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The J gate is $J(\alpha )=HP(\alpha )$, with $P(\alpha )=\mathrm{diag}(1,e^{i\alpha })$. Prepare a fresh qubit in $|+⟩$ and apply the Controlled-Z gate $E$ between it and the input $|\psi ⟩=u|0⟩+v|1⟩$. The resulting state isMeasure the input in the equatorial qubit measurement basis $|{\alpha }_s⟩=(|0⟩+(-1{)}^s{e}^{-i\alpha }|1⟩)/\sqrt{2}$. The unnormalized output isEach outcome has probability $1/2$. Thus one-bit teleportation realizeson the new qubit. Apply the known Pauli X gate correction $X^s$ for the literal $J(\alpha )$ output, or keep the correction in a Pauli frame and adapt later measurements. The old qubit is measured, so this is not cloning the input.

Direct multiplication of the given matrices gives $J(\alpha )X^s=e^{+is\alpha }{Z}^{s}J((-1{)}^s\alpha )$. The displayed negative exponent in the supplied relation has the wrong sign for exact matrix equality. The discrepancy is only a global phase in a fixed measurement branch, so it does not change this measurement implementation or its outcome probabilities. The positive-sign identity is used when tracking exact matrices.

Use superdense coding. For the two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies $Z^a{X}^b$ to her half of the shared Bell state, where $X,Z$ are the Pauli X gate and Pauli Z gate. The encoded states areThese four Bell states are orthonormal. Alice then transmits her one qubit to Bob, who now holds both qubits. A Bell-basis measurement identifies the message with probability one.

More explicitly, Bob can apply a CNOT gate with the received qubit as control and his original qubit as target, followed by a Hadamard gate on the control. The CNOT gate sends the encoded state toand the Hadamard gate sends this to $|a⟩\otimes |b⟩$. Computational-basis projective measurements therefore read out $a,b$ exactly. One transmitted qubit, assisted by one previously shared Bell pair, communicates two classical bits.

The basic one-bit teleportation primitive uses the input as qubit 1 and a fresh $|+⟩$ quantum ancilla as qubit 2. Apply the Controlled-Z gate $E_{12}$ and measure qubit 1 in the equatorial qubit measurement basis with angle $\alpha$. If $s\in \{0,1\}$ is the result, the normalized post-measurement state on qubit 2 isBoth results have probability one half. For example, writing $|\psi ⟩=a|0⟩+b|1⟩$, the unnormalized branch is $(a|+⟩+(-1{)}^s{e}^{i\alpha }b|-⟩)/\sqrt{2}=X^{s}J(\alpha )|\psi ⟩/\sqrt{2}$. Thus $J(\alpha )$ is applied deterministically as a logical gate with known Pauli frame $X^s$; a permitted conditional Pauli X gate would remove this byproduct physically.

The literal resource restriction does not need an unstated direct Pauli X gate. A Pauli Z gate can be enacted by applying $E$ to the data and a fixed $|1⟩$ quantum ancilla. Two consecutive angle-zero one-bit teleportations, with results $p,q$, map an arbitrary current state $|\chi ⟩$ to $X^{q}H{X}^{p}H|\chi ⟩=X^q{Z}^p|\chi ⟩$. Apply the available $Z^p$ to remove the latter factor, up to an irrelevant global phase. This implements a heralded $X^q$: if $q=0$, the input is unchanged; if $q=1$, the required Pauli X gate has been applied.

If the original result is $s=1$ and a physically corrected output is required, repeat this heralded Pauli X correction using controlled-Z and measurements until $q=1$. Each attempt succeeds with probability one half independent of the input, so it terminates with probability one, using two attempts on average. Only $E$, fixed ancillary quantum states and single-qubit quantum measurements are used. This exact physical correction has no finite worst-case measurement bound; the standard finite deterministic realization is the logical Pauli frame version, which suffices for the later output-simulation parts.

Let the ideal two-qubit output before final quantum measurements beThe PDF places $J(\alpha )$ only on the upper wire, followed by the Controlled-Z gate and a quantum measurement in the computational basis on each output wire. Measure vertex $0$ of the square in the computational basis, obtaining $r$, then vertex $1$ of the surviving path in the fixed equatorial qubit measurement basis at angle $\alpha$, obtaining $s$. Put $t=r\oplus s$. The previous part gives $E_{23}{X}_2^{t}J(\alpha {)}_2{Z}_3^r|++⟩$.

