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Write the Steane code basis as uniform quantum superpositions over a binary linear code $D$ and a disjoint coset $t+D$, where $D$ is the even-weight part of the length-seven Hamming code. For $Z(w)={⨂}_{j=1}^7{Z}_j^{w_j}$, diagonal Pauli Z gates cannot connect the two cosets. Within coset $t_a+D$, the overlap isIf $w\notin D^{\perp }$, translation by a word $x_0\in D$ with $w\cdot x_0=1$ negates this sum, so it vanishes. Since the dual code $D^{\perp }$ is the length-seven Hamming code of minimum Hamming distance three, every nonzero $w$ of Hamming weight one or two has vanishing overlap. Taking $u,v$ from $\{0,e_1,\dots ,e_7\}$ proves the displayed identity. The eight two-dimensional error images are orthogonal, so their projective measurement determines the error syndrome without revealing the logical state; applying the identified phase flip again recovers every code state. This proves nondegenerate quantum error-correcting code behaviour for the identity and all single-site phase flips, including their coherent linear span.

Apply one CNOT gate from each input qubit to the same target. Their actions commute because controls remain unchanged and each contribution is added modulo two. This reversible circuit computes the parity bit with exactly one two-qubit gate per input, and its inverse is the same circuit. Unlike measuring the inputs to compute parity classically, the unitary preserves quantum superposition and can be uncomputed after phase application.

Use the ordered basis $(|1⟩,|2⟩)$ and ${\sigma }_z=\mathrm{diag}(1,-1)$, withA pure quantum superposition $|+x⟩=(|1⟩+|2⟩)/\sqrt{2}$ has $\mathbf{R}=(1,0,0)$ on the Bloch sphere. In contrast, a mixed state $\rho =p|1⟩⟨1|+(1-p)|2⟩⟨2|$ has $\mathbf{R}=(0,0,2p-1)$, inside the sphere for $0<p<1$. At equal populations its vector is zero. Equal populations alone therefore do not distinguish a coherent superposition from a mixture; their off-diagonal density matrix elements differ.

Choose the drive phase so the rotating-frame Hamiltonian is $H_{\mathrm{rot}}=\hslash (\Omega {\sigma }_x+\delta {\sigma }_z)/2$, where $\delta =\omega -{\omega }_0$ as specified. The Rabi vector is the angular-velocity vectorHere $\Omega$ is the on-resonance population oscillation angular frequency, not the unhalved coefficient of a cosine coupling. From $\mathbf{R}(0)=\hat{z}$, rotation about $\mathbf{W}$ at $W=\sqrt{{\Omega }^2+{\delta }^2}$ givesFor $\Omega \gg |\delta |$, the rotation axis is nearly transverse, the trajectory nearly a great circle, and transfer is almost complete. For $\Omega =\delta >0$, the axis bisects $+x$ and $+z$, the orbit lies in $R_x+R_z=1$, and $p_2=\frac{1}{2}{\sin }^2(\sqrt{2}\Omega t/2)$ never exceeds $1/2$. Each member remains a pure state in this coherent evolution.

For the Ramsey free interval, let atom $j$ have detuning ${\delta }_j=\delta +{\eta }_j$. Instantaneous ideal pulses give the Ramsey ensemble coherence factor $\chi (T)=⟨e^{i{\eta }_{j}T}⟩$. Averaging the free-evolution vector obtained in step (ii) givesDifferent atoms fan out around the equator, shortening the ensemble Bloch vector even though every individual's vector remains unit length. A broad smooth detuning distribution usually makes the average spiral towards the centre and damps the Ramsey fringes towards $1/2$. The dephasing time is of order $1/\Delta {\omega }_0\sim 5/\delta$. inhomogeneous ensemble dephasing is not necessarily irreversible decoherence of individual atoms; an appropriate spin echo can rephase static frequency offsets.

