Slide 1 Chapter 5 Molecular Vibrations and Time-Independent Perturbation Theory Part A: The Harmonic...

Chapter 5

Molecular Vibrations andTime-Independent Perturbation Theory

Part A: The Harmonic Oscillator and Vibrations of Molecules

Part B: The Symmetry of Vibrations + Perturbation Theory + Statistical Thermodynamics

• The Classical Harmonic Oscillator

• Math Preliminary: Taylor Series Solution of Differential Eqns.

• The Vibrations of Diatomic Molecules

• The Quantum Mechanical Harmonic Oscillator

• Vibrational Spectroscopy

• Harmonic Oscillator Wavefunctions and Energies

• Properties of the Quantum Mechanical Harmonic Oscillator

• Vibrational Anharmonicity

• The Two Dimensional Harmonic Oscillator

• Vibrations of Polyatomic Molecules


The Classical Harmonic Oscillator

Re

V(R

)

0

A BR

V(R) ½k(R-Re)2

Harmonic Oscillator Approximation

or V(x) ½kx2 x = R-Re

The Potential Energy of a Diatomic Molecule

k is the force constant

m1 m2

x = R - Re

Hooke’s Law and Newton’s Equation

Force:21

2

dV df kx kx

dx dx

Newton’s Equation:2

2

d xf a

dt 1 2

1 2

m m

m m

where

Reduced Mass

Therefore:2

2

d xkx

dt

Solution

2

2

d xkx

dt

2

2

d x kx

dt

Assume: ( ) s i n ( ) c o s ( )x t A t B t 2

2 22

sin ( ) cos( )d x

A t B td t

2 x

2 kx x

k

2

or1

2

k

or1

2

k

c

Initial Conditions (like BC’s)

Let’s assume that the HOstarts out at rest stretchedout to x = x0.

( ) s i n ( ) c o s ( )x t A t B t

x(0) = x0

0

0t

dx

dt

cos( ) sin( )dx

A t B tdt

0

0t

dxA

dt

( ) c o s ( )x t B t

0( 0 )x x B and 0( ) c o s ( )x t x t

1

2 2

k

Conservation of Energy

0 sin ( )dx

x td t

0( ) c o s ( )x t x tk

Potential Energy (V) Kinetic Energy (T) Total Energy (E)

21

2V kx

2 20

1cos ( )

2V kx t

21

2

dxT

dt

2 2 20

1sin ( )

2T x t

2 20

1sin ( )

2T kx t

E T V

20

1

2E kx

20

1

2V kx 0T 2

0

1

2E kx

0V 20

1

2E kx2

0

1

2T kx

t = 0, , 2, ... x = x0

t = /2, 3/2, ... x = 0

Classical HO Properties

Energy: E = T + V = ½kx02 = Any Value i.e. no energy quantization

If x0 = 0, E = 0 i.e. no Zero Point Energy

Probability: E = ½kx02

x

P(x)

0 +x0-x0

Classical Turning Points

Math Preliminary: Taylor Series Solution of Differential Equations

Taylor (McLaurin) Series

( )

0

1( ) (0)

!n n

n

f x f xn

Example: ( ) xf x e

Any “well-behaved” function canbe expanded in a Taylor Series

2 3 41 1 11

2 6 24x x x x

0

1

!n

n

xn

0 1

x ff f2 f3( )

( ) xf x e2

2 ( ) 12

xf x x

2 3

3 ( ) 12 6

x xf x x

x ff f3 f6( )

0 3

( ) xf x e2 3

3 ( ) 12 6

x xf x x

2 3 4 5 6

6 ( ) 12 6 24 120 720

x x x x xf x x

Example: ( ) xf x e 2 3 41 1 11

2 6 24x x x x

0

1

!n

n

xn

A Differential Equation

( )( )

df xf x

dx

Solution: ( ) xf x A e

Example: ( ) xf x e

Taylor (McLaurin) Series

( )

0

1( ) (0)

!n n

n

f x f xn

Any “well-behaved” function can

be expanded in a Taylor Series

2 3 41 1 11

2 6 24x x x x

0

1

!n

n

xn

Taylor Series Solution

( )( )

df xf x

dx

2 30 1 2 3

0

( ) nn

n

f x a a x a x a x a x

Our goal is to develop a “recursion formula” relatinga1 to a0, a2 to a1... (or, in general, an+1 to an)

SpecificRecursionFormulas:

21 2 3

( )2 3

d f xa a x a x

dx

2 21 2 3 0 1 2

( )2 3 ( )

d f xa a x a x a a x a x f x

d x

Coefficients of xn must be equal for all n

a1 = a0

2a2 = a1 a2 = a1/2 = a0/2

3a3 = a2 a3 = a2/3 = a0/6

1

1

mm

m

ma x

a1 = a0

a2 = a0/2

a3 = a0/6

2 30 1 2 3( )f x a a x a x a x 2 3

0

1 11 ...

