1
Chapter 7. A Quantum Mechanical Model for the
Vibration and Rotation of Molecules
Harmonic oscillator:
Hooke’s law:
kxF , x is displacement
Harmonic potential:
2
2
1)( kxFdxxV
k is force constant:
02
2
xdx
Vdk (curvature of V at equilibrium)
Newton’s equation:
kxFdt
xdm
2
2
(diff. eqn)
Solutions:
x = Asint, (position)
21)( /k/m (vibrational frequency)
2
Verify:
rhskxtAm
km
tmAtdt
dmA
dt
tAdmlhs
sin
sincossin 2
2
2
Momentum:
tAmdt
dxmp cos
Energy:
22
2
)sin(2
1
2
)cos(
2
tAkm
tAm
Vm
pE
22
)(sin)(cos2
222
2222
kAAm
ttAm
When t = 0, x = 0 and p = pmax
= mA
When t = , p = 0 and x = xmax
= A
Energy conservation is maintained by oscillation between
kinetic and potential energies.
3
Schrödinger equation for harmonic oscillator:
Ekx
dx
d
m
2
2
22
2
1
2
Energy is quantized:
2
1E ... 1, 0,
where the vibrational frequency
2/1
m
k
Restriction of motion leads to uncertainty in x and p, and
quantization of energy.
Wavefunctions:
2/2
)()( yeyHNx
where
4/1
2,
mkxy
Normalization factor
4
2/1
!2
N
)(yH is the Hermite polynomial
1)(0 yH
yyH 2)(1
24)( 22 yyH
...
11 22 HyHH (recursion relation)
Let’s verify for the ground state
2/
00
22
)( xeNx
2/22
242
0
2/222/222/2
0
2/222/2
0
2/2
2
22
0
22
222222
2222
22
222
2
1
2
2
1
2
2
1
2
x
xxx
xx
x
em
xm
kN
ekxexem
N
ekxxedx
d
mN
ekxdx
d
mNlhs
Note that
5
4/1
2
mk
It is not difficult to prove that the first term is zero, and the
second term as 2/ .
2/0
22
2
1 xeNrhs
So, lhsrhs , it is a solution.
One can also show normalization:
14
222
0
20
2
0
22
dxeNdx x
where we have taken advantage of the even symmetry of the
Gaussian
22axxa ee
and used the integral formula
a
dxe ax
40
2
For higher states, the wavefunctions are essentially the product
of a Gaussian and a polynomial:
6
The Gaussian is always an even function, but the polynomial
can either be even )()( xfxf or odd )()( xfxf . As a
result, the wavefunctions are either even or odd, depending on .
Properties of the oscillator
i. Zero point energy ( 0 )
2
10 E (quantum effect)
Even at 0 K, there is still vibration.
ii. Orthonormality:
dx*
iii. Average displacement
0)()(*
dxxxx
7
No matter if )(x is even or odd, 2
)(x is even, and thus the
integrant is odd. Integration of an odd function in , is
always zero.
iv. Average momentum
0)()(*
dxxdx
d
ix
Because the derivative of an even function is an odd function,
and vice versa.
v. Nodes
The state has 1 nodes (due to Hermite polynomial)
vi. Energy separation
)2/1(2/111 EEE
(equal separation).
vii. Tunneling
Wavefunction penetrates into classically forbidden regions.
A quantum phenomenon.
Example: A harmonic oscillator has a force constant of 475 N/m
and a mass of 1.61×10-27 kg. Calculate the ZPE and the photon
frequency needed to bring the oscillator from ground state to the
first excited state.
8
114
271043.5
1061.1
/475
s
kg
mN
m
k
( 2// skgmN )
J
sJsE
20
114340
1086.2
1043.510054.15.02
1
JEE 200 1072.52
nm
sJs
J
h
E
3476
1063.810626.6
1072.5 113
34
20
( c )
Example: The IR spectrum of H35Cl is dominated by a peak at
2886 cm-1. Calculate the force constant of the molecule.
Reduced mass
kgamukgamu 2727 1061.1/1066.197.0351
351
114110 1043.52886/100.314.32
~22
scmscm
c
mNskg
skgk
/475/475
1043.51061.1
2
2114272
9
What about DCl?
HCl
kgamukgamu
2
1014.3/1066.189.1352
352 2727
HClskg
skgk
2
11089.3
1014.3
/475 114
27
2
HClcmscm
s
c
~
2
12063
/100.314.32
1089.3
2~ 1
10
114
Isotope effect is important to assign spectral lines.
