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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 19
The variational Method to estimate the ground- state energy of a
helium atom
- We will use the variational method to estimate the groundstate energy of a helium atom.
- The Hamiltonian operator for a helium atom is
2
2
2
2
e
1
1
e
1
Hˆ  
(   ) 
(  )
2m e
4 0 r1 r2 4 0 r12
2
1
2
2
(1)
- The Schrödinger equation cannot be solved exactly for this
system because of the term involving r12 .
- Equation (1) can be written in the form
2
e
1
Hˆ  Hˆ H (1)  Hˆ H (2) 
4 0 r12
(2)
where
Hˆ H (1)  
2
2e 2 1
 
2m e
4 0 r1
2
1
and
Hˆ H (2)  
2
2e 2 1
 
2m e
4 0 r2
2
2
So, we can write it as
2
2
2
e
1
Hˆ H ( j )  
 
2m e
4 0 r j
2
j
[1]
j = 1and 2 (3)
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 19
Hˆ H ( j ) is
the Hamiltonian operator for a single electron around a
helium nucleus. Thus, Hˆ H (1) and Hˆ H (2) satisfy the equation
Hˆ H ( j ) H (rj , j ,  j )  E j H (rj , j ,  j ) j = 1 or 2 (4)
- where  H (rj , j ,  j ) is a hydrogen like wave function with Z = 2
and where the
E j are
given by
me e 4
mee 4
En   2 2 2  
8 0 h n
32 2 02 2 n 2
Z 2mee 4
E j  
32 2 02 2 n 2j
j = 1 or 2
(5)
- with Z = 2. If we ignore the inter electronic repulsion term
(e2/4πε0r12), then the Hamiltonian operator is separable and
the ground – state wave function would be
0 (r1 , r2 )   1s (r1 ) 1s (r2 )
(6)
from (Table in hydrogen lecture)we take  1s value
1/2
 Z3 
 Zr / a
 1s (r j )   3  e j 0 j = 1 or 2
  a0 
(7)
Where a0 = 4πε0ħ2/mee2. We can use Equations (6) and (7) as
a trial function using Z as a variational constant. Thus, we
must evaluate
E ( Z )   0 (r1 , r2 )Hˆ 0 (r1 , r2 )dr1dr2
[2]
(8)
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 19
 Z 3   Z ( r1  r2 )/a0
0 (r1 , r2 )   3  e
  a0 
(9)
From equation (1). The Hamiltonian operator of the helium
atom is
Hˆ  
2
2
2e 2
2e 2
e2
 
 


2m e
2m e
4 0 r1 4 0 r2 4 0 r12 (10)
2
1
2
2
We substitute equation 9 and 10 in equation 8. The result is
mee 4
E (z ) 
16 2 02
(Z 2 
2
27
Z)
8
(11)
Equation (11) suggests that it is convenient to express E in
unite of
mee 4
16 2 02
2
, and so we can write Equation 11 as
E (z )  ( Z 2 
27
Z ) (12)
8
If we minimize E(z) with respect to Z, we find that Zmin =
27/16. We substitute this result in equation 11 to obtain
mee 4
E (z )  2.8477
16 2 02
2
(13)
The experimental value is
mee 4
E  2.9037
16 2 02
[3]
2
(14)
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 19
Thus, we achieve a fairly good result, considering the
simplicity of the trial function.
The value of Z that minimize E can be interpreted as an
effective charge. That fact that Z comes out to be less than
2 reflects the fact that each electron partially screens the
nucleus from the other , so that net effective nuclear charge
is reduced from 2 to 27/16 (1.68).
[4]