2 6a x x x

Power SeriesExpansion for ex

General Recursion Formula

( )( )

df xf x

dx

1

1 0

m nm n

m n

ma x a x

Coefficients of the two series may be equatedonly if the summation limits are the same.

Therefore, set m = n+1 n=m-1and m=1 corresponds to n = 0

10 0

( 1) n nn n

n n

n a x a x

1( 1 ) n nn a a or 1 ( 1)n

n

aa

n

The Quantum Mechanical Harmonic Oscillator

Schrödinger Equation

Particle vibrating against wall

m

x = R - Re

2 22

2

1

2 2

dkx E

m dx

Two particles vibrating against each other

2 22

2

1

2 2

dkx E

dx

m1 m2

x = R - Re

1 2

1 2

m m

m m

Boundary Condition: 0 as x

Solution of the HO Schrödinger Equation

Rearrangement of the Equation2 2

22

1

2 2

dkx E

dx

2 22

2

10

2 2

dkx E

dx

22

2 2

2 10

2

dkx E

dx

22

2 2 2

20

d E kx

dx

Define

2

2 E

22

k

2 2

2

k

22 2

20

dx

dx

2

2 E

22

k 2

22

22 2

20

dx

dx

Should we try a Taylor series solution?0

nn

n

a x

No!! It would be difficult to satisfy the Boundary Conditions.

Instead, it’s better to use:0

nn asymp

n

a x

( ) a s y m pH x

That is, we’ll assume that is the product of a Taylor Series, H(x),and an asymptotic solution, asympt, valid for large |x|

The Asymptotic Solution

2

2 22

0d

xdx

For large |x|2

2 22

0asymptasympt

dx

dx

We can show that theasymptotic solution is

2 / 2asympt xdxe

dx

2 22

/ 2 2 2 / 22

asym pt x xde x e

dx

22

2 2 / 22

asympt xdx e

dx

2 22 2 / 2 2 2 / 2 0x xx e x e

2 / 2xa s y m p t e

Note that asympt satisfied the BC’s;

i.e. asympt 0 as x

Plug in

A Recursion Relation for the General Solution

2

2 22

0d

xdx

2 / 2

0 0

n n xn asymp n

n n

a x a x e

Assume the general solution is of the form

2 / 2( ) xH x e

If one: (a) computes d2/dx2

(b) Plugs d2/dx2 and into the Schrödinger Equation

(c) Equates the coefficients of xn

Then the following recursion formula is obtained:

2

2 10, 1, 2, 3,

( 1)( 2)n n

na a n

n n

2

2 10, 1, 2, 3,

( 1)( 2)n n

na a n

n n

Because an+2 (rather than an+1) is related to an, it is best to considerthe Taylor Series in the solution to be the sum of an even and odd series

2 2/ 2 2 4 3 5 / 20 2 4 1 3 5

0

n x xn

n

a x e a a x a x a x a x a x e

Even Series Odd Series

From the recursion formula: 0 2 4a a a Even Coefficients

1 3 5a a a Odd Coefficients

a0 and a1 are two arbitrary constants

So far: (a) The Boundary Conditions have not been satisfied

(b) There is no quantization of energy

Satisfying the Boundary Conditions: Quantization of Energy

2

2 E

2

2E

So far, there are no restrictions on and, therefore, none on E

Let’s look at the solution so far:

Does this solution satisfy the BC’s: i.e. 0 as x ?

It can be shown that:

2 2/ 2 2 4 3 5 / 20 2 4 1 3 5

0

n x xn

n



2 / 2lim 0n xx x e unless n

Therefore, the required BC’s will be satisfied if, and only if, boththe even and odd Taylor series terminate at a finite power.

