4.1.1 Separation of Variables
1 Spherical coordinates
Cartesian coordinates:
z y
x ,
,
φ θ
, ,
r
Spherical coordinates:
, sin
2
φ θ
θ τ
d drd
r d
=
P
x y
z
o
, ,
z y
x M
θ
r
•
•
ϕ ϕ
ϕ ϕ
z
y x
A
= =
, cos
sin
φ θ
φ θ
r x
In spherical coordinates the Laplacian takes the form
∂ ∂
+ ∂
∂ ∂
∂ +
∂ ∂
∂ ∂
= ∇
2 2
2 2
2 2
2 2
sin 1
sin sin
1 1
φ θ
θ θ
θ θ
r r
r r
r r
, sin
2
φ θ
θ τ
d drd
r d
=
= =
. cos
, sin
sin
θ φ
θ
r z
r y
The time-independent Schrodinger equation in Cartesian coordinates
In spherical coordinates
ψ ψ
ψ ψ
ψ
E V
z y
x m
= +
∂ ∂
+ ∂
∂ +
∂ ∂
−
2 2
2 2
2 2
2
2
ψ ψ
φ ψ
θ θ
ψ θ
θ θ
ψ
E V
r r
r r
r r
m =
+ ∂
∂ +
∂ ∂
∂ ∂
+ ∂
∂ ∂
∂ −
2 2
2 2
2 2
2 2
sin 1
sin sin
1 1
2
We begin by looking for solutions that are separable into products:
, ,
,
φ θ
φ θ
ψ
Y r
R r
=
Putting this into above equation, we have
ERY VRY
Y r
R Y
r R
dr dR
r dr
d r
Y m
= +
∂ ∂
+ ∂
∂ ∂
∂ +
−
2 2
2 2
2 2
2
sin sin
sin 2
φ θ
θ θ
θ θ
2 2
2mr −
Dividing by RY
and multiplying by
2 sin
1 sin
sin 1
1
2 2
2 2
2 2
= −
− ∂
∂ +
∂ ∂
∂ ∂
+ E
r V
mr Y
Y Y
Y dr
dR r
dr d
R
φ θ
θ θ
θ θ
∂ ∂
+ ∂
∂ ∂
∂ −
= −
−
2 2
2 2
2 2
sin 1
sin sin
1 2
1
φ θ
θ θ
θ θ
Y Y
Y Y
E r
V mr
dr dR
r dr
d R
The term on the left hand depends only on r
, whereas the right depends only on ; accordingly, each must be a constant, which is in the form
ll+1 :
; 1
2 1
2 2
2
+ =
− −
l l
E r
V mr
dr dR
r dr
d R
. 1
sin 1
sin sin
1
2 2
2
+ −
= ∂
∂ +
∂ ∂
∂ ∂
l l
Y Y
Y Y
φ θ
θ θ
θ θ
r V
V =
Spherically symmetric
potential
4.1.2 The Angular Equation
Solution of Y : Equation 4.17 determines
Y function as
Multiplying above equation by Ysin
2
, it becomes:
. 1
sin 1
sin sin
1
2 2
2
+ −
= ∂
∂ +
∂ ∂
∂ ∂
l l
Y Y
Y Y
φ θ
θ θ
θ θ
. sin
1 sin
sin
2 2
2
Y l
l Y
Y
θ φ
θ θ
θ θ
+ −
= ∂
∂ +
∂ ∂
∂ ∂
As always, we try separation of variables:
,
φ θ
φ θ
Φ ⋅
Θ =
Y
Plugging this in, and dividing by Y,
we find
. 1
sin 1
sin sin
1
2 2
2
= Φ
Φ +
+ +
Θ Θ
φ θ
θ θ
θ θ
d d
l l
d d
d d
The first term is a function only of , and the second is a function only of , so each must be a constant. This time I will call the separation constant
m
2
:
φ
; sin
1 sin
sin 1
2 2
m l
l d
d d
d =
+ +
Θ Θ
θ θ
θ θ
θ
d d
Θ
θ θ
. 1
2 2
2
m d
d −
= Φ
Φ
φ
1 The equation is easy to solve:
.
2 2
2
φ
φ φ
im
e m
d d
= Φ
Φ −
= Φ
φ
Now, when advances by 2
, we return to the same point in space, so it is natural to require that
φ
. 2
φ φ
π
= +
In other words,
1
2 2
= =
+ m
i im
im
e e
e
π φ
π φ
From this it follows that m must be an integer
:
, 3
, 2
, 1
, ±
± ±
= m
2 The equation is not simple.
θ
[ ]
. sin
1 sin
sin
2 2
= Θ
− +
+ Θ
m l
l d
d d
d
θ θ
θ θ
θ
Turn into x
by:
θ
θ
cos =
x
. 1
1 2
1
2 2
2 2
2
= Θ
− −
+ +
Θ −
Θ −
x m
l l
d d
x d
d x
θ θ
Above equation is l
th-order associated Legendre equation .
