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Rigged Hilbert space, separable space, domain of CSCO, mapping

  1. Jan 30, 2013 #1
    Suppose that we have rigged Gilbert space Ω[itex]\subset[/itex]H[itex]\subset[/itex]Ω[itex]\times[/itex] (H is infinite-dimensional and separable).
    1. Is the Ω a separable space?
    2. Is the Ω[itex]\times[/itex] a separable space?
    3. Consider the complete set of commuting observables (CSCO) which contain both bounded and unbounded operators. Eigenvectors of CSCO span the space of states. Is this space of states Ω or H?
    4. I know that unbounded closed operators have domain Ω and map Ω into Ω. And what about bounded (compact) operators: Ω into Ω, Ω into H or somewhat else?
    5. In what space do CSCO form algebra?
    Thanks
     
  2. jcsd
  3. Jan 30, 2013 #2

    dextercioby

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    In short.

    3. Ωx.
    4. H into H, because a compact operator is bounded everywhere hence continuous, so it can be uniquely extended to all vectors of H.
    5. Both Ω and Ωx, but for different mappings of course. The (anti)dual mappings in case of Ωx.

    For 1. and 2. I'll have to check.
     
  4. Jan 30, 2013 #3
    3. You say Ω[itex]\times[/itex]. It means that I can decompose any vector from Ω[itex]\times[/itex] over the eigenvectors of CSCO. Is this true for vectors from H\Ω? I don't think so, because nuclear spectral theorem allow us to decompose only vectors from Ω over the eigenvectors of unbounded operator (suppose CSCO contain unbounded operator).
    4-5. Ok. Bounded operators have domain H, and unbounded have domain Ω. I am not sure but if they form CSCO algebra they must have one common invariant domain.
     
  5. Jan 30, 2013 #4

    dextercioby

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    3. Yes I agree with your last statement, vectors from H\Ω bear no significance to physics and are not interesting mathematically.
    4-5. What are you not sure about ?
     
  6. Jan 30, 2013 #5
    3. So, answer is Ω?
    4-5. Operators forming CSCO algebra must have one common invariant domain. You said both Ω and Ω[itex]\times[/itex]. Which of them? I think must be one domain.
     
  7. Jan 30, 2013 #6

    dextercioby

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    3. The eigenvectors must necessarily be in the topological (anti)dual space, if there's at least an operator in the CSCO with (partially/totally) continuos spectrum.
    4-5. The CSCO has Ω for all the operators as a common (dense everwhere in the Hilbert space topology) domain. However, bringing in the commutation condition for these operators, we are 'thown' again in the Ωx, because the commutativity is realized through generalized projectors onto spectral subspaces of Ωx. Again, the need to go to topological (anti)duals stems from the 'continuous spectrum' condition.

    4-5. Example: For the 3D free Galilean particle, x, y, z actually form a set/algebra of operators for which any commutator of any 2 operators vanishes strongly. However, x,y,z have generalized eigenvectors (Dirac distributions from Sx(R3)), so they achive the CSCO status only when 'promoted' as mappings in the distributional space, as per the definition of a CSCO.
     
  8. Jan 30, 2013 #7
    3. Ok, I am aware that eigenvectors of unbounded (continuous, continuous+discrete spectrum) lie in Ω[itex]\times[/itex]. But I think that only vectors from Ω can be decomposed over the CSCO basis.

    Thanks, I'm waiting for 1, 2 )
     
  9. Jan 30, 2013 #8

    strangerep

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    A "Gilbert" space, huh? :-)
    From Wikipedia, i.e., http://en.wikipedia.org/wiki/Separable_space ....
    So: 1: Yes.

    As for 2: I can think of cases in which ##\Omega^\times## has a countable dense set. Not sure if that's true in general, though.
    I believe that vectors from ##\Omega^\times## can also be decomposed, else Fourier transforms wouldn't be anywhere near as useful as they are...
     
  10. Jan 31, 2013 #9
    Sorry for "Gilbert". I am not so good in English ).

    1. I agree. Thanks. I must be considerate towards "wikipedia" ).

    2 and 3 are still actual.
     
