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Kramers theory of reaction kinetics Winter School for Theoretical Chemistry and Spectroscopy Han-sur-Lesse, 7-11 December 2015 Hans Kramers Hendrik Anthony Kramers 1894 – 1952

1911 HBS, 1912 staatsexamen, 1916 doctoraal wis- en natuurkunde 1916 Arrived unannounced in Kopenhagen to do PhD with 1919 promotion in Leiden (Ehrenfest), married in 1920 Lector in Kopenhagen 1926 Professor theoretical in Utrecht 1934 Professor in Leiden (chair of Ehrenfest), professor in TU-Delft

Physics 7(4), April 1940, Pages 284-304 - 8 references

Outline

• Intro Kramers theory • Transition State Theory • The Langevin equation • The Fokker-Planck equation • Kramers reaction kinetics • high-friction vs low-friction limit • memory, Grote-Hynes theory

• Bennet-Chandler approach • Reactive Flux • Transmission coefficient calculation • Free energy methods • Summary • Bibliography Molecular Transitions are Rare Events chemical reactions phase transitions protein folding

defect diffusion self-assembly stock exchange crash Rare Events • Molecular transitions are not rare in every day life • They are only rare events with respect to the femto- second timescale of atomic • Modeling activated transitions by straightforward simulation would be extremely costly (takes forever) • Advanced methods: Transition state theory (TST), Bennet-Chandler (Reactive Flux) approach, Transition Path Sampling (TPS), Free energy methods, Parallel Tempering, String Method,....

A very long trajectory • A and B are (meta-)stable states, i.e. A B attractive basins

• transitions between A and B are rare

• transitions can happen fast

• system looses memory in A and B Free energy landscape Rare Events Transition State

The free energy has local minima separated by barriers

Many possible transition paths via meta-stable states

Projection on reaction coordinate shows FE profile P.G. Bolhuis, D. Chandler, C. Dellago, P.L. Geissler with transition state barrier - G/kT k = k0 e Δ Annu. Rev. Phys. Chem 2002 Reaction rate depends Arrhenius equation exponentially on the barrier height A long trajectory Reaction coordinate Transition State Theory • Derived by Henry Eyring and by Meredith Gwynne Evans and Michael Polanyi in 1935 E /RT • Explains Arrhenius equation (1889) k = Ae a (van’t Hoff 1884) Henry Eyring • Activated complex is in quasi equilibrium with reactants • All reactants that reach the activated complex state, continue to products

[AB]‡ probability to reach A + B [AB]‡ C K‡ = the transition state $ ! [A][B]

kBT G‡/RT - rate is frequency of unstable k = k‡K‡ =  e h mode times the TS probability - transmission factor is κ=1 in TST frequency (correction for barrier recrossings) Transition State Theory

• what is a transition state? activated structure, critical nucleus, entropic bottleneck,… • where does the refactor (h dependence) come Henry Eyring from? • TST rate is an upper limit for the true rate • The actual rate is slower that the TST estimate due to recrossing • recrossing occur due to a bad estimate for the TS (dividing surface) / hidden slow variables (aka the reaction coordinate problem) and/or due to noisy dynamics from the environment • is kappa just a cheating factor or does it have a physical meaning? Separation of time scales • Suppose a coordinate X describes the reaction progress • X is coupled to environment • the coupling makes the dynamics of X stochastic (not deterministic) • to escape from the reactant well, X (=random walker) must acquire enough energy • The escape time depends on from the bath the size of the fluctuations • this energy must be lost (X(t)-)2 again after passing the • Enoise << Ebarrier => Ea/RT >> 1 barrier relaxation time τs escape time τe = 1/k 2 d V a ⌧s 2⇡ me↵ / 2 << ⌧e = ⌧s exp (E /kBT ) ⇡ r dX ⇣ ⌘ Diffusion Density of particles: ⇢(r,t) Current density: j(r,t)=⇢(r,t)v(r,t)

A gradient in the density leads to a current: Diffusion of particles due to a concentration gradient j = D ⇢ Fick’s law r diffusion coefficient Mean square displacement:

