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Consider a depth-$D$ circuit whose gates have range at most $r$. Under backward Heisenberg evolution, a one-site observable has a backward light cone of a local quantum circuit of radius at most $rD$. Choose sites $i,j$ separated by $L>2rD$. Their two backward light cones are disjoint. Since the input is a product state, expectations factorize, and hence every connected correlation function between the two output observables vanishes.

For the GHZ statehowever,at every separation. The light cones must therefore overlap, which forcesFor opposite ends of a one-dimensional chain, $L=\Theta (N)$, so $D=\Omega (N)$ and no constant-depth local circuit can prepare the GHZ state. The continuous-time version follows directly from the Lieb-Robinson bound, with preparation time at least $L/(2v_{\mathrm{LR}})$ up to exponentially small tails.

This is the GHZ-state circuit-depth lower bound. Finite-depth local circuits define equivalence within a gapped phase, so a state with this long-range order cannot lie in the same circuit phase as a product state. The GHZ state is the finite-size cat state associated with spontaneous symmetry breaking; it is not a unique short-range-entangled ground state. The persistent distant correlation is precisely the obstruction.

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

Suppose output qubit $j$ is measured. Its probability of outcome zero isBecause $C$ is a Clifford operation, the Pauli group is preserved under conjugation. Propagating $Z_j$ backward through the $N=\mathrm{poly}(n)$ gates therefore produces, including its sign, a tensor-product PauliThe input is a product state, so the expectation factors:Each one-qubit factor follows directly from the classical description of $|{\alpha }_k⟩$, and conjugating a Pauli through each Clifford gate takes constant classical work. Hence $p_0$, and $p_1=1-p_0$, are computable in polynomial time. This is Heisenberg propagation of a Pauli observable through a Clifford circuit.

For positions ${\mathbf{r}}_1,{\mathbf{r}}_2$, the two-particle Schrodinger equation isNeglecting packet spreading and writing $|a{⟩}_i$ for ${\psi }_{ai}$, branchwise evolution givesup to local kinetic phases. This branch-dependent Newtonian gravitational potential energy produces gravitationally induced entanglement.

Under the stated distance approximation, only the three branches $a=b$ acquire an appreciable common phaseThe coefficient matrix iswhere $J$ is the $3\times 3$ all-ones matrix. The reduced density matrix isIt equals $I/3$ when $\cos \varphi =-1/2$, so the first maximally entangled state occurs at $\varphi =2\pi /3$. Therefore

The state does not remain entangled for every $t>0$. Whenever $\varphi =2\pi k$, all branch phases again agree and the state returns to its initial product state. The revival period is

For equal masses with positions ${\mathbf{r}}_1,{\mathbf{r}}_2$, the two-body Schrodinger equation isNeglecting wave-packet spreading and overlap, write $|a{⟩}_i$ for the packet ${\psi }_{ai}$. The initial state and its branchwise gravitational evolution areup to local kinetic phases. The branch dependence of the Newtonian gravitational potential energy is the source of gravitationally induced entanglement.

Under the approximation stated in the question, only the matching pairs $a=b$ acquire an appreciable common phaseThe coefficient matrix of the resulting bipartite state iswhere $J$ is the all-ones matrix. ThereforeA maximally entangled state would require ${\rho }_1=I/n$. This demandsThat equation has no solution for $n>4$. Thus the requested large-$n$ conclusion does not follow from the assumptions printed in the paper: within the stated approximation, the system never becomes maximally entangled in the large-$n$ limit. For completeness, maximal entanglement is possible for $n=3$ at $\varphi =2\pi /3$ and for $n=4$ at $\varphi =\pi$; for $n=2$ it occurs at $\varphi =\pi /2$ or $3\pi /2$. The $n=4$ time would bebut it is not a large-$n$ answer.

The system also does not remain entangled for every $t>0$. Whenever $\varphi =2\pi k$, the phase matrix again factorizes and the state returns to its initial product state. The revival period isFor $n>2$ it is entangled at all intervening times; for $n=2$ there is the additional product-state revival at $\varphi =\pi$.

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.

Every density operator $\sigma$ with positive partial transpose has squared quantum fidelity at most $1/d$ with a fixed maximally entangled state on $d\otimes d$. The partial transpose of a maximally entangled projector and the trace-adjoint identity for partial transpose express that squared overlap as $\mathrm{Tr}(S{\sigma }^{T_B})/d$. Since ${\sigma }^{T_B}$ is positive of trace one and the swap operator has operator norm one, trace duality gives the bound. A matching product state attains equality.

For a code with product state inputs to $n$ uses of a memoryless quantum channel, Subadditivity of Von Neumann entropy bounds its output Holevo quantity by $n{\chi }^{\ast }(\Lambda )$. The Holevo bound and Fano's inequality then give, for asymptotic rate $R$, average and hence maximum error at least $1-{\chi }^{\ast }(\Lambda )/R$ in the limit inferior. In particular, no code above that rate can have vanishing maximum error, even with a collective decoder measurement.

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