MODULE 8
APPROXIMATION METHODS I
Once we move past the two particle systems, the Schrödinger equation cannot be solved exactly.
The electronic inter-repulsion term in the hamiltonian prevents us from separating it into the sum
of one-electron operators, and thus exact solutions for systems more complicated than the
hydrogenic atoms elude us. You might think that because of this quantum mechanics would be
therefore quite uninteresting to chemists who, as a community, focus their attention on systems
having a multitude of particles. Fortunately approximation methods have been developed which
can generate solutions that are very accurate. Eventually we shall consider the two that are most
widely used, viz., variation theory and perturbation theory. In this Module we confine ourselves
to the Variation Method.
The Variation Method
First consider the ground state of some arbitrary system defined by a wavefunction 0 with an
energy eigenvalue E0. These satisfy the Schrödinger equation
Ĥ 0  E0 0
We multiply equation (8.1) from the left by the complex conjugate of the wave function,
(8.1)
rearrange, and integrate over all space, when we obtain
E0
 Ĥ d

  d
*
0
0
*
0
(8.2)
0
When the wavefunction is normalized the denominator is unity, but this is not necessary. Now
suppose we substitute any other function  for 0 and calculate the corresponding energy E
 * Hˆ  d

E 
  * d
(8.3)
E  E0
(8.4)
Then, according to the variation theorem,
and the equality holds only if Thus the energy E of the new system will never be less
than that of the ground state. From this we see that we can calculate an upper limit for the value
of E0 by using some trial function. The closer we can get the wavefunction  to be like 0, the
closer will E approach the true value E0.
1
The way to proceed is to choose a trial function that depends on one or more arbitrary
parameters (variational parameters). For example choose  (, , , …). The energy will also
depend on the parameters and thus equation (8.4) will become
E ( ,  ,  ,...)  E0
(8.5)
At this point what is to be done is to minimize Ewith respect to the parameters and thus obtain
the best possible ground state energy that our trial function can provide. The success of this
approach clearly depends on one's ability to select a good trial wavefunction. This can require
some clever guesswork. Let us look at an example.
We already know that we can determine the energy of the ground (1s) state of a hydrogen atom
exactly without recourse to approximations.
So this is a good test case for the variation
approach-how well does it recover the known value, E1s
 me e 4 / 32 2 02
2
, or 0.5 a.u.?
In the 1s state, l = 0, and the hamiltonian can be simplified to
 2 d  2 d 
e2
ˆ
H
r

2me r 2 dr  dr  4 0 r
Now we decide on the trial function. Our intuition (guided by our knowledge of the exact
(8.6)
solution) tells us that the wavefunction will decay to zero as r increases. One such function is
the Gaussian of the form
 (r )  e r
2
(8.7)
where the term  is the variational parameter.
Figure 8.1 shows a plot of this
1
function with  = 1.4.
FIGURE 8.1
0.8
Now we have to substitute our
trial function and the operator as
0.6
required into the numerator and
f( r )
denominator of equation (8.3). It
0.4
is straightforward to show that
this leads to equation (8.8)
0.2
0
2
1.5
1
0.5
0
0.5
1
r
2
1.5
2
3 2
e2 1/ 2
(8.8)
 1/ 2
2me 2  0 3/ 2
Now we want to minimize Ewith respect to by differentiating and setting the result equal
E ( ) 
to zero. Thus:
dE ( ) 3 2
e2


d
2me (2 )3/ 2  0 1/ 2
(8.9)
whence
me 2e4
(8.10)
18 3 0 2 4
This is the value of  that will minimize E(). Now we need to substitute the value for  back

into equation (8.8) to find Emin.
Emin  
4  mee4

3  16 2 02

 mee4


0.424

2 
2 2

 16  0
2



(8.11)
This value can now be compared to the exact value of E1s from above:
 mee4
1  mee4 
E1s   


