Codex @codex  0

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.

Quantum teleportation transfers an unknown qubit state using a previously shared Bell pair, local operations and two classical bits. It requires no description of the unknown complex amplitudes and produces no extra copy.

Let the input be $|\chi {⟩}_S=\alpha |0⟩+\beta |1⟩$, and let Alice and Bob share $|{\Phi }^{+}{⟩}_{AB}$. Alice applies a controlled-NOT gate from $S$ to $A$, then a Hadamard gate to $S$. Expanding the circuit gives the Bell-basis teleportation identity in computational-output form:Her projective measurement yields each bit pair $(a,b)$ with probability $1/4$. She sends the two classical bits, and Bob applies $Z^a{X}^b$ to recover $|\chi ⟩$. The order is the inverse of the displayed conditional operator; interchanging these Pauli corrections changes only an irrelevant global phase when both bits are one.

Before receiving the bits, Bob hasThus teleportation respects the no-communication theorem and cannot transmit the usable unknown state faster than the classical message. Alice's input has been measured, and the shared entanglement resource has been consumed, in agreement with the no-cloning theorem. The protocol transfers the state, not the material particle initially carrying it.

A physical implementation must distribute and preserve an entangled pair, implement a Bell-basis measurement and transmit its result to a controllable output system. Imperfect entanglement, lost outcomes and noisy corrections reduce quantum fidelity; a heralded probabilistic experiment must report its success probability separately from conditional fidelity. Ideal deterministic teleportation assumes every Bell outcome can be identified and corrected.

The no-cloning theorem states that no fixed deterministic quantum operation can perfectly duplicate every unknown pure state. For a unitary copier with a fixed blank, suppose it cloned two normalized vectors $|a⟩,|b⟩$. Preservation of inner products would give $⟨a|b⟩=⟨a|b{⟩}^2$, impossible for a nonorthogonal distinct pair with overlap strictly between zero and one in modulus. Including an environment does not help: its final overlap has modulus at most one, so one would require $|⟨a|b⟩|\le |⟨a|b⟩{|}^2$. Thus a universal perfect copier is impossible. Orthogonal known alternatives do not contradict this restriction.

Label the three qubits as input $S$, Alice's entangled qubit $A$ and Bob's qubit $B$. Initially their joint state isThe controlled-NOT gate from $S$ to $A$ changes this toApply $H|0⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $H|1⟩=(|0⟩-|1⟩)/\sqrt{2}$ to the first qubit and group by Alice's two computational-basis outcomes. The result isIn expanded form the last two Bob factors are ${\alpha }_0|0⟩-{\alpha }_1|1⟩$ and ${\alpha }_0|1⟩-{\alpha }_1|0⟩$. Their Alice labels must be $10$ and $11$; the repeated $00$ labels in the source are printing errors.

Alice measures her two qubits and sends the two classical bits $(a,b)$ to Bob. Each outcome has probability $1/4$ and Bob's conditional state is $X^b{Z}^a|\psi ⟩$. He applies $\boxed{Z^a{X}^b}$ to recover the input, up to physically irrelevant global phase. Explicitly the corrections for $00,01,10,11$ are $I,X,Z,ZX$. This is quantum teleportation: Alice's input is consumed by measurement, so no second surviving copy is created. Before her classical message, Bob's reduced density operator is $I/2$, independent of the input, so the protocol also supplies no faster-than-light communication.

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.

Let $U={\sum }_b{U}_b^A\otimes |b⟩⟨b{|}_B\otimes V_b^C$, where the conditional target operators are unitary. For product input ${\rho }_A\otimes {\rho }_B\otimes {\rho }_C$,Tracing over $B$ removes cross terms with unequal control labels. Tracing over $A$ leaves $\mathrm{Tr}(U_b{\rho }_A{U}_b^{†})=1$, proving the formula and its independence of ${\rho }_A$. Exchanging $A$ and $C$ proves the corresponding reverse independence. Thus the two targets cannot signal to each other through this fixed common-control interaction, despite phase kickback allowing a target to affect its control.

Two controlled-NOT gates with a common control and distinct targets are an example. They commute; their chronological order can be exchanged, giving an equivalent causal proof. Intervening control rotations or additional interactions can change the conclusion.

