Ch121a Atomic Level Simulations of Materials and Molecules

© copyright 2012 William A. Goddard III, all rights reservedCh120a-Goddard-L03 1

Ch121a Atomic Level Simulations of Materials and Molecules

William A. Goddard III, [email protected] and Mary Ferkel Professor of Chemistry,

Materials Science, and Applied Physics, California Institute of Technology

Room BI 115Lecture: Monday, Friday 2-3pm

Lab Session: Wednesday 2:30-3:30pmExtra help: Monday 3-3:30pm and Wednesday 2-

2:30pm

Lecture 3, April 8, 2013FF1: valence, QEq, vdw


Ch121a Atomic Level Simulations of Materials and Molecules

William A. Goddard III, [email protected] and Mary Ferkel Professor of Chemistry,

Materials Science, and Applied Physics, California Institute of Technology

Room BI 115Hours: Monday, Wednesday, Friday 2-3pm

Lecture 3, April 8, 2013FF1: valence, QEq, vdw

TA’s Caitlin Scott and Andrea Kirkpatrick


Homework and Research ProjectFirst 5 weeks: The homework each week uses generally available computer software implementing the basic methods on applications aimed at exposing the students to understanding how to use atomistic simulations to solve problems. Each calculation requires making decisions on the specific approaches and parameters relevant and how to analyze the results. Midterm: each student submits proposal for a project using the methods of Ch121a to solve a research problem that can be completed in the final 5 weeks.The homework for the last 5 weeks is to turn in a one page report on progress with the projectThe final is a research report describing the calculations and conclusions


kj kj

k

ijijikl

lk

klki ik k reZ

re

ReZZ

mMH

2222

22

2

22

Born-Oppenheimer approximation, fix nuclei (Rk=1..M), solve for Ψel(1..N)

The full Quantum Mechanics Hamiltonian: Htotal(1..Nel,1..Mnuc)

nuclei electrons nucleus-nucleus

electron-nucleus

electron-electron

ijijikj kj

k

i rrZ 1

21 2

Hel(1..Nel) =

NM M M,NN

Hel(1..Nel) Ψel(1..Nel) = Eel Ψel(1..Nel)

Ψel(1..Nel) and Eel(1..Mnuc) are functions of the nuclear coordinates.

Next solve the nuclear QM problem

kj kj

k

ijijikl

lk

klki ik k reZ

re

ReZZ

mMH

2222

22

2

22

kj kj

k

ijijikl

lk

klki ik k reZ

re

ReZZ

mMH

2222

22

2

22

+Eel(1..Mnuc)Hnuc(1..Mnuc) =

Hnuc(1..Mnuc) Ψnuc(1..Mnuc) = Enuc Ψnuc(1..Mnuc)

EPE(1..Mnuc)FF is analytic fit to this


The Bending potential surface for CH2

3B1

1A1

1B1

3Sg-

1Dg

9.3 kcal/mol

linear

H H

C

Note that E(p) E(p)Symmetric about p180°


Force Field

E Evalence Enonbond (Euser Econstraint)

Involves covalent bonds and the coupling between them (2-body, 3-body, 4-body, cross terms)

Evalence Ebond Eangle E torsion E inversion Eba Ebb' Eaa' Ebt Eat

kj kj

k

ijijikl

lk

klki ik k reZ

re

ReZZ

mMH

2222

22

2

22

= Eel(1..Mnuc)EPE(1..Mnuc)

The Force Field (FF) is an analytic fit to such expressions but formulated to be transferable so can use the same parameters for various molecules and solids

General form:

Ecross-terms

E nonbond = Evdw + Eelectrostatic

Involves action at a distance, long range interactions


Re = 0.9 – 2.2 ÅKe = 700 (Kcal/mol)/Å 2

Units: multiply by 143.88 to convert mdynes/ Å to (Kcal/mol)/Å 2

Simple Harmonic

Bond Stretch Terms: Harmonic oscillator form

R

ignore

Taylor series expansion about the equilibrium, Re, d = R-Re

E(d)=E(Re)+(dE/dR)e[d]+½(d2E/dR2)e[d]2+(1/6)(d3E/dR3)e[d]3 + O(d4)

ignore zero ignoreKb

Some force fields, CHARRM, Amber use k=2Ke to avoid multiplying by 2

Re

Problem: cannot break the bond, E ∞

Ground state wavefunction (Gaussian)0(d) (mw/pЋ) exp[- (mw/2Ћ)d2]

