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5.61 Fall 2007 Lecture # page 1
SPHERICAL HARMONICS
θ, φ
l
θ
Y
l
m
m
m
φ
1
2
m
m
im φ
Y
l
θ, φ
l
π
l
l
m
m
P
l
cos θ
e
l = 0, 1, 2,... m = 0, ± 1, ± 2, ± 3,... ± l
Y
l
m
’s are the eigenfunctions to
H
ψ = E ψ for the rigid rotor problem.
1
Y
0
0
4 π
1 2
Y
2
0
π ⎠
2
3cos
2
θ − 1
1 1
Y
1
0
π ⎠
2
cos θ Y
2
± 1
π ⎠
2
sin θ cos θ e
± i φ
1 1
2
2
Y
1
− 1
8 π ⎠
32 π ⎠
sin
2
θ e
± 2 i φ
sin θ e
i φ
Y
2
± 2
1
2
Y
1
1
− i φ
8 π ⎠
sin θ e
m m ′∗ m
Y
l
’s are orthonormal:
Y
l ′
θ, φ
Y
l
θ, φ
sin θ d θ d φ = δ
ll ′
δ
mm ′
1 if l = l ′
1 if m = m ′ normalization
Krönecker delta δ
ll ′
δ
mm ′
0 if l ≠ l ′
0 if m ≠ m ′
orthogonality
m
Energies: (eigenvalues of HY )
l
m
= E
lm
Y
l
Switch l → J conventional for molecular rotational quantum #
2 IE
Recall β = = l l
≡ J J
J = 0, 1, 2,...
2
20
5.61 Fall 2007 page 2
∴
E
2
E
J
= J J + 1
2 I
2
0 ± 1 ± 2 ± 3
Y
3
, Y
3
, Y
3
, Y
3
7x degenerate
J = 3
E
3
I
2
0 ± 1 ± 2
J = 2 E
2
= Y
2
, Y
2
, Y
2
5x degenerate
I
2
0 0
J = 1 E
1
= Y
1
, Y
1
2x degenerate
I
0
J = 0
E
0
= 0 Y
0
nondegenerate
Degeneracy of each state g
J
= ( 2 J + 1
from m = 0, ± 1, ± 2,..., ± J
Spacing between states ↑ asJ ↑
2
2
E =
J + 1
− E
J
2 I
J + 1
− J J + 1
I
J + 1
J + 2
Transitions between rotational states can be observed through
spectroscopy, i.e. through absorption or emission of a photon
δ+
δ-
hν
Absorption
E
J
δ+
δ-
E
J+
hν
δ+
δ-
E J
Emission
δ+
δ-
E J- 1
or
Lecture # 20
00 E =
01 ν→
12 ν→ 23 ν→ 34 ν→ 45 ν→ 56 ν→
5.61 Fall 2007 page 4
E
E
3
= 12 Bhc
E
2
= 6 Bhc
E
1
= 2 Bhc
J = 1
J = 2
J = 3
Δ E
0 → 1
= 2 Bhc
Δ E
2 → 3
= 6 Bhc
Δ E
1 → 2
= 4 Bhc
This gives rise to a rigid rotor absorption spectrum with evenly spaced lines.
J = 0
ν
ν
0 → 1
ν
1 → 2
ν
2 → 3
ν
3 → 4
ν
4 → 5
ν
5 → 6
2 B
Spacing between transitions is 2 B (Hz) or 2 B (cm
ν
J + 1 → J + 2
−ν
J → J + 1
= 2 B
J + 1
− 2 B J + 1
= 2 B
Use this to get microscopic structure of diatomic molecules directly from
the absorption spectrum!
Get B directly from the separation between lines in the spectrum.
Use its value to determine the bond lengthr 0
!
Lecture # 20
5.61 Fall 2007 page 5
h
2
m
1
m
2
2 B =
4 π
2
cI
I =μ r
0
μ =
m
1
2
1 1
h
2
h
2
( B in cm
) or r
0
∴ r = ( B in Hz)
0
8 π
2
8 π
2
cB μ B μ
Lecture # 20
