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Transcript of Symmetry and Spectroscopy. P. R. Bunker and Per Jensen: Molecular Symmetry and Spectroscopy, 2 nd...
![Page 1: Symmetry and Spectroscopy. P. R. Bunker and Per Jensen: Molecular Symmetry and Spectroscopy, 2 nd Edition, 2 nd Printing, NRC Research Press, Ottawa,](https://reader035.fdocuments.us/reader035/viewer/2022081504/56649f045503460f94c18e69/html5/thumbnails/1.jpg)
Symmetry and
Spectroscopy
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P. R. Bunker and Per Jensen:
Molecular Symmetry and Spectroscopy, 2nd Edition,
2nd Printing, NRC Research Press, Ottawa, 2006
(ISBN 0-660-19628-X). $49.95 for 747 pages.
paperback. BJ1
P. R. Bunker and Per Jensen:
Fundamentals of Molecular Symmetry,
IOP Publishing, Bristol, 2004
(ISBN 0-7503-0941-5). $57.95
paperback. BJ2
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Examples of point group symmetry
H2O
CH3F
C60
C3H4
C2v
C3v
D2d
Ih
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Examples of point group symmetry
H2O
CH3F
C60
C3H4
C2v
C3v
D2d
Ih
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Point group symmetry of H2O
z
y
(-x)
The point group C2v consists ofthe four operations E, C2y, yz, and xy
The word ´group´ is loaded. To see howwe do two operations in succession
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Point groups: Number of rotation axes and reflection planes.
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z
y
(-x)
1 2
σyz
C2
z
y
(-x)
1 2
z
y
(-x)
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z
y
(-x)
σxy = C2 σyz
1 2
σyz
σxy
C2
z
y
(-x)
1 2
z
y
(-x)
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Multiplication Table for H2O Point group
z
y
(-x)C2v = {E, C2, yz, xy }
E C2 yz xy E E C2 yz xy C2 C2 E xy yz yz yz xy E C2 xy xy yz C2 E
Multiplication table (=Rrow Rcolumn, in succession)
Use multiplication table to prove that it is a “group.”
σxy = C2 σyz
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{E, C2, yz, xy } forms a “group“ if it obeysthe following GROUP AXIOMS :
•All possible products RS = T belong to the group
•Group contains identity E (which does nothing)
•The inverse of each operation R1 (R1R =RR1 =E ) is in the group
•Associative law (AB )C = A(BC ) holds
E C2 yz xy E E C2 yz xy C2 C2 E xy yz yz yz xy E C2 xy xy yz C2 E
C2v
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Fermi:
{E, C2, yz, xy } forms a “group“ if it obeysthe following GROUP AXIOMS :
•All possible products RS = T belong to the group
•Group contains identity E (which does nothing)
•The inverse of each operation R1 (R1R =RR1 =E ) is in the group
•Associative law (AB )C = A(BC ) holds
E C2 yz xy E E C2 yz xy C2 C2 E xy yz yz yz xy E C2 xy xy yz C2 E
C2v ‘‘Group theory is just a bunch of definitions‘‘
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•All possible products RS = T belong to the group
•Group contains identity E (which does nothing)
•The inverse of each operation R1 (R1R =RR1 =E ) is in the group
•Associative law (AB )C = A(BC ) holds
E C2 yz xy E E C2 yz xy C2 C2 E xy yz yz yz xy E C2 xy xy yz C2 E
Not a GROUP
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•All possible products RS = T belong to the group
•Group contains identity E (which does nothing)
•The inverse of each operation R1 (R1R =RR1 =E ) is in the group
•Associative law (AB )C = A(BC ) holds
E C2 yz xy E E C2 yz xy C2 C2 E xy yz yz yz xy E C2 xy xy yz C2 E
Is a GROUP(subgroup of C2v)
Rotationalsubgroup
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PH3 at equilibrium
Symmetry operations:C3v = {E, C3, C3
2, 1, 2, 3 }
Symmetry elements:
C3, 1, 2, 3
C3 Rotation axis
k Reflection plane
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C32 = 2 1
Multiplying C3v symmetry operations
Reflection
Reflection
Rotation
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Multiplication table for C3v
C32 = σ2σ1
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Multiplication table for C3v
C32 = σ2σ1
Note that C3 = σ1σ2
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Multiplication table for C3v
Rotationalsubgroup
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Multiplication table for C3v
3 classes
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A matrix group
1 0
0 1
21
23
23
21
21
23
23
21
1 0
0 1
21
23
23
21
21
23
23
21
´M1 = M4 =
´ M5 =
´ M6 =
´M2 =
´M3 =
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M1
M2
M3
M4
M5
M6
M1
M1
M2
M3
M4
M5
M6
M2
M2
M3
M1
M6
M4
M5
M3
M3
M1
M2
M5
M6
M4
M4
M4
M5
M6
M1
M2
M3
M5
M5
M6
M4
M3
M1
M2
M6
M6
M4
M5
M2
M3
M1
Multiplication table for the matrix group
Products are Mrow Mcolumn
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E C3 C32 σ1 σ2 σ3
This matrix group forms a “representation” of the C3v groupThese two groups are isomorphic.
