 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/35b/b/iii/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 35B / b / iii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

a ↓ 0

,

E (a) = - V_{0} a + O (a),

so

E (a) < 0

for all sufficiently small positive

a

. On the other hand,

E (a) ⟶ + \infty (a \to \infty) .

The derivative satisfies

E^{'} (a) \to - \infty (a ↓ 0), E^{'} (a) \to \frac{1}{2} (a \to \infty) .

The extremum equation

a (1 + a)^{3} = V_{0}^{2}

has exactly one positive solution because its left side is strictly increasing. Thus

E

has one global minimum, and that minimum is negative.

The potential tends to zero at spatial infinity, so zero is the threshold of the continuous spectrum. A trial state with negative energy expectation forces the spectrum below zero; the corresponding lowest spectral value is a normalizable bound state. Hence every

V_{0} > 0

produces at least one bound state.

 Read the full article

 New to topics? Read the docs here!