1. Torsion-mediated spin-rotation hyperfine splittings in methanol (at moderate to high J values) Li-Hong Xu – University of New Brunswick 2 expt labs: NNOV (S. Belov, G.Yu. Golubiatnikov, A.V. Lapinov) Kharkov (V. Ilyushin, E.A. Alekseev, A.A. Mescheryakov) 2 others: NIST (J.T. Hougen), UNB (Li-Hong Xu) Part I – Unexpected splittings Part II – Hamiltonian and matrix elements Nuclear-spin -- overall-rotation interaction Torsion-mediated spin-rotation operator for E state O H H H C H Torsional function: e ± i  Torsional angular momentum : P  Part III – Current fit and prediction

2. Part I – Unexpected splittings after instrumental improvements i.e. after NNOV/Russia improved the sensitivity and resolution of the Lamb-dip sub-millimeter spectrometer (range  100 to 500 GHz). E species v t = 0, J K = 27 0  27 -1 71 kHz 20 kHz Lamb-dip doublet splittings of 71 kHz in methanol at 104 GHz for an E species Belov’s 2013 Columbus talk Kharkov t = 0,  J = 0, K a = 0  -1

3. Part II - Physics considerations for CH 3 OH Nuclear spins: I C = I O = 0 and I H = ½ CH 3 OH has no nuclei with electric quadrupoles The four H atoms are all magnetic dipoles Where are the magnets & magnetic fields? Magnets: I 1, I 2, I 3 of CH 3 and I 4 of OH with 12 Cartesian components Magnetic fields: overall rotation J (J x, J y, J z ) internal rotation (torsion) P  Now list all proton-spin hyperfine operators, using x,y,z components in the molecule-fixed axis system (to permit using molecular constants). O H4H4 H2H2 H3H3 C H1H1 How many bilinear operators with at least one I component?

4. Without symmetry, there are 102 (= 54 + 36 + 12) bilinear operators with at least one I component. Spin-Spin: (4  3)  9/2 = 54 bilinear operators of form (I 1x )(I 2x ), (I 1x )(I 3y ), (I 1z )(I 4x ), (I 2x )(I 3z ), (I 2y )(I 4x ), etc. Spin-Rotation: 12  3 = 36 bilinear operators of form (I 1x )(J x ), (I 2x )(J y ), (I 3y )(J z ), (I 4z )(J x ), etc. Spin-Torsion (P  ): 12  1 = 12 bilinear operators of form (I 1x )(P  ), (I 2x )(P  ), (I 3y )(P  ), (I 4z )(P  ), etc. With symmetry how many of these operators are permitted? Need symmetry species in the PI group G 6 for CH 3 OH…. In principle, we need to consider: spin-spin, spin-rotation, and spin-torsion But first try to simplify our work by guessing at what the important physics is, … what measurements are telling us….. Without including torsional variation

5. 16 2228 80 70 50 30  Look at some more experimental data ( t = 0,  J = 0, two Q branches)  NNOV t = 0,  J = 0, K a = 2  -1 Kharkov t = 0,  J = 0, K a = 0  -1 Doublet splitting vs J These doublet splittings increase approximately linearly with J. Thus, keep only operators with one J & one I (I 1x J y, etc.). We keep no spin-spin, no spin-torsion, only spin-rotation. 1734

