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

Transcription

1 1. Born-Oppenheimer approx.- energy surfaces 2. Mean-field (Hartree-Fock) theory- orbitals 3. Pros and cons of HF- RHF, UHF 4. Beyond HF- why? 5. First, one usually does HF-how? 6. Basis sets and notations 7. MPn, MCSCF, CI, CC, DFT 8. Gradients and Hessians 9. Special topics: accuracy, metastable states Jack Simons Henry Eyring Scientist and Professor Chemistry Department University of Utah 0

2 Often, we use so-called atomic units. Digression into atomic units. We let each (e and nuc.) coordinate be represented in terms of a parameter a 0 having units of length and a dimensionless quantity: r j a 0 r j ; R j a 0 R j The kinetic and potential energies in terms of the new dimensionless variables: T = -( 2 /2m e )(1/a 0 ) 2 j 2, V en = - Z K e 2 (1/a 0 ) 1/r j,k, V ee = e 2 (1/a 0 ) 1/r i,j Factoring e 2 /a 0 out from both the kinetic and potential energies gives T = e 2 /a 0 {-( 2 /2m e )(1/e 2 a 0 ) j2 } and V = e 2 /a 0 {-Z K /r j,k +1/r j,i } Choosing a 0 = 2 /(e 2 m e ) = Å = 1 Bohr, where m e is the electron mass, allows T and V to be written in terms of e 2 /a 0 = 1 Hartree = ev in a simple manner: T = -1/2 j 2 while V en = - Z K /r j,k and V ee = 1/r i,j 1

3 Let s now look into how we go about solving the electronic SE H 0 ψ Κ (r R) =E K (R) ψ Κ (r R) for one electronic state (K) at some specified geometry R. 2

4 There are major difficulties in solving the electronic SE. The potential terms V ee = Σ j<k=1,n e 2 /r j, k make the SE equation not separable- this means ψ Κ (r R) is not a product of functions of individual electron coordinates. ψ Κ (r R) φ 1 (r 1 ) φ 2 (r 2 ) φ Ν (r N ) (e.g., 1sα(1) 1sβ(2) 2sα(3) 2sβ(4) 2p1α(5) for Boron). This means that our desire to use spin-orbitals to describe ψ Κ (r R) is not really correct. We need to make further progress before we can think in terms of spin -orbitals, orbital energies, orbital symmetries, and the like. 3

5 The correct ψ Κ (r R) have certain properties that we need to know about so that, when we try to create good approximations to ψ Κ (r R), we can build these properties into our approximate functions. 1. Because P i,j H 0 = H 0 P i,j, the ψ Κ (r R) must obey P i,j ψ Κ (r R) = ± ψ Κ (r R) [-1 for electrons]. 2. Usually ψ Κ (r R) is an eigenfunction of S 2 (spin) and S z 3. ψ Κ (r R) has cusps near nuclei and when two electrons get close a. Near nuclei, the factors (- 2 /m e 1/r k / r k Z A e 2 / r k -R A ) ψ Κ (r R) will blow up unless / r k ψ = -m e Z A e2 / 2 ψ(as r k R A ). b. As electrons k and l approach, (-2 2 /m e 1/r k,l / r k,l +e 2 /r k,l ) ψ Κ (r R) will blow up unless / r k,l ψ = 1/2 m e e2 / 2 ψ(as r k,l 0) 4

6 Cusps / r k ψ = -m e Z A e2 / 2 ψ(as r k R A ) and / r k,l ψ = 1/2m e e2 / 2 ψ(as r k,l 0). Cusp near nucleus Cusp as two electrons approach The electrons want to pile up near nuclei and they want to avoid one another. 5

7 In the electronic kinetic energy, in addition to the terms like (- 2 /m e 1/r k / r k Z A e 2 /r k ) ψ Κ (r R) there are terms involving angular derivatives L 2 /2m e r 2 ψ Κ (r R) = - 2 /(2m e r 2 ) {(1/sinθ) / θ(sinθ / θ +(1/sinθ)2 2 / φ 2 } ψ Κ (r R) These terms will also blow up (for any state with L > 0) unless ψ Κ (r R) ψ Κ (r R) 0 (as r k R A ). So, for L = 0 states, one has / r k ψ = -m e Z A e2 / 2 ψ(as r k R A ), and for L > 0 states, both ψ Κ (r R) 0 (as r k R A ) and / r k ψ = -m e Z A e2 / 2 ψ(as r k R A ) hold, but the latter is useless because ψ Κ (r R) 0 anyway. This is why the cusp condition / r k ψ = -m e Z A e2 / 2 ψ(as r k R A ) is useful only for ground states. 6

