The PLUMED Plugin and Free Energy Methods in Electronic-Structure-Based Molecular Dynamics

The PLUMED plugin and free energy methods in electronic-structure-based molecular dynamics

Davide Branduardi, Theoretical Molecular Biophysics Group Max Planck for Biophysics, Frankfurt am Main, Germany

Tuesday, September 4, 2012 TyOutline

What is free energy and why is important Traditional QC approach Explicit sampling Enhanced sampling PLUMED plugin for free energy calculations

Tuesday, September 4, 2012 TyOutline

What is free energy and why is important Traditional QC approach Explicit sampling Enhanced sampling PLUMED plugin for free energy calculations

Tuesday, September 4, 2012 TyWhat is free energy and why is important

s= state descriptor (bound/unbound, reactant/products, F (s)= kBT ln P (s) phaseA/phaseB) −V (x) k T e− b δ(s(x) s)dx P (s)= − Q conformational transitions and equilibria ligand binding mechanism of transporters or channels chemical reactions and catalysis phase transitions .....

P (out) P (near) P (in)

Tuesday, September 4, 2012 TyOutline

What is free energy and why is important Traditional QC approach Explicit sampling Enhanced sampling PLUMED plugin for free energy calculations

Tuesday, September 4, 2012 TyTraditional QC approach

Rigid rotor/ harmonic approximation is well known [1]

F (s)=H(s) TS(s) − ∂ ln q (s)q (s)q (s)q (s) S(s)=R + R ln (q (s)q (s)q (s)q (s)) + RT t e r v t e r v ∂T trasl, elect, rot, vib partition functions where the partition functions are obtained from the calculated vibrational spectrum (Hessian), moments of inertia, electronic structure

H(s)=Hel(s)+Ht(s)+Hv(s)+Hr(s)

It requires electronic structure and Hessian in the local minima and transition states: 1 single conformation per state.

[1] McQuarrie, Simon “Physical Chemistry, a molecular approach” 6

Tuesday, September 4, 2012 Ty Finding the RC (I) Static approach: look for saddle points by using chemical intuition and TS optimization.

m1 m2

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Tuesday, September 4, 2012 Ty Finding the RC Static approach: look for saddle points by using chemical intuition and TS optimization.

Verify then by IRC: move forward and backward to verify to end in minima.

11108

Tuesday, September 4, 2012 Ty Finding the RC Static approach: look for saddle points by using chemical intuition and TS optimization.

m1 m2

Verify then by IRC: move forward and backward to verify to end in minima.

14109

Tuesday, September 4, 2012 Ty Finding the RC Static approach: look for saddle points by using chemical intuition and TS optimization.

m1 m2

Verify then by IRC: move forward and backward to verify to end in minima.

∆‡G = ∆‡H T ∆‡S − kbT ∆‡G/RT [2] k(T )= e− Rates may be derived from Eyring equation : hc0

[2] Eyring, Chem. Rev. 1935 vol. 17 pp 65 1014

Tuesday, September 4, 2012 TyTraditional QC approach

PRO: one calculation per point (3 single points calculations per rate and free energy differences) CONS: optimizing TS is not trivial (redundant coordinates), the landscape must be very simple (Free) energy ortho coor Reaction coor

Tuesday, September 4, 2012 TyProblems with traditional QC approach

Many problems are not smooth as before: many parallel valleys (conformers), solvent induced roughness

Tuesday, September 4, 2012 TyProblems with traditional QC approach

Many problems are not smooth as before: many parallel valleys (conformers), solvent induced roughness

A TS B

Perfectly parallel basins, Not too bad (small correction over static approaches). Typical case of symmetry in a reaction

Tuesday, September 4, 2012 TyProblems with traditional QC approach

Many problems are not smooth as before: many parallel valleys (conformers), solvent induced roughness

Tuesday, September 4, 2012 TyProblems with traditional QC approach

Many problems are not smooth as before: many parallel valleys (conformers), solvent induced roughness

Perfectly parallel basins but energetically asymmetric, Can be done with static QC but need to sample all the paths

