DFT in practice : Part I. Ersen Mete

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1 plane wave expansion & the Brillouin zone integration Department of Physics Balıkesir University, Balıkesir - Turkey August 13, NanoDFT 09, İzmir Institute of Technology, İzmir

2 Outline Plane wave expansion Kohn-Sham equations for periodic systems Potential terms in plane waves Brillouin zone integration k-point sampling & Monkhorst-Pack grid Integration in irreducible Brillouin zone Smearing methods

3 Kohn-Sham equations to be solved self-consistently [ 1 ] V eff ( r) ψ i ( r) = ε i ψ i ( r) V eff ( r) = υ ext ( r) + n( r) = i ψ i ( r) 2 d r n( r ) r r } {{ } Hartree potential + δe XC [n] δn( r) } {{ } Exchange correlation potential If the exchange-correlation functional is exact the total energy and the density will be exact. However, Kohn-sham single particle wavefunctions and eigenenergies do not correspond to any exact physical value.

4 Self-consistent Kohn-Sham loop

5 Basis set representation of Kohn-Sham orbitals Atomic Orbitals (AO) { ψ α ( r) = ψ α (r)yl m e αr 2 Gaussian (θ, ϕ) where ψ α (r) = e αr Slater molecular structures : slightly distorted atoms small basis sets can give good results easy to represent vacuum regions basis functions are attached to nuclear positions non-orthogonal basis set superposition errors (BSSE)

6 Basis set representation of Kohn-Sham orbitals Plane waves ψ k ( r) = e i k r orthogonal (solids) periodic (finite systems) = PBC with supercell approach practical for Fourier transfom and computation of matrix elements atomic wave functions require large number of plane waves nonlocalized = inefficient for parallelization

7 Bravais lattice, primitive cell & Brillouin zone Direct lattice : a = [ a 1, a 2, a 3 ] Reciprocal lattice derives from confinement and Fourier analysis of periodic functions : e i b i a j = δ ij = b i a j = 2πl (l=integer) Reciprocal latt. : b = 2π(ã) 1 = [ b1, b 2, b 3 ] Cell volume : Ω cell = det(a) BZ volume : Ω BZ = det(b) = (2π) 3 /Ω cell Direct lattice vectors : T (n 1, n 2, n 3 ) = n 1 a 1 + n 2 a 2 + n 3 a 3 T n Reciprocal latt. vectors : G(m 1, m 2, m 3 ) = m 1 b1 + m 2 b2 + m 3 b3 G m 1st BZ (Wigner-Seitz cell) : boundaries are the bisecting planes of G vectors where Bragg scattering occurs.

8 Born-von Karman supercell approach point defect slab molecule ψ i, k ( r + N j a j ) = ψ i, k ( r) Supercell must be sufficiently large to maintain isolation.

9 Plane wave expansion of the single particle wavefunctions Bloch s theorem For a periodic Hamiltonian : ψ i, k ( r) = e i k r 1 Ncell u i, k ( r) { } ψ i, k ( r + a j ) = e i k a j ψ i, k ( r)) where k is in the first BZ and periodic u i, k ( r) can be expanded in a Fourier series : u i, k ( r) = 1 c i,m ( { } k)e i G m r u i, k ( r + a j ) = u i, k ( r)) Ωcell Equivalently, m ψ i, k ( r) = m c i,m 1 Ω e i( k+ G m) r q c i, q q ( q = k + G m )

10 Fourier representation of the Kohn-Sham equations For a crystal, Implications of Bloch states ψ i, k is a superposition of PWs with q wavevectors which differ by G m to maintain the PBC. k is well-defined crystal momentum which is conserved. Potential has the lattice periodicity as u i, k ( r). V eff ( r) = m V eff ( G m )e i G m r Only G m vectors are allowed in the Fourier expansion. The non-zero matrix elements are, q V eff q = m V ( G m ) q e i G m r q = m V eff ( G m )δ q q, G m

11 Substituting these Bloch states into the Kohn-Sham equations, multiplying by q from the left and integrating over r gives a set of matrix equations for any given k. [ d r q 1 ] V eff ( r) c i,m q = ε i c i,m q q q m [ ] 1 2 k + G m 2 δ m,m + V eff ( G m G m ) c i,m = ε i c i,m Matrix diagonalization needed, but still easier to solve. Eigenvectors and eigenenergies for each k are independent in the 1st BZ. In the large Ω = N cell Ω cell limit, k points become a dense continuum and ε i ( k) become continuous bands.