QM eigenstates, En = n + ½ , where n=0,1,2,…

E(R) = (ke/2)(R-Re)2,


Note that E(Re) = 0, the usual convention for bond termsDe is the bond energy in kcal/molRe is the equilibrium bond distancea is the Morse scaling parametercalculate from Ke and De

Bond Stretch Terms: Morse oscillator form

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

2 3 4 5 6

De

Re

RWant the E to go to a constant as break bond, popular form is Morse function

E = 0

where X = exp[-a(R-Re)] = exp[-(g/2)(R/Re -1)]

Evdw (Rij) = De[X2-2X]

a= sqrt(ke/2De), where ke = d2E/dR2 at R=Re (the force constant)


Morse Potential – get analytic solution for quantum vibrational levels

where X = exp[-a(R-Re)] = exp[-(g/2)(R/Re -1)]

EMorse(Rij) = De[X2-2X]

a = sqrt(ke/2De), where ke = d2E/dR2 at R=Re (the force constant)

Ev = hv0(n + 1/2 ) – [hv0(n + ½)]2/4De

n0 = (a/2p)sqrt(2De/m)

Level separations decrease linearly with level

En+1 – En = hv0 – (n+1)(hv0)2/2De

Write

En/hc = we(n + ½ ) - xewe(n + ½ )2

De

D0


Angle Bend Terms: cosine harmonic

I K

J

Expand E() as Fourier series.

Note only cos(n) since must

be symmetric about =0 and p

E() = C0 + C1 cos() + C2 cos(2) + C3 cos(3) + ….

E() ~ ½ Ch [cos - cos e]2

E=0 at = e

E’() = dE()/d = - Ch [cos - cos e] sin = 0 at = e

E”() = d2E()/d2 = - Ch [cos - cos e] cos + Ch (sin )2

k = E”(e) = Ch (sin e)2

Using cos(2) = 2 cos2 - 1


The Bending potential surface for CH2

3B1

1A1

1B1

3Sg-

1Dg

9.3 kcal/mol

linear

H H

C

Note that E(p) E(p)Symmetric about p180°


The energy is a maximum at = p.Thus energy barrier to “linearize” the molecule becomes

Ebarrier 1/2C 1 cos(o) 2 1/2K (1 cos(o)sin(o)

)2

E(p ) E(p )By symmetry the angular energy satisfies

E( ) E()This is always satisfied for the cosine expansion but

Barriers for angle term

A second more popular form is the Harmonic theta expansionE() = ½ K [ - e]2 However except for linear molecules this does NOT satisfy the symmetry relations, leading to undefined energies and forces for = 180° and 0°. This is used by CHARMM, Amber, ..


Angle Bend Terms for linear molecule,

I

KJ

If e = 180° then we write E() = K [1 + cos()]

since for = 180 – d this becomes

E(d) = K [1 + (-1 + ½ d2) ~ 1/2 K d2


For an Octahedral complex The angle function should have the form E() = C [1 - cos4] which has minima for = 0, 90, 180, and 270°With a barrier of C for = 45, 135, 225, and 315°Here the force constant is K = 16C = CP2 where P=4 is the periodicity, so we can write C=K/P2

Simple Periodic Angle Bend Terms


For an Octahedral complex The angle function should have the form E() = C [1 - cos4] which has minima for = 0, 90, 180, and 270°With a barrier of C for = 45, 135, 225, and 315°Here the force constant is K = 16C = CP2 where P=4 is the periodicity, so we can write C=K/P2