The given spread does not specify a probability distribution or whether it denotes an RMS width. An exact envelope is therefore not determined. The figure illustrates, rather than infers, a Gaussian distribution with RMS ${\sigma }_{\delta }=0.2\delta$, for which $\chi =e^{-{\sigma }_{\delta }^2{T}^2/2}$. A uniform distribution would instead give a sinc envelope and side lobes. Both have the same qualitative initial loss of contrast on the quoted inverse-spread scale.

In the rubidium-87 ground-state hyperfine structure, $J=1/2$ and $I=3/2$, so $F=1,2$. Taking $H_{\mathrm{hf}}=A\mathbf{I}\cdot \mathbf{J}/{\hslash }^2$ givesThe zero-field splitting is ${\Delta }_{\mathrm{hf}}=2A\simeq h\times 6.8\mathrm{GHz}$. Ignoring the small nuclear magnetic term, the weak-field Landé g-factors are $g_{F=1}\simeq -1/2$ and $g_{F=2}\simeq +1/2$. The diagram labels all three $F=1$ and five $F=2$ substates, with linear shifts $g_F{\mu }_B{m}_{F}B$.

The pair $|1⟩=|F=1,m_F=0⟩$ and $|2⟩=|F=2,m_F=0⟩$ is connected by a magnetic dipole transition with $\Delta F=1$, $\Delta m_F=0$, driven with its oscillating magnetic component along the quantization axis. Both first-order shifts vanish, giving the rubidium-87 clock transition first-order insensitivity near zero field. Its insensitivity does not imply absence of quadratic shifts.

For these two $m_F=0$ levels the Breit-Rabi formula gives the quadratic Zeeman shift of a clock transition:where the last expression neglects the nuclear term and sets $g_J\simeq 2$. HenceThis uses ${\mu }_B/h\simeq 1.40MHz/G$ and the supplied $A/h\simeq 3.4\mathrm{GHz}$. With physical moments defined by ${\mu }_i=-\partial E_i/\partial B$, the differential magnetic moment is ${\mu }_2-{\mu }_1\simeq -8.2\times 10^{-3}{\mu }_B$. If effective magnetic moment denotes the positive energy slope instead, that slope difference is positive. The magnitude and physical frequency sensitivity are the same. Retaining the full atomic factors gives the familiar quadratic clock-frequency coefficient about $575Hz/{\mathrm{G}}^2$; the simplified supplied parameters give about $576Hz/{\mathrm{G}}^2$.^[ref]

Use a far-detuned optical dipole trap, such as a red-detuned focused-beam trap, to confine both clock states. Their first-order magnetic moments vanish; their residual quadratic magnetic slopes are opposite, so a simple static magnetic-field minimum does not confine both in the same way. An optical potential acts through the electronic polarizability and can trap both ground hyperfine states. It is not perfectly state independent: differential AC Stark shifts, intensity noise and residual magnetic gradients still need control during Ramsey evolution.

For the allowed $n=2k$ states, the normalized standing wave is a quantum superposition of two opposite angular momentum eigenstates:The Born rule therefore predictsThe expectation value is zero, but a single measurement never gives zero for these $k\ge 1$ states. Their uncertainty is $\Delta l_z=k\hslash$. The two components have the same rotor energy ${\hslash }^2{k}^2/(2I)$, so a definite energy does not imply definite angular momentum.

Here $H^2=I$. Both remaining gates fix the all-zero computational vector: the T gate fixes $|0⟩$, and the controlled-NOT gate fixes $|00⟩$. Therefore every finite circuit, on any number of initially zero wires, leaves $|0\cdots 0⟩$ unchanged.

It cannot even approximate $|+⟩\otimes |0\cdots 0⟩$, whose squared overlap with that unchanged vector is $1/2$. Thus the gate set is not universal. The needed quantum superposition cannot be created from this input by basis permutations and diagonal phases alone.

