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Put $c=\beta /2-\alpha$. The change of measure and part (a)(iii) yieldFor a centered normal random variable $G$ of variance $v>0$, a Gaussian integral gives the Gaussian quadratic exponential momentHere $c\le \beta /2$ and $\beta v_t<1/2$ for finite positive $t$, so the required positivity holds, including when $c>0$. ThusExpanding the two exponentials in terms of hyperbolic functions givesAt $t=0$ this is one. As $\beta \downarrow 0$, the denominator tends to $1+2\alpha t$, recovering part (a)(i). At $\alpha =0$ it gives the Laplace transform of the integrated square of Brownian motion. The joint terminal-and-integral transform is the Joint Laplace transform of Brownian square and its time integral.

At the empty Markov chain state, only the constant-rate production chemical reaction is possible. Thus the first-passage time ${\tau }_1$ has an exponential distribution of rate $2$:Its expected value can be obtained by integrating this survival function, and its second moment by integrating $2t$ times the survival function:Consequently the requested expected value and variance are

We use the equilibrium generalized Langevin equation and its generalized fluctuation-dissipation relation, with the factor $5$ in the noise autocorrelation fixing the equilibrium velocity variance. Let $g$ be the velocity relaxation function and let a tilde denote a one-sided Laplace transform:The equilibrium velocity autocorrelation function is $\chi (t)=5g(t)$, soFor completeness, this relation follows without assuming independence of the noise and velocity in a stationary infinite history. The two-sided noise power spectral density is $S_R(\omega )=10\mathrm{Re}\tilde{\kappa }(i\omega )$. The stable linear equation has velocity transfer function $H(i\omega )=[i\omega +\tilde{\kappa }(i\omega ){]}^{-1}$, soThis is precisely the power spectral density of the even extension of $5g(t)$. Stable relaxation removes initial transients with a finite second moment, giving the long-time autocorrelation function defined in the question. All the parameter choices used below have stable relaxation.

Integrating the velocity autocorrelation function twice gives the mean squared displacement from a fixed initial position:A decaying initial transient does not change its asymptotic slope. Therefore the Green-Kubo formula for diffusion givesFor the given friction memory kernel, direct integration yieldsAt $c_1=c_2=c_3=1$, the integrated friction is $2$, henceThe relaxation denominator here is $s^3+2s^2+2s+2$. Its roots have negative real parts by the cubic Routh-Hurwitz criterion, since all coefficients are positive and $2\cdot 2>2$. This justifies the decay and integrability used above. The diffusion coefficient depends on the area under the friction memory kernel, not just its value at zero.

For $c_1=2$, $c_2=4$, $c_3=5$, the friction memory kernel and its Laplace transform areUsing the generalized fluctuation-dissipation relation with equilibrium velocity variance $5$, the velocity autocorrelation function has Laplace transformAll three poles are negative, so the stationary relaxation is stable. A partial fraction decomposition givesTaking the inverse Laplace transform givesThe negative middle coefficient is compatible with an autocorrelation function; individual terms in a partial fraction decomposition need not themselves be separate physical modes with positive weights. Several useful checks follow directly:The Volterra integral equation for $\chi$ predicts exactly ${\chi }^{\prime }(0)=0$ and ${\chi }^{\prime \prime }(0)=-\kappa (0)\chi (0)=-10$. Its integral is also $8-10/9+1/18=125/18=5/\tilde{\kappa }(0)$, agreeing with the Green-Kubo formula for diffusion.

A bosonic coherent state obeys $a|\alpha ⟩=\alpha |\alpha ⟩$ and has expansion $e^{-|\alpha {|}^2/2}{\sum }_n{\alpha }^n|n⟩/\sqrt{n!}$. Its occupation is Poisson distributed, with mean and variance $|\alpha {|}^2$. Its nonzero field amplitude makes phase coherence natural, although physically observable phase comparisons require a reference.

For paired fermions, the corresponding Cooper-pair pseudospin coherent state isIt coherently superposes empty and occupied pair states and has an anomalous average $⟨c_{-\mathbf{k}\downarrow }{c}_{\mathbf{k}\uparrow }⟩\ne 0$. It is not an ordinary complex-eigenvalue coherent state of a single fermion annihilation operator. The broken-symmetry state has indefinite particle number; number projection can restore a fixed-number description without discarding the underlying pairing correlations.

