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Let $|\alpha ⟩=a|0⟩+b|1⟩$ be Alice's unknown qubit, and label her half of the shared Bell state by $A$ and Bob's by $B$. Alice applies a Controlled-NOT gate from the unknown qubit to $A$, then a Hadamard gate to the unknown qubit. The three-qubit state becomesAlice measures her two qubits in the computational basis and sends the two classical outcomes $(j,k)$ to Bob. Bob applies $Z^j{X}^k$, recovering $|\alpha ⟩$ with certainty. This is quantum teleportation; Alice never needs to know the state.

For the shared GHZ state, Charlie first applies a Hadamard gate to his qubit. ThenCharlie measures in the computational basis. Outcome $0$ leaves Alice and Bob with $|{\varphi }^{+}⟩$; outcome $1$ leaves $|{\varphi }^{-}⟩$, which either party converts to $|{\varphi }^{+}⟩$ by applying $Z$ after Charlie reports the outcome. Thus

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.

Across $A|BC$, the GHZ state has the Schmidt decompositionof rank two. Symmetry gives the same rank-two decomposition across $B|AC$ and $C|AB$, so the state is entangled across every bipartition.

Projecting the GHZ state onto Charlie's outcome $0$ leaves the unnormalised state $|00{⟩}_{AB}/\sqrt{2}$. After normalization Alice and Bob have $|00⟩$, a product state.

The Einstein–Podolsky–Rosen criterion of reality says that if a physical quantity can be predicted with certainty without disturbing a system, then an element of physical reality corresponds to that quantity.

Let $X_i,Y_i$ be Pauli measurements on particle $i$. The GHZ state in the question obeyswith certainty. Each local outcome can therefore be predicted by spacelike-separated measurements on the other two particles. The EPR criterion assigns predetermined values $x_i,y_i\in \{-1,1\}$ satisfying those three equations. Multiplying them and using $y_i^2=1$ givesQuantum theory instead predictsso a joint $X$ measurement has product $-1$ with certainty. The EPR elements of reality therefore contradict the quantum prediction, which is the GHZ theorem form of Bell theorem.

Let $z_A,z_B\in \{+1,-1\}$ denote the two $Z$ outcomes and $x_C\in \{+1,-1\}$ the $X$ outcome. Measuring $Z_A$ in the GHZ state projects $B$ onto the same $Z$ eigenstate, while the later $X_C$ outcome is unbiased. Hence the possible triples, in the order $(z_A,x_C,z_B)$, areeach with probability $1/4$.

Choose $(z_A,x_C,z_B)=(+,+,+)$. Immediately before $t=0$, each local readout is $I/2$. The collapse at $(t,x)=(0,0)$ is immediately in $A$'s past light cone, reaches $B$ at $t=1$, and reaches $C$ at $t=2$. ThusHere $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. The $X_C$ collapse at $(3,2)$ reaches $B$ at $t=4$ and $A$ at $t=5$, but the state is already a product after the first collapse, so it does not change their local states. Likewise, the $Z_B$ collapse at $(5,1)$ reaches $A$ and $C$ at $t=6$ without changing their states. For another allowed outcome, replace $|0⟩$ by $|1⟩$ according to $z_A=z_B$ and replace $|+⟩$ by $|-⟩$ according to $x_C$.