Chemistry 342
Spring, 2005
The Hydrogen Atom.
The Φ equation.
The first equation we want to solve is
d 2Φ
dφ 2
  m2Φ
This equation is of familiar form; recall that for the free particle, we had
d2ψ
dx 2
  k 2ψ
for which the solution is
ψ( x)  a 0 cos kx  a 1 / k sin kx
Since
e ix
 cos x  i sin x
a more general solution to equations of this type is
Φ  A e imφ
 B e imφ
In order that
Φ (φ )  Φ (φ  2 π )
The value of Φ at some value of φ
must be the same at φ + 2π, since Φ
is periodic.
it is necessary that
A eimφ
 B e imφ
 A eim( φ 2 π )  B e im( φ 2 π )
 A eimφ eim2 π  B e imφ e im2 π
Since e±im2π = cos (m 2π) ± i sin (m 2π) = 1 only when m = 0, ±1, ±2…, the Φ equation
has solutions
Φ  A eimφ , m  0,  1,  2,
Chemistry 342
Spring, 2005
We can determine A by requiring that the wavefunctions be normalized,

2π
A (2π  0)  1  A
2

2π
0
Φ*Φ dφ 
A
2
2
0
e imφ  eimφ dφ 

A
1
, A 
2π
2

2π
0
dφ  1
1
2π
so
Φm

1
2π
m  0,  1,  2,
e imφ
are the final solutions to the Φ equation.
A postscript.
These wavefunctions are complex. Sometimes it is more useful to have real
wavefunctions. These can be constructed by first defining
Φ

1 i m φ
e
2π

1
(cos mφ  i sin mφ)
2π
Φ

1 i m φ
e
2π

1
(cos mφ  i sin mφ)
2π
and then adding and subtracting Φ  and Φ  …we say, “forming linear combinations”:
Φ symm

1
(Φ 
2
 Φ ) 
1
cos m φ
π
Φ antisymm

1
(Φ 
2
 Φ ) 
1
sin m φ
π
These functions are
also solutions to the
Φ equation. Try it!
each of which is a real function. We cannot associate with these functions a particular m
value, but only with m . The first three of these functions are
Φ0

1
2π
Φ 1 
1
π
cos φ
Φ 1 
1
π
sin φ
etc.
Chemistry 342
Spring, 2005
The Θ equation.
The Θ equation is
1 d 
dΘ 
 sin θ
 
sin θ dθ 
dθ 
m 2Θ
sin 2 θ
 βΘ  0.
Rearranging,
d 2Θ
dθ 2

m2 

 β 
 Θ  0.
sin 2 θ 

cos θ dΘ
sin θ dθ
Now, make the substitutions
x  cos θ
,
d

dθ
  sin θ
d dx
dx dθ
sin 2 θ  1  x 2
d
d2
,
dx
dθ 2
 sin 2 θ
d2
d
 cos θ
2
dx
dx
After some algebra, we get
(1  x 2 )
d 2Θ
dx 2
 2x
dΘ
dx

m2 
 Θ  0.
  β 
(1  x 2 ) 

This equation is identical to the associated equation of Legendre
(1  x 2 )
d 2P
dx 2
 2x
dP
dx

m2 
P  0
  (  1) 
(1  x 2 ) 

if we identify P with Θ and β with ℓ (ℓ + 1).
The solutions P of the associated Legendre equation are called the associated Legendre
functions; these may be expressed in closed form as (since x = cos θ)
P (cos θ)  (1  cos 2 θ)
m
m
2

k 0
(1) k (2  2k )!(cos θ)   m 2 k
2  (  k )! k! (  m  2k )!
Chemistry 342
Spring, 2005
m
Here, P is a polynomial of degree ℓ and order m , where ℓ and m are integers. k is an
(integer) index, and the sum (Σ) runs from k = 0 to an upper limit of
k  (  m ) / 2 if
(  m ) is even
k  (  m  1) / 2 if
(  m ) is odd
Since m is an integer, and since the solutions to the associated Legendre equation are
acceptable only if (  m ) is an integer, it is necessary that
  int eger
,
  m.
with
The solutions P (Θ) must of course be normalized; the requirement that
1 

π
0
Θ*,m Θ,m dθ 
1

1
2
A P* m (cos θ) Pm (cos θ) d (cos θ)
leads to
1

 2  1  (  m !
2
A  



 2  (  m !

which gives
1

 2  1  (  m )!
2 m
Θ ,m (θ)  

 P (cos θ).
2
(


m
)!






