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For a bipartite density operator ${\rho }_{AB}$,Equality holds exactly when ${\rho }_{AB}={\rho }_A\otimes {\rho }_B$.

For a bipartite density operator ${\rho }_{AB}$ and a product reference state,The difference between the two sides is the nonnegative quantum mutual information of ${\rho }_{AB}$.

Ideal corrected quantum teleportation transfers an input system to the receiver while preserving its entire joint density operator with an arbitrary reference. Expand a joint pure state as ${\sum }_j|j⟩|r_j⟩$. Every teleportation branch acts with the same known correction on each system basis vector and does nothing to the reference vectors; undoing that correction restores the complete sum. Linearity extends the result to mixed states. The sender's measured carrier no longer retains the original reference correlations.

The trace distance between density operators isEvery quantum channel, including the partial trace, contracts trace distance.

For two density operators $\rho ,\sigma$, let $P_{+}$ project onto the positive spectral subspace of $\rho -\sigma$. The binary POVM $\{P_{+},I-P_{+}\}$ has output probability difference $(D,-D)$, where $D=D(\rho ,\sigma )$. Its measurement channel therefore preserves this pair's trace distance exactly. The optimizing quantum measurement depends on the pair; a single fixed measurement need not preserve every pair's distance.

The quantum conditional entropy has the relative-entropy representationwhere the minimum ranges over density operators on $B$.

Every tripartite density operator satisfies $S(AC)+S(BC)\ge S(A)+S(B)$. Purify to $ABCD$ and apply Strong subadditivity of Von Neumann entropy to $ACD$, conditioned on $A$. The Schmidt decomposition gives $S(AD)=S(BC)$ and $S(ACD)=S(B)$, proving the inequality. The original tripartite state need not be pure.