Rare Event Simulations Transition state theory 16.1-16.2 - PowerPoint PPT Presentation

Rare Event Simulations Transition state theory 16.1-16.2 Bennett-Chandler Approach 16.2 Transition path sampling16.4

Outline • Part 1 – Rare event and reaction kinetics – Linear Response theory – Transition state theory – Free energy methods – Bennet Chandler approach – Example zeolites • Part 2 – Two ended methods – Transition path sampling – Rate constants – Reaction coordinate analysis – Application to crystallization

catalysis solution reactions crystallisation Rare enzyme reactions complex fluids Events solvent effects folding & binding isomerization

Rare events Interesting transitions in complex systems A – solution chemistry F B – protein folding – enzymatic reactions – complex surface reactions – diffusion in porous media q – nucleation q B These reactions happen on a long time scale compared to the molecular timescale A t dominated by collective, rare events τ mol � τ stable Straightforward MD very inefficient

Example: Diffusion in porous material

Phenomenological reaction kinetics A rare event can be seen as a chemical reaction between reactant A and product B B A A B ↔ The change in population c(t) is (0<c<1) ( ) d c A t d c t ( ) ( ) + k B → A c B t ( ) = − k A → B c A t B k c t k c t ( ) ( ) = + − A B A B A B dt → → dt d c t c t Total number change in population ( ) ( ) ⎡ ⎤ + ⎣ ⎦ = A B 0 dt ( ) =  ( ) = 0 Equilibrium:  c A t c B t c A = k B → A This gives a relation between equilibrium population and reaction rates k A → B c B

Relaxation time Let us make a perturbation of the equilibrium populations, e.g by applying an external field. When releasing the field, the system will relax to the original equilibrium c t c c t c t c c t ( ) ( ) ( ) ( ) For state A For state B: = + Δ = −Δ A A A B B A We can rewrite the kinetics in terms of the perturbation Δ c: 1 d c ( ) t Δ A k c t k c t ( ) ( ) = − Δ − Δ C A (t) A B A B A A → → dt population c t c 0 exp k k t ( ) ( ) ( ) Δ = Δ ⎡ − + ⎤ ⎣ ⎦ A A A B B A C B (t) → → c 0 exp [ t τ ] ( ) = Δ − A time With relaxation time c 1 1 − − 1 B k k ( ) ( ) k 1 c c − τ = + = + = A B B A A B A B → → → k A B → ( ) + c B t ( ) = 1 c A t

Microscopic theory Microscopic description of the progress of a reaction q Reaction coordinate: in this case the z-coordinate of the particle We need to write the kinetics of the reaction in terms of this microscopic reaction coordinate q

Reaction coordinate B A q q * < Reactant A: q q * > Product B: Transition state: q = q * Heaviside θ -function 0 q q * 0 − < ⎧ q q * ( ) θ − = ⎨ 1 q q * 0 − > ⎩ Let us introduce the function g A : g q q * 1 q q * q * q ( ) ( ) ( ) − = − θ − = θ − A With this function we write for the probability c A (t) the system is in state A: ( ) = g A t ( ) c A t

Microscopic theory Is going to give us the macroscopic relaxation in terms of a microscopic time correlation function g 0 g t ( ) ( ) Δ Δ A A exp [ t ] − τ = c c A B

Perturbed Hamiltonian Let us consider the effect of a H H g q q * ( ) = − ε − static perturbation: 0 A This external potential increases the concentration of A For the equilibrium concentration as a function of ε : = g A ε − g A c c c Δ = − A A A 0 0 ε We need to compute the ensemble average in the form of : d A exp [ H ] ∫ Γ − β 0 A = 0 d exp [ H ] ∫ Γ − β 0

Linear Response theory (static) The original Hamiltonian (H 0 ) is perturbed by ε D: H = H 0 − ε D This gives as change in the expectation value of A: A A A Δ = − 0 with ( ) ∫ ⎡ ⎤ d A exp H d Γ A exp − β H 0 − ε D [ ] ∫ Γ − β ⎣ ⎦ 0 A = A = ( ) ∫ ⎡ ⎤ d Γ exp − β H 0 − ε D 0 d exp [ H ] Γ − β ∫ ⎣ ⎦ 0 ∂ A A = A 0 If the perturbation is small we can write 0 + ε ∂ ε

∂ A ε = ∂ A Δ A = 0 For such a small perturbation ε ∂ ε ∂ ε 0 ! # ! # ( ) ( ) d Γ β AD exp − β H 0 − ε D d Γ exp − β H 0 − ε D ∫ ∫ ∂ A ∂ A " $ " $ with = = 2 ∂ ε ∂ ε { ! # } ( ) d Γ exp − β H 0 − ε D ∫ " $ ! # ! # ( ) ( ) d Γ A exp − β H 0 − ε D d Γ β D exp − β H 0 − ε D ∫ ∫ " $ " $ − 2 { } ! # ( ) d Γ exp − β H 0 − ε D ∫ " $ Evaluated for ε = 0 ∫ ∫ ∫ d Γ β AD exp − β H 0 ⎡ ⎤ d Γ A exp − β H 0 ⎡ ⎤ d Γ β D exp − β H 0 ⎡ ⎤ ∂ A ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ = − × { } ∫ ∫ ∂ ε d Γ exp − β H 0 ⎡ ⎤ d Γ exp − β H 0 ⎡ ⎤ ∫ d Γ exp − β H 0 ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 0 { } ∂ A Giving: = β 0 − A AD 0 D ∂ ε 0 0

