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Write the unknown pure photon polarization qubit asThe polarizing beam splitter sends the $H$ component down the lower path and the $V$ component up the upper path. After rotating the upper photon polarization to $H$, both paths have the same internal quantum state and can interfere at BS2. Absorb fixed apparatus phases into a calibrated zero for the tunable quantum phase. With the preceding reflection convention, the quantum state before BS2 is $(\alpha |\ell ⟩+i\beta e^{i\theta }|u⟩)|H⟩$, so the two output amplitudes areConsequentlyRepeated counts while scanning $\theta$ determine the complex quantum coherence $\gamma$. The interference visibility is $2|\gamma |$, and the fringe position determines the relative quantum phase $\arg \beta -\arg \alpha$ whenever both components are nonzero.

A fringe alone is not enough to determine an arbitrary photon polarization quantum state: it fixes $|\alpha \beta |$ but not which of $|\alpha {|}^2$ and $|\beta {|}^2$ is larger. Indeed, $|H⟩$ and $|V⟩$ both give the flat fringe $P_1=1/2$. For a general pure quantum state, the two possible population differences areResolve this ambiguity by temporarily detecting the two paths before BS2. Their relative counts give $h=|\alpha {|}^2$ and $1-h=|\beta {|}^2$ using the same polarizing beam splitter and detectors. Taking $\alpha =\sqrt{h}$ as a global-phase choice, the reconstructed quantum state for $h>0$ isIf $h=0$, it is $|V⟩$ up to a global phase. This completes interferometric polarization tomography. The counts are obtained from many identically prepared photons; they do not identify an arbitrary quantum state from one photon.

Without the upper-arm rotation, the photon polarization still labels the path. Before BS2 the quantum state isAfter the balanced beam splitter, the polarization-valued output amplitudes areThe detectors do not resolve photon polarization. Since $⟨H|V⟩=0$, the cross terms vanish in the Born rule:These probabilities are independent of the tunable quantum phase and the input photon polarization, so there is no usable wave interference fringe. Equivalently, taking the partial trace over photon polarization removes the quantum coherence between the two path quantum states. The which-path information is retained in orthogonal photon polarizations even without anybody measuring it. Rotating the two occupied photon polarization quantum states to the same quantum state removes that distinction and allows the cross term in part (ii).

A central application is quantum order finding, which supplies the quantum step in the Shor algorithm for factoring an integer $M$. Choose an integer $a$ coprime to $M$, and let $r$ be its multiplicative order modulo $M$. On valid residue quantum states define the reversible modular-multiplication unitary operator $U_a|x⟩=|ax\mathrm{mod}M⟩$, extending it reversibly to any unused computational basis quantum states.

The orbit $|1⟩,|a⟩,\dots ,|a^{r-1}⟩$ has the orthonormal eigenstatesfor $s=0,\dots ,r-1$. One need not know $r$ to prepare an eigenstate: the easy input $|1⟩=r^{-1/2}{\sum }_s|u_s⟩$ makes quantum phase estimation sample one of these eigenphases. Its control register estimates $s/r$. With $N=2^t$ chosen at least of order $M^2$, the nearest quantum phase label has rational approximation error small enough for continued-fraction recovery in quantum order finding. For instance, $N>2M^2$ gives $|m/N-s/r|<1/(2r^2)$ on the nearest-label event. If $s$ and $r$ are coprime, the recovered denominator is $r$; otherwise one obtains a divisor and can repeat, combine candidates and verify them by modular exponentiation.

When the verified order is even and $a^{r/2}\not\equiv -1(\mathrm{mod}M)$, the greatest common divisorsgive nontrivial factors: their product is divisible by $M$, while neither factor inside the gcd is zero modulo $M$. Unsuccessful choices of $a$ or quantum phase samples are handled by repetition and classical verification. Controlled powers of $U_a$ are implemented by efficient reversible modular arithmetic, rather than by naively applying $U_a$ exponentially many times.