The inverse Bogoliubov transformation gives $c=ua-v{b}^{†}$ and $d=v{a}^{†}+ub$. Because $(a^{†}{b}^{†}{)}^2=0$ by the canonical anticommutation relations, the exponential truncates:The fermionic signs are fixed byIt follows thatBoth annihilation conditions hold precisely when $B=v/u$. Normalizing the paired fermionic vacuum, using the orthogonality of its two Fock states, gives $A^2(1+B^2)=1$. Choose the irrelevant overall phase so that $A>0$. The result isEquivalently,Its positive pair coefficient follows from the specified order $ab+b^{†}{a}^{†}$ in the Hamiltonian operator. The state has even fermion parity and is a quantum superposition of zero and two original particles. In the limit $\Delta \to 0^{+}$, $A\to 1$ and $B\to 0$, so it reduces to the original vacuum state, as expected for $\epsilon >0$.

For teleportation with an ambiguous Bell-state phase label, write $\rho =|\psi ⟩⟨\psi |$. The Bell-basis measurement outcomes ${\varphi }^{+}$ and ${\varphi }^{-}$ each occur with probability $1/4$. In either case a correct guess gives Bob $\rho$, while the wrong guess differs by the Pauli Z gate and gives $Z\rho Z$. The other two Bell states are identified correctly. Since Alice guesses each ambiguous label with probability $1/2$, the overall wrong-correction probability is $(1/2)(1/2)=1/4$. Averaging over the actual quantum measurement outcome and her guess gives the phase-flip channelConditional only on an ambiguous analyzer result, the output is $(\rho +Z\rho Z)/2$; that conditional channel has weight $1/2$ in the displayed unconditional result. There is no rejection of those rounds.

Use the teleportation fidelity for a pure state target, namely the squared quantum fidelityThis compares the input with the resulting mixed state; averaging state vectors before forming a density matrix would give a different, incorrect answer. Direct substitution givesThus the maximum is $1$, attained by $|0⟩$ and $|1⟩$ up to a global phase, because these are eigenstates of the Pauli Z gate. The minimum is $3/4$, attained by every equal-weight quantum superpositionon the equator of the Bloch sphere. The unsquared quantum fidelity is $\sqrt{F_{\psi }}$ if that alternative convention is used. No average over different input quantum states is needed for the requested state-dependent answer.

Define the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ and write $s=e^{-i\varphi }$. After $U_{\varphi }$ the two-qubit quantum state isApplying the Hadamard gate to the first qubit gives the normalized quantum stateFor $\varphi =n\pi$, $s=(-1{)}^n$, soThe first is a product state. The second is exactly the minus Bell state in the printed phase gate convention; its reduced density matrices are $I/2$.

For three qubits, let $U_{12}$ use the ordered pair $(1,2)$ and $U_{23}$ the ordered pair $(2,3)$. The two diagonal quantum gates commute. For $\varphi =n\pi$ their action before the final Hadamard gates isIf $n$ is even this is $|+++⟩$, and $H_1{H}_3$ gives $|0⟩|+⟩|0⟩$. If $n$ is odd, group the quantum superposition by the middle qubit:Since $H_{\mathrm{Had}}|+⟩=|0⟩$ and $H_{\mathrm{Had}}|-⟩=|1⟩$, the final states areThe odd case is a GHZ state: each individual qubit has reduced density matrix $I/2$, and it is related to $(|000⟩+|111⟩)/\sqrt{2}$ by the local Pauli gates $Z_1{X}_3$. Conversely, applying those gates converts $|G⟩$ to that standard form. They have not been applied in the printed network, so the actual output used in the next part is $|G⟩=(|001⟩-|110⟩)/\sqrt{2}$.

