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Let $\Phi :M_d(\mathbb{C})\to M_D(\mathbb{C})$ be a linear map. It is positive when $X\ge 0$ implies $\Phi (X)\ge 0$, and completely positive whenis positive for every ancillary dimension $r$. In finite dimensions it is enough to check $r=d$.

A finite family of Kraus operators $A^{\alpha }:{\mathbb{C}}^d\to {\mathbb{C}}^D$ defines the Kraus representationThis map is completely positive because, for every positive semidefinite matrix $Y$ on the enlarged space,

Conversely, use the unnormalized maximally entangled vector $|\Omega ⟩={\sum }_{j=1}^d|j⟩|j⟩$. Complete positivity makes the Choi matrixpositive semidefinite. By the spectral theorem for normal operators, $C_{\Phi }={\sum }_{\alpha =1}^R|v_{\alpha }⟩⟨v_{\alpha }|$, where $R=\mathrm{rank}{C}_{\Phi }\le dD$. Reshape each $v_{\alpha }\in {\mathbb{C}}^D\otimes {\mathbb{C}}^d$ into a matrix $A^{\alpha }$ by $|v_{\alpha }⟩={\sum }_{\mu j}{A}_{\mu j}^{\alpha }|\mu ⟩|j⟩$. The Choi matrix inversion formulathen gives $\Phi (X)={\sum }_{\alpha }{A}^{\alpha }X(A^{\alpha }{)}^{†}$. Thus finite Kraus representations characterize finite-dimensional completely positive maps.

The additional normalizationmakes $\Phi$ trace preserving and hence a quantum channel; ${\sum }_{\alpha }{A}^{\alpha }(A^{\alpha }{)}^{†}=I_D$ instead makes it unital.

The average of $z_j{z}_k^{\ast }{z}_{j^{\prime }}^{\ast }{z}_{k^{\prime }}$ vanishes unless the exponent of every independent fourth root is balanced modulo four. The surviving index patterns are $j=k,j^{\prime }=k^{\prime }$ and $j=j^{\prime },k=k^{\prime }$. Their intersection $j=k=j^{\prime }=k^{\prime }$ has been counted twice. Hence, writingthe phase average isSolving for the projector onto the maximally entangled state gives

Entanglement monogamy says that maximal entanglement with one independent system excludes entanglement with another. For three qubits this is quantified by the Coffman--Kundu--Wootters inequalityFor example, if $AB$ is a maximally entangled state, then $C_{AB}=1$ and the global state factorizes as $|{\psi }^{-}{⟩}_{AB}\otimes |\varphi {⟩}_C$ up to local unitaries, so $C_{AC}=C_{BC}=0$. The three-qubit GHZ state instead has entanglement across every one-versus-two cut but no pairwise concurrence after the third qubit is traced out.

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.

For equal masses with positions ${\mathbf{r}}_1,{\mathbf{r}}_2$, the two-body Schrodinger equation isNeglecting wave-packet spreading and overlap, write $|a{⟩}_i$ for the packet ${\psi }_{ai}$. The initial state and its branchwise gravitational evolution areup to local kinetic phases. The branch dependence of the Newtonian gravitational potential energy is the source of gravitationally induced entanglement.

Under the approximation stated in the question, only the matching pairs $a=b$ acquire an appreciable common phaseThe coefficient matrix of the resulting bipartite state iswhere $J$ is the all-ones matrix. ThereforeA maximally entangled state would require ${\rho }_1=I/n$. This demandsThat equation has no solution for $n>4$. Thus the requested large-$n$ conclusion does not follow from the assumptions printed in the paper: within the stated approximation, the system never becomes maximally entangled in the large-$n$ limit. For completeness, maximal entanglement is possible for $n=3$ at $\varphi =2\pi /3$ and for $n=4$ at $\varphi =\pi$; for $n=2$ it occurs at $\varphi =\pi /2$ or $3\pi /2$. The $n=4$ time would bebut it is not a large-$n$ answer.

The system also does not remain entangled for every $t>0$. Whenever $\varphi =2\pi k$, the phase matrix again factorizes and the state returns to its initial product state. The revival period isFor $n>2$ it is entangled at all intervening times; for $n=2$ there is the additional product-state revival at $\varphi =\pi$.

Every density operator $\sigma$ with positive partial transpose has squared quantum fidelity at most $1/d$ with a fixed maximally entangled state on $d\otimes d$. The partial transpose of a maximally entangled projector and the trace-adjoint identity for partial transpose express that squared overlap as $\mathrm{Tr}(S{\sigma }^{T_B})/d$. Since ${\sigma }^{T_B}$ is positive of trace one and the swap operator has operator norm one, trace duality gives the bound. A matching product state attains equality.

A quantum channel is entanglement breaking exactly when its Choi state is a separable quantum state. If $C_{\Lambda }={\sum }_a{p}_a{\sigma }_a\otimes {\tau }_a$, the Choi reconstruction formula gives a measure-and-prepare channel with POVM $E_a=d{p}_a{\tau }_a^T$. Conversely, applying an entanglement-breaking channel to a maximally entangled state produces a separable Choi state. Horodecki, Shor and Ruskai's entanglement-breaking channel paper develops the equivalences.