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Thread: Harmonic Oscillator Finding <x^2>

  1. #1
    Senior Member bugatti79's Avatar
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    Harmonic Oscillator Finding <x^2>

    Dear Experts,

    I am having difficulty undersanding the pattern when working with operators which is part of the work to evaluate <x^2>. See attached jpg.

    Likewise the same with evaluating $\displaystyle a_{+}a_{-}\psi_{n}(x)$

    Thanks
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by bugatti79 View Post
    Dear Experts,

    I am having difficulty undersanding the pattern when working with operators which is part of the work to evaluate <x^2>. See attached jpg.

    Likewise the same with evaluating $\displaystyle a_{+}a_{-}\psi_{n}(x)$

    Thanks
    $\displaystyle a_{+} \sqrt{n+1} \, \psi_{n+1} = \sqrt{n+1} \, a_{+} \psi_{n+1}$ and $\displaystyle a_{+} \psi_{n+1} = \sqrt{n+2} \, \psi_{n+2}$ by definition of the raising operator.
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  3. #3
    Senior Member bugatti79's Avatar
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    On that basis,

    I calculate $\displaystyle a_{+}a_{-} \psi_{n}=\sqrt{n} a_{+} \psi_{n-1}=\sqrt{n} \sqrt{n+1} \psi_{n}$

    $\displaystyle a_{-} a_{+}\psi_{n}=\sqrt{n+1} a_{-} \psi_{n+1}=\sqrt{n+1} \sqrt{n} \psi_{n}$

    $\displaystyle a_{-}^2 \psi_{n}=\sqrt{n} a_{-} \psi_{n-1}=\sqrt{n} \sqrt{n-1} \psi_{n-2}$

    The first 2 calculations are wrong by my lectures notes, so obviously dont understand the definition.

    For example, it there is an adding operator, I will sum it with n. Ie, adding a_+ operator to n+1 gives n+2, subtracting a_- to n gives n-1 etc.....
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  4. #4
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    Quote Originally Posted by bugatti79 View Post
    On that basis,

    I calculate $\displaystyle a_{+}a_{-} \psi_{n}=\sqrt{n} a_{+} \psi_{n-1}=\sqrt{n} \sqrt{n+1} \psi_{n}$

    $\displaystyle a_{-} a_{+}\psi_{n}=\sqrt{n+1} a_{-} \psi_{n+1}=\sqrt{n+1} \sqrt{n} \psi_{n}$

    $\displaystyle a_{-}^2 \psi_{n}=\sqrt{n} a_{-} \psi_{n-1}=\sqrt{n} \sqrt{n-1} \psi_{n-2}$

    The first 2 calculations are wrong by my lectures notes, so obviously dont understand the definition.

    For example, it there is an adding operator, I will sum it with n. Ie, adding a_+ operator to n+1 gives n+2, subtracting a_- to n gives n-1 etc.....
    Use the definitions more carefully. eg. $\displaystyle a_{+} \psi_{n-1} \neq \sqrt{n+1} \psi_{n}$ etc.
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Use the definitions more carefully. eg. $\displaystyle a_{+} \psi_{n-1} \neq \sqrt{n+1} \psi_{n}$ etc.
    I have it now, when I compare
    $\displaystyle a_{+}\psi_{n}=\sqrt{n+1} \psi_{n+1}$ and $\displaystyle a_{+} \psi_{n-1}$ together I know hows it derived.

    Cheers
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