Quantum phase estimation turns modular-multiplication eigenphases into an efficiently usable period, enabling quantum factoring in time polynomial in the bit length of the integer. Factoring is important to public-key cryptography, and this is a major example of a quantum algorithm improving on known classical methods. Another use is extracting energy eigenvalues by applying quantum phase estimation to a controlled time-evolution operator, with care to avoid energy aliasing and to prepare an input having overlap with the desired eigenstate.

Put $N=2^t$ and write $U|u⟩=e^{i\varphi }|u⟩$, with $|u⟩$ normalized. The Hadamard gates prepare the control register in $N^{-1/2}{\sum }_{k=0}^{N-1}|k⟩$. Controlled powers of the unitary operator then give quantum phase kickback:Thus the register quantum state is $N^{-1/2}{\sum }_k{e}^{ik\varphi }|k⟩$, while the target remains $|u⟩$. Define the quantum Fourier transform by $F_N|m⟩=N^{-1/2}{\sum }_k{e}^{2\pi ikm/N}|k⟩$. Its inverse givesWhen $\varphi =2\pi n/N$, the inner finite geometric series equals $N$ if $m\equiv n(\mathrm{mod}N)$ and zero otherwise. Therefore exact quantum phase estimation givesThe quantum phase, rather than an unrestricted integer $n$, is recovered modulo $2\pi$.

The printed circuit has no bit-reversal swaps. Its top control uses $U^{2^{t-1}}$, and its measured top wire gives the least significant output bit. Write $n\mathrm{mod}N={\sum }_{j=0}^{t-1}{2}^j{b}_j$. Before the top wire's final Hadamard gate, its relative quantum phase is $2^{t-1}\varphi =\pi n$, so that gate produces $b_0$. The controlled inverse quantum phase gates then subtract its contribution from lower wires. More generally, before the final Hadamard gate on wire $j$, the relative quantum phase isso that wire gives $b_j$ deterministically. For the three-wire drawing, the correct interpretation isEquivalently, its Fourier-inversion circuit is $R{F}_N^{†}$, where $R$ reverses the register bits. Relabeling the measured bits or appending swaps gives the usual integer output. This bit convention matters, for example, when quantum phase $\pi$ produces top-to-bottom bits $001$, representing the label four rather than one.

Number the physical wires $1,\dots ,5$ from top to bottom and write a computational basis ket in that order. The first four are ancilla qubits and the fifth carries the input. It is enough to follow the basis input $|b⟩$, $b\in \{0,1\}$, because the encoding quantum circuit acts linearly.

The first fan-out consists of four controlled-NOT gates from wire 5 to wires 1–4. It sends $|0000b⟩$ to $|bbbbb⟩$. The three Hadamard gates then giveThe two Pauli X gates on wires 2 and 3, followed by a Controlled-Z gate and the same two X gates, restore the bits but contribute $(-1{)}^{(1-c)(1-d)}$: the sign is negative exactly when the original bits on both wires were zero. Finally, the right-hand controlled-NOT gates are $3\to 4$, $2\to 5$, $1\to 4$ and $1\to 5$. They leave the first three bits unchanged and turn the last two into $b\oplus a\oplus d$ and $b\oplus a\oplus c$. ThusExplicitly, the logical basis quantum states produced by the printed gates areEach has eight orthogonal terms of equal magnitude, and their supports are disjoint, so they are normalized and mutually orthogonal. The arbitrary input is therefore encoded asThis is a realization of the five-qubit error correcting code. The input amplitudes and their relative quantum phase are preserved in the logical subspace; the encoding is an entangled-state linear isometry, not five independent copies of the unknown input. The explicit signs and bit ordering above are fixed by this circuit, rather than by a convention for a different encoder.

