 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-324/3/a/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

A_{1}, \dots, A_{m}

e^{t \sum_{j = 1}^{m} A_{j}} = k \to \infty lim (e^{t A_{1} / k} \dots e^{t A_{m} / k})^{k},

where convergence is in operator norm. In particular, for Hermitian matrices

H_{j}

A_{j} = - i H_{j}

to obtain a product of unitary operators approximating

e^{- i t \sum_{j} H_{j}}

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