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A Pareto distribution is a proper probability distribution for every $\alpha ,\beta >0$, even when its mean is infinite. To see both facts, integrate the probability density function against $x^q$. For real $q\ge 0$,The improper integral converges exactly when $q-\alpha -1<-1$. At $q=\alpha$ it diverges logarithmically; above that it diverges as a positive power of the upper endpoint. Setting $q=0$ gives one, proving normalization regardless of whether higher moments exist.

Taking $q=1,2$ and using $\mathrm{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X]{)}^2$ when both terms are finite givesFor $0<\alpha \le 1$, the nonnegative random variable has expected value $+\infty$; a centered variance about a finite mean is unavailable. For $1<\alpha \le 2$, the mean is finite and the variance is infinite.^[ref] A finite total probability and a finite expected value answer different questions.

At the chosen minimum $\beta =10$, $\alpha =3$ gives $\mathbb{E}[X]=15$, $\mathbb{E}[X^2]=300$ and variance $300-15^2=75$, but an infinite third moment. At $\alpha =3/2$, the mean is 30 and the variance is infinite; at $\alpha =3/4$, even the mean is infinite, while the median remains the finite number $102^{4/3}\simeq 25.198$. At the boundary $\alpha =2$, the earlier example has mean 20 but infinite variance. A finite sample contains finite values almost surely; that fact cannot establish that the underlying mean or variance is finite.

Conditioning a Pareto distribution on exceeding a threshold changes its minimum while preserving its shape parameter. For $u\ge \beta$ and $x\ge u$, the definition of conditional probability givesEquivalently, for $r\ge 1$,The ratio to the threshold has the same Pareto distribution at every admissible threshold.^[ref] For the chosen $\alpha =2$, $\beta =10$, only $1/4$ of outcomes exceed 20, but among those outcomes the probability of exceeding 40 is again $1/4$. When $\alpha >1$, the same identity gives conditional expectation $\mathbb{E}[X\mid X\ge u]=\alpha u/(\alpha -1)$.

This multiplicative scaling differs from the additive memoryless property of an exponential distribution. Indeed $Y=\log (X/\beta )$ has exponential distribution of rate $\alpha$, because $\mathbb{P}(Y>y)=e^{-\alpha y}$ for $y\ge 0$. Thus logarithms turn multiplication into addition and explain the scaling property. Conversely, if $U$ has uniform distribution on $(0,1)$, inverse transform sampling gives $X=\beta (1-U{)}^{-1/\alpha }$.

The excess $Z=X-u$ conditional on $X\ge u$ has survival function $(1+z/u{)}^{-\alpha }$ for $z\ge 0$. In the generalized Pareto distribution convention $\mathbb{P}(Z>z)=(1+\xi z/\sigma {)}^{-1/\xi }$, this means $\xi =1/\alpha$ and $\sigma =u/\alpha$. Here the excess starts at zero; the original Pareto distribution starts at $\beta >0$.

Let $X,Y$ be independent fair bits and set $Z=X\oplus Y$, their exclusive or. The four possible triples are $000,011,101,110$, each with probability $1/4$. Every coordinate is fair, and every pair has all four outcomes with equal probability. Thus each pair's mutual information is zero, as in pairwise independence does not imply joint independence.

The full random vector has joint entropy ${\log }_{2}4=2$ bits, while the marginal information entropies sum to three bits. ConsequentlyThe product of the marginals predicts probability $1/8$ for every triple, including the four impossible ones. On each possible triple, the ratio of true to product probability is $(1/4)/(1/8)=2$; averaging its base-two logarithm gives the same one bit of relative entropy.

The example gives a concrete use for joint information measures. A lossless source coding scheme that treats the three coordinates separately needs three bits here. If the decoder knows the integer parity rule, the encoder can send just $X,Y$ and let the decoder reconstruct $Z$ exactly. Pairwise independence does not reveal that saving. This is a stipulated source law and an explicit two-bit code, rather than a promise that arbitrary data can be shortened.

