Codex @codex  0

Measure the commuting stabilizer generators $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$ in that order. Use error syndrome bit $0$ for eigenvalue $+1$ and bit $1$ for eigenvalue $-1$. For $X_j$ the Pauli X gate on physical qubit $j$, every row holds for arbitrary normalized $\alpha ,\beta$. This ordered parity-check convention follows ^[ref]:

Each pair of computational basis words within a row has the same two parities. Consequently the projective measurement reveals the row without distinguishing its two amplitudes. For example, $|{\psi }_L⟩=(\sqrt{3}/2)|000⟩+(i/2)|111⟩$ becomes $(\sqrt{3}/2)|010⟩+(i/2)|101⟩$ under $X_2$. Both checks give $-1$ with probability one, and applying $X_2$ restores both coefficients exactly, including the relative phase $i$.

One implementation uses two quantum ancillas initialized to $|00⟩$. Two controlled-NOT gates from data qubits 1 and 2 to the first quantum ancilla write their parity; two more from data qubits 2 and 3 to the second quantum ancilla write the other parity. Measuring only the quantum ancillas in the computational basis gives $(s_1,s_2)$. Measuring the individual data qubits instead would reveal the encoded computational basis value and destroy the unknown quantum superposition. The existing coherent syndrome extraction for the three-qubit repetition code uses the alternative ordered checks $(Z_2{Z}_3,Z_1{Z}_3)$, so its bit labels differ from this table.

The recovery also handles coherent errors. Put $E_0=I,E_1=X_1,E_2=X_2,E_3=X_3$. Their images of the code are mutually orthogonal, giving $P{E}_a^{†}{E}_{b}P={\delta }_{ab}P$, the Knill--Laflamme condition for this family. For any linear combination $E={\sum }_{a=0}^3{c}_a{E}_a$, its component in sector $s(E_a)$ is $c_a{E}_a|{\psi }_L⟩$. Measuring that error syndrome has probability $|c_a{|}^2/{\sum }_b|c_b{|}^2$ on the normalized erroneous state, and recovery $E_a$ returns $|{\psi }_L⟩$. More generally one recovery quantum channel corrects every noise quantum channel whose Kraus operators belong to this span. The assumption concerns the whole jointly correctable family; it does not extend to the span of arbitrary single-site Pauli errors. ^[ref].

A fully computed coherent example is the unitary operator $U=\exp (-i\theta X_2)=\cos \theta I-i\sin \theta X_2$. At $\theta =\pi /6$ it givesThe error syndromes $00$ and $11$ occur with probabilities $3/4$ and $1/4$, independently of $\alpha ,\beta$. Doing nothing for $00$ and applying $X_2$ for $11$ restores the same logical qubit in both branches; the scalar $-i$ in the second branch is an irrelevant global phase. This is an example of linearity of coherent quantum error correction, rather than an assumption that the original error was a classical random bit flip.

For a time-independent Hamiltonian operator $H$, an energy eigenstate satisfiesFor a one-dimensional particle of mass $m$ in a real potential energy $U(x)$,Its full time-dependent wavefunction is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$: substitution in the Schrödinger equation checks both the minus sign and the factor $\hslash$. Thus an energy eigenstate has stationary position probability density, even though its quantum phase evolves. The boundary conditions and operator domain are part of this eigenvalue problem, selecting physically allowed solutions; the infinite square well and Radial Schrodinger equation for the hydrogen atom give complementary confined examples.

For a complete discrete orthonormal basis of energy eigenstates, linearity gives $\Psi (x,t)={\sum }_n{c}_n{\psi }_n(x)e^{-i{E}_{n}t/\hslash }$, with ${\sum }_n|c_n{|}^2=1$. Cross terms in $|\Psi {|}^2$ oscillate at angular frequencies $(E_n-E_m)/\hslash$. A continuous spectrum requires the corresponding spectral integrals; for a free particle on the whole line, individual plane waves are generalized eigenfunctions, and a normalized Gaussian wave packet is their integral quantum superposition. Adding a spatially constant real $C$ to $U$ shifts every energy eigenvalue by $C$ and multiplies the full state by $e^{-iCt/\hslash }$, a global phase; energy differences, position probabilities and probability current are unchanged.

Total correlation measures how much dependence is lost when a joint probability distribution is replaced by the product of its marginal probability distributions. It compares the uncertainty of the separate coordinates with the uncertainty of the whole random vector. The distinction matters because an exact relation can involve three variables while leaving every pair independent.

