lecture 8/11 - uni-wuppertal.de
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1 1 3
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The feasibility of an operation depends
on potential energy barriers and
experimental resolution
Should we extend this definition?
H2Te Rigid Rotor Energy Levels [E(JKaKc)-E(JJ0)]
JJ,0
JJ,1
JJ-1,1
JJ-1,2
JJ-2,2
JJ-2,3
JJ-3,3
JJ-3,4
JJ-4,4
JJ,0
JJ,1
JJ-1,1
JJ-1,2
J
Ae=6.26, Be=6.11, Ce=3.09 cm-1
Actual H2Te Energy Levels [E(JKaKc)-E(JJ0)]
J
JJ,0
JJ,1
JJ-1,1
JJ-1,2
JJ,0
JJ,1
JJ-1,1
JJ-1,2
JJ-2,2
JJ-2,3
JJ-3,3
JJ-3,4
JJ-4,4
-6
-5
-4
-3
-2
-1
0
0 10 20 30 40 50 60 70 80
Six-fold enery clusters in the
vibrational ground state of PH3
A1/A2
A2/A1
E
E
Cluster = A1 A2 2E in C3v(M)
1 PCS
= E 1 PCS
3 PCS
= (132)1 PCS
4 PCS
= (23)*1 PCS
2 PCS
= (123)1 PCS
5 PCS
= (12)*1 PCS
6 PCS
= (13)*1 PCS
Cluster = A1 A2 2E in C3v(M)
J
Rotational energy surface
The rotational energy surface is a radial plot of
EJ EJ(min) as a function of J,J
Classical Hamiltonian function
z
x
y
PH3 rotational energy surface, J = 100
Six equivalent maxima
Each cluster state at highest
energy corresponds to stable
angular momentum trajectory
around a maximum
Cluster states are „localized“
at their respective maxima
„No tunneling between
cluster states“
Extension of feasibility definition:
If energy splittings resulting from potential
energy tunneling cannot be resolved,
permutation-inversion operations accompanied
by this tunneling are considered unfeasible
If energy splittings resulting from rotational
energy surface tunneling cannot be resolved,
permutation-inversion operations accompanied
by this tunneling are considered unfeasible
Old:
New = Old together with:
See also Per Jensen and P. R. Bunker: The Molecular Symmetry Group
for Molecules in High Angular Momentum States, J. Mol. Spectrosc. 164,
315-317 (1994).
Application to H2X:
MS group is C2v(M) = { E, (12), E*, (12)* }
1
2
1
1 1
2 2
2
When tunneling between cluster states
is neglected, (12), E*, (12)* are
unfeasible, only E is feasible.
New MS group is C1 = { E} with one irrep A
A
Reverse correlation C1 C2v(M)
A
Application to PH3:
MS group is C3v(M) = { E, (123),(132) , (12)*,(13)*,(23)* }
When tunneling between cluster states
is neglected, (123), (132), (12)*,
(13)*, (23)* are unfeasible,
only E is feasible.
New MS group is C1 = { E} with one irrep A
A
Reverse correlation C1 C3v(M)
A
etc.
A1 A2 2E
The MS group of the ammonia dimer (NH3)2 has 144
elements when the effect of inversion of the two
NH3 moieties is observable
Does (NH3)2 then
have higher
symmetry than C60 ?
A Question To Think About:
Symmetry and molecules - summary:
Symmetry operations (e.g., permutation-
inversion operations) for a molecule leave the
molecular Hamiltonian invariant.
The operations can be organized in symmetry
groups.
The symmetry groups have irreducible
representations.
The wavefunctions for the molecular eigenstates
transform according to the irreducible
representations of the symmetry group.
How does symmetry help us?
A.
In practice, we solve the molecular Schrödinger
by diagonalizing matrix representations of
the Hamiltonian. These calculations are
facilitated by symmetry.
B.
Symmetry imposes selection rules on
molecular transitions (for example, absorption
and emission transitions).
Vanishing integral rule
The quantum mechanical integral
must vanish (i.e., be = 0) unless the integrand
contains a totally symmetric component in the
symmetry group(s) of the Hamiltonian
𝐼 = 𝜓′ 𝑂𝜓′′𝑑𝜏 *
𝜓′ 𝑂𝜓′′ *
Parenthetic remark:
symmetry of a product
Given two sets of wavefunctions
s-fold degenerate,
irrep n
r-fold degenerate,
irrep m
Which representation is generated by the set of
products Φ𝑛𝑖 Φ𝑚𝑗, i= 1, 2, 3, ..., s; j = 1, 2, 3, ..., r ?
