Solving Hartree Fock Equation using a Basis Set HF Roothaan equation

1 HF Roothaan Equation For molecules numerical basis not efficient so use atom centered basis set to describe the N b a s i s  a r1    C ua u a  1, 2 ,...N b a s is u  1

f r1 a r1    aa r1  Nbasis Nbasis f r1   Cua u   a  Cua u  u 1 u 1 Nbasis Nbasis C  * r f r  r dr   C  * r  r dr  ua  v  1   1  u  1  1 a  ua  v  1  u  1  1 u 1 u 1 Nbasis Nbasis  CuaFvu   a  CuaSvu a  1,2,...Nbasis u 1 u 1

푭푪 = 푺푪흐

2 Self Consistent Field

We just have to solve the fock equation: PROBLEM FOCK OPERATOR HAS THE SOLUTION INSIDE F  C  C  SC  So put in a guess Cguess this allows you to get C1 F  C g u e s s  C 1  SC 1  Then put in C1 this allows you to get C2

Continue the cycle until you get convergence on Cinput and Coutput

SELF CONSISTENT FIELD (SCF) method

3 Fock Operator in Basis Representation

푎, 푏 are for spin orbitals 푢, 푣 are for basis sets

Core Hamiltonian Two electron Matrix  n / 2  core  * r  2J r  K r  r dr huv   v * r1 hr1  u r1 dr  v 1  b 1 b 1 u 1 1   b1  N n / 2 1 2 Z I hr1     1    2 vb| ub  vb| bu  2 I 1 r  1I b1 Calculated once in a n / 2 Nbasis Nbasis     C wb *C xb 2 vw| ux  vw| xu  SCF cycle b1 w1 x1 4 Fock Operator in Basis Representation

F   * r f r  r dr   * r h r  r dr vu  v  1   1  u  1   v  1   1  u  1  n / 2 Nbasis Nbasis    Cwb *Cxb 2 vw| ux  vw| xu  b1 w1 x1 Nbasis Nbasis  hvu    Pwx 2 vw| ux  vw| xu  w1 x1 Where we define the density matrix as n / 2 Pwx   Cwb *C xb b1

Since C changes every iteration this has to be recalculated every time

5 Self Consistent Field Method vw| ux   * r  * r r 1 r  r dr dr  v  1  w  2  12 u  1  x  2  1 2

Calculate initially

S  d r  *  F vu   v *  r 1  f  r 1  u  r 1  d r 1 vu  1 v u 

F  C  C  SC  UPDATE

Get newest results 6 Basis Sets

7 Slater Type Orbital vs Gaussian Type Orbital

In the H2 case we used Slater type orbitals for radial part where basis on HA is given as

1 / 2 STO 3   r  R A  1 S   , r  R A     /   e Another tye is a Gaussian type orbital for the radial part where

basis on H Ais given as

2 GTO 3 / 4   r  R A  1 S   , r  R A    2  /   e

8 GTO vs STO STO GTO

STO larger slope near r=0

d STO  , r  R  d GTO  , r  R  1S A  0 1S A  0 dr r  0 dr r  0 9 Solve the hydrogen atom by

 2      e2 2ll 1 Hˆ   r 2      2    2   2r r  r   40r 2r 

1S l=0, using atomic units

 1   2   1  Hˆ  Hˆ   r   trial trial  2  E   2r r  r  r  trial  trial trial

2 using  trial  Expr 

Solve for variational minimum of  and obtain the minimum energy. Compare the value with the exact one.