2 2/ 2 2 4 3 5 / 20 2 4 1 3 5

0

n x xn

n



We cannot let either Taylor series reach x.

2

2 10, 1, 2, 3,

( 1)( 2)n n

na a n

n n

We can achieve this for the even or odd series by settinga0=0 (even series) or a1=0 (odd series).

We can’t set both a0=0 and a1=0, in which case (x) = 0 for all x.

So, how can we force the second series to terminate at a finitevalue of n ?

By requiring that an+2 = 0•an for some value of n.

2

2 10, 1, 2, 3,

( 1)( 2)n n

na a n

n n

If an+2 = 0•an for some value of n, then the required BC willbe satisfied.

Therefore, it is necessary for some value of the integer, n, that:

2 10

( 1)( 2)

n

n n

2 1 0n

This criterion puts restrictions on the allowed values of and,therefore, the allowed values of the energy, E.

Allowed Values of the Energy

22

k

2 2

2

2

2 E

2

2E

Earlier Definitions Restriction on

2 1 0n n = 0, 1, 2, 3,...

(2 1)n

n = 0, 1, 2, 3,...

Quantized Energy Levels

2

2E

2

(2 1)2

n

1

2E n

or 1

2E n h 2

2 1n

2 1

2

n

HO Energies and Wavefunctions

Harmonic Oscillator Energies

1 10, 1, 2 , 3,

2 2nE n n h n

Angular Frequency: k

Circular Frequency: 1

2

k

En

erg

y

1

2

3

2

5

2

7

2

Wavenumbers: 1

2

k

c

where 11

( ) ( / )cm

cm c cm s

E

ner

gy

1

2

3

2

5

2

7

2

Quantized Energies

Only certain energy levels are allowedand the separation between levels is:

E

The classical HO permits any value of E.

Zero Point Energy

The minimum allowed value for theenergy is:

0

1

2E

The classical HO can have E=0.

Note: All “bound” particles have a minimum ZPE. This is a consequence of the Uncertainty Principle.

The Correspondence Principle

The results of Quantum Mechanics must not contradict those ofclassical mechanics when applied to macroscopic systems.

The fundamental vibrational frequency of the H2 molecule is 4200 cm-1. Calculate the energy level spacing and the ZPE, in J and in kJ/mol.

ħ = 1.05x10-34 J•sc = 3.00x108 m/s = 3.00x1010 cm/sNA = 6.02x1023 mol-1

2 2 c 1 0 12 3 .1 4 3 .0 0 1 0 / 4 2 0 0x c m s c m 1 4 17 . 9 2 1 0x s

Spacing: E 3 4 1 4 11 .0 5 1 0 7 .9 2 1 0x J s x s 2 08 . 3 1 0x J

2 3 1 2 06 .0 2 1 0 8 .3 1 0AN E x m o l x J

ZPE: 0

1

2E 34 14 11

1 .05 10 7 .92 102

x J s x s 2 04 . 1 5 1 0x J

45 . 0 0 1 0 / 5 0 /x J m o l k J m o l

2 3 1 2 0 40 6 . 0 2 1 0 4 . 1 5 1 0 2 . 5 1 0 / 2 5 /AN E x m o l x J x J m o l k J m o l

2 4 1 4 12 3 .1 4 1 .0 0 1 0 6 .2 8 1 0x s x s

Spacing: E 3 4 4 1 3 01 .0 5 1 0 6 .2 8 1 0 6 .6 1 0x J s x s x J

ZPE: 0

1

2E 34 4 1 301

1 .0 5 1 0 6 .2 8 1 0 3 .3 1 02

x J s x s x J

ħ = 1.05x10-34 J•sc = 3.00x108 m/s = 3.00x1010 cm/sNA = 6.02x1023 mol-1

Macroscopic oscillators have much lower frequencies thanmolecular sized systems. Calculate the energy level spacing andthe ZPE, in J and in kJ/mol, for a macroscopic oscillator with afrequency of 10,000 cycles/second.

2 3 1 3 0 6 96 . 0 2 1 0 6 . 6 1 0 4 . 0 1 0 / 4 . 0 1 0 /AN E x m o l x J x J m o l x k J m o l

2 3 1 3 0 6 90 6 . 0 2 1 0 3 . 3 1 0 2 . 0 1 0 / 2 . 0 1 0 /AN E x m o l x J x J m o l x k J m o l

Harmonic Oscillator Wavefunctions

Application of the recursion formula yields a different polynomial for each value of n.

