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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

Quantum no-signalling says that, without communicating a measurement outcome, any local trace-preserving operation performed on subsystem $A$ leaves all outcome probabilities for measurements on subsystem $B$ unchanged; equivalently, it leaves the reduced density matrix ${\rho }_B$ unchanged.

The two Bell states are orthogonal. Applying the inverse Bell-state preparation circuit, namely a Controlled-NOT gate from $A$ to $B$ followed by a Hadamard gate on $A$, givesA computational-basis measurement therefore distinguishes them perfectly, soIf only $B$ is accessible, however,Every measurement on $B$ consequently has the same outcome distribution under both hypotheses. With equal prior probabilities no information is available, andthe error probability of a random guess.

The printed sum contains $Z_n$ although the stated labels end at $n-1$. We use the natural periodic convention $Z_n=Z_0$, with $n\ge 3$. If an open chain was intended, omitting the final term gives the same asymptotic bound.

Set $H_j=X_{j-1}{Z}_j$, with indices modulo $n$, and implement the product-formula Hamiltonian simulationEach factor acts on two qubits, so it is a two-qubit unitary operator. More explicitly, if $a=j-1$, $b=j$, and $C_{a\to b}$ is a controlled-NOT gate,where ${\mathsf{H}}_a$ is a Hadamard gate. This gives a constant number of one-qubit and two-qubit gates per factor; one-qubit gates can also be viewed as two-qubit gates tensored with the identity.

The spectral norm obeys the triangle inequality, submultiplicativity of the operator norm, and invariance under multiplication by unitary operators. In particular, the telescoping bound for products of operators gives $\Vert V^k-W^k\Vert \le k\Vert V-W\Vert$ for unitary $V,W$. The first-order unitary product-formula error bound consequently yieldsOnly neighboring terms can have a nonzero commutator, because all other supports are disjoint. There are $n$ neighboring unordered pairs, and $\Vert [H_j,H_l]\Vert \le 2\Vert H_j\Vert \Vert H_l\Vert =2$. Thus $\Vert U-\tilde{U}\Vert \le n/k$. Choosing $k>n/\epsilon$ givesThe coarser bound that counts all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairs also proves the often-used $O(n^3/\epsilon )$ construction. Locality improves that cubic estimate to the quadratic bound above; neither assertion is a lower bound on the best possible circuit.

The controlled-NOT gate sends $|0x⟩$ to $|0x⟩$ and $|1\bar{x}⟩$ to $|1x⟩$. HenceSince the Hadamard gate satisfies $H^2=I$,

The three controlled-NOT gates have directions $1\to 2$, $2\to 1$, and $1\to 2$. On computational basis bits they act asBy linearity, the whole quantum circuit is the SWAP gate. Thereforeso $|\alpha ⟩=|\varphi ⟩$ and $|\beta ⟩=|\psi ⟩$.

Prepare an ancilla qubit in $|0⟩$. To measure $X\otimes X$, apply a Hadamard gate to each data qubit, apply a controlled-NOT gate from each data qubit to the ancilla, measure the ancilla in the computational basis, and apply a Hadamard gate to each data qubit again. The ancilla records the parity of the two rotated computational-basis bits, so outcome $0$ corresponds to eigenvalue $+1$ and outcome $1$ to eigenvalue $-1$. The data register is projected by $[I+(-1{)}^{s}X\otimes X]/2$, so its complete post-measurement state is retained. For $Z\otimes X$, perform the same ancilla-assisted Pauli measurement but apply the basis-changing Hadamard gates only to the second data qubit.

Use $n$ controlled-NOT gates in parallel. For each $j=1,\dots ,n$, the $j$th input qubit is the control and the $j$th output-register qubit is the target. Their combined action iswhich is exactly

For the controlled-NOT gate $U={\mathrm{CNOT}}_{12}$, propagation of the Pauli generators givesThus an $X$ on the control propagates forward to the target, while a $Z$ on the target propagates backward to the control.

Suppose $\tilde{V}$ has the same four conjugation rules and put $W=U^{†}\tilde{V}$. Then $W$ commutes with $X_1,X_2,Z_1,Z_2$. These generators span the full two-qubit operator algebra, so its commutant consists only of scalar multiples of the identity. Hence $W=e^{i\varphi }I$ and

Apply the gates from left to right. The first Hadamard gate and the first controlled-NOT gate produce the Bell stateThe parallel $Z\otimes H$ gates then produceThe final controlled-NOT has the lower qubit as control; it interchanges $|01⟩$ and $|11⟩$, whose amplitudes here are equal. Thus the output isBy quantum measurement in the computational basis, every basis outcome has probability equal to the squared modulus of its amplitude:

Use three controlled-NOT gates with control-target directionsOn computational-basis bits they act asThey therefore implement the Three-CNOT decomposition of the SWAP gate, and linearity proves the identity on every two-qubit state.

Each summand of $J_X$ acts on one qubit, while each summand $Z_i{Z}_j$ of $J_Z$ acts on two. Therefore $J=J_X+J_Z$ is a 2-local Hamiltonian. We haveSplit $t$ into $r$ steps of length $\delta =t/r$ and use the second-order product formulaFor one step, $\Lambda =O(\delta n^2)$ in the stated estimate, so the spectral-norm error is $O({\delta }^3{n}^6)$. The error bound for a product of unitary operators makes the total errorIt is therefore enough to choosewith $r$ also large enough that the small-step estimate applies.

All $Z_i{Z}_j$ terms commute. A factor $e^{-i\delta Z_i{Z}_j}$ uses a constant-size circuit of two controlled-NOT gates and one phase gate, up to a global phase, so one product-formula step costs $O(n^2)$ gates. The complete Hamiltonian simulation consequently has size

Write $|\psi ⟩=\alpha |0⟩+\beta |1⟩$ and $t=e^{i\theta }$. Before measurement, the three controlled-NOT gates map a computational-basis component $|x,y,0⟩$ toThe measured bit is therefore $b=y\oplus x$. For $b=0$, the unnormalized state of the first qubit isFor $b=1$, it isEach branch has probability $1/2$; after normalization and removal of the irrelevant global phase $t$, the two outputs arewith equal probability. This is probabilistic phase-gate injection.

Apply a Hadamard gate to Alice's qubit, followed by a controlled-NOT gate from Alice to Bob and another from Alice to Charlie:Thus this circuit gives the required quantum circuit preparation of a three-qubit GHZ state.

The magic state $|A_{\theta }⟩=(|0⟩+e^{i\theta }|1⟩)/\sqrt{2}$ and a short controlled-NOT gate circuit can apply either $P(\theta )$ or $P(-\theta )$ to an input, with a measured bit identifying the branch.