### A Quantum Mechanical Model for Vibration and Rotation of Molecules

```Translational
Quantum Mechanical Models
Vibration and Rotation of Molecules
Chapter 18
Vibrational
Molecular
Energy
Molecular
Motions
Rotational
Electronic
…
Vibrations of Molecules:
Frequency of vibration – classical approach
Model approximates molecules to atoms joined by
springs. A vibrations (one type of – normal mode of
vibration) of a CH2 moiety would look like;
Water:
CO2
O2
http://en.wikipedia.org/wiki/Molecular_vibration
The motions are considered as harmonic oscillators.
For a molecule of N atoms there are 3N-6 normal
modes (nonlinear) or 3N-5 (linear).
Vibrational motion - harmonic oscillator, KE and PE
– classical approach
Center of mass, does not move.
Center of mass coordinates
Diatomics:
Force constant of spring = k
Whether pulled apart or pushed together from the
equilibrium position, the spring resists the motion
by an opposing force.


Amplitudes are real numbers, b1 and b2 are real or = 0.
Applying the BC, set at t = 0; x(0) = 0, v(0) = v0.
=1
Solutions to the DE will be of the form;
Therefore:
where b1  c1  c2 and
b2  i (c1  c2 )
and to find T:

angular velocity = 
Frequency  = 1/T
k
 2

where   frequency
 = phase angle
Energy terms (KE, PE) are
1 2
1
v
PE  kx 2
2
2
where v and x are velocity and position.
KE 
Harmonic oscillator potential function
Morse curve
 0
No restriction on
E classically.
Study Example Problem 18.1
Similar for low energy situations = ground state
1. Expression for total vibrational energy,
Etot ,vib  KE  PE 

x
,x
1 2 1 2
 v + kx
2
2
p2 1 2
 kx
2 2
2. Construct the total energy Operator, the Hamitonian
2
Equivalent to a particle of mass , vibrating around its
equilibrium distance. How can a particle like that be
described by a set of wavefunctions.
  1 p 2  1 kx 2  1  i d   1 kx 2
H
2
2
2   dx  2
3. Set the Schrodinger equation, eigenequation
  E
H
H n ( 1/ 2 x ) - Hermite polynomials
n = vibrational quantum number
Solutions (normalized wavefunction, eigenfunctions)
of which are;
With the normalization constant
The first few wavefunctions,  n (n = 0,..4)
The first few
wavefunctions
(n = 0,..3)
The respective eigen energies are;
Note the resemblance to 1D box.
zpe
Equal energy gaps ~ contrast to 1D box.
The first few wavefunctions,  n2 (n = 0,..4)
The constraint imposed on the
particle by a spring results in zero point energy.
The energy values of vibrational states are precisely
known, but the position of the particles (as described
by the amplitude is imprecise, only a probability
 ]  0.
density 2(x) of x can be stated, [x, H
Note: the overflow (forbidden) of wavefunction beyond
the potential barrier.
Note the resemblance to 1D box.
At high quantum numbers n (high energy limit) the system
gets closer to a classical system (red probability of x in q.m.
oscillator, blue probability of x classical oscillator.
n = 12
n >>0
high
low high
high
low high
Blue to black, a particle in a barrel
Rotational energies of a classical rigid rotor (diatomic).
In vibrational motion, velocity, acceleration and
momentum are parallel to the direction of motion.
In the center of mass coordinate system, rigid rotor motion
equivalent to a mass of  moving in a circle of radius
r (= bond length) with an angular velocity  and
a tangential velocity v. Rotation in 2D (on a plane)
y
r0
No opposing force to rotation –
no PE (stored of energy).
All energy = KE.
x
p = linear momentum
Angular velocity vector
Angular velocity and acceleration,
r0
Mass with constant  makes  = 0.
RHR
Tangential linear velocity,
Angular velocity (~ momentum)
vector normal to the plane of motion.
(coming out of the plane)

= I, moment of inertia
The rotation on the x,y plane occurs freely. Equivalently
the particle moves on the circle (radius r = r0) freely;
Epot = V(x,y) = 0
QM Angular momentum (2D):
p
2E
I
angular momentum
Etot ,vib  KE  PE  KE 
r
p2
2
2
2
2
2
  p       ( x, y )    ( x, y ) 
H


2
2
2   x
y 2 
Magnitude of l = l
Energy:
  vr
SE:
No restriction on
l (or E) classically.
In polar coordinates
(variable is ):
r2sin d d dr
BC: ( )  (  2 )
Rotational
eigenfunction.
General solution for DE (as seen before):
ml = integer
clockwise
counter-clockwise
ml = rotational quantum number, quantization
Normalization constant A;
2

*
ml
  ( ) 
( ) ml ( )d  1
1
2
e  iml
0
2
A2
e
 iml  iml
e
d  1
0
2
A2

d  1
A2  2   1
0
A 
  ( ) 
1
2
m being an integer,
rotational frequency
 will take finite
values.
