slide 1 chapter 5 molecular vibrations and time-independent perturbation theory part b: the symmetry...

Chapter 5

Molecular Vibrations andTime-Independent Perturbation Theory

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

• Symmetry and Vibrational Selection Rules

• Spectroscopic Selection Rules

• The Vibrational Partition Function Function

• The Symmetry of Vibrational Normal Modes: Application to the Stretching Vibrations in Ethylene

• Internal Energy: ZPE and Thermal Contributions

• Time Independent Perturbation Theory

• Statistical Thermodynamics: Vibrational Contributions to the Thermodynamic Properties of Gases


• Applications to O2(g) and H2O(g)

• QM Calculations of Vibrational Frequencies and Dissoc. Energies

Symmetry and Vibrational Selection Rules

2

??axxe dx

2a xx e

x

+

-

2

0axxe dx

By symmetry: f(-x) = -f(x)

Consider the 1s and 2px orbitals in a hydrogen atom.

1 2 ? ?xs p d V

+-+

+- + x

z

1 2 0xs p d V

By symmetry: Values of the integrand for -x are the negative of values for +x; i.e. the portions of the integral to the left of the yz plane and right of the yz plane cancel.

If you understand this, then you know all (or almost all) there is toknow about Group Theory.

The Direct Product

There is a theorem from Group Theory (which we won’t prove)that an integral is zero unless the integrand either:

(a) belongs to the totally symmetric (A, A1, A1g) representation

(b) contains the totally symmetric (A, A1, A1g) representation

0fd unless 1( )f A

The question is: How do we know the representation of an integrand when it is the product of 2 or more functions?

( ) ?a b a bf d d

The product of two functions belongs to the representationcorresponding to the Direct Product of their representations

a bf

xf a b

How do we determine the Direct Product of two representations?

Simple!! We just multiply their characters (traces) together.

f a a b bf

C2V E C2 V(xz) V’(yz)

A1 1 1 1 1

A2 1 1 -1 -1

B1 1 -1 1 -1

B2 1 -1 -1 1

B1xB2 1 1 -1 -1 A2

A2xB1 1 -1 -1 1 B2

A2xB2 1 -1 1 -1 B1

When will the Direct Product be A1?

A2xA2 1 1 1 1 A1

B1xB1 1 1 1 1 A1

Only when the two representations are the same.(This is another theorem that we won’t prove)

x xf a b c

The representation of the integrand is the Direct Product of therepresentations of the three functions.

i.e.

What is f if one of the 3 functions belongs to the totallysymmetric representation (e.g. if (c) = A1) ?

1x x A xf a b a b

What if the integrand is the product of three functions?

e.g. ( ) ?a b c a b cf d d

Therefore unless1f A a b

When will the Direct Product be A1?

Only when the two representations are the same.(This is another theorem that we won’t prove)

A minor addition.

When considering a point group with degenerate (E or T)representations, then it can be shown that the product of twofunctions will contain A1 only if the two representations are the same.

e.g. E x E = A1 + other

( ) ?a b a bf d d

xf a b

Based upon what we’ve just seen: 1f A

unlessa b

Therefore, the integral of the product of two functions vanishesunless the two functions belong to the same representation.

x xf a b c

The representation of the integrand is the Direct Product of therepresentations of the three functions.

i.e.

What is f if one of the 3 functions belongs to the totallysymmetric representation (e.g. if (c) = A1) ?

1x x A xf a b a b

What if the integrand is the product of three functions?

e.g. ( ) ?a b c a b cf d d

Therefore unless1f A a b

Spectroscopic Selection Rules

When light (of frequency ) is shined on a sample, the light’selectric vector interacts with the molecule’s dipole moment, whichadds a perturbation to the molecular Hamiltonian.

( 0 ) ( 1 ) ( 0 ) ( 0 ) c o s ( )H H H H E H E t

The perturbation “mixes” the ground state wavefunction (0 = init ) withvarious excited states (i = fin). A simpler way to say this is that thelight causes transitions between the ground state and the excitedstates.

Time dependent perturbation theory can be used to show thatthe intensity of the absorption is proportional to the squareof the “transition moment”, M0i.

Consider a transition between the ground state (0 = init ) and thei’th. excited state (i = fin).

2

0 iI n t e n s i t y C M

The transition moment is: *0 ˆ ˆi f i n i n i t f i n i n i tM d

ˆ e r e x i y j z k

is the dipole moment operator:^

0 ˆi f i n i n i t f i n i n i tM e x i y j z k

0 i f i n i n i t f i n i n i t f i n i n i tM e x i e y j e z k

0 0 0 0x y z

i i i iM M i M j M k

0

0

0

xi fin init

yi fin init

zi fin init

M e x

M e y

M e z

Thus, the transition moment has x, y and z components.

The above equation leads to the Selection Rules in various areasof spectroscopy.

The intensity of a transition is nonzero only if at least one componentof the transition moment is nonzero.

That’s where Group Theory comes in.

As we saw in the previous section, some integrals are zero due tosymmetry.

Stated again, an integral is zero unless the integrand belongs to(or contains) the totally symmetry representation, A, A1, A1g...

0 0 0 0x y z


0

0

0

xi fin init

yi fin init

zi fin init

M e x

M e y

M e z

2


Selection Rules for Vibrational Spectra

Infrared Absorption Spectra

0 0 0 0x y z


0

0

0

xi fin init

yi fin init

zi fin init

M e x

M e y

M e z

2


The equation governing the intensity of a vibrational infrared absorption is allowed (i.e. the vibration is IR active) is:

init and fin represent vibrational wavefunctions of the initialstate (often the ground vibrational state, n=0) and final state (oftencorresponding to n=1).