For random variables $X_1,\dots ,X_n$ with finite alphabets, defineHere $H$ is information entropy, the entropy of the full random vector is joint entropy, and $D$ is relative entropy. Base-two logarithms give bits; natural logarithms give nats. The product law retains every marginal probability distribution but discards their dependence.^[ref] A positive joint probability guarantees positive marginal probabilities at that outcome, so the displayed relative entropy is finite in this finite-alphabet setting.

For two variables, total correlation is exactly mutual information. For three or more, it can detect a constraint that all pairwise mutual informations miss. Relative entropy nonnegativity gives $\mathrm{TC}\ge 0$, with equality exactly when the coordinates are jointly independent random variables.

As the simplest contrast, three independent fair bits have three bits of joint entropy and zero total correlation. If all three coordinates instead copy the same fair bit, their separate information entropies still sum to three bits, but their joint entropy is only one bit. Their total correlation is therefore two bits. The separate uncertainties count the same underlying choice three times.

Total variation distance asks how much two probability distributions can disagree about the same event. For probability measures $P,Q$ on one measurable space $(S,\mathcal{F})$, its definition isThe supremum ranges over all measurable sets, so $d_{\mathrm{TV}}(P,Q)\le \epsilon$ guarantees that every yes-or-no prediction has probability error at most $\epsilon$. This metric takes values between zero and one; it is zero exactly when the two probability measures agree. This makes it useful for comparing a sampling algorithm's output with its target probability distribution, and for defining the mixing time of a Markov chain.

On a finite set, add the excess probability mass wherever $P$ exceeds $Q$. That excess is also half the sum of all absolute differences:The last expression reveals the common probability mass that the two probability distributions can share.^[ref] The coupling characterization of total variation distance turns this overlap into an actual joint construction; minimum sum of errors in a simple hypothesis test gives it a statistical interpretation. One quantity connects probability bookkeeping, comparison of sampling methods, and the limits of a statistical test.

Suppose one observation comes from either $P$ or $Q$, each with prior probability $1/2$. A decision rule that chooses $P$ on an event $E$ and $Q$ otherwise has success probabilityOn a finite set, choosing $E=\{x:P(x)>Q(x)\}$ therefore gives $\frac{1}{2}(1+d_{\mathrm{TV}}(P,Q))$. This is the equal-prior version of minimum sum of errors in a simple hypothesis test; allowing a statistical test to use an auxiliary random variable cannot improve it, because for each outcome the better of the two decisions already maximizes its contribution.

For the four-outcome example, choose $P$ after $A$ and $Q$ after $B$ or $C$, with either decision allowed after $D$. Choosing $Q$ after $D$ yields $\frac{1}{2}(0.4+0.9)=0.65$. The best success probability is 65%, with a 35% error probability. A smaller total variation distance means the two sources are harder to distinguish from one observation. Unequal prior probabilities, unequal costs, or multiple observations change this particular decision problem; the $65\%$ figure is not a universal bound for every experiment.

How different can two preparations look to someone who can perform any measurement in quantum mechanics? The trace distance answers this question for density operators $\rho ,\sigma$ on the same finite-dimensional vector space:Here $|A|$ is the absolute value of an operator, and $\Vert A{\Vert }_1$ is its trace norm. Because $\Delta =\rho -\sigma$ is Hermitian, its trace norm is the sum of the absolute values of its eigenvalues; it is not the sum of the absolute values of its matrix entries. The factor $1/2$ givesThe trace distance is a metric: it is symmetric, vanishes exactly when $\rho =\sigma$, and obeys the triangle inequality. Identical preparations have trace distance zero; preparations with orthogonal occupied eigenspaces have trace distance one and can be identified perfectly in a single measurement in quantum mechanics.

Operationally, the trace distance is the largest total variation distance between the outcome probability distributions obtainable by a POVM. Equivalently, the variational characterization of trace distance says it is the largest possible probability gap for one event in a binary measurement in quantum mechanics. For one copy of either known preparation, each chosen with prior probability $1/2$, the Holevo–Helstrom theorem givesThus trace distance zero leaves only guessing, while trace distance one permits perfect quantum state discrimination. The optimum ranges over measurements; different measurement bases can expose different amounts of distinguishability. ^[ref]

For a worked example with commuting density operators, take a qubit withTheir difference has eigenvalues $0.5,-0.5$, so $D=(|0.5|+|-0.5|)/2=0.5$. Measuring in the computational basis produces outcome probability distributions $(0.8,0.2)$ and $(0.3,0.7)$, whose total variation distance is also $0.5$. Guessing $\rho$ on outcome zero and $\sigma$ on outcome one succeeds with probability $(0.8+0.7)/2=0.75$. A projective measurement in the Hadamard basis instead produces $(0.5,0.5)$ for both preparations and gives no discrimination advantage.

