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Hermitian operator-prove product of operators is Hermitian if they commute

  1. Feb 9, 2010 #1
    Hermitian operator--prove product of operators is Hermitian if they commute

    1. The problem statement, all variables and given/known data

    If A and B are Hermitian operators, prove that their product AB is Hermitian if and only if A and B commute.

    2. Relevant equations

    1. A is Hermitian if, for any well-behaved functions f and g,
    [tex]\int f^* \hat{A}g d\tau = \int g (\hat{A}f)^* d\tau[/tex]

    2. If A and B are Hermitian, then (A + B) is Hermitian

    3. The Eigenenfucntions of a Hermitian operator that correpsond to different eigenvalues are orthogonal.

    4. Commuting Hermitian operators have simultaneous eigenfunctions.

    5. The set of eigenfunctions of any Hermitian operator form a complete set.

    3. The attempt at a solution
    Well, it seems I could play around with this
    [tex]\int f^* \hat{A}\hat{B}g d\tau = \int g (\hat{B}\hat{A}f)^* d\tau[/tex]
    [tex]\int f^* \hat{B}\hat{A}g d\tau = \int g (\hat{A}\hat{B}f)^* d\tau[/tex]
    and try and set the different halves equal, or something, but I can't seem to justify moving any terms around.

    Property #4 has the appearance of relevance; however, I also do not think I can arbitrarily turn the operators into functions and still go forward with the proof.

    Besides the above points and a little discussion of how operators distribute/commute, there doesn't seem to be much basis from my course knowledge to figure this out. But hopefully I am missing something simple?
     
    Last edited: Feb 9, 2010
  2. jcsd
  3. Feb 9, 2010 #2
    Re: Hermitian operator--prove product of operators is Hermitian if they commute

    Welcome to Physics Forums Dunhausen!!

    You're not turning an operator into a function. Property #4 means that if [A,B]=0 then [itex]A\psi[/itex] and [itex]B\psi[/itex] can be known simultaneously where [itex]\psi[/itex] is the eigenfunction. The most likely won't have the same eigenvalue, but the eigenfunction will be the same for the two.
    Hope that helps a little.
     
  4. Feb 9, 2010 #3

    marcusl

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    Re: Hermitian operator--prove product of operators is Hermitian if they commute

    Don't overthink the problem, just try writing down the few relevant facts.
    First write down the condition (definition) for the product AB to be Hermitian. I can save writing integrals, etc. by just writing down the operators.
    AB = (AB)+
    Now expand the right hand side, and apply what you know (that A and B are individually Hermitian). You should see the solution from there. Write again if you need more help.
     
  5. Feb 9, 2010 #4
    Re: Hermitian operator--prove product of operators is Hermitian if they commute

    jdwood:
    Well, I'm happy to have a better understanding of the property! I guess I had just been wondering whether thinking in terms of eigenfunctions instead could verify the intended theorem.

    marcusl:
    The only definition I had been given for the Hermitian operator was the form of the integral.

    The expression "AB = (AB)+" is somewhat mysterious to me. Are you use + the way I am using *?
     
  6. Feb 10, 2010 #5
    Re: Hermitian operator--prove product of operators is Hermitian if they commute

    As far as I can tell, he is. Most physicists use a dagger (which is sort of like a +) to mean the hermitian conjugate.

    Figure out how to write (AB)* in terms of A* and B* using the definition of hermitian. The rest should fall out naturally.
     
  7. Feb 10, 2010 #6
    Re: Hermitian operator--prove product of operators is Hermitian if they commute

    Ah! You will have to excuse me, as noted I was not familiar with the notation (and had only seen the integral form!) but from what you just said, I assume that all of these properties apply:

    (A+B)*=A*+B*
    (rA)*=rA*
    (AB)*= B*A*
    A**=A

    And "Hermitian or self-adjoint if A = A*" is what marcus meant when he said the definition of the Hermitian is AB=(AB)*.

    So I can say:
    assume A, B, and AB are Hermitian.
    AB=(AB)*=B*A*=BA*=BA
    therefore, AB commutes.

    Fyi, neither my professor nor my text discussed matrices, but this makes more sense to me than thinking about it in terms of integrals. However, I would love if anyone can think of any introductory references about the matrix method, especially which might help me to see why the formulations are equivalent.
     
    Last edited: Feb 10, 2010
  8. Feb 10, 2010 #7
    Re: Hermitian operator--prove product of operators is Hermitian if they commute

    What text are you using?
     
  9. Feb 10, 2010 #8

    Gokul43201

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    Re: Hermitian operator--prove product of operators is Hermitian if they commute

    J. J. Sakurai, Modern Quantum Mechanics (first chapter)
     
  10. Feb 10, 2010 #9
    Re: Hermitian operator--prove product of operators is Hermitian if they commute

    Atkins and De Paula, Physical Chemistry, 8th edition.

    Also, thank you. Gokul!
     
  11. Feb 10, 2010 #10

    marcusl

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    Re: Hermitian operator--prove product of operators is Hermitian if they commute

    Sorry I couldn't get back to you in a timely fashion, Dunhausen--a family medical emergency required my attention. I indeed used + to mean Hermitian adjoint, and you figured it all out nicely. Good work!
     
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