4.9 Applications and results 77

Table 4.1.Bond lengths in atomic units for three

different molecules. Hartree–Fock (HF) and

experimental results are shown.

Molecule HF Expt.

H 2 1.385 1.401

N 2 2.013 2.074

CO 1.914 1.943

DatatakenfromRef.[6].

The integrals overr 1 andr 2 yield two delta-functions ink 1 andk 2 , andEq. (4.121)
transforms into

〈1s,α,A; 1s,β,B|g|1s,γ,C; 1s,δ,D〉

= 4 πM( 2 π)−^3 (π^2 /ρσ )^3 /^2

∫

d^3 kk−^2 e−k

(^2) (ρ+σ )/( 4 ρσ)
eik(RP−RQ). (4.122)
We have already encountered this integral in the previous subsection. The final
result is now
〈1s,α,A; 1s,β,B|g|1s,γ,C; 1s,δ,D〉

2 π(^5 /^2 )
(α+γ )(β+δ)(α+β+γ+δ)^1 /^2
×exp[−αγ /(α+γ)|RA−RC|^2 −βδ/(β+δ)|RB−RD|^2 ]
×F 0

[

(α+γ )(β+δ)

(α+β+γ+δ)

|RP−RQ|^2

]

. (4.123)

4.9 Applications and results

After having considered the implementation of the Hartree–Fock in a computer
program, we now present some results of HF calculations for simple molecules
[ 6 , 17 ]. As the HF calculations yield an energy for a static configuration of nuclei,
it is possible to find the stable configuration by varying the positions of the nuclei and
calculating the corresponding energies – the lowest energy corresponds to the stable
configuration. In this way, the equilibrium bond lengths of diatomic molecules can
be determined. InTable 4.1, HF results are shown for these bond lengths, together
with experimental results. The table shows good agreement between the two. The
same holds for bond angles, given inTable 4.2. It is also possible to calculate the