  11. Jan 31, 2013 #10

    strangerep

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    Hmm. I thought 3 was already answered.

    As for (2), you might get answers from better-informed people if you ask in the mathematics forums (i.e., functional analysis & topology).

    And BTW, it would help if you tell us which references on rigged Hilbert space you have already studied... (?)

    But let's go back... Question 3 was:
    Summarizing Antoine in "Quantum Mechanics beyond Hilbert space" ...

    We assume ##\Omega## is complete with respect to a locally convex topology ##\tau_\Omega## on ##\Omega## (i.e., a finer topology than the norm topology inherited from ##H##).

    ##\Omega^\times## is the space of ##\tau_\Omega##--continuous antilinear functionals on ##\Omega##.

    We also assume that ##\Omega## is reflexive (with some technical conditions).

    In most cases, ##\Omega## is the intersection of a countable family of Hilbert spaces. That makes it a Frechet space, according to Antoine.

    We also assume that ##\Omega## is "nuclear" which is a technical condition about the respective embeddings of members of the countable family of Hilbert spaces mentioned above.

    Then if ##A## is a closed operator in ##H##, which maps ##\Omega## into itself (continuously,... blah blah blah),.... then ##A## may be extended by duality to a linear operator ##A^\times : \Omega^\times \to \Omega^\times## which is an extension of the usual adjoint ##A^*## ...

    If ##A## is self-adjoint and ##\Omega## is nuclear and complete, the nuclear spectral theorem asserts that ##A## possesses a complete orthonormal set of generalized eigenvectors ##\xi_\lambda \in \Omega^\times##, which means that for any ##\phi,\psi \in \Omega##, we have
    $$
    \def\<{\langle}
    \def\>{\rangle}
    \<\phi|\psi\> ~=~ \int_R \<\phi|\xi_\lambda\> \; \<\xi_\lambda|\psi\> \; d\mu(\lambda)
    $$
    So... I can see why you ask whether only vectors from ##\Omega## can be decomposed in terms of the ##\xi_\lambda##. But every element (functional) of ##\Omega^\times## is only defined in terms of its action on elements of ##\Omega##. ##\Omega^\times## was constructed in terms of a particular operator (or operators), and it's the spectrum of those operator(s) and the associated generalized eigenvectors which matter.

    Hence I believe the answer to 3 is that elements of ##\Omega^\times## can also be decomposed in terms of the generalized eigenvectors of such an operator ##A##.

    [WARNING: Although I have studied these subjects, I am not an expert. ]
     
  12. Jan 31, 2013 #11
    OK, but [itex]\Omega[/itex] is not a subspace of H. The point is that [itex]\Omega[/itex] has a finer topology than the subspace topology. So all we demand is that [itex]\Omega\rightarrow H[/itex] is a injective bounded operator. We don't demand subspace topology.

    I think the answer to 1 is "no", but I can't seem to find a good counterexample. Every example of rigged Hilbert spaces seem to be separable.
     
  13. Jan 31, 2013 #12
    I don't think that reflexive is part of the definition, is it?? Anyway, if it is reflexive, then [itex]\Omega[/itex] is separable if and only if [itex]\Omega^\times[/itex] is separable. So the answer to (1) and (2) are equivalent in that case.
     
  14. Jan 31, 2013 #13
    The triple [itex]\ell^1\subset \ell^2 \subset \ell^\infty[/itex] is a counterexample to (2).
     
  15. Jan 31, 2013 #14
    And it seems to me that [itex]L^\infty([0,1])\subset L^2([0,1])\subset (L^\infty([0,1]))^\times[/itex] is a counterexample to (1).
     
  16. Jan 31, 2013 #15

    strangerep

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    Ah, excellent! Micromass is getting involved. I might actually learn something... :eek:

    OK, thanks. I meant "subspace" in the "sub vector space" sense, which was sloppy terminology.

    Yes, QM people don't like nonseparable spaces since they're more difficult to work with.

    Well, it's certainly part of Antoine's definition, but perhaps different authors use other conventions? Which author(s) did you have in mind?

    Certainly, these spaces in physics are usually (always?) assumed to be reflexive.