Tuesday, September 4, 2012 TyProblems with traditional QC approach

Many problems are not smooth as before: many parallel valleys (conformers), solvent induced roughness

Tuesday, September 4, 2012 TyProblems with traditional QC approach

Many problems are not smooth as before: many parallel valleys (conformers), solvent induced roughness

Rough paths: vibrations are meaningless, paths from IRC are not representative of reactions: USE MD BASED METHODS 12

Tuesday, September 4, 2012 TyProblems with traditional QC approach

Many problems are not smooth as before: many parallel valleys (conformers), solvent induced roughness

Tuesday, September 4, 2012 TyOutline

What is free energy and why is important Traditional QM approach Explicit sampling Enhanced sampling PLUMED plugin for free energy calculations

Tuesday, September 4, 2012 TyThe ingredients of the probability picture

Free energy is connected to the probability of a given state to occur in a single measurement

F (s)= k T ln P (s) − B

Probability State descriptor V (x) k T (bond length, folded/ e− B δ(s(x) s)dx P (s)= V (x) − unfolded descr., liquid/ e− kB T dx crystal/amorphous)

Tuesday, September 4, 2012 TyExplicit sampling via molecular dynamics

F (s)= k T ln P (s) − B a real system a set of independent (cuvette) objects (ensemble) a b a a a a b a a a a a a a a a b b MD: evolve in time with Newton’s eq. of motion use ergodic hypothesis a b a a a a b a a F (s)= k T ln P (s) ∆t 2∆t 3∆t ... − B all the available states can be sampled with t->inﬁnite 15 Tuesday, September 4, 2012 a TyLimitations

Transition of 8 kcal/mol can take up to milliseconds at 300K

ν = ν exp ( β∆V ‡) 0 − 9 1 ν = 5 10 s− ∆V ‡ 1 0 k T 2 B Free Energy React Prod RC

Tuesday, September 4, 2012 TyLimitations

Transition of 8 kcal/mol can take up to milliseconds at 300K

ν = ν exp ( β∆V ‡) 0 − 9 1 ν = 5 10 s− ∆V ‡ 1 0 k T 2 B Free Energy React Prod RC MD (DFT) can reach hundreds of ps. Lucky if you see one single event!

Tuesday, September 4, 2012 TyLimitations

Transition of 8 kcal/mol can take up to milliseconds at 300K

ν = ν exp ( β∆V ‡) 0 − 9 1 ν = 5 10 s− ∆V ‡ 1 0 k T 2 B Free Energy React Prod RC MD (DFT) can reach hundreds of ps. Lucky if you see one single event! property time

Tuesday, September 4, 2012 TyLimitations

Transition of 8 kcal/mol can take up to milliseconds at 300K

ν = ν exp ( β∆V ‡) 0 − 9 1 ν = 5 10 s− ∆V ‡ 1 0 k T 2 B Free Energy React Prod RC MD (DFT) can reach hundreds of ps. Lucky if you see one single event! 100 ps 100 property time

Tuesday, September 4, 2012 TyLimitations

Transition of 8 kcal/mol can take up to milliseconds at 300K

ν = ν exp ( β∆V ‡) 0 − 9 1 ν = 5 10 s− ∆V ‡ 1 0 k T 2 B Free Energy React Prod RC MD (DFT) can reach hundreds of ps. Lucky if you see one single event! 100 ps 100 property time

If you want thermodynamics you must average over events

Tuesday, September 4, 2012 TyLimitations

Transition of 8 kcal/mol can take up to milliseconds at 300K

ν = ν exp ( β∆V ‡) 0 − 9 1 ν = 5 10 s− ∆V ‡ 1 0 k T 2 B Free Energy React Prod RC MD (DFT) can reach hundreds of ps. Lucky if you see one single event! 100 ps 100 property time

If you want thermodynamics you must average over events property time

Tuesday, September 4, 2012 TyLimitations

Transition of 8 kcal/mol can take up to milliseconds at 300K

ν = ν exp ( β∆V ‡) 0 − 9 1 ν = 5 10 s− ∆V ‡ 1 0 k T 2 B Free Energy React Prod This is not RCyour solution MD (DFT) can reach hundreds of ps. Lucky if you see one single event! 100 ps 100 property time