A vibration will be Infrared active if any of the three componentsof the transition moment are non-zero; i.e. if

1 1( ) x ( ) x ( ) ( , , . . . )f i n i n i t gx A A A

1 1( ) x ( ) x ( ) ( , , . . . )f i n i n i t gy A A A or

1 1( ) x ( ) x ( ) ( , , . . . )f i n i n i t gz A A A or

2


0 0 0 0x y z


0

0

0

xi fin init

yi fin init

zi fin init

M e x

M e y

M e z

Raman Scattering Spectra

When light passes through a sample, the electric vector createsan induced dipole moment, ind, whose magnitude dependsupon the polarizability, . The intensity of Raman scattering dependson the size of the induced dipole moment.

Strictly speaking, is a tensor (Yecch!!)

indx xx xy xz xindy xy yy yz yindz xz yz zz z

E

E

E

I have used the fact that the tensor is symmetric; i.e. ij = ji

There are 6 independentcomponents of : xx , yy , zz , xy , xz , yz

indx xx xy xz xindy xy yy yz yindz xz yz zz z

E

E

E

Therefore, a vibration will be active if any of the followingsix integrals are not zero.

u vf i n i n i t u = x, y, z and v = x, y, z

uv has the same symmetry properties as the product uv;

i.e. xx belongs to the same representation as xx, yz belongs to the same representation as yz, etc.

Therefore, a vibration will be Raman active if the direct product:

1 1( ) x ( u v ) x ( ) ( , , . . . )f i n i n i t gA A A

for any of the six uv combinations; xx, yy, zz, xy, xz, yz

The Symmetry of Vibrational Normal Modes

We will concentrate on the four C-Hstretching vibrations in ethylene, C2H4.

C2h E C2 i h

Ag 1 1 1 1 x2,y2,z2,xy

Bg 1 -1 1 -1 xz,yz

Au 1 1 -1 -1 z

Bu 1 -1 -1 1 x,y

Ethylene has D2h symmetry. However,because all 4 C-H stretches are in theplane of the molecule, it is OK to usethe subgroup, C2h.

x

y

C C

H H

H H

1

(3030 cm-1)

C C

H H

H H

2

(3100 cm-1)

C C

H H

H H

3

(3110 cm-1)

C C

H H

H H

4

(2990 cm-1)

1 1 1 1 1 ag

ag

2 1 1 1 1 ag

ag

3 1 -1 -1 1 bu

bu

4 1 -1 -1 1 bu

bu

Symmetry of Vibrational Wavefunctions

One Dimensional Harmonic Oscillator

n = 02 / 2

0 0xN e

n = 222 / 2

2 2 ( 4 2 ) xN x e

n = 12 / 2

1 1 ( 2 ) xN x e

Harmonic Oscillator Normal Mode: k

Normal Coordinate: Qk

n = 02 / 2

0 0k k kQN e

n = 22 / 22

2 2 ( 4 2 )k k kQk kN Q e

n = 12 / 2

1 1 ( 2 )k k kQk kN Q e

0 ( t o t . s y m m . r e p . )kgA

2 ( t o t . s y m m . r e p . )kgA

1k

kQ

31 2

1 2 3 1 2 3, , , . . . 1 2 3, , ,v i bn n n n n nn n n

Total Vibrational Wavefunction of Polyatomic Molecules

31 2

1 2 3 1 2 3, , , . . . x x xv i bn n n n n n

GroundState: 31 2

0 , 0 , 0 , . . . 0 0 00 , 0 , 0 ,v i b

31 20 , 0 , 0 , . . . 0 0 0x x xv i b

x x x . . .g g g ga a a A

SinglyExcited State: 31 2

0 ,1 , 0 , . . . 0 1 00 , 1 , 0 ,v i b

31 20 ,1 , 0 , . . . 0 1 0x x xv i b

2 21 1x ( ) x x . . . ( )g ga a

C-H Stretching Vibrations of Ethylene

C2h E C2 i h

Ag 1 1 1 1 x2,y2,z2,xy

Bg 1 -1 1 -1 xz,yz

Au 1 1 -1 -1 z

Bu 1 -1 -1 1 x,y

Chapter 13: Slide 44

C C

H H

H H1(3030 cm-1)

Ag

C C

H H

H H3

(3110 cm-1)

Bu

C C

H H

H H2(3100 cm-1)

Ag

C C

H H

H H4

(2990 cm-1)

Bu

IR Activity of Fundamental Modes

0 , 0 , 0 , 0 1 , 0 , 0 , 0Ag:

0 1 , 0 , 0 , 0 0 , 0 , 0 , 0x v i b v i biM e x

31 2 41 , 0 , 0 , 0 1 0 0 0x x xv i b

gA

31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b

gA

x xg u g uA B A B

0 1 , 0 , 0 , 0 0 , 0 , 0 , 0y v i b v i biM e y x xg u g uA B A B

0 1 , 0 , 0 , 0 0 , 0 , 0 , 0z v i b v i biM e z x xg u g uA A A A

Ag vibrations are IR Inactive

C2h E C2 i h

Ag 1 1 1 1 x2,y2,z2,xy

Bg 1 -1 1 -1 xz,yz

Au 1 1 -1 -1 z

Bu 1 -1 -1 1 x,y


C C

H H

H H1(3030 cm-1)

Ag

C C

H H

H H3

(3110 cm-1)

Bu

C C

H H

H H2(3100 cm-1)

Ag

C C

H H

H H4

(2990 cm-1)

Bu

IR Activity of Fundamental Modes

0 , 0 , 0 , 0 0 , 0 , 1 , 0Bu:

0 0 , 0 ,1 , 0 0 , 0 , 0 , 0x v i b v i biM e x

31 2 40 , 0 ,1 , 0 0 0 1 0x x xv i b

uB

31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b

gA

x xu u g gB B A A

0 0 , 0 ,1 , 0 0 , 0 , 0 , 0y v i b v i biM e y x xu u g gB B A A

0 0 , 0 ,1 , 0 0 , 0 , 0 , 0z v i b v i biM e z x xu u g gB A A B

Bu vibrations are IR Active and polarized perpendicular to the axis.