These polynomials are called Hermitepolynomials, Hn

2 / 2( ) xn n nN H x e

2 / 2( ) zn n nN H z e

or

where z x

Some specific solutions

n = 02 2/ 2 / 2

0 0 0x zN e N e Even

n = 22 22 / 2 2 / 2

2 2 2( 4 2 ) ( 4 2 )x zN x e N z e Even

n = 12 2/ 2 / 2

1 1 1( 2 ) ( 2 )x zN x e N z e Odd

n = 32 23 / 2 3 / 2 3 / 2

3 3 3( 8 1 2 ) ( 8 1 2 )x zN x x e N z z e Odd

2 / 2

0

n xn

n

a x e

0

1E

2

1

3E

2

2

5E

2

2

E = ½x02

x

P(x)

0 +x0-x0


Quantum Mechanical vs. Classical Probability

Classical

P(x) minimum at x=0

P(x) = 0 past x0

QM (n=0)

P(x) maximum at x=0

P(x) 0 past x0

E = ½x02

x

P(x)

0 +x0-x0


CorrespondencePrinciple

E = ½x02

x

P(x)

0 +x0-x0


-6 -4 -2 0 2 4

vals.1

0.0

0.2

0.4

0.6

0.8

1.0

x 2 V zero

n=9

Note that as n increases, P(x)approaches the classical limit.

For macroscopic oscillatorsn > 106

Properties of the QM Harmonic Oscillator

2

0

1

2xe dx

2

0

1

2xxe dx

22

0

1

4xx e dx

2320

1

2xx e dx

2420

3

8xx e dx

Some Useful Integrals

Remember

If f(-x) = f(x)

0( ) 2 ( )f x d x f x d x

Even Integrand

If f(-x) = -f(x)

( ) 0f x dx

Odd Integrand

Wavefunction Orthogonality

2 / 20 0

xN e - < x < 2 / 2

1 1 ( 2 ) xN x e - < x < 22 / 2

2 2 ( 4 2 ) xN x e - < x <

<0|1>

2 2/ 2 / 20 1 0 1 2

x xN e N x e d x

2

0 1 0 1 2xN N xe d x

Odd Integrand

0 1 0

Wavefunction Orthogonality

2 / 20 0

xN e - < x < 2 / 2

1 1 ( 2 ) xN x e - < x < 22 / 2

2 2 ( 4 2 ) xN x e - < x <

<0|2>

2 2/ 2 2 / 20 2 0 2 4 2x xN e N x e d x

0 2 0 2

1 14 2 2 2

4 2N N

2

0

1

2xe dx

22

0

1

4xx e dx

2 220 2 0 2 4 2x xN N x e d x e d x

0 2 0

Normalization

We performed many of the calculations below on 0 of theHO in Chap. 2. Therefore, the calculations below will beon 1. 2 / 2

1xA x e - < x <

22

/ 21 1 1 1xA xe dx

fromHW Solns.

1/ 4

2A

Positional Averages

22

/ 21 1 1

xx x x A xe d x

fromHW Solns.

<x>

0x

22

2 2 2 / 21 1 1

xx x x A xe d x

fromHW Solns.

<x2>

2 3

2x

Momentum Averages

2

2

/ 21/ 2

1 1 1ˆ

x

xd A xe

p p A xe dxi dx

fromHW Solns.

<p>

0p

fromHW Solns.

<p2>

2 23

2p

2

2

2 / 212 2 / 2 2

1 1 1 2

x

xd A xe

p p A xe dxdx

^

Energy Averages

Kinetic Energy

221

2 2

pT p

21 3

2 2T

fromHW Solns.

3

4T

Potential Energy

2 21 1

2 2V kx k x

1 3

2 2V k

fromHW Solns.

