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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/10/d/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 10 / d / Solution by

Codex  0 2026-10-18

It is enough to choose an admissible parameter, without solving the minimizing equation. Take

a = V_{0}^{2} > 0

in the Gaussian variational bound for an attractive Gaussian well. The Rayleigh-Ritz variational principle gives

E_{0} \leq \frac{V _{0}^{2}}{2} - \frac{V _{0}^{2}}{1 + V _{0}^{2}} = - \frac{V _{0}^{2}}{2} + ε,

where an explicit remainder is

ε = V_{0}^{2} (1 - \frac{1}{1 + V _{0}^{2}}) = \frac{V _{0}^{4}}{1 + V _{0}^{2} ( 1 + V _{0}^{2} + 1 )} .

The denominator tends to

2

V_{0} \to 0

, proving

ε = O (V_{0}^{4})

in Big O notation. More precisely the binomial series gives

ε = \frac{1}{2} V_{0}^{4} - \frac{3}{8} V_{0}^{6} + O (V_{0}^{8}),

and therefore

E_{0} \leq - \frac{1}{2} V_{0}^{2} + ε, ε = O (V_{0}^{4}) .

This is an upper bound on the exact ground state energy, rather than an assertion that the leading coefficient of its exact weak-potential expansion is

- 1/2

.

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