The Pauli X gate interchanges the two basis quantum states. Because the bit-flip outcome is a classical unobserved error rather than a coherent superposition, the quantum bit-flip channel givesFor $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, its explicit density matrix isFor a pure reference quantum state, the square of the paper's quantum fidelity is $F^2=⟨\psi |{\rho }_2|\psi ⟩$. ThereforeWith $|\psi ⟩=\cos (\theta /2)|0⟩+e^{i\varphi }\sin (\theta /2)|1⟩$, the expectation is $⟨X⟩=\sin \theta \cos \varphi$, the $x$ component of the Bloch vector. Hence“All possible quantum states” is interpreted as the rotationally invariant uniform pure-qubit average, with measure $d\mu =\sin \theta d\theta d\varphi /(4\pi )$. Thusso the average pure-qubit fidelity of a bit-flip channel isEquivalently, rotational symmetry gives $⟨r_x^2⟩=⟨r_y^2⟩=⟨r_z^2⟩=1/3$, since a pure-state Bloch vector has unit length. The $X$ eigenstates are unchanged as density matrices and have fidelity one, while a computational basis quantum state has $F^2=1-p$; these provide checks on the angular expression. A uniform distribution in $\theta$ rather than in solid angle would produce the wrong ensemble average. The requested quantity is the mean squared fidelity, not the square of the mean unsquared fidelity.

Write ${\rho }_1=|\psi ⟩⟨\psi |$ and ${\rho }_2=|\phi ⟩⟨\phi |$. A pure-state density matrix is a rank-one orthogonal projector, so ${\rho }_1^{1/2}={\rho }_1$ andTaking its square root and trace yields the unsquared quantum fidelityUse unitary invariance of trace distance and fidelity to choose a basis of the span of the two quantum states in which $|\psi ⟩=|0⟩$ and $|\phi ⟩=c|0⟩+s|1⟩$, with $c=F\ge 0$ and $s=\sqrt{1-c^2}\ge 0$. Choosing state-vector phases to make the coefficients nonnegative does not change either density matrix. On that span,Its eigenvalues are $s$ and $-s$, with zero eigenvalues on the orthogonal complement. The positive square root of ${\Delta }^{†}\Delta$ therefore has two eigenvalues $s$, so the trace distance is $(s+s)/2=s$. Thus the pure-state trace distance and fidelity identity isIdentical pure quantum states give $F=1,D=0$; orthogonal pure quantum states give $F=0,D=1$. If the span is one-dimensional, the identical-state result follows directly, so no second basis vector is required. This formula uses the unsquared fidelity convention in the paper; if fidelity denotes squared overlap instead, the square inside the formula is omitted.

For a positive operator $A$, the spectral theorem writes $A={\sum }_j{a}_j|j⟩⟨j|$ with $a_j\ge 0$. Hence its unique positive square root of an operator is $\sqrt{A}={\sum }_j\sqrt{a_j}|j⟩⟨j|$. Conjugating this spectral decomposition shows $\sqrt{UA{U}^{†}}=U\sqrt{A}{U}^{†}$ for a unitary operator $U$.

Let $\Delta ={\rho }_1-{\rho }_2$ and ${\rho }_i^{\prime }=U{\rho }_i{U}^{†}$. Then ${\Delta }^{\prime }=U\Delta U^{†}$ and $({\Delta }^{\prime }{)}^{†}{\Delta }^{\prime }=U{\Delta }^{†}\Delta U^{†}$. The square-root covariance and the cyclic property of the trace giveFor the quantum fidelity, covariance first gives $({\rho }_1^{\prime }{)}^{1/2}=U{\rho }_1^{1/2}{U}^{†}$. ThereforeApply the same square-root identity and the cyclic property of the trace to obtainThus unitary invariance of trace distance and fidelity givesThe same $U$ acts on both quantum states: these quantities depend on the relative distinguishability of the quantum states, not on a choice of basis.

Write the promised phase as $\varphi =2\pi m/8$. Prepare three logical qubits in $|+⟩$ and attach zero ancillas in three blocks of sizes four, two and one. A block of size $r$ is prepared by fanout CNOT gates asApply one copy of $U_{\varphi }$ to every physical qubit in every block, all in the same time step. Undo each fanout. The logical block output is $(|0⟩+e^{ir\varphi }|1⟩)/\sqrt{2}$ and every ancillary wire returns to zero. This is parallel phase multiplication by coherent fanout; its intermediate entanglement copies basis labels, and does not clone an arbitrary quantum state.

