Quantum Annealing

Quantum annealing is a family of finite-time Hamiltonian processes that returns measured samples. A typical protocol prepares a state favored by a driver Hamiltonian, changes the driver and problem coefficients through a declared schedule, and decodes a terminal measurement into candidate answers. The evolution may be closed or open, nearly adiabatic or deliberately diabatic, monotone or reversed, and interrupted by pauses or quenches. A valid claim therefore needs more than an Ising objective and an anneal time: it needs a complete process, output, resource, and evidence record. This page develops that device-agnostic record. It does not identify quantum annealing with ideal adiabatic quantum computation, an equilibrium Gibbs sampler, classical simulated annealing, a particular hardware platform, or computational speedup.

Required background. Adiabatic Quantum Computation supplies driver–problem paths, accepted-answer and decoder semantics, schedule and energy normalization, closed-system evolution, and the boundary between an adiabatic certificate and a merely finite-time process. Thermal Master Equations supplies detailed balance, Gibbs stationarity, global-versus-local generator caveats, weak-coupling and secular assumptions, and the strong-coupling handoff needed to interpret thermal rates.

Quantum Annealing as a Finite-Time Sampling Model

For an instance $x$, an annealing specification starts with a finite-dimensional logical Hilbert space and a time-dependent Hamiltonian

$$H_x(t), \qquad 0\leq t\leq T.$$

The specification is incomplete until it also states an initial state $\rho_{0,x}$, a dynamical map $\Phi^{(x)}_{T:0}$, a terminal measurement $\{M_y\}_y$, and a classical decoder $D_x$. The resulting sample distribution is

$$p_x(y) := \operatorname{Tr} \!\left[ M_y\, \Phi^{(x)}_{T:0}(\rho_{0,x}) \right].$$

This operational definition includes several regimes without conflating them:

• a closed process can be analyzed by a unitary propagator;
• an open process requires a declared reduced-dynamics model and its validity regime;
• a slow schedule may admit an adiabatic approximation, but only under a licensed theorem or error certificate;
• a fast or nonmonotone schedule may use coherent interference without being adiabatic;
• a thermal population model may relax toward an instantaneous equilibrium law without reaching it; and
• a device may implement only a restricted coefficient range, graph, timing interface, or readout channel.

Quantum annealing is therefore not one universal differential equation. Shared endpoint Hamiltonians do not imply shared output distributions. Temperature alone does not determine an output distribution, and longer runtime need not improve success when relaxation, excitation, freeze-out, control error, or readout are present.

The page’s stopping rule is equally important. The process record can establish a conditional sample distribution or resource count for a declared model. Optimization Case Studies owns complete application evidence and matched solver comparisons, while Algorithmic Benchmarking owns the general end-to-end time-to-accepted-answer and advantage ledger.

The Ten-Field Quantum Annealing Record

Use the following record before calculating a spectrum, simulating dynamics, or interpreting samples. It is intentionally vertical so that assumptions do not disappear on narrow screens.

1. Annealing task, instance family, and licensed claim. State the instance distribution or fixed instance, the requested output, and the narrow conclusion the calculation can support.
2. Logical variables, encoding, basis, and promises. Declare bit, spin, or higher-dimensional variables; basis order; bit-to-spin convention; feasibility promises; and all classical preprocessing.
3. Driver, problem Hamiltonian, catalysts, and decoder. Give every Hamiltonian term, coefficient unit, offset and scale, catalyst, accepted set, and classical decoding map.
4. Initial state, preparation, and endpoint convention. State the prepared state or ensemble, its cost or assumption, endpoint inclusivity, dwell behavior, and when measurement occurs.
5. Schedule, controls, pauses, and runtime. Give the physical coefficient envelopes, total duration, pauses, reversals, quenches, bandwidth assumptions, and timing boundary.
6. Dynamics, environment, temperature, rates, and equilibrium or freeze-out assumptions. Name the closed or reduced open model, initial correlations, bath and coupling assumptions, rates, temperature units, equilibrium hypothesis, and any operational freeze criterion.
7. Measurement, sample distribution, success event, and quality metric. Declare the POVM, readout channel, decoder, accepted event, energy or cost statistic, and whether probabilities are exact, estimated, conditional, or postselected.
8. Embedding, gauges, precision, resources, repeats, and comparator boundary. Separate logical and physical variables, chain policy, gauge transformations, coefficient precision, all resource currencies, repeat assumptions, and the classical comparison boundary.
9. Verification data, uncertainty, and reproducibility. Record analytic checks, numerical methods and tolerances, sampling or fit uncertainty, calibration provenance when applicable, and enough detail to reproduce each value.
10. Conclusion, stopping point, and canonical handoff. State exactly what follows, what does not, and which owner receives the next theorem, physical implementation, optimization, or benchmarking question.

