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Uncountable infinite dimensional Hilbert space

  1. Jul 6, 2005 #1
    Does anybody know an example for a uncountable infinite dimensional Hilbert space?(with reference or prove).i know about Banach space:\L_{\infty} has uncountable dimension(Functional Analysis,Carl.L.Devito,Academic Press,Exercise(3.2),chapter I.).but it is not a Hilbert space.
    thank you.
     
  2. jcsd
  3. Jul 6, 2005 #2

    dextercioby

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    Example 3.2.13, page 125, L.Debnath, P.Mikusinski "Introduction to Hilbert Spaces with Applications", AP, 1990.

    "Let [itex] \mathcal{H} [/itex] be the space of all complex valued functions defined on [itex] \mathbb{R} [/itex] which vanich everywhere, except a countable number of points in [itex] \mathbb{R} [/itex] and the values of those points form a square summable sequence. For such functions the inner product

    [tex] \langle f, g\rangle =\sum_{x\in\mathbb{R}} f(x)\bar{g}(x) [/tex]

    is well defined, because the summation is actually over a countable set. The space [itex] \mathcal{H} [/itex] is not separable, because, for any sequence of functions [itex] f_{n}\in\mathcal{H} [/itex], there are nonzero functions [itex] f [/itex], such that

    [tex]\langle f,f_{n}\rangle =0 \ , \ \forall n\in\mathbb{N}[/tex] "

    Satisfied ?

    As a side note, quantum physics (whether QM or quantum theory of fields) operates ONLY with separable Hilbert spaces.

    Daniel.
     
  4. Jul 6, 2005 #3
    If I may barge in, why is that? Why is separability so important?
     
  5. Jul 6, 2005 #4

    dextercioby

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    It allows to define basis (complete set of mutually orthogonal vectors which span the space).

    Daniel.
     
  6. Jul 8, 2005 #5
    thank you i think this example is true. but nonseparable Hilbert spaces have important role in physics for example Fock space in quantum filed theory(i think it is a nonseparable Hilbert spaces but i don't know prove of it).my aim of this question was to find a simple and straight example.
     
  7. Jul 8, 2005 #6

    George Jones

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    The Fock space constructed from a separable 1-particle state space is itself separable.

    Regards,
    George
     
  8. Jul 8, 2005 #7
    By Haag's theorem in QFT (Local Q.P by Haag)one can obtain two representations from one algebra that are not unitarity equivalent. but all seperable Hilbert spaces(with infinit daimension) are topologically isomorphe( for example: thorem 2.3.3,Mathematical Quantization,Nik Weaver,).it shows Fock space is nonseprable.(i didn't read all of this book and i don't know
    this subject is discucced .)
    Regards,
    reza.
     
  9. Jul 9, 2005 #8

    George Jones

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    I don't understand what you're saying.

    1) an algebra can have representations that are not unitarily equivalent. (OK)
    2) all infinite-dimensional separable Hilbert spaces are homeomorphic. (OK)
    3) Fock space is not separable. (???)

    Are you saying that 3) follows from 1) and 2)? If so, how?

    I still stand by the statement I made in my previous post.

    A topological space is separable if it contains a countable, dense subset. As dextercioby said, for a Hilbert space [itex]V[/itex], this is equivalent to the existence of a complete orthonormal basis [itex]B[/itex], i.e., a countable set of orthonormal vectors iin terms of which every vector in [itex]V[/itex] can be expressed as a strong limit of linear combinations of elements of [itex]B[/itex].

    For an infinite-dimensional Hilbert space [itex]V[/itex] with complete orthonormal basis [itex]B[/itex], let

    [tex]
    T = \mathbb{C} \oplus V \oplus \left( V \otimes V \right) \oplus \left( V \otimes V \otimes V \right) \oplus ...
    [/tex]

    A complete orthonormal basis for [itex]V \otimes V[/itex] is [itex]B \times B[/itex], which is countable, since the Cartesian product of countable sets is countable. By induction, every space [itex]V \otimes ... \otimes V \right[/itex] has a countable orthonormal basis. Therefore a basis for [itex]T[/itex] is [itex]\left\{ \psi_{ij} \right\}[/itex], where the index [itex]i[/itex] labels the space of i-particle states and the the index [itex]j[/itex] labels the j-th basis vector for the space of i-particle states.

    Again, since the Cartesian product of countable spaces is countable, this basis has a countable number of elements. Hence, [itex]T[/itex] is separable. Fock space, which is the appropriate symmetric or anisymmetric subspace of [tex]T[/itex], is also separable.

    Regards,
    George
     
  10. Jul 9, 2005 #9
    False. If it was true, by induction L_infinity([0,1],dx) is a separable hilbert space because L_n([0,1],dx) for any finite n is separable.

    Seratend.
     
    Last edited: Jul 9, 2005
  11. Jul 9, 2005 #10

    George Jones

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    I wasn't very explicit, but I meant ... to represent a finite (but arbitrary) number of tensor products. I think that's all I need for my argument.

    Regards,
    George
     
  12. Jul 9, 2005 #11
    {0,1}^n for any finite n is countable however {0,1}^|N is uncountable (same cardinality as |R) what do you conclude?

    Seratend.
     
  13. Jul 10, 2005 #12
    only two nonseprable Hilbert spaces can be INequivalent.two seprable Hilbert spaces certainly are equivalent.(this is the origin of Unrh efect and Hawking radiation)
    reza.
     
  14. Jul 10, 2005 #13
    I guess you mean isomorphic. This is only true for separable hilbert spaces with the *same* dimension (dim in |N U {+oO}).