{ } ∂ A = β 0 − A AD 0 D If we apply this result for c A : ∂ ε 0 0 H H g q q * ( ) with = − ε − 0 A c g g Since g A =0 or 1: Δ = − A A A 0 ε g A (x) g A (x) =g A (x) c ∂Δ ) ( 2 2 A g g ( ) = β − A A 0 ∂ ε 0 ( ) ( ) g 1 g = β − A A 0 0 ( ) ( ) c 1 c c c = β − = β A A A B 0 0 0 0 Δ c A = β c A 0 ε 0 c B Giving:

Linear Response theory (dynamic) Let us now switch off the perturbation at t=0 H = H 0 − ε D H = H 0 at t>0: Let us see how the system relaxes to equilibrium (dynamical perturbation) We take <A> 0 =0 ( ) = A t ( ) − A ( ) Δ A t 0 = A t Similar as for the static case for small values of ε , we have ( ) D exp − β H 0 ( ) ∫ ⎡ ⎤ d Γ β A t ∂ A t ( ) A t ( ) ⎣ ⎦ = = β D 0 { } ∂ ε ∫ ⎡ ⎤ d Γ exp − β H 0 ⎣ ⎦ 0 ( ) = βε D 0 ( ) A t ( ) Δ A t Giving:

( ) = βε D 0 ( ) A t ( ) Δ A t D = Δ g A and A = Δ g A If we apply this result to ( ) = βε Δ g A 0 ( ) Δ g A t ( ) Δ c A t We obtain: ( ) Δ c A 0 From static perturbation: βε = c A c B ( ) Δ g A t ( ) Δ g A 0 ( ) = Δ c A 0 ( ) Δ c A t c A c B Compare linear response expression with the macroscopic expression ( ) = Δ c A 0 ( ) exp − t τ ⎡ ⎤ Δ c A t ⎣ ⎦

Microscopic rate expression Δ has disappeared because of the g 0 g t ( ) ( ) Δ Δ A A exp t [ ] derivative − τ = c c A B Derivative ( )  ( ) ( ) g A t ( )  g A 0 g A 0 g A t − 1 ⎡ ⎤ = − τ exp − t τ ⎦ = ⎣ c A c B c A c B Stationary (t is arbitrary, only depends on τ ) d ( ) B t + τ ( ) = 0 dt A t ( )  ( ) + ( ) B t + τ ( ) = 0  B t + τ A t A t ( )  ( ) = −  ( ) B τ ( ) B τ A 0 A 0

 ( ) g A t ( ) g A 0 1 We have " $ τ exp − t τ % = # c A c B c B − 1 ( ) Using − 1 τ = k A → B 1 + c A c B = k A → B ( ) g A t ( )  g A 0 ( ) = For sufficiently short t, we obtain k A → B t c A Using the definition of g A we can write ( ) ( ) ∂ g A q − q * ∂ g B q − q * ( ) =  g A q − q * = −   q q ∂ q ∂ q ( ) ( ) − q * ∂ g B q 0 ( ) ( )  q 0 g B t We now have an expression that ∂ q ( ) = relates the macroscopic reaction k A → B t rate to microscopic properties c A

( ) ( ) − q * ∂ g B q 0 ( ) ( )  q 0 g B t Let us look at the ∂ q different terms in this ( ) = k A → B t equation c A ( ) ( ) = θ q t ) ( ) − q * Only when the system is in the g B t product state we get a contribution to the ensemble average ( ) ( ) ( ) − q * ( ) − q * ∂ g B q 0 ∂Θ q 0 Only when the system starts at = the transition state, we get a ∂ q ∂ q contribution to the ensemble ( ) ( ) − q * average = δ q 0 ( )  q 0 Velocity at t=0 ( ) c A = Θ q * − q Concentration of A ( ) θ q t ( ) ( ) δ q 0 ( ) − q * ( ) − q *  q 0 ( ) = k A → B t ( ) θ q * − q

Transition state theory ( ) θ q t ( ) ( ) δ q 0 ( ) − q * ( ) − q *  q 0 ( ) = k A → B t ( ) θ q * − q Let us consider the limit: t → 0 + ( ) = θ  ( ) ( ) − q * ( ) lim t → 0 + = θ q t q t This gives: ( ) θ   ( ) δ q 0 ( ) − q * ( ) q 0 q TST = k A → B ( ) θ q * − q Eyring’s transition state theory

Decay of rate expression ( ) θ q t ( ) ( ) δ q 0 ( ) − q * ( ) − q *  q 0 ( ) = k A → B t C(t) ( ) θ q * − q lower value because of recrossings t k TST k(t) k AB ~exp(-t/ ! rxn ) ! mol t

Transition state theory ( ) θ q t ( ) ( ) δ q 0 ( ) − q * ( ) − q *  q 0 ( ) = k A → B t ( ) θ q * − q We can rewrite this expression as a product ( ) θ q t ( ) ( ) ( ) δ q 0 ( ) − q * ( ) − q * ( ) − q * δ q 0  q 0 ( ) = × k A → B t ( ) ( ) ( ) − q * θ q * − q δ q 0 Ratio of probabilities to find Conditional “probability” to find a particle on the top of the particle on top of the barrier barrier with a positive velocity and in the state A ( ) ( ) − q * δ q 0 ( ) ( ) =  ( ) θ q t ( ) − q * k A → B t q 0 q = q * × ( ) θ q * − q