However if we wanted the minima to be at = 45, 135, 225, and 315° with maxima at = 0, 90, 180, and 270°then we want to use E() = ½ C [1 + cos4] Thus the general form is E() = (K/P2)[1 – (-1)Bcos4]Where B=1 for the case with a minimum at 0° and B=-1 for a maximum at 0°

Simple Periodic Angle Bend Terms


Dihedral barriers

HOOH

1.19 kcal/mol Trans barrier

7.6 kcal/mol Cis barrier

HSSH: φe ~ 92°

5.02 kcal/mol trans barrier

7.54 kcal/mol cis barrier

Part of such barriers can be explained as due to vdw and electrostatic interactions between the H’s.But part of it arises from covalent terms (the pp lone pairs on each S or O)This part has the form E(φ) = ½ B [1+cos(2φ)]Which is E=0 for φ=90,270° and E=B for φ=0,180°

φe ~ 111°


dihedral or torsion terms

Where each kn is in kcal/mol, n = 1, is the periodicity of the potential and

d=+1 the cis conformation is a minimumd=-1 the cis conformation is thea maximumInput data is kn and dn for each n.

E() 1/2[kn (1 dcos(n)]n1

p

IJ K

LE(φ) = C0 + C1 cos(φ) + C2 cos(2 φ) + C3 cos(3 φ) + ….which we write as

I K

Jφ LGiven any two bonds IJ and KL attached

to a common bond JK, the dihedral angle φ is the angle of the JKL plane from the IJK plane, with cis being 0


Consider the central CC bond in ButaneThere are nine possible dihedrals: 4 HCCH, 2 HCCC, 2 CCCH, and 1 CCCC.Each of these 9 could be written asE(φ) = ½ B [1 + cos(3 φ)], For which E=0 for φ = 60, 180 and 300°and E=B the rotation barrier for φ = 0, 120 and 240°.

IJ K

L


Consider the central CC bond in Butane

IJ K

L

For ethane get 9 HCCH terms.The total barrier in ethane is 3 kcal/mol but standard vdw and charges of +0.15 on each H will account for ~1 kcal/mol.Thus only need explicit dihedral barrier of 2 kcal/mol.

There are nine possible dihedrals: 4 HCCH, 2 HCCC, 2 CCCH, and 1 CCCC.Each of these 9 could be written asE(φ) = ½ B [1 – cos(3 φ)], For which E=0 for φ = 60, 180 and 300°and E=B the rotation barrier for φ = 0, 120 and 240°.


Consider the central CC bond in Butane

IJ K

L

Can do two ways. Incremental: treat each HCCH as having a barrier of 2/9 and add the terms from each of the 9 to get a total of 2 (Amber, CHARRM)Or use a barrier of 2 kcal/mol for each of the 9, but normalize by the total number of 9 to get a net of 2 (Dreiding)

There are nine possible dihedrals: 4 HCCH, 2 HCCC, 2 CCCH, and 1 CCCC.Each of these 9 could be written asE(φ) = ½ B [1 + cos(3 φ)], For which E=0 for φ = 60, 180 and 300°and E=B the rotation barrier for φ = 0, 120 and 240°.For ethane get 9 HCCH terms.The total barrier in ethane is 3kcal/mol but standard vdw and charges of +0.15 on each H will account for ~1 kcal/mol.Thus only need explicit dihedral barrier of 2 kcal/mol.


Twisting potential surface for etheneThe ground state (N) of ethene prefers φ=0º to obtain the highest overlap of the pp orbitals on each CThe rotational barrier is 65 kcal/mol. We write this as E(φ) = ½ B [1-cos2 φ)]. Since there are 4 HCCH terms, we calculate each using the full B=65, but divide by 4.

φ = 0° φ = 90°φ = -90°

T

N

The triplet excited state (T) prefers φ =90º to obtain the lowest overlap.


InversionsWhen an atom I has exactly 3 distinct bonds IJ, IK, and IL, it is often necessary to include an exlicit term in the force field to adjust the energy for “planarizing” the center atom I.

