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Hermitian operators

  1. Sep 6, 2007 #1

    T-7

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    1. The problem statement, all variables and given/known data

    Show that if [tex]\Omega[/tex] is an hermitian operator, and [tex]\varphi[/tex] and [tex]\psi[/tex] are (acceptable) wavefunctions, then then

    [tex]\int \phi^{*} \Omega \psi dz = \int \psi (\Omega \phi)^{*} dz[/tex]

    2. Relevant equations

    Consider the wave function [tex]\Psi = \phi + \lambda\psi[/tex]

    3. The attempt at a solution

    In my solution, I eventually arrive at

    [tex]\int \Psi^{*} \Omega \Psi dz = \int \phi^{*} \Omega \phi dz + \lambda\lambda^{*}\int \psi^{*} \Omega \psi dz + \lambda\int \psi^{*} \Omega \phi + \lambda^{*}\int \phi^{*} \Omega \psi [/tex]

    The first two terms on the right, I conclude, are real. (I deduce this because they are both essentially expectation measurements on their respective functions; if the functions are legitimate wave functions, and if the operator is Hermitian, then the result must be real). Likewise for the term on the left, which is a superposition of these two wavefunctions.

    This leaves the last two terms. I'm guessing the next step is to compare real and imaginary parts of both sides -- which implies these two are also real, since there isn't anything imaginary on the left.

    Now the answer would drop out nicely if I equated the conjugate of the fourth term on the right with the third term on the right, but I'm not totally clear on how that step would be justified (I presume this is the thing to do, though). Could someone make the logic of that move very explicit for me. (I'm probably just being obtuse).

    Cheers.
     
  2. jcsd
  3. Sep 6, 2007 #2

    Hurkyl

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    No; it implies their sum is real.


    [tex]
    \left(\int f(x) + i g(x) \, dx\right)^* =
    \left( \left( \int f(x) \, dx \right) + i \left(\int g(x) \, dx \right) \right)^* = \ldots[/tex]


    You could also do it by looking at Riemann sums.
     
    Last edited: Sep 6, 2007
  4. Sep 7, 2007 #3

    dextercioby

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    If [itex] \Omega \subset \Omega^{\dagger} [/itex], then [itex] \forall \psi\in D_{\Omega} \ , \ \Omega\psi=\Omega^{\dagger} \psi[/itex]. Now consider the abstract expression [itex] \langle\phi,\Omega\psi\rangle [/itex] for some [itex] \phi,\psi\in D_{\Omega} [/itex]. Translate it into [itex]L^{2}\left(\mathbb{R},dx\right) [/itex] language. Then translate the folowing formula associated with the definition of the adjoint: if the adjoint exists, then [itex] \forall\phi\in D_{\Omega^{\dagger}}, \forall \psi\in D_{\Omega} ,\langle \phi, \Omega\psi\rangle =\langle \Omega^{\dagger}\phi, \psi\rangle [/itex] also in the same language. Then use the hermiticity of the operator and you're there.
     
  5. Sep 8, 2007 #4

    cks

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    Using bra-ket notation,

    [tex]\int \phi^{*} \Omega \psi dz = \int \psi (\Omega \phi)^{*} dz[/tex]

    Since omega is hermitian, then omega=omega^(dagger)
    <phi|omega|psi>=<phi|omega^(dagger)|psi>=<omega phi|psi>
     
  6. Sep 8, 2007 #5

    T-7

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    Thanks very much folks. :smile:
     
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