Every field must contain a value or a specific reason that the quantity is not applicable. Writing “anneal with transverse field” is not a substitute for this record: it omits the process that connects a Hamiltonian name to observed samples.

Driver, Problem Hamiltonian, and Accepted Answers

Fix the computational convention

$$Z|0\rangle=+|0\rangle, \qquad Z|1\rangle=-|1\rangle.$$

If a classical bit is $x_i\in\{0,1\}$, the corresponding spin eigenvalue is

$$s_i=(-1)^{x_i}, \qquad x_i=\frac{1-s_i}{2}.$$

For a classical objective $C$ and decoder $D$, a safe diagonal encoding contract is

$$H_{\rm P} := \sum_z E(z)|z\rangle\langle z|, \qquad E(z) := a\,C(D(z))+b, \qquad a>0.$$

The positive scale $a$ and offset $b$ preserve the ordering of classical costs, but they do not disappear physically. The scale changes spectral gaps, the dimensionless product $\beta H_{\rm P}$, coefficient ranges, and usually transition rates. The offset changes only a closed-system global phase when it is applied uniformly, but it still belongs in a convention-complete record.

A common two-local Ising problem Hamiltonian is

$$H_{\rm P} := cI + \sum_i h_iZ_i + \sum_{i<j}J_{ij}Z_iZ_j.$$

Penalty terms must be large enough to enforce the stated feasible set under the declared perturbation and precision model. “Large” cannot mean arbitrarily large: coefficient ranges and rescaling can shrink every other physical energy scale. Degenerate ground strings are not an error when the task accepts a set. The record must distinguish an accepted subspace from one preferred representative and must say whether the decoder identifies several physical strings with one logical answer.

The usual transverse driver is

$$H_{\rm D} := -\sum_i\Gamma_iX_i, \qquad \Gamma_i>0,$$

whose product ground state is easy to prepare when the coefficients are independent and positive. It is one driver, not the definition of quantum annealing. Nonuniform transverse terms, multi-spin drivers, catalysts, and longitudinal biases change preparation, symmetry, stoquasticity, gaps, and control requirements. The Transverse-Field Ising Model owns clean-chain phases, criticality, finite-size gaps, and exact many-body physics; this page uses driver and Ising terms only as components of a computation record.

An accepted set $\mathcal A_x$ should be defined in decoded space. If $D_x(z)$ is the answer returned by bit string $z$, then the accepted computational outcomes are

$$\widetilde{\mathcal A}_x := \{z:D_x(z)\in\mathcal A_x\}.$$

This distinction matters when embeddings, broken-chain repair, redundant encodings, or several ground strings represent the same answer.

Schedules, Pauses, Reversals, and Quenches

A general controlled path has the form

$$H(t) = A(t)H_{\rm D} + B(t)H_{\rm P} + H_{\rm cat}(t), \qquad 0\leq t\leq T.$$

The coefficient envelopes are physical controls, not decorative interpolation parameters. A reproducible record gives their units, boundary values, complete time dependence, total runtime, regularity, and permitted bandwidth. It also says whether the terminal measurement occurs immediately, after a dwell, or after an additional control segment.

A forward schedule usually reduces the relative driver strength and increases the problem strength. It need not be linear or monotone. A pause holds one or more controls fixed for a declared duration; in an open model that changes the available relaxation budget, while in a closed model it adds coherent phase evolution. A reverse anneal begins from a declared seed, moves away from the problem-dominated endpoint, and later returns. It is not preparation of the driver ground state in reverse. A quench is an ideal discontinuity or a finite-bandwidth ramp whose duration and approximation error must be stated.

These controls answer different questions. A slow smooth path may license an adiabatic approximation under additional hypotheses. A deliberately diabatic path may exploit coherent transitions. A pause may improve or worsen the accepted probability depending on the incoming population and the instantaneous equilibrium bias. A quench can create interference that makes success oscillate with dwell time. None of these outcomes follows from the endpoints alone.

Closed, Open, Thermal, and Nonadiabatic Regimes

For a closed model, the state obeys

$$i\hbar\frac{d}{dt}|\psi(t)\rangle = H(t)|\psi(t)\rangle.$$

This equation contains no bath temperature. Temperature may enter only through a separately declared initial ensemble. Closed evolution can be adiabatic or nonadiabatic, coherent or dephased only by an explicitly added channel; calling a Hamiltonian path an anneal does not determine which regime applies.

