On to 3 dimensions:
z
(x, y, z) → (r, θ, φ)
r0
•
r
•
θ
motion of particle on
the surface of a sphere
y
x
φ
0 < r < ∞
0 < θ < π
0 < φ < 2π
≡ 3D Rigid rotor
volume element r2 sin θ dr dθ dφ
x = r sinθ cosφ
y = r sinθ sinφ
z = r cosθ
−= 2
2 μ r02
⎡ 1 ∂ ⎛
1 ∂ 2Y ⎤
∂Y ⎞
⎢ sin θ ∂θ ⎜ sin θ ∂θ ⎟ + sin 2 θ ∂φ 2 ⎥ = EY
⎝
⎠
⎣
⎦
Y(θ, φ) = wave function
2 μ r02
β= 2 E
=
∂
sin θ
∂θ
∂Y ⎞
∂ 2Y
⎛
2
⎜ sin θ
⎟ + β sin θ Y = − 2
θ
∂
∂φ
⎝
⎠
separation of variables
⇒ Y (θ , φ ) = Θ(θ )Φ (φ )
spherical harmonics
1
d ⎛
dΘ ⎞
2
θ
sin θ
sin
⎜
⎟ + β sin θ
Θ
dθ ⎝
dθ ⎠
depends only on θ
=
−1 d 2 Φ
Φ dφ 2
depends only on φ
⇒ this must be equal to a constant
Chapter 7, continued
3D Rigid Rotor
1
d ⎛
dΘ ⎞
2
2
θ
β
θ
sin θ
sin
+
sin
=
m
A
⎜
⎟
Θ
dθ ⎝
dθ ⎠
1 d 2Φ
= − mA2
2
Φ dφ
Φ mA = AeimAφ ,
mA = 0, ±1, ±2,...
β = A(A + 1), A = 0,1, 2,...
(but see below)
quantization conditions
mA = −A, −A + 1,...0,..., A − 1, A
A=0
→ mA = 0
s
A =1
→ mA = −1, 0,1
p
A=2
→ mA = −2, −1, 0,1, 2
Y (θ , φ ) = YAmA (θ , φ ) = ΘAmA (θ )Φ mA (φ )
d
H atom
θ,φ equations → two quantum #s
2 μ ro2 E 2 I
β=
= 2E
2
=
=
=2
E = A ( A + 1) , A = 0,1, 2,...
2I
2
=
ˆ mA = A ( A + 1) Y mA
HY
A
A
2I
degeneracy = 2A +1
Aˆ 2YAmA = = 2 A ( A + 1) YAmA
Aˆ and Hˆ obviously commute
2
Â
2 : “angular momentum”
operator
G
A = = A ( A + 1)
components of the angular momentum operator
=⎛ ∂
∂ ⎞
Aˆ x = ⎜ y − z ⎟ =
∂y ⎠
i ⎝ ∂z
=⎛
∂
∂ ⎞
sin
cot
cos
−
−
φ
θ
φ
⎜
⎟
∂θ
∂φ ⎠
i⎝
=⎛ ∂
∂ ⎞ =⎛
∂
∂ ⎞
Aˆ y = ⎜ z − x ⎟ = ⎜ cos φ
− cot θ sin φ
⎟
∂z ⎠ i ⎝
∂θ
∂φ ⎠
i ⎝ ∂x
=⎛ ∂
∂ ⎞ = ∂
Aˆ z = ⎜ x − y ⎟ =
i ⎝ ∂y
∂x ⎠ i ∂φ
⎡Aˆ x , Aˆ y ⎤ = i=Aˆ z
⎣
⎦
⎡Aˆ y , Aˆ z ⎤ = i=Aˆ x
⎣
⎦
⎡Aˆ z , Aˆ x ⎤ = i=Aˆ y
⎣
⎦
Aˆ zYAmA = mA =YAmA
Can simultaneously know the magnitude of the angular momentum and
one of its components
Spherical harmonics
Y00 =
1
4π
Y10 =
3
cos θ
4π
spherically symmetric
1/2
⎛ 3 ⎞
Y1±1 = ⎜ ⎟ sin θ e ± iφ
⎝ 8π ⎠
pz
3
sin θ cos φ
4π
3
sin θ sin φ
4π
1/2
⎛ 5 ⎞
2
Y20 ⎜
⎟ (3cos θ − 1)
⎝ 16π ⎠
1/2
⎛ 15 ⎞
Y2±1 ⎜ ⎟ sin θ cos θ e ± iφ
⎝ 8π ⎠
1/2
⎛ 15 ⎞
2
±2 iφ
Y2±2 ⎜
⎟ sin θ e
⎝ 32π ⎠
s
px
py
d z2
15
sin θ cos θ cos φ
4π
d xz
15
sin θ cos θ sin φ
4π
d yz
15
sin 2 θ cos 2φ
16π
d x2 − y 2
15
sin 2 θ sin 2φ
16π
d xy
Consider a sphere of radius
A ( A + 1)=
For ℓ=2, the allowed solutions can be represented as 4 cones and 1 disk.
Although the z component of the angular momentum is specified for each
cone, the x and y components can take on any value.
A(A + 1) =
A=2
You might find the graphical representations in the supplementary
material to be helpful