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The Ghirardi--Rimini--Weber theory modifies unitary time evolution by random Gaussian position localizations. Each constituent collapses at a small fixed rate, while the rate for an entangled macroscopic body grows with its number of constituents.

For $P_{\alpha }=\mathrm{diag}(1,e^{i\alpha })$, conjugation by the Hadamard gate turns a relative quantum phase into an $x$-axis Bloch vector rotation. In the convention $\rho =(I+\mathbf{P}\cdot \mathbf{\sigma })/2$, the map is $(P_x,P_y,P_z)\mapsto (P_x,P_y\cos \alpha -P_z\sin \alpha ,P_y\sin \alpha +P_z\cos \alpha )$. An $x$-polarized input is unchanged, whereas a $z$-polarized input gives a sinusoidal population difference. A maximally mixed input remains maximally mixed under every unitary time evolution.

The Hamiltonian operator represents the total energy and generates unitary time evolution through the Schrödinger equation.

An impulsive quantum kick is the limit of a pulse with finite integrated Hamiltonian operator, rather than a bounded finite step. For a bounded self-adjoint operator $K$ with units of action, a rectangular pulse $H_{\tau }=K/\tau$ lasting $\tau$ has exact unitary time evolution $e^{-iK/\hslash }$ for every $\tau$. In the limiting notation $H(t)=K\delta (t)$, $\delta$ is the Dirac delta function andThe integral estimate for quantum state continuity at a finite Hamiltonian step no longer tends to zero: the pulse height diverges. A bounded background Hamiltonian operator contributes vanishing evolution during this limit. A more general pulse with noncommuting generators can require time ordering; integrated area alone then need not specify the result.

For example, $K=\hslash \varphi {\sigma }_y/2$, with a $y$ Pauli matrix, sends a qubit in $|0⟩$ to $\cos (\varphi /2)|0⟩+\sin (\varphi /2)|1⟩$. At $\varphi =\pi$, the Born rule predicts $|1⟩$ with probability one, however short the regularized pulse is. Its finite rotation is precisely what a bounded finite step cannot accomplish at the switching instant.

A normalized ground state initially supported on $(0,a)$ is $\sqrt{2/a}\sin (\pi x/a)$. Expanding the infinite square well to $(0,2a)$ leaves this wavefunction unchanged during an ideal sudden switch. Its overlap with the new ground state $a^{-1/2}\sin (\pi x/(2a))$ is $4\sqrt{2}/(3\pi )$, so the Born rule gives $32/(9{\pi }^2)$. This differs from ground-state overlap after sudden expansion of a square well with a fixed midpoint, because translation of the new well changes the overlap. Under a fixed final Hamiltonian operator, all final energy eigenstate probabilities remain constant during subsequent unitary time evolution.

For a bounded linear operator, the series $e^A={\sum }_{n=0}^{\infty }{A}^n/n!$ converges in operator norm. For an unbounded self-adjoint operator $H$, the spectral theorem instead defines the bounded unitary operator $e^{-iHt/\hslash }$ by applying the scalar function $e^{-iEt/\hslash }$ to its projection-valued measure. This yields unitary time evolution without assuming that every vector lies in the domain of every power of $H$. In finite dimensions it is the matrix exponential.

The operator support is the set of subsystems on which an operator may act nontrivially; it acts as the identity outside that set. Unitary time evolution can enlarge the support, while a Lieb-Robinson bound controls the error of replacing the evolved operator by one with smaller support. The support diameter and the number of sites in the support are different quantities.

Fix the qubit polarization convention by $\rho =(I+\vec{P}\cdot \vec{\sigma })/2$, where the Pauli matrices are ordered in the basis $(|1⟩,|2⟩)$ and ${\sigma }_z=|1⟩⟨1|-|2⟩⟨2|$. Thus the ground state is the north pole of the Bloch sphere. In a rotating frame and within the rotating-wave approximation, choose the drive phase and detuning sign so thatFor example, with this ground-state convention one may take $\Delta ={\omega }_L-{\omega }_0$ and choose the transverse coupling positive. The Bloch vector precesses around the effective field $(\Omega ,0,\Delta )$ at angular frequency ${\Omega }_R=\sqrt{{\Omega }^2+{\Delta }^2}$, tracing a circle at fixed projection on that axis. Its length is conserved under coherent unitary time evolution. On resonance the axis is transverse and the Rabi oscillation can fully invert an initially ground-state atom. Away from resonance the axis tilts, andThere is no relaxation towards an equilibrium Bloch vector in this ideal model. An inhomogeneous ensemble can nevertheless lose its averaged qubit polarization through dephasing even without spontaneous emission; the circle description assumes a common drive and detuning.

