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Show linear combination is not Hermitian

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data

    Linear combination is [itex]\hat{A}[/itex] + i[itex]\hat{B}[/itex]. It's given that it is not Hermitian already.


    2. Relevant equations
    ∫ψi * [itex]\hat{Ω}[/itex] ψj = (∫ψj * [itex]\hat{Ω}[/itex] ψi)*


    3. The attempt at a solution

    ∫ψi * ([itex]\hat{A}[/itex] + i[itex]\hat{B}[/itex]) ψj = (∫ψj * ([itex]\hat{A}[/itex] + i[itex]\hat{B}[/itex]) ψi)*

    I chose to work with the right hand side of the equation first.
    ∫ψi * ([itex]\hat{A}[/itex] + i[itex]\hat{B}[/itex]) ψj = {∫ψj * [itex]\hat{A}[/itex] ψi + i(∫ψj * [itex]\hat{B}[/itex] ψi)}*

    So I have to take the complex conjugate of the right hand side (not sure if that's the proper way to say it). What I don't understand is why the operator would become ∫ψj * [itex]\hat{A}[/itex] ψi - i(∫ψj * [itex]\hat{B}[/itex] ψi) and then ∫ψi * ([itex]\hat{A}[/itex] - i[itex]\hat{B}[/itex]) ψj.

    What are the mathematical reasons? The complex conjugate of + i[itex]\hat{B}[/itex] is -i[itex]\hat{B}[/itex]. I would just be replacing the operator with its complex conjugate? That would give me the answer, but it doesn't seem that simple. Clarification at this would help! Thank you!
     
  2. jcsd
  3. Feb 24, 2014 #2
    The reason is the wave functions [itex]\Psi_{i,j}[/itex] also get affected by conjugating. And when you want the hermitian conjugate of a producat of operators, you have to reverse the order of factors, since the hermitian conjugate (for example) of a matrix is the complex conjugate of the transpose of the original matrix: [itex](A^*)_{ij}=\bar{A_{ji}}[/itex]

    If that's what you mean
     
  4. Feb 24, 2014 #3

    hilbert2

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    Gold Member

    Are ##\hat{A}## and ##\hat{B}## hermitian operators? If they are, it's not generally true that ##\hat{A} + i\hat{B}## is not hermitian (consider the case where ##\hat{B}## is the zero operator).
     
  5. Feb 24, 2014 #4
    Ahat and Bhat are hermitian operators, yes. But (Ahat -/+ iBhat) is not hermitian. I think I understand but not sure. We have to take the complex conjugate to show that it is not hermitian. So (iBhat)* is (-iBhat). I was wondering if there was more to that step than what I am thinking.

    Apologies for no code. I am on the iPhone app and am not sure how to do it.
     
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