Supporting Information Part 1. DFTB3: Extension of the self-consistent-charge. density-functional tight-binding method (SCC-DFTB)

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1 Supportng Informaton Part 1 DFTB3: Extenson of the self-consstent-charge densty-functonal tght-ndng method SCC-DFTB Mchael Gaus, Qang Cu, and Marcus Elstner, Insttute of Physcal Chemstry, Karlsruhe Insttute of Technology, Kaserstr. 12, Karlsruhe, Germany, and Department of Chemstry and Theoretcal Chemstry Insttute, Unversty of Wsconsn, Madson, 1101 Unversty Avenue, Madson, Wsconsn E-mal: marcus.elstner@kt.edu γ-functon and Dervatve γ The γ-functon s descred n detal n refs 1,2 and when ntroducng the revatons α = 16 5 U a, β = 16 5 U, r = R R a t can e wrtten as for the γ h -functon see elow 1 r S f r 0,α β γ = 1 r S g r 0,α = β α r = 0 2 Karlsruhe Insttute of Technology Unversty of Wsconsn, Madson 1

2 f α,β,r = S f = e αr f α,β,r + e βr f β,α,r 3 S g = e αr gα,r 4 αβ 4 2α 2 β 2 2 β 6 3α 2 β 4 α 2 β 2 3 r gα,r = αr + 9α 2 r 2 + α 3 r r 5 Thus, we fnd for the dervatve for the dervatve of the γ h -functon see elow: γ = 16 5 γ = 16 5 Sf Sg r 0,α β r 0,α = β 5 16 r = 0 7 S f = e αr f α,β,r re αr f α,β,r + e βr f β,α,r 8 S g = e αr gα,r re αr gα,r 9 f α,β,r = β 6 + 3α 2 β 4 2α 2 β α3 β 4 α 2 β 2 4 r f β,α,r = 2β 3 α 3 β 2 α β 4 α 3 β 2 α 2 4 r gα,r = αr + 3α 2 r Note the symmetry of γ: γ = γ a and γ = γ a. Kohn-Sham Equatons To determne the MO coeffcents c µ of the LCAO ansatz ψ = a µ a c µ φ µ 13 2

3 approxmate Kohn-Sham equatons are derved y fndng the maxmum of the total energy wth respect to the coeffcent c δd, the ass functon δ eng located at atom d. Therefore, takng nto account the constrants ψ ψ d 3 r = 1 14 the respectve dervatve s gven y [ E DFTB3 j n j ε j µ a ν c µ j c ν j S µν 1 ] = 0 d,δ d,. 15 The total energy expresson E DFTB3 = E H0 + E γ + E Γ + E rep = c µ c ν Hµν ν n q a q γ µ a q 2 a q Γ + E rep 16 as derved n the manuscrpt can now e nserted, and the Mullken charge analyss for estmatng the charge fluctuatons q a = q a q 0 a can e employed q a = n j j µ a ν c µ j c ν j S µν 17 where S µν = φ µ φ ν are the overlap matrx elements. Takng advantage of the symmetry of H 0 µν = H 0 νµ and S µν = S νµ eq 15 can e formed to 2 ν n c ν Hδ 0 d ν + Eγ + EΓ 2n ε ν c ν S δd ν = 0 18 E γ E Γ = 1 2 q a q γ = 1 2 = 1 3 q 2 a q Γ = 1 3 q a q γ + γ a 19 q a q 2 q a Γ + q Γ a 20 3

4 The dervatve of the Mullken charge wth respect to c δd can e wrtten as q a = δ ad c ν c n c ν S δd ν + n c µ S δd µ. 21 µ a Thus, eqs 19 and 20 expand to E γ E Γ = 1 2 q γ d + γ d + γ c + γ c n c ν S δd ν 22 c ν c = 1 3 q 2 q d Γ d + q Γ d + 2 q c Γ c + q Γ c n c ν S δd ν. 23 c ν c Dvdng eq 18 y 2n, comnng wth eqs 22 and 23, and renamng ndces gves ν c ν Hµν ε S µν = 0, a, µ a, 24 H µν 1 = Hµν 0 + S µν q c c 4 γ ac + γ ca + γ c + γ c q aγ ac + q Γ c + q c 6 Γ ca + Γ c a,, µ a,ν. 25 Takng advantage of the symmetry γ = γ a note, that Γ Γ a s not symmetrc!, and renamng the ndces results n ν c ν Hµν ε S µν = 0, a, µ a, 26 H µν 1 = Hµν 0 + S µν q c c 2 γ ac + γ c q aγ ac + q Γ c + q c 6 Γ ca + Γ c a,, µ a,ν 27 4

