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The Schmidt decomposition theorem states that a normalized vector in a finite-dimensional tensor product ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$ has the form ${\sum }_{j=1}^r{s}_j|u_j⟩|v_j⟩$, where the two families are orthonormal, $s_j>0$, ${\sum }_j{s}_j^2=1$, and $r\le \min (\dim {\mathcal{H}}_A,\dim {\mathcal{H}}_B)$. Extending each family to an orthonormal basis for two qubits givesThe nonzero Schmidt coefficients can be made positive by absorbing phases into the basis vectors. Both are positive exactly when the pure state is entangled. A product state has Schmidt rank one, so one coefficient is zero; the source's assertion of two positive coefficients for every pure state needs this exception.

Choose each local orthonormal basis independently as the computational basis. In that basis ${\sigma }_z=|0⟩⟨0|-|1⟩⟨1|$. This is a local change of coordinates, implemented by separate unitary matrices, rather than a physical restriction on the original quantum state. The associated Pauli operators supply the other two local axes. Up to an irrelevant common phase, each unitary change of qubit basis corresponds to a rotation of its Bloch sphere.

Put $s=2{\lambda }_0{\lambda }_1$. The Schmidt-basis Pauli correlation tensor is diagonal. The Pauli operators ${\sigma }_x\otimes {\sigma }_x$ exchange $|00⟩$ and $|11⟩$, while ${\sigma }_y\otimes {\sigma }_y$ do so with minus signs, and ${\sigma }_z\otimes {\sigma }_z$ leaves both fixed. HenceMixed components vanish: those containing one $z$ and one transverse Pauli operator map the occupied basis vectors outside their span, while the $xy$ and $yx$ matrix elements are purely imaginary and cancel for real Schmidt coefficients. By bilinearity, for arbitrary real vectors,For a CHSH inequality test take unit quantum measurement axesThese CHSH axes for an entangled pure two-qubit state giveThe local bound is two: for each hidden state, $A(B+B^{\prime })+A^{\prime }(B-B^{\prime })$ is $\pm 2$ when all four outcomes are $\pm 1$, and averaging cannot increase its absolute value. Thus every entangled pure two-qubit state violates a CHSH inequality, the content of Gisin's theorem. A product state has $s=0$ and does not violate it, so the unqualified final claim in the question is false for that case. The maximum $2\sqrt{2}$ occurs for equal Schmidt coefficients. This excludes local hidden-variable theories satisfying measurement independence, but does not permit faster-than-light signalling: the local reduced density matrix and its quantum measurement probabilities are unchanged by the remote choice of axis.

Every separable quantum state has positive partial transpose. Here the partial transpose iswith eigenvalues $1/2,1/2,\alpha /2,-\alpha /2$. Any nonzero $\alpha$ therefore gives a negative eigenvalue, excluding a separable quantum state. At $\alpha =0$, the explicit decompositionis a mixture of product states. ConsequentlyOnly necessity of the positive partial transpose criterion is needed here; sufficiency at zero follows directly from the decomposition.

Let $\Gamma =T_B$ denote the partial transpose. The partial transpose of a maximally entangled projector iswhere $S$ is the swap operator, satisfying $S^{†}S=I$ and $\Vert S{\Vert }_{\mathrm{op}}=1$. If $\sigma$ has positive partial transpose, then ${\sigma }^{\Gamma }$ is a positive operator of trace one, hence $\Vert {\sigma }^{\Gamma }{\Vert }_1=1$.

For a pure state ${\varphi }^{+}=|\Phi ⟩⟨\Phi |$, the unsquared quantum fidelity obeys $F({\varphi }^{+},\sigma {)}^2=⟨\Phi |\sigma |\Phi ⟩$. Using the trace-adjoint identity for partial transpose and part (i),The overlap is nonnegative, so taking its square root proves the PPT maximally entangled overlap bound:The product state $|00⟩⟨00|$ attains the bound, so it is sharp.

Write $|a⟩={⨂}_j|a_j⟩$. Instead of storing the output vector, use Heisenberg propagation of a Pauli observable through a Clifford circuit:Initialize the compact Pauli group representation at $Z_1$ and conjugate successively by $U_N,U_{N-1},\dots ,U_1$, in that order, using $P\mapsto U_j^{†}P{U}_j$. This giveswith $O(N+n)$ local-update work and $O(n)$ storage. The resulting Pauli operator is Hermitian, so we can express it as $P=\eta {⨂}_j{\sigma }_j$, where $\eta \in \{+1,-1\}$ and each ${\sigma }_j$ is $I,X,Y$ or $Z$. Any factors $XZ$ in the representation from part (i) are converted using $XZ=-iY$, with their phases absorbed into $\eta$.