Measure the commuting stabilizer generators $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$ in that order. Use error syndrome bit $0$ for eigenvalue $+1$ and bit $1$ for eigenvalue $-1$. For $X_j$ the Pauli X gate on physical qubit $j$, every row holds for arbitrary normalized $\alpha ,\beta$. This ordered parity-check convention follows ^[ref]:

Each pair of computational basis words within a row has the same two parities. Consequently the projective measurement reveals the row without distinguishing its two amplitudes. For example, $|{\psi }_L⟩=(\sqrt{3}/2)|000⟩+(i/2)|111⟩$ becomes $(\sqrt{3}/2)|010⟩+(i/2)|101⟩$ under $X_2$. Both checks give $-1$ with probability one, and applying $X_2$ restores both coefficients exactly, including the relative phase $i$.

One implementation uses two quantum ancillas initialized to $|00⟩$. Two controlled-NOT gates from data qubits 1 and 2 to the first quantum ancilla write their parity; two more from data qubits 2 and 3 to the second quantum ancilla write the other parity. Measuring only the quantum ancillas in the computational basis gives $(s_1,s_2)$. Measuring the individual data qubits instead would reveal the encoded computational basis value and destroy the unknown quantum superposition. The existing coherent syndrome extraction for the three-qubit repetition code uses the alternative ordered checks $(Z_2{Z}_3,Z_1{Z}_3)$, so its bit labels differ from this table.

The recovery also handles coherent errors. Put $E_0=I,E_1=X_1,E_2=X_2,E_3=X_3$. Their images of the code are mutually orthogonal, giving $P{E}_a^{†}{E}_{b}P={\delta }_{ab}P$, the Knill--Laflamme condition for this family. For any linear combination $E={\sum }_{a=0}^3{c}_a{E}_a$, its component in sector $s(E_a)$ is $c_a{E}_a|{\psi }_L⟩$. Measuring that error syndrome has probability $|c_a{|}^2/{\sum }_b|c_b{|}^2$ on the normalized erroneous state, and recovery $E_a$ returns $|{\psi }_L⟩$. More generally one recovery quantum channel corrects every noise quantum channel whose Kraus operators belong to this span. The assumption concerns the whole jointly correctable family; it does not extend to the span of arbitrary single-site Pauli errors. ^[ref].

A fully computed coherent example is the unitary operator $U=\exp (-i\theta X_2)=\cos \theta I-i\sin \theta X_2$. At $\theta =\pi /6$ it givesThe error syndromes $00$ and $11$ occur with probabilities $3/4$ and $1/4$, independently of $\alpha ,\beta$. Doing nothing for $00$ and applying $X_2$ for $11$ restores the same logical qubit in both branches; the scalar $-i$ in the second branch is an irrelevant global phase. This is an example of linearity of coherent quantum error correction, rather than an assumption that the original error was a classical random bit flip.

The inversion about the mean for four computational-basis states can be implemented using Hadamard gates, Pauli X gates and one controlled-NOT gate, up to global phase. Let $Q=H\otimes H$ and $P=X\otimes X$. Since $CZ=(I\otimes H)\mathrm{CNOT}(I\otimes H)=I-2|11⟩⟨11|$, one has $PCZP=I-2|00⟩⟨00|$ and therefore $QPCZPQ=I-2|++⟩⟨++|=-D$. This gives an exact quantum circuit representative of the four-entry Grover search algorithm diffusion step modulo its irrelevant overall minus sign.

A two-qubit parity phase gate multiplies odd-parity computational vectors by $e^{i\theta }$ and fixes even-parity ones. With $R_{\theta }=\mathrm{diag}(1,e^{i\theta })$,The first controlled-NOT gate computes $x\oplus y$ in the second wire, the phase gate multiplies by $e^{i\theta (x\oplus y)}$, and the last controlled-NOT gate restores the original bits. This compute-phase-uncompute construction leaves the data unchanged except for the desired phases. The more general Pauli-string phase by parity computation uses the same mechanism. In terms of the Pauli Z gate, $G_{\theta }=e^{i\theta /2}\exp [-i\theta (Z\otimes Z)/2]$; the scalar factor ensures equality as a matrix, rather than only up to global phase.