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Approximate quantum counting estimates the number $k$ of satisfying assignments of a Boolean function, using a quantum oracle and quantum measurement. Controlled repetitions of a Grover search algorithm iteration let quantum interference reveal the count. For one input bit, a Hadamard test can give Pauli Z gate expectation value $1-k$: with the complemented phase oracle $U_f=\mathrm{diag}(2f(0)-1,2f(1)-1)$ and $G=-X{U}_f$, use $⟨+|X=⟨+|$ to obtain $⟨+|G|+⟩=-[(2f(0)-1)+(2f(1)-1)]/2=1-k$. Repeated preparations estimate this expectation value; one binary measurement result does not distinguish all three counts $0,1,2$.

The $m$-qubit bit-flip repetition code stores one logical qubit in $|0_L⟩=|0{⟩}^{\otimes m}$ and $|1_L⟩=|1{⟩}^{\otimes m}$. Its stabilizer generators are $Z_j{Z}_{j+1}$ for $j=1,\dots ,m-1$, and its logical operators are $\overline{X}=X_1\cdots X_m$ and $\overline{Z}=Z_1$. Interchanging the Pauli X gate and Pauli Z gate gives the phase-flip repetition code. For odd $m$, majority-vote decoding of a repetition code corrects up to $(m-1)/2$ physical Pauli X gate errors, assuming ideal operations and a noise family containing only these bit flips.

For $m=3$, prepare two quantum ancillas in $|00⟩$ and apply controlled-NOT gates from the input qubit to each of them. By linearity, encoding sendsThis does not produce $(\alpha |0⟩+\beta |1⟩{)}^{\otimes 3}$, which would contain cross terms. For example, encoding $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ produces $(|000⟩+|111⟩)/\sqrt{2}$, whereas three independent copies would have eight equal computational basis amplitudes. The encoding creates quantum entanglement when both coefficients are nonzero and respects the no-cloning theorem.

The three-qubit bit-flip syndrome table corrects the family $\{I,X_1,X_2,X_3\}$ and its entire linear span, as specified by the Knill--Laflamme condition. It cannot correct arbitrary errors on one physical qubit: the phase flipcommutes with both checks and acts as logical $\overline{Z}$. In particular, it takes the encoded $|+⟩$ to the orthogonal encoded $|-⟩$ with error syndrome $00$. Thus its full quantum code distance is one, despite correcting one error of the restricted Pauli X gate type. The Shor code, Steane code and five-qubit error correcting code provide protection against arbitrary one-qubit errors.

Let $1$ be a leaf with neighbour $2$ and let the graph on the other vertices be generated by the Controlled-Z gate product $E_{\mathrm{rest}}$. If the leaf has Pauli Z gate frame $Z_1^r$, measuring it at angle $\alpha$ with result $s$ applies one-bit teleportation and leavesAny known Pauli Z gate frames on other vertices remain as additional factors before those vertices' input states. The identity $J(\alpha )Z^r=X^{r}J(\alpha )$ combines the leaf byproduct with the measurement result.

Let $S$ be the SWAP gate. If the drift Hamiltonian operator and every allowed control Hamiltonian operator commute with $S$, every reachable time-evolution operator also commutes with $S$. Indeed, $SUS$ satisfies the same initial-value Schrödinger equation as $U$ and starts at $I$, so uniqueness gives $SUS=U$. Yet $S(Z\otimes I)S=I\otimes Z\ne Z\otimes I$. An individual Pauli Z gate is consequently unavailable within that symmetric control model. Equal nuclear resonances, isotropic exchange interaction, and identical coupling to a common homogeneous radio frequency field provide an example. An added interaction that breaks this symmetry can remove the obstruction.

The Hadamard gate is the unitary operatorwhere $I$ is the identity operator. It maps the computational basis to the Hadamard basis:

These outputs have equal computational-basis measurement probabilities, but their relative quantum phase differs. Applying another Hadamard gate sends them to different definite outputs. Calling this gate a classical coin toss would discard precisely the information that makes this possible. ^[ref].

Starting from $|0⟩$, compare two quantum circuits, reading matrix products from right to left:Here $Z$ is the Pauli Z gate. After the first Hadamard gate, the probability amplitudes are $(1/\sqrt{2},1/\sqrt{2})$. Inserting $Z$ changes them to $(1/\sqrt{2},-1/\sqrt{2})$ without changing their squared magnitudes. At the final gate, the contributions to output zero cancel and those to output one add. Without $Z$, cancellation occurs at output one instead. This is controlled interference, with the quantum phase determining which answer emerges.