For block weights $4,2,1$, the three logical wires now containApply the specified negative-sign quantum Fourier transform and measure. Part (a) gives label $m$ deterministically. Thus seven parallel phase-gate applications, on three logical wires and four ancillas, identify the phase in one oracle time step. Preparation, uncomputation and the Fourier transform cost no time under the stated model; using serial powers would not meet the time requirement.

Take the labels $r,s\in \{0,\dots ,n-1\}$, with basis indices read modulo $n$. There is a missing hypothesis in the printed claim: the root of unity must be primitive. For example, when $n=2$ and $\omega =1$, the two labels $(0,0)$ and $(1,0)$ produce the identical state $(|00⟩+|11⟩)/\sqrt{2}$, so they are not orthogonal. More generally, if the order of $\omega$ is $d<n$, changing $r$ to $r+d$ repeats a state. The orthonormality claim is false for a nonprimitive root.

For the intended result, choose a primitive root of unity, for example $\omega =e^{2\pi i/n}$. Since $|\omega |=1$, expansion in the orthonormal basis givesIf $r=r^{\prime }$, the sum is $n$. If $r\ne r^{\prime }$ modulo $n$, primitiveness gives $q={\omega }^{r^{\prime }-r}\ne 1$ while $q^n=1$, and the finite geometric series is $(1-q^n)/(1-q)=0$. ThusThere are $n^2$ vectors in an $n^2$-dimensional Hilbert space, so they form the generalized Bell basis. Each has equal Schmidt coefficients $1/\sqrt{n}$ and is a maximally entangled state.

For qudit teleportation, let $C$ be Alice's unknown input, write $|\chi {⟩}_C={\sum }_j{\alpha }_j|j{⟩}_C$, and let Alice and Bob share $|\Phi {⟩}_{AB}=n^{-1/2}{\sum }_k|k{⟩}_A|k{⟩}_B$. Alice holds $C,A$ and Bob holds $B$. Define the qudit shift and phase operatorsBoth are unitary. Alice makes a joint projective measurement of $C,A$ in the generalized Bell basis, with projectors $|{\psi }_{rs}⟩⟨{\psi }_{rs}|$. Completeness follows from the orthonormality just proved.

For outcome $(r,s)$, contracting the corresponding bra with the joint input and resource gives Bob's unnormalized state:The Born rule therefore gives $p_{rs}=1/n^2$, since $X^s{Z}^{-r}$ preserves the input norm. Conditional on this outcome, Bob holds $X^s{Z}^{-r}|\chi ⟩$. Alice sends him the classical label $(r,s)$, and he applies its inverse in the displayed order:The corrected receiver state is exactly the unknown input state for every outcome. Alice needs no knowledge of the amplitudes. The label has $n^2$ possibilities and can be encoded in $\lceil {\log }_2(n^2)\rceil$ classical bits; when $n$ is a power of two this is exactly $2{\log }_{2}n$. The protocol consumes the shared maximally entangled state and the original carrier is measured, so it does not retain an additional copy of the unknown state.

Equivalently, the complete joint quantum state has the teleportation expansionIt follows from the contraction calculation and completeness of the generalized Bell basis. After correction the branch operators are all $I/n$, so their sum defines the identity quantum channel: ${\sum }_{r,s}(I/n)\rho (I/n)=\rho$. Acting on the input part of a joint state with a reference gives the same identity, which proves teleportation as an identity channel on a reference and extends the protocol to mixed inputs.

Before receiving the label, Bob instead has the outcome-averaged stateTo check the equality, averaging over $r$ kills the off-diagonal matrix entries by the same root-of-unity sum; averaging the cyclic shifts $s$ makes every diagonal entry the total trace divided by $n$. This is the Heisenberg-Weyl twirling channel. Thus the classical message is necessary to recover the input, consistently with the no-communication theorem derived in Question 1.

Take any rank-one orthogonal projection $P$ and its generalized Bloch vector $s$. The antipodal vector $-s$ is on the same sphere, but its corresponding trace-one Hermitian matrix isOn the range of $P$, this matrix has eigenvalue $2/N-1$, which is negative for $N>2$. It is therefore not a positive semidefinite matrix and cannot be a density operator. Hence the pure states do not cover the sphere when $N>2$. This also shows directly why the purity bound for generalized Bloch vectors does not by itself guarantee a physical quantum state.

