= Schmidt-tail entanglement monotone
{c}
{title2=$E_2(\chi)=1-\lambda_{\max}(\rho_A)$}
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 $\lambda_{\max}(\rho)=\max_{\|v\|=1}\langle v|\rho|v\rangle$; 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 $\sum_r p_rE_2(\chi_r)\leq E_2(\chi)$. 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 $pE_2(\phi)\leq E_2(\psi)$, since failure branches contribute nonnegative amounts.
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