The helium atom
Consists of a nucleus and
two electrons
Hamiltonian has four kinds
of terms
 Nuclear kinetic energy
 Electronic kinetic
energy
 Nuclear-electron
Coulombic potential
energy
 Electron-electron
Coulombic potential
energy
Physical Chemistry
Lecture 18
The Helium Atom’s Spatial States
H
 Tn

Center of mass separation
Focus on the relative problem
Center of mass is assumed to be at the
nucleus



H
Good approximation as nuclei get heavier
Determine only relative energies of a single atom
Center-of-mass problem again gives a particle-ina-box solution; translational energies
 Te1

pe2
2me
 Te 2

 Vne1
pe22
2me

 Vne2
1
e2
4 0 | r1  r2 |
Hydrogen-like energies and
wave functions
Difference between the
approximate helium atom
and the hydrogen atom
 Z = 1 for hydrogen
 Z = 2 for helium
Nuclear-charge difference
produces scaling of wave
functions and energies
Each electron of helium is
defined by three quantum
numbers, but the energy
depends on only the
principal quantum number

pe21
2me

 Vne1
pe22
2me
nlm (r )  Rnl ( Zr / a0 )Ylm ( ,  )
 
Z 2 Eh 1
2 n2
 Vne2
 Vee
2e 2  1
1 
 


4 0  | r1 | | r2 | 
1
e2
4 0 | r1  r2 |
Consider the problem with neglect of the
electron-electron repulsion
H
 Te1

pe21
2me
 Vne1

 Vne2
 Te 2
2e 2 1

4 0 | r1 |
pe22
2me

2e 2 1
4 0 | r2 |
Decompose into two hydrogen-like problems
H 2 2 (r2 )  E2 2 (r2 )
 (r1 , r2 )  1 (r1 )2 (r2 )
E
 E1
 E2
Approximate wave functions
and energies for helium
In the simplest approximation, wave
functions are products of one-electron wave
functions
Energies are sums of one-electron energies
n1 ,l1 ,m1 ,n2 ,l2 ,m2 (r1 , r2 ) spin
En

Simplification of the heliumatom problem
H11 (r1 )  E11 (r1 )
 Vee
2e 2  1
1 
 


4 0  | r1 | | r2 | 
 Te1  Te 2
pn2
2mn
En1 ,l1 ,m1 ,n2 ,l2 ,m2
 
Z 2 Eh
2
 n1 ,l1 ,m1 (r1 )n2 ,l2 ,m2 (r2 ) spin
 1
1
 n 2  n 2 
2 
 1
1
Independent-electron energies
of the helium atom
Consider the state with
lowest energy
Each electron has a
principal quantum
number of 1


n1 = 1
n2 = 1

  108.8 eV
E1, 2
Next highest

2 2 Eh  1 1 
  
2  12 12 
  4 Eh
E1,1  
n1 = 1 and n2 = 2
n1 = 2 and n1 = 1
 
2 2 Eh
2
1 1
 2  2
1 2 
5
Eh
2
  68.0 eV
 
Checking energies – the
ionization potential
We measure energy
differences
Consider the ionization process
The energy change of this
process can be measured with
a mass-spectrometric
experiment
Experimental first ionization
potential of helium is 24.6 eV
First ionization potential
theoretically predicted to be
much larger by this
E
independent-electron
approximation
Major source of error
Model does not represent the situation accurately
Must incorporate electron-electron repulsion
to give a “better” picture of the energy state
Can be done by several procedures



Exact calculation – very hard
Perturbation theory
Variational calculation

J1s ,1s
Average over a known wave function of the
Coulombic interaction
Can be calculated through tedious calculus
Depends on the kinds of states involved


all
space
 1
e2  *
1s (r2 ) (r2 )d 3r 
1*s (r1 )1s (r1 )
 4 0 | r1  r2 | 

5
Eh
4
 
11
Eh
4
 54.4 eV
*
H = Ha + Hb + Vee
Vee neglected in the
independent-electron
calculation
3
all
space
Use independent-electron
product wave functions
The Hamiltonian must
include all terms

  (r) H (r)d r
E 
Eab
  (r ) (r ) H  (r ) (r )d r

*
a
*
b
1
3
a
2
a
1
b
2
all
space

  (r ) (r ) H  (r ) (r )d r
*
a
1
*
b
3
b
2
a
1
b
2
all
space

  (r ) (r )V
*
a
1
*
b
2
ee
a (r1 ) b (r2 )d 3r
all
space
 Ea( 0 )
 Eb( 0 )
 E (1)
Ionization potential with the
Coulomb repulsion included
The ionization
potential is the

 e
measurement of the He  He
energy difference
With the inclusion of
E  EHe,1s  EHe,(1s )
the electron
  54.4 eV  (74.8 eV )
repulsion, the
 20.4 eV
ionization potential
is much closer to the
measured value
Eexp  24.6 eV
2
5
Eh
4
Total energy is a sum of the unperturbed
energy with the first-order correction
E   4 Eh

First-order perturbation
theory
Coulomb repulsion integral, J

 E(1s ) 2 , He
  54.4 eV   108.8 eV 
With an approximate
wave function, one can
calculate an
approximate energy

Ground-state energy of helium
through first order

 e
Perturbation theory
Neglect of electron-electron repulsion gives
an incomplete picture of the energy situation

 E1s , He
Must reconsider the
approximation

Problems with independentelectron approximation

He  He 
  74.8 eV
2
Variational principle
One can guess any
approximate wave
function



Eapp
Use to calculate an
approximate energy
Approximate will be
always be incorrect
unless the true wave
function is chosen
*
3
 app Hd r
all
space
Eapp
 Etrue
Variational principle


All approximate energies
will be higher than the
exact energy
Can use this to “improve”
the wave function and
the energy
Eapp
c
The approximations to the
helium-atom energies
The improvements of
the wave function give
improvements of the
energies
With modern computerassisted techniques,
one gets approximate
states that are
extremely close to the
exact solution of the
problem
 0
Summary
One cannot solve for the helium atom’s spatial
states exactly
Use first-order perturbation theory


Independent-electron model is solvable
“Correct” the energy by integral over unperturbed states
 Evaluate Coulomb integrals
Still not adequate to model the experimental data


Variational calculation allows a closer approach
Major problem is “correlation energy”
 Electron avoidance
 More sophisticated numerical solutions allow close
approach to ground-state wave functions


Self-consistent field
Configuration interaction
3