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Ch 8. The Vibrational and Rotational
Spectroscopy of Diatomic Molecules
- light interacts with molecules to induce transitions
between states.
- Absorption of electromagnetic radiation in the
infrared and microwave regions of the spectrum.
- Transitions between eigenstates of vibrational and
rotational energy induced by light
MS310 Quantum Physical Chemistry
8.1 An introduction to spectroscopy
What is Spectroscopy?
: see the ‘bond’ → a lot of information about the character of
chemical bond and reactivity
Our focus in this chapter : rotation and vibration
(Atomic spectroscopy : Ch 11, electron spectroscopy : Ch 15,
NMR : Ch 18)
Bond length : rotational spectroscopy
Frequency of characteristic oscillation with chemical bond :
vibrational spectroscopy
Discrete energy level in Q.M →
absorption and emission spectrum make individual peaks
Transition energy is given by
h | E1  E2 |
MS310 Quantum Physical Chemistry
Energy, wavelength of light(photon) and name of wave
Visible light : very small range
MS310 Quantum Physical Chemistry
Used wave in spectroscopy
→ from microwave to X-ray region : nine order of magnitude
1
~
In spectroscopy, we use the wave number   
Therefore, energy is given by | E2
 E1 | h  hc~
Experimentally, only a few transition occur(not arbitrary
chosen state) : selection rule
MS310 Quantum Physical Chemistry
Electric and magnetic fields associated with a traveling light
wave.
→ consider the dipolar diatomic molecule in time-dependent
electric field
MS310 Quantum Physical Chemistry
Interaction between classical
harmonic oscillator in electric
field.
(arrow : direction of mass)
Oscillator absorbs energy in
both the stretching and
compression half cycles.
Real Q.M oscillator : similar
Electric field effects 2 way : permanent and dynamic
Ex) polar HCl molecule has permanent dipole moment, and
dynamic dipole moment can be generated.
MS310 Quantum Physical Chemistry
8.2 Absorption, Spontaneous emission, and
Stimulated emission
Photon-assisted transition
Absorption : photon induces a transition to higher level
Spontaneous emission : excited state relaxes to lower level
Stimulated emission : photon induces a transition from excited
state to lower level
MS310 Quantum Physical Chemistry
Spontaneous emission : random event, related to lifetime of
excited state
Absorption, Stimulated emission : related to radiation density ρ(ν)
In equilibrium, overall transition 1 to 2 and 2 to 1 must be same.
System is described by
B12  ( ) N1  B21 ( ) N 2  A21 N 2
Use blackbody spectral density
function, we can obtain
A21 16 2 3
B12  B21 ,

B21
c3
MS310 Quantum Physical Chemistry
Ex) 8.1
Derive the result by the (1) the overall rate of transition is zero at
equilibrium, (2) ratio of N2 to N1 is governed by the Boltzmann
distribution
Sol) overall transition rate is zero :
B12  ( ) N1  B21 ( ) N 2  A21 N 2
N 2 g2  h / kT

e
 e  h / kT ( g1  g2 )
Boltzmann distribution is given by
N 1 g1
 ( ) 
A21
B12e h / kT
8h 3
1

 B21
c 3 e h / kT  1
A21 8h 3 16 2 3


Two expression must be same → B12  B21 ,
3
B21
c
c3
Lightbulb : incoherent photon source(random direction)
Laser : coherent photon source(all photons are in phase and
same direction)
MS310 Quantum Physical Chemistry
8.3 An introduction to vibrational spectroscopy
Vibrational frequency
→ depends on the 2 vibrating atoms at the end of bond
other atoms affect much less degree
→ group frequency : characteristic frequency of bond
Can caculate the # of first excited state(N1) and # of ground
state(N0) at 300K and 1000K by the Boltzmann distribution
Except the Br2, N1<<N0 acceptable even 1000K
→ absorption of light at characteristic frequency occur at n=0
Experimental result in Q.M harmonic oscillator
: ∆n = nfinal – ninitial = +1
MS310 Quantum Physical Chemistry
MS310 Quantum Physical Chemistry
MS310 Quantum Physical Chemistry
Use more higher sensitive instrument
: ∆n = +2, +3, … can measure, but much weaker than ∆n = +1
→ selection rule ∆n = +1 is not rigorously for anharmonic
potential
Model of anharmonic potential : Morse potential
V ( R)  De [1  e  ( R Re ) ]2
De : dissociati on energy relative to the bottom of potential,   k / 2 De
d 2V
force constant k  ( 2 ) x  xe , 
dx
k