The parity construction of pairwise independent bits turns two independent fair bits into three pairwise independent random variables. With $X=U$, $Y=V$, and $Z=U\oplus V$, their allowed triples are $000,011,101,110$, each of probability $1/4$. Every coordinate has the Bernoulli distribution with parameter $1/2$. Every pair has the same joint probability mass function:All twelve pair cells equal the product of the corresponding probabilities in their marginal distributions. This establishes pairwise independent random variables, a requirement involving every pair of value-events, rather than merely one chosen event per coordinate. For general real-valued random variables $R_i$, the requirement isfor every pair of Borel sets $D,E$. For these binary random variables, checking all four cells of each pair suffices: any value-event is a union of the checked cells. In fact, for binary random variables, independence of their one-value events also gives the other cells by taking complements. Siegrist gives the all-set definition and its connection to indicator random variables in Random Services: independence of random variables.

The full joint probability mass function tells a different story:The integer parity of $X+Y+Z$ is always even. Any two bits determine the third: $Z=X\oplus Y$, $X=Y\oplus Z$, and $Y=X\oplus Z$. Each coordinate is independent of either other coordinate individually, but is dependent on the pair together. For example, the conditional probabilities satisfy $\mathbb{P}(Z=1\mid X=1)=1/2$ and $\mathbb{P}(Z=1\mid X=1,Y=1)=0$. See also pairwise independence does not imply joint independence for the general distinction.

The constraint changes the sum's probability distribution without changing its expected value or variance. Let $S=X+Y+Z$. Only $000$ gives $S=0$; each other allowed triple gives $S=2$. Direct calculation givesThe same variance follows from the variance of a sum: each Bernoulli distribution has variance $1/4$, and pairwise independent random variables have zero cross covariances. Linearity of expectation gives the expected value $3/2$ even without independence.

For comparison, the sum $T$ of three mutually independent fair bits has the binomial distribution $\mathrm{Binomial}(3,1/2)$:Both sums have identical expected values and variances, but $S$ never takes the values one or three. Even their third moments differ: $\mathbb{E}[S^3]=6$, whereas $\mathbb{E}[T^3]=27/4$. Pairwise independence preserves these first two sum summaries while allowing a different joint probability distribution and a different sum law. The calculation is an exact finite example.

For three bits, the Boolean function $p(x)=x_1\oplus x_2\oplus x_3$ outputs one when $x_1+x_2+x_3$ is odd, where $\oplus$ denotes exclusive or. Unlike the three-bit majority decision tree, an exact decision tree computing parity must inspect all three bits on every input. If any coordinate is unread, two completions differing only at that coordinate follow the same observed graph path but have opposite parity. A single leaf cannot correctly label both.

The same argument works for $n$ bits: parity needs all $n$ queries on every input, whatever their order. Thus its decision-tree depth is $n$, its expected query cost is $n$ for every input probability distribution, and each coordinate revealment is one. The algorithm remains required to be correct on the whole Boolean cube, even if the chosen probability distribution assigns some inputs zero probability; a promise restricting permitted inputs would be a different computational problem.

Parity violation means that a physical law fails to be invariant under the spatial reflection represented by the parity operator. A process and its fully inverted counterpart can then have different probabilities. For beta-minus decay with polarized parent atomic nuclei, a nonzero correlation between nuclear spin and emitted electron momentum is such a test. Rotational symmetry alone allows this correlation: a pseudoscalar is unchanged by proper rotations but changes sign under spatial reflection.

The postulates of quantum mechanics assign a pure state to a normalized ray in a complex Hilbert space, and an observable to a self-adjoint operator. A spectral outcome projector $P_a$ has Born rule probability $p_a=⟨\psi |P_a|\psi ⟩$. An ideal projective measurement updates the state to $P_a|\psi ⟩/\sqrt{p_a}$ when $p_a>0$; degeneracy requires the full outcome projector.

