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Completeness of eigenvectors in a complete, commuting set

  1. Dec 24, 2011 #1
    "Completeness" of eigenvectors in a complete, commuting set

    Hi guys, I asked this question on Physics Stack Exchange many days ago and even after substantial discussions and revisions, this has remain unanswered. This is:
    http://physics.stackexchange.com/qu...igenvectors-in-a-complete-commuting-set/18272

    It seems quite fundamental that I want to know what's actually going on. Is Dirac's generalization argument incomplete? Or is there a problem in my concepts of mathematics of ket spaces?

    The discussion on this can go on in this thread as it is usually discouraged on Stack Exchange.

    Hope to find substantial insight into it here.
     
  2. jcsd
  3. Dec 24, 2011 #2

    Hurkyl

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    Re: "Completeness" of eigenvectors in a complete, commuting set

    As you were answered, there isn't a |c,d> in the vector space you describe in the first part of your post.

    For your second question, I don't know the argument you're referring to, so I can't help. But I suspect at least one of these two comments would be relevant:

    • The operators he's considering don't have the analog of repeated eigenvalues. (e.g. a continuous spectrum)
    • Each variable of integration independently parametrizes the entire spectrum
     
  4. Dec 25, 2011 #3
    Re: "Completeness" of eigenvectors in a complete, commuting set

    If you go through the entire spectrum of comments, we settled on the issue that there is no good reason for |c,d> to occur. Why can't it be missing? All these developments are clearly indicated in the edited question which I posted above solid line.

    There is no second question. I don't know where you found it. The question was edited after all the discussion related to first answer and that was then posted above solid line and original question is below solid line. I have mentioned it all in my question. Please read it.
     
  5. Dec 26, 2011 #4
    Re: "Completeness" of eigenvectors in a complete, commuting set

    Hello,

    Maybe a small example will show that |c,d> doesn't have to exist, as claimed by Hurkyl. Define

    [itex] A = \left( \begin{array}{ccc}
    a & 0 & 0 \\
    0 & a & 0 \\
    0 & 0 & c
    \end{array} \right) [/itex] and [itex] B = \left( \begin{array}{ccc}
    b & 0 & 0 \\
    0 & d & 0 \\
    0 & 0 & b
    \end{array} \right) [/itex]

    It is then clear that [itex]|a,b \rangle = \left( \begin{array}{} 1 \\ 0 \\ 0 \end{array} \right)[/itex], [itex]|a,d \rangle = \left( \begin{array}{} 0 \\ 1 \\ 0 \end{array} \right)[/itex] and [itex]|c,b \rangle = \left( \begin{array}{} 0 \\ 0 \\ 1 \end{array} \right)[/itex].

    Also note that no vector corresponds to [itex]|c,d \rangle [/itex].

    As for your continuation
    I can't help, as I'm not yet familiar with what you're describing. (I think that Hurkyl referred to this as your "second question".)
     
  6. Dec 27, 2011 #5
    Re: "Completeness" of eigenvectors in a complete, commuting set

    Exactly that's what I am trying to point out there. That there need not be a |c,d>. So, if |c,d> need not occur. Now consider this:
    There is a complete, commuting set of 'r' observables all of which have only continuous eigenvalues and no discrete eigenvalues. Now, it can happen that an eigenvector of the form [itex] |\xi_{1},\xi_{2}.....\xi_{r}\rangle [/itex] may be "missing" where the different [itex] \xi_{1},\xi_{2}.....\xi_{r} [/itex] are the corresponding eigenvalues of the 'r' observables (these eigenvalues are well-defined in the sense that each lies in corresponding observable's domain). In such a case you can't consider an r-fold integral of a function of these 'r' eigenvalues with the integrals being respect to different eigenvalues. I claim the previous statement as the limits of integration are not quite independent because for some value [itex] \xi_{i} [/itex] of 'i'th eigenvalue the limits to which other eigenvalues can be integrated become a function of this 'i'th eigenvalue as noted in the |c,d> example above. Although the example is of discrete case, it well conveys the idea of what I am trying to say.

    But, Dirac has used this integral in his book and I suspect that this integral is not well-defined by the argument in the previous paragraph. On the other hand, it's hard for me to believe that Dirac would err on this trivial looking thing and it would go unnoticed till date. Hence, I am looking for an argument to make this integral well-defined if I have not made any mistakes in the argument above. If I have made a mistake, please point it out. Thanks. :)

     
  7. Dec 27, 2011 #6

    strangerep

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    Re: "Completeness" of eigenvectors in a complete, commuting set

    I don't really understand what the starting point is here. Having found that there's no [itex]|c,d\rangle[/itex] in a 3D space, we can simply go to a 4D space, e.g.,
    [tex]
    A = \left( \begin{array}{cccc}
    a & 0 & 0 & 0 \\
    0 & a & 0 & 0 \\
    0 & 0 & c & 0 \\
    0 & 0 & 0 & c
    \end{array} \right) ~~~~\mbox{and}~~~
    B = \left( \begin{array}{ccc}
    b & 0 & 0 & 0 \\
    0 & d & 0 & 0 \\
    0 & 0 & b & 0 \\
    0 & 0 & 0 & d
    \end{array} \right)
    [/tex]
    Then
    [tex]
    \def\<{\langle}
    \def\>{\rangle}
    |a,b \> = \left( \begin{array}{} 1 \\ 0 \\ 0 \\ 0 \end{array} \right) ~,~~~~
    |a,d \> = \left( \begin{array}{} 0 \\ 1 \\ 0 \\ 0 \end{array} \right) ~,~~~~
    |c,b \> = \left( \begin{array}{} 0 \\ 0 \\ 1 \\ 0 \end{array} \right) ~,~~~~
    |c,d \> = \left( \begin{array}{} 0 \\ 0 \\ 0 \\ 1 \end{array} \right)
    [/tex]
    (or did I miss something?)
     
