Answers Quantum Chemistry NWI-MOL406 G. C. Groenenboom and G. A. de Wijs, HG00.307, 8:30-11:30, 21 jan 2014

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1 Answers Quantum Chemistry NWI-MOL406 G. C. Groenenboom and G. A. de Wijs, HG00.307, 8:30-11:30, 21 jan 2014 Question 1: Basis sets Consider the split valence SV3-21G one electron basis set for formaldehyde (H 2 CO). 1a. What is the dimension of this basis? Show how you determined the dimension. Answer: This is a split-valence basis set, H has two s-functions, C and O each have three s-functions and two p x, two p y, and two p z functions. The dimension of the basis is (2+2+2) = 22. 1b. How many primitive Gaussians are contained in this basis? Again, show how they were counted. Answer: SV3-21G means that core orbitals use 3 Gaussians and valence shells are split in 2+1, so they also require 3 Gaussians. Thus, an H-atom requires 3 Gaussians, and C and O require 3+3+9=15 Gaussians. The total number of Gaussians is 2 (3+15) = 36 Gaussians. The basis set can be extended by adding polarization functions. This results in the SV3-21G* basis. 1c. List these polarization functions for formaldehyde. Answer: The single star in SV3-21G* means polarization functions are only added to second-row atoms, i.e., C and O. The each get a set of d-functions. In a Cartesian-Gaussian basis that is 6 functions, d xy, d xz, d yz, d xx, d yy, and d zz. In a spherical basis that is 5 functions each, d 2, d 1, d 0, d 1, and d 2. Question 2: Point group symmetry The point group symmetry of the ammonia molecule (NH 3 ) is C 3v. The C 3v character table is: C 3v E 2C 3 (z) 3σ v A A E The nitrogen atom has four valence atomic orbitals (AOs), (2s, 2p x, 2p y, 2p z ). 2a. Which of these AOs belong to the A 1 irrep? Answer: The C 3 (z) axis is taken as the z-axis, so the 2s and the 2p z are invariant under all symmetry operators and belong to the A 1 irrep. Page 1 of 10

2 2b. Consider the effect of the rotation operator and give the irrep label of the remaining orbitals. Answer: The 2p x and 2p y orbitals mix under rotation, so they belong to the only two-dimensional irrep in this group, the E irrep. The general expression for a character projector for irrep Γ (dimension o Γ) of a group G (order o G) of operators ĝ with characters χ (Γ) (ĝ) is ˆP (Γ) o Γ o G χ (Γ) (ĝ) ĝ. ĝ G Basis set B consists of the 1s AOs of the three hydrogen atoms H A, H B, and H C B = {1s A,1s B,1s C }. 2c. Use the character projector formula to adapt basis set B to point group symmetry C 3v. Answer: Take 1s A on the plane of reflection σ 1, 1s B on σ 2, and 1s C on σ 3. The operators acting in the orbitals give Ê Ĉ 3 Ĉ3 1 ˆσ 1 ˆσ 2 ˆσ 3 1s A 1s A 1s B 1s C 1s A 1s C 1s B 1s B 1s B 1s C 1s A 1s C 1s B 1s A 1s C 1s C 1s A 1s B 1s B 1s A 1s C ˆP (A 1) 1s A = 1 6 (1s A +1s B +1s C +1s A +1s C +1s B ) = 2 3 (1s A +1s B +1s C ) ˆP (E) 1s A = 2 6 [2(1s A) 1s B 1s C ] ˆP (A 2) 1s A = 1 6 (1s A +1s B +1s C 1s A 1s C 1s B ) = 0 Page 2 of 10