C2h E C2 i h

Ag 1 1 1 1 x2,y2,z2,xy

Bg 1 -1 1 -1 xz,yz

Au 1 1 -1 -1 z

Bu 1 -1 -1 1 x,y

Raman Activity of Fundamental Modes

0 , 0 , 0 , 0 1 , 0 , 0 , 0Ag:

31 2 41 , 0 , 0 , 0 1 0 0 0x x xv i b

gA

31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b

gA

x xg g g gA A A A x xg g g gA B A B

Ag vibrations are Raman Active

u v 0i u = x, y, z and v = x, y, z0( ) x ( u v ) x ( )i

21 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i bx

21 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i by

21 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i bz

1 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i bx y

1 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i bx z

1 , 0 , 0 , 0 0 , 0 , 0 , 0v i b v i by z

C2h E C2 i h

Ag 1 1 1 1 x2,y2,z2,xy

Bg 1 -1 1 -1 xz,yz

Au 1 1 -1 -1 z

Bu 1 -1 -1 1 x,y

Raman Activity of Fundamental Modes

x xu g g uB A A B x xu g g uB B A A

Bu vibrations are Raman Inactive


20 , 0 ,1 , 0 0 , 0 , 0 , 0v i b v i bx

20 , 0 ,1 , 0 0 , 0 , 0 , 0v i b v i by

20 , 0 ,1 , 0 0 , 0 , 0 , 0v i b v i bz

v i bv i b x y 0,0,0,00,1,0,0

0 , 0 ,1 , 0 0 , 0 , 0 , 0v i b v i bx z

0 , 0 ,1 , 0 0 , 0 , 0 , 0v i b v i by z

31 2 40 , 0 ,1 , 0 0 0 1 0x x xv i b

uB

31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b

gA

0 , 0 , 0 , 0 0 , 0 , 1 , 0Bu:

The Shortcut C2h E C2 i h

Ag 1 1 1 1 x2,y2,z2,xy

Bg 1 -1 1 -1 xz,yz

Au 1 1 -1 -1 z

Bu 1 -1 -1 1 x,yCorky: Hey, Cousin Mookie. Wotcha doing all that work for? There’s a neat little shortcut.

Mookie: Would you please stop looking for shortcuts? They can get you in trouble.

Corky: But look!! If there’s an x,y or z attached to the representation, then the vibration’s IR active; Bu vibrations are IR Active. Ag vibrations are IR Inactive.

If there’s an x2, y2, etc. attached to the representation,then the vibration’s Raman active; Ag vibrations are Raman Active.Buvibrations are Raman Inactive.

Mookie: How do you determine the activity of a combination mode?

Corky: What’s a combination mode?

IR Activity of Combination Modes

0 0 ,1 , 0 ,1 0 , 0 , 0 , 0x v i b v i biM e x

31 2 40 ,1 , 0 ,1 0 1 0 1x x xv i b

31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b

gA

x xu u g gB B A A

0 0 ,1 , 0 ,1 0 , 0 , 0 , 0y v i b v i biM e y x xu u g gB B A A

0 0 ,1 , 0 ,1 0 , 0 , 0 , 0z v i b v i biM e z x xu u g gB A A B

2 + 4 is IR Active and polarized perpendicular to the axis.

C2h E C2 i h

Ag 1 1 1 1 x2,y2,z2,xy

Bg 1 -1 1 -1 xz,yz

Au 1 1 -1 -1 z

Bu 1 -1 -1 1 x,y


C C

H H

H H1(3030 cm-1)

Ag

C C

H H

H H3

(3110 cm-1)

Bu

C C

H H

H H2(3100 cm-1)

Ag

C C

H H

H H4

(2990 cm-1)

Bu

0 , 0 , 0 , 0 0 , 1 , 0 , 1 2 + 4

x x xg g g u ua a a b B

C2h E C2 i h

Ag 1 1 1 1 x2,y2,z2,xy

Bg 1 -1 1 -1 xz,yz

Au 1 1 -1 -1 z

Bu 1 -1 -1 1 x,y

IR Activity of Combination Modes

0 0 , 0 ,1 , 0 0 , 0 , 0 ,1x v i b v i biM e x

31 2 40 , 0 ,1 , 0 0 0 1 0x x xv i b

x xu u u uB B B B

0 0 , 0 ,1 , 0 0 , 0 , 0 ,1y v i b v i biM e y

0 0 , 0 ,1 , 0 0 , 0 , 0 ,1z v i b v i biM e z x xu u u uB A B A

3 - 4 is IR Inactive.


C C

H H

H H1(3030 cm-1)

Ag

C C

H H

H H3

(3110 cm-1)

Bu

C C

H H

H H2(3100 cm-1)

Ag

C C

H H

H H4

(2990 cm-1)

Bu

0 , 0 , 0 , 1 0 , 0 , 1 , 0 3 - 4

x x xg g u g ua a b a B

31 2 40 , 0 , 0 ,1 0 0 0 1x x xv i b


x xu u u uB B B B

C2h E C2 i h

Ag 1 1 1 1 x2,y2,z2,xy

Bg 1 -1 1 -1 xz,yz

Au 1 1 -1 -1 z

Bu 1 -1 -1 1 x,y

x xu g g uB A A B x xu g g uB B A A


20 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i bx

20 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i by

20 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i bz

0 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i bx y

0 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i bx z

0 ,1 , 0 ,1 0 , 0 , 0 , 0v i b v i by z

Raman Activity of Combination Modes

2 + 4 is Raman Inactive.