: also valid for harmonic oscillator
Bond energy D0 : respect to lowest allowed level
1 ( h )2
1 2
E

h

(
n

)

(
n

)
Energy level is given by n
2
4 De
2
: anharmonic correlation
MS310 Quantum Physical Chemistry
MS310 Quantum Physical Chemistry
Material-dependent parameter is given by this table.
B : rotational constant, r : bond length
MS310 Quantum Physical Chemistry
8.4 The origin of selection rule
Transition probability is not zero
→ transition dipole moment follow this condition
 xmn   m* ( x ) x ( x ) n ( x )d  0
In real case, position depends on time.
We can take Taylor expansion at the x=0(equilibrium position)
 x ( x(t ))  0 x  x(t )(
d x
) x 0  ...
dx
First term : permanent dipole moment
Second term : dynamic dipole moment(time-dependent)
We can think absorption occurs state of n=0 : reasonable
MS310 Quantum Physical Chemistry
We can calculate the transition dipole moment


m0
x
 Am A0  0 x  H m (
1/ 2
x )H 0 (
1/ 2
x )e
x 2
dx


d x
1/ 2
1/ 2
x 2
 Am A0 [(
) x  x0 ]  H m ( x )xH 0 ( x )e dx
dx

First integral : 0 by orthogonality, We only see the second
integral
m=even, Hm(α1/2x) : even, m=odd, Hm(α1/2x) : odd
Therefore, if m is even, overall integral becomes integral of
odd function and it becomes zero.
→ value of integral may not zero if the transition of
n=0 → m=2b+1, b=0,1,2,…
MS310 Quantum Physical Chemistry
Figure of m=1,3,5
Except the m=1 case, area of red and
blue region is same.
  xm 0  0
Therefore, only m=1 transition allowed.
→ ∆n=1 : selection rule for absorption
∆n=-1 : selection rule for emission
Vibrational excitation : only dμx/dx≠0
H2, N2, O2 : μx0 =0, dμx/dx=0
→ 99.93% of atmosphere doesn’t
absorb the IR radiation
IR radiation only absorb by CO2, NOx,
and hydrocarbon : they occur
greenhouse effect
MS310 Quantum Physical Chemistry
8.5 Infrared absorption spectroscopy
IR spectroscopy follows the Beer-Lambert Law
Incident light I0(λ) through the distance dl
Absorption related to dl, intensity, and concentration(also it
related to # of molecule), ε(λ) : molar absorption coeffieicnt
dI ( )   ( ) MI ( )dl
dI ( )
I ( )
  ( ) Mdl ,
 e  (  ) Ml
I ( )
I 0 ( )
MS310 Quantum Physical Chemistry
Ex) 8.4
ε(λ) of ethane : 40 (cm bar)-1 at 12μm
Calculate the I(λ)/ I0(λ) when 10cm absorption cell length and
2.0ppm. Also, how long the cell length if I(λ)/ I0(λ)=0.90?
Sol)
I ( )
 e  (  ) Ml  exp(40(cm bar) -1 (2.0  10-6 bar)(10cm) )
I 0 ( )
 0.9992  1.0
Therefore, it is difficult to measure.
Rearrange the equation, we can obtain
1
I ( )
1
3
l
ln

ln
0
.
90

1
.
3

10
cm
1
6
 ( ) M I 0 ( )
(40 cm bar ) (2.0  10 bar )
This order of length can make by the mirror and it uses to
the gas detection.
MS310 Quantum Physical Chemistry
How does ε(λ) depend on the wavelength?
Consider the ketone molecule
Vibration frequency of carbonyl group : determined by the force
constant of C=O bond.
Force constant depends on the chemical bond between C and O
and other alkyl groups affect much less!
Vibrational modes : depends on the degree of freedom
total degree of freedom : 3n
translational freedom : 3
rotational freedom : 3(nonlinear), 2(linear)
vibrational freedom : 3n-6(nonlinear), 3n-5(linear)
Benzene : 30 vibrational modes, but only 20 distinct vibrational
frequencies because of the degeneracy
MS310 Quantum Physical Chemistry
MS310 Quantum Physical Chemistry
See the IR spectroscopy of CH4 and CO
Case of CH4, 9 vibrational frequency expected. However,
there are only 2 peaks and a lot of additional frequencies.
Also, broadening in CO peak. Why?
→ rotational spectra
MS310 Quantum Physical Chemistry
There are many rotational states with only 1 vibrational transition,
n=0 → n=1.
We can analyze the rotational spectra if we use the highresolution instrument.
In CH4 spectra, there are only 2 peaks instead of 9. Why?
Case of CH4, there are only 2 peaks
1306 cm-1 : 3 degenerate C-H bending modes
3020 cm-1 : 3 degenerate C-H stretching modes
Where are other 3 modes?
Use group theory, these modes are symmetric and doesn’t
satisfy the condition dμx/dx≠0. Therefore, these modes are
infrared inactive.
However, all modes of CH4 and CO are raman active.(section 8.8)
MS310 Quantum Physical Chemistry
Consider the CO2 case.
2 C=O bonds are equivalent and expected there are only 1 peaks.
However, there are 2 peaks by the experiment. Why?
→ symmetric and antisymmetric stretching
MS310 Quantum Physical Chemistry
Symmetric stretch : only depends on k1 ,  symmetric 
1
2
k1