Hartree-Fock-Bogoliubov theory approximates a two-body interaction using both the normal density ${\rho }_{ij}=⟨c_j^{†}{c}_i⟩$ and anomalous pairing tensor ${\kappa }_{ij}=⟨c_j{c}_i⟩$. Direct and exchange contractions determine a normal one-particle field, while anomalous contractions determine the pair potential. The resulting quadratic Hamiltonian is diagonalized by a fermionic Bogoliubov transformation. In a homogeneous singlet case its Nambu block isThe fields are determined self-consistently, rather than selected independently of the quasiparticle state. For example,together with the particle-number equation. The pairing field extends ordinary Hartree-Fock theory to states with pair correlations. This mean-field treatment omits fluctuations beyond its self-consistent quadratic description.

For bosons, write the field as a coherent condensate plus fluctuations, $\hat{\psi }=\varphi +\delta \hat{\psi }$. The same normal/anomalous strategy retains $\tilde{n}=⟨\delta {\hat{\psi }}^{†}\delta \hat{\psi }⟩$ and $\tilde{m}=⟨\delta \hat{\psi }\delta \hat{\psi }⟩$, with a bosonic, commutation-preserving Bogoliubov transformation. In the leading weakly depleted contact-gas limit, the condensate density $n_0$ and effective coupling $g>0$ give $E_k=\sqrt{{ϵ}_k({ϵ}_k+2g{n}_0)}$, with ${ϵ}_k={\hslash }^2{k}^2/(2m)$. This Bogoliubov theory connects low-momentum sound to high-momentum particle excitations. More elaborate self-consistent truncations must respect the symmetry constraints needed for a gapless condensate mode.

Two assumptions of the usual normal linear model used for the reported coefficient Student t-tests are:

The second statement includes homoscedasticity. Normality concerns the conditional errors, rather than requiring the weights themselves to have a normal distribution.

Use Poisson regression for the offspring count, with testosterone as a predictor. It models a nonnegative integer response through its conditional mean. If the conditional variance substantially exceeds that mean, a negative binomial regression is a possible alternative for overdispersion.

Each observation has normal distribution probability density functionThe observations are independent and identically distributed, so their joint probability density function, viewed as a function of the unknown mean, is the product:Independence justifies multiplication; the common mean and known unit variance justify using this same density factor for every observation.

The exponential distribution prior has rate $\lambda =1/3$ and support $\mu \ge 0$, so its prior density is $(1/3)e^{-\mu /3}{1}_{\{\mu \ge 0\}}$. By Bayes' theorem,where $C$ is a normalizing constant independent of $\mu$.

Completing the square gives a useful exact reference for the numerical sampling. PutThen this normal mean posterior with an exponential prior is a normal distribution with location $m$ and variance $s^2$, truncated below at zero. With $\varphi$ the standard normal distribution density and $\Phi$ its cumulative distribution function,The exponential rate is $1/3$, not three; the prior mean is three. The support restriction must be retained in both the formula and the sampler.

Square the operator solutions, retaining the symmetrized cross term rather than treating $x^{\prime }(0)$ and $p^{\prime }(0)$ as commuting variables. The initial symmetrized expectation vanishes, soThe means remain zero. These are also the variances. Unless $\alpha =m\omega /\hslash$, the Gaussian state breathes: its position and momentum widths exchange periodically while the mean energy remains constant.

Introduce the dimensionless width $s=\hslash \alpha /(m\omega )>0$. Multiplying the variances givesThe minimum holds at all times precisely when $\alpha =m\omega /\hslash$, the oscillator ground-state width. For any other $\alpha$, it is still attained at the isolated quarter-period times $t=n\pi /(2\omega )$, including zero. At a fixed generic time, minimizing over $\alpha$ again selects $s=1$.

The excess product has a covariance interpretation:Thus the breathing Gaussian state in a harmonic oscillator saturates the Robertson-Schrödinger uncertainty relation throughout, even when its position-momentum product exceeds the lower bound in the Heisenberg uncertainty principle.

First trace the state explicitly specified in (c). orthogonality in the $y$ oscillator removes terms with different occupation numbers:For a thermal oscillator of frequency $\omega$, the ratio of successive Boltzmann weights is $e^{-\hslash \omega /(k_{B}T)}$. Consequently the conditional answer for that two-mode squeezed state isAt $\theta =0$ this is interpreted by the limit $T_{\mathrm{sq}}=0$. The reduced state is mixed for nonzero squeezing although the joint state is pure; its effective thermal statistics arise from entanglement, not from heating by a bath.

