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Ladder operators and harmonic oscillator

  1. Jun 11, 2009 #1
    1. Explain why any term (such as AA†A†A†)with unequal numbers of raising and lowering operators has zero expectation value in the ground state of a harmonic oscillator.

    Explain why any term (such as AA†A†A) with a lowering operator on the extreme right has zero expectation value in the ground state of a harmonic oscillator.

    2. Relevant equations



    3. The attempt at a solution

    Please help as I'm totally stuck with this. Does it have to do with orthonormality of different eigenfunctions?
     
  2. jcsd
  3. Jun 11, 2009 #2

    malawi_glenn

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    Yes it has, and also consider the ground state as the problem tells you to do.

    what is the effect of A on |0> ? And what about it's dual? i.e. <0|A†

    Now give it a try, star with your first operator AA†A†A† and tell us what you get and why
     
  4. Jun 11, 2009 #3
    with AA†A†A† I get [tex]\psi[/tex][tex]_{}n+2[/tex]. The first to terms convert [tex]\psi[/tex][tex]_{}n[/tex] into different eigenfunctions. Because [tex]\psi[/tex][tex]_{}n[/tex] is orthonormal to these eigenfunctions, these terms can be dropped from the integral.

    Correct?


    Also since the lowering operator is first to act on the ground-state harmonic iscillator, this takes it below the ground-state which cannot be allowed.

    On the right track?
     
  5. Jun 11, 2009 #4

    malawi_glenn

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    I have a hard tome to understand your language..

    with AA†A†A† acting on the state to the right (remember that this is sandwiched between [itex]\psi^* _0 [/itex] and [itex]\psi _0 [/itex], the two first will give you [itex]\psi _2 [/itex] so what is left is

    [itex]\psi^* _0 AA^{\dagger} \psi_2 [/itex]

    do the rest of the operation:
    [itex]\psi^* _0 \psi_2 [/itex]
    thus integral gives zero

    And yes, you can't go further down with the ground state, thus A|0> = 0
     
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