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Hermitian Operators

  1. Oct 26, 2012 #1
    Hi,

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

    Let A and B be hermitian operators. Show that C=i[A,B] is hermitian aswell.



    2. Relevant equations

    -

    3. The attempt at a solution

    Well, I tried just to use the definition but I'm not sure if that's enough (my guess would be no lol).

    [tex] C=i[A,B]=iAB-iBA=iA^\dagger B^\dagger-iB^\dagger A^\dagger =i[A^\dagger , B^\dagger ][/tex]


    I'm not used to proof stuff as we don't learn it in my degree program.
    Is that right, or is that a whole wrong approach?

    Thanks for your help
     
  2. jcsd
  3. Oct 26, 2012 #2
    [tex]C=i[A,B]=iAB-iBA = iA^{\dagger}B^{\dagger}- i B ^{\dagger} A^{\dagger}[/tex]
    [tex]C^{\dagger}=(iAB-iBA)^{\dagger}[/tex]
    Can you show that [itex]C[/itex] and [itex]C^{\dagger}[/itex] are the same?
     
  4. Oct 26, 2012 #3
    I'd say:

    [tex] C^\dagger = (iAB)^\dagger - (iBA)^\dagger = iA^\dagger B^\dagger-iB^\dagger A^\dagger [/tex]
     
  5. Oct 26, 2012 #4
    Yup, so I guess you're done ;)
     
  6. Oct 26, 2012 #5
    Thanks for the help, but is that really it?

    I can't believe that actually because it looks so trivial.

    I came to the conclusion that I could show aswell that the diagonal of the operator is real, but that would be much more of an afford I guess... ^^
     
  7. Oct 26, 2012 #6

    Dick

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    You might want to give your reasoning behind that. It's easy to write down the correct answer for the wrong reason. What is [itex] (iAB)^\dagger[/itex]?
     
  8. Oct 27, 2012 #7
    Hm I just used a calculation rule which says:
    [tex] (AB)^\dagger = A^\dagger B^\dagger[/tex]
     
  9. Oct 27, 2012 #8

    Fredrik

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    I think Dick was more interested in how you dealt with the i. It looks like you know what you're doing, but your calculation in #3 doesn't show what you were thinking. In particular, you didn't make it clear that you understand that the first term on the left of the final equality is equal to the second term on the right. So it looked like you might have made two mistakes that cancel each other out.

    Edit: As you can see in posts 13-14 below, I didn't even notice that the operators are in the wrong order on the right-hand side of the quoted equality. I used the correct rule myself, but I didn't see that you (Lindsayyyy) was using the wrong one.
     
    Last edited: Oct 27, 2012
  10. Oct 27, 2012 #9
    Well, I don't know what to say now :blushing: . I thought it's allowed to factor out the i.
     
  11. Oct 27, 2012 #10

    Fredrik

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    If you just take it outside, both terms on the right in #3 would get the wrong sign.
     
  12. Oct 27, 2012 #11
    I guess I did it wrong then. I don't know the right answer.

    edit: hold on a second. I try something

    [tex] C^\dagger = (iAB)^\dagger - (iBA)^\dagger = -iA^\dagger B^\dagger +iB^\dagger A^\dagger [/tex]

    would that be right?
     
  13. Oct 27, 2012 #12

    Fredrik

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    No, you had the right answer the first time. It's true that $$(iAB)^\dagger - (iBA)^\dagger = iA^\dagger B^\dagger-iB^\dagger A^\dagger,$$ but if you want to show that you know what you're doing, you have to do the two terms separately:
    \begin{align}
    (iAB)^\dagger &=-i(AB)^\dagger=-iB^\dagger A^\dagger\\
    (-iBA)^\dagger &=i(BA)^\dagger =iA^\dagger B^\dagger.
    \end{align}
     
  14. Oct 27, 2012 #13

    vela

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    There is no such rule.
     
  15. Oct 27, 2012 #14

    Fredrik

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    Ahh...I didn't even see that the operators were in the wrong order on the right.
     
  16. Oct 27, 2012 #15
    whoops, sorry. I didn't read the rule clearly. I have to drive home now, thanks for your help. I'll take a closer look tomorrow and post again if I need further help. :smile:
     
  17. Oct 27, 2012 #16

    Fredrik

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    It is now clear that you did make two mistakes that canceled each other out: You didn't change the order of the operators, and you didn't take the complex conjugate of the i.
     
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