Rare Event Simulations

Rare Event Simulations

Homogeneous nucleation is a rare event (e.g. Liquid Solid) Crystallization requires supercooling (µ solid < µ liquid ) Crystal nucleus 2r Free - energy gain r 3 4! 3! GBulk = ñr! ì 3 Free - energy loss :! G Surface = 4!r 2 ã s,l s,l > < 0 0 r 2

!G tot!g * ~ r 2 ~ -r 3 0 r r c

Also, in many chemical reactions, a free-energy barrier separates reactants from products.

Classical chemical kinetics: A B

because

Statistical mechanics approach.

reactant product 1 A B 0 q q * g A (q-q * )

Common (not best) choice for g A (q-q * ): If we normalize the total concentration to 1, then <c A > = <g A >, and hence

Linear response theory (next slide) Because:

Intermezzo: linear response theory (ultra short).

Consider the response of an observable A due to an external field f B that couples to an observable B: For simplicity, assume that For small f B we can linearize:

Hence Now consider a weak field that is switched off at t=0. f B ΔA t 0

Using exactly the same reasoning as in the static case, we find: In the present case,

Time dependent linear response theory: We eliminate the external field, using then

The macroscopic equation was: And therefore

More convenient: consider at time derivative

Because in equilibrium: Hence: And so, for t=0:

Specific choice for g A : g q " q * = 1 " g q " q* =! q " q * B ( ) ( ) ( ) " g q # q* B ( ) " q A g q " q * = 1 "! q " q* =! q *" q A ( ) ( ) ( ) =! q # q * ( )

We get With

This can be rewritten as: Now the second term on the right is simply equal to For the first term on the right we have:

Our final result for k A B is Kinetic prefactor

In words: Rate= equilibrium probability to find system at the top of the barrier, multiplied by The cross-correlation of the velocity in the direction of the reaction coordinate at t=0 and the probability that the system has crossed the barrier at time t.

Transition state theory k A! B ( t) = = ( ) ( 0 *) " gb q # q q& ( 0) gb t " q c ( )! ( ) " " ( q* # q) A ( ) ( # ) ( ( )# ) q& 0 q 0 q* q t q * Let us consider the limit: t 0 + k TST A# B ( ( ) ) ( ) ( 0)! ( ( 0 ) $ *)" (&) " ( q * $ q) lim = # * = & " ( t) ( ) +! q t q! q t t 0 = q& q q q

The motion of the particle is ballistic The product or reactant states can equilibrate k TST A# B ( t) q& ( 0)! ( q( 0) $ q1) " ( q& )! ( q( 0) $ q1) = %! ( q( 0) $ q1) " ( q1 $ q) ( q( 0) & q1) ) dq ( q( 0) & q1) exp & U ( q) = ( q1 q) ) dq ( q( 0) & q1) exp & U ( q)!! $ ' " % ( # &! $ ' " % ( ( ) ( )! q( 0) = q 1 ( "# ( ) ( ) ( q1) ( ) exp $ &"! U q1 %' ( ) dq exp $ &"! U q %' ) ( ) ( ) ( ) Transition state theory ( ) ( ) ( ) ( & 1) $ & ( )% dq! ( q( 0) % q1) exp # & %" U ( q) $ ' dq! ( q( 0) % q1) exp #%" U ( q) $ q& 0! q 0 & q1 " q& dqq& 0! q 0 & q1 " q& exp $ ' &# U q % ( = & dq! q 0 q exp ' # U q ( ) ( ) = 0.5 q& 0 ( ) = 0.5 q& 0 ( ( & '

( ) = 0.5 &( 0) k t q TST A" B q kbt = 2" m 1 ) #$ q 1 ) #$ ( ) exp %#! U q & ' 1 ( ( ) dq exp %' #! U q &( ( ) exp %#! U q & ' 1 ( ( ) dq exp %' #! U q &( depends on the choice for q 1.

What happens if you choose the wrong value for q *? Bennett-Chandler: less efficient, but still OK TST: wrong

Example: Ballistic barrier crossing

Transition state theory Requires accurate knowledge of free energy Yields an upper bound to the reaction rate Assumption: no re-crossings

k TST A# B k A# B ( t) ( t) ( $ ) (& 1 ) ( ( ) $ 1) ( $ q1) " ( q1 $ q) ( $ 1) ( ( ) $ 1) ( ( ) $ 1)! ( q( 0) $ q1) " ( q1 $ q) ( ) ( )! q( 0) q& 0! q 0 q " q! q 0 q = % ( ) ( ) q& 0! q 0 q " q t q! q 0 q = % Transmission coefficient! ( t) # k A" B TST ka" B ( t) ( 0) ( 0) ( # ) ( ( )# ) q! q q " q t q = & 0.5 q& 0 1 1 ( ) Bennett Chandler Approach

Bennett-Chandler approach Results are independent of the precise location of the estimate of the transition state, but the accuracy does. If the transmission coefficient is very low Poor estimate of the reaction coordinate Diffuse barrier crossing

Simulating rare events: the diffusive limit reactant product ω A B q * q

How do concentration fluctuations decay? If the concentration increase on the left side of the barrier is ΔC A (0), then the concentration increase on the top of the barrier is, on the left side: And, on the right side: 0 ( absorbing boundary ) The diffusive flux across the barrier is:

Therefore The solution is with

In practice, the expression is only slightly more complex. To compute the rate of an activated process, we must compute: 1. The free energy barrier (ΔG) 2. The diffusion constant at the top of the barrier (D) 3. The effective width of the barrier (for a square barrier, this is W in other cases it is a bit different)

