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Completeness of the Number States

  1. Nov 26, 2008 #1
    All the books I read just say that the set of number states (n) forms a complete space. They just say that the sum of all the ket-bra of states n (over all possible n) is equal to unity.

    I am having trouble proving this. Can anyone direct me to a good proof and/or show me the way?
     
  2. jcsd
  3. Nov 26, 2008 #2
    Just to clarify: Are you asking how to prove that the sum of ket-bra of states n is equal to unity? Or is it more that you know how to show that the sum is equal to unity, but want to know how this proves that the states are complete?

    If you're after the former then Messiah volume 1 has a good section round about chapter XII section 3,4 and 5. There's quite a lot of material to summarise, but if you can't get hold of the book, I'd be happy to do my best if someone else can't come up with something succinct.

    If you're after the latter, then if we assume that the basis u_i is complete we have:

    [tex]|\psi>= \sum_{i=1}^{n}|u_i > <u_i|\psi>[/tex] (decomposition)
    Which is the same thing as
    [tex]|\psi>= \sum_{i=1}^{n}(|u_i > <u_i|)|\psi>[/tex]
    Thus the sum over the ket-bra must equal unity/identity. I'm not sure this proves that the sum of ket-bra can only equal unity for a complete set though. Merely that for a complete set, the sum is unity.


    EDIT: https://www.physicsforums.com/showthread.php?t=173896

    This thread seeks to answer the question, though I'll admit it's a little convoluted.
     
    Last edited: Nov 26, 2008
  4. Nov 27, 2008 #3
    ?

    Actually it's the first question: how can I Prove that the sum of ket-bra of states n is equal to unity?

    I did look up the chapters on the Messiah book, but there I only found a treatment on why the eigenvalues are non-integers and the states are non-degenerate. I am still not clear how you can prove that they form a complete space.

    Thanks DeShark.
     
  5. Nov 27, 2008 #4
    I asked a similar question in the math section. The problem seems to be, that we need to know what spaces we are talking about. If you have a five dimensional space spanned by five functions. Than it is obvious, why the sum over the five projectors equals the identity. But then we have to stay inside this space with our Hamiltonian too.
    So in a way what you are asking is, to show that a set of base vectors is complete in the space that they span. Otherwise you have to ask "complete in what space".

    Usually you just show that a set of functions is dense in some space or that you can approximate certain functions with these base functions as closely as you want. You could probably find some decent examples in lectures about the Fourier transform.

    In general the identity is not true for arbitrary unmeasurable functions, but physics professors usually dodge this issue.
     
  6. Nov 27, 2008 #5
    I think that's essentially what's important for the quantum mechanical framework. Essentially it comes down to measurement. I only know about the N states in terms of harmonic oscillators really, but the fact that the hamiltonian can be expressed using the creation and annihilation operators means, in a way, that the eigenvalues (corresponding to the |n> state vectors) are what matters and you can have any value of the energy or any superposition. This result could just as well come from that assumption that there are an infinite amount of eigenvalues (observed energies). Since this is backed up well enough by experiment, this ought to be a sufficient physical argument..

    Mathematically I'm not so sure though. Re-reading Messiah, you're right that it doesn't nail it on the head and uses more physical arguments, almost assuming that the basis is complete. Perhaps an isomorphic mapping would be a sound mathematical route to try. By proving that another representation has basis vectors which are complete and finding the mapping between the spaces, a proof might be found. But I'm at a loss when it comes to this area.
     
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