Multiplication tableshave the ‘same shape’
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Irreducible Representations
The matrix group we have just introducedis an irreducible representation of the C3v
point group.
The sum of the diagonal elements (character)of each matrix in an irreducible representationis tabulated in the character table of thepoint group.
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The characters of this irreducible rep.
1 0
0 1
21
23
23
21
21
23
23
21
1 0
0 1
21
23
23
21
21
23
23
21
´M1 = E M4 = 1
´ M5 = 2
´ M6 = 3
´M2 = C3
´M3 = C32
C3v C3v
2
-1
-1
0
0
0
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The characters of this irreducible rep.
E (123) (12)
1 2 3
A1 1 1 1
A2 1 1 1
E 2 1 0
The 2D representation M = {M1, M2, M3, ....., M6}of C3v is the irreducible representation E. In thistable we give the characters of the matrices.
E C3 σ1
C32 σ2
σ3
Elements in the same class have the same characters
3 classes
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Character Table for the point group C3v
E (123) (12)
1 2 3
A1 1 1 1
A2 1 1 1
E 2 1 0
The 2D representation M = {M1, M2, M3, ....., M6}of C3v is the irreducible representation E. In thistable we give the characters of the matrices.
E C3 σ1
C32 σ2
σ3
Elements in the same class have the same characters
Two 1Dirreduciblerepresentationsof the C3v group
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The matrices of the E irreducible rep.
1 0
0 1
21
23
23
21
21
23
23
21
1 0
0 1
21
23
23
21
21
23
23
21
´M1 = E M4 = 1
´ M5 = 2
´ M6 = 3
´M2 = C3
´M3 = C32
C3v C3v
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The matrices of the A1 + E reducible rep.
21
23
23
21
21
23
23
21
1 0
0 1
21
23
23
21
21
23
23
21
´M1 = E M4 = 1
´ M5 = 2
´ M6 = 3
´M2 = C3
´M3 = C32
C3v C3v
1 0 00 1 00 0 1
1 0 00
0
1 0 00
0
1 0 00
0
1 0 000
1 0 00
0
‘
‘
‘
‘
‘
‘
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‘
The matrices of the A2 + E reducible rep.
21
23
23
21
21
23
23
21
1 0
0 1
21
23
23
21
21
23
23
21
´M1 = E M4 = 1
´ M5 = 2
´ M6 = 3
´M2 = C3
´M3 = C32
C3v C3v
1 0 00 1 00 0 1
-1 0 00
0
1 0 00
0
1 0 00
0
-1 0 000
-1 0 00
0
‘
‘
‘
‘
‘
‘
‘
‘
‘
‘
‘
‘
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Character table for the point group C2v
E (12) E* (12)*
A1 1 1 1 1
A2 1 1 1 1
B1 1 1 1 1
B2 1 1 1 1
E C2 σyz σxy
Irreducible representations are “symmetry labels”
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Some of Fermi’s definitions• Group
• Subgroup
• Multiplication table of group operations
• Classes
• Representations
• Irreducible and reducible representations
• Character table
See, for example, pp 14-15 and Chapter 5 of BJ1
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Some of Fermi’s definitions• Group
• Subgroup
• Multiplication table of group operations
• Classes
• Representations
• Irreducible and reducible representations
• Character table
See, for example, pp 14-15 and Chapter 5 of BJ1
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Irreducible representationsThe elements of irrep matrices satisfy the„Grand Orthogonality Theorem“ (GOT).