8 This means when we try to approximately solve the electronic SE, we should use trial functions that have such cusps. Slater-type orbitals (exp(-ζr k )) have cusps at nuclei, but Gaussians (exp(-αr k2 )) do not. tight Gaussian orbital with cusp at r = 0 loose Gaussian d/dr(exp(-αr k2 ) = -2αr k (exp(-αr k2 ) =0 at r k = 0, d/dr(exp(-ζr k )= -ζ (exp(-ζr k ) = -ζ at r k = 0. medium Gaussian r So, sometimes we try to fit STOs by a linear combination of GTOs, but this can not fix the nuclear cusp problem of GTOs 7

9 It is very difficult to describe the ee cusp (Coulomb hole) accurately. Doing so is important because electrons avoid one another. We call this dynamical correlation. We rarely use functions with e-e cusps, but we should (this is called using explicit e-e correlation). 5 D T The coulomb hole for He in cc-pvxz (X=D,T,Q,5) basis sets with one electron fixed at 0.5 a 0 8

10 The nuclear cusps / r j ψ Κ = -m e Z A e 2 / 2 ψ Κ (as r j R A ) depend on Z. Given a wave function ψ Κ (r R), one can compute the electron density ρ(r) = N ψ * K (r,r 2,r 3...r N R) ψ K (r,r 2,r 3...r N R)dr 2 dr 3...dr N Using the cusp condition that ψ Κ obeys, one can show that / rρ(r) = -2m e Z A e2 / 2 ρ(r) (as r R A ). So, if we knew the ground-state ρ(r) and could find the locations of its cusps, we would know where the nuclei are located. If we also could measure the strength -2m e Ze 2 / 2 of the cusps, we would know the nuclear charges. If were to integrate ρ(r) over all values of r, we could compute N, the number of electrons. This observation that the exact ground-state ρ(r) can be used to find R, N, and the {Z K } and thus the Hamiltonian H 0 shows the origins of densityfunctional theory. 9

11 Addressing the non-separability problem and the permutational and spin symmetries: If V ee could be replaced (or approximated) by a one-electron additive potential V MF = Σ V MF (r j R) each of the solutions ψ Κ (r R) would be a product (an antisymmetrized product called a Slater determinant) of functions of individual electron coordinates (spin -orbitals) φ j (r R): A spinorbital is product of spin-and spatial function ϕ kα (r j ) = ϕ k (r j )α j ψ ( r, r,..., r ) = 1 2 n 1 = n! P p( P) ( 1 ) P{ ϕ1 α (1) ϕ2 α (2)... ϕnβ ( n) } O{ ϕ1 αϕ2α... ϕnβ} ϕ kβ (r j ) = ϕ k (r j )β j ϕ kα ϕ kβ = dvϕ * k (r)ϕ k (r) α β 10

12 Is there any optimal way to define V MF = Σ V MF (r j R)? Does such a definition then lead to equations to determine the optimal spin-orbitals? Yes! It is the Hartree-Fock definition of V MF. Before we can find potential V MF and the Hartree-Fock (HF) spin-orbitals φ j (r)(α or β), we need to review some background about spin and permutational symmetry. 11

13 A brief refresher on spin S Z α =1/2 α S Z β = 1/2 β S 2 α = 2 1/2(1/2 +1)α = 3/4 2 α S 2 β = 2 1/2(1/2 +1)β = 3/4 2 β Special case of J 2 j,m >= 2 j( j +1) j,m > For acting on a product of spin-orbitals, one uses S S 2 = S S + + S 2 Z = S Z ( j) S = S ( j) Z + S Z j j < α α >=1 < α β >= 0 < β β >=1 S α = 1 2 (1 2 +1) 1 2 (1 1)β = β 2 S β = 0 Special case of J ± j,m >= j( j +1) m(m ±1) j,m ±1> Examples: S Zα(1)α(2) =1/2 α(1)α(2) +1/2 α(1)α(2) = α(1)α(2) S α(1)α(2) = β(1)α(2) + α(1)β(2) 12

14 Let s practice forming triplet and singlet spin functions for 2 e s. We always begin with the highest M S function because it is pure. α(1)α(2) S Z α(1)α(2) =1/2 α(1)α(2) +1/2 α(1)α(2) = α(1)α(2) S Z β(1)β(2) = 1/2 β(1)β(2) 1/2 β(1)β(2) = β(1)β(2) So, M S =1; has to be triplet So, M S =-1; has to be triplet S α(1)α(2) = β(1)α(2) + α(1)β(2) = 1(1+1) 1(1 1) S =1, M S = 0 > So, 1,0 >= 1 2 [α(1)β(2) + β(1)α(2)] This is the M S =0 triplet How do we get the singlet? It has to have M S = 0 and be orthogonal to the M S = 0 triplet. So, the singlet is 0,0 >= 1 [α(1)β(2) β(1)α(2)] 2 13

