Norm Conserving Pseudopotentialsand

The Hartree Fock Method

Eric NeuscammanMechanical and Aerospace Engineering 715

May 7, 2007

Goals

1. Produce a stand alone Hartree Fock code using C++

2. Apply code to generate pseudopotentials

Goals

1. Produce a stand alone Hartree Fock code using C++

2. Apply code to generate pseudopotentials

Success!

Goals

1. Produce a stand alone Hartree Fock code using C++

2. Apply code to generate pseudopotentials

Success!

Very limited success.

1. By removing core electrons

Basic Idea

Carbon n = 6 n3 = 216

Silicon n = 14 n3 = 2744 nv = 4 nv

3 = 64

Pseudopotentials simplify calculations

2. By creating nodeless valence orbitals

Creating Pseudopotentials

Step 1: Solve atomic system exactly.

Step 2: Construct the target valence orbital. It should be exact for large r.

Step 3: Invert the Schrodinger eq. for the V(r) that produces your valence orbital

The Hartree Fock Method

Apply the variational principal to a Slater determinant.

Result

N ... 11

iii

N

aaa KJ

r

Z

2

2

1

iii F

This differential equation has an infinite number of solutions.

Solving the Fock Equation

Introduce a basisiii F

j

M

jjii C j

M

jjii C

j

M

jjii C Assumes all orbitals are doubly occupied

Unrestricted Hartree Fock:

Restricted Hartree Fock:

Solving the Fock Equation

The basis gives a matrix eigenvalue problem

irpq

qrpqrq

qrpqrp CSC FF

εCSCF

Fock MatrixOrbital

Coefficients Overlap Matrix

Diagonal Eigenvalue Matrix

qppq FF

j

M

jjii C qppqS

Solving the Fock Equation

Structure of the Fock operator

sqprPrspqPHF rsr s

Trs

corepqpq ||

)()(

1)()(| 22

211121 rr

rrrrddrspq srqp

)(2

1)( 2 r

r

ZrdH qp

corepq

One Electron

Integrals

Two Electron Integrals

qppq FF

Solving the Fock Equation

Finding the integrals is the hardest part!

)()(

1)()(| 22

211121 rr

rrrrddrspq srqp

)(2

1)( 2 r

r

ZrdH qp

corepq

One Electron

Integrals

Two Electron Integrals

Basis Sets

To simplify integration, choose linear combinations of

gaussian basis functions.

This gives us

i

rcbai

izyxr2

e)(

1

0

222

1

20

221

2

11

1

2222

12

2222

111 e1

e1

e

|

drdrrr

r

rdrr

r

r

rrangularf

rspq

r

r

nrr

n

r

nn

angularfnAn analytic integral We have reduced a 6-dimensional

integral to a 2-dimensional integral

The SCF Algorithm

Calculate the Fock Matrix

sqprPrspqPHF rsr s

Trs

corepqpq ||

Guess the initial density matrices P P

Diagonalize the two systems to find the coefficient matrices

εCSCF εCSCF

Update the density matrices

r

qrprpq CCP r

qrprpq CCP Converged?

SolutionFound

No

Yes

Target Valence Orbitals

Now that we know the exact valence orbital functions, we may construct new valence orbitals that lack nodes.

Questions

)()( rr valpseudo

valexact

1. How do we ensure they match in the valence region?

2. How do we enforce normalization? (Norm conservation)

Answer: Method of Hamman, Schluter, and Chiang

Introduce a cutoff function

Holding the core electrons fixed, re-solve the Schrodinger (Fock) equation, but with a modified potential:

ccmod r

rfcrVrrfrV )(1)(

is a cutoff function such thatf

crrf 1 crrf 0

Outside the core, . Thus solutions to the exact and modified Schrodinger equations with the same eigenvalues will be identical in the valence region.

)()( rVrVmod

Example: 4

e)( xxf

Introduce a cutoff function

modexactmodmod rV

)(

2

1 2

To generate our target valence orbital, we repeatedly solve the modified Fock equation, adjusting c until the eigenvalue of the nodeless solution is the same as for the exact orbital.

ccmod r

rfcrVrrfrV )(1)(

A Potential Problem

Incorporating in the Hartree Fock method means modifying the one and two electron integrals.

)(rVmod

)()(

1)()(| 22

2111121 rr

rrrrrrfddrspq srqpc

One Electron: Not a problem

Two Electron: Breaks symmetry!

)(2

1)( 2 rrrf

r

ZrdH qcp

corepq

Solution: Only modify the OEI

Rather than modifying the nuclear, coulomb, and exchange potentials, only modify the nuclear attraction.

Then only the one electron integrals need modification and the method is tractable again.

Norm Conservation

Due to the homogeneity of the Schrodinger equation, the modified and exact valence functions may differ by a constant multiple.

This is easily fixed by scaling the modified function to match the exact function in the valence region.

Afterwards, however, our function is no longer normalized!

Norm Conservation

To re-normalize our function without altering its valence behavior, we must change it’s form in the core.

This is achieved by using a cutoff function again.

cmodmod rrf ~

The normalization condition is then

12

drrf cmod

Of the two roots for δ, choosing the smaller one will produce a smoother wavefunction.

The Pseudopotential

We have now generated a target valence orbital that is normalized and matches the exact orbital outside the core.

Our pseudopotential is then whatever potential generates our target orbital. To find it, we invert the Fock equation.

modexactmod

N

aaapseudo KJV ~~

2

1 2

Solve for me!

Easier Said Than Done

The integral form of the coulomb and exchange operators, coupled with the fact that not all the core electrons will be in spherically symmetric orbitals, make inverting this equation cumbersome.

modexactmod

N

aaapseudo KJV ~~

2

1 2

Currently, my code can only calculate Vpseudo when all of the occupied orbitals are spherically symmetric.

This limits me to Lithium and Berillium. Yuck!

Results

Pseudopotential method successful for both Li and Be. Results reported here employ the 6-31G basis set.

Hartree Fock code correctly predicts orbital occupations for 2nd row elements (need to check 3rd row).

Small modification would allow d-orbital basis functions to be employed, permitting modeling of transition metals. However, accuracy will degrade as relativistic effects grow.

Substantial work needed to allow calculation of PPs for atoms with p electrons.

Results for Lithium (rc = 2.0)

-20.0

-15.0

-10.0

-5.0

0.0

5.0

0 1 2 3 4 5 6

Radius (a.u.)

Pseudopotential -Z/r

Results for Lithium (rc = 2.0)

-300

-250

-200

-150

-100

-50

0

0.00 0.02 0.04 0.06 0.08 0.10

Radius (a.u.)

Pseudopotential -Z/r

Results for Lithium (rc = 2.0)

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5 6

Radius (a.u.)

Exact Wavefunction Wavefunction from PP

Results for Beryllium (rc = 0.9)

-20.0

-15.0

-10.0

-5.0

0.0

5.0

10.0

15.0

20.0

0.0 0.5 1.0 1.5 2.0 2.5

Radius (a.u.)

Pseudopotential -Z/r

Results for Beryllium (rc = 0.9)

0

30

60

90

120

150

180

0.00 0.02 0.04 0.06 0.08 0.10

Radius (a.u.)

Pseudopotential

Results for Beryllium (rc = 0.9)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.5 1.0 1.5 2.0 2.5

Radius (a.u.)

Exact Wavefunction Wavefunc from PP

Conclusion

Although implementing it was an excellent educational tool, the Hartree Fock is ill-suited for pseudopotentials

Even if my code’s current shortcomings could be removed, relavistic effects would prevent the method from applying to heavier atoms where pseudopotentials can greatly reduce the number of electrons.

Questions

?