5 Total Energy The total energy eq 16 as derved n the manuscrpt can also e express the energy n terms of ε usng eqs 13, 14, 24, 25, the symmetry S µν = S νµ, and the defnton of q a eq 17: n ε = µ a ν = µ a ν n c µ c ν H µν n c µ c ν Hµν 0 + µ a ν n c µ c ν 1 S µν q c c 4 γ ac + γ ca + γ c + γ c q aγ ac + q Γ c + q c 6 Γ ca + Γ c = µ a ν 1 n c µ c ν Hµν 0 +q a q c ac 2 γ ac + γ ca q aγ ac q cγ ca 28 Thus, nsertng eq 28 nto eq 16 and usng q a = q a q 0 a, the total energy can e wrtten as E DFTB3 = 1 n ε q a q c ac 2 γ ac + γ ca q aγ ac q cγ ca q a q γ q 2 a q Γ + E rep = n ε 1 2 q q a γ a + q 0 aγ 1 3 q a q Γ q a + q 0 a 1 3 q a q 2 Γ a + E rep 29 Consderng the symmetry γ = γ a the total energy can e smplfed to E DFTB3 = n ε 1 2 q a + q 0 a q γ 1 3 q a + q 0 a q a q Γ 1 3 q a q 2 Γ a + E rep. 30 Forces An analytcal force equaton s derved for a Cartesan coordnate system usng the dervatve of the total energy wth respect to the atomc coordnates R, whle sujectng to the constrants eq 14. Wth k, the respectve atom ndex, and an ndex x {1,2,3} for the Cartesan coordnate we can 5

6 wrte usng eq 13: F = [ E DFTB3 n ε µ a ν c µ c ν S µν 1 ] k, x 31 The energy depends explctely on the atomc coordnates va S µν, H 0 µν, and γ and mplctly va the coeffcents c µ : E DFTB3 = E DFTB3 c µ R,R 32 de = dr c µ R,R dc µ + EDFTB3 µ dc µ dr 33 de DFTB3 Because of the varatonal prncple see eq 15 the mplct dependence va the coeffcents c µ,.e., the frst term of eq 32 s equal to zero. The second term,.e., the explct dependence of the energy wth respect to the coordnates EDFTB3 Insertng needs to e carred out and s shown n the followng. E DFTB3 = E H0 + E γ + E Γ + E rep 34 nto eq 31 gves F = [ = F H0 + Fγ + FΓ E H0 + E γ + E Γ + E rep + Frep n ε µ a ν c µ c ν S µν 1 ] + Fnorm k, x, 35 6

7 where F H0 = µ a ν F γ = 1 2 q a q γ = 1 2 n c µ c ν H 0 µν γ qa q + q a q = µ a ν 1 2 γ q a q n c µ c ν H 0 µν 36 = F γ1 + F γ2 37 F Γ = F rep = 1 3 = q 2 a q Γ q a Γ 2 q a q + q a q 1 3 q 2 Γ a q q a q 2 q a Γ + q Γ a 1 3 q 2 Γ a q = F Γ1 + F Γ2 38 = Erep F norm = V = rep 39 n ε µ a ν c µ c ν S µν 1 = µ a ν n ε c µ c ν. 40 Because H 0 µν = H 0 νµ, H 0 µν = H0 νµ, and H 0 µν wth µ a and ν s only dependent on the coordnates R a and R, 1 and for a = t s not dependent on any coordnate 1 F H0 to can e smplfed F H0 = = µ a ν k µ k ν µ k ν k = 2 Hµν 0 n c µ c ν Hµν 0 n c µ c ν Hµν 0 n c µ c ν µ a ν k µ a ν k k µ a ν k Hµν 0 n c µ c ν Hµν 0 n c µ c ν Hµν 0 n c µ c ν 41 7

8 The same apples for F norm whch can e smplfed to F norm = µ a ν n ε c µ c ν = 2 µ a ν k n ε c µ c ν 42 The dervatve of the atomc charge wth respect to an atomc coordnate s gven y q q k = µ a ν k = µ k k ν = n c µ c ν 43 µ k k ν n c µ c ν + µ k ν k n c µ c ν n c µ c ν, 44 where the last recastng n latter equaton s due to the fact that the overlap of two ass functons centered at one atom does not change wth a change n the coordnate of that atom. Further the ass functons centered on one atom are orthogonal to each other. Takng advantage of the symmetry 8

