Transcript 08ColToluene.ppt

Torsion-Rotation Program for
Six-Fold Barrier Molecules
Toluene MW Fit
Vadim V. Ilyushin1, Zbigniew Kisiel2, Lech Pszczolkowski2,
Heinrich Mäder3, Jon T. Hougen4
1Institute
of Radio Astronomy of NASU, Kharkov, Ukraine
2Institute of Physics, Polish Acad. of Sci., Warsaw, Poland
3Institute for Phys. Chem., Kiel University, Kiel, Germany
4Optical Technology Division, NIST, Gaithersburg, MD
H2
H3
PI Group Operations
H1 z
Cf
x
H5
Cb
Cg
Ca
H7
Cd
Ce
Cc
H8
Toluene
H4
Toluene

C3 top C2 frame
(23)* (123)
(ab)(cd)x
(45)(67)
G12  C6v
A1, A2, B1, B2, E1, E2
H6
Methanol

C3 top C2 frame
(23)* (123)
none
G6  C3v
A1, A2, E
Transformation properties of the torsional
and rotational variables under various
operations of the PI group G12 for toluene
torsional
E

(123)
 + 2/3
(ab)()()()

(ab)()()()(123)   2/6
(23)*

rotational
, , 
, , 
 + , , 
 + , , 
  ,   ,  + 
Symmetry species  in G12 and
time reversal symmetry (+) or ()

A1
A2
B1
B2
E1
E2
Torsional
cos6 (+)
sin6 (+)
cos3 (+)
sin3 (+)
exp(i)
exp(2i)
Momenta
Jz (), p ()
Jy ()
Jx ()
|JKaKc
ee
eo
oe
oo
Character table for the subgroup G6 of G12
needed to understand basis set construction
in the two-step diagonalization procedure
basis functions: exp(6k+)i |JKM>
A
B
E1+
E1
E2+
E2
E (ab)(123) (ab)(123)2 (123) (123)2 (ab)
1
1
1
1
1
1
1
-1
-1
1
1
-1
1
-
-*

*
-1
1
-*
-
*

-1
2
*

*

1
2

*

*
1
 = exp(2i/3)

0
3
+1
-1
+2
-2
K
even
odd
Where are we in the fitting procedure?
Answer: In the middle of it.
234 transitions for m = 0, 1, 2, +3, -3
21 parameters
standard deviation = 5.9 kHz
J  25 for m = 0, but J  8 for m = 1,2,+3,-3
Ka, Kc labels are problematic (next slides)
Current understanding of the toluene spectrum:
J = 2 ← 1 region
Observed
(waveguide
FTMW)
Calculated
= |m|≤3 in fit
= |m|>3
assigned,
not yet fitted
Specimen fit:
K=0 levels for m = 0,1,2,3,-3
60
m = -3 B species
m = +3 B species
E - 0.5(B+C)J(J+1) cm-1
50
40
30
m = 2 E2 species
20
10
m = 1 E1 species
Barrier Height V6
m = 0 A species
0
0
20
40
60
J(J+1)
80
100
Observed m = 0 Reduced Energy Levels
6
E - 0.5(B+C)J(J+1) cm-1
5.5
K = 5 B species
5
4.5
K = 4 A species
4
K = 3 B species
3.5
3
K = 2 A species
2.5
2
K = 1 B species
K = 0 A species
1.5
1
0
20
40
60
J(J+1)
80
100
Observed m = 1 Reduced Energy Levels
14
K = 5 E2 species
E - 0.5(B+C)J(J+1) cm-1
13
12
K = 4 E1 species
11
K = 3 E2 species
10
K = 2 E1 species
K = 5 E2 species
9
K=0
8
K = 1 E2 species
K = 4 E1 species
K=3
K=2
K=1
7
0
50
J(J+1)
100
Observed m = 2 Reduced Energy Levels
34
E - 0.5(B+C)J(J+1) cm-1
32
K = 5 E1 species
30
K = 4 E2 species
K = 3 E1 species
28
K = 2 E2 species
26
K = 1 E1 species
K=2
24
K = 0 E2 species
K = 1 E1 species
K = 3 E1 species
22
0
20
40
60
J(J+1)
80
100
Observed m = +3 and -3 reduced energy
levels
E - 0.5(B+C)J(J+1) cm-1
64
62
K = 5 A species
60
K = 4 B species
58
K = 3 A species
56
K = 2 B species
K = 1 A species
K = 0 B species
54
K = 0 B species
K = 1 A species
K = 2 B species
K = 3 and 4
52
50
0
20
40
60
J(J+1)
m = -3
m = +3
80
100
Conclusions:
1. Our 6 kHz fit is excellent, but:
2. Assignments need to be extended into
higher J and K problem regions (see
Zbigniew Kisiel at this meeting).
3. Measure J = 1  0 region (see Heinrich
Mäder at this meeting).
4
Labeling algorithm in program needs to
be able to deal with Ka,Kc (high-barrier)
and K (low barrier) labels.