Between measurements, an isolated system obeys $i\hslash {\partial }_t|\psi ⟩=H|\psi ⟩$, giving unitary evolution. Composite systems use tensor products; identical particles additionally obey the appropriate exchange symmetry. Mixed preparations use a density matrix, with $p_a=\mathrm{tr}(\rho P_a)$. These postulates require a specified Hamiltonian and measurement model; they do not assign those automatically.^[ref]

The WKB approximation describes a wave whose local wavelength changes slowly compared with one wavelength. For the one-dimensional stationary Schrödinger equation, put $p(x)=\sqrt{2m[E-V(x)]}$. In a classically allowed region,The phase is the classical action divided by $\hslash$. The amplitude varies as $p^{-1/2}$ so that a single travelling component has constant probability current. A useful local validity condition is $|\hslash p^{\prime }/p^2|\ll 1$.

For $V>E$, write $\kappa =\sqrt{2m(V-E)}/\hslash$. The wave instead has growing and decaying components proportional to ${\kappa }^{-1/2}\exp (\pm \int \kappa \mathrm{d}x)$. At a simple turning point, $p$ vanishes and neither local WKB expression is valid. Linearizing the potential gives an Airy-function solution that connects the two sides and supplies the turning-point phase shift. It is not legitimate to carry the divergent WKB amplitude straight through the turning point.^[ref]

A useful application is tunnelling through a smooth opaque barrier. With turning points $x_1,x_2$, the leading transmission probability isThe exponent is twice the forbidden-region action in units of $\hslash$, because transmission is a probability rather than a wave amplitude. For an ideal rectangular barrier of height $V_0>E$ and width $d$, the same leading exponential is $\exp [-2d\sqrt{2m(V_0-E)}/\hslash ]$; matching at the abrupt edges supplies an additional prefactor.

For a bound state between two smooth turning points, the related WKB quantization rule is ${\int }_{x_1}^{x_2}p\mathrm{d}x=(n+1/2)\pi \hslash$. In a harmonic potential the integral equals $\pi E/\omega$, recovering $E_n=(n+1/2)\hslash \omega$. This exact agreement of energies for the harmonic oscillator does not make WKB wavefunctions uniformly exact near its turning points.

Spin and spatial exchange symmetry. Write the normalized spin singlet and triplet states asThe singlet is antisymmetric under exchange and has total spin zero; the triplet is symmetric and has total spin one. For orthogonal orbitals defineIdentical spin one-half particles are fermions, so the complete state must be antisymmetric. Thus $C{\chi }_{00}$ is allowed, and $A{\chi }_{1M}$ is allowed for each triplet component. The spatially symmetric same-orbital states $|aa⟩,|bb⟩$ also require the singlet. A same-orbital triplet vanishes by the Pauli exclusion principle.

These products give a spin-adapted basis. More generally the physical state can be writtenA superposition of the singlet and triplet sectors need not factor into one spatial function times one spin function. Factorization holds for a definite-spin sector, as used in the subsequent model. If the original normalized orbitals are not orthogonal, the corresponding $ab\pm ba$ normalization is $\sqrt{2(1\pm |⟨a|b⟩{|}^2)}$; the remainder of this calculation uses the stipulated orthogonal orbitals.

Integral matrix elements and positivity. Let $a(\mathbf{r})$ and $b(\mathbf{r})$ denote the localized spatial orbitals. With fixed proton positions, the one-electron electronic operator isThe proton-proton repulsion is a separate common nuclear energy. The requested one-electron and direct two-electron elements areSince $H_1=h\otimes I$,Likewise,The integrand is nonnegative and is positive on a set of nonzero measure. This is the repulsion cost of double occupancy; the coincidence singularity is integrable for ordinary smooth localized orbitals.

One electron on each center. Suppress the normalized singlet spin factor and use $|\psi ⟩=C$. For each of the spatial products $ab$ and $ba$, the one-body sum has expectation $2E_0$. The cross terms vanish because a one-body hop leaves an orthogonal spectator orbital. The interaction contribution is the direct integral plus the exchange integral, both neglected in the model. ConsequentlyThis state has no same-center double occupancy, but remains an exchange-symmetric spatial superposition rather than assigning a distinguishable electron permanently to each proton.

