 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-81/2/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

Write

b_{k} = ⟨ b_{k} ⟩ + δ b_{k}

. The mean-field approximation drops the quadratic product of the fluctuations:

b_{k}^{†} b_{k^{'}} ≃ b_{k}^{†} ⟨ b_{k^{'}} ⟩ + ⟨ b_{k}^{†} ⟩ b_{k^{'}} - ⟨ b_{k}^{†} ⟩ ⟨ b_{k^{'}} ⟩ .

For a Hermitian pairing interaction, the two linear terms are Hermitian conjugates. Defining the BCS anomalous average and pairing field by

Δ_{k} = - \sum_{k^{'}} V_{k, k^{'}} ⟨ b_{k^{'}} ⟩

ξ_{k} = ϵ_{k} - μ

, gives

K \equiv H - μ N = k, σ \sum ξ_{k} c_{kσ}^{†} c_{kσ} - k \sum (Δ_{k} b_{k}^{†} + Δ_{k}^{*} b_{k}) - k, k^{'} \sum V_{k, k^{'}} ⟨ b_{k}^{†} ⟩ ⟨ b_{k^{'}} ⟩ .

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