15 Slater determinants (P i,j ) in several notations. First, for two electrons. ψ (r 1,r 2 ) = 1 ϕ α (r 1 ) ϕ β (r 1 ) 2 ϕ α (r 2 ) ϕ β (r 2 ) Shorthand = 1 2 ϕ(r 1 )α(1) ϕ(r 2 )α(2) ϕ(r 1 )β(1) ϕ(r 2 )β(2) ψ (r 1,r 2 ) = ϕ α ϕ β ψ (r 1,r 2 ) = 1 ( 2 ϕ(r )α(1)ϕ(r )β(2) ϕ(r )β(1)ϕ(r )α(2) ) = ϕ(r 1 )ϕ 2 (r 2 ) 1 2 { α(1)β(2) β(1)α(2) } Symmetric space; antisymmetric spin (singlet) φ 1 αφ 2 α = 1 2 [φ 1α(1)φ 2 α(2) φ 1 α(2)φ 2 α(1)] = 1 2 [φ 1(1)φ 2 (2) φ 2 (1)φ 1 (2)]α(1)α(2) ψ (r 1,r 2 ) = ψ (r 2,r 1 ) 1 2 ϕ α (r 1 ) ϕ β (r 1 ) ϕ α (r 2 ) ϕ β (r 2 ) = 1 2 Antisymmetric space; symmetric spin (triplet) ϕ α (r 2 ) ϕ β (r 2 ) ϕ α (r 1 ) ϕ β (r 1 ) Notice the P i,j antisymmetry 14

16 More practice with Slater determinants ψ (r 1,r 2,...,r n ) = ϕ 1α,ϕ 2α...,ϕ nβ Shorthand notation for general case ψ (r 1,r 2,...,r n ) = ϕ 1α (1) ϕ 2α (1) ϕ nβ (1) ϕ 1α (2) ϕ 2α (2) ϕ nβ (2) ϕ 1α (n) ϕ 2α (n) ϕ nβ (n) Odd under interchange of any two rows or columns ψ ( r, r,..., r ) = 1 2 n 1 = n! P p( P) ( 1 ) P{ ϕ1 α (1) ϕ2 α (2)... ϕnβ ( n) } O{ ϕ1 αϕ2α... ϕnβ} P permutation operator ( 1) p( P) O = 1 1 p P n! P ( ) ( ) P antisymmetrizer The dfn. of the Slater determinant contains a N -1/2 normalization. parity ( p(p) least number of transpositions that brings the indices back to original order) 15

17 Example : Determinant for 3-electron system { } = P ij O ϕ 1 ϕ 2 ϕ 3 + P ijk ϕ ϕ ϕ i, j i, j,k { } = 1 ϕ 1 (1)ϕ 2 (2)ϕ 3 (3) ϕ 2 (1)ϕ 1 (2)ϕ 3 (3) ϕ 3 (1)ϕ 2 (2)ϕ 1 (3) ϕ 1 (1)ϕ 3 (2)ϕ 2 (3) 6 +ϕ 2 (1)ϕ 3 (2)ϕ 1 (3) + ϕ 3 (1)ϕ 1 (2)ϕ 2 (3) permutations 1, P 12, P 13, P 23, P 231, P 312 transpositions parity

18 The good news is that one does not have to deal with most of these complications. Consider two Slater determinants (SD). ( 1) p P Pϕ 1 (1)ϕ 2 (2)ϕ 3 (3)...ϕ N (N) ψ A = 1 N! ψ B = 1 N! P Q ( 1) qq Qϕ' 1 (1)ϕ' 2 (2)ϕ' 3 (3)...ϕ' N (N) Assume that you have taken t permutations 1 to bring the two SDs into maximal coincidence. Now, consider evaluating the integral j<k=1,n d1d2d3...dnψ A * [ f ( j) + g(i, j)]ψ B where f(i) is any one-electron operator (e.g., -Z A / r j -R A ) and g(i.j) is any two-electron operator (e.g., 1/ r j -r k ). This looks like a horrible task (N! x (N + N 2 ) x N! terms). 1. A factor of (-1) t will then multiply the final integral I 17

