Chapter 7. A Quantum Mechanical Model for the Vibration and Rotation of Molecules

Harmonic oscillator:

Hooke’s law:

F  kx, x is displacement

Harmonic potential:

1 2 V (x)   Fdx  kx 2 k is force constant:

d 2V k  (curvature of V at equilibrium) 2 dx x0

Newton’s equation:

d 2x m  F  kx (diff. eqn) dt 2

Solutions:

x = Asint, (position)

  (k/m)1/ 2 (vibrational frequency)

1 Verify:

d 2 Asin t d lhs  m  mA cost  mA2 sin t dt 2 dt k  m Asin t  kx  rhs m

Momentum:

dx p  m  mAcost dt

Energy:

p2 E  V 2m (mAcost)2 1   k(Asin t)2 2m 2 m 2 A2  (cost)2  (sin t)2  2 m 2 A2 kA2   2 2

When t = 0, x = 0 and p = p = mA max When t = , p = 0 and x = x = A max

Energy conservation is maintained by oscillation between kinetic and potential energies.

2 Schrödinger equation for harmonic oscillator:

2 2   d 1 2   2  kx   E  2m dx 2 

Energy is quantized:

 1  E       0, 1, ...  2 where the vibrational frequency

1/ 2  k       m 

Restriction of motion leads to uncertainty in x and p, and quantization of energy.

Wavefunctions:

 y2 / 2  (x)  N H (y)e where

1/ 4  mk  y  x,      2 

Normalization factor

3 1/ 2    N      2!

H (y) is the Hermite polynomial

H0 (y) 1 H1(y)  2y 2 H2 (y)  4y  2 ... H1  2yH  2H1 (recursion relation)

Let’s verify for the ground state

 2 x2 / 2 0 (x)  N0e

2 2   d 1 2   2 x 2 / 2 lhs  N0  2  kx e  2m dx 2  2   d  2 x 2 / 2 2 1 2  2 x 2 / 2   N0  e  x kx e   2m dx 2  2     2 x 2 / 2 2 2  2 x 2 / 2 2  1 2  2 x 2 / 2   N0  e  x  e    kx e   2m   2  2 4 2 2  k       2 2   2  x / 2  N0   x  e  2 2m  2m 

Note that

4 1/ 4  mk       2 

It is not difficult to prove that the first term is zero, and the second term as  / 2.

1 2 2 rhs  N e x / 2 2 0

So, rhs  lhs, it is a solution.

One can also show normalization:

  2 2  2 x2    0 dx N0 2e dx  2 2 1  0  4 where we have taken advantage of the even symmetry of the Gaussian

2 2 eax  eax and used the integral formula

 ax2  e dx  0 4a

For higher states, the wavefunctions are essentially the product of a Gaussian and a polynomial:

5

The Gaussian is always an even function, but the polynomial can either be even f (x)  f (x) or odd f (x)   f (x). As a result, the wavefunctions are either even or odd, depending on .

Properties of the oscillator i. Zero point energy (  0)

1 E   (quantum effect) 0 2

Even at 0 K, there is still vibration. ii. Orthonormality:

 *   dx    iii. Average displacement

 *   (x)x (x)dx  0 

6 2 No matter if  (x) is even or odd,  (x) is even, and thus the integrant is odd. Integration of an odd function in  , is always zero. iv. Average momentum

 *   d    (x)  (x)dx  0   i dx 

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

E  E1  E   11/ 2 ( 1/ 2)   (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.

7 k 475N / m     5.431014 s1 m 1.611027 kg ( N / m  kg / s2 )

1 E    0.51.0541034 Js  5.431014 s1 0 2

 2.861020 J

20 E    2E0  5.7210 J

E 5.721020 J     8.631013 s1 h 6.6261034 Js  3476nm (  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

135    0.97amu 1.661027 kg / amu 1.611027 kg 1 35

  2  2c~

 23.143.01010 cm/ s  2886cm1  5.431014 s1

2 k   2 1.611027 kg  5.431014 s1    475kg / s2  475N / m

8 What about DCl?

2 35   1.89amu 1.661027 kg / amu  3.141027 kg 2  35

 2HCl

k 475kg / s2 1     3.891014 s1    3.141027 kg 2 HCl

 3.891014 s1 1 ~    2063cm1  ~ 2c 23.143.01010 cm/ s 2 HCl

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 2   , Ir  dt

The rotational energy (classical):

l 2 E  2I

9 Schrödinger eqn. for 2D motion:

2   2  2       E  2 2  2mx y 

Use polar coordinate system ( x  r cos, y  rsin , it becomes for 2D rigid rotor:

