Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 1 / 14C / ii

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- Solution ii

Solution

Before the change, the normalized ground state of the infinite square well is

ψ_{0} (x) = \frac{2}{a} sin \frac{π x}{a} .

For an allowed post-quench energy

E \in (0, U_{0})

satisfying part (i), define the unnormalized eigenfunction

χ_{E} (x) = ⎩ ⎨ ⎧ sin (k x), \frac{sin ( ka /2 )}{sinh ( l a /2 )} sinh (l (a - x)), 0 \leq x \leq a /2, a /2 \leq x \leq a .

Its normalization factor is

N_{E}^{- 2} = \int_{0}^{a /2} sin^{2} (k x) d x + \frac{s i n ^{2} ( ka /2 )}{s i n h ^{2} ( l a /2 )} \int_{a /2}^{a} sinh^{2} (l (a - x)) d x .

The Born rule therefore gives

prob (E) = \frac{2 N _{E}^{2}}{a} \int_{0}^{a /2} sin \frac{π x}{a} sin (k x) d x + \frac{sin ( ka /2 )}{sinh ( l a /2 )} \int_{a /2}^{a} sin \frac{π x}{a} sinh (l (a - x)) d x^{2} .

For a value of

E

that is not an eigenvalue, this probability is zero. The sudden change leaves the wavefunction fixed, while the energy eigenbasis changes.

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