DFT in practice

7/29/2019 DFT in practice

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Plane wave expansion & the Brillouin zone integration

DFT in practice : Part Iplane wave expansion & the Brillouin zone integration

Ersen Mete

Department of PhysicsBalkesir University, Balkesir - Turkey

August 13, 2009 - NanoDFT09, Izmir Institute of Technology, Izmir

Ersen Mete DFT in practice : Part I
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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

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Kohn-Sham equations to be solved self-consistently

122 + Veff(r)i(r) = ii(r)

Veff(r) = ext(r) +

dr

n(r )|r r|

Hartree potential+

EXC[n]

n(r)

Exchangecorrelationpotentialn(r) =

i

|i(r)|2

If the exchange-correlation functional is exact the total energy and the

density will be exact.

However, Kohn-sham single particle wavefunctions and eigenenergies donot correspond to any exact physical value.

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Self-consistent Kohn-Sham loop


Pl i & h B ill i i i
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Basis set representation of Kohn-Sham orbitals

Atomic Orbitals (AO)

(r) = (r)Ym (, ) where (r) =

er

2

Gaussianer 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)


Plane a e e pansion & the Brillo in one integration


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Basis set representation of Kohn-Sham orbitals

Plane waves

k(r) = eikr

orthogonal

periodic (solids) (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




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Bravais lattice, primitive cell & Brillouin zone

Direct lattice : a = a1,a2,a3Reciprocal lattice derives from confinementand Fourier analysis of periodic functions :

eibiaj = ij = bi aj = 2 (=integer)

Reciprocal latt. : b = 2(

a)1 =

b1, b2,b3Cell volume : cell = det(a)

BZ volume : BZ = det(b) = (2)3/cell

Direct lattice vectors : T(n1, n2, n3) = n1a1 + n2a2 + n3a3 TnReciprocal latt. vectors : G(m1,m2,m3) = m1b1 + m2b2 + m3b3 Gm

1st BZ (Wigner-Seitz cell) : boundaries are the bisecting planes of Gvectors where Bragg scattering occurs.


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Born-von Karman supercell approach

point defect slab molecule

i,k(r + Njaj) = i,k(r)

Supercell must be sufficiently large to maintain isolation.


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Plane wave expansion of the single particle wavefunctions

Blochs theorem

For a periodic Hamiltonian :

i,k(r) = e

ikr 1Ncell

ui,k(r)

i,k(r +aj) = eikaj

i,k(r))

where k is in the first BZ and periodic ui,k(r) can be expanded in aFourier series :

ui,k(r) =

1cell

m

ci,m(k)eiGmr

u

i,k(r +aj) = ui,k(r))

Equivalently,

i,k(r) =

m

ci,m1

ei(k+Gm )r

q

ci,q|q (q = k+ Gm)


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p g

Fourier representation of the Kohn-Sham equations

For a crystal,

Implications of Bloch states

i,k is a superposition of PWs with q wavevectors which differ by

Gm to maintain the PBC.

k is well-defined crystal momentum which is conserved.

Potential has the lattice periodicity as ui,k(r).

Veff(r) =

m

Veff(Gm)eiGmr

Only Gm vectors are allowed in the Fourier expansion.The non-zero matrix elements are,

q|Veff|q =

mV(Gm)q|eiGmr|q =

mVeff(Gm)qq,Gm


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p g

Substituting these Bloch states into the Kohn-Sham equations,multiplying by q| from the left and integrating over r gives a set ofmatrix equations for any given k.

drq|

1

22 + Veff(r)

q

ci,m|q = iq

ci,m|q

m

1

2|k+ Gm|2m,m + Veff(Gm Gm)

ci,m = ici,m

Matrix diagonalization needed, but still easier to solve.

Eigenvectors and eigenenergies for each k are independent in the 1stBZ.

In the large = Ncellcell limit, k points become a dense continuumand i(k) become continuous bands.