The eigenvalues $p_j$ of a density operator are nonnegative and sum to one. ThereforeFor a non-pure state at least two eigenvalues are positive, so the last expression is strictly positive. Combining this with the Generalized Bloch representation givesThus every non-pure quantum state lies strictly inside the radius sphere for $N\ge 2$. It need not be an interior point of the entire physical state body: for $N>2$, a rank-deficient mixed density operator can lie on that body's boundary while still lying inside the Euclidean sphere.

For a spin singlet state, quantum spin measurements along unit vectors $\mathbf{a},\mathbf{b}$ predict $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$. Take ${\mathbf{a}}_0=\mathbf{z}$, ${\mathbf{a}}_1=\mathbf{x}$, ${\mathbf{b}}_0=(\mathbf{z}+\mathbf{x})/\sqrt{2}$ and ${\mathbf{b}}_1=(\mathbf{z}-\mathbf{x})/\sqrt{2}$. These giveThe violation therefore excludes the whole conjunction of the sufficient assumptions, for any model reproducing these correlations. It does not identify which premise fails without further commitments about causal explanation, state completeness and measurement setting generation.

A defensible verdict is to retain ordinary independent setting preparation and reject Bell local causality, specifically its classical screening-off factorization. In standard quantum theory, taking the pure entangled state as the complete state preserves parameter independence: each local marginal is $1/2$, regardless of the distant setting. Yet outcome independence fails. For equal spin axes, $P(A=+,B=-)=1/2$, whereas $P(A=+)P(B=-)=1/4$. The distant outcome changes the conditional prediction even after the quantum state is specified. This failure is compatible with quantum no-signalling, since conditioning on that outcome requires learning a result which cannot itself be controlled.

Adding further past variables does not restore the entire Bell conjunction while retaining the violating correlations; the CHSH inequality already allowed arbitrary such $\lambda$. Alternatively, one may reject measurement independence, permit retrocausal setting dependence in the relevant past state, or adopt another non-Bell-local description. Those are logically possible responses, but each must explain its replacement probability and causal model rather than treating the inequality as a failed algebraic theorem. A deterministic theory is not exempt from this requirement.

The assumption of measurement independence is supported by the purpose and design of independently varied settings, but cannot be proved for every conceivable hidden variable merely by inspecting observed setting frequencies. Conversely, a rejection of Bell factorization does not alone establish a controllable superluminal signal or select a unique ontology. The reasoned verdict is against Bell's screening-off locality, conditional on the other stated premises; the unconditional conclusion concerns their conjunction.

Consider two spacelike-separated measurement regions $R_A$ and $R_B$, with locally chosen settings $a,b\in \{0,1\}$ and recorded outcomes $A,B\in \{-1,1\}$. A relevant past region supplies a sufficient specification $\lambda$ of the common preparation and other hidden physical information needed to screen the outcomes. This is not necessarily just a quantum state written at the source, nor a conditioning on arbitrary records which already contain the future choices. The spacetime shielding and sufficiency requirements are substantive causal assumptions.

Use Bell local causality: once the relevant past is sufficiently specified, learning facts in the remote measurement region changes no local outcome probability. In particular, requirewhere conditional probabilities are defined. The probability chain rule givesThis factorization combines parameter independence with outcome independence. Neither probability is required to be zero or one: the argument includes stochastic local hidden-variable theories.

Also assume measurement independence, $\rho (\lambda \mid a,b)=\rho (\lambda )$, so all four setting pairs average over one normalized nonnegative past-variable distribution. This is a statistical condition on the setting procedure and relevant hidden information, not a proof of a metaphysical doctrine of free will. Finally the recorded outcomes must represent the same ensemble of trials. One can include nondetections in a specified bounded-outcome rule; if instead one discards events depending on setting or outcome, a further sampling assumption is required. Setting-dependent postselection is not covered silently by the argument.

