The Hartree-Fock Method for Approximating Wave Functions in Quantum Chemistry
The Hartree-Fock Method
For hydrogen, the exact wave function is known. For lithium and helium, very accurate wave functions have been calculated by including interelectronic distance in the variation function. For atoms of higher atomic number, the best approach to finding a good wave function lies in first calculating an approximate wave function using the Hartree-Fock procedure, which we shall outline in this section. The Hartree-Fock method is the basis for the use of atomic and molecular orbitals in many-electron systems.
The Hamiltonian operator for an n-electron atom is given by:
[Insert mathematical equation here]
where an infinitely heavy point nucleus was assumed.
- The first term sum contains the kinetic energy operator for n-electrons.
- The second term sum is the potential energy for attraction between the electrons and the nucleus of charge Ze, for a neutral atom Z=n.
- The last term is the potential energy of the interelectronic repulsion.
Problems on the Hartree-Fock Method
1. The He SCF Calculation
This calculation used a basis set of two STOs, X1 and X2. For the helium ground state treated with this basis set:
(a) Write down the configuration state functions (CSFs) that are present in the wave function in a full CI treatment.
Since we used two basis functions, the SCF calculation yielded two SCF orbitals, φ1 and φ2. In the ground state of this two-electron atom, only φ1 is occupied, and φ2 is an unoccupied orbital. We found:
[Insert mathematical equation here]
but did not bother to find φ2. Use of the Roothaan equation:
[Insert mathematical equation here]
The normalization condition, we can write:
[Insert mathematical equation here]
C22 = 1.816 and C12 = -1.620
The SCF orbitals are:
[Insert mathematical equation here]
and
[Insert mathematical equation here]
The term that arises from placing two electrons into these two s-orbitals. Since the ground state is an s-orbital, we include only the CSFs which are:
[Insert mathematical equation here]
and
[Insert mathematical equation here]
(b) Carry out a CI calculation that includes only doubly excited CSFs.
The CFS Φ2 is doubly substituted, and Φ3 is singly substituted, so we include only Φ2 in addition to Φ1. The vibrational function is:
[Insert mathematical equation here]
and the secular equation is given by:
[Insert mathematical equation here]
The orthogonality of the orbitals φ2 and φ1 ensures that:
[Insert mathematical equation here]
The wave function Φ1 is the SCF energy -2.862 Hartrees calculated in the previous example.
Evaluating the integrals in the secular determinant, we can omit the spin factor in the wave function since summation over these gives 1.
The He Hamiltonian is H=H(1)+_H(2) and:
[Insert mathematical equation here]
where the values of the integrals were taken from the example. Similarly, we for:
[Insert mathematical equation here]
The H integrals vanish for orthonormality of Φ2 and Φ1.
The secular equation is given by:
[Insert mathematical equation here]
Then the value of E is given by:
E = -2.876, 3.24
The lower root gives:
E = -2.876 Hartrees = -78.25eV
as compared with the SCF energy of -77.87eV and the true energy of -79eV. This CI calculation has recovered 34% of the correlation energy. The CI ground-state Ψ is found to be:
ψ = 0.9989φ1 – 0.0474φ2
One finds that the inclusion of the singly excited CSFs φ3 gives only a very slight further improvement. Significant further improvement requires redoing the SCF calculation with a larger basis set, which will generate more virtual orbitals so that many more electron CSFs can be included in the CI calculation.
2. Do an SCF Calculation for the He Atom Ground State
Use a basis set of two 1S STOs with orbital exponents ε1=1.45 and ε2=2.
To solve the Roothaan equations, we need the integrals Frs and Srs. The overlap integral Srs are:
[Insert mathematical equation here]
where the Appendix integrals were used.
The integrals Frs are given and depend on rs/tu:
[Insert mathematical equation here]
The integrals H are calculated in the same way that similar integrals were evaluated in the variation treatment of He. We find:
[Insert mathematical equation here]
Many of the electron repulsion integrals (rs/tu) are equal to one another. For real basis functions, one can show that:
[Insert mathematical equation here]
The electron repulsion integrals are evaluated using the 1/r12 expansion. One finds:
[Insert mathematical equation here]
To start the calculation, we need an initial guess for the ground state Ao expansion coefficient Csi so that we can get an initial estimate of the density matrix elements Ptu. We saw that the optimum orbital exponent for the helium Ao that consists of one 1S To is 27/16=1.6875. Since the orbital exponent ε1 is much closer to 1.6875 than is ε2, we expect that the coefficient of X1 in:
[Insert mathematical equation here]
will be substantially larger than the coefficient of X2. Let us take as an initial guess:
[Insert mathematical equation here]
The normalization condition:
[Insert mathematical equation here]
gives for real coefficients:
[Insert mathematical equation here]
where k=C11/C21. Substitution of k=2 and S12=0.8366 gives C21≈0.3461 and C11≈2C21=0.6922 with n=2 and b=2.
The initial guess C11≈0.6922, C21≈0.3461 gives as the initial density matrix elements:
P11≈0.9583; P12≈P21≈0.4791; P22≈0.2396
The Fock matrix elements are found from b=2 using P12=P21 for real functions, we get:
[Insert mathematical equation here]
After substitution, we get:
[Insert mathematical equation here]
The initial estimate of the secular equation:
[Insert mathematical equation here]
Substitution of the lower root ε1 into the Rootaan equation with r=2 gives:
[Insert mathematical equation here]
The SCF energy is found from with n=2 and b=2 as:
[Insert mathematical equation here]
A more precise calculation with ε1=1.46363 and ε2=2.91093 gives an SCF energy of -2.8616726 as compared with the limiting Hartree-Fock energy -0.8616799 Hartrees found with five basis functions.
Reference Book:
Quantum Chemistry – IRA. N. LEVINE
Submitted by
S. Rubini
Dept of Physics