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For integer batch size $k\ge 1$ and rates $\lambda ,\beta \ge 0$, a Batch-birth linear-death process has transitionsWriting $P_n(t)=\mathbb{P}(N_t=n)$ and taking $P_j=0$ for $j<0$, its Master equation is

For independent lattice steps of length $a$ at intervals $\tau$, with probabilities $\alpha +\beta =1$, the diffusive scaling $a^2/\tau \to 2D$ and $(\alpha -\beta )a/\tau \to V$ requires a bias of order $a$. Expanding the master equation gives the advection-diffusion equation $p_t+V{p}_x=D{p}_{xx}$. The finite-step variance coefficient is $2a^2\alpha \beta /\tau$; it has the same limit because the squared bias vanishes. Fixed nonzero bias does not give finite drift under this diffusive scaling.

For transverse coupling with equal mode amplitudes $\lambda$, Fermi's golden rule and the oscillator creation operator and annihilation operator givewhere the first index is the final spin and $g$ is the density of states per unit angular frequency. Decay creates a quantum and carries the factor $n_{\Omega }+1$; excitation annihilates one and carries $n_{\Omega }$. For nonuniform resonant occupations, $n_{\Omega }$ is their density-weighted mean.

With a thermal Bose-Einstein distribution, $n_{\Omega }/(n_{\Omega }+1)=e^{-\beta \hslash \Omega }$. The two-state master equation therefore has the Boltzmann distribution as its unique stationary populations when ${\Gamma }_0>0$. The population deviation relaxes at rate ${\Gamma }_0(2n_{\Omega }+1)$. If there is no resonant coupling, both rates vanish and stationarity does not determine the populations.

For transition rate $w_{ij}$ from state $i$ to state $j$, the Master equation is

The thermal oscillator occupation is the Bose-Einstein distribution. With the rate notation ${\Gamma }_{\eta \sigma }$ for a transition from $\sigma$ to $\eta$, the master equation for the up population isAt thermal equilibrium, assuming ${\Gamma }_0>0$, vanishing time derivative is detailed balance between the two spin states. HenceorTogether with normalization, this gives $p_{\uparrow }=1/(1+e^{\beta \hslash \Omega })$ and $p_{\downarrow }=e^{\beta \hslash \Omega }/(1+e^{\beta \hslash \Omega })$, the Boltzmann distribution for the energy eigenvalues $\pm \hslash \Omega /2$. At zero temperature, absorption vanishes and emission relaxes the spin into its down ground state.

The nonzero-rate condition is necessary: if $\lambda =0$ or $g(\Omega )=0$, both rates vanish, so every initial pair of populations is stationary within this approximation. In that case stationarity alone cannot imply a thermal distribution. The equilibrium conclusion therefore presumes coupling to resonant oscillator modes.

Conservation of total probability gives $A(t)=1-B(t)$. The two-state master equation is thereforeSolving this linear ordinary differential equation givesThe middle panel in d sketches this solution for the explicitly illustrative ratio $k_f/(k_f+k_b)=0.6$. The curve starts at zero with slope $k_f$, is increasing and concave down, and approaches the horizontal limit $B_{\infty }=k_f/(k_f+k_b)$ on time constant $(k_f+k_b{)}^{-1}$. Its initial tangent meets the asymptote at that time constant. The complementary limit is $A_{\infty }=k_b/(k_f+k_b)$, so $B_{\infty }/A_{\infty }=k_f/k_b=e^{-\Delta G_{AB}/(k_{B}T)}$, the Boltzmann distribution ratio. These are occupancy probabilities; $B(t)$ is not the cumulative first-passage probability when reverse transitions remain possible.

Detailed balance means that at thermodynamic equilibrium each elementary transition and its reverse carry equal probability flux. For normalized state probabilities $A_{\mathrm{eq}}$ and $B_{\mathrm{eq}}$,With the printed convention $\Delta G_{AB}=G_A-G_B$, the sign is thereforeThe $G$ values are free energies per system, including internal degeneracies; for molar Gibbs free energies, replace $k_{B}T$ by $RT$. If B has lower free energy, its forward-to-reverse rate ratio exceeds one, as it should.