An exact system–environment statement begins instead with

$$\rho_S(t) = \operatorname{Tr}_E \!\left[ U_{SE}(t)\rho_{SE}(0)U_{SE}^{\dagger}(t) \right].$$

Any reduced equation therefore inherits assumptions about the coupling, environmental state, initial correlations, memory, and approximation scale. A time-local model

$$\dot\rho = -\frac{i}{\hbar}[H(t),\rho] + \mathcal D_t(\rho)$$

is meaningful only after $\mathcal D_t$, its domain, and its approximation regime are specified. The Lindblad–GKSL Equation owns the general generator structure, and Strong Coupling owns the boundary where bare-system Gibbs and memoryless descriptions can fail.

Fast driving, unresolved Bohr frequencies, strong coupling, or memory can invalidate an instantaneous global thermal generator. A local dissipator need not make the Gibbs state of the interacting Hamiltonian stationary. Conversely, agreement with a diagonal rate equation does not establish the microscopic bath model that produced it. The annealing record may adopt a reduced model, but it must keep that conditional status visible. Noise in Quantum Information owns the separate taxonomy of coherent, incoherent, leakage, crosstalk, SPAM, drift, and correlated device errors.

Relaxation, Freeze-Out, and Endpoint Distributions

The equilibrium comparison associated with a specified Hamiltonian and inverse temperature is

$$\rho_\beta(t) = \frac{e^{-\beta H(t)}}{\operatorname{Tr}e^{-\beta H(t)}}, \qquad \pi_\beta(z) = \frac{e^{-\beta E(z)}}{\sum_{z'}e^{-\beta E(z')}}.$$

The Thermal Density Operators page owns this equilibrium construction. An observed diagonal state is not necessarily Gibbs, and a stationary state need not be the Gibbs state of the bare problem Hamiltonian. Temperature plus an endpoint spectrum does not determine a finite-time output distribution without dynamics, initial data, runtime, and readout.

A useful phenomenological population model is

$$\dot p(t) = -\gamma(t)\bigl[p(t)-q(t)\bigr], \qquad \Lambda(t) = \int_0^t\gamma(u)\,du.$$

Its exact solution is

$$p(T) = e^{-\Lambda(T)}p(0) + \int_0^T \gamma(u)q(u) e^{-[\Lambda(T)-\Lambda(u)]}\,du.$$

This equation makes schedule memory explicit: when the remaining integrated rate is small, the population retains much of its earlier deviation. The diagnostic

$$\Lambda_{\rm rem}(t) = \int_t^T\gamma(u)\,du$$

supports an operational freeze criterion only after the observable, kinetic model, and numerical threshold are fixed. There is no unique, model-independent freeze-out point.

Suppose, additionally, that the late path commutes and has the special form

$$H(t)=a(t)H_{\rm P}+b(t)I,$$

and that populations equilibrate before freezing at $t_*$. Only under those conditions does an endpoint fit in final-energy units obey

$$\beta_{\rm eff} = \beta\frac{a(t_*)}{a(T)}.$$

Residual transverse terms, a changing eigenbasis, subsequent transitions, readout bias, or another kinetic model break this inference. Relaxation and Thermalization owns the broader distinction between equilibration mechanisms and thermal ensembles.

Stoquasticity, Tunneling, and Interpretation Limits

In a specified orthonormal basis, a real Hamiltonian is stoquastic when

$$\langle z|H(t)|z'\rangle\leq0, \qquad z\neq z',$$

throughout the stated time interval. The basis, sign convention, and full path belong to the claim. A basis change can alter the signs of off-diagonal matrix elements without altering the spectrum, so stoquasticity is not a basis-independent physical label.

The Sign Problem Preview owns the relationship between representations, sign-free weights, and simulation obstacles. A stoquastic path need not mix rapidly or solve an optimization problem easily. A nonstoquastic term need not confer an advantage. The existence of tunneling or entanglement likewise establishes a dynamical mechanism, not a favorable scaling comparison.

Barrier language also needs a declared coordinate and comparator. A narrow barrier in one effective coordinate may still sit inside a hard many-body landscape, and a semiclassical tunneling estimate does not by itself determine end-to-end sample quality. Computational evidence requires an instance family, matched resources, uncertainty, and a competitive classical portfolio.

Embedding, Gauges, Precision, and Resource Accounting

An embedding maps logical variables and couplings to physical controls. Its record includes the number of logical and physical variables, chain or code structure, penalty strengths, broken-chain policy, coefficient rescaling, rejected-sample rule, and the decoder applied before scoring. A logical instance is not the same resource as its physical realization.

For a spin-reversal gauge $g_i\in\{-1,+1\}$,

$$U_gH(h,J)U_g^\dagger = H(g_i h_i,g_i g_jJ_{ij}),$$

and decoding must undo

$$s_i'=g_i s_i.$$

Gauge averaging can expose or reduce selected control biases, but it is not error correction and does not erase correlations between samples. Gauge choice, programming count, reads per gauge, and any aggregation rule are part of the experiment.

