Derivation of Quadratic Response Time Dependent Hartree-Fock (TDHF) Equations

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1 Derivation of Quadratic Response Time Dependent Hartree-Fock (TDHF) Equations Joshua Goings, Feizhi Ding, Xiaosong Li Department of Chemistry, University of Washington Nember 21, Derivation of Equations of Motion through Second Order Exp the Fock matrix, F the density matrix, P, in a perturbative series as a function of arbitrary λ. Let µ ν be arbitrary perturbation indicies. We use Einstein notation, recognizing that λ µ is a sum er all possible µ, which includes λ ν. P = P (0) λ µ P (1) λ µ λ ν P (2) F = F (0) λ µ F (1) λ µ λ ν F (2) In a similar fashion, exp the Fock matrix the density matrix, as a Taylor series in arbitrary λ. P(t, λ) = P (0) P λ µ λ µ λ µλ ν P 2 2 λ µ λ ν F(t, λ) = F (0) F λ µ λ µ λ µλ ν F 2 2 λ µ λ ν This suggests that the following relations hold (with the factor of 2 absorbing into the expressions λ; upon collecting terms later on it cancels.): P (1) = λ µ P (2) = 2 λ µ λ ν F (1) = F λ (t) λ µ F (2) = 2 F λ (t) λ µ λ ν Recalling the definition of the Fourier transform: f(t) = f(ω) e iωt dω And using the postulate that any real observable corresponds to a Hermitian operator (e.g. Fock Density operators are Hermitian), gives, in the frequency domain: P (1) = λ µ = (ω µ )e iωµt ( ω µ )e iωµt 1

2 F (1) = F λ (t) λ µ = F (µ) (ω µ )e iωµt F (µ) ( ω µ )e iωµt P (2) = 2 λ µ λ ν = P µν (ω µ, ω ν )e i(ωµων)t P µν (ω µ, ω ν )e i(ωµ ων)t P µν ( ω µ, ω ν )e i( ωµων)t P µν ( ω µ, ω ν )e i(ωµων)t F (2) = 2 F λ (t) λ µ λ ν = F µν (ω µ, ω ν )e i(ωµων)t F µν (ω µ, ω ν )e i(ωµ ων)t F µν ( ω µ, ω ν )e i( ωµων)t F µν ( ω µ, ω ν )e i(ωµων)t The n-neumann type equation of motion is: Plugging in abe expressions for left h side: i P(t) = F(t), P(t) (1) t i t P(t) = i t P(0) t λ µ (ω µ )e iωµt t λ µ ( ω µ )e iωµt t λ µλ ν P µν (ω µ, ω ν )e i(ωµων)t t λ µλ ν P µν (ω µ, ω ν )e i(ωµ ων)t t λ µλ ν P µν ( ω µ, ω ν )e i( ωµων)t t λ µλ ν P µν ( ω µ, ω ν )e i(ωµων)t Which is to say that i t P(t) = λ µω µ ( (ω µ )e iωµt ( ω µ )e iωµt) λ µ λ ν (ω µ ω ν )P µν e i(ωµων)t (ω µ ω ν )P µν e i(ωµ ων)t ( ω µ ω ν )P µν e i( ωµων)t (ω µ ω ν )P µν e i(ωµων)t Solving for the right h side: 2