2 - also end-to-end distance x (t) =2Dt - ballistic : x2(t) t2 h i ! Velocity auto correlation function: Self-diffusion of a particle seen as a random walk 1 D = d⌧ vx(⌧)vx(0) 0 h i ensemble average ... Z h i Langevin equation

mass x acceleration = force 2 d x dU(x) dx Paul Langevin m = + ⇣(t) 1872-1946 dt2 dx dt

U is external viscous drag random potential γ is friction collisions (dissipation)

Describes Brownian motion random motion of small particles (pollen grains) in Robert Brown a solvent due to thermal fluctuations (Brown 1828) 1773-1858 Illustration with Octave/Matlab script d2x dU(x) dx m = + ⇣(t) dt2 dx dt

% brownian_2d.m simulates a two-dimensional Brownian motion

N=10000; % number of steps to take T=70; % maximum time h=T/N; % time step t=(0:h:T); % t is the vector [0 1h 2h 3h ... Nh] sigma = 1.0; % strength of noise

x=zeros(size(t)); % place to store x locations y=zeros(size(t)); % place to store y locations

x(1)=0.0; % initial x location y(1)=0.0; % initial y location for i=1:N % take N steps x(i+1)=x(i)+sigma*sqrt(h)*randn; y(i+1)=y(i)+sigma*sqrt(h)*randn; end;

plot(x,y); grid on % add a grid to axes % axis([0 T -3 8]); % set axis limits Illustration with Octave/Matlab script d2x dU(x) dx m = + ⇣(t) dt2 dx dt

% brownian_2d.m simulates a two-dimensional Brownian motion

N=10000; % number of steps to take T=70; % maximum time h=T/N; % time step t=(0:h:T); % t is the vector [0 1h 2h 3h ... Nh] sigma = 1.0; % strength of noise

x=zeros(size(t)); % place to store x locations y=zeros(size(t)); % place to store y locations

x(1)=0.0; % initial x location y(1)=0.0; % initial y location for i=1:N % take N steps x(i+1)=x(i)+sigma*sqrt(h)*randn; y(i+1)=y(i)+sigma*sqrt(h)*randn; end;

plot(x,y); grid on % add a grid to axes % axis([0 T -3 8]); % set axis limits Langevin equation

- laminair flow (small velocities) Stokes formula for viscous drag: - spherical particle with radius α - γ is the friction = 6⇡⌘↵ - 1/γ is the mobility - η is the dynamic viscosity f = v - v is the velocity of the particle d Equation of motion with only drag and random forces:

dv(t) - Langevin equation without ∇V m = 6⇡⌘↵v(t)+⇣(t) - stochastic differential equation dt - what do we know about ζ(t)? - ⇣(t) =0 h i 2 - ⇣(t)⇣(t0) =2D (t t0) Taking the ensemble average: - h(white noise,i Wiener process) - (Markovian dynamics vs memory) d v(t) m h i = 6⇡⌘↵ v(t) - since average random force is zero dt h i

t/⌧ - relaxation time of initial perturbation which implies: v(t) = v(0) e h i ⌧ = m/(6⇡⌘↵) strong friction

Particle in a very viscous solvent: d2x dU(x) dx dx - m = + ⇣(t) = ⇣(t) dt2 dx dt dt - particle is driven by fluctuating solvent - no inertial motion -> m=0 - no external potential solving the differential equation: t 1 - since ⇣(t) =0 h i x(t)=x(0) + dt0⇣(t0) - unsurprisingly: x(t) = x(0) - h i h i Z0 consider many paths from x(0) what about the variance? - x(0) to the left and square both sides t t 2 1 x(t) x(0) = dt0 dt0 ⇣(t0 )⇣(t0 ) 2 1 2 1 2 Z0 Z0 ⌦ ↵ ⌦ ↵ 2 - particle gets 1016 (gas) to 1020 (liquid) kicks per sec. - kicks are very fast and considered uncorrelated x(t) x(0) =2Dt 2 - ⇣(t)⇣(t0) =2D (t t0) h i ⌦ ↵ weaker friction