Precision cannot be summarized by the number of printed decimal places. Report coefficient ranges, quantization, rescaling, calibration uncertainty, drift, and whether the accepted set is robust to the permitted perturbations. If ideal accepted and rejected bands have separation $\Delta_{\rm acc}$, the sufficient Weyl guard

$$\|\delta H_{\rm P}\| < \frac{\Delta_{\rm acc}}{2}$$

preserves their spectral separation. It does not certify calibration, dynamical fidelity, or sampling accuracy. Analog Quantum Simulation owns target-to-device correspondence, while Control, Readout, and Calibration owns platform control and measurement practice.

A complete ledger separates setup, programming, physical variables, couplers, controls, anneal duration, pause duration, reads and resets, gauges, decoding, verification, rejected samples, and comparator work. Fixed costs may be amortized only over the batch that actually shares them.

Success Probability, Repetitions, and Evidence Boundaries

For a terminal POVM and classical decoder,

$$p_T(y) = \operatorname{Tr}[M_y\rho(T)], \qquad p_{\rm acc} = \sum_{y:D(y)\in\mathcal A}p_T(y).$$

If measurement is followed by a classical readout channel,

$$p_{\rm obs}(y) = \sum_zR(y|z)p_T(z).$$

Exact optimum probability, feasibility, threshold success, energy or cost quantiles, and approximation quality answer different questions and should be reported separately. Low mean energy does not imply a large probability of an accepted answer.

For a known, fixed, independent per-run probability $0<p_{\rm acc}<1$, the runs required for at least one accepted result with confidence $\eta$ are

$$R_\eta = \left\lceil \frac{\ln(1-\eta)}{\ln(1-p_{\rm acc})} \right\rceil.$$

The cases $p_{\rm acc}=0$ and $p_{\rm acc}=1$ must be handled separately. Drift, correlations, gauge dependence, and uncertainty in an estimated probability invalidate blind use of this formula.

An annealing-specific time ledger may be written

$$T_{{\rm total},\eta} = T_{\rm setup} + T_{\rm program} + R_\eta \left( T_{\rm anneal} + T_{\rm read/reset} + T_{\rm decode} + T_{\rm verify} \right).$$

This form assumes one fixed or batched program term; reprogramming between reads requires the corresponding batch sum. The number of gauges, batches, and amortized reads must be explicit. Algorithmic Benchmarking owns the general confidence, scaling, comparator, and end-to-end advantage framework, and Optimization Case Studies owns application and hardware evidence. Neither a single successful instance nor anneal-only timing establishes scaling or speedup.

Quantum Algorithms for Optimization compares finite-time annealing with other solver families under matched accepted-answer and total-cost conventions; this page retains the process, open-system, embedding, and sample-distribution contract.

Worked Audit: A Coherent Quench–Pulse–Quench Anneal

This audit is an exact one-qubit closed process with $\hbar=1$.

1. Annealing task, instance family, and licensed claim. Minimize the one-bit endpoint energy of $H_{\rm P}=-Z/2$ and return $0$. The only licensed claim is the exact finite-time distribution of this toy process.

2. Logical variables, encoding, basis, and promises. Use $Z|0\rangle=|0\rangle$ and $Z|1\rangle=-|1\rangle$ on one qubit. There is one accepted string and no promise beyond the exact controls below.

3. Driver, problem Hamiltonian, catalysts, and decoder. Use

   $$H_{\rm D}=-\frac{X}{2}, \qquad H_{\rm M}=-\frac{X+Z}{2}, \qquad H_{\rm P}=-\frac{Z}{2},$$

   with no catalyst. A computational-basis read decodes directly.

4. Initial state, preparation, and endpoint convention. Prepare the exact driver ground state $|+\rangle$. Measure immediately after the final quench; a later closed dwell under $H_{\rm P}$ changes phases but not computational-basis probabilities.

5. Schedule, controls, pauses, and runtime. In $H=AH_{\rm D}+BH_{\rm P}$, quench $(A,B):(1,0)\to(1,1)$, hold for time $\tau$, and quench to $(0,1)$. Both switches are ideal zero-duration, unbounded-bandwidth controls.

6. Dynamics, environment, temperature, rates, and equilibrium or freeze-out assumptions. Use exact closed Schrödinger evolution with no bath, temperature, stochastic noise, relaxation rate, equilibrium assumption, or freeze-out interpretation.

