 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-58/4/b/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

W

W W^{†} = I

. Thus for every nonnegative integer

j

(W^{†} H W)^{j} = W^{†} H^{j} W .

Insert this into the convergent matrix exponential series in the finite-dimensional qubit setting:

W^{†} e^{i H} W = \sum_{j = 0}^{\infty} \frac{i ^{j}}{j !} W^{†} H^{j} W = \sum_{j = 0}^{\infty} \frac{i ^{j}}{j !} (W^{†} H W)^{j} .

Consequently unitary conjugation commutes with the exponential:

W^{†} e^{i H} W = e^{i W^{†} H W} .

For an unbounded self-adjoint Hamiltonian, the same identity follows from the spectral theorem for normal operators, with the domain transported by

W

; no unbounded power-series manipulation is needed.

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