31 2 40 ,1 , 0 ,1 0 1 0 1x x xv i b

31 2 40 , 0 , 0 , 0 0 0 0 0x x xv i b

gA


0 , 0 , 0 , 0 0 , 1 , 0 , 1 2 + 4

C2h E C2 i h

Ag 1 1 1 1 x2,y2,z2,xy

Bg 1 -1 1 -1 xz,yz

Au 1 1 -1 -1 z

Bu 1 -1 -1 1 x,y

x xu g u gB A B A x xu g u gB B B B 20 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i bx

20 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i by

20 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i bz

0 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i bx y

0 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i bx z

0 , 0 ,1 , 0 0 , 0 , 0 ,1v i b v i by z

Raman Activity of Combination Modes


0 , 0 , 0 , 1 0 , 0 , 1 , 0

31 2 40 , 0 ,1 , 0 0 0 1 0x x xv i b

3 - 4

x x xg g u g ua a b a B

31 2 40 , 0 , 0 ,1 0 0 0 1x x xv i b


3 - 4 is Raman Active.

Time Independent Perturbation Theory

Introduction

One often finds in QM that the Hamiltonian for a particular problemcan be written as:

( 0 ) ( 1 )H H H

H(0) is an exactly solvable Hamiltonian; i.e. ( 0 ) ( 0 ) ( 0 ) ( 0 )H E

H(1) is a smaller term which keeps the Schrödinger Equation from being solvable exactly.

One example is the Anharmonic Oscillator:

2 22 3 4

2

1

2 2

dH kx x x

d x

H(0)

Exactly SolvableH(1)

Correction Term

( 0 ) ( 1 )H H H ( 0 ) ( 0 ) ( 0 ) ( 0 )H E where

( 0 ) ( 1 ) ( 2 ) ( )nE E E E E

In this case, one may use a method called “Perturbation Theory” toperform one or more of a series of increasingly higher order correctionsto both the Energies and Wavefunctions.

( 0 ) ( 1 ) ( 2 ) ( )n

We will use the notation: ( 0 )E E E ( 0 ) and

Some textbooks** outline the method for higher order corrections.However, we will restrict the treatment here to first order perturbationcorrections

e.g. Quantum Chemistry (5th. Ed.), by I. N. Levine, Chap. 9

First Order Perturbation Theory

( 0 ) ( 1 )H H H ( 0 ) ( 0 ) ( 0 ) ( 0 )H E where

( 0 )E E E ( 0 ) Assume: and

H E

( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 0 )H H E E

( 0 ) ( 0 ) (1) ( 0 ) ( 0 ) (1)

( 0 ) ( 0 ) ( 0 ) ( 0 )

H H H H

E E E E

One can eliminate the two terms involving the product of two small corrections.

( 0 ) ( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 0 ) ( 0 )H H H H E E E E

One can eliminate two additional terms because: ( 0 ) ( 0 ) ( 0 ) ( 0 )H E

( 1 ) ( 0 ) ( 0 ) ( 0 ) ( 0 )H H E E

( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 )* * * *H d H d E d E d

Multiply all terms by (0)* and integrate:

H(0) is Hermitian. Therefore:

( 0 ) ( 0 ) ( 0 ) ( 0 )* *H d H d ( 0 ) ( 0 ) *E d ( 0 ) ( 0 ) *E d

( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 )* * * *H d E d E d E d

Plug in to get:

Therefore: ( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 0 )* *H d E d E

( 0 ) ( 1 ) ( 0 )*E H d is the first order perturbation theorycorrection to the energy.

Applications of First Order Perturbation Theory

PIB with slanted floor

Consider a particle in a box with the potential:

( ) 0 ,V x x x a

0( ) 0V

V x x x aa

V0For this problem:

2 2( 0 )

20

2

dH

m dx

2 2( 0 )

28n

n hE

ma

( 0 ) 2sinn

n x

a a

The perturbing potential is: (1) 0VH x

a

x

V(x

)

0 a0

We will calculate the first order correction to the nth energy level.In this particular case, the correction to all energy levels is the same.

( 0 ) ( 1 ) ( 0 )*E H d 0

0

2 2sin sin

a Vn x n xx dx

a a a a a

20

0

2sin

aV n xx dx

a a a

sin 2 sin 2 0n

a aa

cos 2 cos 2 1n

a aa

22

2

sin(2 ) cos(2 )sin ( )

4 4 8

x x x xx x dx

Integral Info

202

0

2s in

aV nx x dx

a a

20

2 2 2

2 sin (2 ) co s(2 ) co s(0 )0 0

4 4 8 8

V a a a a

a

20

2 2 2

2 1 1

4 8 8

V a

a

0

2

VE Independent of n

Anharmonic Oscillator

For this problem:2 2

( 0 ) 22

1

2 2

dH kx

m dx

( 0 )0

1

2E

21/ 4

( 0 ) / 20

xe

The perturbing potential is:( 1 ) 3 4H x x

2 3 41( )

2V x k x x x

Consider an anharmonic oscillator with the potential energy of the form:

We’ll calculate the first order perturbation theory correction to the groundstate energy.

kk

2 21/ 4 1/ 4

/ 2 3 4 / 2x xe x x e dx

( 0 ) ( 1 ) ( 0 )0 0*E H d

21/ 2

3 4 xx x e dx

2 21/ 2 1/ 2

3 4x xx e dx x e dx

2420

3

8

xx e dx

23

4k

Note: There is no First order Perturbation Theory correction due to the cubic term in the Hamiltonian.

However, there IS a correction due to the cubic termwhen Second order Perturbation Theory is applied.

