Jack Simons, Henry Eyring Scientist and Professor Chemistry Department University of Utah

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1 1. Born-Oppenheimer pprox.- energy surfces 2. Men-field (Hrtree-Fock) theory- orbitls 3. Pros nd cons of HF- RHF, UHF 4. Beyond HF- why? 5. First, one usully does HF-how? 6. Bsis sets nd nottions 7. MPn, MCSCF, CI, CC, DFT 8. Grdients nd Hessins 9. Specil topics: ccurcy, metstble sttes Jck Simons, Henry Eyring Scientist nd Professor Chemistry Deprtment University of Uth 0

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3 Two issues: How does one compute g K nd H K,L nd wht do you do with them? Assume you hve g K vilble t some strting geometry X 0 = {X 1, X z, X 3N }. One cn ttempt to move downhill towrd locl-minimum by tking smll displcements δx K proportionl to, but in opposition to, the grdient g K long tht direction δx K = - g K. The energy E is then expected to chnge by δe = - K (g K ) 2. This is the most simple lgorithm for stepping downhill towrd minimum. The prmeter cn be used to keep the length of the step smll. A series of such steps from X 0 to X 0 + δx cn often led to minimum (t which ll 3N g K vlues vnish). 2

4 One problem with this pproch is tht, if one reches point where ll 3N g K vnish, one cn not be certin it is minimum; mybe it is first-, second-, or higher-order sddle point. Minimum: ll 3N g K vnish nd 3N-6 eigenvlues of the H K,L mtrix re positive. First-order sddle (trnsition stte TS): ll 3N g K vnish nd 3N-7 eigenvlues of the H K,L mtrix re positive; one is negtive. 3

5 So, one is usully forced to form H K,L nd find its 3N eigenvlues λ nd eigenvectors V k Σ L=1,3N H K,L V L = λ V k. 3 of the λ hve to vnish nd the 3 corresponding V k describe trnsltions of the molecule. 3 more (only 2 for liner molecules) of the λ hve to vnish nd the corresponding V k describe rottions of the molecule. The remining 3N-6 (or 3N-5) λ nd V k contin the informtion one needs to chrcterize the vibrtions nd rection pths of the molecule. 4

6 If one hs the grdient vector nd Hessin mtrix vilble t some geometry, δe = Σ K g K δx K + 1/2 Σ K,L H K,L δx K δx L Becuse the Hessin is symmetric, its eigenvectors re orthogonl nd they form complete set Σ K V K V K b = δ,b Σ V K V L = δ K,L. This llows one to express the tomic Crtesin displcements δx K in terms of displcements δv long the eigenmodes δx K = Σ L δ K,L δx L = Σ V K (Σ L V L δx L ) = Σ V K δv. 5

7 Inserting into gives where δx K = Σ V K δv. δe = Σ K g K δx K + 1/2 Σ K,L H K,L δx K δx L δε = Σ {g δv + 1/2 λ (δv ) 2 } g = Σ L V L g L This wy of writing δε llows us to consider independently mximizing or minimizing long ech of the 3N-6 eigenmodes. 6

8 Setting the derivtive of {g δv + 1/2 λ (δv ) 2 } with respect to the δv displcements equl to zero gives s suggested step δv = - g /λ Inserting these displcements into gives δε = Σ {g δv + 1/2 λ (δv ) 2 } δε = Σ {- g 2 /λ + 1/2 λ (-g /λ ) 2 } = -1/2Σ g 2 /λ. So the energy will go downhill long n eigenmode if tht mode s eigenvlue λ is positive; it will go uphill long modes with negtive λ vlues. Once you hve vlue for δv, you cn compute the Crtesin displcements from δx K = Σ V K δv 7

9 If one wnts to find minimum, one cn. Tke displcement δv = - g /λ long ny mode whose λ is positive. b. Tke displcement tht is smll nd of opposite sign thn - g /λ for modes with negtive λ vlues. The energy will then decrese long ll 3N-6 modes. Wht bout finding trnsition sttes? 8

10 Wht bout finding trnsition sttes? If one is lredy t geometry where one λ is negtive nd the 3N-7 other λ vlues re positive, one should. Visulize the eigenvector V k belonging to the negtive λ to mke sure this displcement mkes sense (i.e., looks resonble for motion wy from the desired trnsition stte). b. If the mode hving negtive eigenvlue mkes sense, one then tkes δv = - g /λ for ll modes. This choice will cuse δε = Σ {- g 2 /λ + 1/2 λ (-g /λ ) 2 } = -1/2Σ g 2 /λ to go downhill long 3N-7 modes nd uphill long the one mode hving negtive λ. Following series of such steps my llow one to locte the TS t which ll g vnish, 3N-7 λ re positive nd one λ is negtive. 9