Particle in a less viscous solvent: d2x dU(x) dx t/m - m = + ⇣(t) v(t) = v(0) e dt2 dx dt h i - inertial motion -> m > 0 - still no external potential For the position we use: - τ=m/γ t x(t)=x(0) + dt0v(t) Z0 γ=0.001

take the ensemble average position γ=0.1 t x(t) = x(0) + dt0 v(t) γ=1.0 γ=2.0 h i 0 h i Z time m t/m x(t) = x(0) + v(0) 1 e h i Fluctuation-dissipation dv(t) m = 6⇡⌘↵v(t)+⇣(t) dt conversion of heat dissipation of kinetic into kinetic energy energy into heat of the bath of the particle

Velocity auto correlation in weak friction:

- decays to zero with τ=m/γ D (t2 t1)/m v (t )v (t ) = e - i.e. correlated when t2 – t1 < m/γ h x 1 x 2 i m D - v (t)v (t) = h x x i m Gives also the kinetic energy: 3 - equipartition theorem: E = k T kin 2 B 1 2 3D Ekin = m v(t) = - fluctuation-dissipation theorem: 2 h i 2 there is a relation between the kicking and the friction!

kBT - combine with MSD -> determine kB ! Einstein relation: k T D = x2 = B t h i ⇡⌘↵ - Jean Baptiste Perrin 1926 Nobel prize Fokker-Planck

Integrating the Langevin equation from an initial point (x0,t0) results in a stochastic trajectory. What is the probability to arrive at point x at time t? —> P(x,t | x0,t0) - we could integrate over all possible trajectories conditional probability - or try to predict how the probability evolves

dx d2x dU(x) dx E.g. strong friction case: = ⇣(t) m = + ⇣(t) dt dt2 dx dt 2 x(0) =0, x (t) =6Dt - (now in 3 dimensions) h i h i The distribution that gives this variance is a Gaussian:

2 1 3/2 x /4Dt 3 P (x,t)=( ) e - prefactor from: d xP(x,t)=1 4⇡Dt Z @P Solution of diffusion equation: = D 2P - simplest Fokker- @t r Planck equation Diffusion equation P Fick’s law: j = D P - current probability (or r current density) j j(x,t)=P (x,t)v(x,t)

volume flux in flux out with P @P @j Conservation law: = j j @t @x ∆x @P Combine: = D 2P @t r

Solution of diffusion equation:

1 3/2 x2/4Dt P (x,t)=( ) e 4⇡Dt Illustration with Octave/Matlab script

1 3/2 x2/4Dt P (x,t)=( ) e 4⇡Dt

% Script tp plot diffusion equation clear; clf;

N=1000; % number of steps to take T=50; % maximum time h=T/N; % time step nx=100; % size of xgrid xmin=-5.0; xmax=5.0; dx=(xmax-xmin)/nx; x=zeros(size(nx)); % place to store y locations y=zeros(size(nx)); % place to store y locations sigma = 0.01; % strength of noise D=1.0*4*3.14159265; t=0.00001;

for it=1:T for ix=1:nx x(ix)=xmin + ix*dx; y(ix)= (1.0/(D*t))^(3/2) * exp(-(x(ix)^2)/(4*D*t)); end;

plot(x,y,"linewidth"), hold on; title(num2str(it*h)); xlabel ("X"); ylabel ("P(X)"); axis([xmin xmax 0.0 10.0]); grid on; pause(1);

t=t+h; end; Summary part 1

kBT G‡/RT Transition state theory (Eyring, 1935) k =  e h Diffusion, Brownian Motion, Langevin equation: d2x dU(x) dx m = + ⇣(t) dt2 dx dt

kBT Fluctuation-dissipation theorem, Einstein relation: D = @P Diffusion equation, Fokker-Planck equation: = D 2P @t r