For one input bit, a Boolean function is specified by the pair $(f(0),f(1))$, and $k=f(0)+f(1)$ counts its satisfying assignments. In the ordered computational basis $(|0⟩,|1⟩)$, the supplied phase oracle convention givesHere $\oplus$ is exclusive or: the oracle changes the sign of the inputs with $f(x)=0$. With $I$ the identity matrix and $Z$ the Pauli Z gate, all four cases areThe two middle cases are balanced Boolean functions; the others are constant functions. The specified zero-state sign change and the Hadamard gate are

Use the top qubit as the first tensor product factor. Write the controlled unitary gate as $C(V)=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes V$. Reading the bracketed quantum circuit from left to right, its four operations are $C(U_f)$, $C(H)$, $C(U_0)$, $C(H)$. Thus the target operation in its control-one branch iswhere the rightmost matrix acts first. The control-zero branch is the identity operator. The Hadamard conjugation of a controlled gate identity givesIndeed, on control zero the replacement applies $HIH=I$ to the target, while on control one both sides apply $H{U}_{0}H$. By linearity, the identity holds on every quantum superposition and therefore in every repetition of the bracket. This replaces the two internal controlled Hadamard gates as a pair.

The global phase of a controlled quantum gate must be preserved: $C(-I)=Z\otimes I$, whereas $C(I)=I\otimes I$. Consequently the minus signs in $U_0=-Z$ and the special phase oracles $U_f=-I,-Z$ cannot be discarded as an unobservable global phase of the complete quantum circuit.

A good atomic qubit needs two reproducible, orthogonal internal states whose populations and relative quantum phase survive for many gate times. Both should be long lived, with weak spontaneous emission and little decoherence from fluctuating fields or atomic motion. Their transition must nevertheless permit fast, selective, coherent control with little leakage to other states. Reliable initialization and state-selective quantum measurement are also required; optical pumping and fluorescence readout can supply these functions.

For rubidium-87, use the two ground state hyperfine levels in the $5{}^2{S}_{1/2}$ electronic configuration. The electronic angular momentum is $J=1/2$, and addition of angular momentum to the given nuclear spin $I=3/2$ gives $F=|I-J|,\dots ,I+J=1,2$. A suitable Rubidium-87 clock qubit isThese are coupled electron-nucleus angular momentum eigenstates, rather than two bare nuclear-spin projections. Their hyperfine structure supplies a resolvable splitting ${\nu }_{01}\simeq 6.835\mathrm{GHz}$. Both belong to the electronic ground state, avoiding a short optical excited-state lifetime. Since the weak-field Zeeman shift is proportional to $g_F{m}_{F}B$, both have zero linear shift at $B=0$: their rubidium-87 clock transition suppresses first-order magnetic-field dephasing near zero field. A quadratic Zeeman shift of a clock transition and differential optical shifts remain. A small stable bias field can define the quantization axis and help resolve unwanted transitions, at the cost of some residual magnetic sensitivity.^[ref]

Drive the magnetic dipole transition with a resonant microwave field of suitable polarization, here $\Delta m_F=0$, or use two phase-coherent optical fields for a far-detuned stimulated Raman transition with their frequency difference resonant with ${\nu }_{01}$. The optical excited state is then only virtually populated, reducing spontaneous emission. In the rotating frame and rotating-wave approximation, define the drive phase $\varphi$ by the effective HamiltonianHere $X,Y,Z$ are the Pauli matrices on the chosen qubit, and $\Omega$ is the Rabi angular frequency. Pulse duration controls the rotation gate angle, while phase selects the axis in the equatorial plane of the Bloch sphere. A $\pi$ pulse exchanges the populations; a $\pi /2$ pulse makes an equal quantum superposition from either basis state. Controlled Rabi detuning, a differential energy shift, or compositions of these pulses give $z$ rotations. For example, $R_z(\alpha )R_y(\beta )R_z(\gamma )$ realizes any one-qubit unitary operator up to a global phase. Thus the atomic encoding admits arbitrary single-qubit quantum gates.

