- #1

Moston-Duggan

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## Homework Statement

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"

## Homework Equations

Above

## 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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