Commute its Pauli frame through the Controlled-Z gate. Since $E_{23}{X}_2^t=X_2^t{Z}_3^t{E}_{23}$ and $E_{23}$ commutes with $Z_3^r$, the actual output isFinally measure vertices $2,3$ in the computational basis, with raw results $d_2,d_3$. A Pauli X gate flips a computational basis result, whereas a Pauli Z gate changes only its phase. Thus the purely classical correction isThis four-cycle graph-state simulation of an entangle-and-measure circuit reproduces the full joint output distribution, not just each marginal. All measurement bases are fixed beforehand, and no physical byproduct correction is necessary. As a check, the ideal Controlled-Z gate is diagonal in the output basis, so $P(k,l)=\frac{1}{2}{\cos }^2(\alpha /2)$ for $k=0$ and $\frac{1}{2}{\sin }^2(\alpha /2)$ for $k=1$, independently of $l$. The measurement procedure gives exactly these probabilities after its classical relabelling.

Apply a Pauli X gate and then a Hadamard gate to the third qubit, and apply Hadamard gates to the first two. This preparesOne application of the Boolean quantum oracle now gives, by quantum phase kickback,The answer qubit is unentangled and may be ignored. If all three qubits must be restored to the displayed register form, applying $H$ and then $X$ to the answer qubit maps $|-⟩$ back to $|0⟩$. Thus the first register is exactly $|f⟩$ after one use of $U_f$.

The Hadamard gate conjugates the Pauli X gate into the Pauli Z gate:When the control qubit is $|0⟩$, all three circuits act as the identity on the target. When it is $|1⟩$, the right-hand circuit acts on the target as $HXH=Z$. It therefore agrees with the Controlled-Z gate on every computational basis state, provingConjugating both sides by $I\otimes H$ and using $H^2=I$ immediately gives the reverse identity

Let the unknown qubit be $|\alpha ⟩=a|0⟩+b|1⟩$. Alice holds this qubit and her half of the shared Bell state $|{\Psi }^{+}⟩$, while Bob holds the other half.

Alice performs a measurement of her two qubits in the Bell basis and sends the two-bit outcome to Bob through the classical channel. For the shared state $|{\Psi }^{+}⟩$, the teleportation with the psi-plus Bell state correction table isup to physically irrelevant global phases. Here $X$ and $Z$ are the Pauli X gate and Pauli Z gate. After the indicated correction, Bob's qubit is $|\alpha ⟩$. This is quantum teleportation; no physical copy of the unknown qubit is sent.

Let $D$ denote Alice's unknown input qubit and $A$ her GHZ qubit. Expanding the four-qubit state in the Bell basis of $D,A$ giveswhere a global sign in the last term is immaterial.

For outcomes ${\Phi }^{+}$ and ${\Phi }^{-}$, Bob and Charlie already haveFor either $\Psi$ outcome, both apply the Pauli X gate. Since $X\otimes X$ exchanges $|00⟩$ and $|11⟩$, the ${\Psi }^{+}$ branch becomes $|{\varphi }_1⟩$ and the ${\Psi }^{-}$ branch becomes $|{\varphi }_2⟩$, up to a global phase. This proves the Bell measurement on one qubit and one leg of a GHZ state result.

Finally, tracing out Charlie givesfor either sign. This reduced state has rank two exactly when $a\ne 0$ and $b\ne 0$. By the entanglement criterion for a two-term correlated state,

The spectral norm of an operator $O$ is the induced operator normEvery Pauli X gate $X_i$ is unitary, so $\Vert X_i\Vert =1$. The triangle inequality for the norm therefore givesIn fact equality holds, as the product of $+1$ eigenvectors of the $X_i$ has eigenvalue $n$ under $J_X$.

First construct a controlled unitary gate from the uncontrolled oracle. Keep the supplied $|v_0⟩$ in a reference register. Controlled on an extra qubit, swap the data and reference registers, query $U$ on the register that contains $|v_0⟩$ in one branch, and swap back. Because $U|v_0⟩=|v_0⟩$, that branch is unchanged, while the other branch acquires $U$ on the data. Reversing the control convention with Pauli X gates gives controlled-$U$. The reference state is returned unchanged, and each use costs one query to $U$ and $O(n)$ controlled-SWAP gates.