11 At minimum or TS, one cn evlute hrmonic vibrtionl frequencies using the Hessin. The grdient (g L or g = Σ L V L g L vnishes), so the locl potentil energy cn be expressed in terms of the Hessin only. The clssicl dynmics Hmiltonin for displcements δx K is H = Σ K,L 1/2 H K,L δx K δx L + 1/2 Σ K m K (dδx K /dt) 2 Introducing the mss-weighted Crtesin coordintes δmwx K = (m K ) 1/2 δx K llows the Hmiltonin to become H = Σ K,L 1/2 MWH K,L δmwx K δmwx L + 1/2 Σ K (dδmwx K /dt) 2 where the mss-weighted Hessin is defined s MWH K,L = H K,L (m K m L ) -1/2 10

12 Expressing the Crtesin displcements in terms of the eigenmode displcements δx K = Σ V K δv llows H to become H = Σ {1/2 λ (δv ) 2 + 1/2(dδV /dt) 2 }. This is the Hmiltonin for 3N-6 uncoupled hrmonic oscilltors hving force constnts λ nd hving unit msses for ll coordintes. Thus, the hrmonic vibrtionl frequencies re given by ω = (λ ) 1/2 so the eigenvlues of the mss-weighted Hessin provide the hrmonic vibrtionl frequencies. At TS, one of the λ will be negtive. 11

13 It is worth pointing out tht one cn use mss-weighted coordintes to locte minim nd trnsition sttes, but the sme minim nd trnsition sttes will be found whether one uses Crtesin or mss-weighted Crtesin coordintes becuse whenever the Crtesin grdient vnishes g L = 0 the mss-weighted grdient will lso vnish g = Σ L V L g L = 0 12

14 To trce out rection pth strting t trnsition stte, one first finds the Hessin eigenvector {V K1 } belonging to the negtive eigenvlue. One tkes very smll step long this direction. Next, one re-computes the Hessin nd grdient (n.b., the grdient vnishes t the trnsition stte, but not once begins to move long the rection pth) t the new geometry X K + δx K where one finds the eigenvlues nd eigenvectors of the mss-weighted Hessin nd uses the locl qudrtic pproximtion δε = Σ {g δv + 1/2 λ (δv ) 2 } to guide one downhill. Along the eigenmode corresponding to the negtive eigenvlue λ 1, the grdient g 1 will be non-zero while the components of the grdient long the other eigenmodes will be smll (if one hs tken smll initil step). One is ttempting to move down strembed whose direction of flow initilly lies long V K 1 nd perpendiculr to which there re hrmonic sidewlls 1/2 λ (δv ) 2. 13

15 One performs series of displcements by ) moving (in smll steps) downhill long the eigenmode tht begins t V K 1 nd tht hs significnt grdient component g, b) while minimizing the energy (to remin in the strembed s bottom) long the 3N-7 other eigenmodes (by tking steps δv = - g /λ tht minimize ech {g δv + 1/2 λ (δv ) 2 }. As one evolves long this rection pth, one reches point where λ 1 chnges sign from negtive to positive. This signls tht one is pproching minimum. Continuing onwrd, one reches point where the grdient s component long the step displcement vnishes nd long ll other directions vnishes. This is the locl minimum tht connects to the trnsition stte t which the rection pth strted. One needs to lso begin t the trnsition stte nd follow the other brnch of the rection pth to be ble to connect rectnts, trnsition stte, nd products. 14

16 When trcing out rection pths, one uses the mss-weighted coordintes becuse dynmicl theories (e.g. the rection-pth Hmiltonin theory) re formulted in terms of motions in mss-weighted coordintes. The minim nd trnsition sttes one finds using mss-weighted coordintes will be the sme s one finds using conventionl Crtesin coordintes. However, the pths one trces out will differ depending on whether mss -weighting is or is not employed. 15

17 So, how does one evlute the grdient nd Hessin nlyticlly? For methods such s SCF, CI, nd MCSCF tht compute the energy E s E = <ψ H ψ>/<ψ ψ>, one mkes use of the chin rule to write E/ X K = Σ I E/ C I C I / X K + Σ iµ E/ C iµ C iµ / X K + <ψ H/ X K ψ>/<ψ ψ>. For MCSCF, E/ C I nd E/ C iµ re zero. For SCF E/ C iµ re zero nd E/ C I does not exist. For CI, E/ C I re zero, but E/ C iµ re not. So, for some of these methods, one needs to solve response equtions for E/ C iµ. 16

