Physics Help Forum Hermitian operators

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 Dec 15th 2010, 11:06 AM #1 Junior Member   Join Date: Dec 2010 Posts: 6 Hermitian operators 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: = *, 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.
 Jan 4th 2011, 09:04 PM #2 Junior Member   Join Date: Jan 2011 Posts: 6 It looks like your main problem here is confusion with indices (the m and n inside the sum are not the same across the equality, for example). Regardless, this would be easier for you (and closer to what your Prof intends) if you broke |a> and |b> into the |$\displaystyle /phi$n> basis, rather than breaking down A.

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