Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hilbert space

  1. Sep 25, 2004 #1
    Please forgive this physicist's thread :

    I can define a Hilbert space that is :
    1) [tex]\mathbb{R}^n[/tex] with the euclidian norm, especially on a real field, and which is finite dimensional : is it right ? This is the most stupid question ever.
    2) over the quaternions [tex]\mathbb{H}[/tex] ?
    3) if the dimension is infinite non-countable, it is not separable. There is no need to talk about topological or metrical separability, the two coincide.

    Please some one answer. It is due to questions [thread=44301]here[/thread].
     
  2. jcsd
  3. Sep 25, 2004 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    1) It's partly a matter of definition. Finite dimensional Euclidean spaces and Hilbert (infinite dimensional) are very similar, but usually the terms are kept separate.

    2) I've never seen anyone use quarternions as the scalar field for a vector space. I suppose it's possible, but it would have to be worked out.

    3) The topology of a metric space is determined by the metric. I doubt if there is a distinction in the use of the term separable, but I'm a little rusty here.
     
  4. Sep 25, 2004 #3

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    To me, Hilbert space is distinguished not by tis dimension, but by the rpesence of an inner product. Thus a finite dimensional Hilbert space is finite dimensional euclidean space equipped with its inner product.

    The notion of dimension of a hilbert space can be confusing. for a vector space the dimension is usually the cardinality of an maximal independent set. For any hilbert space, this is never countably infinite.

    Sometimes the dimension of a hilbert space on the other hand is thought of as the cardinality of a maximal orthonormal set, which can be countably infinite, as in the case of "little L2".

    For a hilbert space to be separable as a topological space, i.e. to have a countable dense subset, it is necessary and sufficient to have an at most countable maximal orthonormal set.

    There is in fact only one infinite dimensional, separable hilbert space, namely "little L2", up to isomorphism.
     
  5. Sep 25, 2004 #4
    Since i am french, I always refered to Bourbaki for what matters of semantics. They defined a Hilbert space independently of the dimensionality, finite or infinite. I appreciate the answers. Thanks.

    EDIT : I don't see any reason why dimensionality should occur in the definition of a Hilbert space
     
    Last edited: Sep 25, 2004
  6. Sep 25, 2004 #5

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    the book: foundations of modern analysis, by the fine french writer Dieudonne', was my source for these remarks.
     
  7. Sep 25, 2004 #6
    Thank you mathwonk. I understand you go in the same direction as i do. Indeed Dieudonne is one of the Bourbaki guys.

    Yet mathman seem to imply there is a distinction, and finite dimensional Hilbert spaces are not accepted.
     
  8. Sep 25, 2004 #7

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    i think he just means they are not the hilbert spaces of interest to many people.
     
  9. Sep 25, 2004 #8
    I totally agree. This question is kind of irrelevant anyway. It keeps going on in the other forum, I wanted to make it stop for that reason.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Hilbert space
  1. Hilbert Transform (Replies: 2)

  2. Hilbert Cube (Replies: 3)

  3. Hilbert's Paradox flaw (Replies: 18)

Loading...