Simplification of the hydrogenatom problem
By substitution, the hydrogen-atom
Hamiltonian becomes
Physical Chemistry
H
Lecture 17
The Hydrogen Atom, a CentralForce Problem

1 2
PCM
2M
 H CM
1 2
p
2

 H rel
Decompose into two problems
H CM CM
 ECM CM
H rel rel
  CM rel
The hydrogen atom
Nuclear kinetic energy
Electronic kinetic energy
Nuclear-electron
Coulombic potential
energy


H
 Tn

 Te
pn2
2mn
 Vne
pe2
2me





Center of mass, RCM
Relative position, r
Total mass, M
Reduced mass, 
E  ECM

1 e2
4 0 | rne |
The relative problem
Consider relative motion
of nucleus and electron,
with only Coulombic
interaction
m1
r1 
M
 r1

R CM
r  r2
M
 m1
m2
r2
M
The dependence of 2
on angular variables
simplifies the problem
 m2
 
m1m2
m1  m2
p12
2m1
p22
2m2

 Erel
A particle of mass, M, moving in no
potential
Like the particle-in-a-box problem
Energies and wave functions known
from that model
Translational degrees of freedom of the
hydrogen atom
Center of mass
Hamiltonian as a
function of nuclear
and electronic coordinates is complex
Define
 Erel rel
Center-of-mass problem
Consists of two
particles, a positively
charged nucleus and a
negatively charged
electron
Hamiltonian has three
terms

1 e2
4 0 r



2
PCM
2M

Eigenfunctions of L2 are
well known
H  E


2 2
 
2
 2 1   2  
r
 
2  r 2 r  r 
1 e2
  E
4 0 r
1 2
L 
2 r 2
1 e2

4 0 r
 E
 (r , ,  )  R (r )Ym ( ,  )
p2
2
1
Hydrogen-atom energy levels
The radial equation
Use of the spherical harmonic functions gives an
equation for the radial part

 1   2 R 
 r
 
2  r 2  r r 
2
  1
2
2
R
r2
2
1 e
R  ER
4 0 r

This can be put into dimensionless form by defining
the radial distance in bohrs
r
 a0
a0
 0h2
me e 2

The dimensionless equation is related to Laguerre’s
differential equation
Rn ( ) 
Defined relative to
the “vacuum level”

States are labeled
by n and 

Hydrogenic energies and
eigenfunctions
nm (r , ,  ) 
Anm r / a0  Ln (r / a0 ) exp(r / na0 )Ym ( ,  )

Laguerre polynomials

0
0
1
0
1
2
n
1
2
2
3
3
3
Ln(x)
1
2-x
1
3 – 2x + 2x2/9
4 – 2x/3
1
 
1 e2 1
4 0 2a0 n 2
gn
Addition of spin
The spatial part is incomplete
One incorporates spin as a separate coordinate
Wave function is a product
n ,l ,m , I ,mI (r , ,  ) 
Energies of the state depend only on n
En
States of different 
given by letters
 s ( = 0)
 p ( = 1)
 d ( = 2)
A  Ln ( ) exp  / n 
Products of spherical harmonic functions with
functions related to Laguerre polynomials
Energy level in
which the electron
and the nucleus are
just pulled apart
Energy is unchanged by addition of spin
variables
Increases degeneracy
 n2
Energies
Electron spin
Can give energies in
“macroscopic units”


S=½
Two states
Joules
Ergs


Often convenient to
use “atomic units”



Hartree
Rydberg
Electron volt
A Rnl (r / a0 )Ylm ( ,  ) spin , S ,ms
1 hartree  27.2114 electron volts
1 rydberg  13.606 electron volts
 0.5 hartree
1 electron volt  96.4853 kilojoules / mole
m = ½ or 
m = - ½ or 
Eigenstates of the spin
angular momentum
operators




S2  = ½(½+1)2 
S2  = ½(½+1)2 
Sz  = (/2) 
Sz  = - (/2) 
2
Summary
Central-force (hydrogen-atom) problem
separates


Center of mass movement (translation)
Relative motion
 Angular part is constant-angular-momentum problem
 Radial part is related to Laguerre’s equation
Energy depends on principal quantum
number, n, as predicted by Rydberg
Degeneracy


States with same n but different angular
momentum quantum numbers,  and m, have
same energy
gn = n2
3