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 ${\rho }_A$.

Suppose two density operators on $R\otimes E$ have the same reduced density matrix ${\sigma }_R$. Every measurement confined to $R$ then has the same outcome probabilities for the two preparations: a POVM effect $F_R$ has probability $\mathrm{Tr}(F_R{\sigma }_R)$ in either case. This remains true when $R$ contains several laboratories brought together for a joint measurement.

For independent copies, the accessible states are both ${\sigma }_R^{\otimes n}$. No collective measurement on those copies can distinguish the original preparations. Adaptive local processing also cannot help, since applying the same quantum operation to equal inputs gives equal outputs. Access to an additional environment measurement record is a change of resources, and can reveal differences hidden by the partial trace.

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 $B$.

For local Kraus operators $A_i$ with ${\sum }_i{A}_i^{†}{A}_i=I$, let ${\rho }_{AB}^{\prime }={\sum }_i(A_i\otimes I){\rho }_{AB}(A_i^{†}\otimes I)$. For every remote operator $B$, cyclicity of the trace gives $\mathrm{Tr}[(I\otimes B){\rho }_{AB}^{\prime }]={\sum }_i\mathrm{Tr}[(A_i^{†}{A}_i\otimes B){\rho }_{AB}]=\mathrm{Tr}[(I\otimes B){\rho }_{AB}]$. Hence the remote reduced density matrix is unchanged. In particular every remote POVM has the same outcome probabilities, proving the no-communication theorem from the measurement rule rather than assuming it as a physical postulate.

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.

A nonzero bipartite pure state is a product state if and only if all its partial inner products on the first subsystem lie in a common one-dimensional vector subspace of the second. For a product $|a⟩\otimes |b⟩$, every contraction equals $⟨u|a⟩|b⟩$. Conversely, contract against each vector of an orthonormal basis to write $|\Psi ⟩={\sum }_i|i⟩\otimes |{\chi }_i⟩$. If $|{\chi }_i⟩=c_i|b⟩$ for a common nonzero $|b⟩$, then $|\Psi ⟩=({\sum }_i{c}_i|i⟩)\otimes |b⟩$. Zero conditional vectors are allowed. A nontrivial linear dependence relation requires multipliers not both zero; allowing both multipliers to vanish would give no test at all.

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 two systems of equal dimension $d\ge 2$, no joint pure state guarantees equal outcome labels whenever both systems undergo the same arbitrary basic measurement. Equal outcomes in the computational basis would force $|\Psi ⟩={\sum }_j{c}_j|jj⟩$. On any two coordinates j,k, changing both bases to the real sum/difference pair makes the different-outcome amplitude $(c_j-c_k)/2$, so all coefficients must agree. Changing both to the imaginary-phase pair makes a different-outcome amplitude $(c_j+c_k)/2$, so those coefficients must also be negatives. They must all vanish, contradicting normalization. The one-dimensional case is an exception.

The requirement of all complex bases matters. A maximally entangled vector $d^{-1/2}{\sum }_j|jj⟩$ does produce matching labels if one party uses a basis and the other its coordinatewise complex conjugate basis. Conjugate-state steering from a maximally entangled pair gives this conjugation explicitly. Identical real bases are a special case, not evidence for identical arbitrary complex bases.

Quantum entanglement between electromagnetic polarization degrees of freedom. For example, $(|HV⟩+|VH⟩)/\sqrt{2}$ is a Bell state of two photons; neither photon separately has a definite horizontal polarization or vertical polarization in this description.

A pair of photons whose joint electromagnetic polarization has polarization entanglement.

Entanglement distillation uses local operations and classical communication on many noisy entangled states to produce fewer high-fidelity Bell pairs. It is also called entanglement purification. The protocol may fail or abort; naming a purification step does not guarantee successful output for arbitrary inputs. If the output is an exact pure state of Bell pairs, every purification of a density operator factors between those pairs and the environment, so measuring each pair in the computational basis gives a shared uniform bit independent of an eavesdropper. Approximate privacy follows when the entire retained output is sufficiently close to that pure target. A separable quantum state cannot be distilled into a Bell pair by LOCC, because each local branch preserves separability.

Entanglement dilution uses local operations and classical communication to prepare partially entangled states from maximally entangled states, such as Bell pairs. For many identical bipartite pure targets, the asymptotic cost in Bell pairs per target, allowing a vanishing approximation error, is their entanglement entropy in bits. The quantum typical subspace of the target's Schmidt coefficients has almost all its weight and dimension at most $2^{m(S+\delta )}$. A maximally entangled state of this dimension can be converted to the normalized truncated target by Nielsen's pure-state conversion theorem. The resulting state approaches the full target as the retained weight tends to one. Average monotonicity of pure-state entanglement entropy and entropy continuity give the converse lower bound. This is the reverse asymptotic task to entanglement concentration.