7. Measurement, sample distribution, success event, and quality metric. The pulse propagator and state are

   $$\begin{aligned} U_{\rm M}(\tau) &= \cos\!\left(\frac{\tau}{\sqrt2}\right)I + i\sin\!\left(\frac{\tau}{\sqrt2}\right) \frac{X+Z}{\sqrt2}, \\ |\psi(\tau)\rangle &= \cos\!\left(\frac{\tau}{\sqrt2}\right)|+\rangle + i\sin\!\left(\frac{\tau}{\sqrt2}\right)|0\rangle. \end{aligned}$$

   Therefore

   $$p_0(\tau) = \frac{3-\cos(\sqrt2\,\tau)}{4}, \qquad p_1(\tau) = \frac{1+\cos(\sqrt2\,\tau)}{4}, \qquad \langle H_{\rm P}\rangle = \frac12-p_0.$$

   The exact audit table is:

   $\tau$	$p_0$	$p_1$	$\langle H_{\rm P}\rangle$
   $0$	$1/2$	$1/2$	$0$
   $\pi/(2\sqrt2)$	$3/4$	$1/4$	$-1/4$
   $\pi/\sqrt2$	$1$	$0$	$-1/2$
   $3\pi/(2\sqrt2)$	$3/4$	$1/4$	$-1/4$
   $\sqrt2\,\pi$	$1/2$	$1/2$	$0$

   The return to $p_0=1/2$ after a longer dwell is an exact counterexample to “longer is always better.”

8. Embedding, gauges, precision, resources, repeats, and comparator boundary. The abstract record uses one qubit, two ideal quenches, one dwell, and one terminal read, with no embedding or gauge. Coefficients and arithmetic are exact; physical coefficient precision is N/A. At $\tau=\pi/(2\sqrt2)$, four fixed independent abstract runs give at least one $0$ with confidence at least $0.99$. Physical quench cost and bandwidth are not licensed, so physical time-to-solution and a classical comparison are N/A.

9. Verification data, uncertainty, and reproducibility. Verify the analytic $2\times2$ exponential, normalization, the $\tau=0$ limit, period $\sqrt2\,\pi$, all five rows, and

   $$R_{0.99} = \left\lceil\frac{\ln(0.01)}{\ln(1/4)}\right\rceil =4.$$

   Exact arithmetic carries no sampling uncertainty; an empirical realization would.

10. Conclusion, stopping point, and canonical handoff. This is a valid ideal finite-time coherent annealing schedule and can reach the endpoint ground state by interference. It is not an adiabatic certificate, a hardware implementation, a scalable optimization result, or a speedup claim.

Worked Audit: A Two-Stage Detailed-Balance Rate Model

This audit is an explicitly phenomenological diagonal late-anneal reduction with $k_{\rm B}=\hbar=1$. It is not a microscopic master-equation derivation.

1. Annealing task, instance family, and licensed claim. Track one effective ground/excited pair through two thermal-rate stages and compute the endpoint population. The claim is conditional on the declared scalar rate law.

2. Logical variables, encoding, basis, and promises. The fixed basis has ground energy $0$ and excited energy $\Delta_j$ in stage $j$. Coherences and changes of eigenbasis are excluded by construction.

3. Driver, problem Hamiltonian, catalysts, and decoder. The late-stage Hamiltonian is $H_j=\Delta_j|1\rangle\langle1|$. The transverse driver and all catalysts are negligible in this reduced window; $0$ decodes as success.

4. Initial state, preparation, and endpoint convention. Start with $p_e(0)=1/2$. Read the output immediately after stage 2.

5. Schedule, controls, pauses, and runtime. Use stage 1 with $(\Delta_1,\gamma_1,t_1)=(1,1,1)$ and stage 2 with $(\Delta_2,\gamma_2,t_2)=(3,0.1,2)$. The total modeled duration is $3$ in the declared reciprocal-rate units, and the instantaneous stage change is an idealization.

6. Dynamics, environment, temperature, rates, and equilibrium or freeze-out assumptions. Fix $\beta=2$ and

   $$\dot p_e = -\gamma_j(p_e-q_j), \qquad q_j = \frac{1}{1+e^{\beta\Delta_j}},$$

   with $\Gamma_{\uparrow,j}=\gamma_jq_j$ and $\Gamma_{\downarrow,j}=\gamma_j(1-q_j)$, so

   $$\frac{\Gamma_{\uparrow,j}}{\Gamma_{\downarrow,j}} =e^{-\beta\Delta_j}.$$

   This is a detailed-balance population law, not a claim that an arbitrary driven device has this generator. “Freeze-out” refers only to the quantified remaining relaxation.

