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Operators and Hermiticity

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider two operators, A and B which satisfy:
    [A, B] = B ; B†B = 1 − A

    A. Determine the hermiticity properties of A and B.
    B. Using the fact that | a = 0 > is an eigenstate of A, construct the other
    eigenstates of A.
    C. Suppose the eigenstates of A form a complete set. Determine if eigen-
    states of B can be constructed, and if so, determine the spectrum of the
    eigenstates of B.



    2. Relevant equations

    Condition for Hermiticity:

    [tex] \int (A^{\dagger}\psi)^{*}\psi.dx = \int \psi^{*}A\psi.dx [/tex]


    3. The attempt at a solution[/b

    Completely lost on this one. All I could think of was trying to work to the commutation relation from

    [tex] \int \psi^{*} B^{\dagger}B\psi.dx [/tex]

    but that just leads to a dead end. Any help would be appreciated.
     
  2. jcsd
  3. Aug 24, 2011 #2

    vela

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    It's easy to show A = 1-BB is Hermitian. Just take the adjoint of both sides of the equation. B, I'm pretty sure, is anti-Hermitian, but I'm not sure how to show that.
     
    Last edited: Aug 24, 2011
  4. Aug 24, 2011 #3

    Dick

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    B is a ladder operator for A, isn't it? I don't think it's either Hermitian or anti-Hermitian.
     
  5. Aug 24, 2011 #4

    vela

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    Yup, you're right. :redface:
     
  6. Aug 24, 2011 #5

    Dick

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    Beer-monster, suppose A|a>=a|a>, i.e. |a> is an eigenvector of A with real eigenvalue a. vela already told you how to show A is hermitian. Use [A,B]=B to say something about the eigenvalue of B|a>. Give us some help here.
     
  7. Aug 25, 2011 #6
    Okay, I guess I really need to study this stuff again because I'm lost.

    I've tried taking the adjoint of A but am not quite sure what to do with it. I guess I just need to hash through the maths.

    How did you see that B was a ladder operator, it not really clear to me.
     
  8. Aug 25, 2011 #7

    vela

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    What does it mean for an operator to be Hermitian?
     
  9. Aug 25, 2011 #8
    That is satisfies the condition that I mentioned above.

    [tex]\int (A^{\dagger}\psi)^{*}\psi.dx = \int \psi^{*}A\psi.dx[/tex]

    and also the quantities of the operators are all real.
     
  10. Aug 25, 2011 #9

    vela

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    That's not quite it. If you have two operators [itex]\hat{A}[/itex] and [itex]\hat{A}^\dagger[/itex] that satisfy[tex]\int (\hat{A}^{\dagger}\psi_l)^*\psi_r\,dx = \int \psi_l^*\hat{A}\psi_r\,dx[/tex]for all pairs [itex]\psi_l[/itex] and [itex]\psi_r[/itex] from the Hilbert space, you would say [itex]\hat{A}^\dagger[/itex] is the Hermitian adjoint of [itex]\hat{A}[/itex]. Now how are the operator and its adjoint related if it's Hermitian?
     
  11. Aug 25, 2011 #10
    The operator and its adjoint are equal if it is Hermitian?
     
  12. Aug 25, 2011 #11

    vela

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    That's right. So you want to calculate [itex]\hat{A}^\dagger[/itex] and show it equals [itex]\hat{A}[/itex]. You should find rules in your textbook on how to calculate the adjoint of products and sums, which you will find helpful.
     
  13. Aug 25, 2011 #12
    I see it now. Here I was adding in kets and bras because I've often found it useful with commutator stuff.

    The [tex]B^\dagger[/tex] becomes B and vice versa would give [tex]BB^\dagger[/tex] but because factors swap round in an adjoint you get what you started with.

    I'm still not sure how you worked out B was a ladder operator. Though I had guessed something along those lines as ladder operators were the only way I could think to get an eigenspectrum with no wavefunction. i.e. the question gave it away.
     
  14. Aug 26, 2011 #13

    Dick

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    I guessed B was a ladder operator because I've seen a form like [A,B]=B before. That's the big reason. You can figure it out for yourself if you operate both sides on an eigenfunction of A, like I suggested in post #5. Or have you already figured that out?
     
  15. Aug 27, 2011 #14
    Not yet. But I think it`s just a matter of beating my head against the maths and cracking out a textbook that goes over the mathematical manipulations of operators again.

    Incidently, I noticed there was a mistake in my first post. The relation between B and A is:

    [tex] B^{\dagger}B=1-A^{2} [/tex]

    This shouldn't effect the results too much as I would imagine if A*A is Hermitian so A would be.
     
  16. Aug 28, 2011 #15

    Dick

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    Untrue, if A=i*I where I is the identity operator then A^2 is hermitian, but A isn't. You might have a bit more work to do on this problem.
     
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