Transcript Vibrations of polyatomic molecules
Vibrations of polyatomic molecules
MJ, Feb 7
Outline
* Normal modes * Selection rules * Group theory (Tjohooo!) * Anharmonicity MJ, Feb 7
Describing the vibrations
Molecule with N atoms has 3N-6 vibrational modes, 3N-5 if linear.
V
Find expression for potential energy.
Taylor expansion around equilibrium positions.
V
( 0 )
i
V
x i
0
x i
1 2
i
,
j
2
V
x i
x j
0
x i x j
...
i
,
j
1 ...
3
N
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i i
, ,
j j k ij ij x i q x i q j j
Total energy
k ij K ij
x i
2
k m i ij x m j
j
0 Kinetic energy
T
1 2
i m i x
i
2 1 2
i q
i
2 Introduce mass weighted coordinates:
q i
m i x i
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Total energy
We can now write the total vibrational energy as:
E
1 2
i q
i
2 1 2
i
,
j K ij q i q j
Nasty cross-terms when
i
j
What we want is to find set of coordinates where the cross-terms disappear . Is this at all possible?
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A look at CO
2 Vibrations of the individual atoms Can be broken down as linear combinations of: These modes do not change the centre of mass, and they are independent.
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Normal coordinates
So, we can write the energy as:
E
1 2
i
i
2 1 2
i
i Q i
2 where Q are the so called normal coordinates.
They can be a bit tricky to find, but at least we know they are there.
Before we see how can use this, lets have a look at the normal modes for our CO 2 .
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Normal modes of CO
2 3 x 3 - 5 = 4 vibrational modes Symmetric stretch Anti-symmetric stretch Orthogonal bending MJ, Feb 7
QM
Since the total energy is just a sum of terms, so is the Hamiltonian of the vibrations. We write it as:
i i i
1 2 2 2
Q i
2 1 2
i Q i
2 Also the vibrational wavefunction separates into a product of single mode wavefunctions: 1 (
Q
1 ) 2 (
Q
2 ) ...
3
N
6 (
Q
3
N
6 ) MJ, Feb 7
Schrödinger equation
The Scrödinger equation then becomes: 1 2 2 2
Q i
2 1 2
i Q i
2
i
(
Q i
)
E
i
(
Q i
) … and this we recognise, right?
Harmonic oscillator with unit mass and force constant .
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Energy levels:
Harmonic oscillator
E n i
(
n i
½)
i
;
i
i
;
n i
Wavefunctions:
n i
N n i H n i
(
y i
)
e
y i
2 / 2 ;
y i
i Q i
Total vibrational energy:
E
i
(
n i
½)
i
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Harm. Osc. …
We know the ground state: 0 1 0 2 ...
0 3
N
6 Ground state energy:
E
½
i
i
Ground state wavefunction: 0
N
i e
y i
2 / 2
Ne
y
2 / 2 ;
y
2
i y i
2 All normal modes appear symmetrically, and as squares The ground state is symmetric with respect to all symmetry operations of the molecule.
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Selection rules
Molecular dipole moment depends on displacements of the atoms in the molecule: Taylor expand...
0
i
Q i
Q i
0 ...
Dipole transition matrix element of a particular mode: 000 ...
n i
...
000 ...
n i
...
n
i
0
n i
Q i
0
n
i Q i n i n
i
n i
0 if
n
i
n i
1 MJ, Feb 7
Selection rules
Selection rules for IR absortion :
n i
1 ;
Q i
0 0 Similarly, by observing that
n i
n i
n i
0
Q i
0
Q i n i
we get selection rules for Raman activity:
n i
1
Q i
0 0 It can be hard to see which vibrations are IR/Raman active, but, as we have seen before, Group Theory can come to rescue.
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Group theory and vibrations
First find a basis for the molecule. Let’s take the cartesian coordinates for each Example: H 2 O (the following stolen from Hedén) Water belongs to the C contains the operations
2v
E x 3 group which , C 2 , s v (xz) z 3 x y 2 3 and s v ’ z 2 (yz) .
y 2 x 1 z 1 y 1 The representation becomes G red E 9 C 2 -1 s v 1 (xz) s v ’ 3 (yz) MJ, Feb 7
Continued water example
Character table for C 2v .