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Φ𝑛1, Φ𝑛2, Φ𝑛3, Φ𝑛4, … , Φ𝑛𝑟
Φ𝑚1, Φ𝑚2, Φ𝑚3, Φ𝑚4, … , Φ𝑚𝑠
Effect of symmetry operation R on the
functions Φ𝑛𝑖 and Φ𝑚𝑗:
Effect of R on the product Φ𝑛𝑖 Φ𝑚𝑗:
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Effect of R on the product ni mj:
Short-hand notation:
Elements of transformation „supermatrix“:
Diagonal elements:
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Diagonal elements of transformation „supermatrix“:
Characters of „product representation“ nm:
Notation (direct product):
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Special case:
n1, n2 have E symmetry in C3v(M)
Which representation is generated by the three
products n1n1, n2n2, n1n2= n2n1?
Symmetric product representation:
with characters:
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nth order symmetric product
representation [E]n
has the characters
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Its dimension is 𝑛 + 1 and it is generated, for example,
by the 𝑛 + 1 products:
n1𝑛, n1
𝑛−1n2, n1𝑛−2 n2
2, n1𝑛−3n2
3, …
…n13n2
𝑛−3, n12n2
𝑛−2, n1 n2𝑛−1, n2
𝑛
Vanishing integral rule
The quantum mechanical integral
must vanish (i.e., be = 0) unless the integrand
contains a totally symmetric component in the
symmetry group(s) of the Hamiltonian
𝐼 = 𝜓′ 𝑂𝜓′′𝑑𝜏 *
𝜓′ 𝑂𝜓′′ *
Totally symmetric component?
All groups have a totally symmetric irreducible
representation (s), e.g.
In the totally symmetric
representation, each group
element is represented by
the 11 matrix 1. The group
{1} is homomorphic to any
group.
The representation of must be
= n(s) (s) .... with n(s) 0
in order that the integral can be non-vanishing
𝜓′ 𝑂𝜓′′ *
Diagonalizing the molecular
Hamiltonian
Schrödinger equation
We apply the vanishing integral rule
Eigenvalues and –functions are found by
diagonalization of a matrix with elements
𝜓𝑗 = 𝑐𝑗𝑛
𝑛
𝜓𝑛 0
𝐻𝜓𝑗 = 𝐸𝑗𝜓𝑗 ^
𝐻𝑚𝑛 = 𝜓𝑚∗𝐻𝜓𝑛𝑑𝜏
0 0 ^
Diagonalizing the molecular
Hamiltonian
The Hamiltonian is invariant under symmetry operations
so the integrand in Hmn generates the product
characters
*
The number of times that the totally symmetric
representation occurs is
m
*
𝐻𝑚𝑛 = 𝜓𝑚∗𝐻𝜓𝑛𝑑𝜏
0 0 ^
Diagonalizing the molecular
Hamiltonian
Hmn can only be non-vanishing if and belong to
the same irreducible representation
The Hamiltonian matrix
factorizes, for example for H2O
Computing time for diagonalization
𝑁3 without factorization
4 (𝑁4)3 =
𝑁3
16 with factorization;
m
*
Total matrix
dimension N
𝜓𝑚 𝜓𝑛 0 0
Intensities
A (A = X, Y, Z) is a space-fixed component of the
molecular dipole moment
Cre is the charge, Ar the A
coordinate of particle r 𝜇𝐴 = 𝐶𝑟𝑒𝐴𝑟 𝑟
Selection rules for transitions
So the intensity of a rotation-vibration transition is
proportional to the square of
Vanishing integral rule: For the integral to be non-
vanishing, the integrand must have a totally symmetric
component.
𝐼TM = Φrve 𝜇𝐴 Φrve d𝜏 ´* ´´
Symmetry of A for H2O
Axis system XYZ with origin in molecular center of
mass; the protons are labeled 1, 2
Both protons have the charge +1e
X
Y
Z
(Xi,Yi,Zi)
(12) (X1,Y1,Z1) = (X2,Y2,Z2)
(12) (X2,Y2,Z2) = (X1,Y1,Z1)
(12) (X3,Y3,Z3) = (X3,Y3,Z3)
(12) A = A
𝜇𝐴 = 𝐶𝑟𝑒𝐴𝑟 𝑟
Symmetry of A for H2O
X
Y
Z
(Xi,Yi,Zi)
(12)* (X1,Y1,Z1) = (X2, Y2,Z2)
(12)* (X2,Y2,Z2) = (X1, Y1,Z1)
(12)* (X3,Y3,Z3) = (X3, Y3,Z3)
E* (X1,Y1,Z1) = (X1, Y1,Z1)
E* (X3,Y3,Z3) = (X3, Y3,Z3)
E* (X2,Y2,Z2) = (X2, Y2,Z2)
E* A = A
(12)* A = A
𝜇𝐴 = 𝐶𝑟𝑒𝐴𝑟 𝑟