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Wave mechanics: the adjoint of a hamiltonian

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    The operator Q satisfies the two equations

    [tex]Q^{\dagger}Q^{\dagger}=0[/tex] , [tex]QQ^{\dagger}+Q^{\dagger}Q=1[/tex]

    The hamiltonian for a system is

    [tex] H= \alpha*QQ^{\dagger}[/tex],

    Show that H is self-adjoint

    b) find an expression for [tex]H^2[/tex] , the square of H , in terms of H.

    c)Find the eigenvalues of H allowed by the result from part(b) .

    where [tex]\alpha[/tex] is a real constant

    2. Relevant equations



    3. The attempt at a solution

    [tex]QQ^{\dagger}=1-Q^{\dagger}Q[/tex]

    [tex]H=\alpha*QQ^{\dagger}=\alpha*(1-Q^{\dagger}Q)[/tex]

    self adjoint of an operator

    [tex](Q^{\dagger}\varphi,\phi)=(\varphi,Q\phi)[/tex] Should I take the conjugate of the operator H?

    b)[tex]H^2=(\alpha)*(1-Q^{\dagger}Q)(\alpha)*(1-Q^{\dagger}Q)=(\alpha)^2*(1-Q^{\dagger}Q-Q^{\dagger}Q+Q^{\dagger}Q^{\dagger}QQ.) [/tex] since [tex]Q^{\dagger}Q^{\dagger}=0 [/tex] then the expression for H^2 is : [tex]H^2=(\alpha)^2*(1-Q^{\dagger}Q-Q^{\dagger}Q=(\alpha)^2(1-2*Q^{\dagger}Q)[/tex]. Now what?
     
    Last edited: Sep 15, 2009
  2. jcsd
  3. Sep 15, 2009 #2
    a) any operator of the form [itex]cQQ^\dagger, c \in R[/itex], is self-adjoint. You don't need to use any other equation to prove this.

    b) What does finding an expression for [itex]H^2[/itex] in terms of [itex]H[/itex] mean? [itex]H^2[/itex] is an expression for [itex]H^2[/itex] in terms of [itex]H[/itex].
     
  4. Sep 15, 2009 #3
    a)show [tex] H=H^{\dagger}[/tex]? if so[tex] H^{\dagger}=((\alpha)QQ^{\dagger})^{\dagger}=QQ^{\dagger}(\alpha)[/tex]
     
  5. Sep 15, 2009 #4
    What is the adjoint of a product?
     
  6. Sep 15, 2009 #5
    i dunno [tex](PQ)^{\dagger}=Q^{\dagger}P^{\dagger}[/tex]?
     
  7. Sep 15, 2009 #6
    Well, yeah, so you just apply this rule to the definition of H and use the fact that [itex](Q^\dagger)^\dagger=Q[/itex].
     
  8. Sep 15, 2009 #7
    I did that: [tex]
    H^{\dagger}=((\alpha)QQ^{\dagger})^{\dagger}=QQ^{\dagger}(\alpha)
    [/tex]
     
  9. Sep 15, 2009 #8
    So did I apply the rule correctly? Hope there is no hard feelings
     
  10. Sep 16, 2009 #9
    er... bump! I am having trouble with part b) [tex]H^2=HH=(\alpha)QQ^{\dagger}*(\alpha)QQ^{\dagger}=(\alpha)^2*Q^{\dagger}Q(1-QQ^{\dagger})=(\alpha)^2*(Q^{\dagger}Q-QQ^{\dagger}Q^{\dagger}Q[/tex]. Using the fact that [tex]Q^{\dagger}Q^{\dagger}=0[/tex] [tex]H^2[/tex] the equals: [tex] H^2=(\alpha)^2*(Q^{\dagger}Q)[/tex]; Shouldn't [tex](Q^{\dagger}Q)[/tex] be squared?

    c) [tex]H*\varphi=E*\varphi[/tex] ; Not sure how to finish this problem.
     