7. Measurement, sample distribution, success event, and quality metric. Read the fixed energy basis, accept $0$, and set $p_{\rm acc}=1-p_e$. Each constant stage has the exact update

   $$p_{\rm out} = q+(p_{\rm in}-q)e^{-\gamma t}.$$

   Applying it twice gives

   $$\begin{aligned} q_1 &=0.11920292202211755, & p_1 &=0.2592903382683391, \\ q_2 &=0.0024726231566347743, & p_2 &=0.21273718445380704. \end{aligned}$$

   The actual ground probability is $1-p_2=0.7872628155461929$, while the endpoint Gibbs ground probability is $1-q_2=0.9975273768433652$. The binary total-variation distance is

   $$|p_2-q_2| = 0.21026456129717228.$$

   Only $1-e^{-0.2}=0.18126924692201815$ of the stage-2 deviation relaxes; the retained fraction is $e^{-0.2}=0.8187307530779818$. A two-level endpoint fit gives

   $$\beta_{\rm eff} = \frac13\ln\!\left(\frac{1-p_2}{p_2}\right) = 0.43616820355726066.$$

   This is a fit diagnostic, not the bath inverse temperature $2$ and not proof of a unique freeze point.

8. Embedding, gauges, precision, resources, repeats, and comparator boundary. The model uses one effective population, two exact gaps, two exact total rates, two stage durations, and one read per run; physical embedding, gauges, and coefficient precision are N/A. Conditional on fixed independent $p_{\rm acc}=1-p_2$, $R_{0.99}=3$ abstract reads suffice. Programming, thermalization, measurement, and comparator costs are N/A because the reduced model does not define them.

9. Verification data, uncertainty, and reproducibility. Recompute $q_1,q_2$ from $\beta\Delta=(2,6)$ and apply the exact update twice. Verify

   $$\frac{p_2-q_2}{p_1-q_2}=e^{-0.2},$$

   the total-variation distance, fitted $\beta_{\rm eff}$, and repeat count to $10^{-12}$ absolute tolerance. There is no empirical uncertainty in the toy calculation.

10. Conclusion, stopping point, and canonical handoff. The same endpoint Hamiltonian and bath temperature do not determine the output distribution. Finite relaxation leaves schedule history, and a fitted effective temperature does not establish equilibrium, a physical freeze mechanism, or optimization advantage. Thermal Master Equations owns microscopic validity.

Common Failure Modes and Canonical Handoffs

Equating quantum annealing with adiabatic quantum computation. A finite-time anneal can be open, paused, reversed, quenched, or intentionally nonadiabatic. Shared driver and problem endpoints do not import an adiabatic theorem or a circuit-equivalence claim; Adiabatic Quantum Computation owns those certificates.

Treating “slow” as a theorem. Runtime alone says nothing without a system-size family, gaps or another licensed error bound, schedule regularity, initialization, and a declared dynamical model. In an open process, longer exposure can add helpful relaxation, harmful excitation, or both.

Inferring a distribution from temperature and the endpoint Hamiltonian. The coupling operators, rates or spectral density, initial state, schedule, runtime, and readout are also required. A Gibbs fit does not prove thermalization, and positive rates do not by themselves satisfy thermal detailed balance. Local generators can also have the wrong stationary state for an interacting Hamiltonian.

Naming a unique freeze point without an operational test. Freeze-out depends on the observable, model, and threshold. An effective-temperature fit is not a mechanism, especially when transverse terms remain or the energy basis keeps changing.

Assuming that longer, pausing, or reversing must help. Coherent success can oscillate, and relaxation moves a population toward the instantaneous equilibrium value from either side. Each control claim needs its own before-and-after distribution.

Turning a sign label into a complexity result. Stoquastic does not mean easy, and nonstoquastic does not mean advantageous. Tunneling and entanglement likewise do not establish speedup. Quantum Complexity Classes owns asymptotic promise-problem language.

Discarding offsets, scales, embeddings, gauges, or decoder choices. A positive rescaling preserves an argmin but changes gaps, $\beta H$, ranges, and rates. Embedding and gauge averaging consume resources and are not free error correction. Broken-chain handling and rejected samples change the measured distribution.

Replacing success probability with mean energy. A low average energy can coexist with negligible mass on the accepted set. State the accepted event before computing repeats, and include confidence intervals when its probability is estimated.

Reporting anneal-only time as end-to-end time. Setup, programming, gauges, reads and resets, decoding, verification, rejected samples, and comparator work belong in the ledger. A fixed-instance demonstration does not establish favorable scaling, hardware advantage, or speedup. Claims, Hype, and Evidence Standards owns broad evidence language.

For microscopic dissipative derivations, hand off to Thermal Master Equations and Open-System Simulation. For native controls, calibration, leakage, and device-specific readout, hand off to the Hardware Overview and Control, Readout, and Calibration. For device data, matched classical portfolios, scaling, and speedup vocabulary, use Optimization Case Studies and Algorithmic Benchmarking rather than extending this process record.

