Ground State of the He Atom – 1s State
First order perturbation theory
Neglecting nuclear motion
2
2
2
2
2
Ze
Ze
e
H 
12 
 22 


2mo
2mo
4 o r1 4 o r2 4 o r12
kinetic energies
attraction of electrons
to nucleus
electron – electron repulsion
1 - electron 1
2 - electron 2
r1 - distance of 1 to nucleus
r2 - distance of 2 to nucleus
r12 - distance between two electrons
Copyright – Michael D. Fayer, 2007
Substituting
 o h2
a0 
 mo e 2
Bohr radius
r1  a0 R1
r2  a0 R2
Distances in terms of Bohr radius
r12  a0 R12
2
1 2
 2
, etc.
2
2
 x1 a0  X 1
spatial derivatives in units of Bohr radius
Gives
2


1
Ze 2
Ze 2
e2
2
2
H 
1   2 


2
2mo a0
4 o a0 R1 4 o a0 R2 4 o a0 R12
Copyright – Michael D. Fayer, 2007
In units of
 1 e2



ground
state,
1s,
energy
of
H
atom


 2 4 o a0

e2
4 o a0
H 


1 2
Z
Z
1
1   22 


2
R1 R2 R12
(Nothing changed. Substitutions
simplify writing equations.)
Take
H 
0
H'


1 2
Z
Z
1   22 

2
R1 R2
1
R12
Zeroth order Hamiltonian.
No electron – electron repulsion.
Perturbation piece of Hamiltonian.
Electron – electron repulsion.
Copyright – Michael D. Fayer, 2007
Need solutions to zeroth order equation
H  0  E 0 0
0
Take
and
 0   0 (1) 0 (2)
E 0  E 0 (1)  E 0 (2)
H0 has terms that depend only on 1 and 2. No cross terms. Can separate
zeroth order equation into

1 2 0
Z
1 (1)   E 0 (1)   0 (1)  0
2
R1 

 0
1 2 0
Z  0
 2 (2)   E (2) 
 (2)  0
2
R2 

These are equations for hydrogen like atoms with nuclear charge Z.
Copyright – Michael D. Fayer, 2007
For ground state (1s)
 0 (1) 
 0 (2) 
1

1

Z 3 / 2 e  ZR1
Z 3 / 2 e  ZR2
Hydrogen 1s wavefunctions for electrons 1 and 2
but with nuclear charge Z.
The zeroth order solutions are
3
Z
 (1, 2)   (1) (2) 
e  ZR1 e  ZR2
0
0
0

E 0  E 0 (1)  E 0 (2)  2 Z 2 E1s ( H )
product of 1s functions
sum of 1s energies with
nuclear charge Z
Copyright – Michael D. Fayer, 2007
Correction to energy due to electron – electron repulsion
  H1s ,1 s
E   H nn
expectation value of perturbation piece of H
  0* H ' 0 d 1d 2
e 2 ZR1 e 2 ZR2

d 1d 2
2 
4 o a0 
R12
e2
Z6
d 1  sin 1 R12 d1d 1dR1
Electron – electron repulsion depends
on the distance between the two electrons.
spherical polar coordinates
d 2  sin 2 R22 d 2 d 2dR2
Copyright – Michael D. Fayer, 2007
This is a tricky integral.
The following procedure can be used in this and analogous situations.
R12  R12  R22  2 R1 R2 cos 
 is the angle between the two vectors
R1 and R2.
1 e-
R1

+
R2
2 eLet
R> be the greater of R1 and R2
R< be the lesser of R1 and R2
Then
R12  R 1  x 2  2 x cos 
x
R
R
Copyright – Michael D. Fayer, 2007
1
1

R12 R
1
1  x 2  2 x cos( )
Expand in terms of Legendre polynomials (complete set of functions in cos()).
1
1