We do not discuss the GOT here, but we list threeconsequences of it: • Number of irreps = Number of classes in the group.
• Dimensions of the irreps, l1, l2, l3 … satisfy
l12 + l2
2 + l32 + … = h,
where h is the number of elements in the group.
• Orthogonality relation
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• Irreducible and reducible representations
These are used as ‘‘symmetry labels‘‘on energy levels.
Which energy levels can ‘‘interact‘‘and which transitions can occur.
Can also determine whether certain terms are in the Hamiltonian.
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IN SOME CIRCUMSTANCES THERE ARE PROBLEMS IF WETRY TO USE POINT GROUP SYMMETRY TO DO THESE THINGS
BUT
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How do we use point group symmetry if the molecule rotates and distorts?
H3+
D3h C2v
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2
3
113
2
Or if tunnels?
NH3
C3vD3h
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What are the symmetriesof B(CH3)3, CH3.CC.CH3, (CO)2, (NH3)2,…?
Nonrigid molecules (i.e. moleculesthat tunnel) are a problemif we try to use a point group.
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What should we do if we study transitions (or interactions) between electronic states that have different point group symmetries at equilibrium?
Also
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Point groups used for classifying:
The electronic states for any moleculeat a fixed nuclear geometry (see BJ2 Chapter 10), and
The vibrational states for molecules,called “rigid” molecules, undergoing infinitesimal vibrations about a unique equilibrium structure(see BJ2 Pages 230-238).
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Rotations andreflections
Permutationsand the inversion
J.T.Hougen, JCP 37, 1422 (1962); ibid, 39, 358 (1963)H.C.Longuet-Higgins, Mol. Phys., 6, 445 (1963)P.R.B. and Per Jensen, JMS 228, 640 (2004) [historical introduction]
To understand how we use symmetrylabels and where the point groupgoes wrong we must studywhat we mean by “symmetry”
See also BJ1 and BJ2
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Symmetry not from geometry since molecules are dynamic
• Centrifugal distortion
eg. H3+ or CH4 dipole
moment
• Nonrigid molecules: eg. ethane, ammonia, (H2O)2, (CO)2,…
• Breakdown of BOA: eg. HCCH* - H2CC
Also symmetry appliesto atoms, nuclei and fundamental particles. Geometrical point group symmetry is not possible for them.
We need a more general definition of symmetry
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Symmetry Based on Energy Invariance
Symmetry operations are operations that leave the energy of the system (a molecule in our case) unchanged.
Using quantum mechanics we define a symmetry operationas follows:
A symmetry operation is an operation that commutes with the Hamiltonian:
(RH – HR)n = [R,H]n = 0
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• Uniform Space ----------Translation• Isotropic Space----------Rotation • Identical electrons------Permute electrons• Identical nuclei-----------Permute identical nuclei • Parity conservation-----Inversion (p,q,s) (-p,-q,s) P(E*) • Reversal symmetry-----Time reversal (p,q,s) (-p,q,-s) T • Ch. conj. Symmetry-----Particle antiparticle C
Symmetry Operations (energy invariance)
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• Uniform Space ----------Translation
Symmetry Operations (energy invariance)Separate translation…Translational momentum
Ψtot = Ψtrans Ψint
int = rot-vib-elec.orb-elec.spin-nuc.spin
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• Uniform Space ----------Translation• Isotropic Space----------Rotation • Identical electrons-------Permute electrons• Identical nuclei-----------Permute identical nuclei • Parity conservation-----Inversion (p,q) (-p,-q) P(E*) • Reversal symmetry-----Time reversal (p,s) (-p,-s) T • Ch. conj. Symmetry-----Particle antiparticle C
Symmetry Operations (energy invariance)
K(spatial) group,J, mJ or F,mF labels