For constant rates, conservation gives $A=1-B$, so the master equation is $\dot{B}=k_1(1-B)-k_{2}B$. Its steady state is $B_0=k_1/(k_1+k_2)$, and separation of the decaying deviation yieldsfor the specified initial state A. The relaxation time constant is $(k_1+k_2{)}^{-1}$.

For each state, add its incoming reaction rates and subtract its outgoing reaction rates. This gives the master equationSumming gives ${\sum }_{i,j}{\dot{S}}_{ij}=0$, verifying the conservation law for total channel density. Nonnegative initial densities stay nonnegative because every loss term vanishes when its own density reaches zero.

With total density one, the open probabilities for the individual gates are $m=S_{10}+S_{11}$ and $h=S_{01}+S_{11}$. Adding the relevant master equations cancels transitions of the other gate:These marginal ordinary differential equations hold even for correlated initial gates. Their product gives $S_{11}$ when the independence assumption in part (c) holds. Solving the first-order linear differential equations from the all-closed initial state givesPut $\lambda =\alpha +\beta =\gamma +\delta >0$. ThenFor positive opening rates this increases strictly from zero, so its half-equilibrium time is uniquely determined by $(1-e^{-\lambda t}{)}^2=1/2$:If an opening rate is zero, the all-open equilibrium density is zero and a nontrivial half-rise time is undefined.

The pair-birth rate here is the constant immigration rate $\lambda$, while deaths occur at total rate $\beta n$. This interpretation is fixed by the requested mean formula. With $P_n(t)=\mathrm{Pr}(N_t=n)$ and $P_j=0$ for $j<0$, the master equation isThe Markov jump-process generator acts as $Lf(n)=\lambda [f(n+2)-f(n)]+\beta n[f(n-1)-f(n)]$. Taking $f(n)=n$ gives $m^{\prime }=2\lambda -\beta m$, henceFor the second moment $s=\mathbb{E}{N}_t^2$, the two jumps give $s^{\prime }=4\lambda m+4\lambda -2\beta s+\beta m$. Differentiating $V=s-m^2$ therefore yields $V^{\prime }=4\lambda +\beta m-2\beta V$. Substitution of the mean and solution of this scalar linear equation giveThus $V(t)\to 3\lambda /\beta$, assuming $\beta >0$ and finite initial second moment. This is pair immigration with linear deaths; births at rate $\lambda n$ would be a different model and would not give the printed mean.

Let $p_j^n$ be the probability of occupying site $ja$ after $n$ steps, with $p_j^0=1_{\{j=0\}}$ and $\alpha +\beta =1$. The master equation isWrite $p_j^n\approx ap(ja,n\tau )$ so that the limiting $p$ is a probability density. A Taylor expansion of this master equation givesThe weakly biased diffusion limit of a weakly biased random walk keeps $a^2/\tau$ and $(\alpha -\beta )a/\tau$ finite as $a,\tau \to 0$, so $\alpha -\beta =O(a)$. The limiting advection-diffusion equation isIts initial condition is the Dirac delta function at the origin. At finite spacing, the exact variance gained per step is $a^2[1-(\alpha -\beta {)}^2]=4a^2\alpha \beta$. Thus a finite-step central limit theorem gives the variance-matched coefficient $2a^2\alpha \beta /\tau$; its difference from the displayed $D$ vanishes in the stated weak-bias diffusion limit of a weakly biased random walk. Keeping a fixed nonzero bias while taking diffusive space-time scaling would instead make $V$ diverge.

Multiply the master equation by $s^n$ and sum over $n\ge 0$. For the immigration terms,For the death terms, shifting the first sum givesTherefore the probability generating function obeys

The additional transition is $n\to n+k$ at rate $ϵ(n)$. Its gain term at state $n$ comes from state $n-k$, so the new master equation iswhere the gain term is zero when $n<k$. In the generating-function equation the new contribution isFor a general state-dependent $ϵ$, this is not closed in $\varphi$. Even when it is closed, $s^k-1$ is not proportional to $s-1$ for $k>1$, so the Poisson ansatz used in part (a) is not preserved.

