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For any normalized pure state on finite-dimensional Hilbert spaces ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$, the Schmidt decomposition givesThe $u_j$ and $v_j$ form orthonormal sets in their respective spaces. The positive numbers $\sqrt{p_j}$ are the Schmidt coefficients, and $r$ is the Schmidt rank. Taking a partial trace gives the reduced density matricesThus the nonzero eigenvalues of the two reduced density matrices agree.

One construction is a singular value decomposition of the coefficient matrix $M$ in $|\psi ⟩={\sum }_{a,b}{M}_{ab}|a⟩|b⟩$. If $M=U\Sigma V^{†}$, the columns of $U$ give $u_j$, the complex conjugates of the columns of $V$ give $v_j$, and the singular values give $\sqrt{p_j}$. This also explains why the Schmidt coefficients are real and nonnegative.

The Schmidt decomposition states that every finite-dimensional bipartite pure state haswhere $s_j>0$, ${\sum }_j{s}_j^2=1$, and the two displayed families are orthonormal. The integer $r$ is the Schmidt rank.

To prove it, choose product bases and writeApply the singular value decomposition $C=US{V}^{†}$. Absorbing the columns of $U$ and the complex conjugates of the columns of $V$ into new orthonormal bases gives the stated sum, with the nonzero singular values as the Schmidt coefficients. Equivalently, $s_j^2$ are the common nonzero eigenvalues of the two reduced density matrices, so $r=\mathrm{rank}{\rho }_A=\mathrm{rank}{\rho }_B$.

The Bell state $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$ has two nonzero Schmidt coefficients. Since the density operator is pure, every ensemble decomposition uses vectors in the same one-dimensional support, and hence

For particle positions ${\mathbf{r}}_1,{\mathbf{r}}_2$, the two-particle Schrödinger equation isWrite $|a{⟩}_i$ for the localized wave packet ${\psi }_{ai}$ and $d_{ab}=|x_{a1}-x_{b2}|$. Neglecting packet spreading and branch overlap, the initial product state evolves branchwise asup to phases generated independently on the two particles. These branch-dependent phases generally cannot be separated into one phase depending only on $a$ and one depending only on $b$, so the Newtonian gravitational potential energy creates gravitationally induced entanglement.

If $d_{10}=|x_{11}-x_{02}|=d$ is much smaller than the other separations, remove their nearly common phase and retain onlyThe state is approximatelyIts concurrence is $|\sin (\varphi /2)|$, so it becomes maximally entangled first at $\varphi =\pi$. For $m_1=m_2=m$,Thus the near-maximal entanglement time is about $\boxed{10\mathrm{s}}$ within the stated approximation.

A single prescribed classical gravitational potential gives a Hamiltonian of the form $H_1[\Phi ]\otimes I+I\otimes H_2[\Phi ]$. Its evolution factorizes as $U_1\otimes U_2$ and preserves every initial product state, so it cannot generate this entanglement. A semiclassical mean field sourced only by expectation values likewise gives each particle a local one-body potential and does not provide a quantum mediator carrying branch correlations.

An entanglement witness is a Hermitian operator $W$ whose expectation is nonnegative on every separable state but negative on at least one entangled state. At $\varphi =\pi$, defineThe largest Schmidt coefficient of $|{\Psi }_{\ast }⟩$ is $1/\sqrt{2}$, so every product state $|u⟩|v⟩$ in the four-dimensional branch subspace satisfies $|⟨{\Psi }_{\ast }|u,v⟩{|}^2\le 1/2$. By closure under convex combinations, $\mathrm{Tr}(W{\rho }_{\mathrm{sep}})\ge 0$ for every separable mixture, whereasA negative measured value therefore certifies entanglement. Under the assumptions that the masses began unentangled and interacted only through gravity, such certification would show that the mediator can transmit quantum coherence; it would be evidence against a purely classical gravitational channel and for the quantum nature of gravity.

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.

For the pure state ${\lambda }_0|00⟩+{\lambda }_1|11⟩$ with real normalized Schmidt coefficients, the two-body Pauli correlators form $T=\mathrm{diag}(s,-s,1)$, $s=2{\lambda }_0{\lambda }_1$. Thus arbitrary local axes give $⟨(\mathbf{a}\cdot \sigma )\otimes (\mathbf{b}\cdot \sigma )⟩=s(a_x{b}_x-a_y{b}_y)+a_z{b}_z$. The negative $yy$ component follows from the phases in the Pauli operators.

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 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.

Let $c,s\ge 0$ and $c^2+s^2=1$. Starting from the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, Alice applies the measurement in quantum mechanics with Kraus operators $M_0,M_1$ displayed above. Completeness follows from $M_0^{†}{M}_0+M_1^{†}{M}_1=I$. Each outcome has probability $1/2$. The normalized outputs are $c|00⟩+s|11⟩$ and $s|00⟩+c|11⟩$. Communicating the outcome allows both parties to apply a Pauli X gate in the second branch, making both outputs equal to the first. Thus this LOCC protocol deterministically converts one Bell pair into any two-qubit pure state with these Schmidt coefficients, after appropriate local unitary operations. It includes the product state endpoint without postselection.