Asymmetric stretch : C moves → opposite direction of each
oscillator, F=-(k1+2k2)x :
1 k1  2k 2
 antisymmet ric 
2

Symmetric and antisymmetric O-H stretching modes
MS310 Quantum Physical Chemistry
8.6 Rotational spectroscopy
Rotational selection rule : ∆J=Jfinal – Jinitial=±1
We see the Example 8.1
Ex) 8.1
Use these eigenfunctions of rigid rotor, J=0 →J=1 transition is
allowed and J=0 → J=2 is forbidden.
1 12
Y ( ,  )  ( )
4
3 12
0
Y1 ( ,  )  ( ) cos 
4
1
5
Y20 ( ,  )  (
) 2 ( 3 cos 2   1)
16
0
0
MS310 Quantum Physical Chemistry
Sol) assume the electromagnetic field : z direction → μz=μcosθ
2

0
0
 zJ 0    d  YJ0 ( ,  )(cos  )Y00 ( ,  ) sin d
J=0 → J=1
2

3 cos 3   
3
 
d  cos  sin d  
[
]  0  
0

4 0 0
2
3
3
3
10
z
2
J=0 → J=2
 z20  
5
8
2

0
0
2
d

(
3
cos
  1) cos  sind
 
3 cos 4  cos 2   
5 1 1

[

]  0  
[  ]  0
8
4
2
8 4 4
5
Therefore, J=0 → J=1 transition is allowed, J=0 → J=2
transition is forbidden. Also, μzJ0 is zero unless ∆MJ=0.
MS310 Quantum Physical Chemistry
Consider the charged two particles in rotational motion.
Rotational absorption : nonzero
‘permanent’ dipole moment!
(by contrast, in vibrational absorption,
nonzero dynamic dipole moment)
We use the angular quantum number
J instead l.(l used for orbital angular
momentum)
Energy is given by the
2
h2
E
J ( J  1) 
J ( J  1)  hcBJ ( J  1)
2r02
8 2 r02
B : rotational constant
MS310 Quantum Physical Chemistry
Simulated rotational spectroscopy
MS310 Quantum Physical Chemistry
calculate the energy change of the transition ∆J=1 and ∆J=-1
E  E ( J final )  E ( J initial )
2
2
J ( J  1)  2hcB( J  1)
( J  1)( J  2) 
E  
2 r02
2 r02
2
2
J ( J  1)  2hcBJ
( J  1)J 
E  
2 r02
2 r02
| ∆E+ | ≠ | ∆E- | : energy level not
spaced equally
Rotational energy not depends on mJ :
2J+1 fold degenerate
Difference between ∆ν, but ∆(∆ν) is
same and its value is 2cB
MS310 Quantum Physical Chemistry
In Real case : rotational and vibrational change simultaneously
∆Erotational << kT : many rotational peaks
observe
Calculate the ratio of ∆Erot and ∆Evib
E rot  2 / r02


 2
Evib  k /  r0 k
This ratio is 0.028(H2) and 0.00034(I2).
moment of inertia, force constant large
→ smaller ratio
MS310 Quantum Physical Chemistry
Relative ratio of given J state
nJ gJ ( J  0 ) / kT
  2 J ( J 1) / 2 IkT