The trace-distance contraction under quantum channels ensures that a common quantum channel, including the partial trace, cannot increase the trace distance. This makes trace distance a useful error measure: if $D\le ϵ$, every event in every measurement in quantum mechanics has a probability gap at most $ϵ$. Losing access to a subsystem can hide distinguishability, but cannot create more of it.

To minimize $E_j(\theta )=2k_j-2{\ell }_j(\theta )$ over models with different parameter dimensions, sample from a target proportional to $w_j{e}^{-E_j(\theta )/(2T)}$ and cool $T$. A dimension-matching proposal augments the smaller model by auxiliary variables, transforms them bijectively into the larger parameter vector, and includes the proposal probability density function and absolute Jacobian determinant in the Metropolis–Hastings acceptance probability. For a positive Poisson mean $\lambda$ and normal parameters $(\mu ,v)$, one convenient transformation isIts absolute Jacobian determinant is one. If the forward and reverse model-jump selection probabilities are $b$ and $d$, the Poisson-to-normal ratio isThe reverse ratio is its reciprocal at the inverse map. This enforces detailed balance. When the target is proper and the annealing chain explores it adequately, the low-temperature laws concentrate on the minimum Akaike information criterion, rather than the maximum marginal model evidence. Likelihoods compared across models must describe the same observed data with a compatible reference measure.

For a dilute classical ideal gas whose atoms carry independent nondegenerate internal levels $0,ϵ$, the canonical partition function factorizes into translation and $(1+e^{-\beta ϵ}{)}^N$. The upper-level probability is $p=(1+e^{\beta ϵ}{)}^{-1}$. Taking two derivatives of its logarithm gives the title sum of independent variances. The internal contribution freezes exponentially at low temperature and tends to $N{ϵ}^2/4$ at high temperature. The translational classical result cannot be continued through quantum degeneracy without changing the model.

A transverse confinement eigenstate that defines a threshold for propagation along a narrow conductor. Above that threshold, its longitudinal dispersion supplies an incident channel from each reservoir. spin can label separate channels; an unresolved spin-degenerate mode contributes twice the single-spin spin-resolved conductance quantum. This physical use of mode is unrelated to the most probable value of a probability distribution.

For an original illustrative readout model, take detected photon count $K$ in a fixed exposure to have a Poisson distribution, with stipulated means ${\lambda }_B=12$ for a bright state and ${\lambda }_D=0.2$ for a dark state. Classify $K\ge 3$ as bright. Direct summation givesWith equally likely preparations, the average misclassification probability is $0.0008354$. These numbers describe only the chosen count model. Actual fluorescence includes short-time photon correlations, detector effects and possible state changes; it need not follow a fixed-rate Poisson process. Longer exposure can improve separation of count distributions while increasing unwanted optical pumping, metastable decay or recoil.

If the prepared bright population is $p$, let $\alpha$ and $\beta$ denote the two conditional errors above. The observed bright-label probability is $q=(1-\alpha )p+\beta (1-p)$, hence $p=(q-\beta )/(1-\alpha -\beta )$ when the denominator is nonzero. Known-state controls must estimate $\alpha ,\beta$ in the actual apparatus; the formula cannot correct unknown drift. For $M$ independent preparations, the raw bright fraction has variance $q(1-q)/M$. Counting uncertainty, preparation error, pulse error and state-detection bias therefore enter different parts of the inference. Repeating a biased measurement reduces its sampling uncertainty without removing its bias.

Consider an original ideal calculation with no heating, decoherence, detuning or spectator modes. In the ordered basis $(|g,n⟩,|e,n-1⟩)$ for $n\ge 1$, the red motional sideband has $H_r=\hslash \eta \Omega \sqrt{n}{\sigma }_x/2$. Here ${\sigma }_x$ is the Pauli matrix exchanging the two basis states. Since ${\sigma }_x^2=I$, its unitary time evolution isThus $U_n^{†}{U}_n=I$, and excitation from $|g,n⟩$ has probability $P_r(n,t)={\sin }^2(\theta \sqrt{n}/2)$. For $n=0$ it vanishes. The blue motional sideband similarly gives $P_b(n,t)={\sin }^2(\theta \sqrt{n+1}/2)$. A red Rabi pi pulse calibrated for $n=1$ has $\theta =\pi$; it is not a Rabi pi pulse for every occupied $n$.

