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An Introduction To: Density Functional Theory (DFT) Xavier Blase CNRS

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An Introduction To: Density Functional Theory (DFT) Xavier Blase CNRS An introduction to: Density Functional Theory (DFT) Xavier Blase CNRS and Laboratoire de Physique de la Mati`ere Condens´ee et Nanostructures (LPMCN) CECAM Tutorial Lyon, November 2006 0-0 Outline: Empirical vs ab initio calculations • The many-body problem • Hartree-Fock (Exercise: the Helium atom) • Density Functional Theory (DFT) • { The energy is a functional of the density { Variational principle for the density { The Kohn-Sham approach and the "eigenvalues" { The local density approximation (LDA) { Self-consistency { A few identified problems Bibliographie • Appendix: a few practical aspects • 0-1 Empirical vs ab initio calculations Empirical approaches: interaction between two atoms described by an explicit function of their positions (ex: Lennard-Jones, 1922): V (r) = 4[(σ=r)12 (σ=r)6] − where (σ; ) are adjusted to reproduce experimental interatomic distances, vibrational frequencies, etc. There are also empirical approaches to the elec- tronic problem (tight-binding, Huc¨ kel, etc.) Question: can the same parameters describe diamond, graphite, nanotubes and fullerenes ? In the case of inaccessible systems (earth center, new mate- rials, etc.) which parameters do we use ? ab initio approaches : can we calculate structural, dynamical, electronic, response properties without any adjustable parameter ? 0-2 In principle, one can calculate everything from first-principles: 1 N N 1 ions Z H = 2 + vext(r ) + ; vext(r ) = I −2ri i r i − R r i i=1 i<j ij I i X X X XI j − j Ground-state energy (without E ): E =< H > with: ionic 0 0j j 0 = [r σ ; :::; r σ ] antisymmetric • 0 0 1 1 N N [r σ ; :::; r σ ) 2dr ::dr probability of finding electron i in [r ,r +dr ]. • j 0 1 1 N N j 1 N i i i n(r) = ρ(r) =< δ(r r ) > charge density • 0j i − i j 0 ρ (r; r0) =< Pδ(r r )δ(r0 r ) > density of pairs, etc. • 2 0j i6=j − i − j j 0 Problem: We don'tP know 0 ! We could use variational principle. However with = [r1σ1; :::; rN σN ], calculat- ing < H > would require (ngrid)3N 1030 opera- tions (forj smallj molecule). With 1012 flops∼machine = centuries. ) 0-3 Hartree-Fock: Variational approach with exact Hamiltonian but approxi- mate wavefunction: a single Slater determinant (Pauli principle). φ1(r1)σ1 :::: φN (r1)σN 1 φ1(r2)σ1 :::: φN (r2)σN (x1; x2; :::; xN) = 2 . 3 pN! . .. 6 7 6 φ (r )σ :::: φ (r )σ 7 6 1 N 1 N N N 7 1 (P ) 4 5 = ( 1) φP (1)(r1)...φP (N)(rN )σP (1)...σP (N) pN! − XP Ground-state energy: EHF = φ 1 2 + vext φ + 1 N (J K ) 0 ih ij − 2 r j ii 2 i;j=1 ij − ij 2 0 2P ∗P 0 ∗ 0 φ (r) φ (r ) φ (r)φj(r)φi(r )φ (r ) J = drdr0 j i j j j j and K = δ drdr0 i j ij r r0 ij σiσj r r0 Z j − j Z j − j with Jij and Kij the (classic) Coulomb (Hartree) and (quantum) exchange 2 6 (Fock) terms. One now has N integrals in (ngrid) dimension (instead of 3N (ngrid) dimension). The φi will be given by variational principle. 0-4 The exchange hole (X-hole) 0 0 Density of pairs: ρ2(x; x ) = i6=j δ(x xi)δ(x xj) ; x = (r; σ) gives probability to have an electronh j with spin− σ in r−withj ani electron with spin σ0 in r0. In HF, electrons withP opposite spins uncorrelated = energy exp HF ) penality. Correlation energy: E - E =EC < 0 Normalized distribution function: 0 0 0 g(x; x ) = ρ2(x; x )=n(x)n(x ): 1 ρ (r; r0) 1 n(r)n(r0) 1 n(r)h (r; r0) Eee = 2 drdr0 = drdr0 + X drdr0 2 r r0 2 r r0 2 r r0 Z j − j Z j − j Z j − j h (r; r0) = n(r0)(g(r; r0) 1) the X-hole which verifies: h (r; r0)dr0 = 1. X − X − R 0-5 The helium atom: effect of correlations Ground state: L=S=0 (singulet) symmetric through radial part f(r1; r2). Ground-state (GS) energy: Eexp = -5.8075 Rydb. Trial 1: We take 1s state to build GS wavefunction : 3=2 1=2 −ar f(r1; r2) = φa(r1)φa(r2); φa(r) = (a /π )e where a is a variatiobal parameter. E(a) is minimum for a0=(2-5/16) with E(a ) = -5.695 Ry (effective ionic charge: (2- 5/16)e screening). 0 Trial 2: We provide more variational freedom: f(r1; r2) = φa(r1)φb(r2) + φb(r1)φa(r2) Minimum energy (Taylor,Parr): (a0; b0) = (2:183; 1:1885) and E(a0; b0) = - 5.751 Ry in better agreement with experiment. 