Matlab/Octave scripting and plotting Fokker-Planck d2x dU(x) dx m = + ⇣(t) More general: with an external potential: dt2 dx dt - strong limit —> m=0 v = V + ⇣(t) r t+t 1 1 - after a small displacement x = vt = V t + dt0⇣(t0) - integrate over the random r t kicks during this small time Z interval The expectation value of a small displacement: 1 x = V t - since ⇣(t) =0 h i r h i The correlation between displacements is also not difficult: t+t t+t t+t 1 2 x x = @ V @ V t t dt0 @ V ⇣ (t0)+@ V ⇣ (t0) + dt0 dt00 ⇣ (t0)⇣ (t00) h i ji 2 h i j i h i j j i i h i j i Zt - i, j are different directions Zt Zt t+t t+t t+t - one integral drops in last 1 2 term because ⇣(t) =0 xixj = @iV @jV t t dt0 @iV ⇣j(t0)+@jV ⇣i(t0) + dt0 dt00 ⇣i(t0)⇣j(t00) h i h i 2 h i h i h i - 1st and 2nd terms: order (δt)2 Zt Zt Zt Fokker-Planck

t+t t+t t+t 1 2 x x = @ V @ V t t dt0 @ V ⇣ (t0)+@ V ⇣ (t0) + dt0 dt00 ⇣ (t0)⇣ (t00) h i ji 2 h i j i h i j j i i h i j i Zt Zt Zt 2 - ignore terms of order (δt)2 xixj =2ijDt + (t ) 1 h i O - average: x = V t h i r So, which probability distribution has this average and variance? Let’s start from the conditional probability: probability to be at x at t+dt, if we were at x’ at time t P (x,t+ t x0,t)= x (x0 + x) = | ensemble average over all possible displacements δx ⌦ ↵ such that go from x’ to x

We bravely Taylor expand the delta-function at δx=0: @ 1 @2 P (x,t+ t x0,t)= 1+ xi + xixj + ... (x x0) | h i@xi0 2h i@xi0 @xj0 ⇣ ⌘ - (what are the derivatives of the delta function???) Fokker-Planck @ 1 @2 P (x,t+ t x0,t)= 1+ xi + xixj + ... (x x0) | h i@xi0 2h i@xi0 @xj0 ⇣ ⌘ at t’, in-between t and t0 the Use the Chapman-Kolmogorov equation: particle had to be somewhere! so, integrate over all possible x’ 1 3 P (x,tx ,t )= d x0P (x,tx0,t0)P (x0,t0 x ,t ) | 0 0 | | 0 0 Z1 fill in our Taylor expansion: @ 1 @2 P (x,t+ t x0,t0)=P (x,tx0,t0) xi P (x,tx0,t0) + xixj P (x,tx0,t0)+... | | @xi0 h i | 2h i@xi0 @xj0 | @ 1 ⇣ @2 ⌘ P (x,t+ t x0,t0)=P (x,tx0,t0) xi P (x,tx0,t0) + xixj P (x,tx0,t0)+... | | @xi0 h i | 2h i@xi0 @xj0 | ⇣ ⌘ 1 - use: x = V t and x x =2 Dt h i r h i ji ij 1 @ @V @2 P (x,t+ t x0,t0)=P (x,tx0,t0)+ P (x,tx0,t0) t + D 2 P (x,tx0,t0)t + ... | | @xi0 @xi | @x | 1 @ @V @2 ⇣ ⌘ P (x,t+ t x0,t0)=P (x,tx0,t0)+ P (x,tx0,t0) t + D 2 P (x,tx0,t0)t + ... | | @xi0 @xi | @x | ⇣ ⌘ Fokker-Planck 1 @ @V @2 P (x,t+ t x0,t0)=P (x,tx0,t0)+ P (x,tx0,t0) t + D 2 P (x,tx0,t0)t + ... | | @xi0 @xi | @x | ⇣ ⌘

Left hand side is also: - Taylor expansion in dt 1 P (x,t+ t x ,t )=P (x,tx ,t )+ P (x,tx ,t )t + ... | 0 0 | 0 0 @t | 0 0

Combining with previous result gives: D here independent of position; otherwise D is inside of ∇ 2 @P (x,t) 1 = (P (x,t) V (x)) + D 2P (x,t) @t r · r r drift or transport diffusion

