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Vibrational Spectroscopy
Outlines
- Quantum harmonic oscillator
- Vibrational transitions
- Selection Rules
- Vibration of polyatomic molecules
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Quantum (Mechanical) Harmonic Oscillator
A simple form of Schrödinger equation
Hamiltonian Operator
Eigenvalue Eigenfunction
For One Dimensional Schrödinger Equation
VTHH total ˆor
EH
Kinetic Operator Potential Operator
ExVdx
d
m
2 2
22
ExV
dx
d
m
2
22
2
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1D Schrödinger equation of a diatomic molecule for any states
xExxVdx
xdnnn
n
)(
2 2
22
For solving the Schrödinger Equation, we need to define the Eigen functions associated with the state.
It is commonly known that the exponential function is a good one to work with.
Let see a simple example
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Consider a given function:
Ex. Show that the function satisfies the Schrödinger equation for the 1D quantum harmonic oscillator. What conditions does this place on ? What is E?
2 xe
2 xe
Schrödinger equation
1D Schrödinger equation
EH
ˆ 22 xxtotal EeeH
2
22 2
22xx EeexV
dx
d
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For the left side of Eq.:
2
2
2
2
2 2
2
22
2
1 2
2
2
xx
xx
ekxdx
xed
exVdx
ed
ax
ax
axedx
de
2
udvvduuvd )( 22 2d ,
,1d ,xx xevev
uxu
22
22
22
2
2
2
2
2
) 2(
xx
xx
xxx
exe
eex
exexdx
xed
1st derivative :
2nd derivative :
2
2
1kxxV
2 2
22
2xexV
dx
d
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222
222
2
2
2 222
2
2 2 2
2 2
2
14
2)2(
2
2
1) 2(2
2
2
1 2
2
xxx
xxx
xx
ekxexe
ekxexe
ekxdx
xed
The last two terms must be
cancelled in order to satisfy the
Schrödinger eigenfunction
22
2
2
2 xxx
exedx
xed
ˆ 22 xxtotal EeeH
22
)2(2
E
Thus from the product:
Therefore, the energy eigenvalue:
) 2(2
ˆ 22 2
xxtotal eeH
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To find β, we use the cancellation of the last two terms
02
14
2
22 2 222
xx ekxex
2 ,
4 22
kk
2
1
22
22
kkE
Substitute β, we get the energy :
Therefore, the condition for β that makes the function satisfies the Schrödinger equation
k
From the classical treatment :
hvvh
E2
1)2(
22
1
2
1
hvE2
1
the energy
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Unlike classical treatment, quantum harmonic oscillation of diatomic
molecule exhibits several discrete energy states.
They can be better described using Hermite polynomial wave
functions :
,...2,1,0 and ,2/21 2
nexHAx x
nnn
Eigenfunctions
Vibrational energy states
4/1
n ! 2
1
nAn
2/ k
Normalization constant
xH n
21
Herrmite Polynomials 2/2xe
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HnHH nnn )()1(2)(2 21
Hermite polynomials
.
.
.
124816
128
24
2
1
244
33
22
1
0
H
H
H
H
H
The first few Hermite polynomials
The recurrence relation for Hermite polynomials
For n > 2
n = 0
n = 1
n = 2
n = 3
n = 4
xH n
21
x21
;
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2
2
2
2
2/13
4/13
3
2/124/1
2
2/1
4/13
1
2/14/1
0
329
124
4
x
x
x
x
exxx
exx
xex
ex
The first four Hermite polynomial wave functions, (n= 0,1, 2, 3)
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2
2
1
21 x
nnn exHAx
22
2
1
21
2
1
21
total x
nnn
x
nn exHAEexHAH
hvE2
1
2 xe ˆ 22 xxtotal EeeH
hvnE
hvnE
n
n
2
1)12(
)2
1(
The solution for the eigenstate is:
If
Solving the1D Schrödinger equation
If
The solution for the n eigenstate is:
or
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hvnEn 2
1)12(
Plot of the first few eigenfunctions of the quantum harmonic oscillator (red) together with the potential energy
Zero point energy
ground vibrational state
1st exited vib. state
The energy of a system cannot have zero energy.
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The Probability (ψ2(x))
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hvnhvnEn )2
1(
2
1)12(
The solution of quantum harmonic oscillation gives the energy:
n vibrational state
Evib
(harmonic)En-En-1
0 Ground state 1/2 h -
1 1st excited st. 3/2 h h
2 2nd 5/2 h h
3 3rd 7/2 h h
Where the vibrational quantum number (n) = 0, 1, 2, 3 …
Noted that the constant Evib= h is good for small n. For larger n, the
energy gap becomes very small. It is better to use anharmonic
approximation.