Use the two-pair component of polarization-entangled down-conversion to account for identical bosons in the same collected modes. Write $a_H^{†}$ for a creation operator in mode $a$ with horizontal polarization, and similarly for the other modes. For the plus Bell state, the pair-creation operator is $K^{†}=a_H^{†}{b}_V^{†}+a_V^{†}{b}_H^{†}$. The normalized four-photon state isThese are normalized Fock states. In particular, $(a_H^{†}{)}^2|0⟩=\sqrt{2}|2H_a⟩$; omitting this factor would misnormalize the four-photon input.

The first polarizing beam splitter sends $a_H^{†}$ to $T_H^{†}$ and $a_V^{†}$ to $c_V^{†}$. Only the middle term can give exactly one trigger photon and leave another photon in $c$. In the first term both $a$ photons reach $T$ and none reaches $c$; in the last there is no trigger. Thus the relevant, still unnormalized component is

For the selected paths of the half-wave plate and beam splitter, takeHere $e$ labels the output towards $D_3$. An immaterial common reflection phase has been absorbed in the mode definitions. Keeping one photon at $D_3$ leavesAt the final polarizing beam splitter, the $H_{c^{\prime }}{H}_d$ alternative yields one $H$ at each of $D_1,D_2$, while $V_{c^{\prime }}{V}_d$ yields one $V$ at each. The two opposite-polarization alternatives send both photons to the same detector and fail fourfold coincidence counting. The accepted component is thereforeAfter conditioning and normalization, the three detected output modes haveThe optical mode matching must erase any timing, frequency or other record that distinguishes the two alternatives. Otherwise the same polarization populations could describe an incoherent mixed quantum state, rather than this quantum superposition. The Bouwmeester three-photon GHZ experiment uses this principle. The photonic GHZ preparation by fourfold coincidence is a measurement in the coincidence basis: detecting all four photons projects and consumes them. A click at $T$ by itself does not herald a guaranteed freely available three-photon state.

The two $D$ states are metastable, so a quantum superposition can retain its relative phase much longer than one involving a rapidly decaying $P$ state. In the stipulated transition diagram they have no allowed downward electric-dipole transition; real weaker decay channels limit, rather than eliminate, this lifetime advantage. Their distinct energies permit resolved state-selective optical control.

Both $D$ levels couple to $P_{3/2}$. Apply coherent beams near 850 nm and 854 nm, detuned from that common intermediate level by a large Rabi detuning ${\Delta }_R$. Absorption from one beam and stimulated emission into the other transfer population between the two $D$ states without appreciably populating the short-lived intermediate level. Tune the frequency difference to their splitting, including differential AC Stark shifts. In the far-detuned limit,for an initially pure $D_{3/2}$ state on two-photon resonance. The controllable relative optical phase sets the rotation axis; the large Rabi detuning suppresses spontaneous-scattering errors. Suitable polarizations and a chosen pair of magnetic sublevels are required in an actual experiment.

Discarding the photon-number result gives a classical mixture of the two outcome branches. The atomic state, in the basis $(|e⟩,|g⟩)$, isIt is also the reduced state before measuring the field; a nonselective operation on the field does not change the atomic reduced density matrix. The atom is not in a coherent superposition with these populations.

For an atomic energy gap $\hslash \omega$, the thermal population ratio is $p_e/p_g=e^{-\hslash \omega /(k_{B}T)}$. Matching it givesA finite positive temperature exists when $0<p_e<p_g$, with $T\to 0^{+}$ at $p_e=0$ and $T\to +\infty$ as $p_e\to 1/2$ from below. Equal populations correspond to zero inverse temperature. If $p_e>1/2$, the expression is a negative effective temperature for a bounded two-level system, not an equilibrium state at any ordinary positive temperature. For a concrete counterexample to an unrestricted positive-temperature interpretation, at $t=0$ the atom is certainly excited, whereas every positive-temperature Gibbs distribution has $p_e\le 1/2$. The fully excited endpoint is the formal limit $T\to 0^{-}$. Matching one density matrix does not establish genuine thermalization with a bath.

