SCF calculation on HeH +

Transcription

1 SCF calculation on HeH + Markus Meuwly Department of Chemistry, University of Basel, Basel, Switzerland Abstract This document describes the main steps involved in carrying out a SCF calculation on the two-electron system HeH +. The text follows the procedures outlined in Szabo and Ostlund in Modern Quantum Chemistry and is directly related to the computer code hehp.f provided in the lectures. 1

2 Procedures The calculations are carried out with a minimal basis set: STO-3G. For an SCF calculation one first chooses a geometry of the nuclei which are located at R 1 and R 2 with an internuclear distance R = R 1 R 2. In the program R = a 0 is used. Next, the basis functions for the H and He atoms need to be specified. Here, a contraction (sum) of 3 Gaussian orbitals approximating the Slater orbital of H and He are used. The two Slater functions are φ 1 (ξ 3 1 /π)1/2 exp ξ 1 r R 1 φ 2 (ξ 3 2 /π)1/2 exp ξ 2 r R 2 (1) (2) where φ 1 describes the electron of He and φ 2 that of H. The standard Slater exponent ξ 2 for H in a molecule is ξ 2 = 1.24 a 1 0 and for He a variational calculation together with a correction for the positive charge of the molecule yields ξ 1 = a 1 0. Both Slater functions in Eqs. 2 are now represented as a sum (i.e. contraction ) over 3 Gauss functions φ CGF = 3 p=1 d pµφp GF (α pµ, r R a ) (therefore STO-3G) with φp GF (α pµ, r R a )=(2α pµ /π) 3/4 exp α pµ r R a 2. They are for H atom (including the normalization of (2α/π) 3/4 ): φ CGF = φ GF and for the He atom: (α = ) φ GF GF (α = ) φ (α = ) (3) φ CGF = φ GF (α = ) φ GF GF (α = ) φ (α = ) In a next step the necessary integrals over the basis functions are evaluated. Here the formulae involving functions are summarized. What is required are a) overlap integrals (2- center overlap), b) electron kinetic energy integrals, c) nuclear attraction and d) two-electron (4) 2

3 repulsion. The reason why Gaussian functions are used is the fact that the product of two Gaussian functions centered at R A and R B is again a Gaussian centered at R P. With g ( r R A ) = exp α r R A 2 one obtains g ( r R A )g ( r R B ) = exp α r R A 2 exp β r R B 2 = Kg ( r R P ) (5) with K = exp [ αβ/(α+β) R A R B 2 ] (6) and R P = (α R A + β R B )/(α + β) (7) Integrals a) and b) above can be evaluated using standard integrals over Gauss functions, c) and d) use Fourier transform techniques. The standard integral required is 0 drr 2 exp pr2 = (π/p) 3/2 /(4π) (8) With this, the overlap integral a) can be evaluated as (A B) = d r 1 g ( r 1 R A )g ( r 1 R B ) = [π/(α + β)] 3/2 exp [αβ/(α+β) R A R B 2 ] (9) and the kinetic integral b) yields (A B) = αβ 2αβ [3 (α + β) (α + β) R A R B 2 π ][ (α + β) ]3/2 exp [αβ/(α+β) R A R B 2 ] (10) For integrals c) and d) the following expressions are obtained: (A Z C B) = 2π r 1C α + β Z C exp [ αβ α+β R A R B 2 ] Fo [(α + β) R P R C 2 ] (11) 3

4 where F o is the error function F o (t) = 1 2 (π t )1/2 er f(t 1/2 ) (12) and (AB CD) = 2π 5/2 /[(α + β)(γ + δ)(α + β + γ + δ) 1/2 ] (13) αβ [ exp α+β R A R B 2 ] [ γδ γ+δ R C R D 2 ] (14) F 0 [(α + β)(γ + δ)/(α + β + γ + δ) R P R Q 2 ] (15) with R P = (α R A + β R B )/(α + β) (16) and R Q = (γ R C + δ R D )/(γ + δ). (17) These are all integrals required for the STO-3G calculation and the corresponding procedures in the program are functions S(A, B, RAB2), T(A, B, RAB2), V(A, B, RAB2, RCP2, ZC) and TWOE(A, B,C, D, RAB2, RCD2, RPQ2). From these integrals the overlap matrix S, the kinetic energy matrix T, the matrix of nuclear attraction energy to nucleus 1 (the He nucleus), V (1), and the matrix of nuclear attraction to nucleus 2 (the H nucleus), V (2), are calculated. Adding H core = T +V (1) +V (2) (18) the core Hamiltonian is computed. This Hamiltonian is the correct one for a single electron in the field of the nuclear point charges. The core Hamiltonian only needs to be evaluated once during the iterative calculations. In the next step, the overlap matrix S is diagonalized and the basis functions are orthonormalized. Here, canonical orthogonalization is used. The basis functions φ µ are transformed to 4

5 φ µ by means of an orthogonal transformation matrix X φ µ = ν X µν φ ν (19) For a canonical transformation X = U/ s (20) where U is the matrix that diagonalizes S, and s is a diagonal matrix with the eigenvalues of S. A matrix element of X is therefore X i j = U i j / s j. For the present problem we obtain φ 1 = φ φ 2 (21) φ 2 = φ φ 2 (22) Now the SCF procedure can be started. First, an initial guess for the density matrix P is required. Here P = 0 and F H core. This may be a poor choice and in more complicated situations a better guess may be needed (for example from a semiempirical calculation - this is what Gaussian will do for you). Next, the Fock matrix is transformed to the orthonormalized basis set F = X FX (23) Diagonalizing this matrix (solving F C = C ε) gives a unitary matrix of coefficients C and eigenvalues ε. From C the coefficients of the original basis functions φ µ are calculated via C = XC (24) Remember: eigenvalues remain unchanged under orthogonal transformations. In the present case the coefficient matrix is C = (25) 5

