 Solution

ID: past-exam-of-the-introduction-to-quantum-information-course-of-the-university-of-oxford/2021/2/b/solution

⟨ Ω∣ A \otimes B ∣Ω ⟩ = \frac{1}{2} j, k = 0 \sum 1 ⟨ j ∣ A ∣ k ⟩ ⟨ j ∣ B ∣ k ⟩ = \frac{1}{2} j, k \sum A_{jk} B_{jk} = \frac{1}{2} tr (A^{T} B) .

The transpose is not an adjoint: the formula is complex-linear in both operators. The equality follows since

(A^{T})_{kj} = A_{jk}

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