1. Theory of microwave 3-Wave mixing of chiral molecules
Kevin K. Lehmann
Depts. of Chemistry & Physics
University of Virginia

2. Where does optical rotation come from?
In chiral molecules, we can have both electric dipole and magnetic dipole transitions between the same pairs of states.
Forbidden by parity selection rules in achiral mol.
There is an interference of these, of opposite sign for two enantiomers
Dispersion part gives optical rotation, absorption part circular dichroism.
These are weak because magnetic dipole transition matrix elements are typically 2-5 orders of magnitude weaker than electric dipole matrix elements.

3. Other ways to detect?
A solution of chiral molecules can have a c3 response, if it has optical rotation (Phys. Rev. Lett. 85, 4253 (2000).
Cannot be phase matched – three E fields must be perpendicular to each other!
Very weak effect for sum-freq. generation
Only seen when resonance enhanced (Phys. Rev. Lett. 85, 4474 (2000).
Only seen in condensed phases.

4. May 23, 2013 issue of Nature contained

6. It is worth noting that the signs of the individual directions of the dipole moment projects have no intrinsic meaning – One is free to pick the “positive” directions of the inertial axes at random.
If one takes the inertia axes directions to give positive components of the dipole projections, it is well defined if this choice gives a right or left handed coordinate systems of x,y,z = A,B,C
Another way to say the same thing is that
ma . (mb x mc) is well defined and opposite sign for two enantiomers.

7. Some Questions I had…
Does the method work when higher J transitions (which have multiple M components) are used?
How large a signal (relative to the regular FID) can be generated?
Why does the signal not appear until the Stark Field is turned off?

8. Rotation of Rigid Asymmetric top
A,B,C, called rotational constants and can be calculated from molecular structure,
Proportional to the inverse of the principle moments of inertia. Wave functions can
be expanded in eigenstates of symmetric top:
It is useful to consider how energy levels change as B varies from C -> A, each limit of
which is a symmetric top with definite K value. We label states:

9. The components on the dipole moment along the inertial axes determine which transitions are allowed
Only a chiral molecule can have all three components nonzero.
The product of the three dipole components is opposite sign for a pair of enantiomers.

11. Calculate the Molecular Polarization parallel to the Y axis
Before Stark Field is turned off:
The interference terms in the Stark Field induced detection matrix element cancel
those in the excitation step, leaving only X polarized emission.

12. After Stark Field is turned off adiabatically:
We not have a net Y polarized molecular polarization and this results in a Y polarized
emission. It is proportional to the product to the three dipole moment projections and
thus of opposite sign for R and S molecules. Thus, it is zero for a racemate.
This chiral emission is phase shifted by +/- 90° relative to the X polarized emission.
For generally for weak Stark Field, the Chiral signal is proportional to the Change in the
Stark Electric field, so Es -> -Es would give stronger signal.

13. The general case has been worked out, get same qualitative behavior with
Polarization perpendicular to DC and AC excitation fields only after DC field is
changed. The optimal is to invert the Stark Field giving Twice the signal.
Without Stark Marking, the Y polarized dipole = 0 due to cancellation of the +M and
-M terms for a sample without orientation ( < cos(q) > = 0.

15. If we can neglect Stark Mixing between states that have different J quantum numbers, the results depend only on k = (2B – A – C)/(A-C).
k goes from -1 (prolate symmetric top) to +1 (oblate symmetric top)
There is a symmetry of k -> -k if we interchange A and C in the expressions.

16. Effect of Finite Excitation Pulse
To first order in Es:

17. Three wave mixing version
Apply p/2 pulse on |0> -> |1> transition, polarized along X.
Then apply p pulse on |1> -> |2> transition, polarized along Z.
Sample will generate Free induction decay polarized along Y at frequency of |0> -> |2> transition with amplitude proportional to product of ma . (mb x mc)
Larger signals generated – still have phase matching limitation
Signal proportional to population difference of states |0> and |1>, does not depend on |0> - |2> or |1> -|2> difference.

18. After First X polarized Excitation Pulse on J’’, t’’ -> J t transition
Then, after 2nd , Z polarized ``twist’’ pulse Pulse on J t -> J’, t’ transition

19. We generate a Y polarized dipole amplitude on on J’’, t’’ -> J’ t’ transition
This is optimized for p/2 and then p pulse excitation. Because of M dependence of transition moments, we cannot get complete coherence transfer by second pulse, but we can ~75% as large a PY as PX.
The challenge is that emission from different points will have destructive interference due to phase mis-match. To phase match, all three waves should propagate parallel to one another, but then the three electric fields cannot be perpendicular. To have a chiral signal we need:

20. If we cross the second excitation beams at angle f, the emission will be maximized
At direction:
If we have uniform sample of length L, the emitted E field will be proportional
to L sinc(Dk L/2) . If we have a Gaussian spatial distribution of PY, say to mode shape,
the emitted signal will be ~exp ( - (Dk L/2)2). Molecules from Effusive source have a
a density ~ 1/ ( z2 + h2) a distance h from the nozzle and this gives emission proportional
to exp(- | Dk h|). Dephasing will severely reduce signal levels is sample become
larger in extent than l(max)/2

21. Quasi-Phase Matching Geometry
Two of the 3 fields free space propagate through loops. The Strip line
will propagate third field parallel to the other two, but allowing E to have
Strong component parallel to its propagation direction.

22. Conclusions
Three wave mixing (of which Stark modulation is special case) can only measure the enantiomeric excess of sample (difference in concentrations of two enatiomers). The emission signal can only be quantitatively converted to an e.e. value one has sample of known e.e. value of the same molecule to calibrate response.
Method is potentially valuable to determine small enatiomeric excess but NOT to quantitatively determine a small racemic fraction, i.e. e.e. values only slightly less than unity.

23. Acknowledgements
Brooks Pate for introducing me to this field and many stimulating Conversations.
Simon Lobsiger, Cristobal Perez, Evangelisti, Nathan Seifert, David Patterson, and John Doyle for helpful conversations.