For $|a{|}^2+|b{|}^2=1$, the stated quantum state givesThe diagonal elements are the two state populations. The off-diagonal elements are quantum coherences: they retain the relative quantum phase and determine interference under a subsequent pulse. This is a pure state, even if both populations are nonzero: ${\rho }^2=\rho$, $\mathrm{Tr}{\rho }^2=1$, and $|\vec{P}|=1$. A mixed state averages projectors over preparation probabilities and generally has $\mathrm{Tr}{\rho }^2<1$. Equal populations alone do not distinguish an equal coherent superposition from an unpolarized mixture.

For use in all three cases, multiplication of the gates gives the Hadamard-phase-Hadamard qubit rotationThe global phase cancels in ${\rho }^{\prime }=U\rho U^{†}$. Thus the Bloch vector map is

The final unlabelled Ramsey spectroscopy comparison applies to the three cases as follows. In an atomic clock, an initial coherent pulse prepares a superposition, a free interval $T$ accumulates a relative quantum phase proportional to detuning times $T$, and a final coherent pulse converts that phase to a population. Here $\alpha$ plays the role of that accumulated phase, with sign and fringe offset set by pulse conventions. In (a), the first Hadamard gate creates the needed superposition and the final gate reveals full-contrast Ramsey interferometry. In (b), the first Hadamard gate instead puts the state in $|1⟩$, so the intervening phase gate has no relative quantum phase to imprint: the output is independent of $\alpha$. In (c), the unpolarized density matrix remains unchanged under every gate, so the ensemble has no Ramsey interferometry contrast. A coherent unitary gate alone cannot initialize a maximally mixed ensemble into a pure clock state.

The Ramsey interferometry analogy concerns preparation, phase accumulation, and readout; a Hadamard gate is not literally a transverse $\pi /2$ rotation on every input. In this basis $H=R_y(\pi /2){\sigma }_z$, and its Bloch sphere action is a $\pi$ rotation about $(\hat{x}+\hat{z})/\sqrt{2}$, up to a global phase. The phase conventions of physical Ramsey spectroscopy pulses must be specified to compare their outputs for arbitrary initial states.

Treat $|a⟩=|0⟩$ and $|b⟩=|1⟩$ as an isolated ground-state hyperfine qubit. Their transition angular frequency is ${\omega }_0=2\pi \times 320.7\mathrm{MHz}$. The radiofrequency magnetic field couples them through an allowed magnetic dipole transition. After choosing its phase reference, take the two-level laboratory Hamiltonian operator to bewhere $X,Z$ are Pauli operators. Here $\Omega$ is defined by the on-resonance rotating-frame off-diagonal matrix element $\hslash \Omega /2$.

Use a quantum rotating frame and the rotating-wave approximation, valid for $\Omega ,|\delta |\ll {\omega }_0$, with Rabi detuning $\delta ={\omega }_0-\omega$. Ignoring off-resonant coupling to other levels givesAt resonance $\delta =0$, the unitary time evolution isThus the Born rule gives coherent Rabi oscillations,The stated amplitudes are in the rotating frame; the laboratory carrier phases are tracked by the drive reference.

These logical states are orthogonal, individually resolvable hyperfine levels in the same electronic ground state. They avoid fast optical decay, can be coherently coupled by radiofrequency control, and have a state-selective optical readout through $|c⟩$. The excited level $|c⟩$ is useful for fluorescence detection rather than for the storage superposition in this model. Long lifetimes alone do not guarantee a useful qubit: control isolation and adequate quantum coherence are also required.

For short times, survival under unitary time evolution is $1-(\Delta H{)}^2{t}^2/{\hslash }^2+O(t^3)$. Dividing a fixed time into $N$ intervals and projecting back after each gives survival approaching one as $N\to \infty$. This suppression of departure by repeated projective measurements is the quantum Zeno effect.

For measurement-guided Zeno transfer, rotate the measured basis gradually to transport a known state. LetMeasure in these orthonormal bases successively for $j=1,\dots ,N$. Starting in $|0⟩=|b_0⟩$, the probability to follow the $b_j$ outcome at every step isIts final state is $|b_N⟩=|1⟩$. If every outcome is retained without postselection, the final probability for $|1⟩$ is $[1+{\cos }^N(\pi /N)]/2$, also approaching one; histories with two switches can return to that branch. These two finite-$N$ probabilities should not be confused.

