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For a bipartite density operator ${\rho }_{AB}$,

Applying the amplitude damping channel to the Bloch vector form of a qubit density operator gives an affine map. Transverse components contract by $\sqrt{1-p}$; the longitudinal component contracts by $1-p$ and shifts toward the ground-state north pole. Reading the entries of the output density operator proves the map. The displacement distinguishes this channel from centered depolarization.

A channel's Choi state is its normalized Choi matrix,It is a density operator and satisfies ${\mathrm{Tr}}_{\mathrm{out}}{C}_{\Lambda }=I_{\mathrm{in}}/d$. Some authors use an unnormalized Choi matrix of trace $d$, so reconstruction factors depend on the convention.

For bipartite density operators,This is a consequence of Strong subadditivity of Von Neumann entropy.

For density operators ${\rho }_1,{\rho }_2$ and $0\le p\le 1$, Von Neumann entropy is concave:

For a bipartite input density operator ${\rho }_{AC}^x$ and a POVM $\{E_y\}$ on $A$, let $q(y|x)=\mathrm{Tr}[(E_y\otimes I){\rho }^x]$. For $q(y|x)>0$, the conditioned input on $C$ isIts positivity follows from the positive sandwich before the partial trace. A quantum channel on $C$ acts on these conditional density operators even if the original input is an entangled state.

For every quantum channel $T$,For a partial trace, this says that discarding a subsystem cannot make two density operators more distinguishable in quantum relative entropy.

For real $-1\le \alpha \le 1$, this two-qubit density operator has equal $00$ and $11$ populations and off-diagonal coherences $\alpha /2$. It is a pure state exactly when $\alpha =\pm 1$, and a separable quantum state exactly when $\alpha =0$. Its partial transpose has eigenvalues $1/2,1/2,\alpha /2,-\alpha /2$, directly witnessing entanglement for every nonzero admissible coherence.

For a positive-definite Hermitian operator $A={\sum }_j{t}_j|u_j⟩⟨u_j|$ and a unit vector $v$, the weights $|⟨v|u_j⟩{|}^2$ form a probability distribution. Scalar logarithmic concavity bounds their average of $\log t_j$ by the logarithm of their average of $t_j$. Positive semidefinite operators follow by a limiting convention. Using the eigenbasis of a density operator $\rho$ gives $S(\rho \Vert \sigma )\ge D(r\Vert q)$, where $r$ is the spectrum of $\rho$ and $q$ the diagonal of $\sigma$ in that basis. This yields nonnegativity of quantum relative entropy without commutativity.

For an ensemble of density operators with average $\overline{\omega }={\sum }_j{p}_j{\omega }_j$,Expanding the trace definitions cancels the $\log {\omega }_j$ terms, and the remaining sum combines to a trace against $\overline{\omega }$. Positive-weight ensemble states have their support contained in that of the average, so the first sum on the right is finite in finite dimension. If the reference-state support condition fails, both sides are $+\infty$. The identity is associated with Donald's work on relative entropy.

A bipartite density operator is entangled when it is not a separable quantum state. For a pure state, this is equivalent to not being a product state, or to either reduced density matrix having rank greater than one.

If a density operator $\sigma$ on an $m\ge 2$ dimensional Hilbert space has overlap $t=⟨\psi |\sigma |\psi ⟩$ with a fixed pure state, then $S(\sigma )\le h(t)+(1-t){\log }_2(m-1)$. Dephase in a basis containing $|\psi ⟩$ and apply the entropy bound with one prescribed probability. The abstract state bound is attained by $\sigma =t|\psi ⟩⟨\psi |+(1-t)(I-|\psi ⟩⟨\psi |)/(m-1)$.

If encoding and decoding quantum channels factor through a $k$-dimensional space, their composite Kraus operators all have matrix rank at most $k$. The rank bound for a weighted operator trace and the Kraus formula for entanglement fidelity implyThus high operation fidelity requires the leading $k$ eigenvalues of the source density operator to carry almost all its probability. The bound applies to arbitrary completely positive trace-preserving encoders and decoders.

A linear map $V:\mathcal{H}\to \mathcal{K}$ is an isometry when it preserves inner products, equivalently $V^{†}V=I$. It need not be surjective. It maps an orthonormal basis to an orthonormal family, and $\rho \mapsto V\rho V^{†}$ preserves the nonzero eigenvalues and the Von Neumann entropy of a density operator.

The maximally mixed density operator on an $N$-dimensional Hilbert space assigns equal weights to every vector of any orthonormal basis. It is invariant under every unitary operator. Its purity is $\mathrm{Tr}({\rho }_{\ast }^2)=1/N$, the minimum possible because the nonnegative eigenvalues sum to one and obey the Cauchy-Schwarz inequality. For $N>1$ it is a mixed state; for $N=1$ it is the sole pure state. A rank-$k$ projector has probability $k/N$ in this state.

A mixed state is a density operator that is not a rank-one projector. In finite dimension it is characterized by $\mathrm{Tr}({\rho }^2)<1$.

For density operators, Klein's inequality gives $D(\rho \Vert \sigma )\ge 0$, with equality exactly when $\rho =\sigma$. If the support of a positive operator $\rho$ is not contained in that of $\sigma$, the quantum relative entropy is $+\infty$.

Let $P$ be a rank-$k$ projector with $0<k<N$. No unitary operator can map every pure state whose $P$ probability equals $k/N$ to a state with the same fixed smaller probability. Average $\sqrt{k/N}|g⟩\pm \sqrt{1-k/N}|b⟩$ over good and bad basis vectors and both signs. The average density operator is the maximally mixed state, so every unitary preserves its average $P$ probability. This contradicts a strictly smaller probability for all states in the ensemble. A preparation tailored to one known starting state is a different resource and is used in exact amplitude amplification.