 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016/11/a/iv/solution

A

and

B

(A B)^{†} = B^{†} A^{†} = B A

. Therefore

A B is Hermitian ⟺ A B = B A ⟺ [A, B] = 0.

The required condition is that the matrices commute.

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