Download Approximation Methods

Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 18
A trial function that depends linearly on the variational
parameters leads to a secular determinant
- as another example of the variational method, consider a
particle in one dimensional box. We should expect it to be
symmetric about x = a/2 and to go to zero at the walls.
- one of the simplest functions with this properties is xn ( a-x)n ,
where n is a positive integer , consequently , let’s estimate Eo
by using :
  c1x (a  x )  c 2 x 2 (a  x ) 2
(1)
Where c1 and c2 are to be the variational parameters. We find
after the energy of particle in one dimensional box exactly
equal to:
E exact
h2
h2

 0.125000
2
8ma
ma 2
(2)
Now we will use a trial function  to calculate Emin to particle
in one dimensional box.
A trial function can be generally written as:
N
  C n f n
n 1
(3)
Where the Cn are variational parameters and fn are arbitrary
known functions
[1]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 18
Consider
  c1f 1  c 2f 2
Then :
ˆ d   (c f  c f )Hˆ (c f  c f )d 

H

 11 22 11 22
ˆ d  c c f Hf
ˆ d  c c f Hf
ˆ d  c 2 f Hf
ˆ d
 c12  f 1Hf
1
1 2 1
2
1 2 2
1
2 2
2
 c12 H 11  c1c 2 H 12  c1c 2 H 21  c 22 H 22 (4)
Where
ˆ d
H ij   f i Hf
j
(5)
We will note that :
f
i
ˆ d   f Hf
Hf
j
 j ˆ id
(6)
So Hij=Hji
Using this result, equation (4) becomes
ˆ d   c 2 H  2c c H  c 2 H

H
1
11
1 2
12
2
22 (7)

Similary we have
[2]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 18
2
2
2

d


c
S

2
c
c
S

c
1 11
1 2 12
2 S 22

(8)
Where
S ij  S ji   f i f j d 
(9)
The quantities Hij and Sij are called matrix elements .
By substituting equations 7,8 into equation 10
E 


 0 Ĥ 0d 


0
 0d 
(10)
We find that :
c12 H 11  2c1c 2 H 12  c 22 H 22
E (c1 , c 2 )  2
c1 S 11  2c1c 2S 12  c 22S 22
(11)
Before differentiating E(c1,c2) in equation 11 with respect to
c1and c2 , it is convenient to write equation 11 in the form:
E (c1 , c 2 )(c12S 11  2c1c 2S 12  c 22S 22 )  c12H 11  2c1c 2 H 12  c 22H 22 (12)
If we differentiate equation 12 with respect to c1 we find that
(2c1 S 11  2c 2S 12 )E 
E 2
(c1 S 11  2c1c 2S 12  c 22S 22 )  2c1 H 11  2c 2 H 12 (13)
 c1
Because we are minimizing E with repect to c1 ,
equation 13 becomes
[3]
E
 c1
=0 and so
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 18
(2c1 S 11  2c 2S 12 )E  2c1 H 11  2c 2 H 12
c1 (H 11  ES 11 )  c 2 (H 12  ES 12 )  0
(14)
Similarly by differentiating E(c1,c2)with repect to c2 instead of
c1 we find
c1 (H 12  ES 12 )  c 2 (H 22  ES 22 )  0
(15)
Equations (14) and (15) constitute a pair of linear algebraic
equations for c1 and c2
This equation is not simply solved but if c1 = c2
 H 11  ES 11

 H 12  ES 12
H 12  ES 12 
0
H 22  ES 22 
(16)
To illustrate the use of equation 16 let us go back to solving the
problem of a particle in a one – dimensional box variationally
using equation (1)
  c1x (a  x )  c 2 x 2 (a  x ) 2
We will set a = 1. In this case ,
f1 = x (1- x)
and f2 = x2 (1- x) 2
So, we will solve H11 , H12, H22 , S11, S12 and S22 from the
equations (5) , (9)
[4]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 18
ˆ d
H ij   f i Hf
j
S ij  S ji   f i f j d 
2


d
ˆ
 H 11   f 1Hf 1d    x (1  x ) 
x (1  x ) dx
2
 2m dx

0
1

2
1
x (1  x )  0  2dx

2m
0

2
2m
1
2

2
x

2
x
)dx

0
2 1
2
 2 2 3

x  x  
2m 0 
3  6m
- As the same you can get H12 and H22
2
2
H12 = H21=
and
30m
H22 =
1
2
f
dx
1
- Also we can get S11 = 
0
2
1
   x (1  x )  dx
0
[5]
105m
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 18
1
  (x 2  2x 3  x 4 )dx
0
2
1  1
1
10  x 3  x 4  x 5  
4
5  30
3
- As the same we can get S12 and S22
1
S12 = S21 =
140
1
and S22 =
630
- Substituting the matrix elements Hij and Sij into the
secular determinant gives
1 E 
 1 E

 6  30
30 140 

0
1
E 
 1  E



 30 140 105 630 
where
E  = E m/
2
. The corresponding secular equation is
E 2 - 56 E  +252=0
whose roots are
E
56  2128
 51.065and 4.93487
2
We choose the smaller root and obtain
[6]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 18
2
E min
h2
 4.93487  0.125002
m
m
Compare with Eexact when a =1
E exact
h2
 0.125000
m
The excellent agreement here is better than should be expected
normally for such a simple trial function. Note that E min
E exact , as it must be.
Example
Using Equation f1=x (1-x) and f2= x2(1-x) 2, show explicitly
that H12=H21
Solution: using the Hamiltonian operator of a particle in a box,
we have
1
2

d
2
2
ˆ
H 12   f 1Hf 2dx   x (1  x ) 
x
(1

x
)
dx
2
2
m
dx


0

2
2m
1
2

dx
x
(1

x
)
2

12
x

12
x


0
 1 

 
2m  15  30m
2
2
[7]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 18
2

d
2
2 
ˆ
H 21   f 2 Hf 1dx   x (1  x ) 
x
(1

x
)
dx

2
 2m dx

0
1

2
2m
1
2
2
x
(1

x
)
 2dx

0
 1 

 
2m  15  30m
2
2
[8]