If you want thermodynamics you must average over events property time

Tuesday, September 4, 2012 TyOutline

What is free energy and why is important Traditional QM approach Explicit sampling Enhanced sampling PLUMED plugin for free energy calculations

Tuesday, September 4, 2012 TyThe ancestor: Torrie & Valleau[3]

Free energy: Free energy for a biased system:

use biasing potential to overcome barriers and retrieve free energies

[3] Torrie, Valleau J. Comp. Phys, 1977 vol. 23, pp 187 18

Tuesday, September 4, 2012 TyLessons from Torrie & Valleau

Biasing potentials:

1 k T 2 B Free Energy React Prod RC

[3] Torrie, Valleau J. Comp. Phys, 1977 vol. 23, pp 187 19

Tuesday, September 4, 2012 TyLessons from Torrie & Valleau

Biasing potentials:

biasing pot

1 k T 2 B Free Energy React Prod RC

[3] Torrie, Valleau J. Comp. Phys, 1977 vol. 23, pp 187 19

Tuesday, September 4, 2012 TyLessons from Torrie & Valleau

Biasing potentials: Metastabilities are removed biasing pot 1 k T 2 B resulting potential

1 k T 2 B Free Energy React Prod Free Energy React Prod RC RC

[3] Torrie, Valleau J. Comp. Phys, 1977 vol. 23, pp 187 19

Tuesday, September 4, 2012 TyLessons from Torrie & Valleau

Biasing potentials: Metastabilities are removed biasing pot 1 k T 2 B resulting potential

1 k T 2 B Free Energy React Prod Free Energy React Prod RC RC

allow to measure free energies: useful allow to overcome barriers: cheap need only an additional term to standard DFT forces: easy

[3] Torrie, Valleau J. Comp. Phys, 1977 vol. 23, pp 187 19

Tuesday, September 4, 2012 TyAdaptive sampling methods

Adaptive Biasing Force Darve & Pohorille, J. Chem. Phys. 2001, vol. 115, pp. 9169.

Adaptive Umbrella Sampling Mezei, M. J Comput Phys 1987, vol. 68, pp. 237.

Self Healing Umbrella Sampling Marsili et. al J. Chem. Phys. B vol. 110, pp. 14011.

MetadynamicsLaio & Parrinello, PNAS 2002, vol. 20, pp. 12562.

Conformational Flooding Grubmüller, Phys Rev E 1995, vol. 52, pp. 2893. ...

Tuesday, September 4, 2012 AdaptiveTy sampling methods (e.g. metadynamics) without ES with ES

Φ Ψ from milli secs... free energy is faster ᄔ not black box to nano secs!

Tuesday, September 4, 2012 TyThermodynamic-Integration-like approaches

The ancestor is mean force calculation through harmonic potential F (s)= k T ln P (s) − B ∂F (s) 1 ∂ V (x) kB T = kBT V (x) e− δ(s(x) s)dx ∂s s0 − k T ∂s − s0 e− B δ(s(x) s)dx − Dirac delta function ! Gaussian function (I like this trick!)

V (x) k 2 ∂F (s) 1 ∂ (s(x) s) k T e− kB T e− 2kB T − dx B V (x) k 2 ∂s s0 − k T 2k T (s(x) s) ∂s s0 e− B e− B − dx It is an average of a biased simulation!

V (x)+k/2(s (x) s )2 ∂F (s) 1 − 0 kB T 2 k(s(x) s0)e− dx = k s(x) s0 bias,s0 V (x)+k/2(s (x) s0) ∂s s0 − − − − − e− kB T dx V (x)+k/2(s (x) s )2 ∂F (s) 1 − 0 kB T 2 k(s(x) s0)e− dx = k s(x) s0 bias,s0 V (x)+k/2(s (x) s0) ∂s s0 − − − − − (opposite of the “mean force”) e− kB T dx 22 Tuesday, September 4, 2012 TyThermodynamic Integration " 101

Make a constrained simulation (go over barrier!)