Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 14
The Schrödinger Equation for the Hydrogen Atom can be
solved exactly
- As our model, we shall picture the hydrogen atom as a
proton fixed at the origin and an electron of reduced mass
µ interacting with the proton through the Coulombic
potential
U (r )  
e2
(1)
4 0 r
where e is the proton charge and ε0 is the permittivity of
free space , and r is the distance between the electron and
the proton.
The Schrödinger Equation for the Hydrogen Atom is

2
2 (r , ,  )  U (r ) (r , ,  )  E (r , ,  ) (2)
2
where 𝛁2 written out in full equation 2 is
 1   2 

r
2   r 2 r  r
2
1
 



sin



2

 r sin   
1
 2 

 2 2
2 
 r sin   
 U (r ) (r , ,  )  E  (r , ,  )
- At first sight, this partial differential equation looks
complicated.
[1]
(3)
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 14
- To solve it first multiply through by 2µr2 to obtain

2
  2 
r
r  r



2
 1  
 
1  2 
 sin    sin     sin 2   2 




 2  r 2 U (r )  E   0
- The second term in equation can write it as L̂2ψ. Thus we
can write the Schrodinger equation in the form

2

r
 2   ˆ2
2
r
  L   2 r U (r )  E   0 (4)
 r 
The solution of the Schrodinger equation for the
hydrogen atom is a formidable mathematical
problem, but is of such fundamental importance that
it will be treated in outline here. The solution is
managed by separating the variables so that the wave
function is represented by the product:
(5)
Spherical polar coordinates
The separation leads to three equations for the three
spatial variables, and their solutions give rise to three
quantum numbers associated with the hydrogen
energy levels.
[2]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 14
- When we substitute Lˆ by its value and ψ by R(r) so,
equation 4 become
d  2 dR

r
2 r 2 dr  dr
2
2

  (  1)


U
(
r
)

E
 
 R (r )  0 (6)
2
2

r
 

- Equation (6) is called the radial equation for the hydrogen
atom.
- Equation (6) is an ordinary differential equation in r and must
be solved as the wave function , the energy must be
quantized according to
En  
-
e 4
8 02 h 2 n 2
n= 1, 2,……
(7)
2
2
a


h
/

e
If we introduce the Bohr radius, 0
, then
0
Eq.(7) becomes
e2
En  
8 0a0 n 2
n=1,2,….
(8)
- It is surely remarkable that these are the same energies
obtained from the Bohr model of the hydrogen atom.
- Of course, the electron now is not restricted to sharply
defined orbits but is described by its wave function.
[3]
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 14
- The complete hydrogen wave function are in equation (5)
-Table(1)
The complete Hydrogen like atomic wave functions for n=
1,2, and 3. The quantity Z is the atomic number of the
nucleus, and σ = Z r/a0, where a0 is the Bohr radius.
[4]