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• Uniform Space ----------Translation• Isotropic Space----------Rotation • Identical electrons-------Permute electrons• Identical nuclei-----------Permute identical nuclei • Parity conservation-----Inversion (p,q) (-p,-q) P(E*) • Reversal symmetry-----Time reversal (p,s) (-p,-s) T • Ch. conj. Symmetry-----Particle antiparticle C
Symmetry Operations (energy invariance)
Symmetric group Sn
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For the BeH molecule (5 electrons) Ψorb-spin transforms as D(0) of S5
PEP
Slater determinant ensures antisymmetry so do not need S5
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• Uniform Space ----------Translation• Isotropic Space----------Rotation • Identical electrons-------Permute electrons• Identical nuclei-----------Permute identical nuclei • Parity conservation-----Inversion (p,q) (-p,-q) P(E*) • Reversal symmetry-----Time reversal (p,s) (-p,-s) T • Ch. conj. Symmetry-----Particle antiparticle C
Symmetry Operations (energy invariance)
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• Uniform Space ----------Translation• Isotropic Space----------Rotation • Identical electrons-------Permute electrons• Identical nuclei-----------Permute identical nuclei • Parity conservation-----Inversion (p,q) (-p,-q) P(E*) • Reversal symmetry-----Time reversal (p,s) (-p,-s) T • Ch. conj. Symmetry-----Particle antiparticle C
Symmetry Operations (energy invariance)
CNPI group = Complete Nuclear Permutation Inversion Group
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• Uniform Space ----------Translation• Isotropic Space----------Rotation • Identical electrons-------Permute electrons• Identical nuclei-----------Permute identical nuclei • Parity conservation-----Inversion (p,q) (-p,-q) P(E*) • Reversal symmetry-----Time reversal (p,s) (-p,-s) T • Ch. conj. Symmetry-----Particle antiparticle C
Symmetry Operations (energy invariance)
CNPI group = Complete Nuclear Permutation Inversion Group
EXAMPLE:The CNPI group for H2O is C2v(M) = {E, (12), E*, (12)*}
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The CNPI Group for the Water Molecule
The Complete Nuclear Permutation Inversion (CNPI) group
for the water molecule is C2v(M) = {E, (12), E*, (12)*}
H1 H2
O e+
H2 H1
O e+ H1 H2
Oe-(12) E*
(12)*
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We compare C2v and this CNPI groupMultiplication table (Rrow Rcolumn)
E C2 yz xy E E C2 yz xy C2 C2 E xy yz yz yz xy E C2 xy xy yz C2 E
E (12) E* (12)* E E (12) E* (12)*
(12) (12) E (12)* E* E* E* (12)* E (12)
(12)* (12)* E* (12) E
C2v and CNPI are isomorphic!
C2v
CNPI
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We compare C2v and this CNPI group
E C2 yz xy E E C2 yz xy C2 C2 E xy yz yz yz xy E C2 xy xy yz C2 E
E (12) E* (12)* E E (12) E* (12)*
(12) (12) E (12)* E* E* E* (12)* E (12)
(12)* (12)* E* (12) E
Rotationalsubgroup
Permutationsubgroup
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CNPI group of water: Character table
E (12) E* (12)*
A1 1 1 1 1
A2 1 1 1 1
B1 1 1 1 1
B2 1 1 1 1
This group is called C2v(M)
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Why RH = HR used to Define Symmetry?
RH = RE
HR = ER
H = E
Thus R = c since E is nondegenerate.However R2 = E, so R(RΨ) = Ψ, but R(RΨ) = c2Ψ. Thus c2 = 1,c = ±1 and R = ±
For the water molecule ( nondegenerate and R2 = E for all R) :
Symmetry of H restricts symmetry of eigenfunctions Ψ
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+ Parity - Parity
x
Ψ1+(x)
x
Ψ3+(x)
x
Ψ2-(x)
Ψ+(-x) = Ψ+(x)
Ψ-(-x) = -Ψ-(x)
Eigenfunctions of Hmust satisfyE*Ψ = ±Ψ
R=E*
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”Why RH = HR used to Define Symmetry?”continued……
RH = RE
HR = ER
H = E
Thus R = c. However R2 = E, so c = ±1 and R = ±
Allows us to SYMMETRY LABEL the energy levels using the irreps of the symmetry group
For the water molecule (with nondegenerate states):
Symmetry of H restricts symmetry of eigenfunctions Ψ
RH=HRimplies
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There are four symmetry types of H2O wavefunction
(12) E* 1 1
1 -1 -1 -1 -1 1
E 1111
(12)* 1 -1 1-1
A1
A2
B1
B2
R = ±
A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1
∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.
