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Orthonormality of a complete set of eigenvectors

  1. Jan 22, 2014 #1
    hello

    How to you rigorously express the orthonormality of a complete set of eigenvectors [itex](|q\rangle)_q[/itex] of the position operator given that these are necessarily generalized eigenvectors (elements of the distribution space of a rigged hilbert space)?
    The usual unformal condition [itex]\langle q|q'\rangle=\delta(q-q')[/itex] does not make sense as inner product is not defined for a pair of these vectors.

    thank you
     
  2. jcsd
  3. Jan 22, 2014 #2

    ChrisVer

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    you have some eigenvectors, and you want an orthonormality relation between them.
    To do that, you have to define an inner product.
    Once you define the inner product, you want it to be equal to 1.
    Once you impose that, you'll get the delta of Kroenicker or Dirac's delta...
    Is there something unclear?
     
  4. Jan 22, 2014 #3

    George Jones

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    If you want to do this rigourously, you should look at rigged Hilbert spaces (also called Gelfand triples).
     
  5. Jan 22, 2014 #4

    dextercioby

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    Burakumin, the topological dual of a Hilbert space is a Hilbert space. That's where the whole rigged Hilbert space story (or if you prefer enchilada) starts.
     
  6. Jan 22, 2014 #5

    strangerep

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    The inner product is "extended" to be what you wrote (so-called "delta normalization"), in which case it becomes "distribution-valued" on those elements -- although such an extension does not necessarily encompass arbitrary elements of the distribution space. The primary purpose of this is so that arbitrary vectors in the small space (Schwartz space) can be expanded in terms of the ##|q\rangle##'s.

    In that sense, the "delta normalization" is already rigorous provided one recognizes the nature (and limitations) of this extended meaning of "inner product".
     
  7. Jan 23, 2014 #6
    As you might have noticed, my question already mentions rigged hilbert spaces. So yes I have already read docs about this topic (for example Quantum Mechanics beyond Hilbert space). But I haven't find anything about orthonormality in this context. Do you have references?

    Thanks strangerep. I can perfectly accept that a notion is extended so that it encompasses new situations. But this requires a general definition. Even if it might not be possible for all elements in the distribution space, it must at least be defined for a subset. And we agree that formula [itex]\langle q|q'\rangle=\delta(q-q')[/itex] is certainly not a definition, right? The only reference Google gives me for "distribution-valued inner product" is a discussion on physicsforums. "delta normalization" gives more results but the few I've read only seems to describe the formal aspect, sweeping the mathematical aspect under the carpet. Do you already know better (=mahematically rigorous) references?

    Thank you
     
  8. Jan 23, 2014 #7

    vanhees71

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    A good book, dealing with the formal aspects of quantum theory is

    A. Galindo and P. Pascual. Quantum Mechanics. Springer Verlag, Heidelberg, 1990. 2 Vols.
     
  9. Jan 23, 2014 #8
    Thanks vanhees71, but does it specifically deal with the orthonormality issue ?

    I've found this document : Dirac-orthogonality in the space of tempered distributions. At least its existence proves that the topic requires investigations.
     
  10. Jan 23, 2014 #9

    strangerep

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    Not exactly. I think you should switch to reading about the Schwartz theory of distributions, since this gives a more rigorous treatment of the underlying ideas.

    E.g., if the bra and ket are the functions ##e^{-iqx}## and ##e^{iq'x}## and ##\langle q|q'\rangle## is understood to mean:
    $$
    \int e^{-iqx} e^{iq'x} dx
    $$
    then this is a well-known tempered distribution known as the Dirac delta distribution.

    The Dirac bra-ket notation tends to obscure the fact that one is really working in a framework of dual pairs here, and not necessarily a true inner product. (For the finite-dimensional case, this distinction is often lost, of course.)

    Rigged Hilbert space can be regarded as a generalization of Schwartz distribution theory, so one should acquire proficiency in the latter first. I like Appendix A of Nussenzveig's book "Causality & Dispersion Relations", but this is more of an extended summary than an extensive rigorous treatment.