3 F(t), P(t) = F (0), P (0) λ µ F (0), λ µ F (µ), P (0) λ µ λ ν F (0), λ µ λ ν F (µν), P (0) λ µ λ ν F (µ), λ µ λ ν F (ν), Therefore, to second order F(t), P(t) = F (0), P (0) λ µ {F (0), F (µ), P (0) λ µ λ ν {F (0), F (µν), P (0) F (µ), F (ν), Exping the expression in terms of the frequency dependent response yields F(t), P(t) = F (0), P (0) λ µ F (0), (ω µ )e iωµt ( ω µ )e iωµt λ µ F (µ) (ω µ )e iωµt F (µ) ( ω µ )e iωµt, P (0) λ µ λ ν F (0), (ω µ, ω ν )e i(ωµων)t λ µ λ ν F (0), (ω µ, ω ν )e i(ωµ ων)t λ µ λ ν F (0), ( ω µ, ω ν )e i( ωµων)t λ µ λ ν F (0), ( ω µ, ω ν )e i(ωµων)t λ µ λ ν F (µν) (ω µ, ω ν ), P (0) e i(ωµων)t λ µ λ ν F (µν) (ω µ, ω ν ), P (0) e i(ωµ ων)t λ µ λ ν F (µν) ( ω µ, ω ν ), P (0) e i( ωµων)t λ µ λ ν F (µν) ( ω µ, ω ν ), P (0) e i(ωµων)t λ µ λ ν F (µ) (ω µ ), (ω ν )e i(ωµων)t λ µ λ ν F (ν) (ω ν ), (ω µ )e i(ωµων)t λ µ λ ν F (µ) (ω µ ), ( ω ν )e i(ωµ ων)t λ µ λ ν F (ν) ( ω ν ), (ω µ )e i(ωµ ων)t λ µ λ ν F (µ) ( ω µ ), (ω ν )e i( ωµων)t λ µ λ ν F (ν) (ω ν ), ( ω µ )e i( ωµων)t λ µ λ ν F (µ) ( ω µ ), ( ω ν )e i(ωµων)t λ µ λ ν F (ν) ( ω ν ), ( ω µ )e i(ωµων)t Setting the left right h sides equal, we can collect terms with like λ. After the perturbative orders have been collected, we further collect based off of the exponential terms. Justification for collection based off of the exponential factors can be illustrated by considering the first order terms: ω µ ( (ω µ )e iωµt ( ω µ )e iωµt) = F (0), (ω µ )e iωµt ( ω µ )e iωµt F (µ) (ω µ )e iωµt F (µ) ( ω µ )e iωµt, P (0) 3

4 Multiplying the abe expression by e iωµt yields ω µ ( (ω µ ) ( ω µ )e 2iωµt) = F (0), (ω µ ) ( ω µ )e 2iωµt F (µ) (ω µ ) F (µ) ( ω µ )e 2iωµt, P (0) Which is true if only if the time dependent terms are independent of the time independent terms. Thus the equation is separable into two first order response equations: ω µ (ω µ ) = F (0), (ω µ ) F (µ) (ω µ ), P (0) Plus the complex conjugate. The same procedure of collecting terms based off of exponential is valid for any number of exponential terms. Thus collecting the second order terms yields, in summary: 0 = F (0), P (0) (2) ω µ = F (0), F (µ), P (0) (3) (ω µ ω ν ) = F (0), F (µ), F (ν), F (µν), P (0) (4) Plus their complex conjugates. Frequency arguments have been dropped for clarity. These are the equations of motion up to second order. 2 Derivation of second order response working equations Assuming a converged solution to the Hartree-Fock equations in the canonical molecular orbital basis, P (0) F (0) are the density Fock/Kohn-Sham matrices. The density matrix is given as I 0 P (0) = (5) F (0) = Σ o 0 0 Σv with I equal to the identity matrix of dimension number of occupied orbitals, Σ O Σ V are the diagonal matrices containing occupied orbital virtual orbital eigenvalues, respectively. The total dimension of each matrix is (number occupied number virtual). The first order idempotency constraint = P (0) P (0) (7) (6) Results in = I 0 P (µ) P (µ) I 0 = 0 P (µ) 0 (8) The second-order density matrix idempotency constraint is = P (0) P (0) (9) 4