2 d 2   E 2I d 2

General solution:

1/ 2 1/ 2  1  im  2IE  m ()    e , m     2   2 

Cyclic boundary condition:

()  (  2)

So, (  2 )  Neim( 2 )  Nei2m eim

 N(ei )2m eim  (1)2m ()

2m has to be even integers, i.e.

m  0, 1,  2, ...

10 Quantization

m22 E  (2D) m 2I

Angular momentum:

lz  m (quantized angular momentum)

Angular momentum operator

ˆ Lz  xˆpˆ y  yˆpˆ x          x  y   i  y x  i 

 is a common eigenfunction of both operators:

 1/ 2  1/ 2 ˆ    1  im   1  im Lz    e   im   e  m i   2   i  2 

2 2  1/ 2  2 ˆ    1  im (m) H   2   e    2I   2   2I

Uncertainty principle (just like p and x)

lz   / 2

11 3D rotors:

Spherical coordinates:

x  rsin cos y  rsin sin zr cos d  dxdydz  r2 sindrdd

Hamiltonian:

2 Hˆ   2 2m 2  2 2 2       2 2 2  2m x y z  2 1 2 Lˆ2   r  2m r r2 2mr2

The differential operator 2 is called Laplacian.

If r is fixed, the 3D rigid rotor Hamiltonian:

Lˆ2 Hˆ  (rigid rotor) 2I where

12  2  ˆ2 2 1  1     L    2 2  sin  sin   sin    

Schrödinger eqn.

Lˆ2   E 2I

The solution is the spherical harmonics:

m m (,)  Yl (,)  Nlm Pl (cos)m () where

(2l 1) (l  m)! N  lm 2 (l  m)!

1/ 2  1  im m ()    e (2D rotor)  2 

m and Pl (cos) 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)

13 Rotational energy (3D rotor)

l(l 1)2 E  l 2I

Since E is independent of m, each l level is 2l + 1 degenerate, which can be removed by an external field (Zeeman effect).

m When m  0, Pl (cos) becomes the Legendre polynomial:

P0 (cos) 1 P1(cos)  cos 2 P2 (cos)  3cos  1/ 2 …

For m  0

m m m d Pl (cos)  sin  Pl (cos) d cos m

Examples of spherical harmonics:

0 Y0 1 0 Y1  cos 1 Y 1  sin ei 1 2 1 Y 0  3cos2  1 2 2

14 ˆ2 ˆ They are eigenfunctions for both L and Lz :

ˆ2 m 2 m L Yl (,)  l(l 1) Yl (,) ˆ m m LzYl (,)  mYl (,)

0 Verification for   Y1  cos

 2  ˆ2 2 1  1     L     2 2  sin cos sin   sin    

2  1  2    0   sin    sin   2  2sin cos   22 sin

Lˆ2 22 Hˆ     2I 2I

  Lˆ   cos  0 z i 

The spherical harmonics are orthonormal

 2 m * m sind  dYl (,) Yl (,)  llmm 0 0

Note the volume element and ranges.

15 Spherical harmonics have nodes due to Legendre functions.

0 For Y1  cos , the angular node can be determined by

cos  0,   90o

Vector model:  The angular momentum can be considered as a vector L . In quantum mechanics, it is quantized in

 i. magnitude: L  l(l 1)

ii. orientation: Lz  m Only certain directions of 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.

16 L  l(l 1)  2 1.0551034 Js 1.491034 Js

34 Lz  ml  0, 1.05510 Js (3-fold degen.)  The angle between L and the z-axis for ml 1:

L 1 cos z ,   45o L 2

L2 (1.491034 Js)2 E    2.41021 J 2I 2(4.61048 kgm2 )

Example. A 3D rotor absorbs at 11 cm-1. Calculate its moment of inertia.

(2  0)2 E  E  E  1 0 2I

E  h  hc~ so

2  I   hc~ 2c~ 1.051034 Js  2 3.14 3.01010 cm/ s 11cm1  5.061047 kgm2

17