  8. Dec 27, 2011 #7
    Re: "Completeness" of eigenvectors in a complete, commuting set

    @strangerep: Indeed you're missing the point. Lakshya wants a case where |c,d> doesn't exist. Why? To ask if Dirac's integral (as described by Lakshya; I'm unfamiliar with it myself) is still well-defined in such a case.
     
  9. Dec 28, 2011 #8
    Re: "Completeness" of eigenvectors in a complete, commuting set

    The scenario he has in mind is: we have an observable A with eigenvalues a and c. We have another observable B with eigenvalues b and d. A and B are commuting observables. Normally you'd say that the eigenstates of the system would be {[itex]|ab\rangle, |ad\rangle, |cb\rangle, |cd\rangle[/itex]}. i.e. the state space is 4 dimensional. BUT, he has a constraint - when the eigenvalue of A is c, then the eigenvalue of B MUST be b, it is not allowed to be d. In this way he has reduced the dimensionality of the state space from 4 to 3.

    My question would be - is this allowed ? Mathematically, it's OK as mr Vodka has explicitly shown. His A and B are commuting and the state space is still only 3 dimensional.

    But from the point of view of physics rather than mathematics - according to the postulates of quantum mechanics, if A and B are commuting operators, doesn't it mean that we MUST be able to prepare the system with all allowed eigenvalues of each operator? So if I prepare it with the value c for A, I can take the resulting state and prepare it with the value d for B without screwing up the c eigenvalue (because A and B commute).
     
  10. Dec 30, 2011 #9

    strangerep

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    Re: "Completeness" of eigenvectors in a complete, commuting set

    Er,... guys,... none of the above has anything to do with what's happening on Dirac p65 -- which was the central point of the original question over on TPSE.

    For those who don't have Dirac's textbook at hand, he's simply constructing a tensor product space from the individual Hilbert spaces, each spanned by the eigenvectors of one observable. This is standard QM, though perhaps presented in Dirac's sometimes-unorthodox style.

    Phrased in different words...

    Suppose we have 2 observable physical quantities A & B and we determine their spectrum by experimental measurements. For argument's sake, let's say we always find that, over a large range of prepared system states, measuring A yields either a or c, while measuring B yields either b or d, and that we also find that it makes no difference whether we measure A then B or B then A. Hence we conclude that A & B are independent observables.

    To construct a QM description of this system, the standard procedure is to take one Hilbert space corresponding to span{A's eigenvectors} and another corresponding to span{B's eigenvectors}, and form the tensor product Hilbert space. This construction ensures that [itex]A\otimes 1[/itex] and [itex]1\otimes B[/itex] commute as operators on the tensor product space -- faithfully reproducing the physical property that these quantities are physically independent. This tensor product space is isomorphic to the 4D space I mentioned in my previous post.

    Dirac p65 simply extends this construction to the case of several independent quantities whose spectra are all continuous.

    Afaict, all the confusion about "missing eigenvalues" was just confusion about how tensor product spaces work in QM, and how they correspond to physical cases. There ain't no "missing eigenvalues/eigenvectors" in the construction.
     
    Last edited: Dec 30, 2011
  11. Dec 30, 2011 #10
    Re: "Completeness" of eigenvectors in a complete, commuting set

    If that's the case, then yes I agree there isn't an issue. My knowledge of quantum mechanics is VERY rusty, but I remember that when you want to construct the Hilbert space which represents a combined system consisting of two subsystems, then you take the tensor product of the two subsystem Hilbert spaces as you describe. But here we're looking at constructing the Hilbert space of a single system, and labelling its basis vectors with eigenvalues of a complete set of commuting operators. Is it true in this case that the Hilbert space must be the tensor product of the Hilbert spaces corresponding to each of the operators in the set, i.e. there cannot be constraints as Lakshya is descibing ?

    It certainly sounds reasonable, but I had a quick google around and couldn't find this in any of the descriptions of the postulates of quantum mechanics. Do you have a reference where it is stated in this way ?
     
  12. Jan 1, 2012 #11

    strangerep

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    Re: "Completeness" of eigenvectors in a complete, commuting set

    I, too, had to think about this a bit when I read Dirac's stuff.

    Consider what characterizes a single instance of a (class of) system. The algebra of observables, right? One then demands that the algebra be represented unitarily and eventually we derive the spectra of the observables -- or spectra in certain combinations of observables (such as [itex]J^2[/itex] and [itex]J_z[/itex] in the standard quantum theory of angular momentum).

    Dirac simply decomposes this idea down to its basics: figure out the joint spectra for the observables in a maximal set of commuting quantities in the (enveloping) algebra. (This set corresponds to a maximal set of mutually-independent dynamical variables.) Then deduce the size of the necessary Hilbert space from the details of the joint spectra. He does this by tensoring the spaces of the eigenvectors for each observable in the maximal commuting set. Finally we must check that the entire algebra is indeed representable on this space.

    I think this is really cool -- one builds up a model of a single instance of a (class of) system by tensoring some basic 1D spaces together, then we extend this idea to multiple copies of the system by tensoring again -- like you said.

    I believe so, or else it's not QM, or else it's the wrong Lie algebra for the system being studied.

    If I'm wrong, I hope someone will give a physically realistic counterexample.

    I haven't seen it done this way anywhere else except in Dirac's book -- so the latter is the only reference I know. The group-theoretic approach in (say) Ballentine's textbook is kinda related, but he starts from the entire Galilean algebra so it's not quite the same -- although it ends up in the same place.
     
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