For a general unknown input, the same slowly turning measurements first resolve its amplitudes into the two branches, then drag each branch separately. In the large-$N$ limit, after discarding the records,The first basis approaches the computational one as $N$ grows. Inserting an initial computational measurement makes this branch decomposition exact at the outset, without changing the limiting result. Relative phase is lost, so this procedure transports a known basis state with high probability rather than implementing a coherent Pauli X gate on an arbitrary qubit.

Expand the wavefunction in the complete orthonormal energy eigenstates. The Time-dependent Schrödinger equation makes each coefficient satisfy $i\hslash {\dot{a}}_n(t)=E_n{a}_n(t)$, soThe series is interpreted in the Hilbert space norm. Its norm is conserved because the phase factors have modulus one and the energy eigenstates are orthonormal. For initial data in the domain of the Hamiltonian operator, this is a strong solution of the equation; for any square-summable coefficients it gives the corresponding unitary time evolution.

In the two-dimensional span of the first two energy eigenstates, the measured observable has matrix $S=\left(\begin{array}{cc}7& 24\\ 24& -7\end{array}\right)$. Its square is $625I$, and its trace is zero, giving eigenvalues $25,-25$. The orthogonal complement has eigenvalue zero. Thus the full spectrum is $\boxed{\{-25,0,25\}}$.

The normalized eigenstates for $-25$ and $25$ are respectivelyThe lowest measurement outcome prepares ${\chi }_{-}$ up to an irrelevant overall phase. The subsequent unitary time evolution givesBy the Born rule, the probability of measuring the lowest eigenvalue again is the squared projection onto its one-dimensional eigenspace:This is the two-level quantum return probability with $q=9/25$. At $t=0$ the probability is one; if the two energies happen to be equal it remains one for every time.

Use units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum of the real scalar field is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. Canonical quantization replaces the classical fields by operator-valued distributions and imposes the equal-time canonical commutation relationsThese relations are imposed on an initial slice, with smeared fields understood. In the Heisenberg picture, simultaneous unitary time evolution of both operators preserves their commutators. Since the momentum is $\dot{\varphi }$, the requested equal-time commutator is $[\varphi (t,\mathbf{x}),\dot{\varphi }(t,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$.

The Legendre transform in mechanics, applied pointwise, gives the canonical Hamiltonian density of a real scalar field. ThusTo obtain its dynamics rather than assume them, use the Heisenberg equation of motion $\dot{O}=i[H,O]$. The momentum-square term gives $[H,\varphi ]=-i\pi$. The field-square term gives $[H,\pi ]=i{m}^2\varphi$, while differentiating the Dirac delta function in the gradient term and integrating by parts gives $[H,\pi ]=-i{\nabla }^2\varphi +i{m}^2\varphi$. ConsequentlyBoundary terms vanish for suitably decaying smeared fields. This is the Klein-Gordon equation as an operator identity.

A spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, where $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches and the Hermitian adjoint condition giveThe invariant-normalized scalar mode extraction can be written particularly simply asSubstituting the mode expansion of a free field proves these inversion formulas: the opposite-frequency coefficient is proportional to $E_p-E_q$ and vanishes after spatial Fourier inversion. Alternatively they are the corresponding Klein-Gordon inner products. Taking their commutators using the equal-time fields gives $(E_p+E_q)(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$ in the mixed case. In the two equal-type cases the delta function enforces $\mathbf{q}=-\mathbf{p}$, multiplying $E_p-E_q=0$. ThereforeConversely these mode commutators reproduce $[\varphi ,\dot{\varphi }]=i{\delta }^{(3)}$: the two frequency contributions each supply half the spatial Dirac delta function.

For the covariant oscillator Hamiltonian of a real scalar field, insert the mode expansion of a free field into $H$. Spatial integration sets the two momenta equal in the mixed terms and opposite in the two-creator or two-annihilator terms. The latter have coefficient $-E_p^2+{\mathbf{p}}^2+m^2=0$. The mixed terms giveThe constant $E_0$ is the divergent zero-point energy; a finite box and a momentum cutoff make its meaning explicit. Normal ordering sets the free vacuum energy to zero and leaves all commutators unchanged. Directly applying the mode commutators to the mixed expression yieldsChoose the Fock vacuum by $a(p)|0⟩=0$. Each creation operator adds a boson of energy $E_p$ and four-momentum $p$, while each annihilation operator removes one. Products of commuting creation operators give bosonic statistics from commuting creation operators and unrestricted occupation numbers. The real scalar field has one spin-zero species, identical to its antiparticle. Its one-particle states obey $⟨q|p⟩=(2\pi {)}^{3}2E_p{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$, precisely the relativistic normalization of a one-particle state.

