Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 60 / 3

Fix an orthonormal basis $\{|i⟩{\}}_{i=1}^d$ of the input Hilbert space, where $d=\dim \mathcal{H}$, and a reference copy $R\simeq \mathcal{H}$. Use the normalized maximally entangled vector $|{\Omega }_d⟩=d^{-1/2}{\sum }_i|i{⟩}_R|i{⟩}_{\mathcal{H}}$. The normalized Choi–Jamiołkowski state isComplete positivity makes $J\ge 0$, and trace preservation gives ${\mathrm{Tr}}_{\mathcal{K}}J=I_R/d$, hence $\mathrm{Tr}J=1$. It is therefore a genuine density operator. In the unnormalized Choi matrix convention the factor $1/d$ is omitted and the trace is $d$; specifying normalization distinguishes a Choi state from that matrix convention. The reference basis also fixes the transpose in the Choi reconstruction formula, $\Lambda (X)=d{\mathrm{Tr}}_R[(X^T\otimes I)J]$.

Set $K_y=|y⟩⟨y|$. The map is in Kraus representation:For any ancillary Hilbert space $R$ and positive operator $M$ on $R\otimes \mathcal{H}$,This verifies complete positivity directly, not merely positivity on unextended states. Cyclicity of the trace gives $\mathrm{Tr}\Lambda (X)=\mathrm{Tr}[X{\sum }_y{K}_y^{†}{K}_y]=\mathrm{Tr}X$. Thus $\Lambda$ is a CPTP map, the completely dephasing channel in the given basis.

Let $q_y=⟨y|\rho |y⟩$. Positivity and normalization of the density operator imply $q_y\ge 0$ and ${\sum }_y{q}_y=1$. The completely dephasing channel produceswhich is already a spectral decomposition. Its Von Neumann entropy is thereforeThe quantum output entropy is exactly the Shannon entropy of the basis-measurement probabilities, including possible zero eigenvalues via $0{\log }_{2}0=0$.

Use Klein's inequality, or equivalently nonnegativity of quantum relative entropy. First check the support needed for the logarithm. If $q_y=0$, positivity gives $|{\rho }_{yz}{|}^2\le {\rho }_{yy}{\rho }_{zz}=0$ for every $z$; the corresponding row and column vanish. Thus support inclusion under rank-one dephasing gives $\mathrm{supp}\rho \subseteq \mathrm{supp}\sigma$, and the logarithms may be evaluated on this support.

Since $\log \sigma$ is diagonal in the dephasing basis,The relative-entropy identity for rank-one dephasing follows:Klein's inequality gives $D(\rho \Vert \sigma )\ge (\mathrm{Tr}\rho -\mathrm{Tr}\sigma )/\ln 2=0$. ThereforeEquality holds precisely when $\rho =\sigma$, meaning that the input was already diagonal in the chosen basis. This quantifies why rank-one dephasing removes coherence without reducing the entropy.