Two electrons in the bonding orbital. The normalized one-electron bonding orbital is $g=(a+b)/\sqrt{2}$. The trial spatial state is $|\varphi ⟩=|gg⟩$, whose one-body expectation isExpanding the product gives $|\varphi ⟩=(|aa⟩+|ab⟩+|ba⟩+|bb⟩)/2$. The total probability of double occupancy is therefore $1/2$. Keeping only on-site Coulomb repulsion gives the intended interaction truncation,and the interaction expectation is $U/2$. HenceThis truncation must be stated. In particular, mixed Coulomb elements such as $X=⟨aa|H_{1,2}|ab⟩$ do not vanish merely by orbital orthogonality or inversion symmetry. The printed list does not explicitly name them. If they are retained and are equal and real by site symmetry, their contribution to $E_{\varphi }$ is $2X$; the requested expression uses their neglect along with the other off-site interaction terms. No numerical value of $X$ is supplied or invented.

Energy comparison and the bond. Put $x=|t|/U$ with $U>0$ and $t=-|t|$. ThenThe straight line starts at $1/2$ and crosses zero at $x=1/4$. The purely bonding-orbital trial state is better for $x>1/4$, while the singly occupied covalent trial state is better for $x<1/4$.

For localized orbitals of approximately fixed atomic size, the hopping magnitude $|t|$ decreases rapidly, typically exponentially times a polynomial, as the proton separation grows. The on-site integral $U$ remains of atomic Coulomb scale and tends to the isolated-orbital value; it does not decay as the intersite repulsion $1/R$. If the orbitals themselves are allowed to relax, $U$ also changes, so no universal monotonic law for $U(R)$ follows from the data.

At small $|t|/U$, electron correlation suppresses the ionic configurations $aa$ and $bb$, favoring a singlet with roughly one electron on each center. A single doubly occupied bonding orbital has an artificial finite ionic weight even as the centers separate. At large $|t|/U$, delocalization gains enough kinetic energy that the bonding-orbital description becomes useful and charge fluctuations are substantial.

The crossing of two restricted trial energies is not a sharp transition of the actual two-site ground state. To see this without adding empirical data, include both $C$ and the symmetric ionic state $I_{+}=(|aa⟩+|bb⟩)/\sqrt{2}$. This two-site Hubbard dimer reduces toIts ground state is $\cos \theta C+\sin \theta I_{+}$, with $\tan 2\theta =4|t|/U$ and $0\le \theta \le \pi /4$. The ionic probability isIt changes smoothly from zero to $1/2$. At large separation, the electronic lowering is $-4t^2/U$ to leading order and vanishes as hopping vanishes. An absolute molecular binding curve would additionally require $E_0(R)$, the nuclear repulsion and the neglected intersite terms; this simplified comparison alone does not determine an equilibrium bond length.

Uniform system and dispersion. A plane wave reduces the differential Hamiltonian to a two-by-two matrix acting on its spinor:The Pauli anticommutation relations imply $H(\mathbf{k}{)}^2=|\mathbf{d}{|}^{2}I$. Choosing a spinor in one of its eigenspaces therefore givesAn arbitrary constant spinor multiplying a plane wave is not necessarily an energy eigenstate; it must be an eigenvector of this matrix. For example, for the upper branch and $\epsilon =|\mathbf{d}|$, one can useaway from the coordinate singularity $\epsilon +V=0$, where another normalized eigenvector chart is used.

For the usual $c>0$ convention, the massless branches are $E_{\pm }=\pm \hslash ck$. For nonzero $V$, the two branches have a gap $2|V|$ and are quadratic near $k=0$:Changing the sign of $V$ does not change these energies. It reverses the out-of-plane spin polarization of a given band at the origin. The plots for equal positive and negative mass magnitudes therefore have identical dispersion curves, despite their different spinors.

Spin direction in the massless case. For $k>0$ put $k_x+i{k}_y=k{e}^{i\phi }$. The normalized eigenspinors can be chosen asDirect multiplication gives $⟨{\sigma }_x⟩=s\cos \phi$, $⟨{\sigma }_y⟩=s\sin \phi$ and $⟨{\sigma }_z⟩=0$. Since the Pauli matrices represent physical spin in this question,The spin is parallel to momentum on the positive-energy branch and antiparallel on the negative-energy branch. Thus the branch-independent statement is collinearity; the printed parallel claim needs the positive-branch qualification. At $k=0,V=0$, the Hamiltonian is zero and spin is not fixed by a wavevector direction.

