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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]

    then,

    [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?

    Thanks
     
  2. jcsd
  3. Nov 2, 2011 #2

    Matterwave

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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:

    [tex]\hat{X}\psi=y\psi[/tex]

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

    Therefore

    [tex]x\psi=y\psi[/tex]

    And

    [tex](x-y)\psi=0[/tex]

    Let's make the notation better still:

    [tex](x-y)\psi_y(x)=0[/tex]

    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:

    [tex]\psi_y(x)=\delta(x-y)[/tex]

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

    [tex]\int_{-\infty}^{\infty}\psi_y(x)\psi_z(x)dx=\int_{-\infty}^{\infty}\delta(x-y)\delta(x-z)dx=\delta(y-z)[/tex]

    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

    dextercioby

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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

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    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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