Write $ϵ=0.08$ and $v=1-2ϵ=0.84$. The ensemble is the mixture $(1-ϵ)|\mathrm{GHZ}⟩⟨\mathrm{GHZ}|+ϵ|{\mathrm{GHZ}}^{\prime }⟩⟨{\mathrm{GHZ}}^{\prime }|$, rather than a coherent superposition of the two phases. The ZZZ probabilities stay unchanged. Reversing the relative phase interchanges the allowed parity sector in every equatorial basis, so for XXY, XYX and YXX,In YYY the quantum prediction is the opposite:For a concrete continuation of the ideal local analysis, apply the same phase-flip event to a predetermined local assignment: it reverses X and Y at the affected photon. All three mixed-basis products and the YYY product then reverse together. With unbiased marginals in each parity sector this shared-error local model predicts negative-product YYY outcomes at $0.23$ each and positive-product outcomes at $0.02$ each. The histograms compare this specified model with the quantum result.

Imperfect mixed-basis statistics alone do not force that particular local histogram. A model-independent constraint follows because a positive YYY product requires at least one of the three ideal mixed-product constraints to fail. Each fails with probability $ϵ$, so the union bound givesEquivalently, local assignments satisfy $|⟨XXY⟩+⟨XYX⟩+⟨YXX⟩-⟨YYY⟩|\le 2$, whereas this noisy quantum state gives $4v=3.36$. Thus the violation survives the specified phase error without assuming a unique noisy local histogram or independent errors in the three measurement contexts.

Track the $-i$ factors only when the corresponding transition is addressed. They are common within the addressed two-state subspace, but become physically relevant relative phases between computational basis states. Using a semicolon to separate the shared motional number state, the complete paths areThe middle arrow comprises both Y pulses. In the last row those individual steps are $-i|ge;1⟩\to -|gf;0⟩\to +i|ge;1⟩$. The two X pulses contribute $(-i{)}^2=-1$ for either initially excited control; the Y pair supplies a second minus sign only when the second ion is also excited.

The shared motion returns to its ground state for all four inputs, and the auxiliary internal state is empty. Quantum superposition therefore gives the two-qubit gateDiscarding the X-pulse factors would incorrectly identify $G$ with a Controlled-Z gate itself.

The extra Pauli gate on the control is removed by $Z_{a}G=C_Z$. Conjugating the target by a Hadamard gate then givesThus, from left to right in the network, apply $H_b$, then $G$, then $Z_a$ and $H_b$; the last two gates commute.

For a direct proof, the control-zero block of $G$ is $I_b$ and its control-one block is $-Z_b$. Multiplication by $Z_a$ changes the latter into $Z_b$. Since $H_b{I}_b{H}_b=I_b$ and $H_b{Z}_b{H}_b=X_b$, the result iswhich is exactly the CNOT gate on all inputs, including quantum superpositions.

An AND gate maps four possible pairs to one bit and loses information, so it is not reversible. A reversible embedding retains both inputs and uses a target bit:This is the Toffoli gate; initializing $c=0$ computes $ab$ while preserving enough information to invert the operation.

Classical reversible gates on one or two bits alone cannot synthesize every such operation. Every permutation of two bits is an invertible affine map over the finite field ${\mathbb{F}}_2$: there are $4\cdot |\mathrm{GL}(2,2)|=4\cdot 6=24$ of them, exactly the $4!$ permutations. Their extensions to a larger register and their compositions remain affine. The nonlinear term $ab$ in the Toffoli gate therefore cannot be generated. Three-bit reversible gates remove this obstruction, with ancillary bits initialized appropriately.

