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Apply a Hadamard gate to the first qubit, then a controlled-NOT gate from the first to the second. This maps a computational-basis state $|ij⟩$ to the Bell state with phase bit $i$ and parity bit $j$. Its inverse uses the same gates in reverse order because each gate is self-inverse but they do not commute. Applying the forward sequence twice is not generally a decoding operation.

The controlled-Z gate is $CP(\pi )=\mathrm{diag}(1,1,1,-1)$. Conjugating the target of a controlled-NOT gate by Hadamard gates turns controlled-NOT into controlled-Z.

For $n\ge 2$, apply a Hadamard gate to one qubit, then controlled-NOTs from it to all others. Replacing each controlled-NOT gate by a target Hadamard, a Controlled-Z gate, and another target Hadamard uses $1+3(n-1)$ allowed gates. Each one-qubit reduced state is $I/2$, proving entanglement across each qubit-versus-rest bipartition.

The inverse of a quantum circuit applies the adjoints of its gates in reverse temporal order. The Hadamard gate, Pauli Z gate, and controlled-NOT gate are each self-inverse.

Apply a Hadamard gate to qubit $A$, followed by a controlled-NOT gate with $A$ controlling $B$. The resulting Bell-basis conversion circuit is $U=\mathrm{CNOT}(H\otimes I)$ and givesThus $00,01,10,11$ map respectively to ${\Phi }^{+},{\Psi }^{+},{\Phi }^{-},{\Psi }^{-}$. The first input bit becomes the phase bit and the second becomes the parity bit.

Both gates are self-inverse, but inversion of their product reverses the order:So the same two gates convert the Bell basis back to the computational basis when applied in reverse order: first the controlled-NOT, then the Hadamard. Using the forward order twice does not generally work, since these gates do not commute. For example, $U|{\Phi }^{+}⟩=(|00⟩+|01⟩-|10⟩+|11⟩)/2$, which is not a computational-basis state.

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 $n\ge 2$.

For $|\psi ⟩=a|0⟩+b|1⟩$, the Controlled-NOT gate givesA cloned product would instead beEquality up to a physically irrelevant global phase requires $ab=0$. ThusFor a genuine superposition it creates quantum entanglement instead of two independent copies.