Proposal for Chiral Detection by the AC Stark-Effect Kevin K. Lehmann Departments of Chemistry & Physics, University of Virginia, Charlottesville VA, 22904-4319∗ (Dated: January 22, 2015) Recently,tworelatedthreewavemixingexperimentshavebeendemonstratedthatuseallowedro- tationaltransitionstoproduceafreeinductiondecaysignalwithanamplitudelinearlyproportional to the enantiomeric excess of a chiral molecule. In the present work, a formally five wave mixing experimentisproposedthatwillexploitnear-resonantACStarkshiftstodifferentiallysplitarota- tionaltransitionofRandSversionsofamoleculeandthuswillallowfortheseparatemeasurement of the densities of both enantiomers. 5 1 Molecular Chirality has long been a subject of intense interest to chemists for both esthetic and practical reasons.1 0 2 The homochirality of the principle biological components (sugars, amino acids, and nucleic acids) is fundamental to the ability to grow large polymers of defined structures and to much of molecular recognition.2 Chiral molecules n come in enantiomeric pairs (often designated R and S) that can be mapped into one another by a reflection in a a J plane. Sans tiny parity violating interactions, a pair of enantiomers have identical thermodynamic properties, and most physical and spectroscopic ones as well. One-well known exception is that they interact differently with circular 1 2 polarized light (which itself has a handedness and thus is chiral) leading to optical rotation and circular dichroism which are proportional to the differences in the respectively real and imaginary dielectric constants of chiral samples. ] Thesedifferencearisesfromaninterferenceofcontributionsofelectricandmagneticdipolecomponentsoftransitions. h Sincemagneticdipolematrixelementsaretypicallymuchsmallerthanelectricdipoleones,theseeffectsareweakand p - difficult to observe in the gas phase. As one goes to longer wavelengths (corresponding to slower motions), allowed m magnetic dipole transition moments become even smaller relative to electric dipole transition moments, and as a e result, more difficult to observe. In particular, optical activity has not been observed in the rotational spectroscopy h ofmolecules. Thisisunfortunateasrotationalspectroscopyprovidethegreatestselectivityindistinguishingcomplex c molecules of any method that can be applied in the gas phase. . s Kral et. al.3 pointed out that chiral molecules have a unique spectroscopic property. As a consequence of their c lack of definite parity, they have closed cycles of three electric dipole allowed transitions of the form a→b→c→a. i s Furthermore, the product of the three transition dipole moments will be of opposite sign for a pair of enantiomers. y Following publications of the same group4,5 examined use of such cycles to achieve enantiomeric selection, though h they explicitly assumed that two of the levels were such that they could be accessed by both L and R forms of the p [ molecule, which requires that they be in a state with no or a low barrier to isomerization. Hirota6 exploited the same principle of existence of cycles of three transitions but for excitation of Chiral, C symmetry asymmetric top 1 1 molecules in their ground state. In this case, the three allowed transition moments are proportional to the projection v of the permanent dipole moment (in the molecular frame) projected onto a different inertial axis of the molecule. 2 Even though the individual transition dipole matrix elements depends upon choices for directions for molecular axis 8 2 and state phase conventions, the triple product of matrix elements is invariant and of opposite sign for a pair of 5 enantiomers. The sign of this triple product can be determined by the interference of amplitude transferred between 0 states a and c by the direct a → c transition and the indirect a → b → c pair of transitions. Hirota proposed that 1. such a scheme can be used to spectroscopically distinguish between a pair of enantiomers. Hirota’s theoretical work 0 was followed by an experiment by Patterson, Schnell, and Doyle7 where they demonstrated chiral sensitivity using 5 rotational spectroscopy. In the initial experiment, a chiral molecule was excited with pulsed X polarized radiation in 1 the presence of a Z polarized DC Stark Field. When the Stark Field was rapidly shorted to zero after the excitation v: pulse, the sample produced Y polarized emission that was of opposite sign for the R and S enantiomers and zero (by i interference of the emission) for a racemic mixture. This effect can be understood as arising from the interference of X two components of the electric transition dipole, with one induced by Stark mixing. This work was soon followed by r a second paper by Patterson and Doyle,8 where the DC Stark field was replaced by an RF field that was tuned on a resonance, producing coherent electric dipole emission at the frequency of the a−c transition induced by two photon excitation of the a↔b and b↔c transitions. The emitted field was polarized perpendicular to the fields driving the a ↔ b and b ↔ c transitions. Again, there was observed a π phase shift between the emissions from the R and S enantiomers and thus zero emission intensity for a racemic mixture. Both of these experimental approaches produce a signal that is proportional to the enantiomeric