= Pure-state negative conditional entropy criterion
{title2=$S(B|A)=-S(\rho_A)<0\iff|\psi_{AB}\rangle\text{ is entangled}$}
For a bipartite <pure state> with finite <Von Neumann entropy>, the joint <Von Neumann entropy> is zero, so its <quantum conditional entropy> is $-S(\rho_A)$. The <Schmidt decomposition> makes the nonzero <eigenvalues> of $\rho_A$ the squared <Schmidt coefficients>. Their <Shannon entropy> vanishes exactly when there is one coefficient, equivalently the joint <pure state> is a <product state>. Thus negative <quantum conditional entropy> detects exactly the <entangled> bipartite <pure states>; it is only a sufficient criterion for <entanglement> of general mixed <density operators>.
Back to article page
