The assumed lower bound gives $|a|=|Q{|}^{-1}{\lesssim }_K⟨\lambda {⟩}^{-M}$ at large frequency. Differentiating $1/Q$ repeatedly expresses every derivative as a finite sum of products of derivatives of $Q$ divided by powers of $Q$. Since $D_{\lambda }^{\beta }Q$ has polynomial order at most $M-|\beta |$, induction and symbol calculus giveThe interpolation region $R\le |\lambda |\le R+1$ is compact in frequency and causes no problem. Hence

The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^n;N)=S_{1,0}^N$ consists of $c\in C^{\infty }(X\times {\mathbb{R}}^n)$ such that, for every compact $K\subset X$ and all multi-indices $\alpha ,\beta$,The defining estimate givesThe Leibniz rule givesand the triangle inequality givesThese are the basic rules of symbol calculus.

Combining parts (iii) and (iv), then multiplying by $1+\epsilon$, givesThe dimension bound for quantum conditional entropy isIndeed, Subadditivity of Von Neumann entropy gives $H(A|B{)}_{\tau }\le S({\tau }_A)\le \log d_A$, while the Araki–Lieb inequality gives $H(A|B{)}_{\tau }\ge -S({\tau }_A)\ge -\log d_A$. HenceInterchanging $\rho$ and $\sigma$ proves the continuity bound for quantum conditional entropy:

The other convex decomposition is $\omega =q\sigma +(1-q)\Delta$. Applying the concavity of quantum conditional entropy, which follows from the Strong subadditivity of Von Neumann entropy, yields

Set $q=(1+\epsilon {)}^{-1}$, so $1-q=\epsilon /(1+\epsilon )$ and $\omega =q\rho +(1-q){\Delta }^{\prime }$. For every state ${\xi }_B$, the entropy bound for a binary mixture givesTaking the minimum over ${\xi }_B$ and using the variational characterization of quantum conditional entropy on each term givesSince binary entropy satisfies $H(q)=H(1-q)$, this is

The two forms of ${\omega }_{AB}$ found in part (i) giveTherefore, in the Löwner order,

If $\epsilon =0$, then ${\rho }_{AB}={\sigma }_{AB}$ and the desired continuity bound is immediate, so assume $\epsilon >0$. Apply the positive-negative decomposition of a Hermitian operator toBecause $\mathrm{Tr}X=0$ and $\Vert X{\Vert }_1=2\epsilon$, one hasThus ${\Delta }_{AB}=X_{+}/\epsilon$ is positive with trace one, hence is a density operator. DefineIt is a convex combination of states. The equation $\epsilon {\Delta }_{AB}^{\prime }=(1+\epsilon ){\omega }_{AB}-{\rho }_{AB}$ giveswhich is likewise positive and has trace one.

The Holevo optimality conditions say a POVM is optimal whenis Hermitian and $\Gamma -p_k{\rho }_k\ge 0$ for every $k$. Hereandwhose eigenvalues are $0$ and $2cs/3$. The pretty good measurement is therefore optimal. Its success probability is

Put $c=\cos (\theta /2)$, $s=\sin (\theta /2)$, and $\zeta =e^{2\pi i/3}$. Averaging the three states cancels the off-diagonal phases:For $c,s>0$, the pretty good measurement isbecause ${\bar{\rho }}^{-1/2}|{\psi }_k⟩=|0⟩+{\zeta }^k|1⟩$. The matrices are positive and ${\sum }_k{M}_k=I$. At the endpoint values of $\theta$, the same formula is understood on the support of $\bar{\rho }$ and may be completed arbitrarily on its kernel.

Let $\Delta =p{\rho }_0-(1-p){\rho }_1$. The success probability of $Q$ isWrite the spectral decomposition $\Delta ={\Delta }_{+}-{\Delta }_{-}$. For every effect $0\le Q\le I$,with equality when $Q$ projects onto the positive spectral subspace, with arbitrary action on the kernel. Since $\mathrm{Tr}\Delta =2p-1$ and $\Vert \Delta {\Vert }_1=\mathrm{Tr}{\Delta }_{+}+\mathrm{Tr}{\Delta }_{-}$, the Holevo–Helstrom theorem follows:

In quantum binary hypothesis testing, hypothesis zero supplies ${\rho }_0$ with prior $p$ and hypothesis one supplies ${\rho }_1$ with prior $1-p$. A two-outcome POVM $\{Q,I-Q\}$ decides zero on outcome $Q$. The conditional errors areSymmetric testing minimizes the prior-weighted average error $p\alpha +(1-p)\beta$, equivalently maximizing the average success probability.

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

For a bipartite input,Each second factor is positive and has trace equal to the probability of measurement outcome $k$. Dividing nonzero factors by their traces therefore writes the output as a convex combination of product states. Hence every measure-and-prepare channel is an entanglement-breaking channel.

Partial transposition interchanges the positions occupied by $x$ and $y$. Positivity of ${\rho }^{T_B}$ therefore requiresBecause the system is $2\otimes 2$, the positive partial transpose criterion is necessary and sufficient. Combining these inequalities with validity of the original state gives

After permuting the computational basis, $\rho$ is the direct sum ofThe stated diagonal and trace conditions already give Hermiticity and trace one. Each $2\times 2$ block is positive semidefinite exactly when its determinant is nonnegative. Thus $\rho$ is a density matrix precisely when

The positive partial transpose criterion says separability implies ${\rho }_{AB}^{T_B}\ge 0$. A nonpositive partial transpose therefore proves entanglement. Positive partial transpose is also sufficient for separability in dimensions $2\otimes 2$ and $2\otimes 3$, but not in general higher dimensions.

A bipartite density operator is a separable quantum state whenfor probabilities $p_j$ and local states. If no such convex decomposition exists, it is an entangled state.

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.

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

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.