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Hartree potential in plane waves

VH(G) =1

cell dr ei

Gr

dr

G n(

G)eiGr

|r

r

|

=1

cell

G

n(G)

dr ei(GG)r

du

eiGu

|u|

=1

cell G

n(G)cellG,G2 011 e

iGucos

u u2d(cos )du

= 2n(G)

0

eiGu eiGuiG

du = 4n(G)

G

0

sin(Gu)du

= 4n(G)

Glim0

0

eu sin(Gu)du

= 4n(G)

Glim0

G

2 + G2= 4

n(G)

G2(G = 0)


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Exchange-correlation potential in plane waves

VXC(G) =1

cell dr eiGr

nn(r)xc([n],r)=

1

cell

dr ei

Gr

G

xc([n],G)ei

Gr +G,G

n(G)eiGrxc([n],G

)n

eiGr

=

1

cell

G

xc([n],G)

dr ei(GG)r+

G,G

n(G)xc([n],G

)n

dr ei(

GG+G)r

=

1

cell G

xc([n],G)cellG,G +

G,Gn(G)

xc([n],G)

n cellG,GG= xc([n],G) +

G

n(G G)xc([n],G)

n


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Fourier representation ofEH and EXC terms

By similar arguments,

EH =

1

2 dr dr n(r)n(r)

|rr| = 2cellG=0

n(G)2

G2

EXC =

dr n(r)xc(r) = cell

Gn(G)xc(G)


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External potential in terms of structure and form factors

Ionic potential as a superposition of isolated atomic potentials

ext(r) =nsp=1

nj=1

T

V(r ,j T)

nsp species, n atoms at ,j for , the set of translation vectors T

Vext(G) =1

drext(r)eiGr =

1

nsp=1

nj=1

duV(u)eiG(u+,j)

T

eiGT

Ncell

=

1

cell

nsp

=1

duV(u)eiGu V(G)

n

j=1

eiG

,j

S(G)

nsp

=1

cell S

(G)V

(G)

This form is particularly useful when V(r) = V(|r|)!


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Form factor for a spherically symmetric ionic potential

V(G) =1

0

11

20

V(r)eiGrr2 d d(cos ) dr

=2

V(r)

eiGr(cos )

iGr 1

-1

r2 dr

=2

0

eiGr e-iGriGr

r2 dr

=

4

0 j0(|G|r)V(r)r2dr= V(|G|)


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Cutoff : finite basis set

Computationally, a complete expansion in terms of infinitely many plane

waves is not possible.The coefficients, cm(k), for the lowest orbitals decrease exponentially

with increasing PW kinetic energy (k+ Gm)2/2.

A cutoff energy value, Ecut,

determines the number of PWs (Npw)

in the expansion, satisfying,

(k+ Gm)2

2 < Ecut (PW sphere)

Npw is a discontinuous function of the PW kinetic energy cutoff.

Basis set size depends only on the computational cell size and the cutoffenergy value.


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Electron density in the plane wave basis

n(r) =1

Nk

occk,i

ni,k(r) where ni,k(r) = |i,k(r)|2

Then, in terms of Bloch states,

ni,k(r) =

1

m,m

ci,m(k)ci,m(k)ei(GmGm)r

and

ni,G(r) = 1 m

ci,m(k)ci,m(k)

where Gm = Gm + G = sphere of n(G) has a double radius.


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Brillouin zone integration

Properties like the electron density, total energy, etc. can be evaluated by

intergration over k inside the BZ.

for the ith band of a function fi(k)

1

BZ BZdk fi(k) = fi = 1

Nk kfi(k)

Example :

a 2D square lattice

25 k-points in the BZ

fi = fi(k1) + . . . + fi(k25)


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k-point sampling : uniform vs non-uniform

Moreno and Soler [Phys.Rev.B 45,24] :

A mesh with uniformly distributedk-points is preferred.

Example : A rectangular lattice

They have nearly the same number ofk-points.

(a) : Isotropic sampling

(b) : finer sampling vertically

poor sampling horizontally



M kh P k id
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Monkhorst-Pack grid

A uniform mesh ofk-points can be generated by Monkhorst-Pack

procedure

kn1,n2,n3 =

31

2ni Ni 12Ni

Gi

where Ni is the number ofk-points in each direction and ni = 1, . . . ,Ni.