3

4V

Total Energy

1 3

2 2E n

T V

The Vibrations of Diatomic Molecules

Re

V(R

)

0A B

R

The Potential Energy

Define: x = R - Req

Expand V(x) in a Taylor series about x = 0 (R = Req)

0 0

1( )

!

nn

nn x

d VV x x

n dx

2 3 4

0 1 2 3 42 3 40

0 0 0 0

1 1 1 1 1( )

0 ! 1! 2 ! 3! 4 !xx x x x

dV d V d V d VV x V x x x x x

dx dx dx dx

2 3 4

2 3 42 3 40

0 0 0 0

1 1 1( )

2 6 24xx x x x

dV d V d V d VV x V x x x x

dx dx dx dx

Re

V(R

)

0

2 3 4

2 3 42 3 40

0 0 0 0

1 1 1( )

2 6 24xx x x x

dV d V d V d VV x V x x x x

dx dx dx dx

00

xV

By convention

0

0x

dV

dx

By definitionof Re as minimum

2

2

0x

d Vk

dx

3

3

0

1

6x

d V

dx

4

4

0

1

24x

d V

dx

2 3 41( )

2V x kx x x

Re

V(R

)

0

2 3 41( )

2V x kx x x

21( )

2V x kx

The Harmonic Oscillator Approximation

2

2

0x

d Vk

dx

where

Ignore x3 and higher order terms in V(x)

The Force Constant (k)

The Interpretation of k

Re

V(R

)

0

2

2

0x

d V d dV dk Slope

dx dx dx dx

i.e. k is the “curvature” of the plot, and represents the “rapidity” with which the slope turns from negative to positive.

Another way of saying this is the k represents the “steepness”of the potential function at x=0 (R=Re).

ReV(R

)

0

Do

ReV(R

)

0

Do

Do is the Dissociation Energy of the molecule, andrepresents the chemical bond strength.

There is often a correlation between k and Do.

Small kSmall Do

Large kLarge Do

Correlation Between k and Bond Strength

0 2 4 6 8 10 12

0

500

1000

1500

2000

2500

HIHBr

H2

O2

NO

CO

N2

Cl2

k [

N/m

]

Do [eV]

Note the approximately linear correlation betweenForce Constant (k) and Bond Strength (Do).

Vibrational Spectroscopy

Energy Levels and Transitions

1 1

2 2nE n n h

1

2 2

k

Selection Rule

n = 1 (+1 for absorption,-1 for emission)

IR Spectra: For a vibration to be IR active, the dipole moment must change during the course of the vibration.

Raman Spectra: For a vibration to be Raman active, the polarizability must change during the course of the vibration.

Transition Frequency

n n+1

1 11

2 2E n h n h h

Wavenumbers: 1Ecm

hc c

Note:1

2

k

c

The Boltzmann Distribution

/nE k Tn nN g e

The strongest transition corresponds to: n=0 n=1

/nE k Te

Calculation of the Force Constant

Units of k

2

2

0x

d Vk

dx

2

Jk

m

2

2

2

kg ms

m

2

kg

s

2kg m

sm

N

m

Calculation of k

1

2

k

c

22k c

The IR spectrum of 79Br19F contains a single line at 380 cm-1

Calculate the Br-F force constant, in N/m. h=6.63x10-34 J•sc=3.00x108 m/sc=3.00x1010 cm/sk=1.38x10-23 J/K1 amu = 1.66x10-27 kg1 N = 1 kg•m/s2

Br F

Br F

m m

m m

79 19

79 19

amu amu

amu amu

1 5 . 3 2 a m u 2 7 2 61 . 6 6 1 0 / 2 . 5 4 1 0x k g a m u x k g

1

2

k

c

22k c 2

1 0 1 2 62 3 .1 4 3 1 0 / 3 8 0 2 .5 4 1 0x cm s cm x k g

21 3 0 /k k g s 21 3 0 / /k g m s m 1 3 0 /N m

Calculate the intensity ratio, ,for the 79Br19F vibration at 25 oC.10

21

II

R

h=6.63x10-34 J•sc=3.00x108 m/sc=3.00x1010 cm/sk=1.38x10-23 J/K

= 380 cm-1~

1 2 1

0 1 0

I N

I N

1

0

/

/

E kT

E kT

e

e

1 0 /E E k Te /h c k Te

34 10 1

23

(6.63 10 ) 3.00 10 / 380

1.38 10 / 298

x J s x cm s cmhc

kT x J K K

1.841 2

0 1

0.16I

eI

= 1.84

1 2

0 1

IR

I

The n=1 n=2 transition is called a hot band, because itsintensity increases at higher temperature

Dependence of hot band intensity on frequency and temperature

Molecule T R~

79Br19F 380 cm-1 25 oC 0.16

79Br19F 380 500 0.49

79Br19F 380 1000 0.65

H35Cl 2880 25 9x10-7

H35Cl 2880 500 0.005

H35Cl 2880 1000 0.04

Vibrational Anharmonicity

Re

V(R

)

0

2 3 41( )

2V x kx x x

Harmonic OscillatorApproximation: 21

( )2

V x kx

The effect of including vibrational anharmonicity in treating the vibrationsof diatomic molecules is to lower the energy levels and decrease thetransition frequencies between successive levels.

