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Equivalent vectors in a Hilbert space

  1. Sep 20, 2012 #1
    In Griffith's intro to QM it says on page 95 (in footnote 6) :

    "In Hilbert space two functions that have the same square integral are considered equivalent. Technically, vectors in Hilbert space represent equivalence classes of functions."

    But that means that if we take for example

    f(x) = 1 | 0 < x < 1


    g(x) = [itex]\sqrt{1/10}[/itex] | 0 < x < 10

    both f(x) and g(x) have the same square integral ∫|f(x)|^2 dx = ∫|g(x)|^2 dx = 1

    But how can they be considered "equivalent" - if they are wavefunctions for a particle, for example, then they represent completely different probability predictions! f(x) says you can find a particle with equal probability in [0,1], while for g(x) it's [0,10]
  2. jcsd
  3. Sep 20, 2012 #2
    Either Griffith is wrong or you must have misunderstood something, but that statement is clearly untrue.

    Vectors in the hilbert space describe the same state if they are related by multiplication of a nonzero complex number. That means physical states are really rays in the hilbert space. Mathematically this is the projective space H/~ where ~ is the equivalence relation that identifies identical states.
  4. Sep 20, 2012 #3


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    It's a very unfortunate statement, to say the least. It should be a transcription of Jazzdude's 2nd paragraph in post #2 in terms of a specific Hilbert space (L^2(over what??)), but it's wrongly worded.

    Scratch that and get a better book.

    P.S. There's no footnote on page 95 of my version of Griffiths (1995, 1st edition I guess).
  5. Sep 20, 2012 #4
    Here it is :

  6. Sep 20, 2012 #5
    So...A vector space has to have a null vector. If L2(-inf,+inf) functions form the vector space, the only null vector is the f(x) = 0 function.

    But also for any vector space, <f|f> has to be >0 for any non null vector. And yet, here
    for a function like

    g(x) = 1 | x=1
    g(x) = 0 | everywhere else

    (and other such functions)

    <g|g>=0 even though clearly g is not a null vector.

    So either we exclude functions like g from the vector space, or we define the vector space to be the set of equivalence classes of functions (where all functions like g would be the same class). But what is the precise formulation/definition of such an equivalence relation?
  7. Sep 21, 2012 #6


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    The functions are "equivalent" if they differ only on a set of Lebesgue measure 0.
    (Isolated points in the domain of a function are of measure 0.)

    See also:

    P.S., The wording of Griffith's footnote
    is wrong. But,
    is ok.
  8. Sep 21, 2012 #7


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    The correct statement should be: f is equivalent to g if ∫|f-g|2dx = 0.
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