Exercises

Exercise 1: Map a Two-Bit QUBO to an Ising Hamiltonian

Using $x_i=(1-Z_i)/2$, map

$$C=x_1+2x_2-3x_1x_2$$

to a diagonal Ising Hamiltonian. Enumerate the four bit strings, identify the accepted minimum set, and explain which affine changes preserve that set.

Solution

Substitution gives

$$H = \frac34I + \frac14Z_1 - \frac14Z_2 - \frac34Z_1Z_2.$$

In the order $(00,10,01,11)$, the objective values are

$$(0,1,2,0).$$

Thus the accepted set is $\{00,11\}$. Adding a constant and multiplying by a positive number preserve this argmin. They do not preserve every physical statement: a positive scale changes gaps, $\beta H$, control ranges, and generally transition rates.

Exercise 2: Apply and Decode a Spin-Reversal Gauge

For

$$H = 0.4Z_1-0.7Z_2+1.2Z_1Z_2,$$

apply the gauge $g=(-1,+1)$, find the transformed Hamiltonian, and decode the transformed ground spin.

Solution

The coefficient rule $h_i'=g_ih_i$, $J_{ij}'=g_ig_jJ_{ij}$ gives

$$H' = -0.4Z_1-0.7Z_2-1.2Z_1Z_2.$$

The original ground spin is $(-,+)$, with energy

$$0.4(-1)-0.7(+1)+1.2(-1)(+1)=-2.3.$$

It maps to the measured transformed spin $(+,+)$. Decoding with $s_i=g_is_i'$ recovers $(-,+)$. The gauge is a unitary relabeling, so the ideal spectrum is invariant; averaging over gauges is not error correction.

Exercise 3: Show That Stoquasticity Depends on the Basis

Compare $-X$ with its transform under $Z$. Decide whether each matrix is stoquastic in the computational basis and state what remains invariant.

Solution

In the computational basis,

$$-X = \begin{pmatrix} 0&-1\\ -1&0 \end{pmatrix}$$

has nonpositive off-diagonal entries. Since $ZXZ=-X$,

$$Z(-X)Z=+X,$$

whose off-diagonal entries are positive in that same basis. The two Hamiltonians are unitarily related and have the same spectrum. Physical predictions agree only when states and observables are transformed as well. The example licenses neither “sign-free means easy” nor “nonstoquastic means advantageous.”

Exercise 4: Solve an Ideal One-Qubit Reverse Anneal

Let $H_{\rm P}=-Z/2$ and seed the system in $|1\rangle$. Quench $(A,B):(0,1)\to(1,0)$, hold $H_{\rm D}=-X/2$ for time $\tau$, and quench back. Compute the probability of returning the accepted state $|0\rangle$.

Solution

During the dwell,

$$e^{i\tau X/2}|1\rangle = \cos\!\left(\frac\tau2\right)|1\rangle + i\sin\!\left(\frac\tau2\right)|0\rangle.$$

The final quench does not change the state, so

$$p_{\rm acc}(\tau) = \sin^2\!\left(\frac\tau2\right).$$

For $\tau=(0,\pi/2,\pi,2\pi)$, the accepted probabilities are $(0,1/2,1,0)$. This ideal result depends on the declared seed and unbounded-bandwidth quenches. It is not a speedup statement.

Exercise 5: Audit a Pause under Thermal Relaxation

For a constant-stage equilibrium excited probability $q=0.1$, rate $\gamma=0.5$, and pause duration $t_p=4$, propagate incoming excited probabilities $0.4$ and $0.05$. Does the same pause always improve ground-state success?

Solution

The constant-stage update is

$$p_e(t_p) = q+[p_e(0)-q]e^{-\gamma t_p}.$$

For the first input,

$$0.4 \longmapsto 0.1+0.3e^{-2} = 0.1406005849709838,$$

so the ground probability increases. For the second,

$$0.05 \longmapsto 0.1-0.05e^{-2} = 0.09323323583816936,$$

so the ground probability decreases. Relaxation moves the population toward $q$ from either side; the same pause can help or hurt according to the incoming state.

Exercise 6: Audit a Two-Site Chain Embedding

For $h>0$, $\kappa>0$, and

$$H_{\rm emb} = -\kappa s_as_b-hs_a+hs_b,$$

enumerate the two-spin energies, find when unbroken chains are the only ground configurations, and test $(h,\kappa)=(1,1.5)$ and $(1,0.75)$.

Solution

In the order $(++,--,+-,-+)$, the energies are

$$(-\kappa,-\kappa,\kappa-2h,\kappa+2h).$$

The two unbroken configurations are the only ground configurations exactly when $\kappa>h$; at $\kappa=h$, one broken configuration becomes degenerate. For $h=1$ and $\kappa=1.5$, the ledger is

$$(-1.5,-1.5,-0.5,3.5),$$

so an unbroken chain wins. At $\kappa=0.75$, it is

$$(-0.75,-0.75,-1.25,2.75),$$

and the $+-$ broken configuration wins. Majority vote ties on any broken two-site chain, so the record must choose a deterministic energy decoder or reject those samples.

