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Number the three data qubits from top to bottom. It suffices initially to follow a pure input $|\psi ⟩=a|0⟩+b|1⟩$; linearity will extend the result to every density operator. The first two controlled-NOT gates create $a|000⟩+b|111⟩$. The first layer of Hadamard gates then encodes this asthe three-qubit phase-flip repetition code. Let $e_i$ indicate whether a Pauli Z gate occurred on data qubit $i$ during storage. The next layer of Hadamard gates uses $HZH=X$ and givesThe two initially zero syndrome quantum ancillas receive the adjacent parities. In the diagram's top-to-bottom order, their error syndrome isBoth terms of the logical superposition have these same parities, so the syndrome is independent of $a,b$. Measuring it reveals no logical-state information.

At the marked position of $R$ in the printed circuit, there has already been another layer of Hadamard gates on the data. There is also a layer immediately after $R$. Thus a desired correction $X_i$ in the parity-extraction frame must be implemented as $R=Z_i$ at the actual recovery location, since $H^{\otimes 3}{Z}_i{H}^{\otimes 3}=X_i$. Choose the recovery according toThe two patterns on each row share a syndrome because complementing all three bits does not change either parity. For a weight-zero or one error, the recovery removes the entire error in the parity-extraction frame. The remaining state is $a|000⟩+b|111⟩$, and the two final controlled-NOT gates give $|\psi ⟩|00⟩$.

For a weight-two or three error, minimum-weight recovery instead leaves all three bits flipped, giving $a|111⟩+b|000⟩$. The final controlled-NOT gates send this to $(a|1⟩+b|0⟩)|00⟩=X|\psi ⟩|00⟩$. Consequently, for every input density operator,The logical failure is a Pauli X gate after decoding, even though the stored physical errors were Pauli Z gates: $Z_1{Z}_2{Z}_3$ exchanges the two encoded codewords. This is the logical failure of the three-qubit phase-flip repetition code.

Take a normalized environment quantum stateThe Controlled-Z gate is symmetric between the two qubits, so it can be written with the environment as the controlling system:Expanding the joint density operator after the interaction givesIn the partial trace, the two off-diagonal environment operators have trace zero, while each diagonal projector has trace one. HenceEquivalently, the Kraus operators are $K_0=\sqrt{1-ϵ}I$ and $K_1=\sqrt{ϵ}Z$, and $K_0^{†}{K}_0+K_1^{†}{K}_1=I$. This explicit Stinespring dilation describes the system's dephasing by discarding information in the environment. A fresh environment in this state at each storage interval also realizes the memoryless iteration used in part (a).

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.

Write the normalized qubit as $|\varphi ⟩=c_0|0⟩+c_1|1⟩$, with $|c_0{|}^2+|c_1{|}^2=1$. Only the $|1⟩$ amplitude changes between the two phase gates, soSince $|e^{i\delta }-1|=2|\sin (\delta /2)|\le |\delta |$,The phase-gate discretization error is uniform over all input states; its exact operator norm is $2|\sin (\delta /2)|$. To choose an approximating gate, round $N\theta /(2\pi )$ to the nearest integer and reduce it modulo $N$. Taking the shortest circular angle difference gives $|\delta |\le \pi /N$, and hence a uniform state-vector error at most $\pi /2^n$. The circular choice handles the identified endpoints $0$ and $2\pi$ correctly.

Neither monotonic decrease alone nor equality of the initial spectra guarantees convergence. A nonnegative Lyapunov function can approach a positive limiting value, and the set where $\dot{V}=0$ can contain trajectories other than the target. Different initial spectra already obstruct exact tracking: unitary time evolution preserves the eigenvalues of each density operator, while two matrices approaching one another in norm must have the same limiting eigenvalues.

Even the spectral obstruction can be absent while the Lyapunov quantum control stalls. Take a qubit with $H_0=Z/2$, $H_1=X$, ${\rho }_d(0)=|0⟩⟨0|$ and $\rho (0)=|1⟩⟨1|$. Both density operators commute with $H_0$, and the commutator of $X$ with $\rho$ has zero diagonal, so $f=\mathrm{Tr}({\rho }_d[-iX,\rho ])=0$. Both states remain constant. They have the identical spectrum $(1,0)$, but $\boxed{\Vert \rho (t)-{\rho }_d(t){\Vert }_{\mathrm{HS}}=\sqrt{2}\text{ for all }t}$. Moreover $iZ,iX$ generate the full special unitary Lie algebra $\mathfrak{su}(2)$, so the failure here is a limitation of this particular feedback law, not a lack of available state-transfer controls.