1
1
  ( ) 
2
Stationary state
1
e  iml
2
1
2
1
2
Non-stationary state
unacceptable
2
real part
e  iml =
1
2
(cos m l + i sin m l )
1
cos 
2
1
cos 3
2
1
cos 5
2
cos 2
cos 4
cos 6
Energy of rotational states:

Note; No non-zero ZPE.
ZPE appears if the potential
energy confinement exists,
not here (V()=0)
Degenerate states
- two fold
(equal energy)

2 Eml
I
=
2E
I
2
2  2 ml2 ml
Eml 

I
I 2I
I
m being an integer, rotational frequency  will take
finite values. – discrete set of rotational frequencies!!
The wavefunctions, …. of a rigid rotor are similar
to the wavefunctions of a free particle restricted to
a wall of a circle (particle in a ring)!
 ,l z ]  0
Note: for 2D rigid rotor [ H
Angular Momentum – 2D rigid rotor.
Operator:
p
l z  i 

l z  i 

Both operators has the same eigenfunctions.
angular analogue
Hence angular momentum (z component) for this case can
be precisely measured.
But position, here , cannot be measured precisely;
,l z ]  0
probability for any interval of d is the same [
r
Angular momentum (in z direction) is quantized!!
 ,l z ]  0
Note: for 2D rigid rotor [ H
QM Angular momentum (3D):
Particle on a circle.
r
r0
r0
2D
3D
Particle on the surface of a sphere.
Classical 3D rotor
KE operator in polar coordinates =
Just the steps in solving the SE
a. collect the constants:
b. multiply by sin2, rearrange to get
PE confinement – none; V(,) = 0
PE operator set to zero. E = KE + PE = KE
SE:
Differentiation only w.r.t. 
Therefore
Wavefunction;

;separation of variables
possible.
spherical harmonic functions
c. substitute for Y(,) divide by ()(), we get
solutions 
Because each side of the equation depends only one of
the variables and the equality exists for all values of the
variables, both sides must be equal to a constant.
spherical harmonic functions written in more detail
would take the form:
solutions 
Energy of degenerate states of q. n l :
For a given l there are (2l +1), ml values. That is the state
described by a l quantum number is comprised of (2l+1)
sub-states. Degeneracy of state with l, is 2l+1.

2  r02 E 2 I
 2 E  l (l  1)
2

El 
2
l (l  1)
2I
quantization of
rotational energy
Eigen-energy:
El 
2
l (l  1)
2I
Quantization of Angular Momentum
As will be seen later, the shapes, directional properties
and degeneracy of atomic orbitals are dependent on the
l and ml vales associated with these orbitals.
Verifiable in SE:
Spherical harmonics are eigen functions of the total
energy operator.
Eml :in a 3D rotor is undefined; the q. n. ml here determines
the z-component of the angular momentum vector I.
For 3D rotor; Etot 
2
l2
 l
H
2I
2I
constant
Therefore both operators have a common set of
eigenfunctions. The operators should commute!
 ,l 2 ]  0
[H
Components of the angular momentum in x, y and z.
Now;
The respective operators are (Cartesian):
and therefore
Note that the above operators differ by a multiplication
constant (1/2I) thus the eigen values differ by (1/2I).
Also note the calculated (and measurable) angular
momentum quantity (of precise (eigen) value) is that of | l2 |
(therefore I).
I  l (l  1)
The respective operators are in spherical coordinates
would be:
The following commutator relationships exist.
 f ( ,  )
The component operators do not commute with one another
 f ( )
The direction of I cannot be specified (known) ‘precisely’
for rotations in 3D in QM systems as opposed to classical
rigid rotors. Need all vectors to specify direction of I.
However, angular momentum wavefunction is an
eigenfunction of the Hamiltonian and therefore,
Conclusion – spherical harmonics are eigenfunctions of
l2 and lz hence magnitudes of both |I| and Iz can be known
simultaneously, but not the x and the z component values
of the angular momentum.
Examining;
!!!
lz  ml 
 , l2 and l commute and l , l and l do not commute.
H
z
x
y
z
Spatial quantization
with one another. Example l  2 case.
ml = +2
ml = +2
Vector model of angular momentum
Vector model of angular momentum
Example l  2 case.
Spherical harmonic wavefunction
shapes:
l=0
l=1
l=2
Complex functions
- visualization –
not possible.
Under such situations alternate wave functions with
same eigenvalues can be constructed from the ‘complete set’
of wave functions available, i. e. proper linear combinations
from the complete set of eigenfunctions.
px =
 
< 90 180


<90 0

90 0

surface on x-z plane,
for simplicity
all p
all d
```