Hermitian operators

  • #1
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Let

[tex]\mathcal{L} = \frac{d}{dx} p(x) \frac{d}{dx} + q(x)[/tex]

be a self-adjoint operator on functions [tex]f : [a,b] \rightarrow \mathbb{C}[/tex]. Under what circumstances is the operator Hermitian with

[tex]<u|v> = \int_a^b u^*(x) v(x) dx[/tex]
?

Can someone give me a hint on this one? I know that hermitian operators satisfies

[tex]<u|\mathcal{L}v> = <\mathcal{L}u|v>[/tex]

but I don't really get the question.
 
Last edited:

Answers and Replies

  • #2
OlderDan
Science Advisor
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Logarythmic said:
Let

[tex]\mathcal{L} = \frac{d}{dx} p(x) \frac{d}{dx} + q(x)[/tex]

be a self-adjoint operator on functions [tex]f : [a,b] \rightarrow \mathbb{C}[/tex]. Under what circumstances is the operator Hermitian with

[tex]<u|v> = \int_a^b u^*(x) v(x) dx[/tex]
?

Can someone give me a hint on this one? I know that hermitian operators satisfies

[tex]<u|\mathcal{L}v> = <\mathcal{L}u|v>[/tex]

but I don't really get the question.
My guess is it is something very close to what is done here.

http://www.math.sdu.edu.cn/mathency/math/h/h215.htm [Broken]

What does [tex]f : [a,b] \rightarrow \mathbb{C}[/tex] mean? Is this a completeness condition?
 
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  • #3
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21
Not a completeness condition, just specifying the domain (from a to b) and the range (complex numbers)...

But as for the question, since you know the Hermitian condition, plug in your actual operator L and then simplify both sides, possibly integrating by parts to shift derivatives around. Then, when you have the two sides looking somewhat similar, you should be able to come up with conditions under which the two sides are equal.

(Yeah, I know that's kind of vague, but it should get you started...)
 
  • #4
OlderDan
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TMFKAN64 said:
Not a completeness condition, just specifying the domain (from a to b) and the range (complex numbers)...
Thanks. I always was weak on those funny letter math symbols :rolleyes:
 

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