6. Group-theoretically allowed spin-rotation interaction operators are made from with sr A 1 = s A 1  r A 1 or s A 2  r A 2 products Rotation: J y is r A 1 and J x, J z are r A 2 Spin: I 1, I 2, I 3, I 4 12 proton spin operators I 4y is s A 1 and I 4x, I 4z are s A 2 5 operators Easy part = non-degenerate A 1 and A 2 operators: I y is s A 1 and I x, I z are s A 2 (I  I 1 + I 2 + I 3 ) 5 operators k 1 I y J y + k 2 I x J x + k 3 I x J z + k 4 I z J x + k 5 I z J z + k 6 I 4y J y + k 7 I 4x J x + k 8 I 4x J z + k 9 I 4x J z + k 10 I 4z J z (10 terms ) Hard part = doubly degenerate s E spin operators: (I E+ ) x  (I 1 + e -2  i/3 I 2 + e +2  i/3 I 3 ) x (I E  ) x  (I 1 + e +2  i/3 I 2 + e -2  i/3 I 3 ) x (I E+ ) y  (I 1 + e -2  i/3 I 2 + e +2  i/3 I 3 ) y (I E  ) y  (I 1 + e +2  i/3 I 2 + e -2  i/3 I 3 ) y (I E+ ) z  (I 1 + e -2  i/3 I 2 + e +2  i/3 I 3 ) z (I E  ) z  (I 1 + e +2  i/3 I 2 + e -2  i/3 I 3 ) z This is where we introduce the torsional variation, because the torsional function e +i  is t E +, e -i  is t E  t E  s E = ts A 1 + ts A 2 + ts E one ts A 1 and one ts A 2 torsion-spin operator from each s E pair (9 terms ) Cannot get sr A 1 O H4H4 H2H2 H3H3 H1H1 C

7. Summary on symmetry: spin-spin and spin-torsion operators are ignored spin-rotation operators are divided into two types: sr A 1 from s A 1  r A 1 or s A 2  r A 2 10 termsSpin-Rotation str A 1 from ts A 1  r A 1 or ts A 2  r A 2 9 terms Torsion-Mediated Spin-Rotation (  6 for  J = 0 hf mixing) The red part gives us hope. It involves t E functions, which don’t exist in “ordinary” molecules, and which therefore might lead to some unexpectedly large splittings in tr E states in methanol.

8. Part III – Current fit and prediction Skip many details (group theory, operator equivalents, lab-fixed axes  molecule-fixed axes) and jump to the quantitative treatment, fit and prediction. First, a schematic display of the 3 (out of 6) torsion-mediated spin-rotation operators that we want to use. Manuscript in preparation Small: J z 2 low K {J x,J y } small D ab {J y,J z } = - I EA1 : ts A 1 I EA2 : ts A 2

9. One of our 3-parameter fits of the doublet splittings we have available at present obs: symbols calc: lines NNOV t = 0,  J = 0 K a = 2  -1 Kharkov t = 0,  J = 0 K a = 0  -1 NNOV t = 0,  J = 0 K a = -2  1 NNOV t = 0,  J = 1 K a = -1  -2 3 parameters 5 K states (-2, -1, 0, 1, 2) 32 data pts e i2  kHz A xx (2) 0.179(43) A yy (2) 2.21(12) A xy (2) 1.15(11)  0.93 Another fit e - i  kHz A xx (1) 0.242(51) A yy (1) 3.015(91) A xy (1) 2.435(80)  1.44

10. Similar to the last slide, but with new data added, no new fits yet NNOV t = 0,  J = 0 K a = 2  -1 Kharkov t = 0,  J = 0 K a = 0  -1 NNOV t = 0,  J = 0 K a = -2  1 NNOV t = 0,  J = 1 K a = -1  -2 NNOV t = 0,  J = 0 K a = 3  -2

11. Small doublet splittings predicted for six branches 20 kHz HFS is not resolved J 1  J 0 Some measurements in these branches have been reported as “no HFS or broad lines”

12. Summary & Next Our initial fits (30-70 kHz doublet splittings) and predictions are very encouraging This explanation is based on three torsion-mediated spin-rotation hyperfine interaction operators, which connect the two components of a doubly degenerate E torsion-rotation level in methanol The fits are not unique at the present time, because we have more interaction parameters than can be determined from the present data set. It would be nice to have more data to further test the model. (We have more fitting parameters if necessary.) There are some lines that are not clear doublets, look more like triplets (see in the next slide from Belov’s 2013 Columbus talk) We don’t understand these non-doublets yet. From preliminary calculations, we think that they have something to do with the I = ½ spin of the OH proton, which would make them a doublet of doublets, with the two middle components overlapping : | || |

13. E- methanol triplet, (24, 1)  (24, 0), V t = 0 From Sergey Belov’s talk at Columbus 2013