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Proof that the set of stationary states are orthonormal?

  1. Mar 30, 2015 #1
    Hello there,

    I am just starting quantum physics with the textbook by griffiths. My lecturer has told me that the set of functions representing stationary states in Hilbert space forms an orthogonal set. He was however unable to prove it. Furthermore he said that it is not always the case, but didn't know when it was true, just that it often was. From the way Griffiths is writing, it seems perhaps that he isn't sure either.

    So my question: When is the set of stationary states orthogonal in with respect to the ordinary relevant inner product
    $$
    \langle f | g \rangle = \int f^*g\,dx
    $$
    Can I spot it from the underlaying physics? Or math? It would be good if I could work out when I work with an orthonormal set without having to explicitly integrate to find out.

    Hope you are able to help.

    Thanks.

    Marius
     
  2. jcsd
  3. Mar 30, 2015 #2

    atyy

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  4. Mar 31, 2015 #3

    Demystifier

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    In mathematics, this result is known as a part of the Sturm-Liouville theory.
    See e.g.

    http://en.wikipedia.org/wiki/User:Dnessett/Sturm-Liouville/Orthogonality_proof
    https://proofwiki.org/wiki/Orthogon...-Liouville_Equation_with_Distinct_Eigenvalues
    http://www.public.asu.edu/~hhuang38/pde_slides_sturm-liouville.pdf
     
  5. Mar 31, 2015 #4
    Thanks for that. However, what are you trying to say? Are you trying to say that ##\psi_n## is solution to the Strum-Louville equation if and only if it induces an orthonormal set -- or something else? Thanks.
     
    Last edited: Mar 31, 2015
  6. Mar 31, 2015 #5

    stevendaryl

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    If you have a Hermitian operator, [itex]\hat{O}[/itex], and a set of states [itex]|\psi_i\rangle[/itex] such that

    [itex]\hat{O} |\psi_i\rangle = \lambda_i |\psi_i\rangle[/itex]

    then whenever [itex]\lambda_i \neq \lambda_j[/itex], it will also be true that [itex]\langle \psi_i | \psi_j \rangle = 0[/itex]

    The proof is pretty simple:

    1. [itex]\hat{O}| \psi_j \rangle = \lambda_j |\psi_j\rangle[/itex]
    2. [itex]\langle \psi_i | \hat{O} = \lambda_i \langle \psi_i | [/itex]
    3. From 1., [itex]\langle \psi_i | \hat{O} | \psi_j \rangle = \lambda_j \langle \psi_i | \psi_j \rangle [/itex]
    4. From 2., [itex]\langle \psi_i | \hat{O} | \psi_j \rangle = \lambda_i \langle \psi_i | \psi_j \rangle [/itex]
    5. 3. & 4. are only possible if [itex]\lambda_i = \lambda_j[/itex] or [itex]\langle \psi_i | \psi_j \rangle = 0[/itex]

     
  7. Mar 31, 2015 #6

    dextercioby

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    The standard proof is valid for an operator with pure point spectrum (e.g. a compact operator). The set of stationary states of a general (time-independent) Hamiltonian are not orthogonal for distinct energy levels.
     
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