Reduced density matrix (${\rho }_A$)

For a bipartite density operator ${\rho }_{AB}$, the reduced state of subsystem $A$ is the partial trace ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$. Every measurement performed only on $A$ has outcome probabilities determined entirely by ${\rho }_A$.

A trace-preserving operation on subsystem $B$ cannot change the reduced density matrix of a disjoint subsystem $A$ when its unobserved outcomes are averaged. Local measurements on $A$ therefore cannot reveal which operation was chosen on $B$.

A bipartite density operator is a product state when it factors as ${\rho }_A\otimes {\rho }_B$. A pure product state factors as $|{\psi }_A⟩\otimes |{\psi }_B⟩$; its reduced density matrices have rank one and zero entanglement entropy. Mixed product states can have nonzero marginal Von Neumann entropy.

Write a two-qubit pure state as ${\sum }_{i,j=0}^1{c}_{ij}|ij⟩$ and form its coefficient matrix $C=(c_{ij})$. The state is a product state exactly when $C$ has rank one, equivalently when

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.

For a normalized pure state of two qubits with coefficient matrix $C$, the concurrence is $2|\det C|$. It ranges from zero for a product state to one for a maximally entangled state.

The state $a|00⟩+b|11⟩$ is entangled exactly when $ab\ne 0$. Its reduced density matrix has nonzero eigenvalues $|a{|}^2$ and $|b{|}^2$.

Entanglement concentration probabilistically converts partially entangled pure states into fewer maximally entangled states by local operations and classical communication.

The Bell states $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$ and $|{\Phi }^{-}⟩=(|00⟩-|11⟩)/\sqrt{2}$ both have reduced state $I/2$ on either qubit. No measurement on one qubit alone can distinguish their relative sign.