Past exam of the theoretical physics major option course of the University of Oxford / 2022 / 3

Here $p_n(t)$ is the probability of occupying state $n$, and $W_{nm}$ for $n\ne m$ is the transition rate from $m$ to $n$. More precisely, over $dt>0$ that transition has probability $W_{nm}dt+o(dt)$. This is a time-homogeneous continuous-time Markov chain: the present state contains all information needed for the next transition, the rates do not depend on time, and finite rates make multiple jumps have probability $o(dt)$. In a physical rate description, untracked coherent dynamics or memory must be negligible on the timescale being modeled.

The master equation separates incoming and outgoing terms:Columns, rather than rows, sum to zero with this indexing. These conditions preserve normalization and nonnegative probabilities. For infinitely many states one also needs a well-defined nonexplosive process. Detailed balance is an additional equilibrium condition, not a general requirement on a master equation.

Write the probability column vector as $\mathbf{p}=(p_0,p_A,p_B{)}^T$, with $0$ denoting the uncharged state. The arrows in the original figure give transitions $0\to A$ at rate $f_1$, $A\to B$ at rate $f_2$, $B\to 0$ at rate $f_3$, and $B\to A$ at rate $f_4$. Exclusion of double occupation means injection occurs only from state $0$. Thus the master equation isFor example, ${\dot{p}}_B=f_2{p}_A-(f_3+f_4)p_B$. Every allowed jump adds to one probability and subtracts from another, which verifies the zero column sums.

Set the time derivatives in the master equation to zero. The first and third rows give $f_1{p}_0=f_3{p}_B$ and $f_2{p}_A=(f_3+f_4)p_B$. Normalization then gives, with $\mathcal{N}=f_2{f}_3+f_1(f_3+f_4)+f_1{f}_2$,These expressions assume positive rates, or a limiting choice for which $\mathcal{N}\ne 0$ and the stationary distribution is unique. If rates vanish so that multiple closed classes appear, the initial condition can matter.

The word “equilibrium” in this part must mean a stationary distribution. With positive $f_1,f_2,f_3$ the one-way contact transitions support a nonzero stationary electric current, so the process is not in thermodynamic equilibrium and does not satisfy detailed balance. In particular, $0\to A$ carries positive probability flow while the direct reverse rate is zero.

Count net transmitted charges per unit time, taking left-to-right as positive. The contact on the left, the internal bond, and the contact on the right give respectivelyThe backward rate $f_4$ must be subtracted in the internal count. Substitution of solution yieldsThe equalities also follow directly from ${\dot{p}}_0=I_R-I_L=0$ and ${\dot{p}}_B=I_{AB}-I_R=0$: there can be no stationary accumulation of electric charge. If each carrier has charge $q$, the electric current is $q{I}_R$. During a stationary interval of duration $t$, the expected net number crossing any one bond is $I_{R}t$. In particular, counting both directions of internal hopping as positive would count traffic, not transmitted charge.

For $f>0$ set $\tau =ft$ and $y(\tau )=p_0(t)$. The master equation becomesUsing $p_A=1-y-p_B$ and $p_B=(y^{\prime }+y)/2$ givesInitially the molecule occupies $B$, so $y(0)=0$ and $y^{\prime }(0)=2$. The characteristic roots are $-2,-3$; the constant particular solution is $1/3$. ConsequentlywhenceIt has the required initial slope $2f$, tends to the stationary uncharged probability $1/3$, and remains nonnegative. As a further check, $p_B=1/6-e^{-2ft}/2+4e^{-3ft}/3$ and $p_A=1/2-e^{-2ft}/2$ are nonnegative and sum with $p(t)$ to one. If $f=0$, no jump occurs and $p(t)=0$, also obtained directly from the boxed expression.