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Hermitian Operators

  1. Mar 3, 2006 #1

    James R

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    My memory is fading. Can somebody please remind me how I would go about determining in each of the following cases whether the operator A is Hermitian or not?

    Case 1.

    [tex]A\psi(x) = \psi(x+a)[/tex]

    Case 2.

    [tex]A\psi(x) = \psi^*(x)[/tex]

    where the star indicates complex conjugation.
     
  2. jcsd
  3. Mar 3, 2006 #2

    vanesch

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    take two vectors, f and g. In your case, these are functions of x.
    Now calculate ff = A f and gg = A g.
    If A is hermitean, then <gg,f> = <g,ff> for all f and g.
     
  4. Mar 3, 2006 #3

    dextercioby

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    Assume for simplicity that [itex] \hat{A} [/itex] is an linear operator continuous on the Hilbert space [itex] L^{2}\left(\mathbb{R}\right) [/itex] in which the scalar product between 2 arbitrary vectors is

    [tex] \langle \psi,\phi \rangle =\int_{-\infty}^{+\infty} dx \ \psi^{*}(x) \phi (x) [/tex].

    If an operator described above is symmetric, then

    [tex] \langle \psi, \hat{A}\phi\rangle =\langle \hat{A}\psi, \phi\rangle , \forall \psi,\phi \in \mathcal{H} [/tex]

    So check both operators now.

    Daniel.
     
  5. Mar 5, 2006 #4

    James R

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    Thanks..................
     
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