Diffusive barrier crossing Consider the following membrane TST: Flux = P(0) <v>/ω Bennett Chandler Flux= P(0) <v>/ω κ Diffusive barrier crossing Flux= P(0) D eff /ω 2

k A! B ( t) =! 0 ( ( 0 ) *) " gb q # q q& ( 0) gb t " q d = $ dt d t! = $ dt 0 q& c A ( ) ( ( 0 ) *) " gb q # q q& ( 0) gb t " q ( 0) c A ( ) ( ( 0 ) *) q& ( t ) " gb q # q " g " q " q " 0 reactant: q<0.5! # gb ( q) = $ 0.5 + q! on the barrier # % 1 product: q>0.5! k " 1 A# B ( t) = 2! $ 0 dt q& c A ( 0) q& ( t) * c A B Diffusive barrier crossing " 1 # g B $ barrier = %! # q $ & 0 elsewhere

Reaction coordinate F(q) r ( ) ( ) t : θ=1 q& 0 and! q& 0 For both ( ) ( ) ( # ) ( ( )# ) q& 0! q 0 q * " q t q * Low value of κ q

Computing free-energy landscapes Suppose we have an order parameter Q(r N ). The probability that Q(r N ) has a value Q is given by From this probability we can derive the variation of the free energy with Q:

This indicates a route to determine F(Q): just make a histogram of the spontaneous fluctuations of Q. P(Q) F(Q) Q Problem: F(Q) is very noisy, except near its minimum.

Application of biased MC simulation to determine P w (Q) But this we can rewrite as

Clearly, or We can choose the bias w(q) such that any desired range of Q-values is sampled. And we can correct for the bias.

The only remaining problem is that the different parts of F(Q) are shifted with respect to each other.

Solutions: 1. Fit to a single curve (not very elegant, but effective) 2. Use Ferrenberg-Swendsen scheme to combine different parts of the histogram (more elegant, but more sensitive to noise) Result:

SOLUTION 2: Use path sampling techniques The advantage of the latest path-sampling techniques (R.J. Allen et al. Phys. Rev. Lett. 94, 18104(2005)) is that they can be used for NON-EQUILIBRIUM systems.

Basic idea: 1. Define a reaction coordinate that distinguishes START from FINISH. 2. Compute the rate Γ 0 (low, but not very low) to reach surface 1. 3. Generate many trajectories from first crossing points and compute the fraction P(12) that reach 2 (without first returning to START). 1 2 3 4 5 4. Repeat for all subsequent planes - and for many starting points. 5. Rate= Γ 0 P(12)P(23)..P(5F)

Illustration: protein crystallization.

!G tot!g * ~ r 2 ~ -r 3 0 r r c

Study nucleation events: 1. By experiment DIFFICULT BECAUSE NUCLEATION EVENTS ARE INFREQUENT, BUT FAST 2. By simulation IMPOSSIBLE WHAT TAKES LONG IN EXPERIMENTS, TAKES FOREVER [O(10 29 ) MD STEPS] IN A SIMULATION

SOLUTION 1: 1. Compute height of the free-energy barrier ΔG * (MC/MD) using biased sampling 2. Compute the transmission coefficient Γ ( using MD) Rate = " exp(#$g % /kt) Kinetic Prefactor (usually weak function of T) Probability of critical fluctuation (strong function of T)

PROTEIN STRUCTURE Problem: HUMAN GENOME PROJECT 3 10 4 (?) protein sequences BUT WHAT IS THEIR 3D STRUCTURE? X-RAY CRYSTALLOGRAPHY REQUIRES GOOD CRYSTALS Only a small fraction of proteins have been successfully crystallized.

Crystallization of globular proteins QUOTE:... mainly trial and error... much like prospecting for gold... (McPherson Preparation and Analysis of Protein Crystals ) N

T Fluid F + S Solid Typical phase diagram of globular proteins T c ρ concentration Meta-stable fluid-fluid

M. Broide et al., PNAS 88,5660(1991) Phase diagram of GLOBULAR PROTEINS (γ-crystallin) T F F + S T c ρ

Protein crystallization conditions T F T c F + S F+F Protein concentration

# of crystalline particles Crystallization Condensation # of particles in a dense cluster (e.g. a droplet)

Crystal nucleation barriers

Nucleation speedup: 10 10

Experimental Evidence?? Experiments by Galkin and Vekilov PNAS 97, 6277 6281(2000) Lyzosyme crystallization

Liquid-liquid phase boundary Nucleation Rate (s -1 cm -3 ) Temperature (ºC)

Crystallization on structured templates If the crystal fits well on the template, it is easy to nucleate

SURPRISE:

RANDOM (!) SUBSTRATES ACT AS UNIVERSAL NUCLEATION SEEDS (Experiments: Prof. Naomi Chayen, Imperial College, London) Figure 1: Electron micrograph of micro-porous silicon (left) shows that the sample contains a network of pores with sizes between 5 and 10 nanometers. The figure on the right hand shows a number of protein crystals (phycobili protein) that have formed on the micro-porous silicon. (N.E. Chayen et al., Ref.. [1])

Explanation: For every Cinderella there is a shoe that fits

Theory: Richard Sear D. Frenkel, NATURE, 443, 641, 2006

Pore-facilitated vapour-liquid nucleation Off-lattice simulations (J.A. van Meel - 2009)

Pore-facilitated ``protein crystal nucleation. Almost epitaxial: crystallises too easily.

but with defects.

Pore-facilitated ``protein crystal nucleation. Far from epitaxial: no crystal at all.

Pore-facilitated ``protein crystal nucleation. Disordered substrate yields perfect crystal that nucleates in wetting layer.