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ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2019-resit/11/a/i/solution

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Past exam of the mathematical methods I course of the University of Oxford / 2019 resit / 11 / a / i / Solution by

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(A B)^{†} = B^{†} A^{†}

and the Hermitian assumptions. The commutator satisfies

[A, B]^{†} = - [A, B]

, so

(i [A, B])^{†} = - i (- [A, B]) = i [A, B] .

(1)

Thus $i [A, B]$ is Hermitian. These adjoint identities are the intended matrix-level setting; for unbounded operators, domains and self-adjointness require additional analysis.

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