E() 1/2C(cos(w) cos(wo))2

I

K

LJ

Kw C sin2wo

Where the force constant in kcal/mol is

LJ

I

K

w Umbrella inversion:ω is the angle betweenThe IL axis and the JIK plane


Amber describes inversion using an improper torsion.

E() 1/2K cos[n( o)]

n=2 for planar angles (0=180°) and n=3 for the tetrahedralAngles (0 =120°).

Improper Torsion: is the angle between the JIL plane and the KIL planeL

J

I

K

AMBER Improper Torsion JILK


CHARMM Improper Torsion IJKLIn the CHARMM force field, inversion is defined as if it were a torsion.

E() 1/2K [ o]2

Improper Torsion: φ is the angle between the IJK plane and the LJK plane

L

JI

K

For a tetrahedral carbon atom with equal bonds this angles is φ=35.264°.


Bond Stretch

Angle bend

Torsion

Inversion

Typical ExpressionsDescription example

2

00 )]cos())][cos(sin(2/[ kEHarmonic-cosine

dihedral

2

00 )]cos())][cos(sin(/2/1[ wwwww kEUmbrella inversion

( )22

)( ob RRKRE Harmonic Stretch

Summary: Valence Force Field Terms

E(φ) = ½ B [1 – cos(3 φ)],


Coulomb (Electrostatic) Interactions

Most force fields use fixed partial charges on each nucleus, leading to electrostatic or Coulomb interactions between each pair of charged particles. The electrostatic energy between point charges Qi and Qk is described by Coulomb's Law as

EQ(Rik) = C0 Qi Qk /(e Rik)

where Qi and Qk are atomic partial charges in electron units

C0 converts units: if E is in eV and R in A, then C0 = 14.403If E is in kcal/mol and R in A, then C0 = 332.0637e = 1 in a vacuum (dielectric constant)

TA check numbers


How estimate charges?Even for a material as ionic as NaCl diatomic, the dipole moment a net charge of +0.8 e on the Na and -0.8 e on the Cl not the idealized charges of +1 and -1

We need a method to estimate such charges in order to calculate properties of materials.

A side note about units.

In QM calculations the unit of charge is the magnitude of the charge on an electron and the unit of length is the bohr (a0)

Thus QM calculations of dipole moment are in units of ea0 which we refer to as au. However the international standard for quoting dipole moment is the Debye = 10-10 esu A

Where m(D) = 2.5418 m(au)


QEq Electrostatics energy

ci

si

ci

oi

oi qRRJ &,,,Three universal parameters for each element:

1991: used experimental IP, EA, Ri; ReaxFF get all parameters from fitting QM

Keeping: i

i Qq

• Self-consistent Charge Equilibration (QEq)• Describe charges as distributed (Gaussian)• Thus charges on adjacent atoms shielded (interactions constant as R0) and include interactions over ALL atoms, even if bonded (no exclusions)• Allow charge transfer (QEq method)

( )

ji iiiiiijjiiji qJqrqqJqE

21),,(}{

Electronegativity (IP+EA)/2

Hardness (IP-EA)

interactions atomic

( ) lk

kir

rErflj

kiij QQQQrE

ij

ijklijklij

,,intJij

rij

1/rij

ri0

+ rj0

I

2


Charge dependence of the energy (eV) of an atom

E=0

E=-3.615

E=12.967

Cl Cl-Cl+

Q=0 Q=-1Q=+1

Harmonic fit

= 8.291 = 9.352

Get minimum at Q=-0.887Emin = -3.676

The 2nd order expansion in Q is ok


QEq: the optimum charges lead to Equilibrationthat is, equal chemical potential

Charge Equilibration for Molecular Dynamics Simulations;

A. K. Rappé and W. A. Goddard III; J. Phys. Chem. 95, 3358 (1991)

Including terms through 2nd order leads to

(2) (3)

Expand the energy of an atom in a power series of the net charge on the atom, E(Q)