19 I = d1d2d3...dnψ A * [ f ( j) + g(i, j)]ψ B j<k=1,n ψ A = 1 ( 1) p P Pϕ 1 (1)ϕ 2 (2)ϕ 3 (3)...ϕ N (N) ψ B = 1 N! N! P 1. The permutation P commutes with the f + g sums, so Q ( 1) qq Qϕ' 1 (1)ϕ' 2 (2)ϕ' 3 (3)...ϕ' N (N) I = 1 N! dτ ( 1) p P ϕ * 1 (1)ϕ * 2 (2)ϕ * 3 (3)...ϕ * N (N)[ f ( j) + g(i, j)]pψ B P j<k=1,n 2. I = N! N! Pψ B = ( 1) p P ψ B and dτ ϕ * 1 (1)ϕ * 2 (2)ϕ * 3 (3)...ϕ * N (N)[ f ( j) + g(i, j)]ψ B P j<k=1,n = dτϕ * 1 (1)ϕ * 2 (2)ϕ * 3 (3)...ϕ * N (N)[ f ( j) + g(i, j)] ( 1) qq Qϕ' 1 (1)ϕ' 2 (2)ϕ' 3 (3)...ϕ' N (N) Now what? P ( 1) p P ( 1) p P = N! j<k=1,n Q so 18

20 I = dτϕ * 1 (1)ϕ * 2 (2)ϕ * 3 (3)...ϕ * N (N)[ f ( j) + g(i, j)] ( 1) qq Qϕ' 1 (1)ϕ' 2 (2)ϕ' 3 (3)...ϕ' N (N) j<k=1,n Four cases: the Slater-Condon rules you should memorize. Recall to multiply the final I by (-1) t ψ A and ψ B differ by three or more spin-orbitals: I = 0 ψ A and ψ B differ by two spin-orbitals-φ Ak φ Al ;φ Bk φ Bl I = dkdlϕ * Ak (k)ϕ * Al (l)g(k,l)[ϕ Bk (k)ϕ Bl (l) ϕ Bl (k)ϕ Bk (l)] Q ψ A and ψ B differ by one spin-orbital-φ Ak ;φ Bk I = j A,B ψ A and ψ B are identical I = k< j A dkdjϕ * Ak (k)ϕ * j ( j)g(k, j)[ϕ Bk (k)ϕ j ( j) ϕ j (k)ϕ Bk ( j)] + dkϕ * Ak (k) f (k)ϕ Bk (k) dkdjϕ * k (k)ϕ * j ( j)g(k, j)[ϕ k (k)ϕ j ( j) ϕ j (k)ϕ k ( j)] + k A dkϕ * k (k) f (k)ϕ k (k) 19

21 In HF theory, we approximate the true ψ(r R) in terms of a single Slater determinant. We then use the variational method to minimize the energy of this determinant with respect to the spin-orbitals appearing in the determinant. Doing so, gives us equations for the optimal spin-orbitals to use in this HF determinant. They are called the HF equations. A single Slater determinant ψ A = 1 ( 1) p P Pϕ 1 (1)ϕ 2 (2)ϕ 3 (3)...ϕ N (N) = ϕ 1 (1)ϕ 2 (2)ϕ 3 (3)...ϕ N (N) N! P can be shown to have a density ρ(r) equal to the sum of the densities of the spin-orbitals in the determinantρ(r) = ϕ * j (r)ϕ j (r) j 20

22 The fourth of the Slater-Condon rules allows us to write the expectation value of H 0 for a single-determinant wave function ψ A = 1 ( 1) p P Pϕ 1 (1)ϕ 2 (2)ϕ 3 (3)...ϕ N (N) = ϕ 1 (1)ϕ 2 (2)ϕ 3 (3)...ϕ N (N) N! P <ψ A * H 0 ψ A >= k< j A dkdjϕ * k (k)ϕ * j ( j) 1 [ϕ k (k)ϕ j ( j) ϕ j (k)ϕ k ( j)] r k, j + k A dkϕ * k (k)[ 1/2 2 (k) M Z M ]ϕ k (k) r k R M = < ϕ * k [ 1/2 2 k A + Z M r R M ] ϕ > k 1 < ϕ * k ϕ * j [ϕ k ϕ j ϕ j ϕ k ] > r k< j A 1,2 M 21

23 E = < ϕ * k [ 1/2 2 k A Z M r R M ] ϕ > + k M 1 < ϕ * k ϕ * j [ϕ k ϕ j ϕ j ϕ k ] > k< j A r 1,2 The integrals appearing here are often written in shorthand as Z M < ϕ * k [ 1/2 2 r R M ] ϕ >=< k f k > k M < ϕ * k ϕ * j 1 r 1,2 [ϕ k ϕ j ϕ j ϕ k ] >=< k, j k, j > < k, j j,k > (Dirac) < ϕ * k ϕ * j 1 r 1,2 [ϕ k ϕ j ϕ j ϕ k ] >= (k,k j, j) (k, j j,k)(mulliken) When we minimize E keeping the constraints <φ k φ j >=δ k,j, we obtain the Hartree-Fock equations fφ k (1) + [ φ * j (2) 1 φ j (2)d2φ k (1) φ * j (2) 1 φ k (2)d2φ j (1)] = ε k φ k (1) = h HF φ k (1) r 1,2 r 1,2 = fφ k (1) + [J j K j ] φ k (1) = fφ k (1) + V HF φ k (1) 22