The ground-state identification must still be qualified. For the true coordinate-coupled ground state, direct normal-mode averages giveTheir ratio is $m^2{\omega }_{+}{\omega }_{-}$, rather than $m^2{\omega }^2$. Therefore this reduced state is not thermal with respect to the original bare frequency $\omega$ for $k\ne 0$. It does have an exact thermal representation for an effective oscillator of mass $m$ and frequencyThe corresponding occupation basis is that of $\nu$, not that of the original $a$. This follows by matching the Gaussian variances to $⟨x^2⟩=\hslash \coth [\hslash \nu /(2k_{B}T)]/(2m\nu )$ and its momentum counterpart. A direct reduced-Gaussian eigenfunction calculation gives the same effective frequency and geometric eigenvalues.^[ref]

On periodic ring states the angular-momentum index $m$ is an integer. At the stated resonance,so $U_T$ is the identity on every orbital state. After $N$ steps from the normalized ground state,Its phase varies in space but its position probability density remains uniform. Periodicity permits integration by parts, givingConsequentlyEquivalently, the angular-momentum probabilities are $P_m=J_m(N\eta {)}^2$, with ${\sum }_m{m}^2{J}_m(z{)}^2=z^2/2$. The mean angular momentum remains zero while its variance grows as $N^2$. This quantum resonance of the kicked rotor gives coherent ballistic energy growth. The Bessel second-moment identity connects the momentum spreading to that quadratic energy law; an incoherent random walk would instead give linear energy growth.

A normalized oscillator coherent state is an eigenstate of the annihilation operator, $a|\alpha ⟩=\alpha |\alpha ⟩$. Expanding it in Fock states givesSince $x=\sqrt{\hslash /(2M\omega )}(a+a^{†})$, $⟨a^2⟩={\alpha }^2$, and $⟨a{a}^{†}⟩=|\alpha {|}^2+1$, the variance is $⟨x^2⟩-⟨x{⟩}^2=\hslash /(2M\omega )$. Henceindependently of the complex amplitude $\alpha$.

For $A=\alpha a^{†}$ and $B=-{\alpha }^{\ast }a$, the commutator is the scalar $[A,B]=|\alpha {|}^{2}I$. The Baker--Campbell--Hausdorff formula therefore terminates and gives the displacement operatorBecause $a|0⟩=0$, applying it to the Fock vacuum yields exactly the Fock state expansion above:For a general complex $\alpha$, its complex conjugate is required in the annihilation term. The expression with $-\alpha a$ printed in the paper is valid when $\alpha$ is real.

The quantum state is continuous through a sudden finite-force quantum quench, so it is still $|0⟩$ at $t=0^{+}$. DefineThe new Hamiltonian operator is $H=\hslash \omega (b^{†}b+1/2)-\hslash \omega {\lambda }^2$. The old Fock vacuum obeys $b|0⟩=-\lambda |0⟩$, so the coherent state after shifting an oscillator equilibrium rotates with amplitude $-\lambda e^{-i\theta }$ relative to the new equilibrium. In the original oscillator basis, the full state isOne way to obtain the phase is to writeand apply the displacement operator composition rule. The coherent-state oscillator expectation values giveThe position uncertainty stays $\sqrt{\hslash /(2M\omega )}$. These mean values obey the forced classical equation, while the nonzero variance retains the quantum width.

For the Poisson process, $q_n=g$ and $r_n=0$, soThis is an ordinary differential equation in time for each fixed $z$, with solutionConsequentlyThe initial condition is interpreted as $P_0(0)=1$, with all other initial probabilities zero. Normalization follows from $F(1,t)=1$. The mean and variance are both $gt$, as expected for independent events arriving at constant rate.

Brownian motion arises from many microscopic collisions. A one-dimensional Langevin description isThe fluctuation-dissipation relation ties random forcing to drag, ensuring thermal equilibrium. For a free particle set $\kappa =0$. With initially thermal velocities, its mean-square displacement isIt is ballistic, $(k_{B}T/m)t^2$, at short times and diffusive, $2Dt$, for $t\gg {\tau }_v$. In three dimensions the corresponding long-time result is $6Dt$. At diffusive times, the position propagator from a specified origin is Gaussian with variance $2Dt$ per coordinate. A small sphere in a viscous liquid supplies a physical example; in the slow unbounded-fluid limit its drag is $6\pi {\eta }_{f}a$.

For a harmonic trap, overdamped relaxation has time ${\tau }_{\kappa }=\zeta /\kappa$, equilibrium variance $k_{B}T/\kappa$ and stationary correlationConsequently stationary trapped Brownian displacement obeysConfinement makes this displacement saturate, unlike free diffusion. A trajectory initialized exactly at the centre instead has variance $(k_{B}T/\kappa )(1-e^{-2t/{\tau }_{\kappa }})$; confusing that conditional variance with the stationary displacement gives an erroneous factor of two. The equilibrium position distribution is proportional to $\exp [-\kappa x^2/(2k_{B}T)]$. More general finite confinement has a geometry-dependent equilibrium distribution and bounded displacement, not necessarily a Gaussian one.