    Seratend.
     
  15. Jul 10, 2005 #14

    George Jones

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    I conclude that this example is not relevant for the construct of Fock space. :smile:

    For [itex]V[/itex] a separable Hilbert space, set

    [tex]
    T = \sum_{n=0}^\infty \otimes^n V,
    [/tex]

    with [itex]\otimes^0 V = \mathbb{C}[/itex]. Just as an operational definition has to be given to the infinite sum

    [tex]
    S = \sum_{n=0}^\infty a_n,
    [/tex]

    an operational definition has to be given to the expression for [itex]T[/itex]. The infinite sum of real numbers is defined by

    [tex]
    S = \lim_{n\rightarrow\infty}s_n,
    [/tex]

    where

    [tex]
    s_n = \sum_{i=0}^n a_i.
    [/tex]

    Note that an infinite sum does directly add up an infinite number of terms; infinity only appears in the limit process.

    [itex]T[/itex] is the set of sequences [itex]\left\{ \left( w_0 , w_1 , ... , w_i , ... \right) \right\}[/itex] such that:

    1) [itex]w_i \in \otimes^i V[/itex];
    2)
    [tex]
    \sum_{i=0}^\infty \|w_i\|^2 <\infty.
    [/tex]

    Condition 1) gives [itex]T[/itex] as a countably infinite direct sum of vector spaces, while condition 2) allows an inner product to be defined on [itex]T[/itex].

    Let [itex]\left\{\psi_{ij} | j \in \mathbb{N} \right\}[/itex] be an orthonormal basis for [itex]\otimes^i V[/itex]. Then [itex]B_T = \left\{\psi_{ij} | \left(i ,j \right) \in \mathbb{N}_0 \times \mathbb{N} \right\}[/itex] is an orthnormal basis for [itex]T[/itex]. Since [itex]\mathbb{N}_0 \times \mathbb{N}[/itex] is countable, [itex]B_T[/itex] is countable, and [itex]T[/itex] is separable.

    In order for your example to be relevant, [itex]i[/itex] would have to equal [itex]\mathbb{N}[/itex], which is not possible since [itex]\mathbb{N}[/itex] is not an element of [itex]\mathbb{N}[/itex].

    It is well-known that the Fock space based on a separable Hilbert space is itself separable.

    Regards,
    George
     
  16. Jul 10, 2005 #15
    You are claiming this while the OP thinks the contrary. I am just saying, with some adequate examples, that you have not given any valid proof for the separability property. And I still cannot see any valid demo in your last answer (I do not understand the meaning of your |No symbol nor the following claim).
    Countability/uncountability demonstrations need some care in order to be valid (it is easy to make wrong demos such as |R is countable).

    So I give another hint: Is [itex]\otimes^\mathbb{N} V[/itex] included in the fock space T?

    Seratend.
     
  17. Jul 11, 2005 #16

    Equivalence means:an unitary map exists between two spaces(isometric topological isomorphism )(unitary equivalence )
    reza.
     
  18. Jul 11, 2005 #17
    : ) An equivalence relation is very general. Therefore it is sometimes usefull to give some precisions.

    Seratend.
     
  19. Jul 11, 2005 #18

    vanesch

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    The problem I see is that although the basis constructed by George (psi_ij) is countable, this constructs a Fock space which, in my opinion, is not closed, and seratend's example of {0,1}^N is relevant, in that {0,1}^N is NOT the same as the union of all {0,1}^n: you need to impose CLOSURE under Cauchy series.

    Now, I have to admit that I don't know if this closure requirement is part of the definition of a Fock space. In any case, it doesn't make sense physically to include states with an infinite number of particles.

    If the closure requirement is not part of the Fock space, then I think George is right, in that there is a countable basis ; but only, the space is not closed (and is, if I understand correctly, not a true Hilbert space). If the closure requirement is part of the Fock space definition, then I tend to agree with seratend in that this closure makes the space non-separable in the same way as the closure of the union of all {0,1}^n (which are countable) make us have something with same cardinality as the real numbers.

    cheers,
    Patrick.
     
  20. Jul 11, 2005 #19

    George Jones

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    Sorry, I should have defined my symbols. [itex]\mathbb{N} := \left\{1, 2, 3, ... \right\}[/itex] and [itex]\mathbb{N}_0 := \left\{0, 1, 2, 3, ... \right\}[/itex]. For each [itex]i[/itex] in [itex]\mathbb{N}_0[/itex], [itex]\psi_{ij}[/itex] is in [itex]\otimes^i V[/itex], and [itex]\left\{ \psi_{ij} \right\}[/itex] is a countable orthonormal basis, labeled by [itex]j[/itex] in [itex]\mathbb{N}[/itex], for [itex]\otimes^i V[/itex]. Therefore, [itex]\psi_{ij}[/itex] = [itex]\psi_{kl}[/itex] iff [itex]\left( i , j \right) = \left( k , l \right)[/itex].

    This establishes a bijection between [itex]B_T = \left\{\psi_{ij} | \left(i ,j \right) \in \mathbb{N}_0 \times \mathbb{N} \right\}[/itex] and [itex]\mathbb{N}_0 \times \mathbb{N}[/itex].

    I agree.

    No.

    Fock space is a formalism for handling an arbitrary, but finite, number of particles, as well as for dealing with a changeable number of particles.

    Regards,
    George
     
  21. Jul 11, 2005 #20

    vanesch

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    With reference to what I posted, does that mean then, that Fock space is not closed (and hence, not a Hilbert space) ?

    cheers,
    Patrick.
     
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