Therefore, the solution of is
, cos
θ θ
m l
m l
AP x
AP =
= Θ
where P
m l
is the associated Legendre function, defined by
, 1
| |
2 |
| 2
x P
dx d
x x
P
l m
m m
l
− ≡
and P
l
x is the
l th Legendre polynomial
, defined by the Rodrigues formula
:
, 1
2 1
2 l
l
l l
x dx
d l
x P
− ≡
, 1
3 1
, ,
1
2 1
− =
= =
x x
P x
x P
x P
3 30
35 8
1 ,
3 5
2 1
, 1
3 2
2 4
4 3
3 2
2
+ −
= −
= −
=
x x
x P
x x
x P
x x
P
P
l
x is a polynomial of degree
l in x
, and is even or odd according to the
parity of
l .
is not, in general, a polynomial——if m
is odd it carries a factor of
, 1
| |
2 |
| 2
x P
dx d
x x
P
l m
m m
l
− ≡
But for associated Legendre function P
m l
:
. 1
2
x −
l m
≤
, 1
= =
x P
x P
,
1 1
x x
P x
P =
=
, 1
1
2 1
2 1
2 1
1
x x
P dx
d x
x P
− =
− =
, 1
3 2
1
2 2
2
− =
= x
x P
x P
, 1
3 1
2 2
2 1
2 1
2
x x
x P
dx d
x x
P −
= −
=
, 1
3 1
2 2
2 2
2 2
2
x x
P dx
d x
x P
− =
− =
not a polynomial
As
, cos
θ
= x
, 1
= r
, cos
θ
= r
P
x y
z
o
, ,
z y
x M
θ
r
•
•
ϕ ϕ
ϕ ϕ
z
y x
A
, sin
θ
= r
cos
θ
m l
P r
=
Plot:
, 1
= x
P ,
cos
1
θ
= x
P
, sin
1 1
θ
= x
P
, 1
cos 3
2 1
2 2
− =
θ
x P
, cos
sin 3
1 2
θ θ
= x
P
, sin
3
2 2
2
θ
= x
P
cos
θ
m l
P r
=
Plot:
P
x y
z
o
, ,
z y
x M
θ
r
•
•
ϕ ϕ
ϕ ϕ
z
y x
A
, 1
cos 3
1
2
− =
θ
r
, cos
sin 3
θ θ
= r
, sin
3
2
θ
= r
, 1
cos 3
2 −
=
θ
r
, cos
sin 3
θ θ
= r
, sin
3
θ
= r
, cos
3 cos
5 2
1
3 3
θ θ
− =
x P
, 1
cos 5
sin 2
3
2 1
3
− =
θ θ
x P
, cos
sin 15
2 2
3
θ θ
= x
P
, cos
1 sin
15
2 3
3
θ θ
− =
x P
cos
θ
m l
P r
=
Plot:
, cos
3 cos
5 2
1
3
θ θ
θ
− =
r 1
, 1
− =
=
π
r r
1 Notice that l
must be a nonnegative integer
, for the Rodrigues formula to make any sense.
Some Notes:
, 1
| |
2 |
| 2
x P
dx d
x x
P
l m
m m
l
− ≡
l m
≤
If |m|l
, then P
l m
=0. Therefore, for any given
l , there are
2l+1 possible
values of m
: values of
m :
, 2
, 1
, =
l
l l
l l
m ,
1 ,
1 ,
, 1
, ,
1 ,
− −
+ −
− =
, ,
, ,
1 1
x P
x P
x P
x P
l l
l l
l l
± −
± ±
2 Now, the volume element in spherical coordinates is
so the normalization condition becomes
1 sin
sin
2 2
2 2
2
= ⋅
=
φ θ
θ φ
θ θ
ψ
d d
Y dr
r R
d drd
r
, sin
2 3
φ θ
θ
d drd
r r
d =
It is convenient to normalize R and Y separately: It is convenient to normalize R and Y separately:
, 1
2 2
=
∞
dr r
R ,
1 sin
2 2
=
π π
φ θ
θ
d d
Y
where R
determined by
Vr
and Y can be obtained.
; 1
2 1
2 2
2
+ =
− −
l l
E r
V mr
dr dR
r dr
d R
The normalized
angular wave functions are called
spherical harmonics
:
where
, cos
| |
| |
4 1
2 ,
θ π
ε φ
θ
φ
m l
im m
l
P e
m l
m l
l Y
+ −
+ =
≤ ≥
− =
. for
, 1
, for
, 1
m m
m
ε
Now we list here some few spherical harmonics: See book for more
φ
θ θ
π φ
θ
i
e Y
± ±
− =
1 cos
5 sin
64 21
,
2 1
3
Actually, the Y
s are automatically orthogonal, so
[ ]
, sin
, ,
2 m
m l
l m
l m
l
d d
Y Y
′ ′
′ ′
=
δ δ
φ θ
θ φ
θ φ
θ
π π
Notice that
, 1
,
φ θ
φ θ
m l
m m
l
Y Y
− =
−
For historical reasons, l
is called the azimuthal quantum number, and m
the magnetic quantum number.
Notice that the angular part of the wave function, Y ,
, is the same for all spherically symmetric potentials; the actual shape of the potential,
Vr , affects
only the radial part of the wave function, Rr
, which is determined by Equation 4.16:
4.1.3 The Radial Equation