Mass-domain-wall eigenstate. Assume $c>0$ for the stated normalizable wall state and writeThen $F^{\prime }=-VF/(\hslash c)$, ${\sigma }_x{u}_{-}=-u_{-}$, and $(i{\sigma }_y+{\sigma }_z)u_{-}=0$. Applying the full operator givesThe transverse derivative and mass terms cancel pointwise. Hence the specified state is an eigenstate with $E=-\hslash c{k}_x$.

For a step mass $V(y)=V_{+}>0$ above the interface and $V(y)=V_{-}<0$ below it,The Localization length at a step Dirac mass wall sets the two exponential tails. The amplitude is continuous, peaks at the interface and decays on both sides, with generally unequal decay lengths. The probability density decays with lengths ${\xi }_{\pm }/2$. A normalized transverse function is $f(y)=\sqrt{2/({\xi }_{+}+{\xi }_{-})}F(y)$ when multiplying the unit spinor $u_{-}$. The longitudinal plane wave can independently be box-normalized. A step in $V$ allows a step in $F^{\prime }$; the first-order Dirac equation does not impose continuity of that derivative.

Propagation. The interface dispersion is the straight lineEvery component on this branch travels in the negative $x$ direction. A packet has the formThis dispersionless chiral interface wavepacket therefore translates rigidly with speed $c$ in that direction, without dispersion in this exactly linear model. It remains confined in $y$; there is no free transverse group velocity, because the bound-state family is not labelled by a transverse travelling-wave momentum.

The continuity equation gives the current components $j_x=c{\Psi }^{†}{\sigma }_x\Psi =-c|\Psi {|}^2$ and $j_y=c{\Psi }^{†}{\sigma }_y\Psi =0$. This explicitly confirms absence of probability transport away from the interface. For the step-mass case the profile is normalizable because its tails have positive decay constants and $c>0$. A general sign-changing profile also requires $F\in L^2(\mathbb{R})$; nonzero asymptotic masses suffice, whereas signs alone do not guarantee confinement if their integrated magnitudes grow too slowly. If $c<0$, the supplied exponential instead grows at infinity; a normalizable state requires the opposite spinor and envelope sign. Reversing the wall orientation also reverses the propagation chirality.

For the quantum harmonic oscillator, the canonical commutation relation $[x,p]=i\hslash$ gives $[a,a^{†}]=1$ and $H_0=\hslash {\omega }_0(a^{†}a+1/2)$. Inverting the definition of the annihilation operator givesWith normalized energy eigenstates and consistent phases, $a|0⟩=0$ and $a^{†}|0⟩=|1⟩$. Their orthogonality therefore givesMore generally, $⟨n|x|0⟩=x_{01}{\delta }_{n1}$, the oscillator electric-dipole selection rule out of the ground state.

To derive the transition amplitude, insert the expansion in stationary energy eigenstates into the Schrödinger equation with $H=H_0+V_1(t)$. The derivatives of the explicit phases cancel the $H_0$ terms. Projection onto $⟨n|$ leaves the exact coefficient equationsThe initial conditions are $c_0(-\infty )=1$ and $c_{n\ne 0}(-\infty )=0$. To first order in the perturbation, replace the amplitudes on the right by these initial values. Since $E_n-E_0=n\hslash {\omega }_0$, integration gives time-dependent perturbation theory in the formThis neglects depletion and higher-order transitions; it is justified when the total transition probability is small. Here $⟨0|x|0⟩=0$, so there is also no first-order diagonal amplitude change from this particular perturbation.

For the exponential electric field, the electric-dipole interaction is $V_1=-qEx{e}^{t/\tau }$, with $\tau >0$. Only $n=1$ couples directly to the initial state, andThe Born rule givesThe real part $1/\tau >0$ makes the lower-limit contribution vanish. ThusSet $u={\omega }_0\tau$ and $\eta =qE{x}_{01}/(\hslash {\omega }_0)$. The dimensionless form is $p_1/{\eta }^2=u^2/(1+u^2)$, increasing monotonically from zero to one, with half the asymptotic value at $u=1$. A sufficient weak-field condition for the entire curve is $|\eta |\ll 1$.

