Approximate Eigenvalues via Variation Method.
If we use approximate wavefunction f , then the variation theorem states:
Where E0 is the ground state energy. We set up the f trial function via the valence bond method or the molecular orbital method.
Molecular Orbital Theory
Molecular orbital method - A molecular orbital (trial function f ) is composed of 1 electron coordinate with several nuclei = it is the linear combination of H-atom like atomic orbitals for which we are dealing with the same (1) electron. The resulting molecular orbital is delocalized over several nuclei, with 1 electron. We finally fill all the molecular orbitals with electrons. LCAO - Linear Combination of atomic orbitals: f = cAj A + cBj B one particular 1-electron molecular orbital (as we add electrons we,in essence, multiply these orbitals), where j 1 & j 2 are atomic orbitals.
The actual wavefunction (for all electrons and nuclei) is the product of the molecular orbitals (due to the orbital approximation): Y = f 1 *f 2 * f 3 ...
Total Wavefn = Product of MOs for each electron, each M.O. = LCAO
Example: H2+
Find the molecular orbital approximate eigenvalues for H2+ (represented by nuclei Ha Hb and one electron).
1) B.O. Approx'n drop Nuc.
K.E., drop Nuc-Nuc repulsion as constant term
2) Treat R as a parameter, such that we will solve H approximately for
each value of R.
3) Basis set for H2+ are 1s atomic orbitals centered
on nuclei a and b, such that the molecular orbital y ± is a L.C. of
these: y ± = ca(1sa) ± cb(1sb)
A linear combination of atomic orbitals LCAO
4) Take spin and Pauli
Exclusion Prinicple into account and obtain the total
wavefunction Y (spin-included) MOs by making use of Slater Determinants.
5) Now compute the energy eigenvalues of the system by variational
principle
utilizing Hele+nuc_rep and total wavefn Y
If we let ca = cb =1 we have two functions: y ± =
1sa ± 1sb
Let’s compute the normalization for y ± = 1sa ± 1sb,
we must first discuss overlap integral: 
Thus we have two cases:
Plus case: {ca(1sb)+cb(1sa)}
= {ca(1sa)+cb(1sb)}
ca(1sa-1sb) - cb(1sa-1sb)
= 0, ca = cb = cg
Minus case {ca(1sb)+cb(1sa)}
= - {ca(1sa)+cb(1sb)}
ca(1sa+1sb) + cb(1sa+1sb)
= 0, ca = - cb = cu
Thus our two possible M.O.s are: s g = cg(1sa+1sb) , s u = cu(1sa-1sb)
4) Normalization of each ò
s g*s gdt =1 ò s u*s udt
=1, leads to values for cu&cg
cg = 1¸ [2(1+sab)]½, cu= 1¸
[2(1-sab)]½ Where sab = ò sa*sbdt
5) Now apply variation
principle to compute Eg and Eu for wavefns s g
& s u
Use fact Haa=Hbb, and Hab=Hba
due to Hermiticity (Problem #2 of Homework
#8) Eg = <s g|H|s g> = [Haa+Hab]¸
[1+Sab], Eu = <s u|H|s u>
= [Haa-Hab]¸ [1-Sab]
6) Evaluation of Haa and Hab
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