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Label a four-cycle $0-1-2-3-0$. Measure vertex $0$ in the computational basis with result $r$, then vertex $1$ in the equatorial qubit measurement basis at angle $\alpha$ with result $s$. Graph-state vertex deletion and one-bit teleportation leaveFinal computational basis measurements with raw results $d_2,d_3$ therefore simulate the ideal circuit $E_{23}J(\alpha {)}_2$ on $|++⟩$ after the classical correction $k=d_2\oplus r\oplus s$, $l=d_3$. The output probability distribution is correct in every prior branch; no physical Pauli frame correction is needed.

A path graph state simulates a sequence of Clifford gates $J({\alpha }_j)$, ${\alpha }_j\in \{0,\pi /2\}$, using fixed equatorial qubit measurement bases on successive path vertices. Let $s_j$ be raw outcomes, $t_j=2{\alpha }_j/\pi \in \{0,1\}$, and start the Pauli frame at $a_0=b_0=0$. Matrix commutation and $J(-\pi /2)=XJ(\pi /2)$ giveInduction shows the final unmeasured state differs from the ideal circuit output by $X^{a_m}{Z}^{b_m}$, up to global phase. A final computational basis result $d$ is corrected to $d\oplus a_m$. All bases are fixed, and projectors on distinct vertices commute, so all measurements, including the final one, can occur in a single layer. The frame recurrence is classical outcome processing, not quantum feed-forward.

Apply a Controlled-Z gate to input $a|0⟩+b|1⟩$ and a fresh $|+⟩$ quantum ancilla. Measuring the input in the equatorial qubit measurement basis at angle $\alpha$, with outcome $s$, leaves the unnormalized outputEach branch has probability $1/2$, independently of the input. The logical output is $J(\alpha )|\psi ⟩$ with known Pauli frame $X^s$; the input qubit has been measured, so this does not clone it.

The Pauli Z gate exchanges the two Hadamard basis states: $Z|+⟩=|-⟩$ and $Z|-⟩=|+⟩$. Multiplying the displayed gate by the Pauli X gate givesOn the other hand,The two operators agree, provingAlso $U(\theta )=H\mathrm{diag}(1,e^{-i\theta })=J(-\theta )$ in the J gate convention. Keeping this angle sign explicit avoids confusing the two conventions for equatorial qubit measurements.

Prepare the three-vertex graph-state wire: start all three qubits in $|+⟩$ and apply Controlled-Z gates on edges $1-2$ and $2-3$. In this paper's equatorial qubit measurement convention the basis vectors can be chosen asThe given identity means that measuring an input qubit at angle $\theta$, with outcome $r$, teleports its state to the next qubit as $X^{r}U(\theta )|\psi ⟩$. Each outcome has probability $1/2$, since $X^{r}U(\theta )$ is unitary. This is one-bit teleportation with a known Pauli frame.

Measure vertex 1 at angle $\alpha$ and record $r$. Then measure vertex 2 at angle $\gamma =(-1{)}^r\beta$, taken modulo $2\pi$, and record $s$. The controlled-Z on edge $2-3$ commutes with the first measurement, so the two teleportation steps can be applied successively. The remaining normalized state isFor $r=0$ this is $X^{s}U(\beta )U(\alpha )|+⟩$. For $r=1$, the relation proved above gives $U(-\beta )X=e^{i\beta }ZU(\beta )$. Thus in every branch the remaining state isThe scalar is a global phase and has no effect on probabilities.

Finally measure vertex 3 in the computational basis, obtaining $z$, and reportThe $Z^r$ byproduct changes only computational-basis phases; $X^s$ changes the output label by $s$. Hence the corrected bit has exactly the ideal circuit's probability distribution in every branch, not merely on average. The procedure uses only single-qubit measurements after preparing the graph state, adaptive classical angle selection, and deterministic classical output processing.

Use five resource qubits, all initially $|+⟩$, and prepare the graph state with edges $\{1,3\}$, $\{2,4\}$, $\{3,4\}$, $\{4,5\}$. The labels here identify resource vertices; logical wire 1 travels from vertex 1 to 3, and logical wire 2 travels from 2 to 4 to 5. Each edge is a Controlled-Z gate, and these entangling gates commute, so all can be applied when preparing the resource.

Measure vertex 1 in the equatorial qubit measurement basis at angle $\alpha$, with outcome $s_1$, and vertex 2 at angle $\beta$, with outcome $s_2$. These measurements can be simultaneous. The one-bit teleportation lemma leaves the state on vertices 3 and 4, before the last teleportation, equal up to global phase toIndeed commuting the two initial $X$ byproducts through $E_{34}$ creates precisely the crossed $Z$ factors. The unused edge from 4 to the fresh vertex 5 is the last one-bit teleportation link.

Measure vertex 4 at angle $\theta =(-1{)}^{s_2}\gamma$, with outcome $s_4$. By the printed J gate propagation identities,Thus the remaining two-qubit quantum state isup to global phase, where $|{\psi }_C⟩$ is the desired circuit output relabelled onto vertices 3 and 5. The sign choice cancels the unwanted angle reversal caused by the $X$ frame on vertex 4.

Finally measure vertices 3 and 5 in the computational basis, obtaining raw results $t_3,t_5$. The Pauli Z gate affects only phases, while the Pauli X gate flips a computational result. The five-vertex graph-state circuit simulation therefore outputsThis deterministic classical correction reproduces the ideal joint probability distribution for every preceding measurement branch. The final fixed-basis measurements may share the second layer with the adaptive measurement of vertex 4, because their projectors act on different vertices; measuring them afterwards is also valid.

