Entanglement monotone (${\sum }_r{p}_{r}E({\rho }_r)\le E(\rho )$)

An entanglement monotone quantifies entanglement without increasing on average under local operations and classical communication. The average is over recorded measurement in quantum mechanics outcomes: individual successful branches can have more entanglement, provided their probabilities compensate for that increase. This distinction makes an entanglement monotone useful for bounding probabilistic state conversion. A pure-state entanglement monotone need only satisfy this inequality on pure input states and pure conditional branches, with local measurement outcomes refined to individual Kraus operators.

The relative entropy of entanglement is the smallest quantum relative entropy from a density operator to a separable quantum state, with the separable set specified for the relevant division into parties. Nonnegativity of quantum relative entropy implies $E_R\ge 0$. For a separable quantum state, choosing $\sigma =\rho$ gives $E_R=0$. For finite-dimensional bipartite states, compactness of the separable set and lower semicontinuity of the relative entropy give the converse. The support convention is $D(\rho \Vert \sigma )=+\infty$ unless $\mathrm{supp}\rho \subseteq \mathrm{supp}\sigma$.

For a bipartite pure state, the squared Schmidt coefficients are the nonzero eigenvalues of either reduced density matrix. Their sum except the largest defines $E_2=1-{\lambda }_{\max }({\rho }_A)$. The largest eigenvalue is a convex function, since ${\lambda }_{\max }(\rho )={\max }_{\Vert v\Vert =1}⟨v|\rho |v⟩$; consequently $E_2$ is concave. If Alice measures locally and records outcome $r$, Bob's conditional reductions satisfy ${\sum }_r{p}_r{\rho }_{B,r}={\rho }_B$. Concavity gives ${\sum }_r{p}_r{E}_2({\chi }_r)\le E_2(\chi )$. For Bob's measurements use Alice's unchanged average reduction instead. Iteration proves average monotonicity under LOCC. For Schmidt rank at most two, $E_2$ is the smaller squared Schmidt coefficient. A conversion succeeding with probability $p$ therefore obeys $p{E}_2(\varphi )\le E_2(\psi )$, since failure branches contribute nonnegative amounts.