The Fokker-Planck equation or Smoluchowski equation or Kolomogorov’s forward equation Adriaan Daniël Fokker Max Karl Ernst Ludwig Planck, 1887 – 1972 1858 – 1947 Dutch and musician. German theoretical physicist Nobel Prize in Physics (1918)

Andrey Nikolaevich Kolmogorov Marian Ritter von 1903 – 1987 Smolan Smoluchowski Soviet mathematician 1872 –1917 Polish physicist Fokker-Planck @P (x,t) 1 = (P (x,t) V (x)) + D 2P (x,t) @t r · r r

FP as a continuity equation: - J is probability current or flux @P (x,t) 1 = J J = (P (x,t) V (x)) + D P (x,t) @t r driftr diffusionr

Helps to see that probability is conserved:

3 @ 3 3 @P 3 - d xP =1 d xP = d x = d x J =0 Z @t @t r - the particle should be Z Z Z somewhere

FP tells us how a system evolves: - for V=0 (diffusion equation), the system spreads out forever k T - for generic V, stationary solutions with ∇J=0: - D = B

V (x)/D Einstein relation V (x)/kB T - Boltzmann P (x) e P (x) e ⇠ ⇠ distribution! Escape over a barrier density ~ 0 Thermal escape from 1D potential equilibrium 1 distribution V (x) !2 (x x )2 ⇡ 2 min min

With initial distribution in the left well: source sink 2 !min ! (x x )2/2k T P (x, t = 0) = e min min B - initial distribution is not s2⇡kBT the equilibrium distribution - there is no equilibrium - assume small flux Steady state flux: 1 J = (P (x,t) V (x)) + D P (x,t) k T @ r r B V (x)/kB T V (x)/kB T k T J = e e P D = B @x

V/k T @ V/k T 1 @V @P e B P e B = P + @x · k T @x @x kBT @ B JeV (x)/kB T = eV (x)/kB T P @x

now integrate both sides from xmin and x* Escape over a barrier x x ⇤ V (x)/k T ⇤ kBT @ V (x)/k T Je B = e B P first integrate the right side @x Zxmin Zxmin x 2 ⇤ V (x)/kB T kBT V (xmin)/kB T V (x )/kB T kBT !min J e = e P (xmin) e ⇤ P (x ) = x ⇤ s2⇡kBT Z min ⇥ at xmin P is at equilibrium at x* P is practically zero⇤

Next, the integral at the left, which is 1 dominated by the potential at V(xmax): V (x) V !2 (x x )2 ⇡ max 2 max max x ⇤ 2⇡kBT J eV (x)/kB T = JeVmax/kB T !2 Zxmin s max Combining the to pieces gives the rate over the barrier as:

!max !min V /k T - J = e max B Kramers results for the reaction rate for 2⇡ strong friction! (no inertial motion; m=0) - transmission coefficient:  = !max/ kappa TST rate Escape over a barrier

Kramers rate equation:

Strong friction (over-damped) limit

!max !min V /k T k = e max B 2⇡

Including inertial motion (m>0): 1 2 ! 2 min Vmax/kB T k = + + !max e !max 2 r 4 2⇡ ⇣ ⌘ kappa (correction to transition state theory rate) TST rate

Weak friction limit: I(Emax) !min V /k T k = p e max B kBT 2⇡ Escape over a barrier Transmission coefficient as a function of the reactant-solvent coupling

Kramers cross-over region

weak coupling: over-damped coupling: ! ! I(Emax) !min V /k T max min Vmax/kB T k = p e max B k = e kBT 2⇡ 2⇡ 1   ⇠ ⇠ if 0 then  0 if then  0 ! ! !1 ! Langevin dynamics in 2D barrier crossing

% Octave script for velocity Verlet algorithm + Langevin for % dynamics in the 2D Muller-Brown potential Muller-Brown potential % Langevin dynamics using PRE 75, 056707 (2007) Bussi & Parrinello clear; clf;