∫ΨaμΨbdτ = 0 if symmetry of product is not A1
Possible labels would be (1,1), (1,-1), (-1,-1), and (-1,1). However. More generally systematic are the irreducible representation labels (or symmetry labels) from the symmetry group.
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The Symmetry Labels of the C2v(M) Group of H2O
(12) E* 1 1
1 -1 -1 -1 -1 1
E 1111
(12)* 1 -1 1-1
A1
A2
B1
B2
R = ±
A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1
∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.
∫ΨaμΨbdτ = 0 if symmetry of product is not A1
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The Symmetry Labels of the C2v(M) Group of H2O
(12) E* 1 1
1 -1 -1 -1 -1 1
E 1111
(12)* 1 -1 1-1
A1
A2
B1
B2
R = ±
because of:The vanishing integral theoremBJ1 pp114-117, BJ2 pp 136-139
Labeling is not just bureaucracy…it is useful
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The Symmetry Labels of the C2v(M) Group of H2O
(12) E* 1 1
1 -1 -1 -1 -1 1
E 1111
(12)* 1 -1 1-1
A1
A2
B1
B2
R = ±
because of:The vanishing integral theoremBJ1 pp114-117, BJ2 pp 136-139
Labeling is not just bureaucracy…it is useful
But first let’s look at three things we overlooked:
Rn=E with n>2, degenerate states, symmetry of a product
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Suppose Rn = E where n > 2.
C3(M) E (123) (132)
1 1 1
A 1 1 1
Ea 1 *
Eb 1 *
= ei2/3
We still have RΨ = cΨ for nondegenerate Ψ, but now Rn Ψ = Ψ.
Thus cn
= 1 and c = n√1
If n = 3 we introduce and c = 1,ε or ε2 (=ε*)
eiπ = -1 C3 C3
2
ei2π = 1
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Suppose Rn = E where n > 2.
C3(M) E (123) (132)
1 1 1
A 1 1 1
Ea 1 *
Eb 1 *
= ei2/3
We still have RΨ = cΨ for nondegenerate Ψ, but now Rn Ψ = Ψ.
Thus cn
= 1 and c = n√1
If n = 3 we introduce and c = 1,ε or ε2 (=ε*)
C3 C32
A pair of separablydegenerateirreps. Degeneratebecause of T
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For nondegenerate states we hadthis as the effect of a symmetry operation on an eigenfunction:
RH = RE
HR = ER
H = E
Thus R = c since E is nondegenerate.
For the water molecule ( nondegenerate) :
What about degenerate states?
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R Ψnk = D[R ]k1Ψn1 + D[R ]k2Ψn2 + D[R ]k3Ψn3 +…+
D[R ]kℓΨnℓ
For each relevant symmetry operation R, the constants
D[R ]kp form the elements of an ℓℓ matrix D[R ].