e
 ( 2J  1)e
n0 g0
Degeneracy (2J+1) dominant for small J
and large T
nJ/n0 decrease to 0 rapidly J increase
Moment of inertia increase
: upper value of J increase
HD case : 4 for 300K, 7 for 700K
CO case : 13 for 100K, 23 for 300K, 33
for 700K
MS310 Quantum Physical Chemistry
Higher frequency(∆J=+1) : R branch
Lower frequency(∆J=-1) : P branch
Center of spectrum : ∆J=0 : forbidden
transition
For raman spectroscopy, selection rule
becomes to ∆J=0, ±2
MS310 Quantum Physical Chemistry
Higher resolution IR spectra of CO molecule
MS310 Quantum Physical Chemistry
8.7 Fourier transform infrared spectroscopy
FTIR : one pulse → same as many single-wavelength
experiment(multiplex advantage)
→ short time!
Instrument : Michelson interferometer
MS310 Quantum Physical Chemistry
At first, we consider the one incident wave
Incident light : amplitude A0ei(kx-ωt), intensity I0
Through the beam splitter S : 50% of light transmitted, and
other 50% of light reflected and go to M2
Transmitted light : reflect by M1 and 50% is reflected by S
Reflected light : reflect by M2 and 50% is transmitted through S
These 2 waves make interference.
A
A
A( t )  0 [e i ( kyD t )  e i ( k [ y D  d ]t ) ]  0 (1  e i ( t ) )e i ( kyD t )
2
2
δ(t) : phase difference from the path difference, Δd
2
2
 (t ) 
( 2 SM1  2 SM 2 ) 
d ( t )


intensity I : related to A*(t)A(t)
I0
I0
2d ( t )
I ( t )  (1  cos  (t ))  (1  cos
), I 0  A02
2
2

MS310 Quantum Physical Chemistry
Δd=nλ : constructive, measured
Δd=(2n+1)λ/2 : destructive, not measured
Measured signal : interferogram, single sine wave
MS310 Quantum Physical Chemistry
After, consider the many incident waves(different frequencies)
Amplitude is given by
A( t )  
j
Aj
2
(1  e
i
2d ( t )
j
i(
)e
Measured intensity is given by I ( t ) 
2
j
y D  j t )
1
2v
I
(

)(
1

cos(

t ))

j
j
2 j
c
FTIR gives all frequency
data simultaneously
→ use for determine of
fraction of air
MS310 Quantum Physical Chemistry
8.8 Raman spectroscopy
Consider the oscillating electric field E  E0 cos 2 t
and characteristic frequency of molecule is νvib
Electric field induce the induced magnetic moment, μinduced and
it related to polarizability
induced (t )  E0 cos 2 t
Polarizability related to bond length xe+x(t) and we can expand
by the Taylor-Mclaurin series
 ( x )   ( xe )  x (
d
) x  xe  ...
dx
Consider the vibration of molecule
x(t )  xmax cos 2 vibt
MS310 Quantum Physical Chemistry
We can calculate induced dipole moment by these results
d
 induce ( t )  E  E 0 cos 2 t[ ( xe )  ( ) x  xe xmax cos 2 vib t ]
dx
1 d
  ( xe ) E 0 cos 2 t  ( ) x  xe xmax E 0 [cos(2   2 vib )t  cos(2   2 vib )t ]
2 dx
Therefore, allowed frequency is ν, ν-νvib, ν+νvib
ν : Rayleigh frequency
ν-νvib : Stokes frequency
ν+νvib : anti-Stokes frequency
Unless the dα/dx≠0, stokes and anti-stokes peak : zero.
It means raman active bond satisfies dα/dx≠0.
However, it is not related to dμx/dx≠0
→ can be raman active although IR inactive!
MS310 Quantum Physical Chemistry
Stokes : n=0 to n=1
Anti-stokes : n=1 to n=1
Intensity of stokes and antistokes peak same?
I anti  stokes nexcited e 3 h / 2 kT

  h / 2 kT  e  h / kT
I stokes
nground e
Range of 1000cm-1 to 3000cm-1 ,
ratio is 8x10-3 to 5x10-7 at 300K
Therefore, 2 peaks are quite
different.
Raman and IR spectroscopy are
complementary.
MS310 Quantum Physical Chemistry
8.9 How does the transition rate between states
depend on frequency?
Solution of time-independent Schrödinger equation : constant
energy → it cannot describe transition state
Consider the 2-level system and E2>E1, wavefunction is written
by
iE t
iE t
 1
 2
( x , t )  a1 1 ( x )e   a2 2 ( x )e   a11  a2 2
At t=0, system in ground state : a1=1, a2=0
Write the initial hamiltonian as H0
Light turns on : electric field applied → permanent or dynamic
dipole moment generate
Assume the electric field along the x axis, time-dependent
potential energy is given by
 E
Hˆ int     E    x E0 cos 2 t   x 0 (e 2it  e  2it )
2
: dipole approximation
MS310 Quantum Physical Chemistry
We must solve time-dependent Schrödinger equation
( x, t )
( Hˆ 0  Hˆ int )( x, t )  i
, ( x, t )  a1 (t )1  a2 (t )2
t
ˆ   E  ,H
ˆ   E  : trivial
H
0 1
1 1
0 2
2 2
Therefore, equation changes to
da ( t )
da ( t )
a1 ( t ) Hˆ int 1  a2 (t ) Hˆ int 2  i1 1  i2 2
dt
dt
Multiply Ψ2* left side and integration
da1 (t )
da2 (t )
* ˆ
*
*
ˆ
a1 (t )  H int 1dx  a2 (t ) 2 H int 2dx  i