Nsteps=2000; % number of stepps h=0.01; % timestep pos=[0.5 0.0]; % initial position vel=[10.0 0.0]; % initial velocity force=[0.0 0.0]; % initial force friction=2.0; % langevin friction temp=30.0; mass=1.0; c1 = exp(-0.5*friction*h) c2 = sqrt( (1-c1*c1) * temp * mass )

% initiate force [Vpot(1), force] = Muller_potential(pos); Ekin(1)=0.5*norm(vel)^2; Etot(1)=Ekin(1) + Vpot(1); t(1)=1;

% main loop for i=1:Nsteps % velocity verlet half step for velocities and positions t(i)=i; vel = c1*vel + c2*randn; x(i)=pos(1); vel += 0.5*force*h; y(i)=pos(2); pos += vel*h;

% plot trajectory and energy % force calculation figure(1); [Vpot(i), force] = Muller_potential(pos); plot(x,y,'b-','linewidth',2,pos(1),pos(2),'ko',0,0,'ro') title(num2str(i*h)) % velocity verlet second half step for velocities axis equal; vel += 0.5*force*h; axis([-2 1.5 -0.5 2.5]); vel = c1*vel + c2*randn; grid on; figure(2); % compute energies plot(Etot,'g-'), hold on; Ekin(i)=0.5*norm(vel)^2; plot(Vpot,'r-'), hold on; Etot(i)=Ekin(i) + Vpot(i); plot(Ekin,'b-'); title("Total energy"); end Rare event simulation Macroscopic phenomenological theory

Chemical reaction: A B

dcA(t) = kA BcA(t)+kB AcA(t) dt

dcB(t) =+kA BcA(t) kB AcA(t) dt

Total number of molecules: Equilibrium:

d[cA(t)+cB(t)] dcA(t) dcB(t) cA kB A =0 = =0 = dt dt dt cB kA B ⇥ ⇥ Rare event simulation Macroscopic phenomenological theory

Make a small perturbation:

cA(t)= cA + cA(t) cB(t)= cB + cA(t) ⇥ ⇥ dcA(t) = kA BcA(t) kB AcA(t) dt

cA(t)=cA(0) exp[ (kA B + kB A)t] = c (0) exp[ t/] A

1 = kA B + kB A ⇥ ⇥ 1 cB 1 ⇥ = kA B 1+ cA / cB = ⇥ k⇤A ⌅B ⇤ ⌅ ⇤ ⌅⇥ ⇥ Rare event simulation

q* Microscopic linear response theory !F 0 if q q < 0 (Reactant A) (q q)= 1 if q q > 0 (Product B)

A B Perturbation add bias to increase concentration cA

H = H g (q q) q 0 A g (q q)=1 (q q)=(q q) A c = c c A A A 0 g probability to be in state A c = g ⇥ g ⇥ A A A A 0 ⇥ ⇥ ⇥ Linear response theory: static

A = A A H = H B 0 0 ⇥ ⇥ ⇥ dA exp[ (H )] dA exp[ (H ⇥B)] A = 0 A = 0 0 d exp[ (H )] d exp[ (H ⇥B)] ⇤ 0 ⇤ 0 ⇥ ⇥ ⇤ ⇤

⇤⇥A dAB exp[ (H0 ⇥B)] d exp[ (H0 ⇥B)] = 2 ⇤⇥ d exp[ (H ⇥B)] ⇧ ⌃ ⌥ ⌥0 dA exp[ ⌥(H0 ⇥B)] dB exp[⇥ (H0 ⇥B)] 2 d exp[ (H ⇥B)] ⌥ ⌥ 0 = AB A B 0 ⌥ 0 0 ⇥ ⇤ ⌅ ⇤ ⌅ ⇤ ⌅ ⇥ Very small perturbation: linear response theory

cA = gA gA 0 H = H g (q q) 0 A ⇥ ⇥ How does the response (!c) depend on the perturbation (!")?

