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Proof that the eigenfunctions of a self-adjoint operator form a complete set.

  1. Sep 9, 2008 #1
    I know this is a common and important fact, so I've been willing to accept it, but this has always been something that has been "outside the scope" of my quantum lectures. Does anyone have reference for a proof?
     
  2. jcsd
  3. Sep 10, 2008 #2
    IMO, this is a postulate. The physical observables (which are often originated from symmetries) are described by the self-adjoint (or Hermitian) operators in the Hilbert space, and the eigenfunctions of the self-adjoint operators span the Hilbert space.

    As in the popular text by Sakurai, the completeness of the set of eigenstates of a Hermitian operator is a postulate.

    Cheers
     
  4. Sep 10, 2008 #3
    The excellent book Mathematical Physics by Hassani has a proof of the spectral decomposition theorem for finite-dimensional vector spaces. As a corollary, a normal operator (one that commutes with its adjoint - more general than simply Hermitian) has eigenvectors that span the space.

    To be honest, I don't know what additional assumptions/definitions you need for this to hold for infinite-dimensional spaces (if it even can at all). I think general self-adjoint operators on a Hilbert space don't even need to have eigenvalues. I'm not a mathematician though, so this is farther than I can go...
     
  5. Sep 10, 2008 #4

    George Jones

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    If you want to go a bit beyond Hassani (which I like) in terms of mathematical rigour, then look at the book Mathematics for Physics and Physicists by Appel. Chapter 14, Bras, kets, and all that sort of thing, uses a distributional rigged Hilbert space approach to treat the infinite-dimensional case.
     
  6. Sep 10, 2008 #5
    Thanks!
     
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