Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 17
Approximation Methods
- We showed before that the Schrödinger equation cannot be
solved exactly for any atom or molecule more complicated
than the hydrogen atom.
- Approximation methods can be used to solve the Schrödinger
equation to almost any desired accuracy.
Approximation methods
The Variational Method
Perturbation Theory
1- The Variational Method Provides an Upper Bound to
the Ground – State Energy of a System
- We will first illustrate the variational method. Consider
the ground state of some arbitrary system.
- The ground state wave function ψ0 and E0 satisfy the
Schrödinger equation
Ĥ  0  E 0 0
Multiply Equation (1) by  0
 0Ĥ  0   0E 0 0
And integrate over all the space to obtain
[1]
(1)
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 17
E0 


 0 Ĥ  0d 


 0 0d 
(2)
where d  represents the appropriate volume element. We
have not set the denominator equal unity in equation (2)
to allow for the possibility ψ0 is not normalized.
- If we substitute any other function ϕ for ψ0 in equation
2 and calculate the corresponding energy according to
E 


 0 Ĥ 0d 


0
 0d 
(3)
then Eϕ will be greater than the ground- state energy E0.
In an equation, we have the variational principle
E   E o (4)
Where the equality holds only if    o , the exact wave
function.
The variational principle says that we can calculate an
upper bound to E0 by using any trial function we wish.
The closer
Eo .
 is to  o
in some sense , the closer
We can choose a trial function

E  to
such that it
depends upon some arbitrary parameters, … called
Varaitational parameters. the energy also will depend
upon these varitional parameters , and equation (4)will
read:
E  ( ,  ,  ,....)  E o (5)
[2]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 17
E
Now we can minimize
with respect to each of
varitional parameters and there by determine the best
possible ground-state energy that can be obtained from
our trial wave function.
As a specific example, consider the ground state of the
hydrogen atom. Although we know that we can solve
this problem exactly , let’s assume that we cannot and
use the variational method. We will compare our
varitional result to the exact result. Because l=0 in the
ground state, the Hamiltonian operator is:
d  2 d
Hˆ  
r
2
2m e r dr  dr
2
e2


 4 o r
(6)
Even if we did not know the exact solution, we would
expect that the wave function decays to zero with
increasing r. consequently, as a trial function, we will
try a Gaussian of the form  (r )  e
where  is a
variational parameter. By a straight forward calculation
we can show.
 r 2

2 32
2
3

e
4   (r )Hˆ  (r )r dr 

12
4 o 
4 2m e 
0
*
2
and that

  
4   * (r ) (r )r 2dr  

2



0
Therefore from equation 3:
[3]
32
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 17
3 2
e 2 1 2
E ( ) 
 12
2m e 2  o  3 2
(7)
We now minimize E () with respect to by
differentiating with respect to  and setting the result
equal to zero. We solve the equation:
dE ( ) 3 2
e2

 12
0
32
d
2me   o (2 )
For  to give
me2e 4

18 o2 3
4
As the value of  that minimize
equation 8 back in equation 7
E min
4

3
 me e 4

2 2
16

o

(8)
E(). Substituting

 me e 4
 0.424 
2 
2 2
16

o



2  (9)

Compared with exact value
1  me e 4
Eo   
2  16o2 2

 me e 4
 0.500 
2 
2 2
16

o



2  (10)

Note that Emin> Eo as the variational theorem assures us.
[4]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 17
Example (1)
Use a trial function of the form e-αr to calculate the
ground state energy of a hydrogen atom.
Solution:
The Hamiltonian operator for the ground state energy of
hydrogen atom is given by equation
Hˆ  
d  2 d
r
2
2m e r dr  dr
2
 r ˆ   r 2
e
 He r dr
E0 
ˆ
He
 r

numerator = 4π  e

 r

=
(2r   r )e
2
2
 r

e2
4 0 r
e  r
ˆ  r r 2dr
He
2 
2
= m
e
2
=
me
2 r 2
e
 r dr
2
2m e r
e2


 4 o r
 (2r   r )e
2
0
2
 r
dr 
e2
0
 2
2  e 2 1
 (2 ) 2  (2 )3    (2 ) 2

 0
2
2m e 

e2
4 0 2
[5]

2 r
e
 rdr
0
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 17
Denominator =
4  e
2 r
8

r dr 
 3
3
(2 ) 
2
and so
2
e 2
E 

2m e 4 o
2
Setting dE/d =0
mee 2

4 o 2
And substituting this result back into E() gives
E min
1  m 2e 4
 
2  16 2 o2

2 

This happens to be the exact ground state energy of a hydrogen
atom.
[6]