The product state input now makes the expectation factorize:Each factor is a two-by-two matrix calculation. More explicitly, for $|a_j⟩={\alpha }_j|0⟩+{\beta }_j|1⟩$, the three nontrivial expectations are $2\mathrm{Re}({\overline{\alpha }}_j{\beta }_j)$, $2\mathrm{Im}({\overline{\alpha }}_j{\beta }_j)$, and $|{\alpha }_j{|}^2-|{\beta }_j{|}^2$ for $X,Y,Z$, respectively.

Thus this single-output process has a strong classical simulation in polynomial time, even when the individual input states are not stabilizer states. As usual, the given state identities must provide efficiently computable amplitudes. With finite-precision inputs, each factor can be evaluated to error $\delta /n$ and its approximation clipped to $[-1,1]$ to obtain the final expectation to error at most $\delta$, because all factors lie in $[-1,1]$. The required extra precision is only $O(\log (n/\delta ))$ bits. Exact evaluation applies when the input descriptions support exact arithmetic. This avoids silently assigning a finite exact-computation cost to arbitrary unspecified real numbers.

Once the probabilities are known, a classical randomized computation can reproduce this measured bit. The argument concerns the specified one-qubit output; it does not assert that every joint measurement distribution for arbitrary product inputs can be simulated by this same single-observable calculation.

Use GHZ preparation with Hadamard and controlled-Z gates for $n\ge 2$. Apply a Hadamard gate to the first qubit, then a controlled-NOT gate from that qubit to each of the other $n-1$ qubits. Every controlled-NOT can be written in the allowed gate set asIndeed conjugating the target $Z$ by $H$ makes the controlled phase into a controlled $X$. The rightmost gate acts first. Consequently the Clifford circuit uses $1+3(n-1)$ allowed gates and givesFor every chosen qubit $j$, tracing out the other qubits yields ${\rho }_j=\frac{1}{2}(|0⟩⟨0|+|1⟩⟨1|)=I/2$. Equivalently, across that qubit versus the rest, the displayed state is a Schmidt decomposition with two nonzero coefficients $1/\sqrt{2}$. It is therefore entangled across every such bipartition.

Entanglement does not prevent the polynomial single-output simulation from part (iii). The simulation follows the measured observable rather than claiming that the intermediate states remain product states. When $n=1$, there is no remaining subsystem and the requested entanglement is impossible; the construction and claim require $n\ge 2$.

A pure bipartite state is entangled when it is not a product state $|u{⟩}_A\otimes |v{⟩}_B$.

Sincewe haveIts coefficient matrix is proportional towhose determinant is $-4\ne 0$. By the two-qubit product-state determinant criterion, a product state would have determinant zero. Therefore $|\alpha ⟩$ is entangled.

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.

Assume a finite-dimensional memoryless quantum channel, no shared entanglement, and a code of $M_n$ messages with $n^{-1}{\log }_2{M}_n\to R$. Use a uniform message $M$ to relate average and maximum error. Each permitted input is a product state ${\rho }_m^{(n)}={⨂}_{i=1}^n{\rho }_{m,i}$, giving output ${\sigma }_m^{(n)}={⨂}_i\Lambda ({\rho }_{m,i})$. The factors may depend on both the message and the channel use.

Let ${\chi }^{\ast }(\Lambda )$ denote the one-use Holevo capacity in the question. The Subadditivity of Von Neumann entropy for the average output, and entropy additivity for each product output, implyThis is the key step of the product-input classical-capacity converse; it does not assume that the average output itself is a product.

Bob may measure all outputs jointly. The Holevo bound still gives $I(M:\hat{M})\le {\chi }_n$. Let $P_e$ be his average message-error probability. Fano's inequality givesSince the maximum error $P_{\max }$ is at least $P_e$,when $R>{\chi }^{\ast }(\Lambda )$. Thus the maximum error cannot vanish asymptotically.

For $N=4$, the binary label $10$ is the integer $2$, sowhere $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. It is a product state and hence has no quantum entanglement.

