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Rigged Hilbert space

  1. Jul 25, 2011 #1
    Rafael de la Madrid writes:

    - de la Madrid (2005): "The role of the rigged Hilbert space in Quantum Mechanics"

    Could the second paragraph be restated as: "The elements of [itex]\Phi[/itex], the vectors, regarded as equivalence classes of functions differing only on sets of zero Lebesgue measure, can be represented by smooth (and thus continuous) functions (although these smooth functions may be equivalent to functions which are not smooth), whereas a general element of [itex]\cal{H}[/itex], considered as such an equivalence class, will not necessarily contain a smooth function."
     
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  3. Jul 25, 2011 #2

    dextercioby

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    Why would they be equivalence classes ?
     
  4. Jul 25, 2011 #3

    Fredrik

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    Because [itex]\Phi[/itex] is a subspace of [itex]L^2(\mathbb R^3)[/itex], which is a vector space whose members are equivalence classes of functions?

    I don't understand that quote from the article. If the vectors in [itex]\mathcal H[/itex] are equivalence classes of functions, then so are the vectors in any subspace. It sounds like he's not considering a subspace of [itex]L^2(\mathbb R^3)[/itex], but a subspace of the semi-inner product space [itex]\mathcal L^2(\mathbb R^3)[/itex] whose members are square-integrable functions. That sounds pretty weird to me, but I don't know much about this stuff.
     
  5. Jul 25, 2011 #4

    tom.stoer

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    Two functions f(x) and g(x) which differ on a set of measure zero are 'identical w.r.t. the relevant functionals F' like integration with test functions etc. For the difference f-g the results for these operations is exactly zero, i.e. F[f-g] = 0. Therefore f and g should be considered as identical, which means that the single elements are not functions but equivalence classes of functions: f ~ g.
     
  6. Jul 25, 2011 #5
    Equivalence classes because otherwise wavefunctions wouldn't comprise a vector space.

    Let [itex]f[/itex] and [itex]g[/itex] be square integrable functions from [itex]\mathbb{R}[/itex] to [itex]\mathbb{C}[/itex]. Consider the inner product defined by

    [tex]([f],[g])=\int_{-\infty}^\infty \overline{f(x)}\cdot g(x) \; dx.[/tex]

    Suppose that, there exists a function [itex]f[/itex] such that [itex](f,f)=0[/itex]. Then, by the inner product axioms, [itex]f=0[/itex]. But if [itex]g[/itex] differs from [itex]f[/itex] only on a set of Lebesgue measure zero, then we also have [itex](g,g)=0[/itex]. So [itex]g=0[/itex]. A group, such as a vector space with addition, contains at most one identity element. Therefore, [itex]f[/itex] must represent the same vector as [itex]g[/itex].
     
  7. Jul 25, 2011 #6

    Fredrik

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    FYP.

    It's a vector space with a semi-inner product, so it's a semi-inner product space, but not an inner product space, and therefore not a Hilbert space.
     
  8. Jul 25, 2011 #7
    Ah, I see. Thanks for the correction.
     
  9. Jul 25, 2011 #8

    BWV

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    More basic question - when the paper refers to the RHS as "the Hilbert space equipped with distribution theory" can someone elaborate or provide a link to what the author is referring to as "distribution theory"?

    Also a link or explanation of "spectra" would be much appreciated. Ballentine refers to the term without (unless I missed it) ever really defining it
     
  10. Jul 25, 2011 #9

    Fredrik

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    Wikipedia is an OK place to start: http://en.wikipedia.org/wiki/Distribution_(mathematics).

    The spectrum of a bounded linear operator A is the set of all complex numbers λ such that A-λ is not invertible. Consider an eigenvalue equation Ax=λx, rewritten as (A-λ)x=0. If A-λ is invertible, we have x=0, so A doesn't have any eigenvectors. The spectrum is therefore a subset of the complex plane that contains all the eigenvalues. It can be shown to be a non-empty compact set, and to only contain real numbers when A is self-adjoint.

    There's a definition for unbounded operators too, but I don't remember it. Check out a book on functional analysis, e.g. "Functional analysis: spectral theory" by V.S. Sunder. It can be downloaded legally here.
     
  11. Jul 25, 2011 #10
    I haven't read these, but they're on my list of things to read:

    http://en.wikipedia.org/wiki/Distribution_(mathematics)
    http://en.wikipedia.org/wiki/Generalized_function

    Likewise, I need to learn more about the idea of spectra. As I currently understand it, the spectrum of a linear operator (linear function from a vector space to itself) is a generalization of the concept of eigenvalues. When the vector space is finite-dimensional, the spectrum of a linear operator is its set of eigenvalues. When the vector space is infinite-dimensional, the spectrum may include other kinds of values. There's a definition here:

    Functional Analysis for Quantum Mechanics (Tillmann Berg)
     
  12. Jul 25, 2011 #11

    BWV

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    Cool, thanks
     
  13. Jul 26, 2011 #12

    dextercioby

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    @all,

    Yes, I overlooked the necessity of building an inner product in the context of Lebesgue integration. I was thinking about the completion procedure where one passes from vectors in a pre-Hilbert space to equivalence classes, as well.
     
  14. Jul 26, 2011 #13

    Hurkyl

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    The whole thing might make more sense when given some more context.

    When we are considering negligible functions (functions that are zero almost except on a set of measure zero) to be actually zero, most equivalence classes of functions don't come with a good choice of representative -- it is intellectually more efficient to think of such things as equivalence classes of normal functions.

