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Hermitian Operators and Inner Products

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the vector space of square-integrable functions [tex]\psi(x,y,z)[/tex] of (real space) position {x,y,z} where [tex]\psi[/tex] vanishes at infinity in all directions. Define the inner product for this space to be

    [tex]<\phi|\psi> = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \phi^* \psi dx dy dz [/tex]

    Show that the operator

    [tex]\hat{L_z} = -i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right)[/tex]

    is a self-adjoint (Hermitian) operator on this space.
    2. Relevant equations



    3. The attempt at a solution

    So I know that for an operator to be Hermitian that [tex]\hat{H} = \hat{H}^\dagger[/tex]

    so just starting from what I know...

    [tex]<\hat{L} \phi|\psi> = <\phi|\hat{L}^\dagger|\psi>[/tex]

    [tex]<\hat{L} \phi|\psi> = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \left(-i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right) \phi \right)^* \psi dx dy dz = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right) \phi^* \psi dx dy dz[/tex]

    and

    [tex]<\phi|\hat{L}^\dagger|\psi> = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \phi^* \left(i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right)\psi \right) dx dy dz = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right) \phi^* \psi dx dy dz[/tex]

    But I guess these are my two main questions:

    For a non-matrix operator

    [tex]\hat{L_z} = -i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right)[/tex]

    does

    [tex]\hat{L_z}^\dagger = i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right)[/tex]

    ??

    and, can I move around my partials in the second equation to match that of my first or do I have to carry them out?

    Thanks, and I'm sorry if what I said is complete junk seeing as how I'm not THAT strong in linear algebra.... yet.
     
  2. jcsd
  3. Oct 19, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Actually, [itex]L_z^\dagger = L_z[/itex] -- that is what you are trying to show (it's precisely what being Hermitian means).

    So the idea is that you start from
    [tex]\langle L_z \phi \mid \psi \rangle[/tex]
    where L works on the first function, and use partial integration to show that it is equal to
    [tex]\langle \phi \mid L_z \psi \rangle[/tex]
    where L works on the second function.

    (Then, of course, since you know that it is also equal to [itex]\langle \phi \mid L_z^\dagger \psi \rangle[/itex] you can conclude that L is equal to its Hermitian conjugate and you are done).
     
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