Exercise 7: Build an Honest Repeat and Time-to-Solution Ledger

Assume known independent $p_{\rm acc}=0.08$ and target confidence $\eta=0.99$. There is one $5\,{\rm ms}$ setup/program cost. Each run uses $20\,\mu{\rm s}$ to anneal, $180\,\mu{\rm s}$ to read and reset, and $50\,\mu{\rm s}$ to decode. Compute the repeat count, end-to-end time, anneal-only time, and their ratio.

Solution

The repeat count is

$$R = \left\lceil \frac{\ln(0.01)}{\ln(0.92)} \right\rceil =56.$$

Each full run costs $0.25\,{\rm ms}$, so

$$T_{\rm total} = 5\,{\rm ms}+56(0.25\,{\rm ms}) = 19\,{\rm ms}.$$

Anneal-only accounting gives $56(0.020\,{\rm ms})=1.12\,{\rm ms}$. It underestimates the declared end-to-end time by

$$\frac{19}{1.12} = 16.96428571428571.$$

This calculation assumes fixed independent trials and a known success probability. Estimation uncertainty, drift, correlation, reprogramming, verification, or comparator work would require a larger ledger.

Exercise 8: Write a Complete Three-Spin Annealing Record

Complete and audit the following toy process record.

Solution

1. Annealing task, instance family, and licensed claim. Return any antiferromagnetic ground string of one three-spin triangle. License only the exact distribution of a declared toy endpoint channel.

2. Logical variables, encoding, basis, and promises. Use three computational-basis spins with $Z|0\rangle=|0\rangle$. The two aligned strings are rejected and the other six are accepted; there is no additional promise.

3. Driver, problem Hamiltonian, catalysts, and decoder. Use

   $$H_{\rm D} = -(X_1+X_2+X_3), \qquad H_{\rm P} = J(Z_1Z_2+Z_2Z_3+Z_3Z_1), \qquad J>0,$$

   with no catalyst and direct spin decoding.

4. Initial state, preparation, and endpoint convention. Prepare $|+++\rangle$ for the nominal ramp and read in the computational basis at $s=1$. The output channel below replaces the endpoint state, so this preparation does not derive the sample law.

5. Schedule, controls, pauses, and runtime. Declare $H(s)=(1-s)H_{\rm D}+sH_{\rm P}$ with $s=t/T$ and symbolic $T$, with no pause, reversal, or quench. No physical runtime claim follows from symbolic $T$.

6. Dynamics, environment, temperature, rates, and equilibrium or freeze-out assumptions. The only licensed output model is the ideal replacement channel

   $$\mathcal E_T(\rho) = \rho_\beta(H_{\rm P})\operatorname{Tr}\rho$$

   at $\beta J=\ln3/4$. It is a mathematical Gibbs assumption, not a derived bath, thermalization, rate, or freeze-out mechanism.

7. Measurement, sample distribution, success event, and quality metric. The six frustrated strings have energy $-J$ and the two aligned strings have energy $3J$. With

   $$Z = 6e^{\beta J} + 2e^{-3\beta J},$$

   each accepted string has probability $3/20$ and each aligned string has probability $1/20$. Therefore

   $$p_{\rm acc} = \frac9{10}, \qquad \langle H_{\rm P}\rangle = -\frac{3J}{5}.$$

8. Embedding, gauges, precision, resources, repeats, and comparator boundary. The abstract record has three logical spins, three $ZZ$ couplings, three driver terms, one nominal ramp, one ideal replacement-channel call, and one read. Embedding, gauges, and coefficient imprecision are N/A in the exact toy model. Two independent abstract calls give success confidence $0.99$ exactly. Physical thermalization cost, physical time-to-solution, and a comparator are N/A because the replacement channel has no licensed implementation.

9. Verification data, uncertainty, and reproducibility. Enumerate all eight strings, verify the $6+2$ degeneracies, and use $e^{-4\beta J}=1/3$. Then

   $$6\left(\frac3{20}\right) + 2\left(\frac1{20}\right) =1,$$

   and direct summation reproduces the accepted probability and mean energy. The exact toy distribution has no empirical uncertainty.

10. Conclusion, stopping point, and canonical handoff. The equilibrium combinatorics are correct conditional on the replacement channel. The record does not claim that the nominal anneal prepares Gibbs, identify a bath or freeze point, supply a physical runtime, sample ground states uniformly on hardware, or establish optimization advantage.

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