10 Hint: Gaussian Integrals   Hˆ   exp  r 2 Hˆ exp  r 2 r 2 dr trial trial      0     exp  r 2 exp  r 2 r 2 dr trial trial      0

 r n   r n exp 2r 2 dr 0 1 r1  4 1  r 2  8 2

4 3  r  2 32 2 11 Why GTO? 4 orbital basis integral: if the basis are on 4 different atoms vw| ux   * r  * r r 1 r  r dr dr  v  1  w  2  12 u  1  x  2  1 2 GTO GTO GTO  1 S  , r  R A  1 S  , r  R B   K AB 1 S  , r  R D 

 R A   R   3 / 4   2  R D  ; K AB  2 /     exp  R A  R B           

1.4

1.2 GTO1 GTO2 1 GTO1GTO2

0.8

0.6

0.4

0.2

0 -3 -2 -1 0 1 2 3 12 Linear Combination of GTO Contracted GTO: One GTO is not enough to describe the STO type orbital so lets use more than one GTO and add with correct coefficient (d) and exponent (a) value

L CGTO GTO   r  R A    d p  p , r  R A  p 1

13 Basis Sets • Minimal Basis: only those in so one 1S orbital for hydrogen, one 1S, 2S, 2P for carbon, oxygen • STO-3G • Split Valence: valence orbital has two, for hydrogen two 1S orbital, one 1S and two 2S, 2P for carbon oxygen • 6-31G, 3-21G • Diffuse: large version of valence orbital • +, ++ • Polarization: higher add 2P for hydrogen, add 3D for carbon, oxygen • *,**, (d), (d,p)

14 Dunning Correlation Consistent Basis Set T. Dunning decided on defining the contraction coefficient and exponential coefficient to maximize electron correlation aug-cc-pVDZ (X=2) aug-cc-pVTZ (X=4) aug-cc-pVQZ (X=4) aug-cc-pV5Z (X=5)

2 E  X   E CBS  B e xp   X  1   C e xp   X  1  

15 Basis Set Effective Core Potential • Most Chemistry occurs between valance electrons, electrons in the core do not contribute to important reactions use EFFECTIVE POTENTIAL FOR THE CORE ELECTRONS: Effective Core Potential (ECP) used for many electron systems. Relativistic Effect become important for heavy transition metal systems so use relativistic ECP (RECP) Hay-Wadt: LANL2DZ LANL2TZ, Dolg In bulk simulation like VASP it is called Psuedo potential

16 Basis Set Library (https://bse.pnl.gov/bse/portal)

17 Pick Basis Set Convergence Dipole moment of H2O Method # of basis Debye HF/STO-3G 7 1.7076 HF/6-31G 13 2.5006 HF/6-31+G 17 2.5824 HF/6-31+G(d,p) 29 2.2339 HF/6-311G 19 2.4881 HF/6-311++G 25 2.5536 HF/6-311++G(d,p) 37 2.1959 HF/6-311++G(3d,3p) 61 1.9744 HF/6-311++G(3df,3pd) 83 1.9681 HF/aug-cc-pVTZ 105 1.9394 HF/aug-cc-pVQZ 215 1.9361 Exp 1.8550 Computational time ~ (# of basis)2 18  1     1  1   2  1 Hˆ   r 2    2r  r 2    2     2  2    2r r  r  r   2r  r r  r   1  1 2 1    2    r r 2 r r   1  1 2 1  Hˆ   exp r 2    exp r 2 r 2dr trial trial    2     r r 2 r r   expr 2  2r expr 2  r 2 expr 2   2  4 2r 2 expr 2  r 2

ˆ 2  2 2 1 2 2  trial H trial  expr  2   2 r  expr r dr   r   3 r 2  2 2 r 4  r1 19 3  3  1  Hˆ   3 r 2  2 2 r 4  r1    trial trial 8 2 16 2 4 3  1   16 2 4 3  1  Hˆ   trial trial 16 2 4 2 1    trial trial  r    1  8 2 trial trial 8 2 3 2     2 r n   r n exp 2r 2 dr 2  0 Derivative with respect to  1 dE 3 2 1 2 2 r1     0    4 d 2   3 

2 1   8  24 16 r  E     2 8 2  9  18 9 2

4 3  r  (ans -0.424, 0.076 hartree off from the true value of -0.5) 32 2 2 20