These wavefunctions, though they appear to be complicated, are not, at least for small ℓ.
For example,
  0, m  0.
Θ 0, 0 (θ) 
2
2
(s)
  1, m  0.
Θ1, 0 (θ) 
6
cos θ
2
(p)
  1, m  1.
Θ1, 1 (θ) 
3
sin θ
2
(p)
Chemistry 342
Spring, 2005
  2, m  0.
Θ 2, 0 (θ) 
10
(3 cos 2 θ  1)
4
(d)
  2, m  1.
Θ 2, 1 (θ) 
15
sin θ cos θ
4
(d)
  2, m  2.
Θ 2, 2 (θ) 
15
sin 2 θ
4
(d)
You have already met these functions before, though possibly not in this form. These are
the angular functions describing the probability amplitudes in s, p, d orbitals!
Some postscripts.

The associated Legendre functions are derivatives of the Legendre polynomials Pℓ
(cos θ)
1  x 2  2
m
Pm ( x) 
dm
P ( x)
dx m
The L. polynomials
P ( x ) 

k 0
(1) k (2  2k )! x  2 k
2  (  k)! k! (  2k )!
upper limit on Σ: ℓ/2 if ℓ
even. (ℓ-1)/2 if ℓ odd.
are, in turn, solutions of the Legendre equation
(1  x 2 )

The functions
d2z
dx 2
  12
 2x
dz
dx
 (  1) z  0
(z  P).
m
P (cos θ) and P (cos θ) form an orthonormal set in the
interval -1 ≤ cos θ ≤ 1.

The L. functions are symmetric or antisymmetric as ℓ is even or odd
P ( cos θ)  (1)  P (cos θ)
P ( cos θ)  (1)  m P (cos θ)
m

m
The functions do not exceed 1 in absolute value
Chemistry 342
Spring, 2005
P (cos θ)  1 ; e.g. P (1)  1 , P (1)  (1)  .

Since the Pℓ (x) are polynomials, there exist ℓ roots, or ℓ values of cos θ, for which Pℓ
(x) changes sign. The sign of Pℓ (x) is often indicated by a circular diagram,
+1
θ=0
θ=
0
π
2
θ=π
-1
At the north pole in this diagram, θ = 0 and x = cos θ = +1; at the equator, x = cos π
2
= 0; at the south pole, x = cos π = -1. We then use lines on the circle to indicate the
values of θ at which the polynomial is zero:
P1 = cosθ
P0 = 1

Recurrence relations exist for both the P
(2  1) (cos θ) Pm

+
+
+
+
 Pm11
m
P2 = ½ (3 cos2 θ - 1)
and Pℓ, e.g.
 Pm11
The product functions Ym (θ, φ)
Ym (θ, φ)  Θ  ,m (θ) Φ m (φ)
are called spherical harmonies. These are given by the formula
1
2

 (2  1) (  m )!
 m
imφ
Ym (θ, φ)  
 P (cos θ) e .
4
π
(


m
)!




“nodes”
Chemistry 342
Spring, 2005
The R equation.
The radial equation for the electron “in orbit” about the nucleus of the hydrogen atom is
d2R
dr 2
2 dR
r dr

 2m 
Ze 2 
(  1) 
  2 E 
R  0
 
r 
r 2 
 
If we consider bound states (E < 0) only, and introduce the new variables n and ρ, where
E  
r 
m Z2 e 4
2 n2 h2
 
1  n2 2 

ρ 
2  m Z e 2 
Z2 e 2
2 n2 a0

 a 0


2 

m e 2 
1  n a0 

ρ
2 Z 
the radial equation becomes
d 2R
d ρ2

2 dR
ρ dρ
 1 n
(  1) 
 R  0.
    