12 Hartree potential in plane waves V H ( G) = = 1 Ω cell d re i G r 1 n( G Ω ) cell G ( d r G n( G )e i G r ) d re i( G G) r r r d u ei G u u = 1 Ω cell G n( G )Ω cell δ G, G 2π 1 e ig u cos θ 0 1 u u 2 d(cos θ)du = 2πn( G) 0 = 4π n( G) G = 4π n( G) G lim η 0 lim η 0 e igu e igu 0 ig du = 4π n( G) G e ηu sin(gu)du G η 2 + G 2 = 4π n( G) G 2 (G 0) 0 sin(gu)du

13 Exchange-correlation potential in plane waves V XC ( G) = 1 Ω cell = 1 d re i G r Ω cell = 1 Ω cell = 1 Ω cell d re i G r { } n( r)ɛ xc ([n], r) n n( G )e i G r ɛ xc([n], G ) n e i G r ɛ xc ([n], G )e i G r + G G, G ɛ xc ([n], G ) d re i( G G) r + n( G ) ɛ xc([n], G ) n G G, G ɛ xc ([n], G )Ω cell δ G, G + n( G ) ɛ xc([n], G ) Ω cell δ G, n G G G G, G = ɛ xc ([n], G) + G n( G G ) ɛ xc([n], G ) n d re i( G G+ G ) r

14 Fourier representation of E H and E XC terms By similar arguments, E H = 1 2 E XC = d rd r n( r)n( r ) r r = 2πΩ cell G 0 n( G) 2 G 2 d rn( r)ɛ xc ( r) = Ω cell n( G)ɛ xc ( G) G

15 External potential in terms of structure and form factors Ionic potential as a superposition of isolated atomic potentials υ ext ( r) = n sp n κ κ=1 j=1 V κ ( r τ κ,j T ) T n sp species, n κ atoms at τ κ,j for κ, the set of translation vectors T V ext ( G) = 1 Ω = Ω d rυ ext ( r)e i G r = 1 Ω 1 Ω cell κ=1 n sp n κ κ=1 j=1 n sp d uv κ ( u)e i n κ G u e i G τ κ,j Ω } {{ } j=1 } {{ } Ω κ V κ ( G) Ω S κ ( G) d uv κ ( u)e i G ( u+ τ κ,j ) T n sp κ=1 This form is particularly useful when V κ ( r) = V κ ( r )! e i G T } {{ } N cell Ω κ Ω cell S κ ( G)V κ ( G)

16 Form factor for a spherically symmetric ionic potential V κ ( G) = 1 Ω κ = 2π Ω κ 1 2π 0 = 2π Ω κ 0 = 4π Ω κ 0 = V κ ( G ) 1 V κ (r) 0 V κ (r)e i G r r 2 dφ d(cos θ) dr e igr(cos θ) igr e igr e -igr r 2 dr igr j 0 ( G r)v κ (r)r 2 dr 1 r 2 dr -1

17 Cutoff : finite basis set Computationally, a complete expansion in terms of infinitely many plane waves is not possible. The coefficients, c m ( k), for the lowest orbitals decrease exponentially with increasing PW kinetic energy ( k + G m ) 2 /2. A cutoff energy value, E cut, determines the number of PWs (N pw ) in the expansion, satisfying, ( k+ G m) 2 2 < E cut (PW sphere) N pw is a discontinuous function of the PW kinetic energy cutoff. Basis set size depends only on the computational cell size and the cutoff energy value.

18 Electron density in the plane wave basis n( r) = 1 N k occ Then, in terms of Bloch states, k,i n i, k ( r) where n i, k ( r) = ψ i, k ( r) 2 n i, k ( r) = 1 c Ω i,m( k)c i,m ( k)e i( Gm Gm) r m,m and n i, G ( r) = 1 c Ω i,m( k)c i,m ( k) where G m = G m + G = sphere of n( G) has a double radius. m

19 Brillouin zone integration Properties like the electron density, total energy, etc. can be evaluated by intergration over k inside the BZ. for the i th band of a function f i ( k) 1 d k f i ( k) = fi = 1 N k Ω BZ BZ k f i ( k) Example : a 2D square lattice 25 k-points in the BZ f i = f i ( k 1 ) f i ( k 25 )

20 k-point sampling : uniform vs non-uniform Moreno and Soler [Phys.Rev.B 45,24] : A mesh with uniformly distributed k-points is preferred. Example : A rectangular lattice They have nearly the same number of k-points. (a) : Isotropic sampling (b) : finer sampling vertically poor sampling horizontally

21 Monkhorst-Pack grid A uniform mesh of k-points can be generated by Monkhorst-Pack procedure 3 2n i N i 1 kn1,n 2,n 3 = Gi 2N i 1 where N i is the number of k-points in each direction and n i = 1,..., N i. Example : MP grid for 2D square lattice G 1 = b 1 = π a ˆk x, G2 = b 2 = π a ˆk y, G3 = 0 k1,1 = ( -3 8, ) -3 8 k1,2 = ( -3 8, ) -1 8 k1,3 = ( -3 8, ) 1 8 k1,4 = ( -3 8, ) 3 8

22 Shifted-grids Example : a 2D square lattice Centered on Γ Centered around Γ = 25 k-points [shifted by (1/8,1/8,0)] 16 k-points They both have the same k-point density in the reciprocal space!

