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Quantum Harmonic Oscillator ladder operator

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the effect of the sequence of ladder operators acting on the ground eigenfunction [itex]\psi_0[/itex]

    2. Relevant equations
    [itex]\hat{A}^\dagger\hat{A}\hat{A}\hat{A}^\dagger\psi_0[/itex]


    3. The attempt at a solution
    I'm not sure if I'm right but wouldn't this sequence of opperators on the ground state result in zero?
     
  2. jcsd
  3. Apr 2, 2013 #2

    vela

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    Yeah.
     
  4. Apr 2, 2013 #3
    Thanks, I thought so. This is part of a larger question concerning the expectation value of [itex]\left\langle p^4_x \right\rangle[/itex]. We are asked to show that

    [itex]\left\langle p^4_x \right\rangle=\displaystyle\frac{ \hbar^4}{4 a^4}\left[ \displaystyle\int^\infty _{-\infty} \psi^*_0 \left(\hat{A}\hat{A}\hat{A}^\dagger\hat{A}^\dagger + \hat{A}\hat{A}^\dagger\hat{A}\hat{A}^\dagger + \hat{A}^\dagger\hat{A}\hat{A}\hat{A}^\dagger\right) \psi_0 dx \right]=\displaystyle\frac{3 \hbar^4}{4 a^4}[/itex]

    Where [itex]\psi_0=\left( \displaystyle\frac{1}{\sqrt{\pi}a} \right)^\frac{1}{2}e^{-x^2/2a^2}[/itex]

    This is not what I get. Due to [itex]\hat{A}^\dagger\hat{A}\hat{A}\hat{A}^\dagger\psi_0=0[/itex] I get [itex]\displaystyle\frac{2 \hbar^4}{4 a^4}[/itex] am I missing something?
    Thanks
     
  5. Apr 2, 2013 #4

    vela

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    What are you getting for the other terms? It works out as it should for me. The first term should give you a 2 by itself.
     
  6. Apr 2, 2013 #5
    As the first two sequences of operators preserve the ground eigenfunction and the last sequence zero, I get

    [itex]\left\langle p^4_x \right\rangle=\displaystyle\frac{ \hbar^4}{4 a^4}\left[ 2 \displaystyle\int^\infty _{-\infty} \psi^*_0 \psi_0 dx \right]=\displaystyle\frac{2 \hbar^4}{4 a^4}[/itex]

    Thanks
     
  7. Apr 2, 2013 #6

    vela

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    Remember that the raising and lowering operators don't simply change the state but introduce a multiplicative factor as well.
     
  8. Apr 2, 2013 #7
    Right so I should be looking at [itex]\hat{A}\hat{A}^\dagger-\hat{A}^\dagger\hat{A}=1[/itex]?
     
  9. Apr 3, 2013 #8
    Hi

    Thanks for your help what I needed was [itex]\sqrt{n}\psi_{n-1}[/itex] and [itex]\sqrt{n+1}\psi_{n+1}[/itex]. I now get the correct answer.
     
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