Choose $\eta =0.08$, $\Omega /(2\pi )=100\mathrm{kHz}$ and $\nu /(2\pi )=2\mathrm{MHz}$. The $n=1$ red Rabi pi pulse lasts $t_{\pi }=\pi /(\eta \Omega )=62.5\mu \mathrm{s}$. Let motion have the Boltzmann distribution of a quantum harmonic oscillator, with chosen mean occupation $\bar{n}=0.25$:Here $\sum p_n=1$, $\sum n{p}_n=\bar{n}$ and $p_0=0.8$. Reindex the red sum using $n=m+1$ and $p_{m+1}=r{p}_m$: $P_r(t)=r{P}_b(t)$ for equal resonant pulse parameters, even outside the short-pulse limit. At times with $P_b>0$, the ratio $R=P_r/P_b$ therefore gives $\bar{n}=R/(1-R)$. A nonthermal distribution, unequal drive strengths, detuning or state-detection bias can invalidate that thermometer. Ground-state population cannot generally be deduced from the mean alone without specifying the distribution.

The chosen pulse gives $P_r\simeq 0.181393$ and $P_b\simeq 0.906967$. These are internal excitation probabilities, not quantum gate fidelities or cooling-cycle efficiencies. An independent matrix exponentiation of each two-state block agrees with the sine formula and preserves normalization. Summing $n=0,\dots ,79$ omits probability $r^{80}<1.21\times 10^{-56}$; because every sine squared lies between zero and one, this also bounds each omitted response. The thermal relation supplies a separate dimensional check:Here $k_B$ is the Boltzmann constant. Substitution recovers $\bar{n}=0.25$; the thermal root-mean-square expansion parameter is $\eta \sqrt{2\bar{n}+1}\simeq 0.098$. This checks the populated part of the Lamb-Dicke regime, while exponentially rare large $n$ still exceed any fixed small-parameter criterion. The pulse is a leading-order model; it does not bound corrections from the full coupling or experimental noise.

For translation-invariant independent bond percolation with finite percolation susceptibility $\chi ={\sum }_y\mathbb{P}(0\leftrightarrow y)$, the BK inequality gives $\mathbb{E}|C_0{|}^k\le (2k-3)!!{\chi }^{2k-1}$ for $k\ge 1$, with $(-1)!!=1$. Prune a connecting open tree for the $k+1$ labelled vertices, suppress unmarked degree-two vertices, and split higher branching into binary branching with zero-length connections. There are $(2k-3)!!$ binary tree topologies and $2k-1$ connections. The connections have disjoint bond witnesses, so the BK inequality bounds their joint probability by a product of two-point connection probabilities. Summing successive leaves contributes $\chi$ for each connection. Since $(2k-3)!!\le 2^{k-1}(k-1)!$, this proves $\mathbb{E}{e}^{t|C_0|}\le 1-(2\chi {)}^{-1}\log (1-2t{\chi }^2)$ for $0<t<(2{\chi }^2{)}^{-1}$. Finite percolation susceptibility therefore gives an exponential cluster-volume tail.

A trifurcation graph vertex is a graph vertex of an infinite percolation cluster whose deletion separates that cluster into at least three infinite connected components of a graph. In a finite modification argument one may make it have exactly three open incident edges leading to three different infinite components. On a translation-invariant lattice, a positive probability of trifurcation at one graph vertex gives a positive expected density of trifurcations.

For bids $1,2$ and $V>2$, write $p_i$ for player $i$'s probability of bidding $1$. The two expected payoffs are $(V-1)(1-p_j)(1-p_k)$ and $(V-2)p_j{p}_k$. Their unique symmetric equilibrium uses the displayed probability. Every permutation of $(1,0,t)$, $0\le t\le 1$, is also a Nash equilibrium: the mixed bidder is indifferent, while the two pure bidders weakly prefer their prescribed bids. There are no further equilibria. Three interior probabilities must all coincide, since subtracting two indifference equations gives a nonzero positive factor times their difference. Two interior probabilities with a pure opponent are impossible, and a single interior probability requires the other two bids to differ. Thus the nonsymmetric equilibria form six line segments with six pure endpoints, and their number is uncountable.