ρ 2 
 4 ρ
(A)
We seek solutions of the form
R  c u(ρ) ρ eρ / 2
(B)
If (B) is substituted into (A), we find that u (ρ) must satisfy the differential equation
ρ
d 2u
dρ 2
 (2  2  ρ)
du
dρ
 (n    1) u  0
(C)
Eq. (C) is of the same form as the associated equation of Laguerre,
x
d2L
dx 2
 (β  1  x )
dL
dx
 (α  β) L  0.
(D)
(D) has solutions, known as the associated Laguerre polynomials, which are of the form
α β
Lβα ( x )    (1) k
k 0
(α!) 2
xk
(α  β  k )! (β  k )! k!
where α and β are integers, k is an index running from 0 to (α - β), and (α - β) is an
integer greater than zero. Thus, the solutions u(ρ) of Eq. (C) are of the form Lβα ( x ) ,
providing one makes the identifications
Chemistry 342
Spring, 2005
x  ρ , (β  1)  (2  2) , (α  β)  (n    1)
Combining these relations, one finds
β  2  1 , α  n  .
and
L2n1 (p) 
n   1

(1) k 1
k 0
n   !2
ρk
(n    1  k )! (2  1  k )! k!
Eigenvalues.
Since the condition for solution is
(α  β)  (n    1)  0
and since ℓ = 0, 1, 2, …, n may take the values
n = 1, 2, 3, …
with the restriction that
n≥ℓ+1
This gives the allowed (negative) values of the energy
En
 
m Z2 e 4
2 n 2 2
,
n  1, 2, 3, (independen t of , m).
0
This result is identical with the
values obtained by means of the
Bohr theory. The resulting energy
level diagram is shown on the right.
+
30
31
32
20
21

10
nL

me 4
( Z  1)
2 2
me 4
8 2
Chemistry 342
Spring, 2005
Eigenfunctions.
The radial wavefunctions for the hydrogen atom are of the form
R (ρ)  c ρ  e ρ / 2 L2n1 (ρ)
To determine the normalizing constant c, we require that


0

R (r) r 2dr  c2  ρ 2 eρ L2n1 (ρ) r 2dr  1
2
2
0
Substituting r = (na0/2Z)ρ, this becomes
 na 
1  c  0
 2Z 
3
2


0
2
ρ 2   2 e ρ L2n1 (ρ) dρ
 na  2nn   !
 c  0
 2Z  (n    1)!
3
3
2
(EWK , p. 66).
so that
 2Z  (n    1)! 

c   
3
na
 0  2nn   ! 
3
1
2
We choose c < 0 to make the (total) wavefunction positive, so
1
3
2

 2Z  (n    1)! 


R nL (r )   

na 0  2n n   !3 





 2Zr   Zr / na0 2 1  2Zr 

 e

L n   
 na 0 
 na 0 
The first few RnL(r) are, expressed in terms of ρ = 2Zr/na0,
3
R 10
 Z 2
 2   e ρ / 2
 a0 
3
R 30

3
R 20

R 21 
2  Z 2
  ( 2  ρ) e ρ / 2
2 2  a 0 
3
R 31 
3
2
1 Z
  ρ e ρ / 2
2 6  a 0 
1  Z 2
  (6  6ρ  6ρ 2 ) e ρ / 2
9 3  a 0 
R 32

1  Z 2
  ( 4  ρ) ρ e ρ / 2
9 6  a 0 
1
9 30
3
2
 Z  2 ρ / 2
  ρ e
 a0 
Chemistry 342
Spring, 2005
Note the very important “structure” of these wavefunctions. Each function consists of a
constant, times a polynomial in ρ, times an exponential factor in -ρ/2. The last factor
looks, of course, like
R10
e-/2

0
=2
R21
R20

so R10 is a simple exponential. But R20, which contains, in addition, the factor (2-ρ), has
a node at ρ = 2, as shown above. And R21, which contains the factor ρ, goes to zero at
the origin, also as shown above.
Also note that, as n increases, the number of nodes increases as (n - 1)…this structure
being dictated by the highest power of ρ appearing in the polynomial!
Chemistry 342
Spring, 2005
Chemistry 342
Spring, 2005