5.61 Fall 2007
Lecture # 20
page 1
SPHERICAL HARMONICS
( )
m
() ()
Yl m θ , φ = Θ l θ Φ m φ
(
(
1
2
)
)
⎡ ⎛ 2 l + 1⎞ l − m ! ⎤
⎥ Pl m cos θ eimφ
Yl θ , φ = ⎢⎜
⎟
⎢⎝ 4π ⎠ l + m ! ⎥
⎣
⎦
l = 0, 1, 2,...
m = 0, ± 1, ± 2, ± 3,... ± l
m
( )
(
)
Yl m ’s are the eigenfunctions to Ĥψ = Eψ for the rigid rotor problem.
Y00 =
⎛ 5 ⎞
Y20 = ⎜
⎝ 16π ⎟⎠
1
( 4π )
12
1
1
2
(3cos θ − 1)
2
1
⎛ 3 ⎞2
Y10 = ⎜ ⎟ cos θ
⎝ 4π ⎠
Y2±1
1
2
⎛ 15 ⎞ 2
= ⎜ ⎟ sin θ cos θ e± iφ
⎝ 8π ⎠
1
2
⎛ 3⎞
Y1−1 = ⎜ ⎟ sin θ eiφ
⎝ 8π ⎠
⎛ 15 ⎞
Y2±2 = ⎜
sin 2 θ e±2iφ
⎟
⎝ 32π ⎠
1
2
⎛ 3⎞
Y11 = ⎜ ⎟ sin θ e− iφ
⎝ 8π ⎠
∫∫ Y (θ ,φ ) Y (θ ,φ ) sin θ dθ dφ = δ
Yl m ’s are orthonormal:
⎧1
if l = l′
⎩0
if l ≠ l′
Krönecker delta δ ll′ = ⎨
Energies:
Switch
Recall
m′∗
l′
m
l
⎧1
δ mm′ = ⎨
⎩0
ll′
δ mm′
if m = m′
normalization
if m ≠ m′
orthogonality
ˆ m = E Y m)
(eigenvalues of HY
l
lm l
l → J conventional for molecular rotational quantum #
2IE
β = 2 = l l +1 ≡ J J +1
J = 0, 1, 2,...
!
(
)
(
)
5.61 Fall 2007
page 2
Lecture # 20
∴
E
!2
EJ =
J J +1
2I
(
)
J=3
6!2
E3 =
I
Y30 , Y3±1 , Y3±2 , Y3±3 7x degenerate
J=2
3!2
E2 =
I
Y20 , Y2±1 , Y2±2 5x degenerate
(
(
!2
E1 =
I
E0 = 0
J = 1
J=0
(
Y10 , Y10 2x degenerate
(
Y00 nondegenerate
(
)
)
)
)
g J = 2J + 1
Degeneracy of each state
from m = 0, ± 1, ± 2,..., ± J
Spacing between states ↑ as J ↑
!2
!2
⎡ J +1 J + 2 − J J +1 ⎤ =
E J +1 − E J =
⎦ I J +1
2I ⎣
(
)(
) (
)
(
)
Transitions between rotational states can be observed through
spectroscopy, i.e. through absorption or emission of a photon
δ+
hν
δ+
Absorption
δ-
EJ
δδ+
or
EJ+1
δ+
Emission
δ-
EJ
δ-
EJ-1
hν
)
Lecture # 20
5.61 Fall 2007
page 3
Molecules need a permanent dipole for rotational transitions.
Oscillating electric field grabs charges and torques the molecule.
Strength of transition
2
dμ
dx
I JJ
(ξ ⋅ μ)
J
electric field
of light
J
dx
dipole moment
of rotor
Leads to selection rule for rotational transitions:
J = ±1
Recall angular momentum is quantized (in units of ).
Photon carries one quantum of angular momentum.
J = ±1
Conservation of angular momentum
Angular momentum of molecule changes by 1 quantum upon absorption or
emission of a photon.
2
E photon = h
J
J +1
photon
J J +1
= Erot = E J +1
( J + 1)
I
EJ =
photon
J J +1
h
=
4
Define
B
and
B
J
J +1
h
8
2
rotational constant (Hz)
I
h
rotational constant (cm-1)
2
8 cI
(
)
(Hz) = 2B J + 1
J
J +1
(
)
(cm -1 ) = 2B J + 1
2
I
( J + 1)
5.61 Fall 2007
page 4
Lecture # 20
E
J=3
E3 = 12Bhc
ΔE2→3 = 6Bhc
J=2
E2 = 6Bhc
ΔE1→2 = 4Bhc
J=1
E1 = 2Bhc
ΔE0→1 = 2Bhc
This gives rise to a rigid rotor absorption spectrum with
00E= evenly spaced lines.
J=0
2B
νν
01 0→1
→
νν
121→2
→
νν
23 2→3
→
νν
34 3→4
→
νν
νν
45 4→5
→
56 5→6
→
ν
2B (Hz) or 2B (cm -1 )
Spacing between transitions is
(
)
(
)
ν J +1→ J +2 − ν J → J +1 = 2B ⎣⎡ J + 1 + 1⎤⎦ − 2B J + 1 = 2B
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 length r0!
5.61 Fall 2007
2B =
h
4π 2 cI
1
∴
page 5
Lecture # 20
I = µ r02
µ=
m1 m2
m1 + m2
1
⎡
h
⎤
2
⎡
h
⎤
2
-1
r0 =
⎢ 2
(B
in
cm
)
or
r
=
⎥
⎢ 2
⎥ (B in Hz)
0
8
π
cB
µ
8
π
B
µ
⎣
⎦
⎣
⎦