    Edit 1: Take those papers by David Carfi with a grain of salt. I don't think his other papers achieve what he claims.

    Edit 2: This paper by Rafael de la Madrid gives another introduction to RHS. Searching Google (Scholar?) should also lead you to his PhD thesis which contains a lot more detail. But... my advice is to become familiar with distribution theory first.
     
    Last edited: Jan 23, 2014
  11. Jan 23, 2014 #10

    George Jones

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    I don't have these books with me right now, and I can't remember if they do what you want, but, in the past, I wrote

     
  12. Jan 28, 2014 #11
    Ok, I'll try to find these references.

    Thanks but I'm not sure there's a problem on this aspect. I studied distribution theory far before I learned any basic concept of quantum mechanics (the former was part of my engineering student curriculum). And as far as I can remember there is no notion of orthogonality within the distribution space itself.

    I don't think this expression is valid in distribution theory. The fact that an expression looks meaningful does not prove it actually is. If we consider these two objects as functions, this integral does not converge. As distributions, this integral is meaningless because distribution theory does not define any scalar-valued product of two distributions. And I do not think you can define spaces where this precise pair can be interpreted as a distribution and a test function as neither is "nice enough" (rapidly decreasing) to be a test function.

    Now I was wondering if you were thinking of the product of distributions, the result of which is a distribution again. This product can sometimes be defined and it is a well know fact that it is impossible in general. But as Mr Carfi's article points it, we cannot be looking for the product of two distributions in this context. We must look for the product of two families (which is in fact already the case for the loose formula ⟨q|q′⟩=δ(q−q′) ). It is arguably a rather different concept from the standard inner product or even from the "duality" product of a distribution and a test function.

    Thank you for the warning but I don't intend to read anything else from Mr Carfi. So my only question is "do you consider this article wrong/invalid?" I may be wrong but his approach looks rigorous to me (but we agree formal appearance proves nothing).
     
  13. Jan 28, 2014 #12

    strangerep

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    OK, good. But did you also study the theory of Fourier transforms in the more general context of distributions? From your other remarks, it seems not.

    Yes -- that's sort-of what I've been saying. Although one can define a "Dirac delta normalization" for particular distributions, it fails to extend sensibly to the entire space of distributions.

    Then you need to study distribution theory at a more advanced level. The integral I gave above is ##2\pi \delta(q-q')##.

    It does have meaning as a distribution. (Sorry, but this is kinda basic.)

    (Sigh.) It has the properties of the Dirac delta distribution. Again, this is elementary.

    I know this, of course. But any regular function (of x, say) which grows at infinity no faster than a polynomial, (i.e., a so-called function of "slow growth"), qualifies as a distribution (since when integrated against a Schwartz function, the result is finite). Such functions can of course be multiplied, and the product also grows no faster than a polynomial, hence also qualifies as a distribution. However, if one takes their respective Fourier transforms (which are still distributions), these cannot be simply multiplied (but rather must be convolved).

    Do you have access to a University library? If so, try Appendix A of H.M. Nussenzveig's textbook "Causality and Dispersion Relations". It presents a lot of useful material on distributions quite efficiently.

    It's been a while since I studied them, and I'm disinclined to put any more time into that. The reason I used the phrase "grain of salt" earlier is that his treatment of generalized spectral theorems don't actually show that every operator on the space of distributions can be decomposed in the way he describes.
     
  14. Jan 30, 2014 #13
    Hi strangerep,

    I now understand more precisely what you are talking about. And all I can say is that IMHO you are mislead. I'm going to try to explain why.

