Solution of the Schrödinger Equation for an LCAO-MO
H (i )  (i )    (i )
mo

mo
mo
mo
mo
(i )H (i )  (i )d    (i )   (i )d
mo
mo
mo
mo
mo
    (i )  (i )d
mo
mo
  c 
where
(
mo
mo
  c 
mo
m
m
n
n
and
n
  is the basis set
n
is an identical summation)
m
substitute:

(i ) H (i )  (i )d     (i )  (i )d
  c  (i ) H (i )  c  (i ) d
mo
m
mo
m
m
mo
mo
mo
n
mo
mo
n
    c   (i )  c  (i )d
mo
n
m
m
m
n
n
n
c
n
m
c   (i ) H (i ) (i )d     c c   (i ) (i )d
c
n
m
c H   c c S
c
m
c H   S
m
m
m
n
n
m
n
mo
mn
n
n
mo m
n
mo
mn
mn
mo
m
n
m
mo
n
S    (i )  (i )d
mn
m
n
m
n
m
n
mn
0
H    (i ) H (i ) (i )d
mn
m
Hamiltonian integrals
Overlap integrals
n
are integrals over the basis set  .
n
c
m
n
m
c H   S
n
mn
mo
mn
0
is the Schrödinger Equation, integrated and reorganized in an LCAO-MO form.
The problem is to solve this for
c  and 
n
mo
.
The Variation Principle: The ground state of the problem will be the one with the lowest
possible energy. For any variable in the wavefunction
, the best choice of its value is

 /   0 .
given by:
mo
For the LCAO-MO problem this means that:
 / c  0 for all n.
mo
n
Differentiate the equation
c
m
n
m
c H   S
n
noting that
mn
H ,S
mn
mo
mn
mn
  0 wrt c
n
are constants
 / c   c c H   S
  0
 2c H   S     c c  S  / c   0
  c H   S   0 for all n
n
m
m
m
n
m
n
mn
m
mn
mo
mn
m
mn
mo
mn
mo
m
n
mn
m
n
mn
mo
n
The SECULAR EQUATIONS
For these equations to have a non-trivial solution requires:
det H   S  0
mn
mo
mn
The SECULAR DETERMINANT
Solving the Secular Determinant give (many) values of the molecular orbital energy
 
a
mo
then substituting a mo energy into the Secular Equations gives the
corresponding mo wavefunction
c 
a
n
Methodology:
1)
Choose a geometry for the molecule (Born Oppenheimer approx.)
2)
Choose a Basis Set of N atomic orbitals
 
n
3)
Obtain values of the Hamiltonian integral
4)
From where??
Set up the N x N Secular Determinant
det H   S  0
mn
mo
mn
and solve for N values of
5) For each value of 
a
  c 
mo
, i.e.
mo
mn
 
and overlap integrals
S
a
mo
solve the Secular Equations for
c  which gives the
a
n
mo
a
LCAO-MO

H
a
n
n
n
This effectively solves the molecular orbital Schrödinger equation for N mo.s of the
molecule.
mn
.