The same calculation works for any phase gate $P_{\varphi }=\mathrm{diag}(1,e^{i\varphi })$:For example, $\varphi =\pi /3$ gives $P(1)=1/4$, while $\varphi =\pi /2$ gives $P(1)=1/2$. The output-one amplitude at $\varphi =\pi /2$ is $(1-i)/2$: its two complex contributions add before the Born rule is applied. A phase gate alone leaves the populations unchanged; recombination converts its relative quantum phase into observable probabilities.

If quantum decoherence completely removes the computational basis coherence between the gates, the intermediate density matrix becomes $I/2$. Both the phase gate and final Hadamard gate leave $I/2$ unchanged, so $P(1)=1/2$ for every $\varphi$. The disappearing fringe explains why preserving quantum coherence matters: it enables the gate sequence to carry phase information into the final measurement.

For $H=J(XX+YY+ZZ)/2$, $U=e^{-itH}$, and an individually applied Pauli Z gate $P=Z\otimes I$, conjugation gives $PHP=J(-XX-YY+ZZ)/2$. The paired Pauli operators $XX,YY,ZZ$ commute with one another, soTwo equal free-evolution intervals, each of duration $t$, therefore cancel exchange while retaining the Ising coupling. This spin echo is exact for instantaneous ideal gates; a finite pulse requires accounting for the interaction during the pulse. A simultaneous Pauli Z gate on each spin commutes with $H$ and cannot perform the same cancellation. The physical availability of the individual gate is a separate quantum control requirement.

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.

The phase-flip repetition code with codewords $|+++⟩,|---⟩$ corrects any one physical Pauli Z gate error. Its two adjacent-pair error syndromes cannot distinguish an error pattern from its three-bit complement. Minimum-weight recovery therefore leaves no residual error at weight zero or one, and leaves $Z_1{Z}_2{Z}_3$ at weight two or three. This residual operator exchanges the codewords, so inverse encoding gives a logical Pauli X gate. For independent physical phase errors of probability $ϵ$, the decoded quantum channel is $(1-p_L)\rho +p_{L}X\rho X$, where $p_L=3{ϵ}^2(1-ϵ)+{ϵ}^3$. For $0<ϵ<1/2$, $ϵ-p_L=ϵ(1-ϵ)(1-2ϵ)>0$.

The codewords $|+++⟩,|---⟩$ have residual logical Pauli X gate after an uncorrectable two- or three-site phase error. Precede their encoding by a logical Hadamard gate and follow decoding by its inverse. Since $HXH=Z$, the same error now acts as a logical Pauli Z gate. Explicitly the encoded coefficients are $(\alpha +\beta )/\sqrt{2}$ and $(\alpha -\beta )/\sqrt{2}$ for an input $\alpha |0⟩+\beta |1⟩$. This convention matters when translating physical phase errors into a claimed decoded phase-error channel.

For the Hadamard test with a density operator $\rho$ in the auxiliary register, the output control qubit has Pauli Z gate expectation $\mathrm{Re}m$ and Pauli Y gate expectation $-\mathrm{Im}m$, where $m=\mathrm{Tr}(\rho U)$. The minus sign comes from the final Hadamard gate, since $HYH=-Y$. Measuring in $(|0⟩\pm i|1⟩)/\sqrt{2}$ therefore estimates the imaginary part using the minus-outcome frequency minus the plus-outcome frequency.

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.

The columns are $i|+⟩$ and $-i|-⟩$. Their norms are one and their inner product is zero, since $|+⟩,|-⟩$ are orthonormal. Thus $M_1$ is a unitary matrix. Equivalently,using the Hadamard gate and Pauli Z gate. The different column phases do not destroy orthonormality.

The Pauli Z gate gives a rotation through $\pi$ about the $z$ axis of the Bloch sphere. In Bloch vector coordinates,For example, this follows from $ZXZ=-X$, $ZYZ=-Y$ and $ZZZ=Z$ in the Pauli matrix expansion $\rho =(I+r_{x}X+r_{y}Y+r_{z}Z)/2$.

