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Raising and lowering operators

  1. Jun 16, 2011 #1
    1. The problem statement, all variables and given/known data

    The quantum simple harmonic operator is described by the Hamiltonian:

    [itex]\hat{H}[/itex] = -[itex]\frac{h^{2}}{2m}[/itex][itex]\frac{d^{2}}{dx^{2}}[/itex] + [itex]\frac{1}{2}[/itex]m[itex]\omega^{2}[/itex]x[itex]^{2}[/itex]

    Show how this hamiltonian can be written in terms of the raising and lowering operators:

    [itex]\widehat{a}[/itex][itex]_{+}[/itex] = -[itex]\frac{h}{\sqrt{2m}}[/itex][itex]\frac{d}{dx}[/itex] + [itex]\sqrt{\frac{m}{2}}\omega[/itex]x

    [itex]\widehat{a}[/itex][itex]_{-}[/itex] = [itex]\frac{h}{\sqrt{2m}}[/itex][itex]\frac{d}{dx}[/itex] + [itex]\sqrt{\frac{m}{2}}\omega[/itex]x

    The "h" in the above eqns are actually "h-bars"

    2. Relevant equations

    Above

    3. The attempt at a solution

    [itex]\widehat{a}[/itex][itex]_{+}[/itex][itex]\widehat{a}[/itex][itex]_{-}[/itex] = (-[itex]\frac{h}{\sqrt{2m}}[/itex][itex]\frac{d}{dx}[/itex] + [itex]\sqrt{\frac{m}{2}}\omega[/itex])( [itex]\frac{h}{\sqrt{2m}}[/itex][itex]\frac{d}{dx}[/itex] + [itex]\sqrt{\frac{m}{2}}\omega[/itex]x) = [itex]\hat{H}[/itex]

    But the solution is in the picture with a red highlight of where my solution differs and i cannot work out how that extra highlighted part is added
     

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    Last edited: Jun 16, 2011
  2. jcsd
  3. Jun 16, 2011 #2

    George Jones

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    Let the second line of your expression for [itex]\widehat{a}[/itex][itex]_{+}\widehat{a}[/itex][itex]_{-}[/itex] operate on a function [itex]f\left(x\right)[/itex].
     
  4. Jun 21, 2011 #3
    Oh that's so simple haha thankyou
     
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