    If I’m correct, you are interpreting the expression ##\int \exp(2 \pi i q x) \cdot \exp(-2 \pi i q' x) \cdot dx## as ##\mathcal{F}1(q'-q)## with ##\mathcal{F}## the Fourier transform and 1 the constant function equal to 1. This interpretation is, in a way, possible but you seem to forget that this is an abuse of notation. How do you define the Fourier transform of a tempered distribution ##T##? By showing that there is one and only one distribution (##\mathcal{F}T##) such that:

    ##\mathcal{F}T(\phi)= T(\mathcal{F}\phi)##

    for all rapidly decreasing functions ##\phi##. There exists no integral definition for it. Now we both agree that it is correct that if ##T## is regular (associated with the function I will call ##\hat{T}## even if generally we intentionally do not make the distinction) and if the function ##\hat{T}## has a fourier transform, ##\mathcal{F}T## is also a regular distribution, and is associated with the function ##t \mapsto \int \exp(-2 \pi i t x) \cdot \phi(x) \cdot dx##. This is not the case for function 1 !

    We tend to use abuses of notations for their evocative properties but we all know that they are always tricky somewhere. One may write :

    ##\delta(t) = \int \exp(-2 \pi i t x) \cdot 1 \cdot dx##

    but one should keep in mind that in distribution context, ##\int \exp(-2 \pi i t x) ... dx## is purely symbolic. There is no « actual » exponential function involved here. Just as we all know that ##\frac{\partial ...}{\partial ...}## is a single symbol and certainly not a fraction.

    Now if we leave Fourier transforms and choose some ##t##, what can we say about the function ##E_t : x \mapsto \exp(2 \pi i t x)##? It is not even a member of ##\mathcal{L}_2(\mathbb{R})## so it cannot be a test function. Can we interpret it as a tempered distribution ? We know that we can, as the distribution :

    ##E_t(\phi) = \int \exp(2 \pi i t x) \cdot \phi(x) \cdot dx##

    But this expression only makes sense if ##\phi## is rapidly decreasing function. So according to distribution theory, an expression like :

    ##E_q(x \mapsto \exp(-2 \pi i q’ x)) = \int \exp(2 \pi i t q) \cdot \exp(-2 \pi i q’ x) \cdot dx##

    just does not make sense and it has nothing to do with any Fourier transform. Similarly if one considers ##|q\rangle## to be an abstract generalized vector, distribution theory says ##\langle q|q’\rangle## is not a meaningful expression and Fourier transform is of no help here.

    I said and I repeat that if we want a definition of orthonormality we cannot rely on a particular family of generalized eigenvectors (unlike you do when considering only Dirac deltas or only complex exponentials). We must think of a undefinite family, potentially containing weird elements (for example Dirac combs or whatever you want) and define concepts that can apply to all families (or at least enough families). This is precisely what Mr Carvi is trying to do and (I suppose) why he has felt necessary to write an article about it. This article would be totally pointless if the problem could be solved as easily as you pretend, right? It might be, but I seriously doubt it and the previous fact should incite to caution.
     
  15. Jan 30, 2014 #14

    strangerep

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    That's not what I said. Hence no need to reply to most of the rest of your post.

    Again, you create here a straw man. I don't respond to straw men.
     
  16. Feb 3, 2014 #15
    I'm very sorry if I have misinterpreted your comments. That was not intentional. But in that case I sincerely do not understand your point.
    I do not have access to a university library but I could finally find this book and read the appendix, just in case I would have ignored some parts of the theory. It is true that I didn’t know of ##\mathcal{E}’##, the space of distribution with compact support and that Fourier transform can be expressed slightly more easily than in ##\mathcal{S}’##. But except that, it contains the standard framework of distribution theory and I cannot see how these usual concepts can be trivially used to express orthonormality, except again by identifying distinct object by means of abuses of notations.
    I may be too stupid to understand, who knows? But at least contrary to what you seem to pretend, I’m not dishonest.
     
  17. Feb 3, 2014 #16

    strangerep

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    You're doing the straw man thing again: you formed an incorrect extrapolated interpretation of something I said, and then you react to that interpretation as if it was fact. But in reality I never thought that you are dishonest.

    Take another look at the "structure" section of the Wikipedia link I gave, i.e., straw man.

    One way to avoid building straw men is, instead of reacting defensively to your interpretation of something I say, you could ask further questions to clarify what I said. This gives the conversation a better chance of proceeding constructively.

