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Kinetic energy of harmonic oscilator

  1. Dec 6, 2007 #1
    [SOLVED] Kinetic energy of harmonic oscilator

    1. The problem statement, all variables and given/known data
    Find the expectation value of the kinetic energy of the nth state of a Harmonic oscillator


    2. Relevant equations
    [tex] <T> = \frac{<p^2>}{2m} [/tex]
    [tex] p_{x} = \frac{1}{i} \sqrt{\frac{m\hbar\omega}{2}} (\hat{a} -\hat{a}^\dagger)[/tex]
    [tex] a^\dagger \psi_{n} = \sqrt{n+1} \psi_{n+1} [/tex]
    [tex] a\psi_{n} = \sqrt{n-1} \psi_{n-1} [/tex]
    3. The attempt at a solution
    So [tex] p_{x}^2 = \left(\frac{1}{i} \sqrt{\frac{m\hbar\omega}{2}} (\hat{a} -\hat{a}^\dagger)\right)^2[/tex]


    So to calculate the <T> do i just do this:
    [tex] <T> = <\Psi_{n}|\frac{p^2}{2m}|\Psi(n)> [/tex]
    [tex] <T> = -\frac{\hbar\omega}{4} \int \psi_{n}^* (aa - aa^\dagger - a^\dagger a + a^\dagger a^\dagger) \psi_{n} dx [/tex]

    [tex] <T> = -\frac{\hbar\omega}{4} \int\psi_{n}^* \sqrt{n(n-1)} \psi_{n-2} + \sqrt{n(n+1)}\psi_{n} + \sqrt{n^2} \psi_{n} + \sqrt{(n+1)(n+2)} \psi_{n+2} dx [/tex]

    only one term will survive, the nth state ones because the wave functions are orthogonal
    [tex] <T> = -\frac{n\hbar\omega}{4} [/tex]

    Is this correct??

    Thanks for your help!
     
  2. jcsd
  3. Dec 6, 2007 #2

    malawi_glenn

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    Science Advisor
    Homework Helper

    Negative Kinetic energy is not allowed.

    In your final integral, two of four terms will survive.

    And also:
    [tex] a\psi_{n} = \sqrt{n-1} \psi_{n-1} [/tex]
    should be:
    [tex] a\psi_{n} = \sqrt{n} \psi_{n-1} [/tex]
     
  4. Dec 8, 2007 #3
    thanks i got the required answer!
    where do i mark this solved?
     
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