A two-state system is first modelled as a discrete-time Markov chain (Chapter 8)

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Thermal, Statistical & Complex Systems · Non-equilibrium statistical mechanics and stochastic thermodynamics

Error, Chance and the Machine, Ch Random processes in continuous time

Problem

A two-state system is first modelled as a discrete-time Markov chain (Chapter 8), sampled every Δt=10 μs\Delta t = 10\,\mu{\rm s}, with one-step transition probabilities P12=0.006P_{12} = 0.006 and P21=0.014P_{21} = 0.014 (and P11=1−P12P_{11}=1-P_{12}, P22=1−P21P_{22}=1-P_{21}). For a two-state chain, the eigenvalue other than 11 of the transition matrix is λ2=1−P12−P21\lambda_2 = 1 - P_{12} - P_{21}, and the chain's population difference from equilibrium decays as λ2n\lambda_2^n after nn steps, so λ2\lambda_2 sets the relaxation rate (Chapter 8). ((a)) Find the continuous-time rates k12=P12/Δtk_{12} = P_{12}/\Delta t and k21=P21/Δtk_{21} = P_{21}/\Delta t, and hence the master equation's relaxation rate and relaxation time for the corresponding two-state generator. ((b)) Compute the number of steps n1/e=−1/ln⁡λ2n_{1/e} = -1/\ln\lambda_2 for the discrete chain's population difference to fall by a factor ee, and compare n1/eΔtn_{1/e}\Delta t with the continuous relaxation time of part (a). ((c)) Find the equilibrium probabilities π1\pi_1 and π2\pi_2 of the two states. ((d)) Treating the two states as two potential wells with πi∝e−Ei/kBT\pi_i \propto e^{-E_i/k_{\rm B}T} at T=300 KT = 300\,{\rm K}, the discrete analogue of the Boltzmann distribution that solves the stationary Fokker-Planck equation, find E2−E1E_2 - E_1 in electronvolts (1 eV=1.602×10−19 J1\,{\rm eV} = 1.602\times10^{-19}\,{\rm J}).

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