  11. Sep 16, 2009 #10

    gabbagabbahey

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    Shouldn't this be [itex]H^2=\alpha^2QQ^{\dagger}(1-Q^{\dagger}Q)[/itex]?
     
  12. Sep 16, 2009 #11

    Hurkyl

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    noblegas: you need to pay more careful attention to non-commutativity. You seem to be randomly swapping [itex]Q Q^\dagger[/itex] with [itex]Q^\dagger Q[/itex] in your expressions... and in general that is simply not allowed when working with operators.


    You should not be surprised to see Q behaving degenerately -- that's one of the big things that the identity [itex]Q^\dagger Q^\dagger = 0[/itex] tells you!

    I don't know if it will help your intuition or not, but that means [itex]Q^\dagger[/itex] will, heuristically speaking, behave something like an "infinitessimal" number whose square is negligible.


    Now, rather than doing things haphazardly, you can try and systematically rewrite equations. The original identities say that [itex]Q^\dagger Q^\dagger[/itex] is trivial, and that [itex]\{ Q Q^\dagger, Q^\dagger Q \}[/itex] is a linearly dependent set. (What about [itex]QQ[/itex]?) It might help if you can decide upon a standard form for any products of Q and [itex]Q^\dagger[/itex], and use these relations to convert any product into the standard form.
     
  13. Sep 16, 2009 #12
    yes. Thats what I meant to write. Nevertheless the second term in the impression goes away.
    . How should I approach the eigenvalue problem? I know that [tex] H*\varphi =E*\varphi ==> \alpha*QQ^{\dagger}*\varphi=E*\varphi[/tex]
     
    Last edited: Sep 16, 2009
  14. Sep 16, 2009 #13

    gabbagabbahey

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    Right, the second term goes away (since [itex]Q^{\dagger}Q^{\dagger}=0[/itex])....and the first term is____?
     
  15. Sep 16, 2009 #14
    Sorry yahiko, I know matrices don't Commute. I mistakenly wrote down the wrong expression
     
  16. Sep 16, 2009 #15
    [itex]
    H^2=\alpha^2QQ^{\dagger}

    [/itex]

    that can't be right because the [tex] QQ^{\dagger} [/tex] is not squared
     
  17. Sep 16, 2009 #16

    gabbagabbahey

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    Why would the [itex]QQ^{\dagger}[/itex] term have to be squared?

    Anyways, [itex]H^2=\alpha^2QQ^{\dagger}[/itex]....and [itex]\alpha^2QQ^{\dagger}=\alpha(\alpha QQ^{\dagger})=\alpha[/itex]__?
     
  18. Sep 16, 2009 #17
    Because [tex] HH=(\alpha)QQ^{\dagger}(\alpha)QQ^{\dagger}[/tex]
     
  19. Sep 16, 2009 #18

    gabbagabbahey

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    Yes, but the conditions [itex]Q^{\dagger}Q^{\dagger}=0[/itex] and [itex]QQ^{\dagger}+Q^{\dagger}Q[/itex] put restrictions on [itex]H[/itex], which allow you to simplify [itex](QQ^{\dagger})^2[/itex] to [itex]\alpha QQ^{\dagger}[/itex]....which is exactly what you've just shown.

    It's no different from saying [itex]x^2=1[/itex] when [itex]x=1[/itex].

    Anyways, [itex]H^2=\alpha^2QQ^{\dagger}=\alpha(\alpha QQ^{\dagger})=\alpha[/itex]__?
     
  20. Sep 16, 2009 #19
    ah! [tex] [itex]
    H^2=\alpha^2QQ^{\dagger}(1-Q^{\dagger}Q=(\alpha*H)
    [/itex]
     
  21. Sep 16, 2009 #20

    gabbagabbahey

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    Right, [itex]H^2=\alpha H[/itex]....Now use that for part (c).... operate on both sides of your eignvalue equation with [itex]H[/itex]...what do you get?
     
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