Self-adjoint operator

  • #1
302
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
gel
533
5
  • #3
302
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But how do you know?
 
  • #4
gel
533
5
But how do you know?
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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