Example : 441 MP grid for 2D square lattice

G1 = b1 =a

kx, G2 = b2 =a

ky, G3 = 0

k1,1 = -38 , -38 k1,3 = -38 , 18k1,2 =

-38 ,

-18

k1,4 =

-38 ,

38

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Ti l


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Time-reversal symmetry

Kramers theorem

Let the hamiltonian, H, be invariant under time-reversal,THT1 = H (in the absence of magnetic fields!)

then H =

TH = HT = T

T (r,t) is also a solution of the same Schrodinger equation withthe same eigenvalue.

The solutions and T are orthogonal, |T = 0

Bloch states, i,-k and

i,k

satisfy the condition (r + Tn) = eikTn(r)

= i,-k = i,k

Kramers theorem implies inversion symmetry in the reciprocal space!



L tti t d th I d ibl B ill i Z


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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 :

(0, 0) : full symmetry of the point group.X 12 , 0 : E, C2, x, yM

12 ,

12

: full symmetry of the point group.

Then, we can unfold the IBZ by the symmetry operations to get thesolution for the full BZ.



P t
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Preserve symmetry

Example : Hexagonal cell

Shift thek-point mesh to preserve hexagonal symmetry!

In this case, even meshes break the symmetry

A mesh centered on preserves the symmetry.



Integration in the IBZ


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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 thek-points

Select the k-points which fall into the IBZ

Calculate weights, k, for each of the selectedk-points

k =number of symmetry connected k-points

total number ofk-points in the BZ

IBZ integration is then given by

fi =IBZk

kfi(k)




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Example : 441 MP grid for 2D square lattice

441 MP grid = 16 k-points in the BZ

4 equivalent k4,4 =

38 ,

38

= wk = 14

4 equivalentk3,3 = 18 , 38 = wk = 14

8 equivalent k4,3 =

38 ,

18

= wk = 12

Then, the BZ integration reduces to,

fi =14 fi(

k4,4) +14 fi(

k3,3) +12 fi(

k4,3)



Smearing methods : fractional occupation numbers


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Smearing methods : fractional occupation numbers

Fermi level : the energy of the highest occupied band.

IBZ integration over the filled states : fi =IBZk

kfi(k)(i(k) 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 (i(k) F) function with a smootherf(

{i(k)

}) function allowing partial occupancies at the Fermi level.



Fermi Dirac smearing
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Fermi-Dirac smearing

In order to overcome the discontinuity of the functions at the Fermi level,

f{i(k)} = 1e(i(k)F)/ + 1

where = kBT

T 0 : Fermi-Dirac distribution approaches to the step function.

Finite temperature T introduces entropy to the system of non-interactingparticles,

S(f) = [f ln f + (1 f)ln(1 f)]and the Free energy is given by,

F = E i

S(fi)

Drawback : reduced occupancies below F are not compensated newoccupancies above F.



Gaussian smearing
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Gaussian smearing

An approximate step function is obtained by integration of aGaussian-approximated delta function.

f{(ik

)} = 12

1 erf

ik F

Smearing parameter, , has no physical interpretation.

Entropy and the free energy cannot be written in terms of f.

S F

= 12

exp F

2



Method of Methfessel-Paxton
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Method of Methfessel-Paxton

To overcome the drawback introduced by the Fermi-Dirac smearing,

Expand the step function in a complete set of orthogonal Hermitefunctions. (Hermite polynomials multiplied by Gaussians)

(x) DN =N

nAnH2ne

x2 (H2n : (x) is even)

(x) SN = 1

DN(t)dt

S0 =1

2(1 erf(x)) (Gaussian smearing)

SN = S0(x) +

Nn=1

AnH2n1(x)ex2

Yields negative occupation numbers!



Marzari-Vanderbilt : cold smearing
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Marzari Vanderbilt : cold smearing

Amends the negative occupation numbers introduced byMethfessel-Paxton.

The delta function is approximated by a Gaussian multiplied by a firstorder polynomial,

(x) =1

e[x(1/

2]2 (2

2x)

where

x =

ik

F



Linear tetrahedron method
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Linear tetrahedron method

Divide the BZ into tetrahedra

Interpolate the function Xiwithin these tetrahedra

Integrate the interpolated functionto obtain Xi


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DFT in practice