There is no conflict with quantum universality: one- and two-qubit gates may pass through quantum superpositions and use relative phases at intermediate steps. They are not restricted to permutations of computational basis states at every step. For example the Hadamard gate, T gate and CNOT gate can decompose a Toffoli gate exactly. Thus the classical affine obstruction does not apply to their intermediate unitary operators.

Choose a real unitary matrix and undo it at the second stage:After the first rotation the photon qubit is $\sqrt{q}|0⟩+\sqrt{p}|1⟩$. A dud leaves this quantum superposition unchanged, so the inverse rotation restores $|0⟩$. Its final projective measurement can never return $1$.

For a live bomb the absorbing component explodes with probability $p$. If the photon survives, the normalized state is $|0⟩$, and the second rotation makes it $\sqrt{q}|0⟩-\sqrt{p}|1⟩$. The unconditional probabilities of the three possible live-bomb outcomes are consequentlyThey sum to one. The nonexplosive result $1$ certifies a live bomb because its dud probability is exactly zero. The inconclusive result $0$ resets the qubit to its original state, allowing another identical trial.

The repeated weak-rotation bomb-test efficiency follows by summing the geometric series over the first successful trial:This is conditional on testing a live bomb, not an average over an unspecified prior fraction of duds. To make the finite-run success probability exceed $1/2-\epsilon$, choose positive $p$ so that $p/[2(2-p)]<\epsilon /2$, then choose $N$ so that $q^{2N}<\epsilon /2$. The first condition controls the limiting deficit and the second controls the remaining tail. Exactly $p=0$ yields no explosion but also no information, so the limit must be approached with nonzero weak rotations.

The half-efficiency limit is consistent with the structure of this Elitzur-Vaidman bomb tester. Suppose generally $U|0⟩=a|0⟩+b|1⟩$. A final outcome that certifies a live bomb must be orthogonal to the dud output $U^{\prime }U|0⟩$. Pulling this outcome backwards through $U^{\prime }$ gives the normalized vector orthogonal to $(a,b)$. Its squared overlap with $|0⟩$ is $|b{|}^2$. Thus the safe-certification probability is $|a{|}^2|b{|}^2$, whereas the explosion probability is $|b{|}^2$. The former never exceeds the latter. Summing these mutually exclusive terminal events over repeated, possibly adaptive trials gives $P_{\mathrm{safe}}\le P_{\mathrm{explosion}}$ and hence $P_{\mathrm{safe}}\le 1/2$. This concerns the specified rotate–interact–rotate–measure protocol.

Let $D={\mathcal{C}}_{\mathrm{ev}}$, and write ${\mathcal{C}}_{\mathrm{odd}}=t+D$ for any odd-weight word $t\in {\mathcal{C}}_H$. The parity functional on the four-dimensional binary linear code ${\mathcal{C}}_H$ is nonzero, so its kernel $D$ has dimension three and its other fibre is the coset $t+D$. For example $t=(1,\dots ,1)$ belongs to the standard length-seven Hamming code and has odd Hamming weight. Denote the two logical basis states by $|{\psi }_0⟩$ and $|{\psi }_1⟩$, corresponding to these two cosets. The eight computational orthonormal basis vectors in each sum are distinct, so both logical states have norm one; their disjoint supports make them orthogonal.

For a binary vector $w\in {\mathbb{F}}_2^7$, define the product of Pauli Z gateswhere the dot product is computed modulo two. The key diagonal overlap, with $t_0=0$ and $t_1=t$, isIf $w\in D^{\perp }$, every summand is one. If $w\notin D^{\perp }$, choose $x_0\in D$ with $w\cdot x_0=1$. Translation $x\mapsto x+x_0$ permutes $D$ and negates every summand. The sum must then be zero. ThusOff-diagonal logical overlaps vanish for every $w$:because a diagonal Pauli operator cannot move a computational basis word from one coset to the disjoint other coset.