Re

V(R

)

0

2 3 41( )

2V x kx x x

A treatment of the vibrations of diatomic molecules which includesvibrational anharmonicity [includes higher order terms in V(x)]leads to an improved expression for the energy:

21 1

2 2n

e

En n x

hc

is the harmonic frequency and xe is the anharmonicity constant.~

Measurement of the fundamental frequency (01) and firstovertone (02) [or the "hot band" (1 2)] permits determination of and xe.

~See HW Problem

Energy Levels and Transition Frequencies in H2

E/h

c [c

m-1]

HarmonicOscillator

Actual

2,2000

6,6001

11,0002

15,4003

2,1700

6,3301

10,2502

13,9303

14 4 0 0~ c m

14 4 0 0~ c m

14 4 0 0~ c m

14 1 6 0~ c m

13 9 2 0~ c m

13 6 8 0~ c m

Anharmonic Oscillator Wavefunctions

-0.5 -0.3 -0.1 0.1 0.3 0.5 0.7

vals.1

0.0

0.2

0.4

0.6

0.8

1.0

xang V zero

-0.5 -0.3 -0.1 0.1 0.3 0.5 0.7

vals.1

0.0

0.2

0.4

0.6

0.8

1.0

xang 2 V zero

2n = 0

-0.5 -0.3 -0.1 0.1 0.3 0.5 0.7

vals.1

-1.0

-0.5

0.0

0.5

1.0

xang V zero

2

-0.5 -0.3 -0.1 0.1 0.3 0.5 0.7

vals.1

0.0

0.2

0.4

0.6

0.8

1.0

xang 2 V zero

n = 2

n = 10

-0.5 -0.3 -0.1 0.1 0.3 0.5 0.7

vals.1

-1.0

-0.5

0.0

0.5

1.0

xang V zero

2

-0.5 -0.3 -0.1 0.1 0.3 0.5 0.7

vals.1

0.0

0.2

0.4

0.6

0.8

1.0

xang 2 V zero

The Two Dimensional Harmonic Oscillator

The Schrödinger Equation

2 21 1( , )

2 2x yV x y k x k y

2 2 22 2

2 2

1 1( , ) ( , )

2 2 2x yk x k y x y E x yx y

2 2 2 22 2

2 2

1 1( , ) ( , )

2 2 2 2x yk x k y x y E x yx y

The Solution: Separation of Variables

The Hamiltonian is of the form: ( , ) ( ) ( )x yH x y H x H y

Therefore, assume that is of the form: ( ) ( )x y( x , y ) x y

2 2 2 22 2

2 2

1 1( ) ( ) ( ) ( )

2 2 2 2x y x y x yk x k y x y E x yx y

222 22 2

2 2

1 1( ) ( )

2 2 2 2yx

y x x x y y x y

ddk x k y E x y

dx dy

222 22 2

2 2

1 1 1 1

2 2 2 2yx

x x y yx y

ddk x k y E

dx dy

The above equation is of the form, f(x) + g(y) = constant.Therefore, one can set each function equal to a constant.

E = Ex + Ey

=

Ex

=

EyEx

222

2

1 1

2 2x

x x xx

dk x E

dx

222

2

1 1

2 2y

y y yy

dk x E

dx

222

2

1

2 2x

x x x x

dk x E

dx

22

22

1

2 2y

y y y y

dk x E

dx

The above equations are just one dimensional HO Schrödinger equations.Therefore, one has:

2 / 2( ) x xxn x n x n xN H x e

1

2x x xE n

xx

k

xx

2 / 2( ) y yyn y n y n yN H y e

1

2y y yE n

yy

k

yy

Energies and Degeneracy

1 1

2 2x y x x y yE E E n n

yxx y

kk

Depending upon the relative values of kx and ky, one may havedegenerate energy levels. For example, let’s assume that ky = 4•kx .