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Ex. for the Hermite polynomial wave functions at n =0 for
HCl
4/1
00! 02
1
A4/1
n ! 2
1
nAn
212 1076.8/ k
)cm 2991 kg, 1064.1( -127
2)2( ck
229822
1076.84/121
0
A
What is the normalization constant at this state?
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Ex. Using Hermite polynomials to show that the ground and first exited states wave function (n =0, n=1) of a diatomic molecule is
2/21 2x
nn exHAx
4/14/1
00! 02
1
A 121
0
xH
2
2
2/1
4/13
1
2/14/1
0
4 x
x
xex
ex
For n = 0
The general form of Hermite polynomials
22 2/14/1
2/121
0
4/1
00! 02
1 xx eexHx
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4/14/1
112
1
! 12
1
A
22 2/1
4/132/12
14/1
1
4)2(
2
1 xx xeexx
x21
2/21 2x
nn exHAx
For n = 1
xH 21
1 22
The recurrence relation for Hermite polynomials
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HnHH nnn )()1(2)(2 21
Hermite polynomials
.
.
.
124816
128
24
2
1
244
33
22
1
0
H
H
H
H
H
The first few Hermite polynomials
The recurrence relation for Hermite polynomials
For n > 2
n = 0
n = 1
n = 2
n = 3
n = 4
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Ex. Plot the ground state wave function (n =0) of HCl where observed frequency 2991 cm-1.
amu 9801.0
212 1076.8/ k
N/m 8.520k
2/4/1
0
2xex
-1cm 2991
20 2
1
2
1kxhvE
hvnEn
2
1
k
Ex 02
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Ex. Draw the vibrational energy diagram of the first five levels of a diatomic molecule using the solution of quantum harmonic oscillation :
hvnhvnEn
2
1
2
1)12(
The first five levels: n = 0, 1, 2, 3, and 4
The energy of a system cannot have zero energy, even at n=0. It is
called “Zero point energy”.
n En
0 1/2 h
1 3/2 h
2 5/2 h
3 7/2 h
4 9/2 h0 1
2
3 4
h
E
at n=0, E0 0
Zero point energy : the energy of the ground vibrational state
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Vibrational transitions
h0
1
2
34 n0 ni
Transitions
01 Fundamental
02 First overtone
03 Second overtone
04 Third overtone
- The fundamental transitions, n=1, are the most commonly occurring. - The overtone transitions n=2, 3, … are much weaker (low intensities).
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Ex. The harmonic vibrational frequency of HCl in wavenumbers is 2991 cm–1.
a) Calculate the energies of the first three vibrational levels in Joules and calculate zero point energy.
1-13
1-1-
10
s 10967.8
cm 2991s
cm10998.2
c
JhvE 200 1097.2
2
1
The ground state vibrational energy, n=0
JhvE 201 1091.8
2
3
The 1st excited state vibrational energy, n=1
JhvE 192 1049.1
2
5
The 2st excited state vibrational energy, n=2
hvnEn
2
1
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JhvE 200 1097.2
2
1
Zero point energy = The ground state vibrational energy
Zero point energy
b) Determine the energy and wavenumber for the fundamental, first-
overtone transitions
1-
1-1301
2001
cm 2991
s 10967.8)(1
10942.5
c
vv
EEh
v
JEEE
fundamental transition : n=0 n=1
1st overtone transition : n=0 n=21-
1902
cm 5982
1019.1
v
JEEE
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Determine the energy and wavenumber for the fundamental, first-
overtone transitions using the vibrational energy function for the
anharmonic correction below:
22
2
1
4
)(
2
1
n
D
hvhvnE
en
Ex. The vibration potential of HCl can be described by a Morse
potential with De=7.41 X 10-19 J, k = 516.3 N m-1 and = 8.967 X 1013
s-1 .
