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Self-adjoint operator

  1. Apr 9, 2008 #1
    Hello.

    I have a linear operator, [tex]L[/tex], and its adjoint [tex]L^a[/tex]. [tex]L[/tex] is self-adjoint, so [tex]L=L^a[/tex]. I'm being told that the following is true:

    [tex]\langle f,Lh\rangle=\langle Lf,h\rangle[/tex].

    But what if the scalar product is not the symmetric product? What if

    [tex]\langle f,h\rangle=\langle h,f\rangle^*[/tex]

    where [tex]^*[/tex] is complex conjugation ? Then my first equation tells me that

    [tex]\langle f,Lf\rangle=\langle Lf,f\rangle[/tex].

    and the second one says that

    [tex]\langle f,Lf\rangle=\langle Lf,f\rangle^*[/tex].

    But which is true?
     
  2. jcsd
  3. Apr 9, 2008 #2

    gel

    User Avatar

    Both, <f,Lf> is real.
     
  4. Apr 9, 2008 #3
    But how do you know?
     
  5. Apr 9, 2008 #4

    gel

    User Avatar

    You just showed that it equals its complex conjugate, <f,Lf> = <Lf,f>* = <f,Lf>*. So it is real.
    Quite a standard result (eg, self adjoint operators represent real valued observables in quantum mechanics).
     
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