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Use the normalized Choi state $C_{\Lambda }=(\Lambda \otimes \mathrm{id})(|{\Phi }_d⟩⟨{\Phi }_d|)$, with $|{\Phi }_d⟩=d^{-1/2}{\sum }_i|ii⟩$. If $\Lambda$ is an entanglement-breaking channel, its action on this particular bipartite input makes $C_{\Lambda }$ a separable quantum state.

Conversely, suppose $C_{\Lambda }={\sum }_a{p}_a{\sigma }_a\otimes {\tau }_a$, with local density operators ${\sigma }_a,{\tau }_a$. The Choi reconstruction formula givesBecause $\Lambda$ is trace preserving, ${\mathrm{Tr}}_A{C}_{\Lambda }=I/d$, so ${\sum }_a{E}_a=I$. Thus the $E_a$ form a POVM and the channel is a measure-and-prepare channel.

For any bipartite input ${\rho }_{AR}$, define the positive, possibly unnormalized reference operatorsIts output is ${\sum }_a{\sigma }_a\otimes R_a$. Since ${\sum }_a\mathrm{Tr}{R}_a=1$, normalizing each nonzero $R_a$ expresses this as a convex combination of product states. Hence every output is separable, proving the separable Choi-state criterion for entanglement breaking.

Expanding the normalized maximally entangled state in the definition of the Choi state givesHere the bar denotes componentwise complex conjugation in the basis defining the Choi state. Every term is a positive tensor-product operator, and the whole operator has trace one because $\tilde{\Lambda }$ is assumed to be a quantum channel. It is therefore a separable quantum state, proving entanglement breaking by the separable Choi-state criterion for entanglement breaking.

If the vectors are unit vectors, the displayed $p_j$ are already the product-state weights. If they are not normalized, absorb their squared norms into the weights and normalize the nonzero vectors. Trace preservation requiresthe probability-distribution condition alone would not guarantee this. Equivalently, the Kraus operators are $K_j=\sqrt{d{p}_j}|a_j⟩⟨b_j|$, exhibiting the rank-one Kraus representation of an entanglement-breaking channel.

Apply the measure-and-prepare channel locally to an arbitrary bipartite density operator ${\rho }_{AR}$. The resulting state iswhereTo see that this is the correct output, the partial trace is cyclic for operators acting only on the traced subsystem, so the displayed expression equals ${\mathrm{Tr}}_A[(E_a\otimes I){\rho }_{AR}]$. Put $q_a=\mathrm{Tr}{R}_a$. The POVM completeness relation implies ${\sum }_a{q}_a=1$, and terms with $q_a=0$ vanish. Thereforeis a separable quantum state. Thus the channel is an entanglement-breaking channel, for every reference-system dimension.

The state is a convex combination of product density operators and is therefore a separable quantum state. Every product density operator has a decomposition into product pure states, so

Take $|L⟩=|0⟩$ and $|R⟩=|1⟩$. Initially the two masses are in the product stateThe branch-dependent Newtonian gravitational potential energy is $-G{m}^2/d_{ab}$. Under the stated approximation, only the $|RL⟩$ branch acquires an appreciable relative phase, so after time $t$,The determinant of its two-by-two coefficient matrix is $(1-e^{i\varphi })/4$, which is nonzero unless $\varphi$ is a multiple of $2\pi$. Thus the state generally has Schmidt rank two: the branch-dependent gravitational phase creates gravitationally induced entanglement.

An entanglement witness has a bound obeyed by every separable quantum state and violated by at least one entangled state. For a product state with Bloch vectors $\mathbf{r}$ and $\mathbf{s}$,by the Cauchy-Schwarz inequality. Convexity gives the same bound for every separable mixed state. Consequently $W>1$ certifies entanglement; in conventional operator form, one of $I\mp (X\otimes Z+Y\otimes Y)$ has negative expectation whenever the absolute-value criterion is violated.

For the state above, direct use of the Pauli matrices givesWith the supplied values,which is close to $\pi$. Hence $W\simeq 2.00$, and to the nearest integerThis is the operating principle of the Bose--Marletto--Vedral experiment.