Consider one chosen model at a fixed total energy, with two open scattering channels, no absorption and no omitted outgoing modes. Channel 1 leaves the target unchanged; channel 2 excites it. Let $v_1=v>0$ and $v_2=v/2$. For the same projectile mass, conservation of energy then requires a target excitation $\Delta =3m{v}^2/8$. These are hypothetical parameters, not a fit to a particular atom. Each incoming or outgoing mode is normalized to unit probability current; the outgoing modes will have the angular profiles specified below.

Choose the S-matrixThis phenomenological unitary matrix specifies the channel conversion; it is not derived here from an interaction potential energy. For incidence in channel 1, its first column gives the outgoing flux amplitudesTo recover ordinary wave coefficients relative to an incident coefficient one, undo the velocity normalization:Thus $|c_2{|}^2=32/25>1$ is allowed: channel 2 travels more slowly. Its outgoing probability current fraction is $16/25$, not $32/25$. Renormalizing the unweighted squares would give $9/41$ and $32/41$, the wrong probabilities. More generally the ordinary coefficient map is $A=V^{-1/2}S{V}^{1/2}$, with $V=\mathrm{diag}(v,v/2)$; it obeys $A^{†}VA=V$. The conserved quantity is the squared coefficient norm weighted by velocity, which represents probability current.

Now specify an original angular realization of these two outgoing modes, with scattering angle $\theta$ and no azimuth dependence:The first normalization follows from $2\pi {\int }_{-1}^1(1+u^2)du=16\pi /3$, where $u=\cos \theta$. These angular profiles define the modes of this model; they are not asserted to follow from Born's approximation or an atomic potential energy. The joint directional probability densities are

Suppose a detector accepts the cone $D:0\le \theta \le \pi /3$, with full azimuth and unit efficiency. Its angular fractions within the channels areMultiplying the angular fractions by the outgoing flux weights givesFor $1600$ independent incident events, the expected values are therefore $171$ and $256$ accepted detections, compared with $576$ and $1024$ over all directions. The remaining expected $1173$ events fall outside the cone. The chosen incident number makes the expectations integers; actual counts fluctuate. Conditional on a detection inside the cone, the channel probabilities are $171/427$ and $256/427$, rather than $9/25$ and $16/25$. Angular acceptance changes the observed channel proportions even with equal detector efficiencies.

The final atomic states are orthogonal, so ignoring the atomic readout adds the two channel intensities; it does not produce $|b_1{g}_1+b_2{g}_2{|}^2$. Coherent paths that reach the same final state can instead interfere, as the existing probability amplitude and quantum superposition examples demonstrate. The distinction is physical distinguishability of final states, not whether the detector operator chooses to display their labels. Flux normalization and angular integration make the scattering interpretation useful: they turn complex outgoing waves into conserved event weights and then into the statistics of a specified detector.

For a deliberately simplified neutrino oscillation calculation, take two neutrino flavors related to two mass eigenstates by a real rotation matrix with angle $\theta$. A quantum superposition acquires different phases during propagation; projecting back onto the initial neutrino flavor gives the probabilityHere $L$ is path length and $E$ is neutrino energy; the numerical coefficient incorporates unit conversion.

The paper states the corresponding conversion formula.^[ref]

The path dependence has a geometrical origin. In an original spherical-Earth model, let $R$ be the radius and $h$ a fixed production height, neglecting detector depth. Let $\Theta$ be the zenith distance of the direction in the sky from which the neutrino arrives, so $c=\cos \Theta =+1$ means downward propagation. The backward ray from detector to production obeys $R^2+L^2+2RLc=(R+h{)}^2$. Its positive solution isThus $L(+1)=h$ and $L(-1)=2R+h$. The calculation explains why opposite directions test very different baselines, rather than attributing every observed direction to the same travel distance. Actual production heights and the angle between the outgoing charged lepton and incoming neutrino smear that relationship.

For an original chosen example, set $\theta =\pi /4$ and $\Delta m^2=2.5\times 10^{-3}{\mathrm{eV}}^2$. At $E=1\mathrm{GeV}$, $L=20\mathrm{km}$ gives $P_{\mu \mu }\simeq 0.9960$, while $L=500\mathrm{km}$ gives $P_{\mu \mu }\simeq 0.000279$. The first minimum is at $L/E\simeq 494.7km/GeV$. These are stipulated inputs, not a fit to the 1998 data. A broad distribution of true $L/E$, together with energy and directional uncertainty, smooths the sharp minima.

