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The hermitian conjugate/adjoint -Quantum Physics

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data

    The hermitian conjugate of an operator, [itex]\hat{F}[/itex], written [itex]\hat{F}[/itex][itex]^{\tau}[/itex] satisfies the condition:

    ∫∅*(r)[itex]\hat{F}[/itex][itex]^{\tau}[/itex]ψ(r)dr=(∫ψ*(r)[itex]\hat{F}[/itex]∅(r)dr)*

    for any choice of wavefunctions ψ and ∅. Show that:

    ([itex]\hat{F}[/itex]+i[itex]\hat{G}[/itex])[itex]^{\tau}[/itex]=[itex]\hat{F}[/itex][itex]^{\tau}[/itex] -i[itex]\hat{G}[/itex][itex]^{\tau}[/itex]

    (10 marks)



    2. The attempt at a solution

    I feel like I'm missing something here, either that or the question's stupidly easily and isn't worth ten marks.
    As ((A + B)* = A* + B*) and as with all complex conjugates (x+iy)*=(x-iy), it can be applied to the above as it's still just a complex conjugate. I know I'm supposed to use the condition above somehow so without obviously telling me the answer, could someone point me in the right direction for how I'm supposed to SHOW it please.
     
    Last edited: Dec 5, 2011
  2. jcsd
  3. Dec 5, 2011 #2
    did you try using (F+iG) as an operator and then using the
    ∫∅*(r)Fˆτψ(r)dr=(∫ψ*(r)Fˆ∅(r)dr)* and seperating them while they are in (..... )* and then apply
    ∫∅*(r)Fˆτψ(r)dr=(∫ψ*(r)Fˆ∅(r)dr)* in reverse way ? not sure but that may help.
     
  4. Dec 5, 2011 #3
    Hmmm. Yes, I can sort of see what you mean. I've tried it but I'm still confused with how the τ's are allowed to be separated. I think it's due to the question being worded in a confusing way. If I knew how operators could be manipulated algebraically then I'd probably get somewhere....I'm assuming they don't function like normal variables?
     
  5. Dec 5, 2011 #4
    ∫∅*(r)(Fˆ+iGˆ)τψ(r)dr=(∫ψ*(r)(Fˆ+iGˆ)∅(r)dr)*
    =(∫ψ*(r)(Fˆ)∅(r)dr+i∫ψ*(r)Gˆ(r)∅(r)dr)*
    =((∫∅*(r)Fˆτψ(r)dr)*+i(∫∅*(r)Gˆτψ(r)dr)*)*
     
  6. Dec 5, 2011 #5
    Brilliant... I understand it now thanks crimsonidol!
     
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