2y x

12 1

2x x y xE n n

32

2x y xE n n

E [

ħ

]

3/20 0

g = 1

5/21 0

g = 1

7/22 0 0 1

g = 2

9/23 0 1 1

g = 2

11/24 0 0 2 2 1

g = 3

Vibrations of Polyatomic Molecules

I’ll just outline the results.

If one has N atoms, then there are 3N coordinates. For the i’th. atom,the coordinates are xi, yi, zi. Sometimes this is shortened to: xi. = x,y or z.

The Hamiltonian for the N atoms (with 3N coordinates), assumingthat the potential energy, V, varies quadratically with the change incoordinate is:

2 2

,2

1( )( )

2 2eq eq

i j i i j ji i ji i

H F x x x xm x

2

,

,eq eqi j

i ji j x x

VF

x

After some rather messy algebra, one can transform the Cartesiancoordinates to a set of “mass weighted” Normal Coordinates.

, ( )eqk i i k i i

i

Q m L x x

Each normal coordinate corresponds to the set of vectors showingthe relative displacements of the various atoms during a given vibration.

There are 3N-6 Normal Coordinates.

For example, for the 3 vibrations of water, the Normal Coordinatescorrespond to the 3 sets of vectors you’ve seen in other courses.

H H

O

H H

O

H H

O

In the normal coordinate system, the Hamiltonian can be written as:

2 23 62 2

21

1

2 2

N

k kk k

H QQ

2 2 2 2 2 22 2 2 2 2 21 1 2 2 3 32 2 2

1 2 3

1 1 1

2 2 2 2 2 2H Q Q Q

Q Q Q

Note that the Hamiltonian is of the form:3 6

1 1 2 2 3 31

( ) ( ) ( ) ( )N

i ii

H H Q H Q H Q H Q

Therefore, one can assume:3 6

1 1 2 2 3 31

( ) ( ) ( ) ( ) ...N

i ii

Q Q Q Q

and simplify the Schrödinger equation by separation of variables.

One gets 3N-6 equations of the form:22

2 21 12

1

1

2 2i

i i iQ EQ

The solutions to these 3N-6 equations are the familiar HarmonicOscillator Wavefunctions and Energies.

As noted above, the total wavefunction is given by:3 6

1

( )N

i ii

Q

The total vibrational energy is:

3 6

1 2 3 1 1 2 2 3 31

1 1 1 1( )

2 2 2 2

N

i ii

E n n n n n n n

3 6

1 2 31

1( )

2

N

i ii

E n n n n h

Or, equivalently:

3 11 2 3

1

( ) 1

2

N

i ii

E n n nn

hc

In spectroscopy, we commonly refer to the “energy in wavenumbers”,which is actually E/hc:

The water molecule has three normal modes, with fundamental

frequencies: 1 = 3833 cm-1, 2 = 1649 cm-1, 3 = 3943 cm-1.~ ~ ~

What is the energy, in cm-1, of the (112) state (i.e. n1=1, n2=1, n3=2)?

1 2 3

(11 2 ) 1 1 11 1 2

2 2 2

E

hc

1 1 13 3 53 83 3 16 49 3 94 3

2 2 2cm cm cm

11 8 , 0 8 1 c m

The water molecule has three normal modes, with fundamental

frequencies: 1 = 3833 cm-1, 2 = 1649 cm-1, 3 = 3943 cm-1.~ ~ ~

What is the energy difference, in cm-1, between (112) and (100),i.e. E(112)/hc – E(100)/hc ?

1(11 2 )18, 081

Ecm

hc

1 2 3

(100 ) 1 1 11 0 0

2 2 2

E

hc

1 1 13 1 13 83 3 16 49 3 94 3

2 2 2cm cm cm 18 , 5 4 6 c m

1 1 1(11 2 ) (1 0 0 )1 8 , 0 8 1 8 , 5 4 6 9 , 5 3 5

E Ecm cm cm

h c h c

This corresponds to the frequency of the combination band in whichthe molecule’s vibrations are excited from n2=0n2=1 andn3=0n3=2.

Slide 1 Chapter 5 Molecular Vibrations and Time-Independent Perturbation Theory Part A: The Harmonic...

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Transcript of Slide 1 Chapter 5 Molecular Vibrations and Time-Independent Perturbation Theory Part A: The Harmonic...