 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-81/4/c/solution

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

The bosonic coherent-state resolution of identity gives the ordinary trace by insertion into a number-state basis. For fermions, the fermionic coherent-state trace requires a sign in the bra. In one mode, for an even operator

A

⟨ - ψ ∣ A ∣ ψ ⟩ = A_{00} - \overset{ˉ}{ψ} ψ A_{11}, \int d \overset{ˉ}{ψ} d ψ e^{- \overset{ˉ}{ψ} ψ} ⟨ - ψ ∣ A ∣ ψ ⟩ = A_{00} + A_{11} .

Without that sign the result would be the supertrace

A_{00} - A_{11}

. Applying this identity mode by mode to the even operator

A = e^{- β (H - μ N)}

Z = \int d [\overset{ˉ}{ψ}, ψ] e^{- \sum_{n} \overset{ˉ}{ψ}_{n} ψ_{n}} ⟨ ζ ψ ∣ e^{- β (H - μ N)} ∣ ψ ⟩ .

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