21/ 2

3 4 xE x x e dx

21/ 2

4

0

0 2 xx e dx

1/ 2 1/ 2

2

32

8

2

3

4E

2

3

4k

Brief Introduction to Second Order Perturbation Theory

As noted above, one also can obtain additional corrections to the energyusing higher orders of Perturbation Theory; i.e.

The second order correction to the energy of the nth level is given by:

( 0 ) ( 1 ) ( 2 )n n n nE E E E

En(0) is the energy of the nth level for the unperturbed Hamiltonian

En(1) is the first order correction to the energy, which we have called E

En(2) is the second order correction to the energy, etc.

2( 0 ) (1) ( 0 )

( 2 )( 0 ) ( 0 )

1

k n

nk n k

HE for k n

E E

( 0 ) ( 1 ) ( 0 ) ( 0 ) ( 1 ) ( 0 )*k n k nH H d where

If the correction is to the ground state (for which we’ll assume n=1), then:

2( 0 ) (1) ( 0 )

1( 2 )1 ( 0 ) ( 0 )

2 1

k

k k

HE

E E

2 2 2( 0 ) (1) ( 0 ) ( 0 ) (1) ( 0 ) ( 0 ) (1) ( 0 )2 1 3 1 4 1( 2 )

1 ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 )1 2 1 3 1 4

H H HE

E E E E E E

Note that the second order Perturbation Theory correction is actuallyan infinite sum of terms.

However, the successive terms contribute less and less to the overallcorrection as the energy, Ek

(0), increases.










• QM Calcs. of Vibrational Frequencies and Dissoc. Energies

The Potential Energy Curve of H2

0.0 0.5 1.0 1.5 2.0 2.5

-1.2

-1.1

-1.0

-0.9E

nerg

y (h

artr

ees)

Bond Length (Angstroms)

Calculation at QCISD(T)/6-311++G(3df,3pd) level

Re(cal) = 0.742 Å

Re(exp) = 0.742 Å

Emin(cal) = -1.17253 hartrees (au)

E(cal) =2(-0.49982) hartrees = -0.99964 hartrees (au)

0.0 0.5 1.0 1.5 2.0 2.5

-1.2

-1.1

-1.0

-0.9

Ene

rgy

(ha

rtre

es)


De(cal) = E(cal) – Emin(cal) = 0.17289 au

The Dissociation Energy

•2625.5 kJ/mol / au

De(cal) = 453.9 kJ/mol

D0(exp) = 432 kJ/mol

De: Spectroscopic

Dissociation Energy

D0: Thermodynamic Dissociation Energy

De = D0 + Evib = D0 + (1/2)h

0.0 0.5 1.0 1.5 2.0 2.5

-1.2

-1.1

-1.0

-0.9

Ene

rgy

(ha

rtre

es)


Relating De to D0

De: Spectroscopic

Dissociation Energy


versus (exp) = 4395 cm-1~

H2: (cal) = 4403 cm-1 [QCISD(T)/6-311++G(3df,3pd)]~

The experimental frequency is the “harmonic” value,corrected from the observed anharmonic frequency.

De(cal) = 453.9 kJ/mol

D0(exp) = 432 kJ/mol

0.0 0.5 1.0 1.5 2.0 2.5

-1.2

-1.1

-1.0

-0.9

En

erg

y (h

art

ree

s)


De: Spectroscopic

Dissociation Energy


De = D0 + Evib = D0 + (1/2)h

1 1

2 2v ibE h h c

3 4 1 0 116 .6 3 1 0 3 .0 0 1 0 / ( 4 4 0 3 )

2x J s x cm s cm

2 0 2 3 13

14 .3 7 9 1 0 6 .0 2 1 0

1 0

k Jx J x m o l

J

2 6 . 4 /v i bE k J m o l

D0(cal) = De(cal) –Evib

= 453.9 – 26.4

= 427.5 kJ/mol

Calculated Versus Experimental Vibrational Frequencies

H2O

(Harm)a

3943 cm-1

3833

1649

(a) Corrected Experimental Harmonic Frequencies

(cal)b

3956 cm-1

3845

1640

(b) QCISD(T)/6-311+G(3df,2p)

(cal)c

4188 cm-1

4070

1827

(c) HF/6-31G(d)

Lack of including electron correlation raises computed frequencies.

Harmonic versus Anharmonic Vibrational Frequencies

H2O

(Harm)a

3943 cm-1

3833

1649

(a) Corrected Experimental Harmonic Frequencies

(cal)b

3956 cm-1

3845

1640

(b) QCISD(T)/6-311+G(3df,2p)

(cal)c

4188 cm-1

4070

1827

(c) HF/6-31G(d)

Lack of including electron correlation raises computed frequencies.

(exp)z

3756 cm-1

3657

1595

(z) Actual measured Experimental Frequencies

Differences between Calculated and Measured Vibrational Frequencies

There are two sources of error in QM calculated vibrationalfrequencies.

(1) QM calculations of vibrational frequencies assume that the vibrations are “Harmonic”, whereas actual vibrations are anharmonic.

Typically, this causes calculated frequencies to be approximately5% higher than experiment.

(2) If one uses “Hartree-Fock (HF)” rather than “correlated electron” calculations, this produces an additional ~5% increase in the calculated frequencies.

To correct for these sources of error, multiplicative “scale factors”have been developed for various levels of calculation.

An Example: CHBr3

(exp)a

3050 cm-1

1149 (E)

669 (E)

543

223

155 (E)

(a) Measured “anharmonic” vibrational frequencies.

