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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.

Let $\lambda =e^{2\pi i\varphi }$. In terms of the Pauli X gate, the given unitary operator is $U_{\varphi }=aI+bX$, with $a=(1+\lambda )/2$ and $b=(1-\lambda )/2$. Since $X|+⟩=|+⟩$ and $X|-⟩=-|-⟩$,Thus the normalized eigenvectors are $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, with eigenvalues $1$ and $e^{2\pi i\varphi }$ respectively. At $\varphi =0$ they coincide as eigenvalues, and the whole two-dimensional Hilbert space is the eigenspace; the same chosen eigenbasis remains valid.

Prepare $|-⟩$ independently of $\varphi$ and use the exact quantum phase estimation circuit of the preceding parts on the controlled-$U_{\varphi }$ black box. It returns $\varphi$ exactly usingThe chosen eigenstate contains all the unknown phase information, unlike $|+⟩$, whose eigenvalue is independent of $\varphi$.

Take ${\alpha }^2+{\beta }^2=1$ and label the first vector of each measurement basis by outcome $0$. Assign the Pauli measurement value $+1$ to outcome $0$ and $-1$ to outcome $1$. The four observables areHere $X$ and $Z$ are the Pauli X gate and Pauli Z gate matrices. These follow by subtracting the two rank-one basis projectors; a real basis rotated through $\gamma$ has observable $\sin (2\gamma )X+\cos (2\gamma )Z$.

The Schmidt-basis Pauli correlation tensor of the state givesConsequently $E_{11}=1/2$, $E_{21}=E_{22}=\sqrt{3}\alpha \beta$ and $E_{12}=-1/2$, where $E_{jk}=⟨A_j\otimes B_k⟩$. For binary outcomes, $\mathbb{E}([A-B{]}_2)=P(A\ne B)=(1-E_{AB})/2$, whereas the final offset term has $\mathbb{E}([B-A-1{]}_2)=P(A=B)=(1+E_{AB})/2$. The chained modular Bell inequality left side is thereforeThe local bound is $I\ge 1$, so the exact violation condition isEquality saturates the bound. The maximal violation for these fixed measurements occurs at $\alpha =\beta =1/\sqrt{2}$ or their common negative, giving $I=(3-\sqrt{3})/2$. Opposite signs do not violate this particular inequality with these fixed bases, although other measurement choices can reveal the state's entanglement. If unnormalized real amplitudes are used, replace $\alpha \beta$ throughout by $\alpha \beta /({\alpha }^2+{\beta }^2)$.

For $n\ge 2$, the Pauli X gate and Pauli Z gate act on different qubits in $X_0{Z}_1$, so they commute. Their product is a Hermitian matrix and squares to the identity operator. Its eigenvalues have modulus one; equivalently it is a unitary operator. Therefore its spectral norm isThe $n\ge 2$ qualification is necessary because the qubit labelled one does not exist for $n=1$.

Write $Q=2^n$, $t=(p-1)(q-1)$ and $a=t/Q$. The easy starting state is uniform over all $n$-bit labels, so its good fraction is $a$, not $t/N$. For distinct primes one has $a\ge 1/4$. If both primes are odd, $t/N\ge (1-1/3)(1-1/5)=8/15$ and $N/Q\ge 1/2$, giving $a\ge 4/15$. If one prime is two, the other is an odd prime $q$; since $2q$ has $n$ bits, $q>2^{n-2}$, and $t=q-1\ge 2^{n-2}=Q/4$.

The uniform coprime state for a semiprime can now be prepared by one exact Grover rotation. SetPrepareUse the good subspace spanned by $|k⟩|1⟩$ with $1\le k<N$ and $\gcd (k,N)=1$. Its squared overlap with $|\Psi ⟩$ is exactly $a\lambda =1/4$. The Euclidean algorithm supplies a reversible computation of the membership predicate, including the range check; condition a sign flip on membership and the extra qubit being one, then uncompute the workspace. This implements $S_G=I-2{\Pi }_G$.

The starting-state reflection $D_{\Psi }=2|\Psi ⟩⟨\Psi |-I$ uses the inverse of its known preparation and a reflection on the all-zero state. Applying $D_{\Psi }{S}_G$ once invokes exact amplitude amplification at $\theta =\pi /6$, and givesThe extra qubit factors off, and the arithmetic workspace is returned to zero. Although the stated range includes $N$, that label is not coprime to itself, so the predicate $k<N$ is equivalent and avoids admitting labels outside the intended range.

The supplied $p,q$ determine $t$ in polynomial time; no factoring procedure is needed. Binary arithmetic, range comparison, the reversible greatest common divisor, the two starting-state reflections and the controlled sign flip all have polynomial-size circuits. Thus the construction takes polynomial time in $n$ in the ideal model allowing the specified one-qubit state preparation. Exactness uses the rotation with known amplitudes $\sqrt{\lambda },\sqrt{1-\lambda }$; with a fixed finite approximate gate library one obtains arbitrary accuracy with precision overhead, rather than an automatic promise of exact state preparation. In the ideal model the preparation succeeds with certainty, without rejection sampling. If desired, the extra $|1⟩$ can be reset by a Pauli X gate.

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 $U^{†}=U^{2^m-1}$.