In the classical equilibrium circuit, the capacitor stores $C{V}^2/2$. Equipartition assigns this quadratic term mean energy $k_{B}T/2$, soThe mean voltage is zero and its variance is $k_{B}T/C$. Resistance determines the equilibration time and noise spectrum, not this full-band equilibrium variance. A finite-bandwidth or appreciably loading readout can measure less than the ideal result.

Equipartition gives mean energy $k_{B}T/2$ for each independent quadratic term in a classical canonical Hamiltonian. For example, $p^2/(2m)$ and $k{x}^2/2$ each contribute this amount in a harmonic oscillator. It is a thermal average, not an instantaneous equality; constraints, nonquadratic terms or quantum level spacings large compared with $k_{B}T$ require a different calculation.

The fluctuation-dissipation theorem connects equilibrium fluctuations with the dissipative part of linear response. If an applied force couples as $-f(t)x$, use response to $e^{-i\omega t}$ and define the two-sided angular-frequency spectrum $S_{xx}(\omega )=\int e^{i\omega t}⟨\delta x(t)\delta x(0)⟩dt$. The classical relation isFor a viscously damped particle, $m\dot{v}=-\zeta v+\xi (t)$, the corresponding white-noise force satisfies $⟨\xi (t)\xi (t^{\prime })⟩=2\zeta k_{B}T\delta (t-t^{\prime })$. Noise replenishes energy lost by friction; their strengths cannot be varied independently at fixed equilibrium temperature.

Electrically, the one-sided voltage-noise spectrum per hertz of an equilibrium resistor is $4k_{B}TR$. A capacitor in parallel filters it toThe same equilibrium voltage variance follows from equipartition. Increasing R makes the spectrum taller and narrower without changing its integrated area. This thermodynamic connection between dissipative conductance and thermal electrical fluctuations is the content of the classical noise calculation.^[ref]

Write $z=e^{\beta (\mu -ϵ)}$. For one nondegenerate fermionic quantum state, the allowed occupations 0 and 1 give ${\Xi }_F=1+z$. For a bosonic state, summing all nonnegative occupations gives ${\Xi }_B=(1-z{)}^{-1}$, requiring $z<1$ or $\mu <ϵ$. Differentiating the logarithm,The interacting point defect has two distinct spin states, not one fermionic state. Its four microstates have grand-canonical weights $1,z,z,q{z}^2$, with $q=e^{-\beta U}$. The first two single-electron weights are separate up/down spin possibilities; the double state has one electron of each spin. ThusFor $U=0$, $\Xi =(1+z{)}^2$ and $\bar{n}=2z/(1+z)$. The sketch is a single smooth step from 0 to 2, centred on $\mu =ϵ$, where $\bar{n}=1$. As $T\to 0$ its width tends to zero.

For positive U with $\beta U\gg 1$, the lowest grand energies are successively $0$, $ϵ-\mu$ and $2ϵ+U-2\mu$. Away from the two crossings,Finite temperature smooths each step over an energy range of order $k_{B}T$. Near the lower step double occupation is exponentially suppressed, so $\bar{n}\simeq 2z/(1+2z)$. Near the upper step put $y=e^{\beta (\mu -ϵ-U)}$ and neglect the empty state, giving $\bar{n}\simeq 1+y/(2+y)$. Spin degeneracy shifts the half-step positions to $ϵ-k_{B}T\ln 2$ and $ϵ+U+k_{B}T\ln 2$.

For a grand-canonical probability $p_s\propto e^{-\beta (E_s-\mu n_s)}$, differentiation gives ${\partial }_{\mu }{p}_s=\beta (n_s-\bar{n})p_s$. Multiplying by $n_s$ and summing provesFor this defect, direct state counting givesAt $\mu =ϵ$, $z=1$ and thereforeThe same limit gives $\bar{n}=2/3$: empty, up and down states have equal weights, while the double state is suppressed. Replacing this degeneracy by a two-outcome equal-weight model would incorrectly give variance $1/4$ at this chemical potential.

Across the full chemical-potential range the variance has two well-separated narrow peaks and tends to zero on the 0, 1 and 2 plateaux. The limiting shapes are $2z/(1+2z{)}^2$ near the lower step and $2y/(2+y{)}^2$ near the upper one. Each peak has height $1/4$, at the shifted half-step position; the variance at the unshifted crossings is instead $2/9$.