Let ${\alpha }_a(\lambda )={\sum }_{A}AP(A\mid a,\lambda )$ and ${\beta }_b(\lambda )={\sum }_{B}BP(B\mid b,\lambda )$. They lie in $[-1,1]$, and the factorization givesFor each $\lambda$,Averaging with the same $\rho$ proves the CHSH inequalityNo separate assumption of simultaneous predetermined values for incompatible quantum observables was needed. The common probability model and its screening relations did the work.

A Reichenbach common cause principle can motivate the same calculation: a complete classical common cause in the past screens the two outcomes, while each setting affects only its own wing. Together with setting-independent sampling this gives the displayed factorization. It is not enough to provide a different screening variable for each pair of settings: those constructions need not form one compatible common-cause model for the four correlations.

A stochastic Einstein locality formulation instead constrains objective chances by facts in the appropriate causal past. To use it as a sufficient premise here, specify both local dependence on settings and a joint screening condition for the two outcomes. Merely requiring local marginal chances not to depend on a remote setting gives parameter independence, which is weaker than factorization. Thus the precise stochastic locality version matters; one should not infer a Bell inequality from every condition loosely described as no superluminal influence.

Primitive causality is an initial-region completeness requirement. A state's full restriction to a suitable region $O$ determines expectations of observables in its future domain of dependence. The domain of dependence is not the whole causal future: it contains points whose past-inextendible causal curves all meet the specified initial data region. Merely being reachable by one causal curve from $O$ is not enough.

A standard local time-slice axiom makes this precise when $O$ contains a neighbourhood of a Cauchy slice of its causal development:with the regions and algebra identifications understood in the specified local net. Definitions applying primitive causality to all bounded open regions can be stronger than a time-slice property stated only for suitable causally convex Cauchy neighbourhoods; these formulations should not be conflated without extra hypotheses.

For a local hyperbolic Klein–Gordon equation, a later field is determined by the field and its normal derivative, or canonical momentum, on a Cauchy slice. Its value can be written using the causal fundamental solution, with integration supported where the past light cone meets the initial slice. If that support lies in the initial patch, the later smeared field is generated by the patch's initial field and momentum operators. Their algebra therefore generates the observables of the patch's domain of dependence. This is the local differential-equation mechanism behind the time-slice property.

The required information is the entire restricted quantum state: it assigns expectations to every observable and product in the initial algebra. Mean values of the initial field and momentum alone do not determine their fluctuations or higher correlations and thus are insufficient. Knowledge of the restricted state fixes later probability distributions and correlations; it does not fix which individual outcome a future measurement will produce.

Finally, primitive causality concerns propagation of information already encoded in initial data, while microcausality controls compatibility of spacelike measurements and quantum no-signalling controls possible interventions. The spectrum condition adds a separate positivity restriction. A causal initial-value evolution, positive energy and no-signalling are complementary requirements, not equivalent definitions of quantum locality.

The deterministic law cannot be transplanted unchanged into measurement-based quantum feedback. Evaluating its state-dependent scalar requires knowledge of the current quantum state. In fact $C=\mathrm{Tr}(i[H_1,{\rho }_d]\rho )$ is the expectation value of a Hermitian observable, but that expectation cannot be obtained exactly from a single system without measurement backaction. Tomography uses many independently prepared copies, not a nondisturbing readout of one evolving state.

Continuous weak measurements can supply a noisy record and a conditional state estimate, but then the state follows stochastic measurement dynamics with additional disturbance and decoherence, rather than the closed unitary equation used in part (a). A new stochastic stability analysis is required. One may instead simulate the corrected Lyapunov quantum control from known initial data and apply its waveform as open-loop control, or learn a pulse across repeated experiments as in Question 3. Those procedures do not constitute real-time state feedback on a single quantum trajectory. The separate source error in part (a) remains: even ideal knowledge of the state would not make its printed overlap law monotone.

The stated $M$ is specifically the transfer from the external input to the internal field $b_1$. Combining $b_1=G{b}_0$ with $b_0=\beta b_{min}-\alpha b_1$ givesFor $\gamma >0$ and a nondegenerate beam splitter $|\alpha |<1$, the closed-loop pole is $-{\gamma }_{meff}/2<0$. The return ratio for the Nyquist stability criterion is $L=\alpha G$. Its frequency locus is a circle of radius $|\alpha |<1$, so it cannot reach or encircle $-1$. The open cavity has no unstable poles; hence the closed loop has none. This agrees with the directly derived state damping rate.