    Ah, of course. Well, that certainly answers those questions if one restricts to the RHS's typically used in QM.

    Could you expand on both of those points, pls? I've always had difficulty coming to grips properly with stuff involving ##\ell^\infty## and ##L^\infty## . :uhh:
     
  17. Jan 31, 2013 #16
    Ah ok, I just looked "rigged Hilbert space" up in wikipedia. There they don't ask anything about being reflexive.

    I'll try to expand on the L-spaces in my next post.
     
  18. Jan 31, 2013 #17
    So, to my understanding (from wikipedia), we are looking for a couple [itex](\Phi,H)[/itex], where H is a Hilbert space and [itex]\Phi[/itex] is a dense subset of H. Furthermore, we equip [itex]\Phi[/itex] with a topology (which is unrelated to the subspace topology of H).

    We must have a map [itex]i:\Phi\rightarrow H[/itex] which is injective and a bounded linear operator. This has to a map [itex]i^*:H=H^*\rightarrow \Phi^*[/itex], with [itex](i^*|v>)(|w>)=<v|w>[/itex].

    So, let's define the real (complex works too) vector space

    [tex]\ell^2=\{(x_n)_n~\vert~\sum|x_n|^2<+\infty\}[/tex]

    We can equip this space with a natural inner product

    [tex]<(x_n)_n,(y_n)_n>=\sum x_ny_n[/tex]

    This is well-defined. Indeed, recall the standard inequality [itex]ab\leq \frac{1}{2}(a^2+b^2)[/itex], this implies

    [tex]|<(x_n)_n,(y_n)_n>|=|\sum x_ny_n| \leq \frac{1}{2}(\sum x_n^2 + \sum y_n^2)[/tex]

    the last two series converge by definition of [itex]\ell^2[/itex]. So the inner product is finite. Checking the properties of the inner product is trivial.

    It is well-known that any inner product defines a norm. In this case, the norm on [itex]\ell^2[/itex] is

    [tex]\|(x_n)_n\|_2=\sqrt{\sum |x_n|^2}[/tex]

    The space [itex]\ell^2[/itex] is complete and thus defines a Hilbert space.

    We can also define

    [tex]\ell^1=\{(x_n)_n~\vert~\sum |x_n|<+\infty\}[/tex]

    This will not turn out to be a Hilbert space, as there is no natural inner product. But we can make a norm by

    [tex]\|(x_n)_n\|_1=\sum |x_n|[/tex]

    This turns out to be a complete space.

    It is true that [itex]\ell^1\subset \ell^2[/itex] (as subsets). And we can prove that [itex]\ell^1[/itex] is dense in [itex]\ell^2[/itex]. Furthermore, we can prove that

    [tex]\|(x_n)_n\|_2\leq \|(x_n)_n\|_1[/tex]

    So the injection [itex]i:\ell^1\rightarrow \ell^2[/itex] is bounded with norm (at most) 1.

    Now, what is the dual of [itex]\ell^1[/itex]. It turns out that it is [itex]\ell^\infty[/itex]. This is defined by

    [tex]\ell^\infty=\{(x_n)_n~\vert~\sup |x_n|<+\infty\}[/tex]

    This is again not a Hilbert space, but we do have a canonical norm

    [tex]\|(x_n)_n\|_\infty=\sup |x_n|[/tex]

    It turns out that [itex]\ell^\infty[/itex] is complete. However, we can prove that [itex]\ell^\infty[/itex] is not separable. The proof of this is essentially some kind of diagonal argument.

    Every element of [itex]\ell^\infty[/itex] induces some bounded functional in [itex](\ell^1)^\times[/itex]. Indeed, if [itex](x_n)_n\in \ell^\infty[/itex], then we define

    [tex]T_{(x_n)_n}:\ell^1\rightarrow \mathbb{R}:(y_n)_n\rightarrow \sum x_ny_n[/tex]

    The map [itex](x_n)_n\rightarrow t_{(x_n)_n}[/itex] defines an isometric bijection between [itex]\ell^\infty[/itex] and [itex](\ell^1)^\times[/itex].