Expand $|b⟩={\sum }_j{b}_j|v_j⟩$ in an eigenbasis of $U$. Exact quantum phase estimation with an $m$-qubit phase register producesApply the available phase gates, controlled by the corresponding bits of $c_j$, to multiply branch $j$ byThen reverse phase estimation. Although no $U^{†}$ oracle was supplied, every eigenvalue obeys ${\lambda }_j^{2^m}=1$, so $U^{-q}=U^{2^m-q}$. The inverse controlled powers can therefore be implemented with forward calls to $U$. Since $2^m=\mathrm{poly}(n)$, phase estimation and its inverse use only $\mathrm{poly}(n)$ queries. The phase register returns to $|0^m⟩$, while the data register isAs a direct check, the same spectral promise implies $U^{†}=U^{2^m-1}$.

A Pauli gate applies a Pauli matrix as a unitary operator to a qubit. The Pauli X gate, Pauli Y gate and Pauli Z gate are common correction operations in quantum teleportation.

A bit-flip channel applies the Pauli X gate with probability $p$ and does nothing otherwise. If the classical error outcome is not recorded, its output is the convex mixture ${\mathcal{E}}_p(\rho )=(1-p)\rho +pX\rho X$, not a coherent superposition of the two alternatives.

// Target: quantum-information-theory.bigb

Quantum error correction protects a logical qubit by encoding its quantum state into a larger Hilbert space, where a specified family of physical errors can be diagnosed and reversed by a common recovery quantum channel. The requirement is to restore every encoded quantum superposition, including its relative phase, rather than only the computational basis states. A quantum error-correcting code specifies which errors have this protection; it does not promise recovery from every possible disturbance.

The beautiful idea is to measure relations between physical qubits that reveal an error syndrome while revealing nothing about the protected logical qubit. For example, the bit-flip repetition code compares neighboring qubits without reading their individual values. Its parity checks can locate one Pauli X gate error while leaving the unknown coefficients intact. Encoding is a linear isometry, consistent with the no-cloning theorem: redundancy in a shared quantum state supplies protection without making independent copies of the unknown input.

Peter W. Shor's 1995 paper ^[ref], published on 1 October, introduced the nine-qubit Shor code to reduce quantum decoherence in memory. Andrew M. Steane's ^[ref], published on 29 July 1996 and ^[ref], connected quantum error correction to classical linear codes; the seven-qubit Steane code protects against arbitrary errors on one physical qubit. These constructions protect relative phase as well as computational basis values, going beyond the elementary bit-flip repetition code.

For an authored account of the mechanism, see John Preskill's ^[ref], especially section 7.1 on collective parity measurements and section 7.9 on stabilizer codes. Andrew Steane's A tutorial on quantum error correction develops the distinction between correcting a stored quantum state and operating with imperfect quantum gates.

The rotation gate about the Pauli X gate axis commutes with $X$. The J gate convention gives $J(\alpha )J(0)=H\mathrm{diag}(1,e^{i\alpha })H=e^{i\alpha /2}{R}_x(\alpha )$, differing from the standard axis rotation only by global phase.

Group the nine physical qubits into three blocks and write $B_r$ for the product of three Pauli X gates in block $r$. The commuting stabilizer generators $B_1{B}_2$ and $B_2{B}_3$ have eigenvalue $+1$ on both logical codewords. A Pauli Z gate in blocks one, two or three gives syndromes $(-,+)$, $(-,-)$ or $(+,-)$ respectively. Apply a Z gate to any one qubit of the indicated block to recover the logical state. Z errors at distinct locations within a block act identically on the code, so this correction is degenerate.

The binary Lüders rule measurement of singlet versus triplet preserves the entire triplet subspace, but its nonselective channel permits signalling. Starting with $|00⟩$, Bob retains $|0⟩$. If Alice instead applies a local Pauli X gate, the input is $|10⟩$ and the channel outputs the equal mixture of $|{\Psi }^{+}⟩$ and $|{\Psi }^{-}⟩$; Bob has reduced density matrix $I/2$. Thus no spacelike local implementation can realize this channel. A Bell-state nondemolition measurement has a finer triplet readout and a different nonselective channel, so this obstruction does not exclude it.

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.

Let $c,s\ge 0$ and $c^2+s^2=1$. Starting from the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, Alice applies the measurement in quantum mechanics with Kraus operators $M_0,M_1$ displayed above. Completeness follows from $M_0^{†}{M}_0+M_1^{†}{M}_1=I$. Each outcome has probability $1/2$. The normalized outputs are $c|00⟩+s|11⟩$ and $s|00⟩+c|11⟩$. Communicating the outcome allows both parties to apply a Pauli X gate in the second branch, making both outputs equal to the first. Thus this LOCC protocol deterministically converts one Bell pair into any two-qubit pure state with these Schmidt coefficients, after appropriate local unitary operations. It includes the product state endpoint without postselection.

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.