5 Which gives = I 0 = 0 (µν) P = P (µν) I P (µ) 0 (10) (11) This implies the following relation = ( ) (12) = (13) which is to say that the occupied-occupied virtual-virtual blocks of the second-order density matrix can be completely specified from the solutions to the first-order density matrix. We will make use of this fact when formulating the final second-order response working equations. The final working equations are specified using the second order equation of motion eq. (4): (ω µ ω ν ) = F (0), F (µ), F (ν), F (µν), P (0) The second-order Fock/KS matrix F (µν) is given by the expression F (µν) = F (1) { F (2) {, (14) The first term, F (1) { is the same as the first order change in the Fock/KS matrix, as we dealt with in the linear response derivation: F (1) { = F P P(µν) (15) where F P or for TDKS, is given, in the TDHF method, as: rs F pq P rs = rs with f xc, called the exchange kernel, defined as: P rs ( H core pq P rs (pq sr) (pr sq) ) = (pq sr) (pr sq) = (pq sr) (16) F pq P rs = (pq sr) (pq f xc sr) (17) f xc (r 1, r 2 ) = 2 E (xc) ρ ρ(r 1 )ρ(r 2 ) (18) where E xc ρ is the adiabatic exchange-correlation energy functional. From this, it is clear that F (1) {P depends linearly on P, such that F (1) { = F (1) { F (1) {. 5

6 The second term F { (2), for the KS matrix depends only on the first-order density matrices, is given as { F (2), = φ p(r 1 )φ q (r 1 )g xc (r 1, r 2, r 3 )ρ (µ) (r 2 )ρ (ν) (r 3 )dr 1 dr 2 dr 3 (19) pq with g xc (r 1.r 2, r 3 ) as the third-order derivative of the adiabatic exchange-correlation functional (compare to f xc ), defined as 3 E (xc) ρ g xc (r 1, r 2, r 3 ) = (20) ρ(r 1 )ρ(r 2 )ρ(r 3 ) Because ρ (µ) ρ (ν) are the first-order changes in the density matrix, given as ρ (µ) (r) = ia φ i(r)φ a(r) ai φ a (r)φ i (r) ia (21) substitution of the abe yields, for real orbitals { F (2), = ( G pq,ia,jb ai ia pq ia,jb ) ( bj ) jb (22) with G pq,ia,jb = φ p (r 1 )φ q (r 1 )φ i (r 2 )φ a (r 2 )φ j (r 3 )φ b (r 3 )g xc (r 1, r 2, r 3 )dr 1 dr 2 dr 3 (23) The important result is that F (2) {, for the KS matrix depends only on the first-order density matrices, which are determined directly from a linear-response TDKS calculation. For TDHF, the result is even simpler: F (2) {, is zero. Plugging eq. (14) into eq. (4), rearranging yields (ω µ ω ν ) F (0), = F (µ), F (ν), F (1) { F (2) {,, P (0) (24) Making use of the linearity of F (1) {P, we substitute the second order density matrix partition, such that { { F (1) {P µν = F (1) F (1) (25) Substituting this result into eq. (24), rearranging gives (ω µ ω ν ) F (0), F (1) {, P (0) = F (µ), F (ν), F (2) {,, P (0) F (1) {, P (0) (26) Because of eqs. (12), (13) (22), the whole right h side of the abe equation can be directly determined from first-order density matrices (of which are obtained from a linear response calculation). We will call the right h side Q, so that Q = F (µ), F (ν), F (2) {,, P (0) F (1) { (ω µ ω ν ) F (0), F (1) {, P (0) (27), P (0) = Q (28) It can be seen that eq. (28) has a similar form to the linear-response working equations, which, upon rearranging, gives: A B 1 0 B A ( (ω ) µ ω ν ) Q 0 1 ( = ) (Q) (29) 6

7 The elements of A B are for Hartree-Fock, A ia,jb = δ ij δ ab (ɛ a ɛ i ) (ia jb) (30) B ia,jb = (ia bj) (31) A ia,jb = δ ij δ ab (ɛ a ɛ i ) (ia jb) (ia f xc jb) (32) B ia,jb = (ia bj) (ia f xc bj) (33) for Kohn-Sham. Q will not necessarily equal zero, so unlike the linear response calculations, Q must be determined first, then the equations are solved iteratively. 7