0-6 Trial 3 We include p-functions (with L=0; see Cohen-Tannoudji p.1017): 2 −1=2 λ f(r1; r2) = (1 + λ ) φa(r1)φb(r2) + φb(r1)φa(r2) + p(r1; r2) f p3 g p(r1; r2) = h10(r1)h10(r2) + h1−1(r1)h11(r2)h1−1(r1)h11(r2) with : 5=2 1=2 −γr h1m(r) = γ =(3π) re Y1m(θ; φ) Minimum energy: (a; b; γ; λ) = (2,176,1.201,2.475,-0.06128) and E(a; b; γ; λ)=- 5.790 Ry Essai 1 Essai 2 Essai 3 Thanks to p orbitals, the pair distribution function: P(r1; r2) = P ( r1 ; r2 ; θ) is angular dependant and P (θ = π) > P (θ = 0) (angular correlations).j j j j 0-7 Trial 4: The Hartree-Fock solution. With the constraint S=0 for the ground-state, the only mono-determinantal solution is: 1 φ(r1)α(1) φ(r2)α(2) α(1)β(2) α(2)β(1) (x1; x2) = = φ(r1)φ(r2) − p2 φ(r1)β(1) φ(r2)β(2) p2 and we are back to the form of (Trial 1). The solution we obtain (numerically) yields: EHF = -5.723 Ry in between (Trial 1) and (Trial 2). We note in particular that the wavefunction in (Trial 2) can be written as a sum of two determinants: A f(r1)α(1) f(r2)α(2) f(r1)β(1) f(r2)β(2) (x1; x2) = p2f g(r1)β(1) g(r2)β(2) − g(r1)α(1) g(r2)α(2) g = by including several determinants containing "excited" single-electron or- bitals,) one can improve the GS wavefunction and energy quantum chem- istry methods (Couple Cluster, Interaction Configurations!(CI), etc.) But CI quickly expensive as well. The approach widely adopted by a large fraction of the community is DFT. 0-8 The density functional theory (DFT) The central idea is to show that the ground-state total energy is a functional of the charge density n(r) (scalar field and physical obervable) instead of the many-body wavefunction: E[ ] E[n]. ! 1 N N 1 Reminder: H = 2 + vext(r ) + −2ri i r i i=1 i<j ij X X X ext ZI v (ri) = = H = 0 − RI ri ) ) XI j − j Theorem HK1 Given n(r) the charge density, there exists only one external potential vext(r) (up to a constant) that can realize n(r) (the reverse statement is obvious!). By absurdum, let's assume that: 0-9 vext(r) n(r) 1 ! 1 ! vext(r) n(r) 2 ! 2 ! ee ext then the variational principle applied to H1 = T + V + v1 yields (non degenerate GS): EGS < H with: 1 h 2j 1j 2i H >=< H + vext vext > h 2j 1j 2 2j 2 1 − 2 j 2 that is: EGS < EGS + dr ρ(r)[vext vext] 1 2 1 − 2 R GS GS ext ext Switching (1) and (2): E2 < E1 + dr ρ(r)[v2 v1 ] and by addition of the two inequalities: − R GS GS GS GS E1 + E2 < E2 + E1 (absurd) 0-10 Variational principle for E=E[n] The density n defines in a unique way vext (and N) that is the Hamiltonian. E is therefore a fonctionnal of n. 3 ext E[ ] = E[n] = d rv (r)n(r) + FHK [n] Zee FHK [n] = T [n] + V [n] ext FHK is a universal fonctionnal of n as does not depend on v . Problem: we do not know FHK . Theorem: Given a "test" density "ntest(r) 0" such that n(r)dr = N, test ≥ then: E[n ] EGS. Variational principle≥ valid for the wavefunctions. Its extensionRto E[n] derives from the bijection between n(r) and the GS many-body wavefunction (HK1): ext test ext;test test nGS v GS and n v so that:()E[ntest(])= E[ test] E[ ()] = E[n ]() ≥ GS GS 0-11 DFT without orbitals: From this variational principle derives a Euler- Lagrange equation which in principle allows to find n(r) by minimization: δ δF [E[n] µ[ d3rn(r) N]] = 0 = vext(r) + HK = µ δn(r) − − ) δn(r) Z with µ the chemical potential associated with conservation of number of par- ticles. However, FHK not known !! The Kohn and Sham (KS) approach Numerous work to find T[n] and Vee[n] (Thomas, Fermi, Slater, Dirac, etc.) However, the most successful approach is to come back to an "exact" expres- sion for the kinetic energy T by re-introducing one-body orbitals. To do that, KS introduce a fictitious equivalent system of non-interacting electrons un- der the action of an effective external potential V eff generating the same density n(r) that the real system: 0-12 N n(r) = 2dr :::dr n(r) = φ (r) 2 j j 1 N () j i j i=1 Z X N 2 T [n] =?? T [n] = < φ −∇ φ > () 0 ij 2 j i i=1 2 X H^ >= E > [−∇ + V eff (r)]φ (r) = φ (r) j j () 2 i i i ee By regrouping: FHK [n] = T0[n]+J[n]+(T T0 +V J)[n], the equivalence between the two systems yields: − − n(r0)dr0 δE [n] V eff (r) = vext(r) + + xc r r0 δn(r) Z j − j Thanks to KS, T0, an "important fraction" of T[n], is calculated exactly.
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