ForT = RS it is straightforward to show that
D[T ] = D[R ] D[S ]
The matrices D[T ], D[R ], D[S ] ….. form an ℓ-dimensional representation that is generated by the ℓ functions Ψnk
The ℓ functions Ψnk transform according to this representation
degenerate energy level with energy Enℓ-fold
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Symmetry of a product:C2v(M) example
E (12) E* (12)*
A1 1 1 1 1
A2 1 1 1 1
B1 1 1 1 1
B2 1 1 1 1
B1 x B2, A1 x A2, B1 x A2, B2 x A2, B1 x B1,… A2 A2 B2 B1 A1
The symmetry of the product of two nondegenerate states is easy:
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Symmetry of a product. Example: C3v
E E: 4 1 0
A1 A1 = A1
A1 A2 = A2
A2 A2 = A1
A1 E = E
A2 E = E
E E = A1 A2 E
Characters of the product representation are the products of the characters of the representations being multiplied.See pp 109-114 in BJ1
Reducible representation
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Symmetry of a product. Example: C3v
E E: 4 1 0
A1 A1 = A1
A1 A2 = A2
A2 A2 = A1
A1 E = E
A2 E = E
E E = A1 A2 E
Characters of the product representation are the products of the characters of the representations being multiplied.See pp 109-114 in BJ1
Reducible representation
We say that E x E A1
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Back to the vanishing
integral theorem
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+ Parity - Parity
x
Ψ+(x)
x
Ψ+(x)
x
Ψ-(x)
Ψ+(-x) = Ψ+(x)
Ψ-(-x) = -Ψ-(x)
∫Ψ+Ψ-Ψ+dx = 0
- parity
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+ Parity - Parity
x
Ψ+(x)
x
Ψ+(x)
x
Ψ-(x)
Ψ+(-x) = Ψ+(x)
Ψ-(-x) = -Ψ-(x)
∫Ψ+Ψ-Ψ+dx = 0
- parity
∫f(τ)dτ = 0 if symmetry of f(τ) does not contain A1
The vanishing integral theorem
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Diagonalizing the molecular Hamiltonian
Schrödinger equation
To apply the vanishing integral rule we look at symmetry of
Eigenvalues and eigenfunctions are found by diagonalization of a matrix with elements
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Diagonalizing the molecularHamiltonian
Hmn vanishes if Γ( ) and Γ( ) are different
The Hamiltonian matrix factorizes, for example for H2O
if Γ(integrand) does not contain Γ(s)
= 0
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F(Ei ) = [ gie-Ei/kT ] / gje-Ej/kT Boltzmann factor∑j
S(f ← i) = | ∫ Φf* μA Φi dτ |2 Line strength∑A=X,Y,Z
Rstim(f→i) = 1 – exp (-hνif /kT ) Stimulated emission
I(f ← i) = ∫
8π3 Na______(4πε0)3hc2
F(Ei ) S(f ← i) Rstim(f→i)
Integrated absorption intensity for a line is:
= νif
~line
Frequency factor
ε(ν)dν~
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Selection rules for transitions
The intensity of a transition is proportional to the square of
For the integral to be non-vanishing, the integrand musthave a totally symmetric component.
μZ = Σ qi Zi
Z
i
Product of symmetries of Φs must contain that of μZ
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Symmetry of Z
Z has symmetry *
* ???
What is
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Symmetry of Z
Z has symmetry *
* has character +1 under all permutations P
1 under all permutation-inversions P*
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Symmetry of Z for H2O
* = A2
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The Symmetry Labels of the C2v(M) Group of H2O
(12) E* 1 1
1 -1 -1 -1 -1 1
E 1111
(12)* 1 -1 1-1
A1
A2
B1
B2
Symmetry of μZ
R = ±
∫Ψa*HΨbdτ = 0 if symmetries of Ψa and Ψb are different.
As a result the Hamiltonian matrix is block diagonal.
∫Ψa*μΨbdτ = 0 if symmetry of product ΨaΨb is not Γ*
Symmetry of H
Using symmetry labels and the vanishingintegral theorem we deduce that:
Γ(μZ) = Γ*
0 0 00
Γ(H) = Γ(s)
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Example of using the symmetry operation (12):
H1
H2
r1´r2´
´(12)
We have (12) (r1, r2, ) = (r1´, r2´, ´)
We see that (r1´, r2´, ´) = (r2, r1, )
Determining symmetry and reducing a representation
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2
3
1
1
3
2
3
12
1
3
23
1 2
1
3
2
1
3
21
3
2
r2´r1´
´
r1r2
r1r2
r1r2
r1r2
r1´r2´
´
r1´r2´´
r2´r1´´
(12)
E
E*
(12)*
r1´ r1 r2´ = r2 ́
r1´ r2 r2´ = r1 ́
r1´ r1 r2´ = r2 ́
r1´ r2 r2´ = r1 ́
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(12)
E
E*
(12)*
r1 r1´ r1 1 0 0 r1 r2 = r2´ = r2 = 0 1 0 r2 ́ 0 0 1
r1 r1´ r2 0 1 0 r1 r2 = r2´ = r1 = 1 0 0 r2 ́ 0 0 1
r1 r1´ r1 1 0 0 r1 r2 = r2´ = r2 = 0 1 0 r2 ́ 0 0 1
r1 r1´ r2 0 1 0 r1 r2 = r2´ = r1 = 1 0 0 r2 ́ 0 0 1
R a = a´ = D[R] a
= 3
= 1
= 3
= 1
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E (12) E* (12)*
A1 1 1 1 1
A2 1 1 1 1
B1 1 1 1 1
B2 1 1 1 1
3 1 3 1
aA1 = ( 13 + 11 + 13 + 11) = 24
1
aA2 = ( 13 + 11 13 11) = 04
1
aB1 = ( 13 11 13 + 11) = 04
1
aB2 = ( 13 11 + 13 11) = 14
1
= 2 A1 B2
A reducible representation
Γ = Σ aiΓi i
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E (12) E* (12)*
A1 1 1 1 1
A2 1 1 1 1
B1 1 1 1 1
B2 1 1 1 1
3 1 3 1
aA1 = ( 13 + 11 + 13 + 11) = 24
1
aA2 = ( 13 + 11 13 11) = 04
1
aB1 = ( 13 11 13 + 11) = 04
1
aB2 = ( 13 11 + 13 11) = 14
1
= 2 A1 B2
A reducible representation
Γ = Σ aiΓi
i
i
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E (12) E* (12)*
A1 1 1 1 1
A2 1 1 1 1
B1 1 1 1 1
B2 1 1 1 1
3 1 3 1
aA1 = ( 13 + 11 + 13 + 11) = 24
1
aA2 = ( 13 + 11 13 11) = 04
1
aB1 = ( 13 11 13 + 11) = 04
1
aB2 = ( 13 11 + 13 11) = 14
1
= 2 A1 B2
A reducible representation
Γ = Σ aiΓi
i
i
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E (12) E* (12)*
A1 1 1 1 1
A2 1 1 1 1
B1 1 1 1 1
B2 1 1 1 1
3 1 3 1
aA1 = ( 13 + 11 + 13 + 11) = 24
1
aA2 = ( 13 + 11 13 11) = 04
1
aB1 = ( 13 11 13 + 11) = 04
1
aB2 = ( 13 11 + 13 11) = 14
1
= 2 A1 B2
A reducible representation
Γ = Σ aiΓi
i
i
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We know now that r1, r2, and generate the representation 2 A1
B2
Consequently, we can generate from r1, r2, and three „symmetrized“ coordinates:
S1 with A1 symmetry
S2 with A1 symmetry
S3 with B2 symmetry
For this, we need projection operatorsPages 102-109 of BJ1
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Projection operators:
General for li-dimensional irrep i
Diagonal element of representation matrix
Symmetry operation
Simpler for 1-dimensional irrep i Character
1
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Projection operators:
General for li-dimensional irrep i
Diagonal element of representation matrix
Symmetry operation
Simpler for 1-dimensional irrep i Character
1
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Projection operators:
General for li-dimensional irrep i
Diagonal element of representation matrix
Symmetry operation
Simpler for 1-dimensional irrep i Character
1
E (12) E* (12)* A1 1 1 1 1
PA1 = (1/4) [ E + (12) + E* + (12)* ]
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Projection operators:
General for li-dimensional irrep i
Diagonal element of representation matrix
Symmetry operation
Simpler for 1-dimensional irrep i Character
1
E (12) E* (12)* A1 1 1 1 1 B2 1 -1 1 -1
PA1 = (1/4) [ E + (12) + E* + (12)* ]
PB2 = (1/4) [ E – (12) + E* – (12)* ]
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Projection operator for A1 acting on r1
= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]
S1 = P11A1r1 = [ E + (12) + E* + (12)* ]r1 4
1
4
1
2
1
= [ + + + ] =
S2 = P11A1 = [ E + (12) + E* + (12)* ] 4
1
4
1