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.

The path graph state is $|{\psi }_3⟩=E_{23}{E}_{12}|+{⟩}_1|+{⟩}_2|+{⟩}_3$. The factor $Z_3^r$ commutes with both the Controlled-Z gates and the quantum measurement of vertex $1$. Treat $Z_1^r|+{⟩}_1$ as the input of one-bit teleportation. The equatorial qubit measurement result $s$ therefore gives the normalized stateHere the subscript on $Z_2^r$ expresses the teleported input factor; it is not an extra operation on the already measured vertex. Use $J(\alpha )Z^r=X^{r}J(\alpha )$ to obtainThe exclusive or exponent is equivalent to the printed sum $r+s$ because $X^2=I$. The conditional probability of $s$ is one half for either $r$, by the one-bit teleportation branch norm. This calculation concerns an equatorial measurement of a graph-state leaf; no third remaining qubit or extra $|-⟩$ factor is present.

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.

The logical depth of a measurement pattern counts sequential layers of quantum measurements forced by dependence of measurement bases on earlier outcomes. Depth one means every basis can be fixed before measurements start, allowing all quantum measurements on distinct qubits to be performed in parallel. This does not require depth-one resource-state preparation or constant-depth classical parity processing.

Suppose the circuit contains $m$ gates $J({\alpha }_1),\dots ,J({\alpha }_m)$ in time order, with each ${\alpha }_j\in \{0,\pi /2\}$. Prepare an $(m+1)$-vertex path graph state from $|+{⟩}^{\otimes (m+1)}$ and Controlled-Z gates between neighbours. Its first qubit supplies the specified input $|+⟩$. Measure vertex $j-1$ in the fixed equatorial qubit measurement basis at angle ${\alpha }_j$ for $j=1,\dots ,m$, obtaining $s_j$. Measure the final vertex in the computational basis, obtaining $d$.

Track the Pauli frame as $X^{a_j}{Z}^{b_j}$, starting with $a_0=b_0=0$. A hypothetical sequential reading of the same quantum measurements gives the step $X^{s_j}J({\alpha }_j)X^{a_{j-1}}{Z}^{b_{j-1}}$. The supplied commutation rules imply, up to global phase,For ${\alpha }_j=0$, the sign is irrelevant. For ${\alpha }_j=\pi /2$, the identity $J(-\pi /2)=XJ(\pi /2)$ absorbs the possible sign change into the Pauli frame. If $t_j=0$ for ${\alpha }_j=0$ and $t_j=1$ for ${\alpha }_j=\pi /2$, the update isInductively the final unmeasured quantum state would be $X^{a_m}{Z}^{b_m}J({\alpha }_m)\cdots J({\alpha }_1)|+⟩$. The final computational basis outcome is corrected by $a_m$ and is unaffected by $b_m$. The induction on a sequential interpretation proves the joint statistics; projectors on distinct vertices commute, so the identical fixed-basis pattern can actually be measured simultaneously, including its final vertex.

The first printed commutation formula has an incorrect exact scalar phase: with the printed matrix definition, $J(\alpha )X=e^{i\alpha }ZJ(-\alpha )$, rather than the negative-exponent prefactor. For example, at $\alpha =\pi /4$ the two versions differ by a factor $i$. They agree up to global phase, so this error does not affect the Pauli frame recurrence or any quantum measurement probability.

Consequently a fixed-basis path-state pattern of logical depth one simulates every such circuit. Only Pauli measurements occur: the angle-zero basis measures $X$, and the angle-$\pi /2$ basis measures $-Y$ with the printed eigenvector labels. Classical exclusive or processing suffices for all output corrections. These gates are Clifford gates, which explains why angle adaptation can be eliminated. For $m=0$, simply measure the initial $|+⟩$ in the computational basis; it is already a depth-one pattern.

With equatorial qubit measurement vectors $(|0⟩+(-1{)}^s{e}^{i\eta }|1⟩)/\sqrt{2}$, the gate in one-bit teleportation is $U(\eta )=H\mathrm{diag}(1,e^{-i\eta })=J(-\eta )$. Direct multiplication gives $U(\eta )Z=XU(\eta )$ and $U(\eta )X=e^{-i\eta }ZU(-\eta )$. Moving a Pauli frame through this gate therefore exchanges its $X$ and $Z$ exponents and reverses the angle if the old $X$ exponent is one, as displayed up to global phase. A measurement outcome $s$ adds a further $X^s$ factor. This calculation supplies adaptive angle choices and the final classical output-bit correction of a linear cluster state computation.

A path graph state on three qubits simulates two successive gates $U(\theta )=H\mathrm{diag}(1,e^{-i\theta })=J(-\theta )$. Measure vertex one in the equatorial qubit measurement basis $|v_r(\alpha )⟩=(|0⟩+(-1{)}^r{e}^{i\alpha }|1⟩)/\sqrt{2}$, then vertex two at angle $(-1{)}^r\beta$, obtaining bits $r,s$. Two one-bit teleportations and $U(\theta )X=e^{-i\theta }ZU(-\theta )$ leave vertex three in $X^s{Z}^{r}U(\beta )U(\alpha )|+⟩$, up to global phase. A computational output bit $z$ is corrected to $z\oplus s$; the $Z$ factor affects only phase. Adaptive sign choice is required for general angles, although the final bit correction is purely classical.