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Homework Help: Expectation value using ladder operators

  1. Nov 18, 2007 #1

    T-7

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    I wonder if someone could examine my argument for the following problem.

    1. The problem statement, all variables and given/known data

    Using the relation

    [tex]\widehat{x}^{2} = \frac{\hbar}{2m\omega}(\widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} )[/tex]

    and properties of the ladder operators, determine the expectation value [tex]<\widehat{x}^{2}>[/tex] for the ground state of the simple harmonic well.

    3. The attempt at a solution

    [tex]<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}\int^{+\infty}_{-\infty}u_{0}^{*}(\widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} )u_{0}.dx[/tex]

    I then argue

    [tex]
    \left[ \widehat{A},\widehat{A}^{+}\right] = \widehat{A}\widehat{A}^{+} - \widehat{A}^{+}\widehat{A} = 1
    [/tex]
    [tex]
    \Rightarrow \widehat{A}\widehat{A}^{+} = 1 + \widehat{A}^{+}\widehat{A}
    \Rightarrow \widehat{A}^{2} + (\widehat{A}^{+})^{2} + \widehat{A}^{+}\widehat{A} + \widehat{A}\widehat{A}^{+} = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 2\widehat{A}^{+}\widehat{A} + 1 = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 2n + 1 = \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 1
    [/tex]

    I am not quite clear if it is right that [tex]\widehat{A}^{+}\widehat{A}[/tex] is equal to n (the eigenfunction number), which is zero here. Could someone comment on that, and on whether or not my treatment above is ok?

    Well, if this is true, then

    [tex]<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}\int^{+\infty}_{-\infty}u_{0}^{*}( \widehat{A}^{2} + (\widehat{A}^{+})^{2} + 1)u_{0}.dx[/tex]

    and by exploiting the orthonormal properties, I argue that the first two integrals are zero (you have one eigenfunction multiplied by another and integrated over infinity), but the third integral is 1, and then

    [tex]<\widehat{x}^{2}> = \frac{\hbar}{2m\omega}[/tex]

    Cheers!
     
  2. jcsd
  3. Nov 18, 2007 #2
    [tex]A^\dagger A[/tex] is indeed the number operator, as
    [tex]A^\dagger A |n> = A^\dagger \sqrt{n} |n-1> = n |n>[/tex]


    ------
    Assaf
    http://www.physicallyincorrect.com/" [Broken]
     
    Last edited by a moderator: May 3, 2017
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