Write $E_n(t)=P(E,n)$ and $C_n(t)=P(ES,n)$. The master equation balances the transitions $E_n\to C_n$ at rate $k_1$, $C_n\to E_n$ at rate $k_1^{\prime }$, and $C_n\to E_{n+1}$ at rate $k_2$:for $n\ge 0$, with $C_{-1}=0$.

The distribution is the geometric distribution on $\{1,2,\dots \}$ with success parameter $q=e^{-\lambda t}$. Its expected value is thereforeEquivalently, multiplying the master equation by $n$ and summing gives $m^{\prime }(t)=\lambda m(t)$ with $m(0)=1$.

This is a Yule process, a pure birth process with state-dependent birth rate ${\lambda }_n=\lambda n$. Probability flows into state $n$ from $n-1$ and leaves it for $n+1$, so the master equation is

Let $w_{ij}$ be the transition rate from $i$ to $j$ and define the probability current $J_{ij}=w_{ij}{P}_i-w_{ji}{P}_j=-J_{ji}$. The master equation is ${\dot{P}}_i=-{\sum }_j{J}_{ij}$. Differentiating the Shannon entropy $S_{\mathrm{sys}}=-{\sum }_i{P}_i\log P_i$ and symmetrizing givesAdding the entropy flow to the environment produces the nonnegative entropy production rate of a Markov chainThe paper's displayed $\sum P\log P$ is the negative of the thermodynamic system entropy, so its derivative has the opposite sign before the environmental contribution is added.

Treat $p_n(t)$ as the probability of the complete coarse-grained state $n$ at time $t$, with $p_n\ge 0$ and ${\sum }_n{p}_n=1$. In the column-vector convention of the master equation, $W_{nm}$ for $n\ne m$ is the transition rate from $m$ to $n$. The short-time transition law isNonnegative transition probabilities and conservation of total probability requireFor a finite state space these conditions define a continuous-time Markov chain. The Markov property means that, conditional on the present state, no further history affects the transition probabilities. A time-independent $W$ additionally assumes time-homogeneous rates; the waiting time in a state is exponential, and multiple jumps in an infinitesimal interval contribute beyond first order. A reduced description that omits relevant slow variables can retain memory and need not obey this closed equation. Quantum coherences are likewise absent from this classical population model. For infinite state spaces one also needs conditions preventing infinitely many jumps in finite time. Neither equal forward/backward rates nor detailed balance is required merely to conserve probability.

Use states $0,A,B$, where $0$ denotes an empty molecule and the other states denote its single occupied site. The source arrows give $0\to A$ at rate $f_1$, $A\to 0$ at rate $f_4$, $A\to B$ at rate $f_2$, and $B\to A$ at rate $f_3$. ThusEach diagonal entry is the negative total escape rate from its column's state. In particular, the empty state cannot make a direct jump to $B$: the reservoir touches $A$ alone. Conversely, $A\to B$ is not a second charge arriving on the molecule; it is an internal hop of its existing charge. Summing the three equations gives $d(p_0+p_A+p_B)/dt=0$, which checks the master equation convention and the single-charge constraint.

For equal rates, the master equation generator is $f$ times the symmetric path-graph matrixIts eigenvectors may be chosen as $(1,1,1)$, $(1,0,-1)$, and $(1,-2,1)$, with eigenvalues $0,-f,-3f$. Resolve the initial empty state asEach nonstationary eigencomponent decays with its own rate, givingThe remaining probabilities are $p_0(t)=1/3+e^{-ft}/2+e^{-3ft}/6$ and $p_A(t)=(1-e^{-3ft})/3$. These satisfy the full equations and sum to one. The limit $p_B(\infty )=1/3$ agrees with the stationary result. At short times $p_B(t)=f^2{t}^2/2+O(t^3)$: two successive jumps, $0\to A\to B$, are necessary, so a term linear in $t$ would signal an incorrectly included direct transition.