Quantum teleportation transfers an arbitrary unknown qubit state to a separated receiver while using local operations and classical communication, rather than sending that input qubit. Alice and Bob first share one maximally entangled state of two qubits, such as a Bell pair; a shared spin singlet state is equivalent after a known local unitary operation.

Alice performs a joint Bell-basis measurement on the input qubit and her half of the pair. There are four possible outcomes. Bob's conditional qubit is the input state transformed by a known Pauli operator determined by that outcome. Alice sends the outcome as two classical bits, and Bob applies the inverse Pauli operator. Thus one shared Bell pair and two classical bits, together with local measurement and correction, transmit one unknown qubit exactly. The entangled resource is consumed.

The procedure does not produce a classical description of the unknown amplitudes and does not leave a second copy at Alice, so it respects the no-cloning theorem. Before receiving the classical outcome, Bob's reduced density matrix is $I/2$, independent of the input; this respects the no-communication theorem. The next part proves the full identity-channel action, including inputs entangled with another system.

Use the BB84 protocol to establish a secret random key, then encrypt the actual message with a one-time pad. Alice chooses independent random bits and independently chooses between the computational orthonormal basis $\{|0⟩,|1⟩\}$ and the diagonal basis $\{|+⟩,|-⟩\}$, where $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. She sends one photon or other qubit in each corresponding state. Bob independently chooses one of the two bases for each measurement.

Over an authenticated public classical channel, they announce their bases but not their bit values. They retain only positions where their bases match; absent disturbance these sifted bits agree. They disclose and discard a randomly selected test sample to estimate the error rate. If it is too large they abort. Otherwise they perform information reconciliation and privacy amplification to produce a shared key about which an eavesdropper has negligible information. Authentication is essential: without it, an active attacker can impersonate the two parties. It may use a previously shared short authentication key, subsequently replenished from the new secret key.

The no-cloning theorem prevents an attacker from copying an unknown nonorthogonal state perfectly: preservation of inner products would require $⟨\psi |\varphi ⟩=⟨\psi |\varphi {⟩}^2$, impossible when its modulus is strictly between zero and one. In a simple intercept-and-resend attack, Eve chooses the wrong basis with probability $1/2$, and then introduces a wrong sifted bit with probability $1/2$. Thus the test error rate is $1/4$, and the probability that $m$ independent tested bits show no error is $(3/4{)}^m$. More general attacks are handled by bounding Eve's information from the observed statistics and shortening the reconciled key through privacy amplification.

Finally Alice sends $C=M\oplus K$, using a uniform independent key of the message's length exactly once. The one-time pad makes the ciphertext independent of the message. The security claim is secrecy with arbitrarily small failure probability after testing and privacy amplification, not that a finite quantum transmission can never be disturbed or intercepted. The scheme assumes trusted devices and an authenticated classical channel; it does not rely on a computational hardness assumption.

Use one shared $|{\Phi }^{+}⟩$ pair to measure $Z_A{Z}_B$, and the other to measure $X_A{X}_B$. The latter entanglement-assisted nondemolition parity measurement is obtained by applying Hadamard gates to both system qubits before and after the $Z$-parity circuit. Since $ZX=-XZ$ at each site, the two minus signs cancel, giving $[Z_A{Z}_B,X_A{X}_B]=0$.

The Bell states have joint eigenvaluesand are therefore distinguished uniquely by the two meter parities. For records $(u,v)$ and $(r,s)$, the combined Kraus operator isEach $P_{z,x}$ is the rank-one projector onto the corresponding Bell state. This is a Bell-state nondemolition measurement: an input Bell state remains exactly that state for every possible local record. The local circuits can be scheduled without communication; the identity of the Bell state is known only when the classical records are brought together. Two parity bits distinguish all four Bell states without disturbing them.

Quantum teleportation transfers the quantum state of an unknown qubit from a sender's carrier to a receiver's carrier. It also transfers correlations with any external reference. It does not determine classical amplitudes or send the original particle.

The standard deterministic protocol consumes one shared maximally entangled Bell pair and two classical bits from Alice to Bob. Alice performs a local Bell-basis measurement on the input and her half of the pair; Bob applies the corresponding local Pauli gate. The classical message is essential: before receiving it, Bob has the outcome-averaged density operator $I/2$ and cannot recover or detect the unknown state. The original input is consumed by the measurement, so quantum teleportation is compatible with the no-cloning theorem.