Rigid rotors
Angular momentum vector
l r p I (r fixed)
where the angular velocity and moment of inertia are
d
dt
, 2I r
The rotational energy (classical):
2
2
lE
I
10
Schrödinger eqn. for 2D motion:
E
yxm 2
2
2
22
2
Use polar coordinate system ( sincos r, yrx , it becomes
for 2D rigid rotor:
Ed
d
I 2
22
2
General solution:
2/1
2
2/12
,2
1)(
IEmeim
m
Cyclic boundary condition:
)2()(
So,
)()1()(
)2(
22
2)2(
mimmi
immiim
eeN
eNeNe
2m has to be even integers, i.e.
, ..., , m 210
11
Quantization
I
mEm
2
22 (2D)
Angular momentum:
mlz (quantized angular momentum)
Angular momentum operator
ixy
yx
i
pypxL xyz
ˆˆˆˆˆ
is a common eigenfunction of both operators:
me
iime
iL imim
z
2/12/1
2
1
2
1ˆ
I
me
IH im
2
)(
2
1
2ˆ
22/1
2
22
Uncertainty principle (just like p and x)
2/ zl
12
3D rotors:
Spherical coordinates:
cossinrx
sinsinry
cosz r
ddrdrdxdydzd sin2
Hamiltonian:
22
2ˆ
mH
2
2
2
22
2
2
2
2
2
22
2
ˆ1
2
2
mr
Lr
rrm
zyxm
The differential operator 2 is called Laplacian.
If r is fixed, the 3D rigid rotor Hamiltonian:
I
LH
2
ˆˆ
2
(rigid rotor)
where
13
sin
sin
1
sin
1ˆ2
2
2
22 L
Schrödinger eqn.
EI
L
2
ˆ2
The solution is the spherical harmonics:
)()(cos),(),( mm
llmm
l PNY
where
)!(
)!(
2
)12(
ml
mllNlm
im
m e
2/1
2
1)(
(2D rotor)
and )(cosm
lP is the associated Legendre function.
The quantization of two angular coordinates leads to two
quantum numbers:
l = 0, 1, 2, ...
m = 0, ±1, ±2, ..., ±l (2l+1)
14
Rotational energy (3D rotor)
I
llEl
2
)1( 2
Since E is independent of m, each l level is 2l + 1 degenerate,
which can be removed by an external field (Zeeman effect).
When 0m , )(cosm
lP becomes the Legendre polynomial:
1)(cos0 P
cos)(cos1 P
2/1cos3)(cos 22 P
…
For 0m
)(coscos
sin)(cos
lm
mmm
l Pd
dP
Examples of spherical harmonics:
100 Y
cos01 Y
ieY
2
1sin1
1
1cos32
1 202 Y
15
They are eigenfunctions for both 2L̂ and zL̂ :
),()1(),(ˆ 22 ml
ml YllYL
),(),(ˆ ml
mlz YmYL
Verification for cos01 Y
22
22
2
2
2
22
2cossin2sin
sinsin
10
cossinsin
1
sin
1ˆ
L
II
LH
2
2
2
ˆˆ
22
0cosˆ i
Lz
The spherical harmonics are orthonormal
mmllm
lm
l YYdd
2
0
*
0
),(),(sin
Note the volume element and ranges.
16
Spherical harmonics have nodes due to Legendre functions.
For cos01 Y , the angular node can be determined by
cos 0 , o90
Vector model:
The angular momentum can be considered as a vector L
. In
quantum mechanics, it is quantized in
i. magnitude:
)1( llL
ii. orientation: mLz
Only certain directions of L
are allowed.
Example (l = 1, ml = 0, ±1):
Example: Calc. amplitude of L, Lz and E for a rotor with I =
4.6x10-48 kg m2 and l = 1.
17
JsJsllL 3434 1049.110055.12)1(
JsmL lz3410055.1 ,0 (3-fold degen.)
The angle between L
and the z-axis for 1lm :
1
cos2
zL
L , o45
Jkgm
Js
I
LE 21
248
2342
104.2)106.4(2
)1049.1(
2
Example. A 3D rotor absorbs at 11 cm-1. Calculate its
moment of inertia.
I
EEE2
)02( 2
01
~hchE
so
247
110
34
2
1006.5
11/100.314.32
1005.1
~2~
kgm
cmscm
Js
chcI