For the Feynman propagator, put $z=x-y$ and contract the two fields in the Fock vacuum. Only $a{a}^{†}$ survives, giving the Wightman functionThe stipulated time-ordered product convention is $i{\Delta }_F(z)=\theta (z^0)W(z)+\theta (-z^0)W(-z)$. In a contour integral over $p^0$, the Feynman i-epsilon prescription places the positive-energy pole just below the real axis and the negative-energy pole just above it. For $z^0>0$ close below, clockwise, obtaining $-i{e}^{-i{E}_p{z}^0}/(2E_p)$; for $z^0<0$ close above, obtaining $-i{e}^{i{E}_p{z}^0}/(2E_p)$. Changing $\mathbf{p}\mapsto -\mathbf{p}$ in the second contribution therefore provesActing with $\square +m^2$ multiplies the integrand by $-(p^2-m^2)$. The remaining factor tends to $-1$ as a distribution, so Fourier inversion givesThe sign also follows from the derivative jump of a free scalar time-ordered two-point function: the first time derivative of $i{\Delta }_F$ jumps by $⟨[\dot{\varphi },\varphi ]⟩=-i{\delta }^{(3)}$, so that of ${\Delta }_F$ jumps by $-{\delta }^{(3)}$.

Expand the initial wavefunction in the complete orthonormal energy eigenstates, with $c_n=⟨{\psi }_n,\Psi (0)⟩$. The unitary time evolution isOn the two-dimensional subspace spanned by ${\psi }_1,{\psi }_2$, the observable has matrix $\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$. Its normalized eigenvectors are the symmetric and antisymmetric combinations. A complete orthonormal eigenbasis is thereforeThe eigenvalues are $1,3,0$, with $0$ infinitely degenerate. The first two are nondegenerate, regardless of the distinct energies $E_n$.

The first measurement in quantum mechanics, with outcome $3$, leaves the state ${\varphi }_2$ up to an irrelevant phase. At time $t$ it has becomeIts probability amplitude for outcome $1$ isThe Born rule gives the two-level observable transition probability

The Hamiltonian operator annihilates every vector orthogonal to both of its defining states:Every positive power also annihilates that vector. The exponential series therefore givesIt is a stationary zero-energy state under this unitary time evolution. The displayed evolution uses $\hslash =1$, as in the question. For $N=1$ the two independent defining vectors already span the whole qubit space, so no nonzero vector satisfies both orthogonality conditions; for larger $N$ their orthogonal complement has dimension $2^N-2$.

The density operators are Hermitian, so their squared Hilbert-Schmidt distance is $2V=\mathrm{Tr}({\rho }^2)+\mathrm{Tr}({\rho }_d^2)-2\mathrm{Tr}(\rho {\rho }_d)$. Both purity of a density operator values are constant under their respective unitary time evolution: cyclicity gives $\mathrm{Tr}(\rho [H,\rho ])=0$ for every Hermitian $H$. ConsequentlyThe terms containing $H_0$ cancel because $\mathrm{Tr}([-i{H}_0,\rho ]{\rho }_d)=-\mathrm{Tr}(\rho [-i{H}_0,{\rho }_d])$. The quantity $F=\mathrm{Tr}({\rho }_d[-i{H}_1,\rho ])$ is real: $[-i{H}_1,\rho ]$ is Hermitian, and the trace of the product of two Hermitian matrices equals its conjugate. The Lyapunov quantum control choice $f=F$ therefore gives $\boxed{\dot{V}=-f^2\le 0}$. For $t_2\ge t_1$, integration yields $V(t_2)=V(t_1)-{\int }_{t_1}^{t_2}f(t{)}^{2}dt\le V(t_1)$. Since the distance is $\sqrt{2V}$, it is monotonically nonincreasing. The decrease need not be strict when $f=0$.

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.

For reachability in control theory, a state $x_1$ is reachable from $x_0$ if an admissible input and a permitted duration drive the dynamical system from $x_0$ to $x_1$. Controllability requires this for every initial-target pair in the specified state space. Restrictions on duration or field amplitudes change the reachable set, so the admissible controls are part of the definition.

For a closed quantum system, $\dot{U}=-iH[f(t)]U$, $U(0)=I$, and states evolve as $\psi (t)=U(t)\psi (0)$ or $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. This gives three notions of quantum controllability. Unitary operator controllability requires every desired $U_d\in U(N)$ to be reachable from the identity; when a global phase is immaterial, one instead asks for every special unitary operation. Density operator controllability requires every pair of density operators with the same spectrum to be connected. The spectral qualification is necessary because unitary time evolution cannot change eigenvalues; the natural state space is a unitary orbit of a density operator. Pure-state controllability requires any normalized initial state to be transferred to any target ray, $U(T){\psi }_0=e^{i\chi }{\psi }_d$.

