mj, feb 7 vibrations of polyatomic molecules. mj, feb 7 outline * normal modes * selection rules *...
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MJ, Feb 7
Describing the vibrations
Molecule with N atoms has 3N-6 vibrational modes, 3N-5 if linear.
Find expression for potential energy.Taylor expansion around equilibrium positions.
...2
1)0(
,
0
2
0
ji
ji
ji
ii
i
xxxx
Vx
x
VVV
Nji 3...1,
Total energy
ji
jiijxxkV
,2
1
0
2
ji
ij xx
Vk; where
Introduce mass weighted coordinates: iiixmq
ji
jiijqqKV
,2
1
ji
ij
ij mm
kK ; where
i
iq 2
2
1 2
2
1ixmT
ii
Kinetic energy
MJ, Feb 7
Total energy
ji
jiiji
iqqKqE
,
2
2
1
2
1
We can now write the total vibrational energy as:
Nasty cross-terms when ji
What we want is to find set of coordinates where the cross-terms disappear.
Is this at all possible?MJ, Feb 7
A look at CO2
MJ, Feb 7
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.
Normal coordinates
MJ, Feb 7
22
2
1
2
1ii
ii
iQQE
So, we can write the energy as:
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 CO2.
Normal modes of CO2
Symmetric stretch
Anti-symmetric stretch
Orthogonal bending
3 x 3 - 5 = 4 vibrational modes
MJ, Feb 7
QM
MJ, Feb 7
Since the total energy is just a sum of terms, so is the Hamiltonian of the vibrations. We write it as:
i
iHH ˆˆ 2
2
2
2
2
1
2
1ˆii
i
iQ
QH
Also the vibrational wavefunction separates into a product of single mode wavefunctions:
)(...)()(63632211
NN
QQQ
Schrödinger equation
MJ, Feb 7
The Scrödinger equation then becomes:
)()(2
1
2
1 2
2
2
2
iiiiii
i
QEQQQ
… and this we recognise, right?
Harmonic oscillator with unit mass and force constant
MJ, Feb 7
Harmonic oscillatorEnergy levels:
iiiiinnnE
i;;½)(
i
i
i
y
innnQyeyHN i
iii ;)( 2/2
Wavefunctions:
Total vibrational energy:
i
iinE ½)(
Harm. Osc. …
We know the ground state:6321
0...00N
i
iE ½Ground state energy:
Ground state wavefunction:
i
y
i
y
i
i yyNeeN 222/2/
0;
22
All normal modes appear symmetrically, and as squares
The ground state is symmetric with respect to all symmetry operations of the molecule.
MJ, Feb 7
MJ, Feb 7
Selection rulesMolecular dipole moment depends on displacements of the atoms in the molecule: Taylor expand...
...0
0
i
ii
Dipole transition matrix element of a particular mode:
iii
i
iiiinQn
Qnnnn
0
0......000......000
0 if , due to orthogonalityiinn 10
iinnif
MJ, Feb 7
Selection rules
Selection rules for IR absortion: 1in 0
0
iQ
;
Similarly, by observing that
ii
i
innnQ
Qn
ii
0
0
we get selection rules for Raman activity:
1in 0
0
iQ
It can be hard to see which vibrations are IR/Raman active, but, as we have seen before, Group Theory can come to rescue.
MJ, Feb 7
Group theory and vibrations
First find a basis for the molecule. Let’s take the cartesian coordinates for each atom. x1
x3
x2
y1
y2
y3
z1
z2
z3
Water belongs to the C2v group which contains the operations E, C2, v(xz) and v’(yz).
The representation becomes E C2 v(xz) v’(yz)
red9 -1 1 3
The details of a normal mode depend on the strength of the chemical bonds and the mass of the atoms. However the symmetries are just a function of geometry.
Example: H2O (the following stolen from Hedén)
Character table for C2v.
yzRyB
xzRxB
xyRA
zyxzA
yzxzCEC
x
y
vvv
,1111
,1111
1111
,,1111
)(')(
2
1
22
2221
22
3119
)(')(22
redvvv yzxzCEC
Now reduce red to a sum of irreducible representations. Use inspection or the formula.
Continued water example
MJ, Feb 7
yzRyB
xzRxB
xyRA
zyxzA
yzxzCEC
x
y
z
vvv
,1111
,1111
1111
,,1111
)(')(
2
1
2
2221
22
The representation reduces to red=3A1+A2+2B1+3B2
trans= A1+B1+B2
rot=A2+B1+B2
vib=2A1+B2
Continued water example
MJ, Feb 7
Modes left for vibrations
3119
)(')(22
redvvv yzxzCEC
MJ, Feb 7
What to use this for?We know that that the ground state is totally symmetric: (A1)
First excited state of a normal mode belongs to the same irred. repr. as that mode because
iiiQyyH )(
1
001 ii
So for , and must span the same irreducible representation for their product to be in A1. transform as translations, so:
i1
For a transition to be IR active, the normal mode must be parallel to the polarisation of the radiation.
MJ, Feb 7
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 x2, y2, 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.
yzRyB
xzRxB
xyRA
zyxzA
yzxzCEC
x
y
z
vvv
,1111
,1111
1111
,,1111
)(')(
2
1
2
2221
22
vib=2A1+B2
MJ, Feb 7
Water again...
A1
A1
B2
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.
MJ, Feb 7
Anharmonicity
At this point overtones like can be allowed, since the matrix element containing the quadratic term Qi2 not necessarily vanishes. This can not be determined from group theory, but must be calculated for every molecule.
ii02
...0
0
i
ii
Electric anharmonicity occurs when our expansion of the dipole moment to first order is not valid.
jiji
ji
ii
i
QQQQ
0
,
2
0
0½
MJ, Feb 7
Anharmonicity
jiji
ji
ii
i
QQQQ
0
,
2
0
0½
We also see from the presence of QiQj cross-terms can cause a mixing of normal modes.
In a perfectly harmonic molecule, energy put into one normal mode stays there. Anharmonicity causes the molecule to thermalise.
MJ, Feb 7
Anharmonicity
...!3
1
2
1,,,
k
kjijiijk
jijiij
xxxkxxkV
Also mechanical anharmonicity can lead to mixing of levels if one needs to add cubic and further terms in the expression for the potential.
0a 0b
1a
2a 1b
bbbaa
b
baanbaQQ
VV
a
a
1002!3
11002 2
2
3
Inversion doublingConsider 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