For $0<p<1$, every bipartition of ${\Psi }_p^{\otimes n}$ has Schmidt rank $2^n$, while $m$ copies of a GHZ state have Schmidt rank $2^m$. Schmidt-rank contraction under product operators therefore forbids any exact successful LOCC branch when $m<n$. Approximate entanglement dilution avoids this obstruction by retaining only a high-weight quantum typical subspace, whose dimension has logarithm $n{h}_2(p)+o(n)$. Exact finite-block conversion and asymptotically faithful conversion are different tasks.

For nonnegative $u,v$ with $u^2+v^2=1$, Alice applies Kraus operators $M_0=\mathrm{diag}(u,v)$ and $M_1=\mathrm{diag}(v,u)$ to one half of a Bell state. Their squared effects sum to the identity, and both outcomes have probability one half. The first branch is the target; the second becomes the target after both parties apply Pauli X gates. Sending the outcome by classical communication makes this exact deterministic entanglement dilution, rather than a postselected conversion.

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.

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.

For a normalized pure two-qubit state written in a magic basis, concurrence is $\mathcal{C}=|{\sum }_j{b}_j^2|$, while $\det {\rho }_A={\mathcal{C}}^2/4$. This bilinear sum uses no complex conjugation inside its square; replacing it by ${\sum }_j|b_j{|}^2$ would incorrectly give maximal concurrence for every state.

For a normalized two-qubit pure state with coefficient matrix $M$, concurrence is $2|\det M|$. The displayed convex roof extension defines its mixed-state version. Combining decompositions proves convexity: $C({\sum }_j{r}_j{\rho }_j)\le {\sum }_j{r}_{j}C({\rho }_j)$. Every separable quantum state has a decomposition into product vectors, each with zero determinant, so its concurrence is zero.

For ${\rho }^{\prime }=(A\otimes B)\rho (A^{†}\otimes B^{†})/p$, with invertible $A,B$, a pure coefficient matrix transforms as $M\mapsto AM{B}^{\mathsf{T}}$. Its unnormalized determinant is multiplied by $\det A\det B$. The invertible filter bijects pure decompositions before and after conditioning, with branch weights rescaled by their success probabilities. Taking both infima in the convex roof extension gives the displayed exact scaling. If either filter has rank one, the retained state is separable across that party.

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.

For one pure rank-two entangled state with descending squared Schmidt coefficients $({\lambda }_0,{\lambda }_1)$, the success operator $F=\mathrm{diag}(\sqrt{{\lambda }_1/{\lambda }_0},1)$ gives a Bell pair with probability $2{\lambda }_1$. The complementary operator $\sqrt{I-F^{†}F}$ makes a valid quantum instrument. Optimality follows from the Schmidt-tail entanglement monotone: a successful Bell branch has smallest weight $1/2$, so its probability cannot exceed $2{\lambda }_1$. Failure branches have nonnegative monotone and cannot loosen this bound.

For a pure state with two descending squared Schmidt coefficients ${\lambda }_0\ge {\lambda }_1$, this quantum instrument acts in one party's Schmidt basis. The first outcome balances the two Schmidt amplitudes when $\tan \theta =\sqrt{{\lambda }_1/{\lambda }_0}$, with probability $2{\lambda }_0{\lambda }_1$. This is not the optimal unrestricted success probability $2{\lambda }_1$ in single-copy entanglement concentration. An embedding in a larger subsystem requires a POVM on the occupied support or a full-space completion.

For $n$ copies of a normalized pure state $\alpha |00⟩+\beta |11⟩$, regroup Alice's and Bob's tensor factors. Alice performs a projective measurement of the total number $k$ of zeroes, preserving all coherence among strings with that count. The outcome has binomial distributionEach retained string has the same complex amplitude ${\alpha }^k{\beta }^{n-k}$, so, up to a common phase, the conditional quantum state isIts equal Schmidt coefficients make it a maximally entangled state of Schmidt rank $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. Alice sends the count to Bob. They can relabel the corresponding local computational basis strings with local unitary operations, and convert the resulting resource to Bell pairs. Measuring the individual bits instead would reveal the entire string and leave a product state.

For $\sqrt{p}|000⟩+\sqrt{1-p}|111⟩$, a projective measurement of the number of ones in one party's $n$ local qubits gives a common type class of size $d_k=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. The conditional pure state is $d_k^{-1/2}{\sum }_{|x|=k}|x,x,x⟩$. It can be converted exactly and deterministically to a uniform common-label state of any dimension $m\le d_k$: one party measures an $m$-element subset $S$ with Kraus operator ${\Pi }_S/\sqrt{\left(\genfrac{}{}{0ex}{}{d_k-1}{m-1}\right)}$, and all parties relabel their common retained labels after receiving $S$. Choosing $m=2^f$ gives $f$ independent GHZ states. Stirling's formula and the weak law of large numbers give $f/n\to h_2(p)$ in probability. Measuring every local bit separately would instead destroy the coherence and leave a product state.

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.