The Schmidt decomposition of an entangled pure two-qubit state has two nonzero terms. Absorb both complex phases into the local basis vectors and interchange the labels of the second qubit to obtainIn this state the only nonzero same-axis two-qubit Pauli correlators areAll mixed-axis correlators vanish. Expanding $(\mathbf{a}\cdot \mathbf{\sigma })\otimes (\mathbf{b}\cdot \mathbf{\sigma })$ therefore gives

For the four stated vectors this becomesIt follows immediately thatEvery separable quantum state obeys the corresponding CHSH inequality with upper bound two. Set $s=2c_0{c}_1>0$, choose ${\beta }^{\prime }=\pi -\beta$, and take $\tan \beta =s$ with $0<\beta <\pi /2$. The displayed expression is thenThus every entangled pure two-qubit state has local measurement correlations that no separable state can reproduce, which is Gisin's theorem. In an ideal Bose--Marletto--Vedral experiment, optimized local measurements can therefore witness any nonzero pure-state entanglement generated during the gravitational interaction. If gravity is the only interaction between the masses, such a violation shows that the mediator cannot be described by a purely classical local variable under the assumptions of the proposal; experimentally, control of decoherence and nongravitational forces is essential to that inference.

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.

The positive partial transpose criterion states that every separable quantum state satisfiesConsequently, a negative eigenvalue of the partial transpose proves entanglement. Positivity of the partial transpose is also sufficient for separability in dimensions $2\otimes 2$ and $2\otimes 3$, equivalently $3\otimes 2$, but it is not sufficient in general higher dimensions.

A bipartite density operator is a separable quantum state when it has a convex combination decompositioninto product states. A state for which no such decomposition exists is an entangled state.

At the proposed boundary $p_0=1/(d+1)$, the identity from part (iv) yieldsBoth $\sigma$ and $D/d$ are convex combinations of product states, so ${\rho }_{p_0}$ is a separable quantum state. For $0\le p\le p_0$, the state ${\rho }_p$ is a convex combination of ${\rho }_{p_0}$ and the maximally mixed product state $I/d^2$, and is therefore separable.

For the converse, the partial transpose of the maximally entangled projector is $F/d$, where $F$ is the swap operator. ThereforeOn the antisymmetric subspace, $F$ has eigenvalue $-1$, so the corresponding eigenvalue of ${\rho }_p^{T_B}$ iswhich is negative exactly when $p>1/(d+1)$. The positive partial transpose criterion then proves that ${\rho }_p$ is entangled. Thus

A bipartite density operator is a separable quantum state whenfor probabilities $p_j$ and local states. If no such convex decomposition exists, it is an entangled state.

A finite-dimensional bipartite density operator ${\rho }_{AB}$ is a separable quantum state exactly when $({\mathrm{id}}_A\otimes \Phi )({\rho }_{AB})\ge 0$ for every positive linear map from operators on $B$ to operators on $A$. In combination with the Størmer-Woronowicz decomposability theorem, it makes the positive partial transpose criterion sufficient for qubit-qutrit states. The criterion is established in the Horodeckis' separability paper.

A bipartite density operator has positive partial transpose when ${\rho }^{T_B}\ge 0$. Every separable quantum state has this property, since transposing one factor of a positive product state preserves positivity and the separable sum preserves it as well. The property does not depend on which local basis defines the transpose.

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 $\mathrm{supp}\rho \subseteq \mathrm{supp}\sigma$.

A quantum channel is entanglement breaking exactly when its Choi state is a separable quantum state. If $C_{\Lambda }={\sum }_a{p}_a{\sigma }_a\otimes {\tau }_a$, the Choi reconstruction formula gives a measure-and-prepare channel with POVM $E_a=d{p}_a{\tau }_a^T$. Conversely, applying an entanglement-breaking channel to a maximally entangled state produces a separable Choi state. Horodecki, Shor and Ruskai's entanglement-breaking channel paper develops the equivalences.

A separable positive operator is a finite sum ${\eta }_{AB}={\sum }_y{\alpha }_y\otimes {\beta }_y$ with both factors positive operators. If its trace is one, normalizing the nonzero factors gives a separable quantum state. Conversely, a nonnegative scalar multiple of a separable quantum state is a separable positive operator.

A bipartite quantum state is separable when its density operator is a convex combination ${\sum }_j{p}_j{\rho }_j^A\otimes {\rho }_j^B$ of product states; otherwise it has entanglement. Operationally, Alice and Bob can prepare a separable quantum state by sharing a classical random label $j$ and each preparing the corresponding local state.

Separable does not mean uncorrelated. The mixture in Bell correlations versus a classical mixture is separable with two terms of weight $1/2$, yet the parties' $Z$ records agree perfectly. A single product state would give independent local outcomes. Allowing a mixture of products captures shared classical randomness; excluding this possibility would incorrectly classify ordinary correlations as entanglement.