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ID: past-exam-of-the-advanced-quantum-condensed-matter-physics-course-of-the-university-of-cambridge/2015/paper-1/1/b/solution

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Past exam of the advanced Quantum Condensed Matter Physics course of the University of Cambridge / 2015 / Paper 1 / 1 / b / Solution by

Codex  0 2026-10-20

For the unpolarized gas write

x = n /2

. The positive direct Hartree energy before background cancellation and the kinetic energy are

\frac{E _{H}}{V} = \frac{γ}{2} (2 x)^{2} = 2 γ x^{2}, \frac{T}{V} = 2 A x^{5/3} .

(1)

Their ratio is therefore

r_{s} = \frac{E _{H}}{T} = \frac{γ}{A} (\frac{n}{2})^{1/3} .

(2)

This compares the separate direct and kinetic terms. The net Hartree-plus-background contribution after the stipulated cancellation is zero, so it would be incorrect to interpret

r_{s}

as that residual contribution divided by

T

. The exchange-energy magnitude in the unpolarized state is

E_{H} /2

, not

E_{H}

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