6 which indicates that the lowest molecular orbital ψ 1 is mainly (0.9291) composed of φ 1 with only a little contribution (0.1398) of φ 2. Without electron repulsion (remember: G = 0 was used up to here) the electrons cluster around the nucleus with the higher charge (He). From equation (25) the first real guess of the density matrix P can be formed N/2 P µν = 2 a C µa C νa (26) which has elements P = (27) here. From P we now form G which contains the two-electron part of the Fock matrix. The explicit expression for a matrix element of G is G µν = P λσ [(µν σλ) 1 (µλ σν)] (28) 2 λσ For the present situation with two basis functions φ 1 and φ 2 there are 2 4 = 16 possible integrals (µν λσ). There are, however, only 6 unique ones here. (φ 1 φ 1 φ 1 φ 1 ) = E h (29) (φ 2 φ 1 φ 1 φ 1 ) = E h (30) (φ 2 φ 1 φ 2 φ 1 ) = E h (31) (φ 2 φ 2 φ 1 φ 1 ) = E h (32) (φ 2 φ 2 φ 2 φ 1 ) = E h (33) (φ 2 φ 2 φ 2 φ 2 ) = E h (34) (35) The one-center integrals (φ 1 φ 1 φ 1 φ 1 ) and (φ 2 φ 2 φ 2 φ 2 ) are the repulsions between an electron in 6

7 φ 1 and another electron in the same orbital φ 1 (ditto for φ 2 ). The two-center integral (φ 2 φ 2 φ 1 φ 1 ) is the repulsion between an electron in φ 1 and one in φ 2. It has an asymptotic value of 1/R for large R. The other three integrals do not have simple classical interpretations. With these integrals the matrix elements G µν in eq. (28) can be evaluated and a new Fock matrix can be formed F = H core + G (36) Now the eigenvalue problem is solved with this new Fock matrix F, and a new guess for C (eq. (24)) and P (eq. (26)) is obtained. I.e. we return to equation 23 and iterate until convergence. Convergence is determined, e.g., with respect to changes in the density matrix P. I.e. if elements P µν in P do not change anymore by more than a pre-specified threshold, the calculation is called converged. The final wavefunctions and orbital energies for the present case are C = (37) and ε = (38) The lowest orbital, ψ 1, is a bonding orbital (same sign of both coefficients). It is mainly composed of the He function φ 1. The virtual orbital, ψ 2, is antibonding (opposite sign of the coefficients). Koopman s theorem allows us to quantify ionization potential and electron affinity: the IP = E h is large as it should be for a cationic species and EA = E h is positive which means that HeH + will bind an additional electron. This does, however, not mean that HeH is stable because the dissociation products (He + H + ) will bind an additional electron much more strongly than HeH +. A Mulliken population analysis yields the partial charges on the two nuclei. For this, one considers the diagonal elements of the matrix PS. The total number of electrons N can be related 7

8 to the molecular orbitals and the matrices P and S in the following way (the second equation follows from ψ a ( r) = K µ=1 C µaφ µ ( r)): N = 2 N/2 a d r ψ a ( r) 2 = µ ν P µν S νµ = µ (PS) µµ = tr(ps) (39) It should be noted that there is no unique definition of the number of electrons associated with a given atom or nucleus. The above analysis is the Mulliken population analysis and associates a charge q A = Z A (PS) µµ (40) µ A where Z A is the charge of nucleus A. Because the trace (tr, sum of diagonal matrix elements) obeys tr(ab) = tr(ba) one also has N = µ (S 1/2 PS 1/2 ) µµ = µ P µµ (41) where P is the density matrix in a symmetrically orthogonalized basis set φ µ( r) = ν (S 1/2 ) νµ φ ν ( r). (42) The diagonal elements associated with P yield the Löwdin population analysis q A = Z A (S 1/2 PS 1/2 ) µµ. (43) µ A 8

9 Short summary 1. Specify a molecule (nuclear coordinates R A ), atomic numbers Z A, the number of electrons N and an electronic basis set φ µ. 2. Calculate all required molecular integrals S µν, H core µν and (µν λσ). 3. Diagonalize overlap matrix S and obtain transformation matrix X (here canonical orthogonalization). 4. Obtain a guess for the density matrix P (here P = H core ; better choice may be required). 5. Calculate the matrix G from P and the two-electron integrals (µν λσ) via G µν = P λσ [(µν σλ) 1 (µλ σν)] (44) 2 λσ 6. Add G to the core Hamiltonian to obtain the Fock matrix F = H core + G 7. Calculate the transformed Fock matrix F = X FX 8. Diagonalize F to obtain C and ε. 9. Calculate C = XC. 10. Form a new density matrix P from C by using N/2 P µν = 2 a C µa C νa (45) 11. Determine whether procedure has converged (e.g. compare old P and new P). If not, return to step (5) with the new P. 12. If procedure has converged, use matrices C, P, F, etc. to determine expectation values and other properties of interest. 9