The externally accessible output is different. From $b_2=\alpha b_{min}+\beta b_1$,It is again an all-pass cavity response. Confusing $b_1$ and $b_2$ would give the wrong output steady state.

For a constant coherent input mean $B=⟨b_{min}⟩$, the steady mean cavity amplitude follows by setting its derivative to zero:The input-output values also follow from $G(0)=-1$ and $M(0)=-\beta /(1-\alpha )$. For any initial mean amplitude, its difference from the stationary mean decays as $e^{-{\gamma }_{meff}t/2}$.

The stability statement needs endpoint qualifications. At $\alpha =1$, normalization forces $\beta =0$, and the cavity obeys $\dot{a}=0$: it is isolated and undamped, not asymptotically stable. Its external output is nevertheless the decoupled input. At $\alpha =-1$, the algebraic elimination divides by zero; the instantaneous feedback connection is singular and is not a well-posed instance of these reduced equations. Thus strict cavity stability holds for $|\alpha |<1$, not for every real pair satisfying only ${\alpha }^2+{\beta }^2=1$.

Finally, $b_{min}(t)$ is a quantum stochastic field, so arbitrary inputs do not approach a literal time-independent operator amplitude. The stationary expressions above describe a constant coherent drive's means, or the formal zero-frequency response. The full solution contains the filtered input-noise convolutionThe stationary cavity density operator also depends on the input noise statistics. For this ideal passive cavity with a coherent drive and vacuum noise, it is the coherent state with the amplitude just calculated; without such input statistics, the gain function alone does not specify a complete stationary quantum state.

Write the normalized input quantum state as $|\psi ⟩=a|0⟩+b|1⟩$. The fresh quantum ancilla is $|+⟩$, so the Controlled-Z gate createsApplying the measurement bra $⟨v_k(\theta )|$ to the first qubit leaves the unnormalized second-qubit vectorSince $|+⟩,|-⟩$ form an orthonormal basis,Also $U(\theta )|\psi ⟩=a|+⟩+e^{-i\theta }b|-⟩$, while the Pauli X gate fixes $|+⟩$ and changes the sign of $|-⟩$. Dividing by the branch norm therefore givesThis is one-bit teleportation with the present positive-exponent measurement vectors; the transferred logical gate is $U(\theta )=J(-\theta )$.

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).

The target specifies the desired output; the controller selects a control input; an actuator realizes it on the system; and the sensor returns a measured output. The environment supplies additional interactions and disturbances. A simple loop is:

For a quantum sensor, a measurement in quantum mechanics has statistical outcomes and generally changes the state through measurement backaction. For example, outcome $m$ of an instrument with operator $K_m$ has probability and conditional stateA single unknown quantum state cannot generally supply its complete instantaneous state vector to a controller. Noncommuting observables cannot be read simultaneously with arbitrary precision, and reliable quantum state tomography normally needs many reproducible preparations. A conditional state estimate can instead be updated from a continuous record, with its disturbance included in the model.

Quantum actuators change a Quantum Hamiltonian through applied electromagnetic fields, tunable interactions or couplings to auxiliary systems. A closed-system actuator implements unitary operators, so it preserves the spectrum and purity of a density operator; it cannot arbitrarily overwrite a state as a classical assignment operation could. Dissipative actuators or measurement in quantum measurements can change these invariants, but their dynamics must be accounted for.

The environment can entangle with the system, causing decoherence and dissipation when unobserved degrees of freedom are traced out. It is consequently part of the dynamical model, rather than just an additive classical disturbance. Conversely, Markovian reservoir engineering uses such coupling constructively to prepare and stabilize states.

In open-loop control the applied waveform is prescribed without using measurements from the current evolution. In closed-loop control it depends on the available measurement record or inferred state. Measurement-based quantum feedback is one realization of the latter; experimental adaptation across repeated preparations is another. The measurement record is classical information, but its production has quantum backaction.