Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 30E / b

Solution

By the evenness of

V

, the semiclassical action integral is twice its integral over the positive half-axis. If

0 \leq E \leq 1

, then

f (E) = 2 \int_{0}^{4 E} E - \frac{x}{4} d x = 8 \int_{0}^{E} u^{1/2} d u = \frac{16}{3} E^{3/2} .

E \geq 1

, split the integral at

x = 4

f (E) = 2 [\int_{0}^{4} E - \frac{x}{4} d x + \int_{4}^{(E + 1)^{2}} E + 1 - x d x] .

\frac{8}{3} (E^{3/2} - (E - 1)^{3/2}) .

In the second, put

r = x

and then

u = E + 1 - r

. It becomes

\frac{4}{3} (E + 1) (E - 1)^{3/2} - \frac{4}{5} (E - 1)^{5/2} .

After simplification,

f (E) = ⎩ ⎨ ⎧ \frac{16}{3} E^{3/2}, \frac{16}{3} E^{3/2} + \frac{16}{15} (E - 1)^{5/2}, 0 \leq E \leq 1, E \geq 1.

The graph begins at the origin, is continuous and strictly increasing, passes through

(1, 16/3)

, and tends to infinity. Therefore for every

n \geq 0

and

ε > 0

f (E) = π ε (n + \frac{1}{2})

has exactly one solution

E = E_{n}

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