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Do the eigenfunctions for the position operator form an orthogonal set?

  1. Nov 1, 2011 #1
    Starting with,

    [itex]\hat{X}\psi = x\psi[/itex]


    [itex]x\psi = x\psi[/itex]

    [itex]\psi = \psi[/itex]

    So the eigenfunctions for this operator can equal anything (as long as they keep [itex]\hat{X}[/itex] linear and Hermitian), right?

    Well, McQuarrie says that "the eigenfunctions of a Hermitian operator are orthogonal", which can be checked with:

    [itex]\int_{-\infty}^{\infty}\psi^*_m \psi_n\, dx = \langle m | n \rangle = 0[/itex]

    But if the eigenfunctions can be anything, then that integral won't always equal zero. What am I missing here?

  2. jcsd
  3. Nov 2, 2011 #2


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    So, you're messing up because you're using x as both the eigenvalue and the operator X. The correct equation should look like:


    We can't know a priori that the eigenfunction [itex]\psi[/itex] has eigenvalue x, so we must use y as the eigenvalue.





    Let's make the notation better still:


    So, our wave function is a function of x, and it has eigenvalue y for the x-operator.

    What "function" can make this true for arbitrary x and y? Well, actually there is no function that can make this true for arbitrary x and y other than the trivial [itex]\psi=0[/itex] function but this is the trivial case. In fact, x has no "eigenfunctions" in the strict sense of the word "function". The only thing that could make that above equality work is:


    Now you notice that this is STILL not square integrable, however, it IS orthogonal in some sense:


    This is the best that we can do. The eigenfunctions of x (and p) are not actually within the Hilbert space.
  4. Nov 2, 2011 #3


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    If X acts on wavefunctions of L2(R) as a multiplication through the real number x, then one can show that X is self-adjoint on his domain of definition. Then by the spectral theorem of Gelfand-Kostyuchenko-Maurin, an extension of it has a complete set of eigenfunctions is a rigged Hilbert space built over L2(R). The eigenfunctions are the Dirac delta distributions.
  5. Nov 2, 2011 #4
    Thanks, but...

    Your explanation


    My head

    I think I'm going to have to find a more mathematically rigorous textbook on the subject as mine doesn't discuss Hilbert spaces, self-adjointness, or that spectral theorem.
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