QEq parameters


Interpretation of J, the hardness

Define an atomic radius as

H 0.84 0.74C 1.42 1.23N 1.22 1.10O 1.08 1.21Si 2.20 2.35S 1.60 1.63Li 3.01 3.08

RA0 Re(A2) Bond distance of

homonuclear diatomic

Thus J is related to the coulomb energy of a charge the size of the atom


The total energy of a molecular complex

Consider now a distribution of charges over the atoms of a complex: QA, QB, etc

Letting JAB(R) = the Coulomb potential of unit charges on the atoms, we can write

or

Taking the derivative with respect to charge leads to the chemical potential, which is a function of the charges

The definition of equilibrium is for all chemical potentials to be equal. This leads to


The QEq equations Adding to the N-1 conditions

The condition that the total charged is fixed (say at 0) leads to the condition

Leads to a set of N linear equations for the N variables QA.

AQ=X, where the NxN matrix A and the N dimensional vector A are known. This is solved for the N unknowns, Q.

We place some conditions on this. The harmonic fit of charge to the energy of an atom is assumed to be valid only for filling the valence shell.

Thus we restrict Q(Cl) to lie between +7 and -1 and

Q(C) to be between +4 and -4

Similarly Q(H) is between +1 and -1


The QEq Coulomb potential lawWe need now to choose a form for JAB(R) A plausible form is JAB(R) = 14.4/R, which is valid when the charge distributions for atom A and B do not overlapClearly this form as the problem that JAB(R) ∞ as R 0In fact the overlap of the orbitals leads to shielding The plot shows the shielding for C atoms using various Slater orbitals

And l = 0.5 Using RC=0.759a0

unshieldedshielded


QEq results for alkali halides


QEq for Ala-His-Ala

Amber charges in

parentheses


QEq for deoxy adenosine

Amber charges in

parentheses


QEq for polymers

Nylon 66

PEEK


Polarizable QEq

ci

si

ci

oi

oi qRRJ &,,,

Four universal parameters for each element:Get from QM

)||exp()()(

)||exp()()(2

2

23

23

si

si

si

si

ci

ci

ci

ci

rrQr

rrQrsi

ci

p

p

Allow each atom to have two charges:A fixed core charge (+4 for Si) with a Gaussian shapeA variable shell charge with a Gaussian shape but subject to displacement and charge transferElectrostatic interactions between all charges, including the core and shell on same atomAllow Shell to move with respect to core, to describe atomic polarizabilitySelf-consistent charge equilibration (QEq)


Noble gas dimers

Ar2

Re

De

s

Most popular form: Lennard-Jones 12-6E=A/R12 –B/R6

= De[12 – 26] where = R/ReNote that E is a min at Re, note the 2

All nonbonded atoms and molecules exhibit a very repulsive interaction at short distances due to overlap of electron pairs (Pauli Repulsion) and a weak attractive interaction scaling like 1/R6 at long R (London Dispersion. Together these are called van der Waals (vdW) interactions

A 2nd form for LJ 12-6 is 4 De[t12 – t6] where tR/sNote that E=0 at R = s, note the 1Here s = Re(1/2)1/6 =0.89 Re


van der Waals interaction

6, 8, 10,6 8 10( ...)ij ij ij

dispi j ij ij ij

C C CE

R R R

What do vdW interactions account for? Short range: Pauli Repulsion of overlapping electron pairs

Dipole-dipole Dipole-quadrupoleQuadruole-quadrupole

We usually neglect higher order (>R6) terms.

(where 0)

1or (where 9)

ijCRrep

i j

rep Ni j ij

E e C

E NR

Long range attraction London dispersion due to Instantaneous fluctuations in the QM charges

Pauli repulsion:

Born repulsion:


Lennard-Jones 12-6E(R)=A/R12 – B/R6

=Dvdw{1/12 – 2/6} where = R/Rvdw

Here Dvdw=well depth,

nd Rvdw = Equilibrium distance for vdw dimer

Alternative form E(R) = 4Dvdw{[svdw/R)12]- [svdw/R)6]}

Where s = point at which E=0 (inner wall, s ~ 0.89 Rvdw)

Dimensionless force constant k = {[d2E/d2]=1}/Dvdw = 72

The choice of 1/R6 is due to London Dispersion (vdw Attraction)However there is no special reason for the 1/R12 short range form (other that it saved computation time for 1950’s computers)LJ 9-6 would lead to a more accurate inner wall.

