 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-1/33c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

With

β = 1/ (k_{B} T)

Z = n \sum e^{- β E_{n}}, p (n) = \frac{e ^{- β E_{n}}}{Z} .

Differentiating

T lo g Z

, with fixed energy levels, gives

S = k_{B} lo g Z + ⟨ E ⟩ / T

. Since

lo g p (n) = - β E_{n} - lo g Z

, normalization yields

S = - k_{B} n \sum p (n) lo g p (n) .

The sums include degeneracy by labeling states rather than distinct energy values, and a zero probability contributes zero by continuity.

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