Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 323 / 1

A quantum channel $T:\mathcal{B}(\mathcal{H})\to \mathcal{B}(\mathcal{K})$ is a linear, completely positive, trace-preserving map. Complete positivity means $T\otimes {\mathrm{id}}_n$ is positive for every ancillary dimension $n$.

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The Kraus representation isThe second identity is precisely trace preservation. Different Kraus families related by an isometry represent the same channel.

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Since $|\varphi ⟩⟨\varphi |\ge 0$ and a channel is completely positive,Writingand tracing over the output system givesThuswhere the factor $1/d$ follows from the normalized maximally entangled state.

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For $T(\rho )={\sum }_k{p}_k{U}_k\rho U_k^{†}$,Each $(U_k\otimes I)|\varphi ⟩$ is a maximally entangled state, so the Choi matrix is a convex combination of maximally entangled pure states.

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Every random unitary channel satisfiesIt is therefore a unital quantum channel.

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The normalized Choi matrix of the trace-to-identity map is $I\otimes I/d$, while that of transposition is the flip operator $F/d$. Hence the Werner–Holevo channel hasthe normalized projector onto the antisymmetric subspace.

If the channel were random unitary, part (i) would express $C_T$ as a mixture of maximally entangled vectors. Every vector in such a mixture must lie in the support of $C_T$, hence in the antisymmetric subspace. Under vectorization, an antisymmetric vector corresponds to a skew-symmetric matrix $A$, while maximal entanglement requires $A{A}^{†}$ to be proportional to $I$. In odd dimension, $\det A=\det (-A^T)=-\det A$, so $\det A=0$; such an $A$ cannot be proportional to a unitary. Thus for odd $d\ge 3$, and in particular $d=3$, this unital channel is not random unitary, disproving the converse. The odd-dimensional qualification matters because antisymmetric maximally entangled vectors can exist in even dimension.

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