In the ordered computational basis $|0⟩,|1⟩$, the Pauli X gate, Pauli Y gate, Pauli Z gate, Hadamard gate and T gate areIn the ordered two-qubit computational basis $|00⟩,|01⟩,|10⟩,|11⟩$, with the first qubit controlling the second, the controlled-NOT gate isHere the T gate uses the conventional representative with first diagonal entry one; another common representative differs only by a global phase.

Initialize two qubits in $|00⟩$. A Hadamard gate on the first followed by a controlled-NOT gate $1\to 2$ prepares the Bell state $|{\varphi }_{+}⟩$. The permitted T gates and Hadamard gates implement the needed Pauli gates exactly:Apply $T_1^4$, followed by $H_2,T_2,T_2,T_2,T_2,H_2$. These are a Pauli Z gate on the first qubit and a Pauli X gate on the second, givingEvery operation is in the allowed gate set; no quantum measurement or probabilistic branch is needed.

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.

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.

Set ${\omega }_1={\omega }_2=\omega$. Define the quantum rotating frame byThe isotropic Heisenberg model coupling commutes with the total $z$ spin angular momentum, so $[H_Z,C]=0$. The transformed Hamiltonian operator is thereforeHere removal of the common Zeeman energy is exact; the exchange is resonant and cannot be discarded using the weak coupling condition for two NMR spins.

Put $a=Jt$, $c=\cos a$, and $s=\sin a$. Exponentiating the parallel-spin entries and the middle $2\times 2$ matrix gives the time-evolution operatorFor example, at $a=\pi /2$ this equals the SWAP gate times $e^{-i\pi /4}$, which checks both the exchange angle and the overall phase.

The printed network reads from left to right as $U$, a Pauli Z gate on the upper spin only, $U$, and another upper-spin Pauli Z gate. LetIts total time-evolution operator, with the first operation on the right, is $W=PUPU$. For this Heisenberg exchange echo,The three paired Pauli operators $X\otimes X$, $Y\otimes Y$, $Z\otimes Z$ commute pairwise: for example both products $(X\otimes X)(Y\otimes Y)$ and $(Y\otimes Y)(X\otimes X)$ equal $-Z\otimes Z$. Thus $P{H}_{\mathrm{r}}P$ commutes with $H_{\mathrm{r}}$, andThe spin echo cancels the $XX+YY$ exchange terms and retains the Ising coupling. Equivalently, over the total free-evolution time $2t$ it produces the effective Hamiltonian operator $JZ\otimes Z/2$ exactly for ideal instantaneous gates. It is generally not the identity. A physical $\pi$ rotation about $z$ is $e^{-i\pi Z/2}=-iZ$; using two such pulses instead of the ideal $Z$ gates changes $W$ only by a physically irrelevant global phase of minus one.

The implementation requires frequency-selective spin control. Under the usual common-radio frequency NMR assumptions, equal resonance frequencies give no way to apply a pulse to the first spin while leaving the second untouched. For equally coupled spins in a spatially uniform control field, the transverse control Hamiltonian operator has the formBoth this control and the drift Hamiltonian operator commute with the SWAP gate $S$. Every reachable time-evolution operator consequently commutes with $S$. But $SPS=I\otimes Z\ne P$: the individual pulse required by the diagram fails this symmetry test. This is the exchange symmetry obstruction to individual spin control. Applying simultaneous $Z$ gates gives $Q=Z\otimes Z$, which commutes with $H_{\mathrm{r}}$ and instead yields $QUQU=U^2$; it does not refocus the exchange. A common change of the RF phase reference also supplies only a collective rotation.

This obstruction assumes the usual common-RF control with no extra handle distinguishing the spins. Equal resonance frequencies do not universally forbid individual addressing: a spatially selective field, unequal control couplings, or another symmetry-breaking interaction could alter controllability. Such resources are absent from the specified equivalent-spin NMR model.

Quantum error correction encodes a logical qubit redundantly, obtains an error syndrome without measuring its unknown logical amplitudes, and applies a recovery that reverses the specified errors. A decoherence-free subspace instead uses a symmetry of a specified noise model: each noise operator acts as a scalar on the entire encoded space, so no syndrome-dependent recovery is required. This is passive protection and can also be viewed as a special case of quantum error correction.

For the two-spin decoherence-free encoding, chooseSimultaneous Pauli Z gates givesoBoth logical components acquire the same global phase; their relative phase and amplitudes survive. The density matrix is exactly unchanged, so even an unknown random choice between applying $I$ and $Z\otimes Z$ causes no logical dephasing.