R12 R
 a P  cos( )
n
n
n
The an can be found.
an  x n
Therefore
1
1

R12 R
x
n
Pn  cos( ) 
n
Copyright – Michael D. Fayer, 2007
Now express the
Pn  cos( )
in terms of the
 1 & 1 ;  2 &  2
the absolute angles of the vectors rather than the relative angle.
The position of the two electrons can be written in terms of the
Spherical Harmonics.
m
Pn
m
Pn
 cos1  e im
1
 cos 2  e im
Complete set of angular functions.
2
The result is
 m  ! R

1
m
m
im 1  2 
 
P
cos

P
cos

e




1
2
1
R12
R

m
!
 
m 
Copyright – Michael D. Fayer, 2007
 m  ! R

1
m
m
im 1  2 
 
P
cos

P
cos

e




1
2
1
R12
R

m
!
 
m 
Here is the trick.
The ground state hydrogen wavefunctions involve
P00 (cos1 )e im1
P (cos 2 )e
0
0
im 2
1s wavefunctions have spherical harmonics with
0
m0
These are constants. Each is just the normalization constant.
A constant times any spherical harmonic except the one with,  0 m  0
which is a constant, integrated over the angles, gives zero.
Therefore, only the  0 m  0 term in the sum survives when doing
integral of each term.
The entire sum reduces to
1
R
for 1s state or any s state.
Copyright – Michael D. Fayer, 2007
Then
e 2 ZR1 e 2 ZR2
E 
d 1 d 2
2 
4 o a0 
R
e2
Z6
The integral over angles yields 162.

e 2 ZR1 e 2 ZR2 2
2
E   16 Z
R
dR
R
dR2
1
1
2


4 o a0 0 0
R
6
e2
This can be written as
R1  R2

16 Z 6 e 2 2 ZR1  1
E 
e


4 o a0 0
 R1
R1
e
0
2 ZR2

R dR2   e
2
2
R1
2 ZR2
 2
R2 dR2  R1 dR1

R2  R1
Copyright – Michael D. Fayer, 2007
Doing the integrals yields
5
e2
E  Z
8 4 o a0
Putting back into normal units
 5 

E    Z  E s (H )
 4 
1 e2
E s (H )  
 13.6 eV
2 4 0 a0
1
1
negative number
Therefore,
5 

E  E 0  E   2Z 2  Z  E s ( H )
4 

1
Electron repulsion raises the energy.
For Helium, Z = 2, E = -74.8 eV
Copyright – Michael D. Fayer, 2007
atom exp. value (eV)
calc. value (eV)
% Error
He
79.00
74.80
5.3
Li+
198.09
193.80
2.2
Be+2
371.60
367.20
1.2
B+3
599.58
595.00
0.76
C+4
882.05
877.20
0.55
Experimental values are the sum of the first and second ionization energies.
Ionization energy positive. Binding energy negative.
Copyright – Michael D. Fayer, 2007
The Variational Method
The Variational Theorem:
If f is any function such that
*
f
 f d  1
(normalized)
and if the lowest eigenvalue of the operator H is E0,
then
f H f   f * H f d  E0
The expectation value of H or any operator for any function is always
great than or equal to the lowest eigenvalue.
Copyright – Michael D. Fayer, 2007
Proof
Consider
f H  E0 f  f H f  f E0 f
 f H f  E0
The true eigenkets of H are
i
H  i  Ei  i
Expand f in terms of the  i
orthonormal basis set
Copyright – Michael D. Fayer, 2007
f   ci  i
Expansion in terms of the eigenkets of H.
i
Substitute the expansion
f H  E 0 f   ci  i  H  E 0   c j  j
i
j
The  j are eigenkets of (H – E0). Therefore, the double sum collapses
into a single sum.
  c j c j  j  H  E0   j
j
Operating H on  j returns Ej.
f H  E0 f   c j c j  E j  E0 
j
Copyright – Michael D. Fayer, 2007
f H  E0 f   c j c j  E j  E0 
j
c jc j  0
A number times its complex conjugate is positive or zero.
E j  E0
An eigenvalue is greater than or equal to the lowest eigenvalue.
Then,
E
and
j
 E0   0
c c  E
j
j
j
 E0   0
i
Therefore,
f H  E0 f  0
f H  E0 f  f H f  E0
Finally
f H f   f * H f d  E0
The equality holds only if
f  0
The lower the energy you
calculate, the closer it is to
the true energy.
, the function is the lowest eigenfunction.
Copyright – Michael D. Fayer, 2007
Using the Variational Theorem
Pick a trial function
f  1 , 2 , 
normalized
Calculate
J   f * H f d
J is a function of the ’s.
Minimize J (energy) with respect to the ’s.
The minimized J - Approximation to E0.
The f obtained from minimizing with respect to the ’s - Approximation to  0 .
Method can be applied to states above ground state with minor modifications.
Pick second function normalized and orthogonal to first function.
Minimize. If above first calculated energy, approximation to next highest energy.
If lower, it is the approximation to lowest state and initial energy is approx.
to the higher state energy.
Copyright – Michael D. Fayer, 2007
Example - He Atom
Trial function
f
Z 3