For ${\omega }_0\tau \ll 1$, the force has acted for too little time to appreciably change the state at the instant of measurement. The short-time amplitude is approximately $iqE{x}_{01}\tau /\hslash$, soFor ${\omega }_0\tau \gg 1$, instead,The nonzero slow-switch limit is a probability in the original $H_0$ basis. It does not contradict the adiabatic theorem. At $t=0$ the instantaneous oscillator is displaced by $d=qE/(m{\omega }_0^2)$:Its ground state is $|0_E⟩=|0⟩+\eta |1⟩+O({\eta }^2)$ in the original basis. Adiabatic following therefore predicts exactly the finite original-basis limit found above.

The exponential switch-on of a driven quantum oscillator also permits an exact check. A linear force preserves a coherent state, with original-basis displacement $\alpha (0)=i\eta u/(1+iu)$. Thus the exact original-basis probability is $|\alpha {|}^2{e}^{-|\alpha {|}^2}$, whose weak-field limit is the expression above. In the instantaneous displaced basis the mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/(1+u^2)$ and does tend to zero for slow switching. The two bases measure different notions of excitation.

For this spin-boson model, take $H_0=H_1+H_2$. Since $[H_0,{\sigma }_z]=0$, the interaction picture observable isThe normalized vectors $|{\varphi }_{\uparrow }⟩$ and $|{\varphi }_{\downarrow }⟩$ belong to different eigenspaces of the Hermitian operator ${\sigma }_z$, and are therefore orthogonal. This remains true even though each vector includes the oscillator degrees of freedom. Hence all cross terms in the expectation value vanish, while the diagonal terms have eigenvalues $+1$ and $-1$:Normalization also gives $|c_{\uparrow \sigma }{|}^2+|c_{\downarrow \sigma }{|}^2=1$, so these are the probabilities of the two spin outcomes.

The orthogonal projection onto the full spin-$\eta$ subspace isHere the identity acts on all oscillator states. The finite-time scattering operator $S(t)$ in the question is the interaction picture time-evolution operator, so $|{\psi }_{\sigma }^I(t)⟩=S(t)|\sigma ,\{n\}⟩$. Projecting its two orthogonal spin components yieldsExpand the right-hand side in the orthonormal basis $|\eta ,\{n^{\prime }\}⟩$. Taking its squared norm, and using the normalization of $|{\varphi }_{\eta }⟩$, givesThus one sums probabilities over unobserved final oscillator states, not their probability amplitudes. Unitarity of $S(t)$ makes the sum over both $\eta$ equal to one.

The Pauli matrix ${\sigma }_x$ flips the spin with unit matrix element. A nonzero oscillator matrix element changes exactly one occupation number, through a creation operator or an annihilation operator:For a spin flip, the spin energy change divided by $\hslash$ is $-s_{\sigma }\Omega$. Thus creation and annihilation in mode $j$ have total frequency changes ${\omega }_j-s_{\sigma }\Omega$ and $-{\omega }_j-s_{\sigma }\Omega$, respectively. Distinct modes or different occupation changes lead to orthogonal final quantum states, so their probabilities add. Fermi's golden rule consequently givesThe Dirac delta function is even, accounting for the sign in the annihilation term. Since all ${\omega }_j,\Omega >0$, an initially up spin can only emit an oscillator quantum, while an initially down spin can only absorb one. Writing the oscillator density of states per unit angular frequency as $g(\omega )={\sum }_j\delta (\omega -{\omega }_j)$, the continuum limit givesThe $+1$ permits emission even from the oscillator vacuum; absorption requires a quantum already present. The notation $n_{\Omega }$ assumes a common or smoothly specified occupation at the resonant frequency. If several resonant modes have different occupations, it denotes their density of states weighted mean, so that $g(\Omega )n_{\Omega }={\sum }_j{n}_j\delta (\Omega -{\omega }_j)$ after continuum smoothing. This is the golden-rule relaxation of a spin coupled to an oscillator bath.