Popular vdW Nonbond Terms: LJ12-6

NonBond

R


Popular vdW Nonbond Terms- Exponential-6Buckingham or exponential-6E(R)= A e-CR - B/R6 which we write as

NonBond

R

where R0 = Equilibrium bond distance; = R/R0 is the scaled distanceDo=well depth z = dimensionless parameter related to force constant at R0

We define a dimensionless force constant as k = {(d2E/d2)=1}/D0 k = 72 for LJ12-6z = 12 leads to –D0/6 at long R, just as for LJ12-6z = 13.772 leads to k = 72 just as for LJ12-6A problem with exp-6 is that E- ∞ as R 0. To avoid this BioGraf, LinGraf, CeriusII calculate the inner maximum and reflect E about this point for smaller R so that E and E’ are continuous. When z>10 this point is well up the inner wall and not important


Converting between X-6 and LJ12-6

Usual convention: use the same R0 and D0 and set z = 12.

I recall that this leads to small systematic errors.

Second choice: require that the long range form of exp-6 be the same as for LJ12-6 (ie -2D0/6) and require that the inner crossing point be the same. This leads to

This was published in the Dreiding paper but I do not know if it has ever been used


Popular vdW Nonbond Terms: MorseNonBond

RE(R) = D0 {exp[-b(R-R0)] - 2 exp[(-(b/2)(R-R0)]}At R=R0, E(R0) = -D0

At R=∞, E(R0) = 0D0 is the bond energy in Kcal/molR0 is the equilibrium bond distanceb is the Morse scaling parameter

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

2 3 4 5 6

D0R0

I prefer to write this asE(R) = D0 [X2 - 2X] where X = exp[(a/2)(1-)]At R=R0, = 1 X = 1 E(R0) = -D0

At R=∞, X = 0 E(R0) = 0k = {(d2E/d2)=1}/D0 = a2/2Thus a = 12 k = 72 just as for LJ12-6Few theorists believe that Morse makes sense for vdw parameters since it does not behave as 1/R6 as R∞ However, the vdw curve matches 1/R6 only for R>6A, and for our systems there will be other atoms in between. So Morse is ok.


vdW combination rules

Generally the vdw parameters are provided for HH, CC, NN etc (diagonal cases) and the off-diagonal terms are obtain using combination rules

D0IJ = Sqrt(D0II D0JJ)

R0IJ = Sqrt(R0II R0JJ)

zIJ = (zIJ + zIJ)/2

Sometimes an arithmetic combination rule is used for vdw radii

R0IJ = (R0II + R0JJ)/2 but this complicates vdw calculations

(the amber paper claims to do this but the code uses geometric combinations of the radii)


1-2 and 1-3 Nonbond exclusions

For valence force fields, it is assumed that the bond and angle quantities include already the Pauli Repulsion and electrostatic contributions so that we do not want to include them again in the nonbond list.

Thus normally we exclude from the vdw and coulomb sums

the contributions from bonds (1-2) and

next nearest neighbor (1-3) interactions


1-4 Nonbond exclusionsThere is disagreement about 1-4 interactions. In fact the origin of the rotational barrier in ethane is probably all due to Pauli repulsion orthogonality effects. Thus a proper description of the vdW interactions between 1-4 atoms should account for the barrier. In fact it accounts only for 1/3 of the barrier. I believe that this is because the CH bond pairs are centered at the bond midpoint not on the H atoms as assumed in the vdw. Thus using atom centered vdw accounts for only part of the barrier necessitating an explicit dihedral term.Thus I believe that the1-4 vdw and electrostatic terms should be included. However some FF, such as Amber, CHARRM reduce the 1-4 interactions by a factor of 2. I do not know of a justification for this.

Ch121a Atomic Level Simulations of Materials and Molecules

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