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Wave mechanics : operator problem

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Two operators , A and B , satisfy the equations

    [tex] A=B^{\dagger}B+3 and A= BB^{\dagger}+1[/tex]

    a)Show that A is self adjoint
    b)Find the commutator of [tex][B^{\dagger},B][/tex]
    c) Find the commutator of [tex][B,B^{\dagger}][/tex]
    d) Suppose [tex]\varphi[/tex] is an eigenfunction of A with eigenvalue a:


    show that if B[tex]\varphi[/tex] =/ 0 then B[tex]\varphi[/tex] is an eigenfunction of A , and find the eigenvalue.

    2. Relevant equations

    3. The attempt at a solution

    I've only worked on the first part of the problem. I will address the remaining 3 parts later.

    [tex](A)^{\dagger}=(B^{\dagger}B+3)^{\dagger}=B^{\dagger}(B^{\dagger})^{\dagger}+(3)^{\dagger}=B^{\dagger}B+(3)^{\dagger}[/tex]. 3 is not an operator so I don't think you can take the adjoint of it.
    Last edited: Sep 20, 2009
  2. jcsd
  3. Sep 20, 2009 #2


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    Hi noblegas! :smile:

    (LaTeX doesn't take any notice of spaces unless you put them inside \text{} :wink:)

    I assume they mean 3 times the identity, I.
  4. Sep 20, 2009 #3
    okak then [tex]B^{\dagger}B+(3I)^{\dagger}=B^{\dagger}B+3*I^{\dagger}=B^{\dagger}B+3*I=[/tex]

    Therefore [tex]A^{\dagger}=A[/tex]? Now proceeding to the next two parts of the problem;

    b) [tex][B,B^{\dagger}]=BB^{\dagger}-B^{\dagger}B[/tex]

    [tex] BB^{\dagger}=A-1, B^{\dagger}B=A-3[/tex], therefore


  5. Sep 21, 2009 #4


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    You seem to have changed the questions :rolleyes:

    but it looks ok :smile:
  6. Sep 21, 2009 #5
    what do you mean? I was writing out my solutions for part b and c of my question; how should I start part d of the problem? Should I start by assuming that [tex]A
    * \varphi=a*\varphi=>A*\varphi-a*\varphi=0
    [/tex]? See my OP
    Last edited: Sep 21, 2009
  7. Sep 21, 2009 #6


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    Hi noblegas! :smile:

    (have a phi: φ and a dagger: † :wink:)
    I meant …
    For d), remember that when the examiner sets you a series of questions, the last one usually comes very easily from the preceding ones …

    in this case, just use c) :smile:
  8. Sep 21, 2009 #7
    oh I see then . [tex][B^{\dagger},B]=-2*I, and [A,B]=-2BI[/tex] correct?
  9. Sep 22, 2009 #8
    did you not understand my latest solution
  10. Sep 22, 2009 #9


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    Yes, that looks ok (except that you needn't write -2BI, you can just write it as -2B). :smile:
  11. Sep 23, 2009 #10
    stuck on part d again: AB-BA=-2BI, ARe they saying [tex]B\varphi[/tex]=> 2BI [tex]\neq[/tex] 0
    Last edited: Sep 23, 2009
  12. Sep 24, 2009 #11


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    (what happened to that φ i gave you? :confused:)

    Sorry, I don't understand what you're asking. :redface:

    An eigenvector cannot be zero.

    d) says that if Aφ = aφ, and if Bφ ≠ 0, then A(Bφ) is a scalar multiple of Bφ …

    prove it using c).​
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