 Solution

ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2019/paper-2/section-a-ii/15/iv/b/solution

Let

(a_{1}, a_{2}, a_{3}, a_{4}) = (0.1, 0.4, 0.3, 0.86)

and

S = 0.9996

. For the time-independent well,

ψ (x, t) = \frac{1}{S} n = 1 \sum 4 a_{n} \frac{2}{L} sin (\frac{nπ x}{L}) e^{- i c n^{2} t /ℏ}, 0 < x < L .

It remains zero outside the well. Each expansion coefficient acquires a phase of unit magnitude, so every energy probability, including the preceding

26.0%

, is time-independent. The position probability density generally changes because cross terms between different energies oscillate.

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