 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/37a/b/solution

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

S = - k_{B} \sum_{n} p_{n} lo g p_{n},

first maximize subject only to

\sum_{n} p_{n} = 1

over

Ω

- (lo g p_{n} + 1) + α = 0,

so every

p_{n}

is equal. Normalization yields the microcanonical ensemble

p_{n} = \frac{1}{Ω} .

For the canonical ensemble, impose both normalization and the mean-energy constraint

\sum_{n} p_{n} E_{n} = E

. Stationarity of

- \sum_{n} p_{n} lo g p_{n} - α \sum_{n} p_{n} - β \sum_{n} p_{n} E_{n}

gives

p_{n} = C e^{- β E_{n}}

C

, so

p_{n} = \frac{e ^{- β E_{n}}}{Z}, Z = \sum_{n} e^{- β E_{n}} .

Thermodynamic consistency identifies

β = (k_{B} T)^{- 1}

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