Write the arbitrary joint pure state of the input $A$ and an external reference $R$ asThe reference vectors need not be normalized or orthogonal. Share $|{\Phi }^{+}{⟩}_{aB}$. Label Alice's Bell states byProjecting onto this Bell basis gives the unnormalized stateEach outcome has probability $1/4$. Alice sends $(r,s)$ and Bob applies $Z_B^r{X}_B^s$, reversing the two Pauli gates. The final state is exactly $|\Omega {⟩}_{BR}$ for every outcome. In particular all input entanglement and other correlations with $R$ now belong to $B$, while the reference marginal is unchanged. This is teleportation as an identity channel on a reference.

Conditioned on the record, Alice's two measured carriers are in $|{\beta }_{rs}⟩$ and factor from $BR$. Thus $A$ no longer retains its original correlations with $R$. There is no second copy, even when the teleported qubit was initially entangled with a larger system.

The equal mixture of $|{\Phi }^{+}⟩$ and $|{\Phi }^{-}⟩$ cancels their off-diagonal coherence, leaving the separable quantum stateA trivial LHV model chooses a shared random bit $j$, uniformly zero or one, and prepares both local qubits in $|j⟩$. Each party uses the local Born rule for its own chosen spin direction, with independent local random seeds if a deterministic description is desired. This reproduces the entire joint measurement distribution, not merely the $z$-axis correlation.

Since the state has an explicit local hidden-variable theory for these local measurements, it cannot violate the CHSH inequality, by the derivation in part (b). Classical correlation of the shared preparation bit is sufficient; the cancellation of coherence removes the entanglement.

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.

Put $M=2^m$ and prepare $m$ control qubits in $|0{⟩}^{\otimes m}$ and the target in the supplied eigenstate $|\psi ⟩$. Apply a Hadamard gate to every control qubit. Let $k={\sum }_{j=0}^{m-1}{2}^j{k}_j$ denote the integer encoded in that quantum register.

For every $j$, apply a controlled unitary gate $U^{2^j}$ with control $k_j$ and the same target. The quantum phase kickback yieldsApplying ${\mathrm{QFT}}_M^{-1}$ to the controls therefore produces $|x⟩$ exactly. A computational-basis measurement returns $x$ with certainty, and the algorithm outputsThis is exact quantum phase estimation; exactness uses the finite dyadic phase promise, not a rounding argument.

Only a controlled-U gate is supplied as a primitive. Its $2^j$ repetitions implement controlled-$U^{2^j}$, so the total number of black-box uses is ${\sum }_{j=0}^{m-1}{2}^j=2^m-1$. Access to powers as unit-cost primitives would be a different quantum query complexity model. This distinction is the cost of exact phase estimation on a dyadic spectrum.

Again put $a=(1+\lambda )/2$, $b=(1-\lambda )/2$. If computational strings $z,y$ differ in $h$ places, the matrix element of $U_{\varphi }^{\otimes n}$ between them is $a^{n-h}{b}^h$: an unchanged bit contributes $a$, and a flipped bit contributes $b$. Hence the given operator isThe printed factored expression $a^n(b/a{)}^h$ is undefined at $a=0$, namely $\varphi =1/2$. Its polynomial form $a^{n-h}{b}^h$ gives the natural continuous extension there, $U_{1/2}^{(n)}=X^{\otimes n}$. For $n\ge 2$ the strict promise $\varphi <1/n$ avoids that singular value; for $n=1$ the extension is needed to include the allowed phase $1/2$. This is a removable defect of the formula, not a failure of unitarity of the tensor-product operator.

Prepare $|-{⟩}^{\otimes n}$. Its eigenvalue is ${\lambda }^n=e^{2\pi in\varphi }$, giving tensor-product eigenphase amplification. Write $n=2^s$. If $s<m$, thenwhere the upper bound is exactly the promise $\varphi <1/n$. Thus the amplified phase has only $m-s$ unknown binary digits and no modulo-one aliasing. Run exact quantum phase estimation with $m-s$ controls on the controlled-$U_{\varphi }^{(n)}$ black box. It returns $x$ exactly, after which division by the original $2^m$ recovers $\varphi$.

The number of joint black-box uses isIf $s\ge m$, the dyadic promise and $\varphi <1/n$ force $x=0$, so output $\varphi =0$ without querying the oracle. This covers the zero-control edge case without an invalid negative number of phase bits. The improvement counts one supplied controlled-$U_{\varphi }^{(n)}$ call as one query. Implementing such a joint call from $n$ individual controlled-$U_{\varphi }$ calls would remove the claimed factor-$n$ primitive-query saving, and preparing the $n$ target qubits is a separate gate cost.