= [ r1 r2 + r1 r2 ] = [ r1 r2 ]
S3 = P11B2r1 = [ E (12) + E* (12)*] r1 4
1
4
1
2
1
= [ + ] = 0
P11B2 = [ E (12) + E* (12)* ] 4
1
4
1
Is „annihilated“ by P11B2
PA1
PA1
PB2
PB2
PB2
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Projection operators for A1 and B2
= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]
S1 = P11A1r1 = [ E + (12) + E* + (12)* ]r1 4
1
4
1
2
1
= [ + + + ] =
S2 = P11A1 = [ E + (12) + E* + (12)* ] 4
1
4
1
= [ r1 r2 + r1 r2 ] = [ r1 r2 ]
S3 = P11B2r1 = [ E (12) + E* (12)*] r1 4
1
4
1
2
1
= [ + ] = 0
P11B2 = [ E (12) + E* (12)* ] 4
1
4
1
Is „annihilated“ by P11B2
PA1
PA1
PB2
PB2
PB2
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Projection operators for A1 and B2
= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]
S1 = P11A1r1 = [ E + (12) + E* + (12)* ]r1 4
1
4
1
2
1
= [ + + + ] =
S2 = P11A1 = [ E + (12) + E* + (12)* ] 4
1
4
1
= [ r1 r2 + r1 r2 ] = [ r1 r2 ]
S3 = P11B2r1 = [ E (12) + E* (12)*] r1 4
1
4
1
2
1
4
1
4
1
PA1
PA1
PB2
Aside: S1, S2 and S3 have the symmetry and form of thenormal coordinates. See pp 269-277 in BJ1, and 232-233 in BJ2
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[H,R] Symmetry and conservation laws(see chapter 14 of BJ2)
iħ ∂Ψ/∂t = HΨ where Ψ is a function of q and t
Does symmetry change with time?
∂<Ψ|R|Ψ>/∂t = <∂Ψ/∂t|R|Ψ> + <Ψ|∂(RΨ)/∂t>
= <∂Ψ/∂t|R|Ψ> + <Ψ|R|∂Ψ/∂t> = [<HΨ|R|Ψ> - <Ψ|R|HΨ>]
iħ__
= <Ψ|[H,R]|Ψ> (H is Hermitian)
= 0
iħ__
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So Far:• Point group (geometrical) symmetry• H2O and PH3 point groups used as examples• Group theory definitions: Irreducible reps and Ch. Tables• Reducible representations and projection operators• Problems using point groups: Rotation, tunneling,…• Use [H,R]=0 to define R as a symmetry operation• Introduce the CNPI group• Explain why [H,R]=0 used to define symmetry• Can label energy levels (the Ψ generate a representation.)• Vanishing integral theorem • Forbidden interactions and forbidden transitions• Conservation of symmetry
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Where are we going?
HΨn = EnΨn
H = H0 + H’
where H0Ψn0 = En
0Ψn0
En0 is -fold degenerate: Eigenfunctions are Ψn1
0,Ψn20,…,Ψn
0
We want to symmetry label the energy levels using theirreducible representations of a symmetry group.
We do this because it helps us to do many things:
Which En0 can be mixed by H’: Block diagonalize H-matrix.
Selection rules for transitions: Only if connected by Γ*.Nuclear spin statistics and intensity ratiosTunneling splittings, Stark effect, Zeeman effect,Breakdown of Born-Oppenheimer approximation…
ℓ ℓ
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The ℓ-fold degenerate eigenfunctions Ψn10,Ψn2
0,…,Ψnℓ0
GENERATE an ℓ-fold irreducible representation of thesymmetry group (this labels the energy level En
0).
The basis of what we do using symmetry is that:
Ψn10
Ψn20
.
.
Ψnℓ0
Ψn10
Ψn20
.
.
Ψnℓ0
R
= D(R)
The matrices D(R) form an irreducible representation
To obtain the matrices D(R), and hence the irreduciblerep. label, we need to know the Ψni
0 and to know how the symmetry ops transform the coordinates in the Ψni
0.
The above follows from the fact that [H,R] = 0.
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But first of all we must decide on the symmetry group that we are going to use. It could be the CNPI group
BUT…
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There are problems with the CNPI Group
Number of elements in the CNPI groups of variousmolecules
C6H6, for example, has a 1036800-element CNPI group,but a 24-element point group at equilibrium, D6h
Huge groups and (as we shall see) multiple symmetry labels