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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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

If $T$ is entanglement breaking, applying it to one half of $|\varphi ⟩⟨\varphi |$ immediately shows that its Choi matrix is separable.

Conversely supposeis separable. The Choi reconstruction formula for the normalized convention isThe trace-preserving condition ${\mathrm{Tr}}_1{C}_T=I/d$ implies ${\sum }_k{M}_k=I$, so $\{M_k\}$ is a POVM. Part (i) now proves that $T$ is entanglement breaking. Thus

Solved by gpt-5.6-sol high.