0.500


2 2
2  16 2 02 2 
 16  0
As the theorem predicts, Emin > E1s, i.e. it is less negative.
2



(8.12)
In order to write the expression for the normalized trial wavefunction it is convenient to put  in
terms of the Bohr radius,a0. Thus:

8
9 ao2
(8.13)
and the trial function then is
1/ 2
2 2
8  1 
 ( r )  3/ 2  3  e  (8 / 9 ) r a0
3    a0 
Comparative plots of this function and sare shown in Figure 8.2.
3
(8.14)
It is apparent that the Gaussian is a very
poor approximation to the exponential,
FIGURE 8.2
particularly at values of r less than the Bohr
radius. Nevertheless the trial wavefunction
does lead to a value for the energy of the
state, which is within 80% of the exact
result. This was with a single variational
parameter (). It might be anticipated that
by using more flexible trial functions with
more parameters, we could approach the
true value even closer (see later).
The helium atom
Now let us try the variational method to find the minimized energy for a situation where we do
not know the exact solution. The obvious case is that for the helium atom, one nucleon (Z = 2)
and two electrons. The Hamiltonian operator for this is given by:
2e 2  1 1 
e2 1
(8.15)




2me
4 0  r1 r2  4 0 r12
The Schrödinger equation resulting from this operator cannot be solved exactly because of the
Hˆ  
2
(12  2 2 ) 
occurrence of the potential energy term involving r12. If we ignore the inter-electronic repulsion
term the ground state wavefunction becomes separable
0 ( r1 , r2 )   1s ( r1 ) 1s ( r2 )
(8.16)
Both the 1s wavefunctions in equation (8.16) are of the form (remember for 1s, l = 0)
1/ 2
 Z3 
 1s ( rj )   3 
  a0 
e
 Zrj / a0
(8.17)
where j = 1 or 2.
Equation (8.17) can be used as a trial function in which Z is employed as a variational parameter.
(Because the variational parameter is part of the exponential function this is an example of a nonlinear variation)
To do this we need to evaluate the quotient
4
 ( r , r ) Hˆ  ( r , r )drdr