First of all, determine the vibrational energies of the ground
state (n=0) and the first and second excited states (n= 1 and 2)
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De=7.41 X 10-19 J = 8.97 X 1013 s-1
h = 5.944 X 10-20 J(h)2/4De= 1.192 X 10-21 JThe vibrational energies of the
ground state and the first and second excited states
22
2
1
4
)(
2
1
n
D
hvhvnE
en
J 1041.1
J 1064.8
J 1094.2
192
201
200
E
E
E
1-
2001
cm 2871
1070.5
v
JEEEFundamental transition : n=0 n=1
1-
1902
cm 5622
1012.1
v
JEEEFirst overtone transition : n=0 n=2
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Selection Rules
• The transition probability from state n to state m is only nonzero if the transition dipole moment satisfies the following condition:
where x = spatial variable μx = dipole moment along the electric field direction
mnx
0)()()(* xxx nxmmnx
...)())((0
0
x
xxx dx
dtxtx
Permanent Dynamics
0mnx
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Assuming transition from the ground state (n = 0) to state m
0mx
n = 0 n = m
0)(...)()( 00
0*
x
dx
dtxx
x
xxm
...)())((0
0
x
xxx dx
dtxtx
0)()()( 0*0 xxx xm
mx
dxxdx
dtxxxx
x
xmxm )()()()()( 0
0
*00
*
Truncate at the first derivative:
dxxtxx
dx
ddxxx m
x
xmx )()()()()( 0
*
00
*0
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2/21 2x
nnn exHAx
dxexHtxexHdx
dAA
dxexHexHAA
xxm
x
xm
xxmxm
mx
22
22
)()()(
)()(
2/10
2/1
00
2/10
2/100
0
Substitute with Hermite polynomial functions
= 0 (Orthogonal)
2/21
000
2xexHAx
2/21 2x
mmm exHAx
n = 0;
n = m;
X Odd functionIf m is odd Odd functionIf m is even Even function
0mnx m must be odd : 1, 3, 5, 7, …
(so permanent dipole is not relevant for IR absorption)
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Possible transitions )0( mnx
m must be odd : 1, 3, 5, 7, … n=0n=1n=0n=3 n=0n=5 …
n=0n=1
n=0n=3
n=0n=5
Integration give an area under the peaks
Integration
n=0n=1 0
n=0n=3 = 0
n=0n=5 = 0
Transitions
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Selection Rules
Selection Rules for IR absortion
n = 1
The only transition from n=0n=1 is allowed.
For the other an integration is zero. The transition is forbidden.
n = +1 absorption
n = -1emission
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How many vibrational modes are observed for a molecule?
A molecule consisting of N atoms
1 atom : 3 coordinates (x1, y1, z1)
N atoms : 3 N coordinates ((x1, y1, z1), (x2,y2,z2), (x3,y3,z3),…..)
* There are only 3 normal modes (coordinates) describing the translational motion.* There are 3 normal modes describing the Rotational motion of the molecules (along 3 axes)
Number of vibrational modes = 3N – 6Number of vibrational modes for linear molecules = 3 N-5
there are only 2 rotational modes
Vibration of polyatomic molecules
Position Representation
Degree of freedoms (Molecular motions)
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Vibration
Stretching vibration Bending vibration
SymmetricStretching
AsymmetricStretching
In plane Bending
Out of plane Bending
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H2O
- non-linear molecule
- N = 3
- number of vibrational normal modes = 3 X 3 –
6 = 3
3685 cm-1 3506 cm-1 1885 cm-1
H2O has three distinct vibrational frequencies.
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CO2
- linear molecule
- N = 3
- number of vibrational normal modes = 3 X 3 – 5
= 4
Because of xz and xy are degenerate, CO2 has three (not four) distinct vibrational frequencies.
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Benzene C6H6
- non-linear molecule
- N = 12
- number of vibrational normal modes = 3 X 12 – 6
= 30
There are 30 vibrational modes but only 20 distinct
vibrational frequencies.
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In the normal coordinate system, the vibrational Hamiltonian of polyatomic system can be written as:
2
1
2
63
1
22
22
N
iii
i
QkQ
H
... 2
1
2
2
1
2
2
1
22332
3
222222
2
222
1121
22
QkQ
QkQ
QkQ
H
...)()()()(63
1332211
N
iii QHQHQHQHH
Therefore, the wave function is of the form :
...)()()( )( 332211
63
1
QQQQN
iii
The total vibrational energy is:
...2
1
2
1
2
1
2
1...)( 332211
63
1331
hvnhvnhvnhvnnnnEN
iii
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Ex. The water molecule has three normal modes, with
fundamental frequencies: 1-bar = 3685 cm-1, 2-bar =
1885 cm-1, 3-bar = 3506 cm-1.
a) What is the energy of the (112) state?
2
12
2
11
2
11)112( 321 hvhvhvE
2
5
2
3
2
3321 vhcvhcvhc
)cm 3506(2
5)cm 1885(
2
3)cm 3685(
2
3 1-1-1- hchchc
J 1040.3 19
(i.e. n1=1, n2=1, n3=2)
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What is the energy difference between (112) and (100)?
J 1040.3)112( 19E
J 1063.1)100( 19E
J 101.77 10)63.14.3(
)100()112(1919
EEE
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The solutions to these 3N-6 equations are the familiar Harmonic Oscillator Wavefunctions and Energies.
hvnEn 2
1)12(
The total vibrational energy is:
The Schrödinger equation is simplified by separation of variables. Therefore, one gets 3N-6 equations of the form
2
1
22
2
22
iiiiii
i EQkQ
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• Coupled system has two vibrational frequencies: the
symmetrical and antisymmetric modes.
