Foundations of Chemical Kinetics Lecture 20: The master equation - PowerPoint PPT Presentation

Foundations of Chemical Kinetics Lecture 20: The master equation Marc R. Roussel Department of Chemistry and Biochemistry

Transition rates ◮ Suppose that P s ( t ) is the probability that a system is in a state s at time t . ◮ The states are members of a set S of allowed states. ◮ Transitions occur randomly between the different states. ◮ For each pair of states r and s , there is a transition rate w rs such that, if the system is in state r at time t , the probability that the system jumps to state s during the subsequent time interval dt is w rs dt .

Transition probabilities Markov property ◮ In general, w rs can depend on the history of the system, i.e. w rs can depend on how long the system has been in state r , which state it came from before that, etc. ◮ In a Markov process, w rs does not depend on the history. ◮ We additionally limit ourselves to homogeneous Markov processes in which w rs does not depend on t .

Derivation of the master equation ◮ The master equation gives us the time evolution of the probabilities P s ( t ). ◮ The probability of being in state s at time t + dt can be written as follows: � � P s ( t + dt ) = P s ( t ) + w rs P r ( t ) dt − w sr P s ( t ) dt r � = s r � = s ◮ The first term is the probability that the system was already in state s at time t . ◮ The terms in the first sum represent the probability that the system was in state r at time t ( P r ( t )) and jumped into state s during the interval dt ( w rs dt ). The sum includes all states from which the system could have come. ◮ The final sum does the same thing as above but for jumps out of state s .

Derivation of the master equation (continued) � � P s ( t + dt ) = P s ( t ) + w rs P r ( t ) dt − w sr P s ( t ) dt r � = s r � = s ◮ Rearrange: P s ( t + dt ) − P s ( t ) � � = w rs P r ( t ) − w sr P s ( t ) dt r � = s r � = s ◮ In the limit as dt → 0, the left-hand side becomes a derivative: dP s � � dt = w rs P r − w sr P s r � = s r � = s This is the master equation. (The argument ( t ) was dropped because all terms are now evaluated at time t .)

Alternative interpretation ◮ If we have a gas containing a large number of molecules, P s might represent the probability that a randomly selected molecule is in state s . ◮ For a large number of molecules, we should have N s = N total P s . ◮ Thus, the master equation could also be written dN s � � dt = w rs N r − w sr N s r � = s r � = s

The master equation and the equilibrium distribution ◮ We rewrite the master equation slightly: dP s � dt = ( w rs P r − w sr P s ) r � = s ◮ At equilibrium, dP s / dt = 0. Thus � � w rs P (eq) − w sr P (eq) � = 0 ∀ s r s r � = s ◮ One solution of this system of equations satisfies detailed balance, i.e. each pair of terms appearing therein individually equals zero: w rs P (eq) − w sr P (eq) = 0 r s

The master equation and the equilibrium distribution (continued) ◮ The detailed balance condition can be rewritten P (eq) = w sr r P (eq) w rs s ◮ At equilibrium, the probabilities should obey a Boltzmann distribution, so � � w sr − E r − E s = exp w rs k B T where r and s label individual quantum states of the system. ◮ Alternatively, if we want r and s to label energy levels, we would have � � w sr = g r − E r − E s exp w rs g s k B T

The master equation and the equilibrium distribution Comments � � w sr = g r − E r − E s exp w rs g s k B T ◮ This equation only fixes the ratio of the transition rates. ◮ The actual values of the transition rates will depend on a number of factors, including the concentrations of collision partners. ◮ Many different functional forms for the transition rates are compatible with this ratio. ◮ Because the transition rates decrease exponentially with increasing energy difference, it is often a good approximation to only consider transitions between adjacent energy levels (Landau-Teller approximation).

Example: A two-level system ◮ For a two-level system, the master equation is dP 1 dt = w 21 P 2 − w 12 P 1 dP 2 dt = w 12 P 1 − w 21 P 2 ◮ Note that dP 1 dt + dP 2 dt = d dt ( P 1 + P 1 ) = 0 so that P 1 + P 2 is a constant, which must be P 1 + P 2 = 1. Thus dP 1 dt = w 21 (1 − P 1 ) − w 12 P 1

Example: A two-level system (continued) dP 1 dt = w 21 (1 − P 1 ) − w 12 P 1 dP 1 w 21 − P 1 ( w 21 + w 12 ) = dt � P 1 ( t ) � t dP 1 dt ′ = t w 21 − P 1 ( w 21 + w 12 ) = ∴ P 1 (0) 0 1 ln [ w 21 − P 1 ( w 21 + w 12 )] P 1 ( t ) ∴ t = − P 1 (0) w 21 + w 12 � w 21 − P 1 ( t ) ( w 21 + w 12 ) � ∴ − t ( w 21 + w 12 ) = ln w 21 − P 1 (0) ( w 21 + w 12 ) w 21 � 1 − e − t ( w 21 + w 12 ) � + P 1 (0) e − t ( w 21 + w 12 ) ∴ P 1 ( t ) = w 21 + w 12

Example: A two-level system (continued) ◮ A slight rewrite of the solution gives P 1 ( t ) = 1 − e − t ( w 21 + w 12 ) + P 1 (0) e − t ( w 21 + w 12 ) 1 + w 12 / w 21 ◮ Now using the Boltzmann detailed balance condition and setting the ground-state energy to zero, we get 1 − e − t ( w 21 + w 12 ) � + P 1 (0) e − t ( w 21 + w 12 ) P 1 ( t ) = � 1 + g 2 − E 2 g 1 exp k B T � 1 − e − t ( w 21 + w 12 ) � = g 1 � + P 1 (0) e − t ( w 21 + w 12 ) � − E 2 g 1 + g 2 exp k B T = g 1 � 1 − e − t ( w 21 + w 12 ) � + P 1 (0) e − t ( w 21 + w 12 ) Q

Example: A two-level system (continued) P 1 ( t ) = g 1 � 1 − e − t ( w 21 + w 12 ) � + P 1 (0) e − t ( w 21 + w 12 ) Q Limits: ◮ As t → 0 ◮ As t → ∞ Relaxation time:

Solving the master equation in general dP s � � dt = w rs P r − w sr P s r � = s r � = s ◮ The master equation is a set of linear differential equation in the P i ’s, so for systems with a finite number of states, it is in principle always solvable. ◮ In practice, it’s not so simple because of the large number of variables.

Solving the master equation in general (continued) dP s � � dt = w rs P r − w sr P s r � = s r � = s ◮ The solution requires the calculation of the eigenvalues and eigenvectors of the matrix of coefficients   − w 11 w 21 . . . w n 1 w 12 − w 22 . . . w n 2   W =  . . .  ... . . .   . . .   w 1 n w 2 n . . . − w nn where w ii = � j � = i w ij .

Solving the master equation in general (continued) ◮ Solution expressible as a sum of terms involving e λ i t , where the λ i ’s are the eigenvalues ◮ For the master equation, λ i < 0 ∀ i . ◮ Solving large eigenvalue problems can be difficult. ◮ There are usually large gaps in the eigenvalue spectrum such that we get a good approximation to the long-term behavior by keeping only a few of the terms (those with the smallest λ i values).

Solving the master equation in general (continued) ◮ Alternative: Direct numerical solution of master equation ◮ Because of the wide range of eigenvalues and potentially large number of variables, these problems can be hard to solve numerically.