Write the system's $z$ basis as $|0⟩,|1⟩$, and use a separate pair of meter qubits in the Bell state $|{\Phi }^{+}⟩$. Alice applies a CNOT gate from system $A$ to her meter qubit; Bob simultaneously applies a CNOT gate from $B$ to his meter qubit. Flipping neither or both meter qubits preserves $|{\Phi }^{+}⟩$, while flipping exactly one gives $|{\Psi }^{+}⟩$. Hence the entanglement-assisted nondemolition parity measurement interaction produceswhere $P_{\mathrm{e}}=|00⟩⟨00|+|11⟩⟨11|$ and $P_{\mathrm{o}}=|01⟩⟨01|+|10⟩⟨10|$. Each party now measures only their meter qubit in the $z$ basis. If their binary records are $u,v$, the system Kraus operator isUnequal records verify zero total $z$ spin, since $S_z^{\mathrm{tot}}=\hslash (Z_A+Z_B)/2$ vanishes precisely on the odd sector. The probability of success is $⟨\psi |P_{\mathrm{o}}|\psi ⟩$, and the successful conditional state is $P_{\mathrm{o}}|\psi ⟩/\sqrt{⟨P_{\mathrm{o}}⟩}$. Every zero-total-$z$-spin state is left unchanged, including any coherent superposition of $|01⟩$ and $|10⟩$. Similarly the even-sector coherence is preserved. This is a quantum nondemolition measurement of the parity, rather than separate measurements of both system spins.

All quantum operations and local meter measurements can finish within the spacelike time window. Nevertheless each local meter record is individually uniform: $K_{u0}^{†}{K}_{u0}+K_{u1}^{†}{K}_{u1}=I/2$. The verification result is obtained only by comparing the records using local operations and classical communication. Thus “instantaneous” refers to the local completion of the joint measurement instrument, not instant access to its nonlocal outcome; quantum no-signalling remains intact.

Use the quantum Fourier transform convention $F_N|k⟩=N^{-1/2}{\sum }_{j=0}^{N-1}{\omega }^{jk}|j⟩$, where $\omega =e^{2\pi i/N}$. Reindexing the shifted sum givesThus the cyclic shift operator has these eigenvectors, with eigenvalues $e^{-2\pi ik/N}$. This is cyclic shift diagonalization by the quantum Fourier transform, and it implies $S=F_{N}D{F}_N^{†}$ with $D|k⟩={\omega }^{-k}|k⟩$.

For $N=4$, the binary encoding is $k=2x+y$, with $x$ the more significant qubit. The required diagonal phase isso $D=P_{-1}\otimes P_{-i}$. The allowed-gate circuit is thereforeIn execution order, apply the inverse QFT, then the two phase gates, then the forward QFT. There is no extra global phase. Reversing the Fourier sign convention would conjugate both phase gate parameters.

This is the Deutsch-Jozsa test with an arbitrary uniform-state unitary, and does not require $N$ to be a power of two. Choose a known unitary operator $F$ such that $F|0⟩=|s⟩=N^{-1/2}{\sum }_i|i⟩$. Prepare the target qubit in $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. One Boolean-oracle call gives quantum phase kickback:Apply $F^{†}$ to the index register. The amplitude on $|0⟩$ isIt equals $+1$ or $-1$ for the two constant strings, and equals zero for a balanced string. Consequently a single query decides the promised problem exactly: measure the index register and report constant for outcome zero, balanced for any other outcome. This is the Deutsch-Jozsa algorithm with the available exact state-preparation operation.

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 $n=1$.

The product of the data Pauli Z gates has computational-basis eigenvalueUse the parity circuit $W$ from part (ii), apply the target unitary gate $R=e^{itZ}=\mathrm{diag}(e^{it},e^{-it})$, and then uncompute the parity with $W^{†}$. For every basis input,This equals $e^{it{Z}^{\otimes n}}$ on the data, with the ancilla qubit returned to zero. The compute-phase-uncompute construction therefore gives an exact linear-size circuit:This is a Pauli-string phase by parity computation, an instance of simulation of a computable diagonal Hamiltonian. Part (i) also explains it by $W^{†}(I\otimes Z)W=Z^{\otimes n}\otimes Z$, whose exponential restricts correctly to the target-$|0⟩$ subspace. If no extra line is desired, accumulate parity into the last data qubit, apply $e^{itZ}$ there, and undo the $n-1$ CNOT gates, for $2n-1$ gates. Both constructions implement the global phase as well as the relative phases exactly.