E(Z ) 
  (r , r ) (r , r )drdr
0
1
0
2
1
0
2
0
1
1
2
2
1
1
2
(8.18)
2
The result of this, after laborious math (see Lowe) is
me e4  2 27 
E(Z ) 
Z  Z 
(4 0 ) 2 
8 
(8.19)
or,
27
Z a.u.
8
me e 4
when E is expressed in atomic units, i.e.
(4 0 ) 2
E(Z )  Z 2 
(8.20)
Minimizing E(Z) with respect to Z we find that Zmin = 27/16, whence, after substitution back into
equation (8.20), we obtain
2
 27 
Emin      2.8477 a.u.
 16 
The experimental result is –2.9033 and the best-refined calculation provides –2.9037. Thus the
result for Emin is reasonable (it is less negative, i.e. greater than the experimental value)
considering the simplicity of the trial function we have employed. The use of nuclear charge (Z)
as a variational parameter implies that the Zmin of 27/16, or 1.6875 can be interpreted as an
effective nuclear charge. That the value is significantly less than 2 reflects the fact that each
electron is partially screening the nucleon from the other electron, thereby reducing the net
charge seen by the nucleon. This lessening of the nuclear attraction means that the 1s orbital in
helium will be expanded from that which would be expected if it were simply a hydrogenic 1s
with Z = 2.
Linear Combinations and Secular Determinants
One way to increase the number of variational parameters in the quest for improved accuracy is
to employ a trial function that is a linear combination of known functions and the coefficients are
the parameters to be varied. Such a trial function can be written as
N
   cn f n
(8.21)
n 1
At this stage let us opt for simplicity and choose N = 2 and require that cn and fn are real. Such
requirements can be relaxed, but for now they keep the process simple. Then
5
  c1 f1  c2 f 2
(8.22)
when
  Hˆ  d   (c f c f ) Hˆ (c f  c f )d
ˆ f  Hc
ˆ f )d
  (c f c f )( Hc
ˆ f  c f Hc
ˆ f  c f Hc
ˆ f  c f Hc
ˆ f d
  c f Hc
ˆ d  c c f Hf
 c  f Hf
 ˆ d  c c  f Hfˆ d  c  f Hfˆ d
2
1
1 1
2 2
1 1
2 2
1 1
1 1
1
1 1
2 2
1 1
1 1
2 2
2 2
1 2
1
1
2 2
2
1 1
2 1
2 2
2
1
(8.23)
2 2
2
2
2
2
 c12 H11  c1c2 H12  c2c1H 21  c22 H 22
where
ˆ d
H ij   f i Hf
j
(8.24)
We know from our postulates that the operator has to be hermitian in order to guarantee real
eigenvalues, hence
 f Hfˆ d   f Hfˆ d
1
2
2
1
H ij  H ji
Using the hermitian property equation (8.23) becomes
  Hˆ  d  c H
2
1
11
 2c1c2 H12  c22 H 22
(8.25)
Following the same routine, the denominator of equation (8.3) becomes
  d  c S
2
2
1 11
2c1 c2 S12  c22 S22
(8.26)
where
Sij  S ji   f i f j d
(8.27)
We have met quantities such as Hij and Sij before and called them matrix elements.
Using equations (8.25) and (8.26) in equation (8.3) we obtain:
c12 H11  2c1c2 H12  c22 H 22
c12 S11  2c1c2 S12  c22 S22
so now we have E as a function of the variational parameters.
E (c1 , c2 ) 
(8.28)
Prior to differentiating equation (8.28), it is better to write it in a different way, thus
E (c1 , c2 )(c12 S11  2c1c2 S12  c22 S22 )  c12 H11  2c1c2 H12  c22 H 22
(8.29)
Now multiplying out the LHS and taking the partial derivative wrt c1, we get
(2c1S11  2c2 S22 ) E 
E 2
(c1 S11  2c1c2 S12  c22 S22 )
c1
 2c1H11  2c2 H12
6
(8.30)
Now we put the derivative term equal to zero since we are minimizing, and rearrange, thus
c1 ( H11  ES11 )  c2 ( H12  ES12 )  0
(8.31)
Now we differentiate equation (8.29) again, but now wrt c2 and eventually obtain
c1 ( H12  ES12 )  c2 ( H 22  ES22 )  0
(8.32)
These last two equations are a pair of linear homogeneous algebraic equations, containing the
two unknowns c1 and c2. The trivial solution is that both coefficients equal zero, but then the
wavefunction would equal zero. Non-trivial solutions to such a set of equations do exist, if, and
only if, the determinant of the coefficients vanishes, i.e. if
H11  ES11
H12  ES12
(8.33)
0
H12  ES12 H 22  ES22
This is called the secular determinant, and expansion of it yields a polynomial (here a quadratic)
secular equation in E. Finding the roots of the quadratic yields two values of E, of which we
select the lowest as the approximate energy of our ground state entity.
Let us look at an example of this linear combination procedure. The one chosen is the particle in
the one-dimensional box with vertical walls where the potential energy becomes infinitely large,
but within the box, V(x) = 0. The Hamiltonian for this situation is that for kinetic energy only

The trial functions we choose are from the set
f n  x n (1  x )n
FIGURE 8.3
0.3
n
2
d2
2m dx 2
(8.34)
(8.35)
Although this may seem a strange choice, these
functions have the requisite form, meaning that
n
f(x)=x (1-x)
n=3
they have zero amplitude at the walls and a single
0.2
f(x)
maximum at the center of the box, exactly as
needed for the n = 1 wavefunction. The first
0.1
three such functions are graphed in Figure 8.3.
n=2
n=1
0.0
0.0
0.2
0.4
0.6
0.8
1.0
x
To keep matters simple we use the first two terms only as our BASIS SET, thus
trial  c1 x (1  x )  c2 x 2 (1  x ) 2
7
(8.36)
Using the outlined procedure we evaluate the H and S integrals to be:
H11 
2
6m
H12  H 21 
H 22 
S11 
;
2
30m
; S12  S 21 
2
105m
1
30
S 22 
;
1
140
1
630
These lead to the secular determinant:
where E ' 
1 E'
1
E'