3 Question 3: Lagrange undetermined multiplier method The Lagrange undetermined multiplier method can be used to minimize a function of n parameters f(x 1,...,x n ), with N constraints g i (x 1,...,x n ) = 0, i = 1,...,N. This method can be used in the derivation of the Roothaan equations in Hartree- Fock theory. 3a. How is the Lagrange undetermined multiplier method used in the derivation of the Roothaan equations? In particular, what is the function f that is being minimized, what are the parameters x 1,...,x n, what is n, and what are the constraints g i and what is N? Answer: The function f that is minimized is the expectation value of the electronic Hamiltonian Ĥ for a single Slater-determinant wave function Φ f = Φ Ĥ Φ The parameters x 1,...,x n are the MO coefficients, i.e., the expansion coefficients of the MOs {ψ j,j = 1,...,n MO } expressed as linear combinations of basis functions {χ 1,...,χ nb }, ψ j = n B i=1 χ i C i,j, j = 1,...,n MO. The number of parameters n = n B n MO. The constraints are the orthonormality of the molecular orbitals so N = n 2 MO. ψ i ψ j = δ i,j Question 4: Slater-Condon rules An n-electron wave function Φ is given by the Slater determinant of orthonormal spin-orbitals χ i, Φ(1,...,n) = 1 n! χ 1 χ 2 χ n. To evaluate the energy expression, E = Φ Ĥ Φ, where Ĥ is the electronic Hamiltonian, the wave function may be written as Φ = n!âχ 1(1)χ 2 (2) χ n (n), Page 3 of 10

4 where the antisymmetrizer  is given by  1 ǫ P ˆP. n! ˆP S n 4a. What are ǫ P, ˆP, and Sn? Answer: The group S n is called the symmetric group. Its elements are all the permutation operators ˆP of n objects. The parity of ˆP is ǫ P, it is 1 for odd and +1 for even permutations. The result is E = ˆP Sn ǫ P χ 1 χ n Ĥ ˆP χ 1 χ n. 4b. Derive this result. For each step in the derivation mention which property of the antisymmetrizer  and hamiltonian Ĥ you are using. It is not necessary to prove the properties you are using. Answer: E (1) = Φ Ĥ Φ = n! Âχ 1...χ n Ĥ Âχ 1...χ n (2) = n! χ 1...χ n ÂĤ Âχ 1...χ n (3) = n! χ 1...χ n Ĥ Â2 χ 1...χ n (4) = n! χ 1...χ n Ĥ Âχ 1...χ n (5) = ˆP Sn ǫ P χ 1...χ n Ĥ ˆPχ 1...χ n Here we use step (1) the definition of  step (2)  is Hermitian:  =  step (3)  commutes with the Hamiltonian: ÂĤ = Ĥ step (4)  is idempotent Â2 =  step (5) the definition of  4c. Simplify the energy expression for a one-electron Hamiltonian Ĥ = n ĥ(i). i=1 Page 4 of 10

5 Answer: E = ˆP Sn ǫ P χ 1...χ n Ĥ ˆPχ 1...χ n = ˆP Sn = n i=1 ǫ P χ 1 (1)...χ n (n) n ĥ(i) χ P(1) (1)...χ P(n) (n) i=1 ˆP S n ǫ P χ 1 (1) χˆp(1)... χ i (i) ĥ(i) χˆp(i)... χ 1 (1) χˆp(1) Because the orbitals χ i are orthormal, this sum only gives nonzero contributions for ˆP(j) = j for all j i, and hence also for i = j, i.e. when ˆP is the identity. This gives n E = χ i ĥ χ i. i=1 Page 5 of 10

6 Question 5: Band-structures Below are three band-structures of silicon (with labels A, B and C ). Silicon crystallizes in the diamond structure, giving 4 valence bands (8 valence electrons per unit cell). The band structures show the (occupied) valence bands and the lowest conduction bands. Three different exchange correlation potentials were used for the three band structures: I: PBE II: HF III: the hybrid functional HSE06 5a. Which functional corresponds to which band structure plot? Briefly justify your choice! No points are given for a mere guess. 10 (A) 5 0 energy (ev) W L Γ X W K Γ K X 10 (B) 10 (C) energy (ev) 5 energy (ev) W L Γ X W K Γ K X W L Γ X W K Γ K X Page 6 of 10

7 Answer: DFT functionals (LDA and GGA) typically underestimate band gaps, whereas HF typically overestimates. Hybrid functionals are somewhere in between. PBE is a GGA functional, HF is Hartree Fock, HSE is a hydbrid. So B = PBE (smallest gap), C = HSE (intermediate), A = HF (largest gap) Page 7 of 10