    It is clear that [itex]\ell^2\subset \ell^\infty[/itex]. We even have

    [tex]\|(x_n)_n\|_\infty\leq \|(x_n)_n\|_2[/tex]

    So the injection [itex]\ell^2\rightarrow \ell^\infty[/itex] is bounded with norm (at most) 1.

    Now, if [itex](x_n)_n\in \ell^2[/itex], then [itex]T_{(x_n)_n}[/itex] is in [itex](\ell^1)^\times[/itex]. Furthermore, we have [itex]T_{(x_n)_n}((y_n)_n)=\sum x_ny_n = <(x_n)_n,(y_n)_n>[/itex]. So we are indeed dealing with a rigged Hilbert space.
     
  19. Jan 31, 2013 #18

    strangerep

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    The Wiki page on "rigged Hilbert space" is sadly minimal. I've been wishing for several years for an expert to flesh it out some more.

    And the "best" textbook I know of, i.e., Gel'fand & Vilenkin vol 4, is quite difficult for the average physics type, including me. It's also growing a bit old.

    Cheers.
     
  20. Jan 31, 2013 #19
    Hmm, I see. There is indeed more to a rigged Hilbert space than the wiki page shows. In particular, we want [itex]\Phi[/itex] to be reflexive. This implies that my example of the [itex]\ell^2[/itex]-space would be wrong, since [itex]\ell^1[/itex] is not reflexive.

    I don't have access to Antoine, sadly, but I'll look on the internet for a more complete definition.
     
  21. Jan 31, 2013 #20

    strangerep

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    Are you sure? Try Google Scholar for Antoine "beyond hilbert space". You should be able to get a pdf from citeseer, or a ps from some other places.

    I'll also dig out some other references and post them shortly.

    Dunno why you say that. You always make me happier when you drop by... :smile:
     
  22. Jan 31, 2013 #21

    strangerep

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    1) Gadella, Gomez,
    "On the Mathematical Basis of the Dirac Formulation of Quantum Mechanics",
    IJTP, vol 42, no 10, 2003, p2225.
    (If you can't access that, send me a PM... ;-)

    2) Gadella, Gomez,
    "Eigenfunction Expansions and Transformation Theory",
    Available as: math/0607548

    and of course, there's always the reliable, but less rigorous:

    3) Rafael de la Madrid,
    "The role of the rigged Hilbert space in Quantum Mechanics",
    Available as: quant-ph/0502053

    Ref (3) is more physicist-oriented, but it does give a more intuitive picture of how the small space ##\Omega## is constructed in terms of an initial Hilbert space, one or more unbounded operators with continuous spectra, and the sequence of seminorms (and hence nested Hilbert spaces) induced by the latter.

    [And now I need to go over your post #17 again more closely... :smile:]
     
  23. Feb 1, 2013 #22

    strangerep

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    So this corresponds to what others call the "canonical embedding map" ?
    So this means ##i## is also a Hilbert-Schmidt operator ?
    What do you mean by "diagonal argument" ?
     
  24. Feb 1, 2013 #23

    strangerep

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    In cases of physical interest, one usually performs the construction in a different sequence...

    For a quantum theory, we must have a space with (+ve-definite) inner product which admits a probability measure. This is the Hilbert space ##H_0## (possibly after a completion in the norm induced by the inner product). The meaning of the "0" subscript will become clear below.

    In a useful physical theory of dynamics, we essentially start from differential equations (of motion), and determine the maximal invariance group of this (system of) differential equation(s). This is the "dynamical group" -- it maps solutions of the dynamics among themselves. Depending on the degree of sophistication in the model, we might be content with restriction to a subgroup, rather than the maximal group.

    So for a quantum-dynamical theory, this group must be represented as operators on the Hilbert space [note #1]. One might use group operators (i.e., in exponential form which are bounded), but more usually one finds the generators of the group and attempts to represent these on the Hilbert space. Typically, these generators are unbounded operators with continuous spectrum, hence cannot be defined everywhere on ##H_0##.

