 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/j/j2/a/i/solution

Since

d / d x = mω /ℏ d / d ξ

ℏ ω /2

yields

- ψ_{ξξ} + ξ^{2} ψ = ϵ ψ, ϵ = \frac{2 E}{ℏ ω} .

For

ψ = f e^{- ξ^{2} /2}

ψ_{ξξ} = e^{- ξ^{2} /2} (f^{''} - 2 ξ f^{'} + (ξ^{2} - 1) f) .

The

ξ^{2} f

terms cancel, leaving

f^{''} - 2 ξ f^{'} + (ϵ - 1) f = 0, ϵ = \frac{2 E}{ℏ ω} .

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