More strongly, the encoding protects against arbitrary collective dephasing:because both basis states have total $Z$ eigenvalue zero. At $\varphi =\pi$, $R_z(\pi )\otimes R_z(\pi )=-Z\otimes Z$, consistent with the previous overall phase. The protection relies on correlated noise: independent $Z$ rotations generally change the logical relative phase, and this two-spin code does not protect a whole logical qubit against arbitrary collective rotations about all three axes or against arbitrary single-spin errors.

Use the ordered computational basis $(|00⟩,|01⟩,|10⟩,|11⟩)$, with qubit 1 written first. The half-time gate requires a branch convention. Even stationary Hamiltonian evolution with $W(2\tau )=\mathrm{SWAP}$ does not uniquely specify $W(\tau )$: different accumulated eigenstate phases can agree at $2\tau$ and differ at $\tau$. In particular, both $S$ and $S^{†}$ below square to SWAP gate and leave $|00⟩,|11⟩$ unchanged. Adopt the conventional continuous isotropic exchange interaction branch, with a relative singlet phase increasing from $0$ to $\pi$ during the full pulse. This is the exchange square-root branch convention.

The symmetric and antisymmetric eigenstateshave SWAP gate eigenvalues $+1,-1$. Let $P_{-}=|{\psi }_{-}⟩⟨{\psi }_{-}|$ and $P_{+}=I_4-P_{-}$ be their antisymmetric and symmetric orthogonal projectors. The adopted exchange evolution isFor effective spin-one-half qubits, $H_{\mathrm{ex}}=\pi \hslash (X_1{X}_2+Y_1{Y}_2+Z_1{Z}_2-I_4)/(8\tau )$, an isotropic exchange interaction plus an energy offset that fixes the common phase. At half time choose $+1$ on the symmetric sector and $+i$ on the antisymmetric sector, the conventional principal square-root value of $-1$. Hence the square root of the SWAP gate isThus $S|00⟩=|00⟩$, $S|11⟩=|11⟩$, $S|01⟩=a|01⟩+b|10⟩$ and $S|10⟩=b|01⟩+a|10⟩$, where $a=(1+i)/2$ and $b=(1-i)/2$. The identities $a^2+b^2=0$ and $2ab=1$ verify $S^2=\mathrm{SWAP}$, and the projector form proves unitarity. The endpoint truth table alone would allow other matrix square roots; the stated exchange branch supplies the additional physical assumption.

For the network, fix the Pauli Z gate, Hadamard gate and S gate conventionsThe letter $P$ distinguishes this local S gate from the two-qubit exchange gate $S$. In particular $P^2=Z$; replacing $P$ by $P^{†}$ changes the network. Subscripts specify the acted-on qubit: $Z_1=Z\otimes I_2$, $P_2=I_2\otimes P$ and $H_2=I_2\otimes H$. Operators act rightmost first, so the printed quantum circuit is

First multiply the central two exchange pulses. Since $a^2-b^2=i$, their off-diagonal terms cancel:This reduces the entire network to $U=H_{2}D{H}_2$. Its action on all four computational basis states, retaining every relative quantum phase, isEquivalently,It flips the second qubit when the first is zero and contributes $-i$ when the first is one. That conditional phase cannot be discarded as a global phase.

To obtain the standard CNOT gate controlled by qubit 1, append an $X$ gate to qubit 2 and $P$ to qubit 1. They commute, andFor control zero, $X_2$ undoes the original flip. For control one, it supplies the desired flip and $P_1$ cancels the phase because $i(-i)=1$. This phase-corrected exchange network for a CNOT gate realizes the standard gate exactly with the declared conventions.

Finally, $S|01⟩=(e^{i\pi /4}|01⟩+e^{-i\pi /4}|10⟩)/\sqrt{2}$ has equal Schmidt coefficients and is maximally entangled: a half exchange creates quantum entanglement although a full SWAP gate does not. More constructively, the network above produces a CNOT gate from two half exchanges and local gates. Since CNOT gates and arbitrary single-qubit unitary operators form a universal quantum gate set, the square root of the SWAP gate and arbitrary single-qubit gates are universal. The alternative inverse exchange branch remains entangling and universal, but its phases and local correction must be recomputed consistently.