Hermitian operators

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  • #1
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Hi, this is actually more a math-problem than a physics-problem, but I thought I'd post my question here and see if anyone can help me.
So I'm writing an assignment in which I have to define, what is understood by a hermitian operator.
My teacher has told me to definere it as:
<ϕm|A|ϕn> = <ϕn|A|ϕm>* , where lϕn> and lϕm> is the n'th and m'th unit operator.
And using this i then have to proof the more general definition:
<a|A|b> = <b|A|a>*, la> and lb> being arbitrary vectors.
I've tried to do so but I have yet not succeeded:
What I've done is to say: Take a hermitian operator. Since it's hermitian it must satisfy:

A = ∑_(m,n)|ϕm>amn <ϕn| = ∑(m,n)|ϕm> anm*<ϕm|
Which when dotted with 2 arbitrary vectors|ψ> and <φ|equals to:
<φ|A|ψ> = ∑(m,n) <φ|ϕm>amn<ϕn|ψ> = ∑(m,n) <ϕm|φ>*amn <ψ|ϕn>* =
∑(m,n) <ψ|ϕn>* amn <ϕm|φ>*
Since A is hermitian this equals to:
∑(m,n) <ψ|ϕn>* amn <ϕm|φ>* = ∑(m,n) <ψ|ϕn>*anm*<ϕm|φ>* = [∑(m,n) <ψ|ϕn>anm<ϕm|φ>]*
Now the proof would work if |ϕn>anm<ϕm| = |ϕm>amn<ϕn|, but that's not right is it?
Can anyone help me how to proove this? I know it's simple, but I'm still finding it a little hard.
 

Answers and Replies

  • #2
Fredrik
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where lϕn> and lϕm> is the n'th and m'th unit operator.
I assume you mean basis vector, not unit operator.

I suggest that you start with <a|A|b>, express |a> and |b> in terms of that basis, and then use the definition your teacher has told you to use.
 

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