dx

i


2 1
2 2 dx


dt
dt
*
2
Ψ1 and Ψ2 are orthonormal, equation can be simply
i
da2 (t )
 a1 (t ) 2* Hˆ int 1dx  a2 (t ) 2* Hˆ int 2dx
dt
MS310 Quantum Physical Chemistry
Assume a1(t) and a2(t) are small change(it means a1=1, a2=0 on
the right side, but not left side!)
Equation is change by
i
( E 2  E1 )t
da 2 ( t )
*
*

i
  2 Ĥ int 1dx  e

2 ( x ) H int 1 ( x )dx

dt
E0 i ( E 2  E1 )t 2it

e
(e
 e  2it )  2* ( x ) x 1 ( x )dx
2
i
( E 2  E 1  h ) t
E0 i ( E 2  E1  h )t


(e
e
)  2* ( x ) x 1 ( x )dx
2
   x21
i
( E 2  E 1  h ) t
E0 i ( E 2  E1  h )t

(e
e
)
2
MS310 Quantum Physical Chemistry
We use dummy variable t’ and integrate it.
t
i
i
( E 2  E 1  h ) t '
( E 2  E 1  h ) t '
i 21 E0


a2   x
(e
e
)dt'


2 0

21
x
i
( E 2  E 1  h ) t

i
( E 2  E 1  h ) t

E0 1  e
1 e
(

)
2
E 2  E 1  h
E 2  E 1  h
MS310 Quantum Physical Chemistry
Unless the μx21 ≠ 0, a2(t) must be zero.
First term of a2(t) : stimulated emission
However, our focus is absorption : second term
period of oscillation becomes zero when E2 - E1 → hν(we
assume E2 > E1, E1 - E2 → hν can neglect)
Use L’Hôpital’s rule
f(x)
f' ( x )
lim
 lim
x 0 g( x )
x  0 g' ( x )
i
 ( E 2  E 1  h ) t

i
 ( E 2  E 1  h ) t

1 e
d( 1  e
) / d ( E 2  E 1  h )
lim a 2 ( t )  lim
 lim
E 2  E 1  h  0
E 2  E 1  h   0 E  E  h
E 2  E 1  h   0 d ( E  E  h  ) / d ( E  E  h )
2
1
2
1
2
1
it  i ( E 2  E1  h )t it
 lim
e

E 2  E 1  h  0 

Resonance condition : E1 - E2 = hν
In this case, a2(t) increase linearly.
MS310 Quantum Physical Chemistry
E1 - E2 is slightly different to hν : do not resonance
→ transition probability almost zero : no transition
Therefore, transition occurs only the hν is equal or extremely
close to E1 - E2.
MS310 Quantum Physical Chemistry
Final goal : find the transition probability, a2*(t)a2(t)
Neglect the first term, we can calculate easily
2
sin
[( E 2  E1  h )t / 2]
a2* ( t )a2 ( t )  E02 [  x21 ]2
( E 2  E1  h )2
→ when resonance condition,
a2*(t)a2(t) increases as t2
Plot a2*(t)a2(t) when 40ps, 120ps,
and 400ps
time increase
→ height increases as t2 and
width decreases as 1/t.
Therefore, there are no transition
without resonance, E1 - E2 = hν.
Why height of peak decrease when time increase?
→ uncertainty principle
MS310 Quantum Physical Chemistry
a2*(t)a2(t) : closely related to observed in an absorption spectra
Then, what is measured by the ‘real’ instrument?
Intrinsic linewidth of vibrational spectra : less than 10-3 cm-1
Resolution of instrument : ~ 0.1cm
In this case, measurement peak broaden and no information
about the intrinsic linewidth by the measurement
→ ‘inhomogeneous broadening’
MS310 Quantum Physical Chemistry
Summary
-Light interacts with molecules to induce transitions between
states and molecular spectroscopy were described.
- It was discussed the absorption of electromagnetic radiation
in the infrared and microwave regions of the spectrum.
- Light of these wavelengths induces transitions between
eigenstates of vibrational and rotational energy.
- The frequency at which energy is absorbed or emitted is
related to the energy levels.
MS310 Quantum Physical Chemistry