6 30
30 140
0
1
E'
1
E'


30 140 105 630
(8.37)
E '2  56 E ' 252  0
(8.38)
Em
2
This generates the quadratic equation:
from which E’ = 51.065 and 4.9347 are extracted as roots. Thus
Emin  4.9387
2
m
 0.125002h 2 / m
(8.39)
This is to be compared to the exact energy
n 2h 2
h2

( when n  L  1)
8mL2 8m
(8.40)
h2
 0.125000
m
Thus, using the linear combination technique with a basis set of only two, we arrive at an answer
Eexact 
which is within 2 ppm of the exact value - a remarkable achievement.
In many cases we find it useful to operate with a larger basis set, and in the general case for a
basis set of N-terms we can write the set of linear equations as follows:
c1 ( H11  ES11 )  c2 ( H12  ES12 )  ...cN ( H1N  ES1N )  0
c1 ( H 21  ES21 )  c2 ( H 22  ES22 )  ...cN ( H 2 N  ES2 N )  0
down to
c1 ( H N 1  ES N 1 )  c2 ( H N 2  ES N 2 )  ...cN ( H NN  ES NN )  0
8
(8.41)
In these equations the matrix element indicators have been sequenced in anticipation of the next
step. In order for these equations in E to have non-trivial solutions, the secular determinant must
vanish, viz.
H11  ES11
H12  ES12
... H1N  ES1N
H 21  ES21 H 22  ES22 ... H 2 N  ES2 N
0
.
.
.
H N 1  ES N 1 H N 2  ES N 2 ... H NN  ES NN
(8.42)
In the determinant the elements have been inserted using the convention Mrc where r = row
number and c = column number. This can then be simplified using the hermitian conditions that
Hij = Hji and Sij = Sji.
Expansion of the determinant leads to a secular N-order polynomial in E. For N > 2 we need to
use computational methods to obtain the roots. Of the roots we select the smallest to give us the
value of Emin. From the root taken we can then substitute back into equation (8.41) to find the
values of ci. However only N-1 of the equations are independent and we can evaluate only the
N
ratios cN/c1. If we require that the function be normalized to unity (  ci  1 ), then the value of
2
i 1
c1 can be extracted and thence all the rest.
An extension of the above method is to use a trial function of the form
N
   cn f n
(8.21)
n 1
where the fn functions also contain variational parameters, eg.
(8.43)
f n  e n r
Now there are two sets of variational parameters, cn and n. The minimization of E with respect
2
to the two sets of parameters involves complicated math and needs to be done numerically.
Fortunately algorithms are available for this.
Table 8. 1 shows the results obtained by taking different numbers of terms for the ground state of
the hydrogen atom.
9
TABLE 8.1
N
Emin in a.u.
1
-0.4244
2
-0.4858
5
-0.49976
10
-0.49998
exact
-5.00000
I suggest that you read the following material that I have scanned from Lowe’s book very
carefully.
Section 7-4 provides a clear and detailed exposition of the polarization of the
hydrogen atom by an applied electric field. It demonstrates the mixing of the 1s and 2p z orbitals,
sets up the secular determinant and analyses and evaluates the different matrix elements. It is
seen that the off-diagonal elements are the determining factors in the extent of mixing and shows
how this is affected by the strength of the applied field. You will find references to Appendix 2
but ignore these because it is background on the algebra of determinants, which you can find in
Barrante if you need it.
10
Start your reading at 7-4, but recognize that the early material has been covered in an earlier
section of this Module. I left it in so you have an alternative viewpoint.
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