dc 2 A = (g )2 g Outside the barrier d⇥ A 0 A 0 ⇤ ⌅ gA = 0 or 1 ⇥ ⇥ = g 1 g gA(x)gA(x)=gA(x) A 0 A 0 ⇤ ⇤ ⌅⌅ ⇥ ⇥ = c 1 c = c c A 0 A 0 A 0 B 0 ⇤ ⇤ ⌅⌅ ⇥ ⇥ ⇥ ⇥ Switch of the perturbation: dynamic linear response

gA(0)gA(t) cA(t)=cA(0) cA cB = c (0) exp[ t/] holds for sufficiently long times A ⇥ ⇥ gA(0)gA(t) exp[ t/]= cA cB ⇥ ⇥ ⇥ 1 gA(0)g ˙A(t) g˙A(0)gA(t) Derivative exp[ t/]= = cA cB ⇥ cA cB ⇥ ⇥ ⇥ ! has⇥ disappeared⇥ because of derivative For sufficiently short t

g˙A(0)gA(t) kA B(t)= cA ⇥ gA(q q) gB(q q) g˙ (q q)=q ˙ ⇥ = q˙ A q q

gB (q(0) q) q˙(0) q gB(t) ka B(t)=⇤ ⌅ ⇥ cA ⇥ Stationary

d A(t)B(t + t) =0 dt ⇥ A(t)B˙ (t + t) + A˙(t)B(t + t) =0

A(t)B˙ (t + t)⇥ = A˙(t)B(t + t⇥) ⇥ ⇥ Eyring’s transition state theory

gB (q(0) q) q˙(0) q gB(t) ka B(t)=⇤ ⌅ ⇤ cA

q˙(0)(q(0) q⇥)⇥(q(t) q⇥) = ⇥ ⇥(q q) ⇥ ⇥ ⇥ Correlation between velocity of states that are at the top of the barrier at t=0 and in the product state B some time t later.

Let us consider the limit t 0+ :

lim = ⇥ q(t) q = ⇥ q˙(t) t 0+ ⇥ ⇥ ⇥ TST q˙(0)(q(0) q)⇥(˙q) ka B(t)= ⇥ ⇥(q q) ⇤ ⌅ ⇤ ⌅ Bennett-Chandler approach

(or Reactive flux method)

q˙(0)(q(0) q)⇥(q(t) q) ka B(t)= ⇥ ⇥(q q) ⇥ ⇥ q˙(0)(q(0) q)⇥(q(t) q) (q(0) q) ka B(t)= ⇥ (q(0) q ) ⇥ ⇥(q q) ⇥ ⇥ Conditional average: q˙(0)(q(t) q) given that we start on top of barrier Probability to find q on barrier top

Computational scheme: • Determine the probability with free energy calculation • Compute conditional average from “shooting” trajectories from barrier top Free energy methods

It is very (very) difficult to compute (or measure) absolute thermodynamic properties, such as free energy and entropy, that depend on the size of the phase space.

But we can compute relative free energies, in particular free energy differences between thermodynamic states.

Reaction equilibrium constants A B

[B[B] ] pBpB KK======[ exp [ (GB(GBGAG)]A)] [A[A] ] pApA Examples: - Chemical reactions, catalysis, isomerization, etc... - Protein folding, ligand binding affinity, protein-protein association - Phase diagrams, coexistence lines, critical points, transitions Free energy perturbation

dsN exp( U ) F = ln(Q /Q )= ln B B A dsN exp( U ) ⇤ A ⇥ ⇤ dsN exp( U ) exp( U) F = k T ln A B dsN exp( U ) ⇤⇧ A ⌅ = k T ln exp( U) B ⇧ A ⇥

F = k T ln exp( U) = k T ln exp(U) B A B B Sampling problems may lead to hysteresis⇥ between the two samples ⇥ Umbrella Sampling Bias the samping along an order parameter q Add and subtract bias potential w(q): N N N dr ⇥(q(r ) q) exp (U(r )+w(q) w(q)) P (q)= drN exp (U(rN + w(q ) w(q )) ⇧ ⇥ N N N dr ⇥(q(r ) q) exp (U(r )+w(q)) exp(w(q)) P (q)= ⇧ ⇥ drN exp (U(rN + w(q ))) exp(w(q )) ⇧ ⇥ N ⇥(q(r⇧ ) q) exp( w(q)) biased ⇥ P (q)= • Let w(q) be a good ⇤ exp(w(q)) biased ⌅ guess of minus the free energy F(q), or exp(⇤ w(q)) ⌅ P (q)= Pbiased(q) • Choose w(q) to confine exp(w(q)) biased sampling to a specific window along q. ⇤ ⌅ • Or do both. F (q)=k T ln P (q)= k T ln P (q) w(q) + const B B biased Constrained MD