(cal)b

3222 cm-1

1208 (E)

691 (E)

546

228

158 (E)

(b) Computed at QCISD/6-311G(d,p) level

(scaled)c

3061 cm-1

1148 (E)

656 (E)

519

217

150 (E)

(c) Computed frequencies scaled by 0.95

Another Example: CBr3•

(cal)b

799 (E) cm-1

324

235

164 (E)

(b) Computed at QCISD/6-311G(d,p) level

(a) Measured “anharmonic” vibrational frequencies.

(exp)a

773 (E) cm-1

???

???

???

(scaled)c

759 (E) cm-1

308

223

156 (E)

(c) Computed frequencies scaled by 0.95

Some Applications of QM Vibrational Frequencies

(1) Aid to assigning experimental vibrational spectra

One can visualize the motions involved in the calculated vibrations

(2) Vibrational spectra of transient species

It is usually difficult to impossible to experimentally measure the vibrational spectra in short-lived intermediates.

(3) Structure determination.

If you have synthesized a new compound and measured the vibrational spectra, you can simulate the spectra of possible proposed structures to determine which pattern best matches experiment.

Statistical Thermodynamics: Vibrational Contributions to Thermodynamic Properties of Gases

Remember these?

2

,

ln

V N

QU kT

T

2

, ,

ln ln

lnV N T N

Q QH kT kT

T V

,V

V N

UC

T

,P

P N

HC

T

lnU

S k QT

l nA U T S k T Q

,

lnln

ln T N

QG H TS kT Q kT

V

The Vibrational Partition Function

1

2 2

k

1, 0 ,1 , 2 ,

2nE n h n

HO Energy Levels:

0

nvib kT

n

q e

The Partition Function

1/ 2

0

n h

kT

n

e

(1/ 2 )

0

nh h

kT kT

n

e e

(1/ 2 )

0

h nh

kT kT

n

e e

2

0

v vnvib T T

n

q e e

v

h

k

where

hc

k

If v/T << 1 (i.e. if /kT << 1), we can convert the sum to an integral.

Let's try O2 ( = 1580 cm-1 ) at 298 K.~

2 2 7 77 .6 4

2 9 8v K

T K

3 4 1 0 1

2 3

(6 .6 3 1 0 )(3 .0 0 1 0 / )(1 5 8 0 )2 2 7 7

1 .3 8 1 0 /v

h c x J s x cm s cmK

k x J K

Sorry Charlie!! No luck!!

Simplification of qvib

v

h hc

k k

2

0

v vnvib T T

n

q e e

Let's consider:0 0

w h erev vn

nT T

n n

e x x e

( )

0

1( ) (0 )

!n n

n

f x f xn

We learned earlier in the chapter that anyfunction can be expanded in a Taylor series:

0

1

1n

n

xx

It can be shown that the Taylor series for 1/(1-x) is:

Therefore:0 0

vnnT

n n

e x

1 1

11

v

Tx

e

v

h hc

k k

2

0

v vnvib T T

n

q e e

0

1

1

v

v

n

T

n T

e

e

and

2

1

v

v

Tvib

T

eq

e

For N molecules, the total vibrational partition function, Qvib, is:

2

1

v

v

N

TNvib vib

T

eQ q

e

Some comments on the Vibrational Partition Function

v

h hc

k k

2

1

v

v

N

TNvib vib

T

eQ q

e

If we look back at the development leading to qvib and Qvib, we see that:

(a) the numerator arises from the vibrational Zero-Point Energy (1/2h)

(b) the denominator comes from the sum over exp(-nh/kT)The latter is the "thermal" contribution to qvib because all terms aboven=0 are zero at low temperatures

In some books, the ZPE is not included in qvib; i.e. they use n = nh.In that case, the numerator is 1 and they then must add in ZPEcontributions separately

2ln ln ln 1v v

vib T TQ N e N e

ZPE Term Thermal Term

In further developments, we'll keep track of the two terms individually.

ln 12

v

v TN N eT

Internal Energy

ln ln 12

vvib v TQ N N e

T

2

,

ln

vibvib

V N

QU kT

T

2 2

, ,

ln 12

vvib v T

V N V N

U kT N kT N eT T T

v i b v i bZ P E t h e r mU U

UZPEvib Utherm

vib

2

,2

vib vZPE

V N

U kT NT T

2 1

2v

dN kT TdT

ZPE Contribution

2

2

1

2vN kT

T

2vNk

2v ibZPE

N k hcU

k

2AN hc

n

This is just the vibrational Zero-Pointenergy for n moles of molecules

or2

v ib AZPE

N hcU

2

,

ln 1v

vib Ttherm

V N

U kT N eT

Thermal Contribution

2 ln 1v

Td

NkT edT

2

1

1

v

v

T

T

NkT de

dTe

2

( 1)

1

v

v

T

T

NkT de

dTe

2

1

v

v

vT

T

NkT de

dT Te

22

1

v

v

Tv

T

eNkT

Te

1

v

v

T

v

T

eNk

e

1 1

v v

v v

T T

A v v

T T

e enN k nR

e e

1

v

v

Tvibtherm v

T

eU nR

e

Although this expression is fine, it's commonly rearranged, as follows:

1

v v

v v

T Tvibtherm v

T T

e eU nR

e e

1v

vib vtherm

T

nRU

e

1v

vib vtherm

T

RU

e

or

Limiting Cases

Low Temperature:(T 0)

1v

vib vtherm

T

RU

e

0

1v

vib vtherm

T

U R

e

Because the denominatorapproaches infinity

High Temperature:(v << T) 1 1

vib vtherm

v

RU

T

R T

The latter result demonstrates the principal of equipartition of vibrationalinternal energy: each vibration contributes RT of internal energy.

However, unlike translations or rotations, we almost never reach thehigh temperature vibrational limit.

At room temperature, the thermal contribution to the vibrational internalenergy is often close to 0.

