Theory of microwave 3-Wave
mixing of chiral molecules
Kevin K. Lehmann
Depts. of Chemistry & Physics
University of Virginia
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.
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.
May 23, 2013 issue of Nature contained
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.
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?
Rotation of Rigid Asymmetric top
H = A× J + B× J + C× J
2
a
2
b
2
c
A >B >C
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:
J,t, M = å At ,K J,K, M
K
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:
JKa,Kc , M
0 £ K a ,Kc £ J
K a + K c = J,J +1
• 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.
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.
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.
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.
• 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.
Effect of Finite Excitation Pulse
To first order in Es:
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.
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
We generate a Y polarized dipole amplitude on on J’’, t -> J’ t transition
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:
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
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.
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.
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.