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Expectation of Position of a Harmonic Oscillator

  1. Jan 17, 2013 #1
    Hey,

    My question is on determing the expectation value of position of the Harmonic Oscillator using raising and lowering operators, the question is part d) below:

    Position.png

    I have determined the position operator to be:

    [tex]\hat{x}=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dagger})[/tex]

    and so the expectation value becomes:

    [tex]\sqrt{\frac{\hbar}{2m\omega}}<E_{1}|(a+a^{\dagger})|E_{0}>[/tex]

    Though this is the bit I'm not sure about, I think only the a-dagger operator acts to increase the E-zero ket and the 'a' operator just causes the state to go to zero thus:

    [tex]\sqrt{\frac{\hbar}{2m\omega}}=<E_{1}|\hat{x}|E_{0}>[/tex]

    Can someone confirm this?

    Cheers,
    SK
     
  2. jcsd
  3. Jan 17, 2013 #2

    vela

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    Looks good.
     
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