Numerical Example

1v

vib vtherm

T

RU

e

2

vib AZPE

N hcU

Let's try O2 ( = 1580 cm-1 ) at 298 K.~

3 4 1 0 1

2 3

(6 .6 3 1 0 )(3 .0 0 1 0 / )(1 5 8 0 )2 2 7 7

1 .3 8 1 0 /v


k x J K

2 3 1 3 4 1 0 1( 6 .0 2 1 0 )( 6 .6 3 1 0 )(3 .0 0 1 0 / ) (1 5 8 0 )

2v ibZ P E

x m o l x J s x cm s cmU

( 2 2 7 7 / 2 9 8 )

8 .314 / 2277

11

v

vib vtherm K K

T

R J m ol K KU

ee

9 4 5 9 / 9 . 4 6 /J m o l k J m o l Independent ofTemperature

9 . 1 / 0 . 0 0 9 /J m o l k J m o l

vs. RT = 2.48 kJ/mol

Thus, the thermal contribution to the vibrational internal energyof O2 at room temperature is negligible.

A comparison between O2 and I2

I2: = 214 cm-1 v = 309 K~

T O2 I2 RT(K) (kJ/mol) (kJ/mol) (kJ/mol)

v ibth e rmU v ib

th e rmU

298 0.009 1.41 2.48

500 0.20 3.00 4.16

1000 2.16 7.10 8.31

2000 8.92 15.4 16.6

3000 16.7 23.7 24.9

Because of its lower frequency (and,therefore, more closely spaced energy levels), thermal contributions to the vibrational internal energyof I2 are much more significant than for O2 at all temperatures.

At higher temperatures, thermal contributions to the vibrational internalenergy of O2 become very significant.

Enthalpy

2

, ,

ln ln

ln

vib vibv ib

V N T N

Q QH kT kT

T V

Qvib independent of V

Therefore: 2

,

ln vibvib vib

V N

QH kT U

T

and: vib vibZPE ZPE

vib vibtherm therm

H U

H U

Similarly:, ,

vib vibvib vibP V

P N V N

H UC C

T T

Heat Capacity (CVvib and CP

vib)1

v

vib vtherm

T

nRU

e

2vib AZPE

nN hcU

, , ,

vibvibvibvib thermZPEV

V N V N V N

UUUC

T T T

Independent of T

1v

vib vV

T

nRdC

dTe

1

1v

Tv

dnR e

dT

2

( 1) 1 1v v

T Tv

dnR e e

dT

2 2

1

v

v

v vT

T

nRe

Te

2

2

1

v

v

Tv

T

enR

Te

It can be shown that in the limit, V << T, CVvib R

2

2

1

v

v

Tvib vib vV P

T

eC C nR

Te

Numerical Example

7 .642

27 .648 .314 7 .64 0 .23 /

1

vib vibV P

eC C J mol K

e

Let's try O2 ( = 1580 cm-1 ) at 298 K (one mole).~

3 4 1 0 1

2 3

(6 .6 3 1 0 )(3 .0 0 1 0 / )(1 5 8 0 )2 2 7 7

1 .3 8 1 0 /v


k x J K

2 2 7 77 .6 4

2 9 8v

T

vs. R = 8.314 J/mol-K

A comparison between O2 and I2

I2: = 214 cm-1 v = 309 K~

298 0.23 7.61

500 1.85 8.05

1000 5.49 8.25

2000 7.47 8.30

3000 7.93 8.31

T O2 I2 (K) (J/mol-K) (J/mol-K)

Vibrational Heat Capacities (CPvib = CV

vib)

Vibrational contribution to heat capacity of I2 is greater because ofit's lower frequency (i.e. closer energy spacing)

The heat capacities approach R (8.31 J/mol-K) at high temperature.

Entropy

lnvib

v ib v ib US k Q

T

1v

vib vtherm

T

nRU

e

2 2v ib v vZ P E

N k nRU

2

1

v

v

N

TNvib vib

T

eQ q

e

2

ln ln

1v

Nv

Tvib

T

ek Q k

e

2ln ln 1vv

T TNk e N k e

/ 2 1ln 1

21

v

v

vib v v vT

T

Nk nR nRS Nk e

T T Te

ln 1

1

v

v

vvib T

T

nR TS nR e

e

ANk nN k nR

Numerical Example

Let's try O2 ( = 1580 cm-1 ) at 298 K (one mole).~

3 4 1 0 1

2 3

(6 .6 3 1 0 )(3 .0 0 1 0 / )(1 5 8 0 )2 2 7 7

1 .3 8 1 0 /v


k x J K

2 2 7 77 .6 4

2 9 8v

T

ln 1

1

v

v

vvib T

T

nR TS nR e

e

7 .6 47 .6 4

8 .3 1 4 7 .6 48 .3 1 4 ln 1

1v ibS e

e

0 . 0 0 4 0 . 0 3 1 0 . 0 3 5 /J m o l K

1 5 1 . 9 4 3 . 8 0 . 0 3 5 1 9 5 . 7 /t r a n r o t v i bS S S J m o l K

"Total" Entropy

Chap. 3 Chap. 4

O2: Smol(exp) = 205.1 J/mol-K at 298.15 K

We're still about 5% lower than experiment, for now.

The vibrational contribution to the entropy of O2 is very low at 298 K.It increases significantly at higher temperatures.