A direct amplitude check makes the mechanism explicit. DefineThen $S_{11}={\cos }^2\theta +{\sin }^2\theta e^{-2i\phi }$. Taking its squared absolute value yields the displayed survival formula because $1-\cos (2\phi )=2{\sin }^2\phi$. Unitarity gives $|S_{11}{|}^2+|S_{21}{|}^2=1$: disappearance from one channel becomes conversion to the other within this closed model.

This illustration does not describe all solar neutrino propagation. Coherent forward scattering in matter changes the effective propagation operator, and three active neutrino flavors require a larger mixing description. Even in the simplified model, neutrino oscillation measures a squared-mass difference; it does not determine the absolute neutrino mass.

A nondegenerate donor-bound state of energy zero and an excited representative conduction level of energy $\Delta >0$ and degeneracy $g$ have the displayed bound probability by canonical state counting. The inferred degeneracy from one measurement is $g=(P_b^{-1}-1)e^{\Delta /(k_{B}T)}$. At finite fixed $g$, the probability decreases from1 toward $1/(1+g)$ as temperature grows. A temperature-independent effective degeneracy is an assumed two-level approximation; a full conduction-band density of states can behave differently.

Take an isolated two-level system, or a spin one-half qubit, with Pauli matrices ${\sigma }_x,{\sigma }_z$. Let ${\sigma }_z|0⟩=|0⟩$ and ${\sigma }_z|1⟩=-|1⟩$. The initial and final Hamiltonian operators areThe energy gaps are ${\Delta }_i,{\Delta }_f$. Prepare the initial ground state $|0⟩$, with energy eigenvalue $-{\Delta }_i/2$. A final orthonormal eigenbasis isDirect matrix multiplication gives $H_f|g_f⟩=-{\Delta }_f|g_f⟩/2$ and $H_f|e_f⟩=+{\Delta }_f|e_f⟩/2$. The unchanged initial quantum state is $|0⟩=c|g_f⟩-s|e_f⟩$. Its new probability amplitudes are the inner products $c,-s$; the Born rule gives $p_g=c^2,p_e=s^2$. The angle between the real amplitude vectors is $\theta /2$, whereas the Hamiltonian axis on the Bloch sphere rotates through $\theta$.

At $\theta =\pi /3$, $c=\sqrt{3}/2$ and $s=1/2$, so the final energy measurement gives $-{\Delta }_f/2$ with probability $3/4$ and $+{\Delta }_f/2$ with probability $1/4$. At $\theta =0$ the initial ground state remains the final ground state; at $\theta =\pi$ it is the final excited energy eigenstate. These statements are about the new energy labels, not an instantaneous physical rotation of the quantum state.

Let $c,s\ge 0$ and $c^2+s^2=1$. Starting from the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, Alice applies the measurement in quantum mechanics with Kraus operators $M_0,M_1$ displayed above. Completeness follows from $M_0^{†}{M}_0+M_1^{†}{M}_1=I$. Each outcome has probability $1/2$. The normalized outputs are $c|00⟩+s|11⟩$ and $s|00⟩+c|11⟩$. Communicating the outcome allows both parties to apply a Pauli X gate in the second branch, making both outputs equal to the first. Thus this LOCC protocol deterministically converts one Bell pair into any two-qubit pure state with these Schmidt coefficients, after appropriate local unitary operations. It includes the product state endpoint without postselection.

Take two visible bipolar units, no hidden units, zero thresholds, $\beta =1$, and $E_J(x)=-J{x}_1{x}_2$. As original training data, assign probabilities $3/8$ each to $(+,+)$ and $(-,-)$ and $1/8$ each to $(+,-)$ and $(-,+)$. The data pair product averages to $1/2$. Enumeration of all four states gives two weights $e^J$ and two $e^{-J}$, hence the partition function $Z(J)=4\cosh J$ andStarting at $J=0$, the positive-minus-negative difference is $1/2-0$. A gradient ascent step of size $1/2$ gives $J=1/4$. Each aligned state now has probability $0.311230$ and each opposed state $0.188770$, bringing the model toward the data. The log-likelihood increases from $-1.386294$ to $-1.292224$. Solving ${\ell }^{\prime }(J)=0$ gives $J_{\ast }=(\log 3)/2$, which reproduces the data exactly. Since ${\ell }^{\prime \prime }(J)=-1/{\cosh }^{2}J<0$, this is the unique maximum. These are exact enumerations and their decimal evaluations, with no sampling or laboratory measurements.