8 Question 6: Exchange-correlation potentials Below are three different expressions (A, B and C). These are parts of three different exchange correlation potentials. In particular, they belong to: I: the Perdew-Zunger LDA correlation energy II: the PBE gradient correction to the correlation energy III: the van der Waals extension to the correlation energy by Dion et al. The (somewhat simplified) expressions in atomic units are: A : n(r)n(r )φ[n(r),n(r ), n(r), n(r )]drdr B 1 B : n(r) dr 1+B 2 rs +B 3 r s } C : n(r)c 1 ln {1+C 2 t 2 1+f[n(r)]t 2 dr. 1+f[n(r)]t 2 +(f[n(r)]t 2 ) 2 Here B 1, B 2, B 3, C 1 and C 2 are constants, and f and φ are known functions. Beware that r s and t are not constants. They change with r via the density n(r), as do k P and k s, according to: n(r) = 3 4r 3 s = k3 P 3π 2, k s = 4k P /π, t = n(r) 2k s n(r). 6a. Which functional corresponds to which expression? Justify your choice! No points are given for a mere guess. Answer: The LDA only depends on the local density, so not on the gradients: I. LDA = B. The PBE also depends on the gradient, could be A or C, but A is nonlocal, so that has to be the van der Waals functional, leaving C for for PBE: II: PBE = C and III: van der Waals = A. Page 8 of 10

9 Question 7: Local Density Approximation The LDA exchange functional can be derived from a calculation on the uniform electron gas (with density n). In this system the electrons are distributed homogeneously over space, and a compensating, uniform positive background charge is present. The starting point is a gas with non-interacting electrons. This has very simple eigenvalue equations: Here k labels the electronic orbitals and eigenvalues. h2 2m e 2 φ k (r) = ǫ k φ k (r). (1) 7a. Explicitly give the eigenfunctions φ k (r) and eigenvalues ǫ k, and show that they indeed are eigenfunctions and eigenvalues. Answer: The eigenfunctions are: φ k (r) = 1 Ω e ik r Let s apply the Hamiltonian: h2 2m e 2 e ik r = h2 k 2 2m e e ik r so ǫ k = h2 k 2 2m e Mind: nothing was said in the exercises to specify the normalization (here assumed to be normalized in a volume Ω). In Hartree-Fock the eigenvalue equations for the homogeneous gas are: ] [ h2 ρ 2 e 2 X k (r,r ) dr φ HF 2m e r r k (r) = ǫhf k φhf k (r). (2) 7b. Explain, in words, why there is no Hartree potential. Answer: Apart from confusion about the word potential, note that the Coulomb potential of the uniform electron gas (including self-term) exactly cancels the Coulomb potential of the compensating positive background charge. 7c. In going from the equations in (1) to those in (2), what happens to the eigenfunctions and eigenvalues? No proof is required. Describe qualitatively the difference between ǫ k and ǫ HF k. Page 9 of 10

10 Answer: The eigenfunctions are unaltered, i.e. they remain the same. The eigenvalues are lowered, by an additional stabilization due to exchange. The dispersion relation ǫ k, i.e. ǫ as a function of k = k is not the free-electron parabola anymore. There is a peculiar, unphysical behaviour at k F where dǫ/dk diverges. 7d. Having obtained ǫ HF k, how do we obtain the exchange energy? Write down the integral, assuming ǫ HF k is given. Give an explicit formula for k F. Answer: The exchange contribution to the eigenvalue is ǫ HF k h2 k 2 2m e We have to integrate this over the volume of the Fermi sphere, counting properly the states ( Ω/(2π) 3 ), the spin degeneracy ( 2), and correct for double counting ( 1/2): ǫ X (n) = 1 N Ω dk 2 ( ) ǫ HF (2π) 3 k h2 k 2 k <k F 2 2m e, k 3 F 3π 2 = N Ω = n Doing the integral of part (7d) will provide the exchange energy per electron as a function of n. Let s call this ǫ X (n). We assume we know this function. 7e. Starting from ǫ X (n), formulate the LDA and give the formula for the LDA exchange energy. Answer: The LDA exchange energy: EX LDA = ǫ X (n) dn {}}{ n(r) dr We assume that locally at r in a volume dr there is as much exchange energy asthereisinauniformgaswithdenstiyn(r), i.e.thelocalexchange energy density is identical to that of the uniform electron gas. Page 10 of 10