The relation $z^{2^n}=1$ makes $z$ a root of unity, so $|z|=1$ and normalization gives $T=2^{n/2}$. Write the binary expansion $y={\sum }_{j=0}^{n-1}{2}^j{y}_j$. Then $z^y={\prod }_j(z^{2^j}{)}^{y_j}$, and henceup to the convention for ordering the binary digits. This explicit tensor product of one-qubit states proves that $|\varphi ⟩$ is a product state.

The pure state $|\Psi {⟩}_{AB}$ is entangled when it cannot be written as a product state $|a{⟩}_A|b{⟩}_B$, equivalently when its Schmidt rank exceeds one. Its subsystem states are the reduced density matriceswhere the partial trace is characterized by $\mathrm{Tr}(M_A{\rho }_A)=⟨\Psi |M_A\otimes I_B|\Psi ⟩$ for every observable $M_A$.

Let Bob's measurement have Kraus operators $M_b$ satisfying ${\sum }_b{M}_b^{†}{M}_b=I_B$. If Alice does not learn Bob's random outcome, her state after the measurement isThis no-communication theorem means that Bob can change Alice's conditional state after she learns his outcome, but cannot change any local outcome distribution available to her alone. Standard nonrelativistic quantum mechanics is therefore operationally compatible with the prohibition of superluminal signalling in special relativity, despite its nonlocal conditional-state updates.

An exact nondisturbing state readout would destroy this protection if it reported the globally collapsed state on an absolute-time slice. For example, Alice and Bob may share Bell states. At a prearranged time Bob encodes a bit by measuring his qubit in either the computational or Hadamard basis. The usual projection postulate assigns Alice respectively one of $|0⟩,|1⟩$ or one of $|+⟩,|-⟩$. A device returning the exact pure state lets Alice identify which basis Bob chose without waiting for his outcome, producing a superluminal signal. Equivalently, Bob may choose whether to measure, and the device distinguishes the resulting proper pure state from the original improper maximally mixed local state.

A causal alternative is a quantum state readout device located at a spacetime point $x$ that reports the local quantum state under objective collapse: take reduced states on spacelike hypersurfaces through $x$ and let those hypersurfaces approach the past light cone of $x$. The result includes localized collapses in $J^{-}(x)$ and excludes spacelike-separated collapses. This assumes a fixed Minkowski spacetime, localized collapse events, ordinary local unitary dynamics between them, and outputs that may control only operations in their causal future. The postulate is logically consistent because it adds a classical record of this local state without changing the state or any standard measurement probability. It also cannot support an indirect signalling algorithm. Inductively through the algorithm's events, every readout at $x$ depends only on operations, collapses, and earlier readouts in $J^{-}(x)$; any operation selected from that output lies in the readout's future light cone. Composing readouts, unitary evolutions, and measurements therefore never carries a controllable dependence outside a future light cone, so relativistic causality is preserved.

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.

A strong classical simulation of a quantum circuit computes any requested output probabilityin polynomial time, to the prescribed inverse-polynomial accuracy. A weak classical simulation of a quantum circuit instead produces classical samples from the circuit's output distribution, with exact or suitably small total-variation error.

The Extended Gottesman--Knill theorem states that a unitary Clifford circuit with an arbitrary product state input and final computational-basis measurements is weakly classically simulable. It is strongly simulable when only $O(\log n)$ output qubits are measured. Indeed, each joint output projector expands into $2^k$ Pauli operators, and Clifford conjugation maps every such operator to another Pauli operator whose expectation factors over the input qubits. The factor $2^k$ is polynomial precisely for $k=O(\log n)$.

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.

Write the spin-one-half Heisenberg antiferromagnet ason a bipartite lattice of coordination number $z$. A one-spin density operator is $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, where $\mathbf{r}$ is its Bloch vector and $|\mathbf{r}|\le 1$. In a two-sublattice product state,The minimum is $-1/4$, attained by pure antiparallel Bloch vectors, so the minimizing product state is a Néel state. In the $z\to \infty$ limit this mean-field approximation becomes exact and the ground-state energy per bond is therefore

The phrase “energy density” requires a coupling convention. For the unscaled Hamiltonian above, every site belongs to $z/2$ bonds andwhich diverges as $z\to \infty$. With the standard Kac normalizationthe finite energy density isIf the convention divides by the spatial dimension $d=z/2$ instead, the answer is $-J/4$ per site. A Hamiltonian written with ${\mathbf{\sigma }}_i\cdot {\mathbf{\sigma }}_j$ rather than ${\mathbf{S}}_i\cdot {\mathbf{S}}_j$ multiplies all these energies by four.