1. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 1 Ch121a Atomic Level Simulations of Materials and Molecules William A. Goddard III, wag@wag.caltech.edu Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology BI 115 Hours: 2:30-3:30 Monday and Wednesday Lecture or Lab: Friday 2-3pm (+3-4pm) Teaching Assistants Wei-Guang Liu, Fan Lu, Jose Mendozq, Andrea Kirkpatrick Lecture 3, April 6, 2011 Force Fields – 1 valence, QEq, vdw

2. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 2 Born-Oppenheimer approximation, fix nuclei (R k=1..M ), solve for Ψ el (1..N) The full Quantum Mechanics Hamiltonian: H total (1..N el,1..M nuc ) nuclei electrons nucleus- nucleus electron- nucleus electron- electron H el (1..N el ) = N M M M,N N H el (1..N el ) Ψ el (1..N el ) = E el Ψ el (1..N el ) Ψ el (1..N el ) and E el (1..M nuc ) are functions of the nuclear coordinates. Next solve the nuclear QM problem +E el (1..M nuc ) H nuc (1..M nuc ) = H nuc (1..M nuc ) Ψ nuc (1..M nuc ) = E nuc Ψ nuc (1..M nuc ) E PE (1..M nuc ) FF is analytic fit to this

3. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 3 Force Field Involves covalent bonds and the coupling between them (2-body, 3- body, 4-body, cross terms) = E el (1..M nuc ) E PE (1..M nuc ) The Force Field (FF) is an analytic fit to such expressions but formulated to be transferable so can use the same terms for various molecules and solids General form: E cross-terms E nonbond = E vdw + E electrostatic Involves action at a distance, long range interactions

4. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 4 Re = 0.9 – 2.2 Å K e = 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, R e,  = R-R e E(  )=E(R e )+(dE/dR) e [  ]+½(d 2 E/dR 2 ) e [  ] 2 +(1/6)(d 3 E/dR 3 ) e [  ] 3 + O(    ignore zero ignore KbKb Some force fields, CHARRM, Amber use k=2K e to avoid multiplying by 2 Re Problem: cannot break the bond, E  ∞ Ground state wavefunction (Gaussian)    m  Ћ) exp[- (m  2Ћ)  2 ] eigenstates, E = + ½, where =0,1,2,… E(R) = (k e /2)(R-R e ) 2,

5. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 5 Note that E(R e ) = 0, the usual convention for bond terms D e is the bond energy in kcal/mol R e is the equilibrium bond distance  is the Morse scaling parameter calculate from K e and D e Bond Stretch Terms: Morse oscillator form D0D0 R0R0 R Want the E to go to a constant as break bond, popular form is Morse function E = 0 where X = exp[-  (R-R e )] = exp[-(  /2)(R/R e -1)] E vdw (R ij ) = D e [ X 2 -2 X ]  = sqrt(k e /2D e ), where k e = d 2 E/dR 2 at R=R e (the force constant)

6. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 6 Morse Potential – get analytic solution for quantum vibrational levels where X = exp[-  (R-R e )] = exp[-(  /2)(R/R e -1)] E vdw (R ij ) = D e [ X 2 -2 X ]  = sqrt(k e /2D e ), where k e = d 2 E/dR 2 at R=Re (the force constant) E v = hv 0 ( + 1/2 ) – [hv 0 ( + ½)] 2 /4D e 0 = (  /2  )sqrt(2D e /m) Level separations decrease linearly with level E +1 – E = hv 0 – ( +1)(hv 0 ) 2 /2D e Write En/hc =  e ( + ½ ) - x e  e ( + ½ ) 2

7. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 7 The Bending potential surface for CH 2 3B13B1 1A11A1 1B11B1 3g-3g- 1g1g 9.3 kcal/mol

8. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 8 Angle Bend Terms: cosine harmonic I K J  E(  ) = C 0 + C 1 cos(  ) + C 2 cos(2  ) + C 3 cos(3  ) + …. ~ ½ C h [cos  - cos  e ] 2 = 0 at  =  e E”(  ) = dE(  )/d  = - C h [cos  - cos  e ] sin  = 0 at  =  e E”(  ) = d 2 E(  )/d  2 = - C h [cos  - cos  e ] cos  + C h (sin  ) 2 = C h (sin  e ) 2 at  =  e Thus k  = C h (sin  e ) 2

9. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 9 The energy barrier to “linearize” the molecule becomes By symmetry the angular energy satisfy This is always satisfied for the cosine expansion but Barriers for angle term A second more popular form is the Harmonic theta expansion E(  ) = 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,..

10. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 10 Angle Bend Terms for linear molecule, I K J  If  e = 180° then we write E(  ) = K  [1 + cos(  )] since for  = 180 –  this becomes E(  ) = K  [1 + (-1 + ½    K   

11. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 11 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= CP 2 where P=4 is the periodicity, so we can write C=K  /P 2 Simple Periodic Angle Bend Terms

12. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 12 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= CP 2 where P=4 is the periodicity, so we can write C=K  /P 2 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  /P 2 )[1 – (-1) B cos4  ] Where B=1 for the case with a minimum at 0° and B=-1 for a maximum at 0° Simple Periodic Angle Bend Terms

13. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 13 Rotational barriers HOOH 1.19 kcal/mol Trans barrier 7.6 kcal/mol Cis barrier HSSH: 5.02 kcal/mol trans barrier 7.54 kcal/mol cis barrier Part of these barriers can be explained as due to vdw and electrostatic interactions between the H’s. But part of it arises from covalent terms (the p  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°

14. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 14 dihedral or torsion terms Where each k n is in kcal/mol, n = 1, is the periodicity of the potential and d=+1 the cis conformation is a minimum d=-1 the cis conformation is thea maximum Input data is K and d for each n. I JK L E(φ) = C 0 + C 1 cos(φ) + C 2 cos(2 φ) + C 3 cos(3 φ) + ….which we write as I K J φ L Given 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

15. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 15 Consider the central CC bond in Butane There are nine possible dihedrals: 4 HCCH, 2 HCCC, 2 CCCH, and 1 CCCC. Each of these 9 could be written as E(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°. I JK L

16. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 16 Consider the central CC bond in Butane I JK L 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. There are nine possible dihedrals: 4 HCCH, 2 HCCC, 2 CCCH, and 1 CCCC. Each of these 9 could be written as E(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°.

17. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 17 Consider the central CC bond in Butane I JK 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 as E(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.

18. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 18 Twisting potential surface for ethene The ground state (N) of ethene prefers φ=0º to obtain the highest overlap with a rotational barrier of 65 kcal/mol for prefers φ =90º to obtain the lowest overlap. 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

19. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 19 Inversions When 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. I K L J Where the force constant in kcal/mol is L J I K Umbrella inversion: ω is the angle between The IL axis and the JIK plane

20. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 20 Amber describes inversion using an improper torsion. N=2 for planar angles (   =180°) and n=3 for the tetrahedral Angles (   =120°). Improper Torsion:  is the angle between the JIL plane and the KIL plane L J I K AMBER Improper Torsion JILK

21. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 21 CHARMM Improper Torsion IJKL In the CHARMM force field, inversion is defined as if it were a torsion. Improper Torsion: φ is the angle between the IJK plane and the LJK plane L J I K For a tetrahedral carbon atom with equal bonds this angles is φ=35.264°.

22. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 22 Bond Stretch Angle bend Torsion Inversion Typical ExpressionsDescriptionIllustration Harmonic-cosine Dreiding dihedral Umbrella inversion Harmonic Stretch Summary: Valence Force Field Terms

23. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 23 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 Q i and Q k is described by Coulomb's Law as E Q (R ik ) = C 0 Q i Q k /(  R ik ) where Q i and Q k are atomic partial charges in electron units C 0 converts units: if E is in eV and R in A, then C 0 = 14.403 If E is in kcal/mol and R in A, then C 0 = 332.0637  = 1 in a vacuum (dielectric constant) TA check numbers

24. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 24 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. We need a method to estimate such charges in order to calculate properties of materials. First a bit more 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 (a 0 ) Thus QM calculations of dipole moment are in units of ea 0 which we refer to as au. However the international standard for quoting dipole moment is the Debye = 10 -10 esu A Where  (D) = 2.5418  (au)

25. GODDARD - MSC/Caltech Berlin 3/22/04 25 ReaxFF Electrostatics energy – Critical element Three universal parameters for each element: 1991: used experimental IP, EA, R i ; ReaxFF get all parameters from fitting QM Keeping: Self-consistent Charge Equilibration (QEq) Describe charges as distributed (Gaussian) Thus charges on adjacent atoms shielded (interactions  constant as R  0) and include interactions over ALL atoms, even if bonded (no exclusions) Allow charge transfer (QEq method) Electronegativity (IP+EA)/2 Hardness (IP-EA) interactions atomic J ij r ij 1/r ij r i 0 + r j 0 I 2

26. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 26 Charge Equilibration Charge Equilibration for Molecular Dynamics Simulations; A.K. Rappé and W. A. Goddard III; J. Phys. Chem. 95, 3358 (1991) First consider how the energy of an atom depends on the net charge on the atom, E(Q) Including terms through 2 nd order leads to (2)(3)

27. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 27 Charge dependence of the energy (eV) of an atom E=0 E=-3.615 E=12.967 ClCl - Cl + Q=0Q=-1Q=+1 Harmonic fit = 8.291= 9.352 Get minimum at Q=-0.887 Emin = -3.676

28. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 28 QEq parameters

29. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 29 Interpretation of J, the hardness Define an atomic radius as H0.840.74 C1.421.23 N1.221.10 O1.081.21 Si2.202.35 S1.601.63 Li3.013.08 RA0RA0 R e (A 2 ) Bond distance of homonuclear diatomic Thus J is related to the coulomb energy of a charge the size of the atom

30. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 30 The total energy of a molecular complex Consider now a distribution of charges over the atoms of a complex: Q A, Q B, etc Letting J AB (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

31. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 31 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 Q A. 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

32. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 32 The QEq Coulomb potential law We need now to choose a form for J AB (R) A plausible form is J AB (R) = 14.4/R, which is valid when the charge distributions for atom A and B do not overlap Clearly this form as the problem that J AB (R)  ∞ as R  0 In fact the overlap of the orbitals leads to shielding The plot shows the shielding for C atoms using various Slater orbitals And = 0.5 Using R C =0.759a 0

33. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 33 QEq results for alkali halides

34. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 34 QEq for Ala-His-Ala Amber charges in parentheses

35. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 35 QEq for deoxy adenosine Amber charges in parentheses

36. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 36 QEq for polymers Nylon 66 PEEK

37. GODDARD - MSC/Caltech Berlin 3/22/04 37 Four universal parameters for each element: Get from QM Polarizable QEq Allow each atom to have two charges: A fixed core charge (+4 for Si) with a Gaussian shape A variable shell charge with a Gaussian shape but subject to displacement and charge transfer Electrostatic interactions between all charges, including the core and shell on same atom Allow Shell to move with respect to core, to describe atomic polarizability Self-consistent charge equilibration (QEq)

38. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 38 Noble gas dimers Ar 2 ReRe DeDe  Most popular form: LJ 12-6 E=A/R 12 –B/R 6 = D e [   12 – 2   6 ] where  = R/Re  D e [   12 –   6 ] where  R/  Here  = R e (1/2) 1/6 =0.89 R e 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/R 6 at long R (London Dispersion. Together these are called van der Waals (vdW) interactions

39. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 39 van der Waals interaction What vdW interactions account for?  Short range repulsion  Overlap of orbitals and the Pauli Principle Dipole-dipoleDipole-quadrupoleQuadruole-quadrupole We often neglect higher order (>R 6 ) terms.  Long range attraction  Instantaneous multipolar interaction (London dispersion) Pauli repulsion: Born repulsion:

40. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 40 Popular vdW Nonbond Terms: LJ12-6 NonBond R Lennard-Jones 12-6 E(R)=A/R 12 – B/R 6 =D vdw {1/  12 – 2/  6 } where  = R/R vdw = 4D vdw {[  vdw /R) 12 ]- [  vdw /R) 6 ]} D vdw =well depth, R vdw = Equilibrium distance for vdw dimer  = point at which E=0 (inner wall,  ~ 0.89 R vdw ) Dimensionless force constant  = {[d 2 E/d  2 ]  =1 }/D vdw = 72 The choice of 1/R 6 is due to London Dispersion (vdw Attraction) However there is no special reason for the 1/R 12 short range form (other that it saved computation time for 1950’s computers) LJ 9-6 would lead to a more accurate inner wall.

41. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 41 Popular vdW Nonbond Terms- Exponential-6 Buckingham or exponential-6 E(R)= A e -CR - B/R 6 which we write as NonBond R where  R 0 = Equilibrium bond distance;  = R/R 0 is the scaled distance D o =well depth  = dimensionless parameter related to force constant at R 0 We define a dimensionless force constant as  = {(d 2 E/d  2 )  =1 }/D 0  = 72 for LJ12-6  = 12 leads to –D 0 /  6 at long R, just as for LJ12-6  = 13.772 leads to  = 72 just as for LJ12-6 A 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  >10 this point is well up the inner wall and not important

42. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 42 Converting between X-6 and LJ12-6 Usual convention: use the same R 0 and D 0 and set  = 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 -2D 0 /  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

43. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 43 Popular vdW Nonbond Terms: Morse NonBond R E(R) = D 0 {exp[-  R-R 0 )] - 2 exp[(-(  R-R 0 )]} At R=R 0, E(R 0 ) = -D 0 At R=∞, E(R 0 ) = 0 D 0 is the bond energy in Kcal/mol R 0 is the equilibrium bond distance  is the Morse scaling parameter D0D0 R0R0 I prefer to write this as E(R) = D 0 [ X 2 - 2 X ] where X = exp[(  /2)(1-  )] At R=R 0,  = 1  X = 1  E(R 0 ) = -D 0 At R=∞, X = 0  E(R 0 ) = 0  = {(d 2 E/d  2 )  =1 }/D 0 =  2 /2 Thus  = 12   = 72 just as for LJ12-6 Few theorists believe that Morse makes sense for vdw parameters since it does not behave as 1/R 6 as R  ∞ However, the vdw curve matches 1/R 6 only for R>6A, and for our systems there will be other atoms in between. So Morse is ok.

44. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 44 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 D 0IJ = Sqrt(D 0II D 0JJ ) R 0IJ = Sqrt(R 0II R 0JJ )  IJ = (  IJ +  IJ )/2 Sometimes an arithmetic combination rule is used for vdw radii R 0IJ = (R 0II + R 0JJ )/2 but this complicates vdw calculations (the amber paper claims to do this but the code uses geometric combinations of the radii)

45. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 45 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 included in the nonbond interactions. Thus normally we exclude from the vdw and coulomb sums the contributions from bonds (1-2) and next nearest neighbor (1-3) interactions

46. © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L03 46 1-4 Nonbond exclusions There is disagreement about 1-4 interactions. In fact the original 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. 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.