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Separability of Hilbert Spaces

  1. Aug 23, 2015 #1
    Why we require the separability of Hilbert spaces in Quantum Mechanics?
  2. jcsd
  3. Aug 23, 2015 #2


    Staff: Mentor

    In reality any actual observation is from a finite dimensional space - but sometimes of a large but unknown dimension. To handle that a limit is taken and you end up with Rigged Hilbert spaces which are separable.

    To be specific the space is of vectors of finite dimension but of any size. Mathematically we take the dual which is the Rigged Hilbert space of this space. It is separable - but convergent only under a very weak topology. Such a large space isn't actually required in practice and part of the art of using Rigged Hilbert spaces is figuring out exactly what subset is needed:

    Last edited: Aug 23, 2015
  4. Aug 23, 2015 #3


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    Gold Member

    Because of results like the Stone-von Neumann theorem. In non-relativistic QM, this theorem asserts that any irreducible unitary realization of the (integrated form of the) canonical commutation relations on a complex Hilbert space ##H## (in principle, not necessarily separable) is unitarily equivalent to the standard realization in the separable space ##L^{2}(R^{3})##, i.e., only in a separable space you can get one such irreducible unitary realizations. Similar results hold for the representation theory of the Poincaré group in relativistic QM.
  5. Aug 23, 2015 #4


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