    In any case, this thread seems to be going nowhere, so if you wish to continue constructively, you could perhaps re-state clearly whatever questions still remain.
     
  18. Feb 4, 2014 #17
    I hope you will admit that it is difficult for me to "re-state clearly whatever questions still remain" as our point of disagreement is that you are pretending something is possible (orthonormality of a family of generalized eigenvectors can be stated within the distribution theory in an elementary way) and I’m pretending it is not. I cannot ask precise questions on your position as according to you I don't understand it.

    So we can try to restart from the very beginning and try to identify where we disagree. I’m a bit worried that we might not be on the same wavelength and that this conversation remains a long mutual misunderstanding, but at least I want to try. First I would like to expose general fact about the context so we can check we are speaking about the same thing. Tell me if you don't consider this Worth it and we will close the discussion.

    We have a hilbert space. As done before, let's consider it is ##\mathcal{L}_2(\mathbb{R})## for simplification. We have a pair of test function space and distribution space associated with it. I think ##(\mathcal{S},\mathcal{S}')=(\mathcal{S}( \mathbb{R} ),\mathcal{S}'( \mathbb{R} ))##, i.e. rapidly decreasing functions and tempered distributions, is the usual choice in this context but feel free to tell me if another pair is more appropriate. I propose, following Mr. Nussenzveig, to use notation ##(T,\phi)## for the duality product of the distribution ##T## with the test function ##\phi##.

    To my knowledge we do not have any inner product on ##\mathcal{S}'## (= function ##\mathcal{S}' \times \mathcal{S}' \mapsto \mathbb{C}## antilinear on first argument and linear on second) but we have some operations:
    • a partial product of distributions ##T\cdot U \in \mathcal{S}'## (partial because it might make no sense for arbitrary pairs of tempered distributions)
    • a partial convolution of distributions ##T * U \in \mathcal{S}'## (partial for the same reason)
    • a Fourier transform that is an isomorphism of ##\mathcal{S}'## and such that ##\mathcal{F}(T * U) = \mathcal{F}T \cdot \mathcal{F}U## and ##\mathcal{F}(T \cdot U) = \mathcal{F}T * \mathcal{F}U## whenever these products make sense
    • a tensor product of distributions ##T\otimes U \in \mathcal{S}'(\mathbb{R}^2)##

    Let’s suppose now that we are given a real-indexed family ##(T_q)_{q\in\mathbb{R}}## of elements of ##\mathcal{S}’##. To remain general I prefer not to suppose this family is a set of eigenvectors of some operator. What I am looking for is a way to extend the usual definition of orthonormality so that one can decide if ##(T_q)_{q\in\mathbb{R}}## is or is not orthonormal. A priori this does not necessarily requires that one must extend the notion of hilbert inner product but it is a possibility.

    I consider this issue to be non-trivial if it must be solved in a mathematically rigorous way. I also consider Mr. Carfi’s article to be a plausible manner of solving it. You claim that it is useless because there exists a trivial manner to solve it. Am I wrong somewhere on all previous statements?

    Now what I understand about your opinion (and that is certainly wrong but I do my best to understand) is that according to you:
    • there exists a sort of partial product between families such that ##(T_q)_{q\in\mathbb{R}}\cdot (U_{q’})_{q’\in\mathbb{R}}## is a distribution again (partial again because it might not exist for arbitrary pairs of families)
    • this operation is just a trivial consequence of previously listed operations
    • the value of ##(T_q)_{q\in\mathbb{R}}\cdot (T_{q’})_{q’\in\mathbb{R}}## can be used to confirm or disconfirm the orthonormality of ##(T_q)_{q\in\mathbb{R}}##.
    • it is what is implied by informal physicist equation ##\langle q|q’\rangle = \delta(q-q’)##
    Is there something misinterpretated here and where?
     
    Last edited: Feb 4, 2014
  19. Feb 4, 2014 #18

    strangerep

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    Yes (to all of the above).

    No. A tensor product would be in ##\mathcal{S}'(\mathbb{R}^2) \otimes \mathcal{S}'(\mathbb{R}^2)##, by definition.

