Let ${\rho }_{n,j}^{\prime }$ be the $j$th one-system marginal of ${\rho }_n^{\prime }$, and put ${\epsilon }_n=\Vert {\rho }_n^{\prime }-{\rho }^{\otimes n}{\Vert }_1$. Contractivity of trace distance under partial trace givesuniformly in $j$. Because the one-system state space is a compact space, continuity of $f(\cdot \Vert \sigma )$ implies uniform continuity. Hence there is a function $\delta (\epsilon )\to 0$ such thatfor every $j$.

Multipartite superadditivity and additivity now implyThereforewhich proves the required lower asymptotic semicontinuity of quantum relative entropy argument.

Apply bipartite superadditivity of quantum relative entropy first to system $1$ and systems $2,\dots ,n$, and then repeat on the remaining joint state. Mathematical induction givesThis is multipartite superadditivity of quantum relative entropy. Repeated application of bipartite additivity similarly gives

The Umegaki relative entropy iswhen the support of $\rho$ is contained in that of $\sigma$, and $+\infty$ otherwise. Its additivity of quantum relative entropy isits superadditivity of quantum relative entropy isand its data-processing inequality for quantum relative entropy is $D(T(\rho )\Vert T(\sigma ))\le D(\rho \Vert \sigma )$ for every quantum channel $T$.

Additivity follows from the logarithm of a tensor product,and the analogous identity for ${\sigma }_A\otimes {\sigma }_B$. For superadditivity, subtract the two marginal relative entropies from the joint one. The reference-state terms cancel, leavingThis is the nonnegativity of quantum mutual information, equivalently Subadditivity of Von Neumann entropy.

Apply the assumed data-processing inequality for quantum relative entropy to the normalized partial trace over $C$, with the two input statesThe channel sends them to ${\rho }_{AB}\otimes I_C/d_C$ and $I_A/d_A\otimes {\rho }_B\otimes I_C/d_C$. Additivity over the common maximally mixed factor reduces data processing toExpanding the Umegaki relative entropy in terms of Von Neumann entropy givesAfter cancelling $\log d_A$, this is precisely the Strong subadditivity of Von Neumann entropy

At the proposed boundary $p_0=1/(d+1)$, the identity from part (iv) yieldsBoth $\sigma$ and $D/d$ are convex combinations of product states, so ${\rho }_{p_0}$ is a separable quantum state. For $0\le p\le p_0$, the state ${\rho }_p$ is a convex combination of ${\rho }_{p_0}$ and the maximally mixed product state $I/d^2$, and is therefore separable.

For the converse, the partial transpose of the maximally entangled projector is $F/d$, where $F$ is the swap operator. ThereforeOn the antisymmetric subspace, $F$ has eigenvalue $-1$, so the corresponding eigenvalue of ${\rho }_p^{T_B}$ iswhich is negative exactly when $p>1/(d+1)$. The positive partial transpose criterion then proves that ${\rho }_p$ is entangled. Thus

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

Each rank-one density operator isTaking their tensor product and averaging over the $4^d$ independent choices of the fourth roots of unity givesThis is the claimed expansion in matrix elements.

The positive partial transpose criterion states that every separable quantum state satisfiesConsequently, a negative eigenvalue of the partial transpose proves entanglement. Positivity of the partial transpose is also sufficient for separability in dimensions $2\otimes 2$ and $2\otimes 3$, equivalently $3\otimes 2$, but it is not sufficient in general higher dimensions.

A bipartite density operator is a separable quantum state when it has a convex combination decompositioninto product states. A state for which no such decomposition exists is an entangled state.

For $p=(d+1)/(2d)$, the weighted difference of the two Werner–Holevo channels isThus the map is the transposition map divided by $d$. The diamond norm of the transposition map and invariance of the trace norm under matrix transpose giveSubstitution into the optimal-error formulas shows that an entangled quantum ancilla permits perfect discrimination, whereas every ancilla-free strategy has

For a fixed input, the Holevo–Helstrom theorem gives the optimal error for the two output states. Optimizing the input and quantum ancilla therefore givesThe stabilization in the diamond norm is exactly the optimization over ancillary systems. Without an ancilla the same argument instead giveswhere $\Vert \cdot {\Vert }_1$ is the induced trace norm.

In quantum channel discrimination, prepare a density operator ${\rho }_{HR}$ on the channel input $H$ and an optional quantum ancilla $R$. Under hypothesis $j\in \{1,2\}$ the output isUse a two-outcome quantum measurement $\{Q,I-Q\}$ and decide for $T_1$ on outcome $Q$. The conditional Type I error and Type II error areWith prior probabilities $p$ and $1-p$, symmetric Bayesian discrimination minimizes the average error $p\alpha +(1-p)\beta$ over the input and measurement.

Let $ϵ=\Omega /(\nu k^2)\ll 1$ and $s=ϵ\Omega s_1+\cdots$. Since $D=\Omega /ϵ+O(ϵ\Omega )$, the leading terms ofgiveThe mode is unstable for every $\boxed{N^2<0}$, a weaker criterion than the inviscid condition. Strong viscosity damps the velocity response and removes the rapid epicyclic restoration, permitting a slow viscous-convective instability; its growth rate nevertheless tends to zero on very small scales.