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Position coordinate

  1. Dec 19, 2006 #1
    Given a state |psi> and an operator A with (non-degenerate) eigenstates |ai> corresponding to real eigenvalues ai, express the wavefunction psi(x) corresponding to the state |psi> in terms of |psi>

    Now this was a question in last year's exam that my lecturer went through, but his answer on the whiteboard was:

    psi(r) = <r|psi>

    But the question asked for psi(x), not psi(r). Are the two then interchangeable?
     
  2. jcsd
  3. Dec 19, 2006 #2

    dextercioby

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    Do you know what a spectral decomposition of a linear selfadjoint operator is ? If so, then write it down and then act with it on |psi>.

    Daniel.
     
  4. Dec 19, 2006 #3
    Okay, well from Wikipedia: "The spectral decomposition of an operator A which has an orthonormal basis of eigenvectors is obtained by grouping together all vectors corresponding to the same eigenvalue."

    So you're referring to a spectral decomposition of the r operator? And the orthonormal basis of eigenvectors would be... x, y and z?
     
  5. Dec 20, 2006 #4

    dextercioby

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    This is the spectral decomposition of A (self-adjoint)

    [tex] A=\sum_{i} a_{i}|a_{i}\rangle \langle a_{i}| [/tex]

    , where the sigma symbol means, as usually, the sum over the discrete spectrum and an integral over parameter space for the continuous spectrum.

    Since you need [itex] \psi (x)= \langle x|\psi\rangle [/itex], and A is self-adjoint, therefore its (possibly generalized) eigenvectors span a basis in the (rigged) Hilbert space of the system, so then [itex] |a_{i}\rangle [/itex] form a complete set, all you need to do is find [itex] \psi (x) [/itex] in terms of [itex] a_{i}(x) [/itex].

    Daniel.
     
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