 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2012/paper-1/1/e/solution

C (β) = [A, e^{- βB}]

C^{'} + BC = [B, A] e^{- βB}, C (0) = 0.

Multiplying by the integrating factor

e^{βB}

[A, e^{- βB}] = e^{- βB} \int_{0}^{β} e^{λ B} [B, A] e^{- λ B} d λ .

Equivalently, differentiate

e^{βB} A e^{- βB} - A

directly; its derivative is the integrand and it vanishes at zero. The argument is exact for finite matrices, and for bounded operators with their norm-convergent exponentials; unbounded linear operators require an appropriate common domain.

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