23 Time-reversal symmetry Kramer s theorem Let the hamiltonian, H, be invariant under time-reversal, THT 1 = H (in the absence of magnetic fields!) then Hψ = εψ THψ = HT ψ = εt ψ T ψ ψ ( r, t) is also a solution of the same Schrödinger equation with the same eigenvalue. The solutions ψ and T ψ are orthogonal, ψ T ψ = 0 Bloch states, ψ i,- k and ψ i, k satisfy the condition ψ( r + T n ) = e i k T n ψ( r) = ε i,- k = ε i, k Kramer s theorem implies inversion symmetry in the reciprocal space!

24 Lattice symmetry and the Irreducible Brillouin Zone We use symmetry of the Bravais lattice to reduce the calculation to a summation over k inside the IBZ. Example : IBZ of the 2D square lattice Special k-points : Γ (0, 0) : full symmetry of the point group. X ( 1 2, 0) : E, C 2, σ x, σ y M ( 1 2, 1 2) : full symmetry of the point group. Then, we can unfold the IBZ by the symmetry operations to get the solution for the full BZ.

25 Preserve symmetry Example : Hexagonal cell Shift the k-point mesh to preserve hexagonal symmetry! In this case, even meshes break the symmetry A mesh centered on Γ preserves the symmetry.

26 Integration in the IBZ Generate a uniform mesh in reciprocal space Shift the mesh if required Employ all symmetry operations of the Bravais lattice to each of the k-points Select the k-points which fall into the IBZ Calculate weights, ω k, for each of the selected k-points ω k = number of symmetry connected k-points total number of k-points in the BZ IBZ integration is then given by f i = IBZ k ω k f i ( k)

27 Example : MP grid for 2D square lattice MP grid = 16 k-points in the BZ 4 equivalent k 4,4 = ( 3 8, 3 8) = wk = equivalent k 3,3 = ( 1 8, 3 8) = wk = equivalent k 4,3 = ( 3 8, 1 8) = wk = 1 2 Then, the BZ integration reduces to, f i = 1 4 f i( k 4,4 ) f i( k 3,3 ) f i( k 4,3 )

28 Smearing methods : fractional occupation numbers Fermi level : the energy of the highest occupied band. IBZ integration over the filled states : f i = IBZ k ω k f i ( k)θ(ε i ( k) ε F ) For insulators and semiconductors, DOS goes to zero smoothly before the gap. For metals, the resolution of the step function at the Fermi level is very difficult in plane waves. Trick is to replace sharp θ(ε i ( k) ε F ) function with a smoother f ({ε i ( k)}) function allowing partial occupancies at the Fermi level.

29 Fermi-Dirac smearing In order to overcome the discontinuity of the functions at the Fermi level, f {ε i ( 1 k)} = e (ε i ( k) ε F )/σ + 1 where σ = k B T T 0 : Fermi-Dirac distribution approaches to the step function. Finite temperature T introduces entropy to the system of non-interacting particles, S(f ) = [f ln f + (1 f ) ln(1 f )] and the Free energy is given by, F = E i σs(f i ) Drawback : reduced occupancies below ε F are not compensated new occupancies above ε F.

30 Gaussian smearing An approximate step function is obtained by integration of a Gaussian-approximated delta function. f {(ε i k )} = 1 2 [ ( )] εi 1 erf k ε F σ Smearing parameter, σ, has no physical interpretation. Entropy and the free energy cannot be written in terms of f. ( ) [ ε εf S = 1 ( ) ] 2 ε σ 2 π exp εf σ

31 Method of Methfessel-Paxton To overcome the drawback introduced by the Fermi-Dirac smearing, Expand the step function in a complete set of orthogonal Hermite functions. (Hermite polynomials multiplied by Gaussians) δ(x) D N = N A n H 2n e x 2 (H 2n : δ(x) is even) n θ(x) S N = 1 D N (t)dt S 0 = 1 (1 erf(x)) (Gaussian smearing) 2 N S N = S 0 (x) + A n H 2n 1 (x)e x 2 n=1 Yields negative occupation numbers!

32 Marzari-Vanderbilt : cold smearing Amends the negative occupation numbers introduced by Methfessel-Paxton. The delta function is approximated by a Gaussian multiplied by a first order polynomial, δ(x) = 1 π e [x (1/ 2] 2 (2 2x) where x = ε i k ε F σ

33 Linear tetrahedron method Divide the BZ into tetrahedra Interpolate the function X i within these tetrahedra Integrate the interpolated function to obtain X i