    Take a typical such generator ##P## (ordinary momentum), represented by (e.g.,) ##-i\hbar\partial_x## on ##H_0##, being a Hilbert space of square-integrable wave functions. Now define
    $$
    H_1 ~:=~ \left\{ \psi\in H_0 : P\psi \in H_0 \right\} ~.
    $$ Clearly ##H_1 \subset H_0##. Further, if the norm on ##H_0## is denoted ##\|\psi\|_0##, we can define another norm on ##H_1## :
    $$
    \|\psi\|_1 ~:=~ \|\psi\|_0 + \|P\psi\|_0 ~,
    $$ so that ##\|\psi\|_1 \le \|\psi\|_0##. Clearly, both norms ##\|\cdot\|_0## and ##\|\cdot\|_1## are well-defined on ##H_1## and one can define a new topology ##\tau_1## on ##H_1## via the 1-norm.

    Continuing in this way we can define a sequence of nested spaces, e.g.,
    $$
    H_{n+1} ~:=~ \left\{ \psi\in H_n : P\psi \in H_n \right\} ~,
    $$ and corresponding ##(n+1)##--norms similarly. Thus, one has an infinite sequence of nested spaces with a corresponding sequence of seminorms and topologies. Gel'fand & Vilenkin call this setup a "countably-Hilbert" space. Then, one puts
    $$
    \Omega ~:=~ \bigcap_{n=0}^\infty H_n ~,
    $$ thus arriving at the "small" space in the triple (which hopefully is not trivially empty).

    Anyway,... the point of posting all this is because I'm not sure whether the extra structure implied by the above might affect what you've been saying about duals...

    [Edit: I also think this construction guarantees that each canonical embedding map from ##H_{n+1}## into ##H_n## is nuclear (Hilbert-Schmidt). It appears (eg Wiki) that "nuclear" requires only trace-class embedding operators, but the nuclear spectral theorem seems to require Hilbert-Schmidt embedding operators. Can you shed any more light on this?]


    [Note #1: Algebraic quantum theory handles this differently: one constructs abstract functionals over a ##C^*##--algebra related to the dynamical group.]
     
    Last edited: Feb 1, 2013
  25. Feb 1, 2013 #24
    Yes.

    The only Hilbert-Schmidt operators I ever encountered were of the form [itex]T:H\rightarrow H[/itex] between Hilbert spaces. I don't see how i is of this form. But I'm sure that generalizations exist.

    The idea is the following. Assume that there is a countably dense subset
    [tex]\{\mathbf{x}_1,\mathbf{x}_2,...\}[/tex] of [itex]\ell^\infty[/itex]. Then we can write this as

    [tex]\mathbf{x}_1: ~x_1^1 ~ x_1^2~ x_1^3 ~ ...[/tex]
    [tex]\mathbf{x}_2: ~x_2^1 ~ x_2^2~ x_2^3 ~ ...[/tex]
    [tex]\mathbf{x}_3: ~x_3^1 ~ x_3^2~ x_3^3 ~ ...[/tex]
    and so on
    But consider the following diagonal sequence: [itex](x_n^n +1)_n[/itex]. This is an element of [itex]\ell^\infty[/itex], but it has distance at least 1 from any of the [itex]\mathbf{x}_k[/itex]. So our set isn't dense.

    I have to admit that this entire post by you is very fascinating. But unfortunately, I am very much out of my comfort zone here, so I can't really help. Gelfand & Vilenkin looks a really nice book, so I will go through it. Maybe I can answer some more after I have read up on the book and the other references you posted.
     
  26. Feb 1, 2013 #25
    Gelfand-Vilenkin seem to define a rigged Hilbert space as a nuclear countably Hilbert space [itex]\Phi[/itex] together with a continuous, injective, linear map [itex]T:\Phi\rightarrow H[/itex] into a Hilbert space.

    However, since [itex]\Phi[/itex] is nuclear countably Hilbert, it is separable. This seems very much nontrivial, but Gelfand and Vilenkin do prove it (in an earlier volume). Furthermore, any nuclear countably Hilbert space is reflexive. This means that [itex]\Phi^\times[/itex] is separable as well.

    So if we use the definition from Gelfand-Vilenkin, then we can answer (1) and (2) with yes.
     
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