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

Table of contents
- Solution c

Solution

For fixed

n

, the right-hand side of the quantization equation tends to zero with

ε

. If

π ε (n + \frac{1}{2}) < f (1) = \frac{16}{3},

the unique energy satisfies

E_{n} < 1

, so the first branch of the action integral for a linear-to-square-root potential applies exactly. Solving

\frac{16}{3} E_{n}^{3/2} = π ε (n + \frac{1}{2})

gives

E_{n} = [\frac{3 π ε}{16} (n + \frac{1}{2})]^{2/3} .

Thus this formula is valid whenever

ε < \frac{16}{3 π ( n + 1/2 )} .

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