9 S µν = S νµ and = S νµ we can rewrte F γ1 and F Γ1 see eqs 37 and 38. F γ1 F Γ1 = 1 2 q a q γ + γ a 1 q k 2 q γ k + γ k = 1 2 q γ + γ a µ a 1 2 q γ k + γ k ν k µ k c k ν c = 1 2 q γ + γ a + γ k + γ k µ a n c µ c ν n c µ c ν ν k n c µ c ν 45 = 1 3 q a q 2 q a Γ + q Γ a 1 q k 3 q 2 q k Γ k + q Γ k = µ k µ a ν k k ν n c µ c ν n c µ c ν c q 2 q a Γ + q Γ a q c 2 q k Γ kc + q c Γ ck = 1 3 q 2 q a Γ + q Γ a + 2 q k Γ k + q Γ k µ a ν k n c µ c ν 46 Takng advantage of the dervatves γ kk R eq 38 to = 0 and Γ kk R = 0 we can also expand F γ2 eq 37 and FΓ2 F γ2 = 1 2 a = 1 2 q k F Γ2 = 1 3 q a q k γ ak 1 2 q a γak + γ ka q 2 Γ ak a q k + k q k q γ k q 2 k q Γ k = 1 3 q Γ k ak Γ ka q a q a + q k Thus, the fnal expresson for the force can e shortly wrtten y nsterng eqs 37, 38, 41, 42, and 9

10 45-48 nto eq 35, usng γ = γ a and reorderng terms as F = µ a ν k n c µ c ν 2 H0 µν 2ε + q c γ ac + γ kc + 1 c 3 2 q aγ ac + q c Γ ca + 2 q k Γ kc + q c Γ ck γ q k ak q a 1 3 q Γ k ak Γ ka q a q a + q k Erep k,x. 49 The dervatves H0 µν and ntegrals H 0 µν and S µν. The force contruton Erep the representaton of E rep can e found n ref 3. are determned y takng the numercal dervatve of the tulated s also calculated analytcally. Detals out In the remander of ths susecton the analytcal expressons for γ k Wth eq 1, the defnton of Γ as dscussed n the manuscrpt Γ = γ q a Γ a = γ q Γ aa = γ aa q a = γ q 0 a q a wth a, q 0 a = γ U q 0 U q wth a, q 0 = 1 γ aa q 0 a 2 q a, q 0 a and Γ k are derved. 50 eq 7, and r k = R R k r ak = R k R a r k = 3 x=1 R x R 2 r ak = 3 x=1 R R ax 2 51 k = R x R r k ak = R R ax r ak 10

11 a,,k,x, we can wrte γ k = γ k k k γ ak = γ ak ak ak 52 Γ k = Γ k k k Γ ak = Γ ak ak ak, 53 and for easer wrtng we use α = 16 5 U a, β = 16 5 U, r = r, and yeld for the dervatve of the γ h -functon see elow γ = γ = 1 r 2 Sf 1 r 2 Sg r 0,α β r 0,α = β 54 0 r = 0 S f S g αr f α,β,r = e αe αr βr f β,α,r f α,β,r + e βe βr f β,α,r 55 αr gα,r = e αe αr gα,r 56 f α,β,r gα,r = β 6 3α 2 β 4 α 2 β 2 3 r 2 57 = 1 r 2 + 3α α3 r Functons f α,β,r and gα,r are defned n eqs 5 and 6, respectvely. Smlarly, Γ = Γ = γ q a = 2 γ + γ 2 U a q a q a = = 2 γ 59 q a The latter term on rght hand sde of the frst lne s equal to zero, ecause the Huard-dervatve s not dependent on any atomc coordnate. Thus, we need eqs 7, 54, and for the dervatve of the 11

12 γ h -functon see elow 2 γ = 2 S f 2 S g r 0,α β r 0,α = β 0 r = S f = 16 S f = 5 = 16 5 [e αr f α,β,rαr 1 α f α,β,r +e βr 2 f β,α,r β f β,α,r ] + 2 f α,β,r r f α,β,r 61 2 S g = 16 5 = 16 5 e αr S g = αr 1gα,r α gα,r + 2 gα,r r gα,r 62 2 f α,β,r 2 f β,α,r 2 gα,r = 12α3 β 4 α 2 β 2 4 r 2 63 = 12α3 β 4 β 2 α 2 4 r 2 64 = 3 8 α α2 r

13 The γ h -Functon The γ h = γh a functon for HX atom pars X {C, H, N, O, P, S} takes the form γ h = 1 r S h 66 Ua +U ζ h = exp r where S = S f when α β and S = S g when α = β. S f and S g as well as α, β, and r are defned n eqs 3 and 4. Eqs 7, 54, and 60 then change to: γ h γ h 2 γ h = S = h + S h = 1 S r 2 h + S h 2 S h + S h + S h + S 2 h wth h = ζ r2 Ua +U ζ 1 h Ua +U ζ h 72 h = 2r 2 h = ζ r 2 Ua +U ζ 1 2 r 2 Ua +U 2 ζ 1 h. 73 References 1 Elstner, M.; Porezag, D.; Jungnckel, G.; Elsner, J.; Haugk, M.; Frauenhem, T.; Suha, S.; Sefert, G. Phys. Rev. B 1998, 58, Elsner, J. Ph.D. thess, Unverstät-Gesamthochschule Paderorn, Gaus, M.; Chou, C..; Wtek, H.; Elstner, M. J. Phys. Chem. A 2009, 113,