Ground State of the He Atom – 1s State
First order perturbation theory
Neglecting nuclear motion
2 2 2 2
Ze 2
Ze 2
e2
H 
1 
2 


2 mo
2 mo
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, 2009
Substituting
 o h2
a0 
 mo e 2
Bohr radius
r1  a0 R1
Distances in terms of Bohr radius
r2  a0 R2
r12  a0 R12
2
1 2
 2
, etc.
2
2
 x1 a0  X 1
spatial
i derivatives
i i
in
i units
i off Bohr radius
i
Gives
2 1
Ze 2
Ze 2
e2
2
2
1   2 


H 
2
2 mo a 0
4 o a0 R1 4 o a0 R2 4 o a0 R12


Copyright – Michael D. Fayer, 2009
In units of
 1 e2



ground
state
state,
1s,
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.
changed Substitutions
simplify writing equations.)
1 2
Z
Z
1   22 

2
R1 R2
Zeroth order Hamiltonian.
No electron – electron repulsion.
p


Take
0
H 
H'
1
R12


Perturbation piece of Hamiltonian.
El t
Electron
– electron
l t
repulsion.
li
Copyright – Michael D. Fayer, 2009
Need solutions to zeroth order equation
H  0  E 0 0
0
Take
and
d
 0   0 (1) 0 (2)
E 0  E 0 (1)  E 0 (2)
H0 has terms that depend
p
only
y on 1 and 2. No cross terms. Can separate
p
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, 2009
For ground state (1s)
 0 (1) 
 0 ((2)) 
1

1

Z 3 / 2 e  ZR1
Hydrogen 1s wavefunctions for electrons 1 and 2
but with nuclear charge Z.
Z 3 / 2 e  ZR2
The zeroth order solutions are
3
Z
e  ZR1 e  ZR2
 (1, 2)   (1) (2) 
0
0
0

E 0  E 0 (1)  E 0 (2)  2 Z 2 E1 s ( H )
product of 1s functions
sum of 1s energies with
nuclear charge Z
Copyright – Michael D. Fayer, 2009
Correction to energy due to electron – electron repulsion
  H 1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 2 dR2
Copyright – Michael D. Fayer, 2009
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 eR1

+
R2
Let
R> be the ggreater of R1 and R2
R< be the lesser of R1 and R2
Then
R12  R 1  x 2  2 x cos 
x
2 e-
R
R
Copyright – Michael D. Fayer, 2009
1
1

R12 R
1
1  x 2  2 x cos(( )
x
R
R
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.
fo nd
an  x n
Therefore
1
1

R12 R
x
n
Pn  cos( ) 
n
Copyright – Michael D. Fayer, 2009
Now express the
Pn  cos( ) 
in terms of the
 1 & 1 ;  2 &  2
the absolute angles
g of the vectors rather than the relative angle.
g
The position of the two electrons can be written in terms of the
Spherical Harmonics, the solutions to the Φ(φ) and Θ(θ) equations in the H atom.
m
Pn
m
Pn
 cos 1  e im
1
 cos 2  e im
Complete set of angular functions.
2
The result is
  m  ! R m

1
m
im 1  2 
 
P
cos
P
cos
e






1
2

 1 
R12
R


m
!
 
 m 
Copyright – Michael D. Fayer, 2009
  m  ! R m

1
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.
1
for 1s state or any s state.
The entire sum reduces to
R
For other
F
th states,
t t li
limited
it d number
b off terms.
t
Group
G
theory.
th
Full
F ll rotation
t ti group.
Integral of product of three functions. Direct product of representations of
function must contain totally symmetric rep. Formulas exist.
Copyright – Michael D. Fayer, 2009
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, 2009
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 o 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, 2009
atom
exp value (eV)
exp.
calc value (eV)
calc.
% Error
He
79.00
74.80
5.3
Li+
198.09
193.80
2.2
Be+2
B
371 60
371.60
367 20
367.20
12
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, 2009
The Variational Method
The Variational Theorem:
If  is any function such that
*

  d  1
(normalized)
and if the lowest eigenvalue of the operator H is E0,
then
 H     * H  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, 2009
Proof
Consider
 H  E0    H    E0 
  H   E0
The true eigenkets of H are
i
H  i  Ei  i
Expand  in terms of the  i
orthonormal basis set
Copyright – Michael D. Fayer, 2009
   ci  i
Expansion in terms of the eigenkets of H.
i
Substitute the expansion
 H  E 0    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.
 H  E0    c j c j  E j  E0 
j
Copyright – Michael D. Fayer, 2009
 H  E0    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
i
j
j
 E0   0
Therefore,
 H  E0   0
 H  E0    H   E0
Finally
 H     * H  d  E0
The equality holds only if
The lower the energy you
calculate, the closer it is to
the true energy.
gy
   0 , the function is the lowest eigenfunction.
Copyright – Michael D. Fayer, 2009
Using the Variational Theorem
c a trial function
u c o
Pick
  1 , 2 ,   
normalized
Calculate
J    * H  d
J is a function of the ’s.
’s
Minimize J (energy) with respect to the ’s.
Th minimized
The
i i i d J - Approximation
A
i ti to
t E0.
The  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.
Mi i i
Minimize.
If above
b
first
fi t calculated
l l t d energy, approximation
i ti to
t nextt highest
hi h t energy.
If lower, it is the approximation to lowest state and initial energy is approx.
to the higher state energy.
Copyright – Michael D. Fayer, 2009
Example - He Atom
Trial function

Z 3

e
 Z  R1  R2 
Th zeroth
The
h order
d perturbation
b i ffunction
i b
but with
i h Z  Z  a variable.
i bl
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, 2009
 1
 1
Z Z 
1  1
H    12   22 
  ( Z  Z  ) 



R1 R2 
 2
 R1 R2  R12


Want to calculate
J    * H  d
using H from above.
The terms in red give
2 Z 2 E1 s ( H )
Zeroth order perturbation energy with
Z  Z
Therefore,
 2

2
2
d 1d 2   
d 1d 2    
d 1d 2
J  2 Z  E1 s  H    Z  Z     
R2
R12
 R1

2
These two integrals have the same value; only difference
is subscript.
Copyright – Michael D. Fayer, 2009
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, 2009
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, 2009
To get the best value of E for the trial function ,
minimize J with respect to Z  .
J 
5
  4 Z   4 Z   E1 s ( H )  0
4
 Z 
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 E1 s ( 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, 2009
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
371.60
369 86
369.86
0 47
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, 2009