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Hilbert and Euclidian space

  1. Oct 23, 2013 #1
    I know that hilbert space is infinite dimension space whereas eucledian is Finite n dimensional space, but what are all other differences between them?
  2. jcsd
  3. Oct 23, 2013 #2


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    There are theorems true in Euclidean space that are not valid in Hilbert space because of the infinite dimension.

    For example: In Euclidean space if a sequence of vectors {xn} converge pointwise to x, then
    ||xn-x|| -> 0.

    This is not necessarily true in Hilbert space.
    Example: kth component of xn = 0, for k≠n, = 1, for k=n. xn -> 0 vector pointwise, but ||xn|| = 1.
  4. Oct 23, 2013 #3
    A Hilbert space is not necessarily infinite dimensional. Some Hilbert spaces are finite dimensional and are isomorphic to Euclidean space.

    So Euclidean space is a special case of a Hilbert space. But of course, there are many Hilbert spaces which are not Euclidean spaces.
  5. Oct 23, 2013 #4
    Hilbert spaces are not necessarily infinite dimensional, I don't know where you heard that. Euclidean space IS a Hilbert space, in any dimension or even infinite dimensional.

    A Hilbert space is a complete inner product space. An inner product space is a vector space with an inner product defined on it. An inner product induces a norm, which induces a metric. A metric space is called complete if every Cauchy sequence is convergent.
  6. Oct 24, 2013 #5
    I guess he's taking a QM course. I have seen QM courses being taught by defining Hilbert space as infinite dimensional. Some even define Hilbert space as ##L^2(\mathbb{R})## or ##\ell^2##. I don't know why they don't teach it the right way though.
  7. Oct 24, 2013 #6
    If hilbert space can be finite or infinite dimension. what about eucledian space can it be infinite dimensional?

    Underwhat special condition a hilber space becomes eucledian space?actually first i thought eucledian space as finite dimension but now I'm confused what it is.
  8. Oct 24, 2013 #7
    Euclidean space is a specific example of a Hilbert space. Euclidean space is the set of all n lists of numbers (n is is the dimension, and may be infinite dimensional), over real numbers, with the dot product as the inner product.
  9. Oct 24, 2013 #8
    An example space which is hilbert but not eucledian space can be set of all continuous function in a given interval..correct ?
  10. Oct 24, 2013 #9
    I don't know who would describe a hilbert space as infinite dimensional only as one of the most important hilbert spaces in QM is two-dimensional. It's hard to talk about Hilbert spaces properly when a very good portion of the students won't even know what completeness means.

    You need to specify an inner product.
    Last edited: Oct 24, 2013
  11. Oct 24, 2013 #10
    Yes, Euclidean space is the same as a finite-dimensional Hilbert space. So the Euclidean space does not have infinite dimension by definition. A Hilbert space can be finite or infinite dimensional.
  12. Oct 24, 2013 #11

    I got to say that this is one of the absolute worst descriptions of Hilbert space that I've ever seen. I'm the last one to say that QM books should do rigorous math, but if they do mention things like Hilbert spaces, it should at the very least be approximately correct. /rant
  13. Oct 24, 2013 #12
    Sorry for the OT, but...

    I think there's some out-of-date terminology that agrees with this. While you and I know a Hilbert space as an inner-product space whose induced metric space structure is complete, I've been told they used to refer to (the) Hilbert space, which is what we would call a separable, infinite-dimensional Hilbert space. Of course the word "the" is validated by there being only one such space (up to unitary equivalence).
  14. Oct 24, 2013 #13
    Yeah. Originally, Hilbert only worked with the ##\ell^2## or ##L^2## space. He didn't consider more general spaces.

    Probably apocryphal, but: once, a mathematician gave a lecture in front of other mathematicians. He said "consider a Hilbert space H and...". He was interrupted as somebody in the audience asked "What exactly is a Hilbert space?" That somebody was David Hilbert.
  15. Oct 24, 2013 #14
    That's a very surprising description.
  16. Oct 24, 2013 #15
    A similar story, about Shizuo Kakutani. He proved a useful generalization of Brouwer's fixed point theorem (to set-valued functions), which is used all over the place in economic theory. Once, he presented at a conference and had a bunch of economists in the audience, so he asked a game theorist buddy why. When the game theorist told him that it was likely because of the Kakutani fixed point theorem, Kakutani replied, "What is the Kakutani fixed point theorem?"
  17. Oct 24, 2013 #16


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    It appears that this question has degenerated into a quibble about definition. The original question was about the difference between (finite dimensional) Euclidean space and (infinite dimensional) Hilbert space. I suggest that replies should try to address that question and not get hung up on definitions.
  18. Oct 24, 2013 #17


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    And again, with Banach, when he was reading a paper asked what a Banach space was (he called them B-spaces IIRC).
  19. Oct 24, 2013 #18
    Uhh, the question is a question about definitions. I don't know what else he's asking for if not for clarification of the definitions.
  20. Oct 25, 2013 #19


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    i agree with mathman.
  21. Oct 25, 2013 #20
    Okay, so rephrasing the question: What makes finite-dimensional Hilbert spaces (a.k.a. Euclidean spaces) special within the class of all Hilbert spaces?
    - Of course, there is a finite spanning set. i.e. For some [itex]n\in \mathbb Z_+[/itex], there is some [itex]\{e_i\}_{i=1}^n[/itex] (which can be taken to be pairwise orthogonal) such that every vector is a linear combination of [itex]\{e_i\}_{i=1}^n[/itex].
    - Every closed, bounded subset is compact.
    - Every bounded sequence converges to something.
    - Every linear subspace is itself a Hilbert space.
    - Every linear map (into any other linear space) is continuous.
    - Carathéodory's theorem: there is a finite number [itex]k\in \mathbb N[/itex] ([itex]k[/itex] can be taken to be [itex]n+1[/itex] with [itex]n[/itex] as above) such that: for every subset [itex]P[/itex] and every [itex]x[/itex] that can be written as a convex combination of points in [itex]P[/itex], there is a way to write [itex]x[/itex] as a convex combination of [itex]k[/itex] or fewer points in [itex]P[/itex].
    - A lot of other stuff...

    Basically, finite-dimensional spaces enjoy a lot of special properties that general Hilbert spaces don't.
  22. Oct 25, 2013 #21
    I think you made a typo here. Of course it should be that any bounded sequence has a convergent subsequence.
  23. Oct 25, 2013 #22
    Hah! Thank you. Yes.

    Every bounded sequence has a subsequence that converges to something.
  24. Oct 25, 2013 #23


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    Isn't this true in any metric space? Consider the bounded sequence as a metric (sub)space S. Then S is a compact (sub)space so that every sequence has a convergent subsequence.
  25. Oct 25, 2013 #24


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    Now, another question:

    Does every Hilbert space have a Schauder basis ?
  26. Oct 25, 2013 #25
    No, it's not true in any metric space. One example is an infinite-dimensional Hilbert space. Or the discrete metric space.

    The failure of your proof is saying that the boundd sequence is compact. Not every bounded and closed space is compact.
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