The derivative of the free energy F(!) with respect to ! can be written as an ensemble average.

⇤F (⇥) drN (⇤U(⇥)/⇤⇥) exp[ U(⇥)] = ⇤⇥ drN exp[ U(⇥)] ⇥NVT ⇧ ⇤U(⇥) = ⇧ ⇤⇥ ⇤ ⌅ The free energy difference between states A and B can then be obtained by thermodynamic integration B ⇥U() F ( ) F ( )= d B A ⇥ ⇤A ⇥ In the case “hard” constraints are used, additional corrections are needed to unbias for sampling in a constraint ensemble (instead of the actual NVT ensemble) Steered MD

Mechanical work to bring the system WA B FA B from state A to state B

Jarzynski’s equality exp[ WA B] = exp[ FA B] A ⇥ Surprisingly, we can obtain the equilibrium free energy difference from a non- equilibrium simulation, in which we force the system in a finite time to move from A to B.

Although, this may sound as a free lunch method, note that it requires sampling an exponential distribution of the work. For infinitely slow switching from A to B, the system is always in equilibrium so that a single simulation gives !F. But the faster the switching the more rare are the important low- work contributions to the average, so that many steered simulations are required for convergence. Metadynamics Escaping free-energy minima, Laio and Parrinello, PNAS (2002)

t′ 2 −(sα − sα ) V (t, s)= Ht′ exp δ2 W 2 t′

The error depends on the height, width, and time HWS interval of the added Gaussians, and on the ⇥ = C diffusion, temperature and order parameter space d D⇤ of the system.

Example Relative solvation free energy of hexane molecule The solvation free energy free difference of a solute in different solvents is here used a target property to parameterized a coarse-grain forcefield.

hexane

dz=30

dz=20

dz=10

dz=0 water Umbrella Sampling

Weighted Histogram Analysis Method (WHAM) to connect the piecewise free energy curves Constrained MD

Spline fit of constraint force measurements

Running average constraint force !F / [kcal/mol] Steered MD Metadynamics

Simulation in two parts with different size of the Gaussian “hills”: • Part 1: H=0.25 K, W=0.4 Å • Part 2: H=0.10 K, W=0.2 Å

Summary

• Intro Kramers theory • Transition State Theory • The Langevin equation • The Fokker-Planck equation • Kramers reaction kinetics • high-friction vs low-friction limit • memory, Grote-Hynes theory

• Bennet-Chandler approach • Reactive Flux • Transmission coefficient calculation • Free energy methods • Summary • Bibliography Bibliography

• Brownian motion in a field of force and the diffusion model of chemical reactions, H. A. Kramers, Physics 7(4), 284-304 (1940) • Kinetic Theory, University of Cambridge Graduate Course, David Tong, Cambridge 2012 (http://www.damtp.cam.ac.uk/user/tong/ kinetic.html) • Reaction-rate theory: fifty years after Kramers, P. Hänggi, P. Talkner, and M. Borkovec, Rev. Mod. Phys. 62, 215 (1990) • A Short Account of RRKM Theory of Unimolecular Reactions and of Marcus Theory of Electron Transfer in a Historical Perspective, F. Di Giacomo, J. Chem. Educ. 92, 476 (2015) • Atkins, P.; De Paula, J. Atkins’ Physical Chemistry, 8th ed.; Oxford University Press: Oxford, 2006; p 821. • Accurate sampling using Langevin dynamics, G. Bussi and M. Parrinello, Phys. Rev. E 75, 056707 (2007)