24 Physical meaning of Coulomb and exchange operators and integrals: J 1,2 = φ* 1 (r)j 2 φ 1 (r)dr= φ 1 (r) 2 e 2 / r-r φ 2 (r ) 2 dr dr K 1,2 = φ* 1 (r)k 2 φ 1 (r)dr = φ 1 (r) φ 2 (r ) e 2 / r-r φ 2 (r) φ 1 (r )dr dr φ 2 (r') Overlap region φ 1 (r) 23

25 What is good about Hartee-Fock? It is by making a mean-field model that our (chemists ) concepts of orbitals and of electronic configurations (e.g., 1s α1s β 2s α 2s β 2p 1 α) arise. Another good thing about HF orbitals is that their energies ε K give approximate ionization potentials and electron affinities (Koopmans theorem). IP -ε occupied ; EA -ε unoccupied 24

26 Koopmans theorem- what orbital energies mean. N-electrons energy E HF = < ϕ * k [ 1/2 2 k=1,n N+1-electrons energy Z M r R M ] ϕ > k M + 1 < ϕ * k ϕ * j [ϕ k ϕ j ϕ j ϕ k ] > k< r 1,2 E HF = < ϕ * k [ 1/2 2 k=1,n +1 Z M r R M ] ϕ > k M + 1 < ϕ * k ϕ * j [ϕ k ϕ j ϕ j ϕ k ] > k< +1 r 1,2 Energy difference (neutral minus anion) if spin-orbital m is the N+1 st electron s ΔE = < ϕ * m [ 1/2 2 M Z M 1 ] ϕ m > < ϕ * m ϕ * j [ϕ m ϕ j ϕ j ϕ m ] > r R M r 1,2 But, this is just minus the expression for the HF orbital energy k=1,n ε m =< φ m h HF φ m >=< φ mk f φ m > + < φ m [J j K j ] φ m > 25

27 The sum of the orbital energies is not equal to the HF energy: So So But E HF = h HF φ k (1) = fφ k (1) + [J j K j ] φ k (1) =ε k φ k ε k =< φ k h HF φ k >=< φ k f φ k > + < φ k [J j K j ] φ k > =< φ k f φ k > + [< k, j k, j > < k, j j,k >] ε k = k=1,n k=1,n {< φ k f φ k > + [< k, j k, j > < k, j j,k >] } < ϕ * k [ 1/2 2 k=1,n Z M r R M ] ϕ > k M + 1 < ϕ * k ϕ * j [ϕ k ϕ j ϕ j ϕ k ] > k< r 1,2 So, the sum of orbital energies double counts the J-K terms. So, we can compute the HF energy by taking E HF =1/2 ε k +1/2 k=1,n < φ k f φ k > k=1,n 26

28 Orbital energies depend upon which state one is studying. So a π* orbital in the ground state is not the same as a π* in the ππ* state. ε k =< φ k h HF φ k >=< φ k f φ k > + < φ k [J j K j ] φ k > p a ε p feels 6 J and 3 K interactions ε a feels 5 J and 2 K interactions p q a b Occupied orbitals feel N-1 others; virtual orbitals feel N others. ε p feels 5 J and 2 K interactions ε p feels 6 J and 3 K interactions ε q feels 6 J and 3 K interactions ε a feels 5 J and 2 K interactions ε b feels 5 J and 2 K interactions ε b feels 6 J and 3 K interactions Occupied orbitals feel N-1 others; virtual orbitals feel N others. This is why occupied orbitals (for the state of interest) relate to IPs and virtual orbitals (for the state of interest) relate to EAs. 27

29 In summary, the true electronic wave functions have P i.j symmetry, nuclear and Coulomb cusps, and are not spin-orbital products or Slater determinants. However, HF theory attempts to approximate ψ(r R) as a single Slater determinant and, in so doing, to obtain a mean-field approximation to Σ j<k=1,n 1/r j,k in the form V HF = [J j K j ] To further progress, we need to study the good and bad of the HF approximation, learn in more detail how the HF equations are solved, and learn how one moves beyond HF to come closer and closer to the correct ψ(r R). 28