    But for the special case of continuous functions, the equivalence class does have a very, very good choice of representative. (the unique one that is continuous) In this case, is often intellectually more convenient to think not in terms of the equivalence class, but instead to think of this choice of representative.

    e.g. while x/x and 1 are not equal as (partial) functions, their difference is negligible, and so they are actually equivalent. Correspondingly, in calculus, when we divide x by x, we usually mentally substitute 1 as the result.
     
  15. Jul 26, 2011 #14

    Fredrik

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    I read section 2.3 where this quote appears. He claims to be talking about a Hilbert space, but the whole section is about the semi-inner product space [itex]\mathcal L^2(\mathbb R)[/itex] of square-integrable functions on ℝ, not about the Hilbert space [itex]L^2(\mathbb R)[/itex] of equivalence classes of square-integrable functions on ℝ.

    It's really weird that he uses a semi-inner product space instead of a Hilbert space to try to justify this claim.
     
  16. Jul 26, 2011 #15

    dextercioby

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    Fredrik, the semi-inner product finesse appears to be overlooked, but he is not really motivated to go into this detail. He's really interested in building the nuclear space, he doesn't care if there are equivalence classes wrt a.e. equality.
     
  17. Jul 28, 2011 #16
    φ=D(G) H=L₂(G) and φˣ=D`(G), where G is a nonempty open set of ℝⁿ with n≥1. D is short for the space of continuously differentiable functions with compact support, and D` is the space dual to D.

    REMARK:
    For u∈D, u:G→ℂ
     
    Last edited: Jul 28, 2011
  18. Jul 29, 2011 #17

    strangerep

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    By "semi-inner product space", I guess you mean a "partial inner product space". I.e., a space in which the inner product is not necessarily well-defined between all pairs of vectors?

    The Hilbert space [itex]\mathcal H[/itex] is meant to be the norm-completion of [itex]\mathcal \Phi[/itex]. The point about [itex]\mathcal \Phi[/itex] is that the usual quantum operators Q and P are well-defined everywhere on [itex]\mathcal \Phi[/itex], which is not generally true on [itex]\mathcal H[/itex].

    Since the inner product on [itex]\mathcal \Phi[/itex] is inherited from that on [itex]\mathcal H[/itex] (i.e., a Lebesgue integral), the usual caveats about equivalence of functions on sets of measure zero apply. But Hurkyl already resolved the question incisively when he said:

    The significance of this remark is that smoothness and continuity are critically important on [itex]\mathcal \Phi[/itex], because we want the P operator (derivative) to be well-defined everywhere.
     
  19. Jul 29, 2011 #18

    Fredrik

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    "Semi-inner product" is the term used in Conway's "A first course in functional analysis" for a sesquilinear form that satisfies all the properties of an inner product except <x,x>=0 only if x=0. The form <,> on the set of square-integrable functions, defined by [tex]\langle f,g\rangle=\int f(x)^*g(x)dx[/tex] for all f,g is a semi-inner product, because the 0 function isn't the only one that satisfies <f,f>=0. Consider e.g. the function f that takes 0 to 1 and everything else to 0. It also satisfies <f,f>=0.

    By "semi-inner product space", I mean a vector space equipped with a semi-inner product, like the vector space of square-integrable functions with the semi-inner product defined above. I don't know if the term is standard. I think the analysis books by Knapp used the term "pseudo-inner product". I'm less fond of that term. Because of how the term "pseudo-Riemannian metric" is used, I kind of expect a "pseudo-inner product" to not satisfy <x,x>≥0, but a semi-inner product does.

    [itex]\Phi[/itex] is supposed to be dense in [itex]\mathcal H[/itex], so [itex]\mathcal H[/itex] can also be described as the closure of [itex]\Phi[/itex]. I consider this description to be slightly simpler. But closure and completion is the same thing when the closure is a complete metric space, so it's certainly OK to describe [itex]\mathcal H[/itex] as the completion of [itex]\Phi[/itex].

    Yes, I believe it makes sense to say that [itex]\Phi[/itex] is the subspace consisting of equivalence classes [f] such that one of the members of [f] satisfies an appropriate "niceness" condition, like continuity or smoothness. But de la Madrid is still wrong to describe a subspace of the Hilbert space [itex]L^2(\mathbb R)[/itex] as not consisting of equivalence classes of functions.

    Edit: What I think he should have done, in order to stay as close as possible to the presentation he chose without saying anything that's wrong, is to start with the semi-inner product space of square-integrable functions, find an appropriate subspace M, and then convert this picture to the Hilbert space of equivalence classes of square-integrable functions, by defining [itex]\Phi[/itex] as the subspace of equivalence classes [f] such that f is in M.
     
    Last edited: Jul 29, 2011
  20. Jul 29, 2011 #19

    Hurkyl

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    At some point, it becomes more convenient to use the term "subspace" to refer to any injective map, rather than ones that are literally inclusions. I confess that I wouldn't have even blinked at a description like
    The space of rapidly decreasing functions is a subspace of [itex]L^2(\mathbb R)[/itex]​
    because of the obvious injective homomorphism from the former to the latter.


    Category theoretically, that's actually how the term "subobject" is defined -- a subobject of X is an (equivalence class) of monomorphisms whose codomain is X. (the details of the equivalence relation can easily be looked up, e.g. on Wikipedia)
     
  21. Jul 29, 2011 #20
    Let S denote the space of "rapidly decreasing functions", then (on top of that already mentioned in the o.p.) we have the following string of set inclusions:

    φ=D(G) ⊂ S ⊂ H=L₂(G) ⊂ φˣ=D`(G).
     
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