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20 When Crtesin Gussins χ,b,c (r,θ,φ) = N',b,c,α x y b z c exp(-αr 2 ) re used, the derivtives / X K χ ν (r) cn be done becuse X K ppers in (x-x K ) nd in r 2 = (x-x K ) 2 + (y-y K ) 2 + (z-z K ) 2. These derivtives give functions of one lower nd one higher ngulr momentum vlue. (from / X K (x-x K ) ) (from / X K exp(-αr 2 )) So, the AO integrl list must be extended to higher L-vlues. More troublesome re < / X K χ ν (r) χ η (r ) (1/ r-r ) χ µ (r) χ γ (r )> becuse there re now 4 times ( the originl plus / X K, / Y K, / Z K ) the number of 2-electron integrls. 19

21 When plne wve bsis functions re used, the derivtives / X K χ ν (r) = 0 vnish (nd thus don t hve to be delt with) becuse the bsis functions do not sit on ny prticulr nucler center. This is substntil benefit to using plne wves. 20

22 The good news is tht the Hellmnn-Feynmn nd integrl derivtive terms cn be evluted nd thus the grdients cn be computed s E/ X K = Σ I E/ C I C I / X K + Σ iµ E/ C iµ C iµ / X K + <ψ H/ X K ψ>/<ψ ψ> = <ψ H/ X K ψ>/<ψ ψ> for SCF or MCSCF wvefunctions. Wht bout CI, MPn, or CC wve functions? Wht is different? 21

23 E/ X K = Σ I E/ C I C I / X K + Σ iµ E/ C iµ C iµ / X K + <ψ H/ X K ψ>/<ψ ψ>. For CI, the E/ C I term still vnishes nd the <ψ H/ X K ψ>/<ψ ψ> term is hndled s in MCSCF, but the E/ C iµ terms do not vnish For MPn, one does not hve C I prmeters; E is given in terms of orbitl energies ε j nd 2-electron integrls over the φ j. For CC, one hs t i,j m,n mplitudes s prmeters nd E is given in terms of them nd integrls over the φ j. So, in CI, MPn, nd CC one needs to hve expressions for These re clled response equtions. C iµ / X K nd for t i,j m,n / X K. 22

24 The response equtions for C iµ / X K re obtined by tking the / X K derivtive of the Fock equtions tht determined the C i,µ / X K Σ µ <χ ν h e χ µ > C J,µ = / X K ε J Σ µ <χ ν χ µ > C J,µ This gives Σ µ [<χ ν h e χ µ >- ε J <χ ν χ µ >] { / X K C J,µ }= Σ µ / X K [<χ ν h e χ µ >- ε J Σ µ <χ ν χ µ >] C J,µ Becuse ll the mchinery to evlute the terms in exists s does the mtrix one cn solve for / X K C J,µ / X K [<χ ν h e χ µ >- ε J Σ µ <χ ν χ µ >] <χ ν h e χ µ >- ε J <χ ν χ µ >, 23

25 A similr, but more complicted, strtegy cn be used to derive equtions for the t i,j m,n / X K tht re needed to chieve grdients in CC theory. The bottom line is tht for MPn, CI, nd CC, one cn obtin nlyticl expressions for g K = E/ X K. To derive nlyticl expressions for the Hessin 2 E/ X K X L is, of course, more difficult. It hs been done for HF nd MCSCF nd CI nd my exist (?) for CC theory. As you my expect it involves second derivtives of 2 -electron integrls nd thus is much more expensive. 24

26 There re other kinds of responses tht one cn seek to tret nlyticlly. For exmple, wht if one dded to the Hmiltonin n electric field term such s Σ k=1,n er k E + Σ =1,M e Z R E rther thn displcing nucleus? So, H is now H + Σ k=1,n er k E + Σ =1,M e Z R E. The wvefunction ψ(e) nd energy E(E) will now depend on the electric field E. de/de = Σ I E/ C I C I / E+ Σ iµ E/ C iµ C iµ / E +<ψ H/ E ψ>/<ψ ψ>. Here, <ψ H/ E ψ>/<ψ ψ> = <ψ Σ k=1,n er k + Σ =1,M e Z R ψ>, is the dipole moment expecttion vlue. This is the finl nswer for HF nd MCSCF, but not for MPn, CI, CC. For these cses, we lso need C I / E nd C iµ / E response contributions. So, the expecttion vlue of the dipole moment opertor is not lwys the correct dipole moment! 25

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