norm conserving pseudopotentials and the hartree fock method eric neuscamman mechanical and...
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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
?