e
 Z  R1  R2 
The zeroth order perturbation function but with Z  Z  a variable.
Writing H as in perturbation treatment after substitutions
H 


1 2
Z
Z
1
1   22 


2
R1 R2 R12
Z Z

Rewrite by adding and subtracting
R1 R2
 1
 1
Z Z 
1  1
H    12   22 
  ( Z  Z  )  


R1 R2 
 2
 R1 R2  R12


Copyright – Michael D. Fayer, 2007
 1
 1
Z Z 
1  1
H    12   22 
  ( Z  Z  )  


R1 R2 
 2
 R1 R2  R12


Want to calculate
J   f * H f d
using H from above.
The terms in red give
2 Z 2 E1 s ( H )
Zeroth order perturbation energy with
Z  Z
Therefore,
 f2

f2
f2
J  2 Z  E1 s  H    Z  Z     
d 1d 2   
d 1d 2    
d 1d 2
R2
R12
 R1

2
These two integrals have the same value; only difference
is subscript.
Copyright – Michael D. Fayer, 2007
For the two integrals in brackets, performing integration over angles gives

6 



Z
2 Z R1
2 Z R2 2
2
2 16
e
R1 dR1  e
R2 dR2   2 Z 
2 
 0
0


In conventional units 2 Z 
e2
4 o a0
The last term in the expression for J 1/ R12 term 
was evaluated in the perturbation problem
except Z  Z  .
The result is (in conventional units)
5
e2
Z
8 4 o a0
Copyright – Michael D. Fayer, 2007
Putting the pieces together yields
e2
2
5
e
J  2 Z 2 E1 s ( H )  ( Z  Z )2 Z 
 Z
4 o a0 8 4 o a0
5 

  2 Z 2  4 Z ( Z  Z )  Z   E1 s ( H )
4 

1 e2
E1 s ( H )  
2 4 o a0
5 

  2 Z 2  4 ZZ   Z   E1 s ( H )
4 

Copyright – Michael D. Fayer, 2007
To get the best value of E for the trial function f,
minimize J with respect to Z  .
J 
5
  4 Z   4 Z   E1 s ( H )  0
 Z 
4
Solving for Z  yields
Z  Z 
5
16
Using this value to eliminate Z in the expression for J
5 

2


J   2 Z  4 ZZ  Z   E1 s ( H )
4 

yields
E  2 Z 2 E1s ( H )
This is the approximate energy of the ground state of the He atom (Z = 2)
or two electron ions (Z > 2).
Copyright – Michael D. Fayer, 2007
atom
exp. value (eV)
calc. value (eV)
% Error
He
79.00
77.46
1.9
Li+
198.09
196.46
0.82
Be+2
371.60
369.86
0.47
B+3
599.58
597.66
0.32
C+4
882.05
879.86
0.15
He perturbation theory value – 74.8 eV
Copyright – Michael D. Fayer, 2007