C
2
v A
1
A
2
B
1
B
2
E
1 1 1 1
C
2 1 1 1 1 s
v
(
xz
) 1 1 1 1 s '
v
(
yz
) 1 1 1 1
z R
2
x
,
R y y
,
R x x
2 ,
y
2 ,
z
2
xy xz yz C
2
v
G
red E
9
C
2 1 s
v
(
xz
) 1 s
v
' (
yz
) 3 Now reduce G red to a sum of irreducible representations. Use inspection or the formula.
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Continued water example
C
2
v A
1
A
2
B
1
B
2
E
1 1 1 1
C
2 1 1 1 1 s
v
(
xz
) 1 1 1 1 s '
v
(
yz
) 1 1 1 1
z R z x
,
R y y
,
R x x
2 ,
y
2 ,
z
2
xy xz yz C
2
v E C
2 s
v
(
xz
) s
v
' (
yz
) G
red
9 1 1 The representation reduces to G red =3 G trans = A 3 1 + B 1 + B 2 A 1 + A 2 +2 B 1 +3 B 2 G rot = A 2 + B 1 + B 2 G vib =2 A 1 + B 2 Modes left for vibrations MJ, Feb 7
What to use this for?
We know that that the ground state is totally symmetric: (A 1 ) First excited state of a normal mode belongs to the same irred. repr. as that mode because
H
1 (
y i
)
y i
Q i
in A 1 1
i
0
i
0 1
i
irreducible representation for their product to be . transform as translations, so: For a transition to be IR active, the normal mode must be parallel to the polarisation of the radiation.
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What
more
to use this for?
By the same argument one can come the the conclution that For a transition to be Raman active, the normal mode must belong to the same symmetry species as the components of the polarisability These scale as the quadratic forms x 2 , y 2 , xy etc.
This also leads to the exclusion rule : In a molecule with a centre of inversion, a mode cannot be both IR and Raman active.
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Water again...
C
2
v A
1
A
2
B
1
B
2
E
1 1 1 1
C
2 1 1 1 1 s
v
(
xz
) 1 1 1 1 s '
v
(
yz
) 1 1 1 1
z R z x
,
R y y
,
R x x
2 ,
y
2 ,
z
2
xy xz yz
A G 1 vib =2 A 1 + B 2 A 1 All three modes are both IR and Raman active, no centre of inversion.
(a) and (b) are excited by z polarised light, and (c) by y polarised.
B 2 MJ, Feb 7
Anharmonicity
Electric anharmonicity occurs when our expansion of the dipole moment to first order is not valid.
i i
Q i i
i i
...
i
,
j
2
Q i
Q j
0
Q i Q j
At this point term
Q i 2
overtones calculated for every molecule.
i
0
i
can be allowed since the matrix element containing the quadratic not necessarily vanishes. This can not be determined from group theory, but must be , MJ, Feb 7
0
i
Anharmonicity
Q i
0
Q i
½
i
,
j
2
Q i
Q j
0
Q i Q j
We also see from the presence of
Q i Q j
can cause a mixing of normal modes. cross-terms In a perfectly harmonic molecule, energy put into one normal mode stays there. Anharmonicity causes the molecule to thermalise .
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Anharmonicity
Also mechanical anharmonicity can lead to mixing of levels if one needs to add cubic and further terms in the expression for the potential.
2 a
V
1 1 b 2
i
,
j k ij x i x j
1 3 !
i j
, ,
k k ijk x i x j x k
...
1 a 2
a
0
b V an
0
a
1
b
1 3 !
3
V
Q a
2
Q b
2
a Q a
2 0
a
0
b Q b
1
b
0 a 0 b MJ, Feb 7
Inversion doubling
Consider ammonia: pyramidal molecule with two sets of vibrational levels: Coupling between the levels lead to mixing of up and down wavefunctions which lifts the degeneracy of the levels
Summary
• Harmonic approximation of energy gives transition rules for IR and Raman activity.
• Group theory can help us figure out which transition are active.
• However, anharmonic terms can come in play and mess everything up.