM that are so un-intuitive and un-historical as possible. That doent make them less good, just less intuitive. $\endgroup$
    – user65081
    Nov 25, 2014 at 4:19
  • $\begingroup$ @Zack: Have a look at the thread Qmechanic linked to: The separability is mostly only fundamental in the sense described in this answer, namely, that the theory evolved around L^2, which happens to be separable. $\endgroup$
    – Martin
    Nov 25, 2014 at 10:48

As showed by Solovay here, in a non-separable Hilbert space $H$ there may be probability measures that cannot be written, for any $M$ closed subspace of $H$, as $\mu (M)=\mathrm{Tr}[\rho \mathbb{1}_M]$, for some positive self-adjoint trace class $\rho$ with trace 1 (density matrix). Here $\mathbb{1}_M$ denotes the orthogonal projection on $M$. [The proof of the existence of such "exotic measures" is undecidable in ZFC, however it is equivalent to the existence of a (real valued) measurable cardinal]

In some sense it means that in non-separable Hilbert spaces there may exist analogues to "normal quantum states" that are not density matrices$^\dagger$.

Remark: For normal quantum state I mean a ultraweakly positive continuous functional on the $C^*$-algebra of bounded operators that is interpreted to give their expectation value.

$^\dagger$: I mean that even if these states are probability measures with the countable additivity of orthogonal closed subspaces property, are not expressed as the trace of density matrices (while on separable Hilbert spaces this is always the case, and these measures are in one to one correspondence with normal states).

  • $\begingroup$ For every infinite dimensional Hilbert space there are states which cannot be represented by a density operator (the states that are not normal). $\endgroup$
    – jjcale
    Nov 25, 2014 at 18:44
  • $\begingroup$ @jjcale : edited to be more clear (in some sense on non-separable spaces there may be "normal states" that are not written as the trace of density operators). $\endgroup$
    – yuggib
    Nov 26, 2014 at 11:28
  • $\begingroup$ This is very interesting to me. Are there any examples of non-normal states on a von Neumann algebra which have a physical interpretation? Are there any of states which can't be represented as density matrices? I imagine this should be the case in QFT. Otherwise, why Thermal Field Theory? $\endgroup$ Jan 9, 2019 at 1:57
  • $\begingroup$ @IvánMauricioBurbano A state is normal w.r.t. a representation of the algebra. Or, due to the GNS construction, w.r.t. another state. The most common representation used in QFT is the Fock representation, that is the GNS representation corresponding to free vacuum vectors. There are indeed physically relevant states that are not Fock-normal, i.e. that cannot be represented as density matrices in the Fock space for the same algebra of observables. The most relevant example are vacuum vectors of interacting relativstic QFTs with the same physical paramteres (mass, charge, spin). $\endgroup$
    – yuggib
    Jan 10, 2019 at 12:48
  • $\begingroup$ In fact, in this case Haag's theorem tells that these vacuum vectors cannot be Fock normal, unless they are the Fock vacuum vector itself (and it is well known that the free vacuum is not a ground state for interacting systems). $\endgroup$
    – yuggib
    Jan 10, 2019 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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