• For symmetrical and asymmetrical, the vibrational
frequency is
1
2
1 kvsymmetric
21 2
2
1 kkv ricqntisymmet
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Ex. The vibration potential of HCl can be described by a Morse
potential with De=7.41 X 10-19 J, k = 516.3 N m-1 and = 8.97 X 1013 s-
1 .
Calculate
ee DxxV )(
Morse function of vibrational potential:
2)(1)( exxe eDxV
At the minimum energy (x=xe), the potential energy:
De = Dissociation energy, Do = bond energy
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en DE
The energy levels for the anharmonic correction:
The highest value of n consistent with the potential that is
ee
DnD
hvhvn
22
2
1
4
)(
2
1
22
2
1
4
)(
2
1
n
D
hvhvnE
en
De=7.41 X 10-19 J = 8.97 X 1013 s-1
h = 5.944 X 10-20 J(h)2/4De= 1.192 X 10-21 J
ee
DnD
hvhvn
22
2
1
4
)(
2
1
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Anharmonicity
2)(1)( exxe eDxV
eD
k
2
exxdx
Vdk
2
2
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Ex. for the Hermite polynomial wave functions at n =0 for
HCl
4/1
00! 02
1
A4/1
n ! 2
1
nAn
212 1076.8/ k
)cm 2991 kg, 1064.1( -127
2)2( ck
28928
102.2
2
14/118
1
A
What is the normalization constant at this state?
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Example for the Hermite polynomial wave functions at n =1
for HCl
4/1
11! 12
1
A4/1
n ! 2
1
nAn
182 102.2/ k
)cm 2991 kg, 1064.1( -127
2)2( ck
28928
102.2
2
14/118
1
A
-110 rads1063.5
What is the normalization constant at this state?
c2 2
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2/21 2x
nn exHAx
To plot the vibrational wave functions, we separate the functions
into three terms, An , Hn and2/2xe
xH n
21
The most complicate one is Herrmite Polynomials
An and are not difficult to calculate 2/2xe
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HnHH nnn )()1(2)(2 21
Hermite polynomials
.
.
.
124816
128
24
2
1
244
33
22
1
0
H
H
H
H
H
The first few Hermite polynomials
The recurrence relation for Hermite polynomials
For n > 2
n = 0
n = 1
n = 2
n = 3
n = 4
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2/21 2x
nn exHAx
4/14/1
00! 02
1
A
22 2/14/1
2/121
0
4/1
00! 02
1 xx eexHx
x21
For n = 0
10 H
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4/14/1
112
1
! 12
1
A
22 2/1
4/132/12
14/1
1
4)2(
2
1 xx xeexx
x21
2/21 2x
nn exHAx
For n = 1
xH 21
1 22
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4/14/1
2222
1
! 22
1
A
x21
2/21 2x
nn exHAx
For n = 2
22 2/124/1
2/124/1
2 124
)24(22
1 xx exexx
242)(424 22212
2 xxH
4/14/1
22
4/1
n 22
1
! 22
1;
! 2
1
An
An
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22/14/1
0xex
Example for the Hermite polynomial wave functions at n =1
for HCl
1) What is the normalization constant at this state?
1 ,2/21 2
nexHAx x
nn
4/1
11! 12
1
A
2/ k
4/1
n ! 2
1
nAn
2/ k
)cm 2991 kg, 1064.1( -127
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Example for the Hermite polynomial wave functions at n =1
for HCl
1) What is the normalization constant at this state?4/1
11! 12
1
A4/1
n ! 2
1
nAn
2/k
)cm 2991 kg, 1064.1( -127
2-
34
27-110-1
s kg 8.520
10626.6
)kg 1067.19801.0)(s cm10998.22)(cm 2991(
k
222 )2( ck
4/14/1
11 2
1
! 12
1
A
)cm 2991 kg, 1064.1( -127
s rad1063.5
)cm 2991s cm10998.22(
2 2
1-10
1-1-10
c
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EXAMPLE
k
tAtx ),sin()(
Answer the following questions.
a) What are the units of A? What role does have in this equation
b) Graph the kinetic and potential energies as a function of time
c) Show that the sum of the kinetic and potential energies is independent of time.
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a) What are the units of A?
Because x(t) has the units of length and the sine function is dimensionless, A must have the units of length.
The quantity sets the value of x at t = 0,because . )sin()0sin()0( AAx
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b) Graph the kinetic and potential energies as a function of timea) Show that the sum of the kinetic and potential
energies is independent of time.
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SolutionFor the J=0 → J=2 transition,
The preceding calculations show that the J=0 → J=1 transition is allowed and that the J=0 → J=2 transition is forbidden. You can also show that is also zero unless MJ=0 .
04
1
4
1
8
5
2
cos
4
cos3
8
5sincos1cos3
8
5
0
242
0
22
0
20
ddz
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Consider the relative motion via the center of mass coordinates.
Angular Momentum (L)
m
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