In first-order time-dependent perturbation theory, the resonant time factor before taking the Fermi's golden rule limit isIts central frequency width is of order $1/t$, and its total area is $2\pi t$. To replace it by $2\pi t\delta (\nu )$, the oscillator density of states and relevant occupations must be approximately constant over that width. Thus $t\Delta \omega \gg 1$. At the same time, the perturbative flip probability must remain much smaller than one.

Define ${\Gamma }_0=2\pi {\lambda }^{2}g(\Omega )/{\hslash }^2$, so the two flip rates are ${\Gamma }_0(n_{\Omega }+1)$ and ${\Gamma }_0{n}_{\Omega }$. A sufficient common time window for both initial spins isFor a specified initial spin, the upper condition is simply its own flip rate times $t\ll 1$. Such a window requires weak enough coupling. The continuum approximation also presumes times shorter than recurrences from a resolved finite oscillator spectrum, and long enough to resolve the positive-frequency resonance from its nonresonant counterpart.

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.

A quantum computing apparatus needs a scalable quantum register of controllable qubits, reproducible initialization, a universal quantum gate set, sufficiently long quantum coherence, and selective readout. A qubit is a two-dimensional state space; several qubits occupy a tensor product space and can develop entanglement. Quantum state preparation must establish a known low-entropy input and supply fresh ancillas during quantum error correction. A universal quantum gate set may use arbitrary single-qubit rotations with an entangling two-qubit gate such as controlled-NOT gate. Quantum coherence and control fidelity must suffice for the intended circuit, and measurement must recover chosen output probabilities without uncontrolled disturbance of other qubits. These are the five DiVincenzo criteria.^[ref]

An atomic implementation can encode qubits in two long-lived internal levels, initialize them by optical pumping, drive coherent transitions with laser or microwave pulses, couple selected qubits through a shared interaction, and distinguish states by state-dependent fluorescence. For trapped ions, a shared vibrational mode can mediate entangling gates, so motional control and cooling are part of the computational apparatus. The motion is a resource only while uncontrolled heating does not erase the required quantum coherence.

The computational resource is coherent control of amplitudes and entanglement, with interference making useful outcomes likely. An $n$-qubit pure state has $2^n$ basis amplitudes, but a single measurement does not reveal those amplitudes or all possible answers. Algorithms exploit structure rather than simply reading out parallel classical computations. The Shor algorithm gives factoring with polynomial resource scaling in the input bit length, compared with the best known classical algorithms' superpolynomial scaling; an exponential classical lower bound has not been proved.^[ref] Grover's algorithm uses order $\sqrt{M}$ oracle queries for unstructured search among $M$ possibilities, a quadratic query improvement.^[ref] Quantum simulation is a further motivation because interacting quantum systems can require exponentially many classical amplitudes for a direct generic representation.

The main challenges are quantum decoherence, fluctuating control fields, imperfect gates, leakage out of the qubit subspace, readout errors, crosstalk, and the difficulty of scaling connectivity and calibration. Faster control can introduce extra errors, while stronger isolation can make preparation and readout harder. Quantum error correction encodes logical information redundantly in several physical qubits and extracts error syndromes without measuring the unknown logical state itself. Fault-tolerant quantum computation prevents the correction apparatus and gates from spreading a few faults into uncorrectable logical errors. It requires error rates below an architecture- and noise-dependent threshold and substantial resources; there is no universal numerical threshold for all implementations. Advantages are task-dependent, while reliable scaling requires both coherent hardware and error control.

For the harmonic-oscillator recoil matrix element, using ${\psi }_1=\sqrt{2}(x/l_0){\psi }_0$ and differentiating the Gaussian transform with respect to $k$,Consequently $|M_{01}{|}^2/|M_{00}{|}^2={\eta }^2$. Reversing the travelling-wave phase changes the sign of the imaginary amplitude. For small $\eta$, the sideband is weaker by $\eta$ in amplitude and ${\eta }^2$ in probability.

With the printed unnormalized Gaussians, the literal integral would be $-2igk{l}_0\exp (-k^2{l}_0^2/2)$. It is distinguished from the physical normalized-oscillator result rather than silently treating inconsistent wavefunctions as normalized.