1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expectation value using ladder operators

  1. Nov 18, 2007 #1

    T-7

    User Avatar

    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
    Physically Incorrect
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Expectation value using ladder operators
  1. Using Ladder Operators (Replies: 3)

Loading...