Thus $\boxed{\text{operator controllability}\Rightarrow \text{density operator controllability}\Rightarrow \text{pure-state controllability}}$, with the operator implication also valid if global phase is discarded. The converse implications need not hold. A pure-state task concerns one vector, whereas an operator task must act correctly on every input vector at once.

Let $B={\sum }_k{g}_k{Z}_k$. In the Zurek spin-bath model, $H=-Z_D\otimes B$, so unitary time evolution with $\hslash =1$ is $U(t)=e^{it{Z}_D\otimes B}$. The bath Hamiltonian terms commute, and device states $|0⟩,|1⟩$ have $Z_D$ eigenvalues $+1,-1$. Hencewhere the two normalized conditional bath states areTake each bath factor normalized, $|{\alpha }_k{|}^2+|{\beta }_k{|}^2=1$, and $|a{|}^2+|b{|}^2=1$. This entails no restriction: if only the product is initially normalized, divide each nonzero factor by its norm; the product of these norms is one.

Taking the partial trace over the bath gives the reduced density matrixThe orientation of this conditional environment overlap fixes the sign of the phase in the upper-right entry. Factorizing the overlap yields the decoherence factorThe populations are conserved because $[H,Z_D]=0$. Only phase coherence can be reduced. In particular,Quantum decoherence here results from distinguishable conditional bath states, although the complete system remains in a pure state under unitary time evolution.

Write $g_{\ast }>0$ for the upper endpoint called $a$ in the question, to distinguish it from the device amplitude. For a single realization of a finite bath,The literal claim of convergence to zero as time tends to infinity is false for every fixed finite $N$. Random-coupling spin-bath decoherence must distinguish individual realizations, ensemble averages and the large-bath limit.

First establish finite spin-bath coherence recurrence without assuming commensurate couplings. Fix a reference time $t_0>0$ and consider $Q^N+1$ points with coordinates $j{g}_k{t}_0/\pi$ modulo $1$, for $0\le j\le Q^N$. Partition the unit cube into $Q^N$ smaller cubes of side $1/Q$. Two points share a cube, so their difference provides an integer $1\le q\le Q^N$ withAt time $q{t}_0$ every cosine is arbitrarily close to $1$. As $Q$ increases, either these $q$ have an unbounded subsequence, or a bounded subsequence supplies a fixed $q$ with all distances exactly zero; its arbitrarily large multiples are then exact recurrences. In both cases there are unbounded times $t_j$ such that $z_N(t_j)\to 1$. This is the simultaneous Dirichlet approximation theorem argument; it applies equally to typical irrational random couplings. Part (i) supplies a particularly simple exact periodic counterexample.

The intended suppression is valid after ensemble averaging. Independence and the uniform density give the ensemble spin-bath coherenceThe continuous value at $t=0$ is $1$. For $t\ne 0$ its magnitude is strictly below $1$ to the power $N$, and for fixed $N$ it has an envelope of order $(2g_{\ast }t{)}^{-N}$ as $t\to \infty$. Thus the ensemble mean tends to zero with time.

The mean square distinguishes actual loss of coherence from cancellation of signs in that mean:At long times this tends to $2^{-N}$, exponentially small for $N\gg 1$, rather than exactly zero for a finite bath. At any fixed $t\ne 0$, the bracket $q(t)$ is strictly less than $1$, since ${\cos }^2(2gt)<1$ on a set of positive measure. The Markov inequality then givesThis proves small coherence for typical large baths at a fixed nonzero time. An even stronger fixed-time formulation uses the strong law of large numbers:The logarithmic singularities at isolated cosine zeros are integrable, and an exact zero has probability zero, so the law applies. Typical coherence consequently decreases exponentially with bath size.

The initial time scale is also explicit. When $g_{\ast }|t|\ll 1$, $\log \cos (2g_{k}t)=-2g_k^2{t}^2+O(g_k^4{t}^4)$, so the short-time Gaussian spin-bath decoherence approximation isHere $N^{-1}\sum g_k^2\to g_{\ast }^2/3$; on the scale $t\sim (g_{\ast }\sqrt{N}{)}^{-1}$ the displayed remainder tends to zero. The coherence is therefore rapidly suppressed for a large bath and is usually tiny at later fixed times, while rare recurrences still prevent a finite-realization long-time zero limit. Ensemble decay or a specified large-$N$ limit is the correct qualification of the printed assertion, consistent with reversible global unitary time evolution.