T Svib

298 K 0.035 J/mol-K

500 0.49

1000 3.07

2000 7.86

3000 10.82

E (Thermal) CV S KCAL/MOL CAL/MOL-K CAL/MOL-K TOTAL 3.750 5.023 48.972 ELECTRONIC 0.000 0.000 2.183 TRANSLATIONAL 0.889 2.981 36.321 ROTATIONAL 0.592 1.987 10.459 VIBRATIONAL 2.269 0.055 0.008 Q LOG10(Q) LN(Q) TOTAL BOT 0.330741D+08 7.519488 17.314260 TOTAL V=0 0.151654D+10 9.180853 21.139696 VIB (BOT) 0.218193D-01 -1.661159 -3.824960 VIB (V=0) 0.100048D+01 0.000207 0.000476 ELECTRONIC 0.300000D+01 0.477121 1.098612 TRANSLATIONAL 0.711178D+07 6.851978 15.777263 ROTATIONAL 0.710472D+02 1.851547 4.263345

Output from G-98 geom. opt. and frequency calculation on O2 (at 298 K)

QCISD/6-311G(d)

0 . 0 5 5 / 0 . 2 3 /v i bVC c a l m o l K J m o l K (same as our result)

2 . 2 6 9 / 9 . 4 9 /v i b v i b v i bZ P E t h e r mU U U k c a l m o l k J m o l

We got 9.46 + 0.009 = 9.47 kJ/mol (sig. fig. difference)

0 . 0 0 8 / 0 . 0 3 3 /v i bS c a l m o l K J m o l K We got 0.035 J/mol-K (sig. fig. difference)

G-98 tabulates the sum ofUZPE

vib + Uthermvib even though

they call it E(Therm.)

Helmholtz and Gibbs Energy

l nv i b v i bA k T Q ,

lnln

ln

vibvib vib

T N

QG kT Q kT

V

Qvib independent of V

v i bA

2

1

v

v

N

TNvib vib

T

eQ q

e

l nNv i b v i b v i bG A k T q l n v i bn R T q l n v i bN k T q

2

ln

1

v

v

T

T

enRT

e

O2 ( = 1580 cm-1 ) at 298 K (one mole).~

3 4 1 0 1

2 3

(6 .6 3 1 0 )(3 .0 0 1 0 / )(1 5 8 0 )2 2 7 7

1 .3 8 1 0 /v


k x J K

2 2 7 7

7 .6 42 9 8

v

T

7.64 / 22

7.640.0219

11

v

v

Tvib

T

e eq

ee

( 8 . 3 1 4 / ) ( 2 9 8 ) l n ( 0 . 0 2 1 9 ) 9 4 6 0 / 9 . 4 6 /v i b v i bG A J m o l K K J m o l k J m o l

Polyatomic Molecules

It is straightforward to handle polyatomic molecules, which have3N-6 (non-linear) or 3N-5 (linear) vibrations.

1, 2 ,i iN

vib kTQ e

1 2 1 1 2 2( 1 / 2 ) ( 1 / 2 ) . . .v i b v i b v i b n h c n h c Energy:

Partition Function

1, 2 ,i iN

kT kTe e

1 , 1 ,i iN N

kT kTe e

Therefore: 1 2v i b v i b v i bQ Q Q

1 2v i b v i b v i bQ Q Q

Thermodynamic Properties

1 2l n l n l nv i b v i b v i bQ Q Q

1 2

, , ,

ln lnln vib v ibv ib

V N V N V N

Q QQ

T T T

Therefore: 1 2

1 2

etc.

vib vib vib

vib vib vib

U U U

S S S

Thus, you simply calculate the property for each vibration separately,and add the contributions together.

Comparison Between Theory and Experiment: H2O(g)

For molecules with a singlet electronic ground state and no low-lyingexcited electronic levels, translations, rotations and vibrations are theonly contributions to the thermodynamic properties.

We'll consider water, which has 3 translations, 3 rotations and 3 vibrations.

Using Equipartition of Energy, one predicts:

Htran = (3/2)RT + RT CPtran = (3/2)R + R = (5/2)R

Hrot = (3/2)RT CProt = (3/2)R

Hvib = 3RT CPvib = 3R

Htran + Hrot = 4RT CPtran + CP

rot = 4R

Htran + Hrot + Hvib = 7RT CPtran + CP

rot + CPvib= 7R

Low T Limit:

High T Limit:

500 1000 1500 2000 2500 3000180

200

220

240

260

280

300

Exp Cal

S

[J/m

ol-K

]

Temperature [K]

Calculated and Experimental Entropy of H2O(g)

Calculated and Experimental Enthalpy of H2O(g)

500 1000 1500 2000 2500 3000

0

20

40

60

80

100

120

140

160

180

High T: 7RT

Low T: 4RT

Cal Exp

H(T

) [

kJ/m

ol]

Temperature [K]

Calculated and Experimental Heat Capacity (CP) of H2O(g)

500 1000 1500 2000 2500 300030

35

40

45

50

55

60

High T: 7R

Low T: 4R

Cal Exp

CP

[J/m

ol-K

]

Temperature [K]

Translational + Rotational + Vibrational Contributions to O2 Entropy

Unlike in H2O, the remaining contribution (electronic) to theentropy of O2 is significant. We'll discuss this in Chapter 10.

0 1000 2000 3000 4000 5000100

150

200

250

300

Expt T TR TRV

S

[J/m

ol-K

]

Temperature [K]

Translational + Rotational + Vibrational Contributions to O2 Enthalpy

There are also significant additional contributionsto the Enthalpy.

0 1000 2000 3000 4000 5000

0

50

100

150

200

Expt T TR TRV

H

[kJ/

mol

]

Temperature [K]

Translational + Rotational + Vibrational Contributions to O2 Heat Capacity

Notes: (A) The calculated heat capacity levels off above ~2000 K. This represents the high temperature limit in which the vibration contributes R to CP.

0 1000 2000 3000 4000 500015

20

25

30

35

40

45

Expt T TR TRV

CP

[J/m

ol-K

]

Temperature [K]

(B) There is a significant electronic contribution to CP at high temperature.

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