The possible errors are $E_0=I$ and $E_j=Z(e_j)$ for $j=1,\dots ,7$. Set $e_0=0$. Each phase flip is its own adjoint and inverse, soIf $i\ne j$, the vector $e_i+e_j$ is nonzero and has Hamming weight one or two. By the given dual code identity $D^{\perp }={\mathcal{C}}_H$ and the minimum Hamming distance three of the Hamming code, it cannot belong to $D^{\perp }$. Hence all its diagonal logical overlaps vanish. If $i=j$, the product is the identity and its logical overlaps are ${\delta }_{ab}$. Combining both cases givesThis proves the Knill--Laflamme condition with an identity error-overlap matrix, not merely a possibly degenerate scalar matrix. In particular the eight two-dimensional error subspaces $E_i{\mathcal{X}}_{\mathrm{Steane}}$ are mutually orthogonal. This is precisely nondegenerate phase-flip correction in the Steane code.

To exhibit recovery rather than only the criterion, let $P$ be the orthogonal projector onto the code and $P_i=E_{i}P{E}_i^{†}$. The projective measurement consisting of these eight mutually orthogonal projectors, completed by $I-{\sum }_i{P}_i$, determines which error occurred. For an arbitrary encoded pure state $|\chi ⟩=\alpha |{\psi }_0⟩+\beta |{\psi }_1⟩$ affected by $E_j$, outcome $j$ occurs with certainty and leaves $E_j|\chi ⟩$ unchanged. Applying $E_j$ again recovers $|\chi ⟩$. The outcome gives the error syndrome and does not reveal $\alpha$ or $\beta$, so recovery preserves logical quantum superpositions.

An equivalent explicit error syndrome uses three commuting X-type stabilizer generators. Choose a parity-check matrix for ${\mathcal{C}}_H$ with the seven different nonzero binary columns:Its rows span ${\mathcal{C}}_H^{\perp }=D$. For a row $h_l$, the operator $X(h_l)={⨂}_j{X}_j^{(h_l{)}_j}$ permutes each of $D$ and $t+D$, and therefore has eigenvalue $+1$ on both logical states. A phase flip at position $j$ changes its eigenvalue to $(-1{)}^{H_{lj}}$, since $X(h_l)Z_j=(-1{)}^{H_{lj}}{Z}_{j}X(h_l)$. Measuring the three commuting stabilizer generators gives the $j$th column of $H$ as a binary syndrome; no error gives the zero column. All eight outcomes are different, and applying the identified $Z_j$ restores the code state. Measuring these stabilizer generators never distinguishes the two logical states.

Finally, if a noise Kraus operator is a coherent linear combination $F={\sum }_j{c}_j{E}_j$, the same syndrome measurement separates its mutually orthogonal components $c_j{E}_j|\chi ⟩$. Conditional recovery returns $|\chi ⟩$ in every nonzero branch. Linearity then also corrects any quantum channel whose Kraus operators lie in this span. The argument therefore corrects arbitrary unknown single-site phase flips nondegenerately, including coherent phase-error amplitudes, rather than only a known error on a basis codeword.

Put $L=2^N$ and define $\theta =\arcsin \sqrt{M/L}$. The normalized good and bad vectors have disjoint computational basis supports, so they are orthogonal. Counting the terms in the uniform quantum superposition givesSince $V_0=I-2|0^N⟩⟨0^N|$, the Hadamard transform gives $-H^{\otimes N}{V}_0{H}^{\otimes N}=2|\varphi ⟩⟨\varphi |-I$, the Grover diffusion operator. On the ordered orthonormal basis $(|{\Psi }_g⟩,|{\Psi }_b⟩)$, the two reflection operators have matricesTheir product isThis is a plane rotation through angle $-2\theta$ in the displayed coordinate convention, with the invariant good-bad plane preserved by both operators. In particular its action sends $(\sin a,\cos a)$ to $(\sin (a+2\theta ),\cos (a+2\theta ))$. Induction therefore gives the Grover rotation angle formula