    Well, this makes things more difficult, but let us continue. (BTW, since we're talking about the specific example of Schwartz space and tempered distributions, the usual operators of position and momentum are applicable here. But ok, let's proceed without mentioning those operators specifically.)

    I'm also surprised that you now wish to exclude this family being a set of eigenvectors of some operator, since your original post in this thread specifically asked about orthnormality of eigenvectors of the position operator.

    Noted.

    I also consider it non-trivial.

    No, I do not say that. In the Carfi article that you linked earlier, he only gets to "Dirac orthonormality" near the end, in section 5. That section contains a definition and a couple of examples, but no theorems or other results. Thus he does not "solve" anything in that section. Nevertheless, I have no problem with Carfi's definition 5.1, since that's nothing new -- it's just a more rigorous statement of the usual physicist's "Dirac-delta orthogonality" concept, (and which the framework of Rigged Hilbert Space generalizes to other sets of operators besides position and momentum).

    More precisely, I say that there might exist a partial product on such families. Carfi's examples show such a case. (If one re-admits operators into the discussion, and the family constitutes a set of generalized eigenvectors of suitably-extended self-adjoint operators, then one could say more. This is the content of the Gelfan-Maurin nuclear spectral theorem.)

    No, I do not claim that.

    I do not claim that in the context of this discussion, since you have excluded discussion of operators.

    I do not claim that in the context of this discussion, since you have excluded discussion of operators.

    If you want to know what I do think about the concept of Dirac-delta orthogonality, I refer you to this paper on the RHS in QM by Rafael de la Madrid, specifically eqns(2.28a-d), and the context leading up to them. If you do a search for his other works on Google Scholar, you can also find his PhD Thesis in which these things are discussed more extensively.
     
  20. Feb 6, 2014 #19
    Would I be misinterpreting you again if I suggested that there is a typo and that you certainly mean ##\mathcal{S}'( \mathbb{R})\otimes \mathcal{S}'( \mathbb{R})## ? That said, to my knowledge ##\mathcal{S}'( \mathbb{R})\otimes \mathcal{S}'( \mathbb{R})## canonically injects in ##\mathcal{S}'( \mathbb{R}^2)## using injection ##i## such that ##(i(T\otimes U), \chi) = (T, x \mapsto (U, y\mapsto \chi(x,y)))## with ##\chi\in\mathcal{S}( \mathbb{R}^2)##. But I guess tensor product of distributions might not be the most important tool for this discussion anyway.
    Actually this was intended because I precisely don’t want that we focus on a specific operator. My initial question refered to the position operator but that was just an example and I never intended to focus on it. I was looking for a general definition and to me [itex]\langle q|q'\rangle=\delta(q-q')[/itex] is just an example. So your answer
    was (and is still) pretty confusing to me.

    If needed it is possible to consider an abstract operator ##X## (densily defined and (essentially) self-adjoint) and the set ##B_X## of all possible complete families of eigenvectors. I suppose that the way to go may be to define the distribution-valued product ##(T_q)_{q\in\mathbb{R}}\cdot (T_{q’})_{q’\in\mathbb{R}}## at least for some families ##(T_q)_{q\in\mathbb{R}} \in B_X## and among them to select the orthonormal ones based on the result of this product. What is still not clear to me is how to define the product in the general case (at least as often as it can makes sense) and what value is expected for orthonormal families.

    I’m not sure to understand how these equations could be an answer. From what I understand after a glance at Mr de la Madrid’s article, it looks more like « we would like objects that formally behave like this » rather than defining them. That said, I’ve downloaded and started to read his phd. His mathematical focus and care for rigor have been very enjoyable so far so I plan to read it entirely. I guess it’s better that I come back with precise questions after I’ve finished.
     
  21. Feb 6, 2014 #20

    strangerep

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    Oops. Yes, it was a typo.

    Yes. It would be better to discuss the content of specific references, rather than my attempted answers that are necessarily brief and incomplete.

    BTW, if you want more completeness and rigor, try Gel'fand & Vilenkin, vol 4.
     
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