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ID: past-exam-of-the-introduction-to-quantum-information-course-of-the-university-of-oxford/2017/3/b/solution

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Past exam of the introduction to quantum information course of the University of Oxford / 2017 / 3 / b / Solution by

Codex  0 2026-10-25

For any column vector

v

and positive semidefinite input

ϱ

, the Kraus representation gives

v^{†} E (ϱ) v = \sum_{k} (A_{k}^{†} v)^{†} ϱ (A_{k}^{†} v) \geq 0.

Thus the output is positive semidefinite. The cyclic property of the trace and completeness of the Kraus operators give

tr E (ϱ) = tr (ϱ k \sum A_{k}^{†} A_{k}) = tr ϱ .

More strongly, tensoring with an arbitrary ancilla replaces

A_{k}

by

A_{k} \otimes I

, so the same argument proves complete positivity rather than only positivity on single-system density matrices.

If

F

has Kraus operators

B_{j}

, the composition of quantum channels has

(F \circ E) (ϱ) = \sum_{j, k} B_{j} A_{k} ϱ A_{k}^{†} B_{j}^{†}, \sum_{j, k} (B_{j} A_{k})^{†} B_{j} A_{k} = \sum_{k} A_{k}^{†} (\sum_{j} B_{j}^{†} B_{j}) A_{k} = I .

This is again a trace-preserving Kraus representation. The composition of completely positive maps is completely positive even without the trace-preserving assumption; in the general case, the product Kraus operators still establish complete positivity.

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