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For a bipartite input density operator ${\rho }_{AC}^x$ and a POVM $\{E_y\}$ on $A$, let $q(y|x)=\mathrm{Tr}[(E_y\otimes I){\rho }^x]$. For $q(y|x)>0$, the conditioned input on $C$ isIts positivity follows from the positive sandwich before the partial trace. A quantum channel on $C$ acts on these conditional density operators even if the original input is an entangled state.

An entanglement witness is a Hermitian operator $W$ such that $\mathrm{Tr}(W\rho )\ge 0$ for every separable quantum state $\rho$, but $\mathrm{Tr}(W\sigma )<0$ for at least one entangled state $\sigma$.

When two masses occupy spatial superpositions, their branch-dependent Newtonian gravitational potential energy produces relative phases. If those phases cannot be separated into one phase for each mass, the interaction turns an initial product state into an entangled state.

Use one-bit remote preparation of real qubit states. Alice measures her qubit in the orthonormal basisBecause this change of basis is real and orthogonal, the shared Bell state has the expansionHer projective measurement gives each outcome with probability one half. She sends $m=0$ for the first outcome and $m=1$ for the second. Bob's conditional pure state is respectively $|{\alpha }_{\theta }⟩$ or $|{\beta }_{\theta }⟩$.

For $m=0$ Bob does nothing. For $m=1$ he applies the fixed unitary operatorwhere $Y$ is the Pauli Y gate. Bob need not know $\theta$, because this correction is the same for all real $\theta$. Both outcomes produce the exact density operator $|{\alpha }_{\theta }⟩⟨{\alpha }_{\theta }|$, using one classical bit and only local operations.

This remote state preparation consumes the shared entangled state. Before receiving the bit, Bob has the mixture $(|{\alpha }_{\theta }⟩⟨{\alpha }_{\theta }|+|{\beta }_{\theta }⟩⟨{\beta }_{\theta }|)/2=I/2$, independent of $\theta$, as required by quantum no-signalling. The real-state restriction is what permits a fixed unitary correction of the orthogonal outcome.

Write $X$ and $Y$ for the classical quantum registers $\tilde{X}$ and $\tilde{Y}$. By the quantum mutual information formula $I(U:V)=S(U)+S(V)-S(UV)$, expanding the right-hand side of the quantum mutual information balance identity givesAll Von Neumann entropies here are evaluated in $\omega$.

We use three facts: quantum mutual information is nonnegative by nonnegativity of quantum relative entropy; a local quantum channel cannot increase it by data processing for quantum mutual information; and the quantum mutual information of a classical-quantum state is its ensemble's Holevo quantity. Write $\chi (\mathcal{T})$ for the Holevo capacity, the supremum of the output Holevo quantity over finite input ensembles.

The $XB$ marginal is an output ensemble for $\mathcal{E}$, with inputs $\rho (x{)}_A={\mathrm{Tr}}_C\rho (x{)}_{AC}$. Consequently $I(X:B{)}_{\omega }\le \chi (\mathcal{E})$. Also ${\omega }_{XBD}$ is obtained from ${\sigma }_{XYD}$ by a quantum channel on $XY$ which retains $X$ and prepares $B$ from $Y$. HenceTo bound the latter even for entangled states $\rho (x{)}_{AC}$, exhibit the conditional input ensemble after a local measurement. Setwhen $q(y|x)>0$; zero-weight outcomes can be omitted. The numerator is a positive operator, its trace is $q(y|x)$, and the $q(y|x)$ sum to one. ThusThis is a classical-quantum state with an output ensemble for $\mathcal{N}$, so $I(XY:D{)}_{\sigma }\le \chi (\mathcal{N})$. Combining these bounds with $I(B:D{)}_{\omega }\ge 0$ yieldsThe last step takes the supremum over all input ensembles on $AC$. The left-hand channel direction is $B\leftarrow A$, as established in part (i); the printed $A\leftarrow B$ in the last inequality is a typographical reversal. Independent product ensembles also give the reverse inequality, so this proves Holevo-capacity additivity for entanglement-breaking channels.

A product state factors as $|{\psi }_1⟩\otimes |{\psi }_2⟩$; an entangled state admits no such factorization. The state $|J,J⟩$ is a product state. When $j_1,j_2>0$, each state with $m=J-1$ displayed in part (iii) has two nonzero terms with orthogonal states in both subsystems, so its Schmidt rank is two and it is entangled. If one subsystem has spin zero, only the $j=J$ multiplet remains and its states factor through the unique state of that subsystem.

For particle positions ${\mathbf{r}}_1,{\mathbf{r}}_2$, the two-particle Schrodinger 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.

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.

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.

Remote state preparation uses a shared entangled state, local quantum measurements and classical communication to prepare a state known to the sender at the receiver. The sender can tailor the measurement to the desired state, so restricted state families can require less communication than quantum teleportation. A real great-circle family of qubit states admits one-bit remote preparation of real qubit states.