excess (ee) of the sampleandthuscanbeexpectedtoproduceweaksignalsforsamplewithsmallee. Inthispaper, Iproposeamethod to produce a differential splitting of R and S rotational transitions produced by a chiral selective AC Stark Shift. This AC Stark shift is produced in third order by the interaction of three mutually perpendicularly polarized electric fields that drive the three transitions in a a→b→c→a cycle. For a simple three state model, one can arrange the experimental conditions such that the second order AC Stark shifts cancel and only the chiral selective third order 2 AC Stark shift remains, leading to opposite sign spectral shifts for an R,S pair. This situation is worked out and the optimal driving condition explored. I then turn attention to a more realistic model that accounts for the 2J +1 degeneracy of rotational states with total angular momentum quantum number J. It is found that one cannot “tune out” the second order AC Stark shift of more than one M component of the transition due to the requirement that the excitation fields be perpendicularly polarized. Despite this, simulation of the absorption of the “dressed states” in the presence of three fields produces substantial chiral differential shifts. Based upon this analysis, I consider this a promising approach to chiral analysis of complex molecules. In particular, one should be able to make accurate relative intensity measurements of the spectrally resolved signals from the R and S isomers and this could lead to accurate measurement of the ee of an enantio-enriched sample, perhaps even when the magnitude of the ee is only slightly less than unity. I. THREE LEVEL MODEL Consider the energy level diagram with three states we will label a,b,c in energy order with respective zero field energy values ¯hω ,¯hω ,¯hω and we use the notation ω =ω −ω . We apply radiation E(t)=E cos(ω t+φ )Xˆ + a b c ij i j 1 1 1 E cos(ω t + φ )Yˆ + E cos(ω t + φ )Zˆ with ω ≈ ω ,ω ≈ ω , and ω = ω + ω ≈ ω . Xˆ,Yˆ,Zˆ are three 2 2 2 3 3 3 1 ba 2 cb 3 1 2 ca orthogonal unit vectors forming a right handed coordinate system. We define Rabi frequencies as Ω = Ω∗ = ba ab (cid:104)b|µ |a(cid:105)E exp(−iφ )/2h¯,Ω = Ω∗ = (cid:104)c|µ |b(cid:105)E exp(−iφ )/2h¯, and Ω = Ω∗ = (cid:104)c|µ |a(cid:105)E exp(−iφ )/2¯h. We X 1 1 cb bc Y 2 2 ca ac Z 1 3 also define detunings ∆ω =ω −ω and ∆ω =ω −ω . The detuning of the b→c transition is ∆ω −∆ω . We b ba 1 c ca 3 c b write the time dependent wavefunction in the Hilbert spaced spanned by these three states as: Ψ(t)=C e−iωat|a>+C e−i(ωa+ω1)t|b>+C e−i(ωa+ω3)t|c> (1) a b c The time dependent Schr¨odinger equation, after making the “rotating wave” approximation of neglecting far-off resonance terms, gives for the time dependence for the amplitudes:        C 0 Ω Ω C C d a ab ac a a i¯hdt Cb =h¯ Ω∗ab ∆ωb Ωbc  Cb =Hds Cb  (2) C Ω∗ Ω∗ ∆ω C C c ac bc c c c which is the same as for the wavefunction coefficients with a time independent “dressed state”9 Hamiltonian matrix, H . Note that if we had not imposed the constraint that ω =ω +ω , we would not have eliminated all the explicit ds 3 1 2 time dependent terms in H . ds If we apply third order perturbation theory to the eigenstates of H (and correct for the energy shifts implied by ds the explicit time dependence in the definition of the C ’s), we get AC Stark shifts of the energy levels: k |Ω |2 |Ω |2 2(cid:60)(Ω Ω Ω ) E /¯h = ω − ab − ac + ac cb ba (3) a a ∆ω ∆ω ∆ω ∆ω b c b c |Ω |2 |Ω |2 2(cid:60)(Ω Ω Ω ) E /¯h = ω + ab + bc + ba ac cb (4) b b ∆ω ∆ω −∆ω ∆ω (∆ω −∆ω ) b b c b b c |Ω |2 |Ω |2 2(cid:60)(Ω Ω Ω ) E /¯h = ω + ac − bc + ca ab bc (5) c c ∆ω ∆ω −∆ω ∆ω (∆ω −∆ω ) c b c c c b where(cid:60)(x)istheRealpartofx. ThelasttermineachcasecomesinthirdorderandtheRealapartarisesfromthetwo time reversed paths through the intermediate states, i.e. for state a we can go a → b → c → a and a → c → b → a. These give the same denominators but numerators that are complex conjugates of each other. If one observes a transition from any of the three levels a,b,c to a forth level, d, one will find this transition shifted by the AC Stark effect of the initial state, assuming that the nonresonant Stark Shifts of state d can be neglected. Except for a chiral molecule of C symmetry (i.e. no symmetry elements except the identity operation), we cannot 1 have a closed loop of three electric dipole allowed transitions such as a → b → c → a. This is due to the fact that each symmetry element besides the identity implies that at least one component of the electric dipole projected onto theinertialaxesmustbezero. Forapairofenantiomers, theproductΩ Ω Ω willbeofoppositesignforthesame ac cb ba sets of transitions. For the case where the three coupled transitions are all rotational transition in a given vibronic state, each transition will have a sign determined by the sign of one of the components of the permanent electrical dipole moment projection onto a different principle axis of the inertial tensor. In assigning the inertial axes to the molecule, one is free to pick the sign of two of the components of the permanent dipole moment to be positive, but then the sign of the third component will be fixed by the requirement that one has a right handed coordinate system. 3 Thus, if we have a mixture of a pair of enantiomers, there will be a different sign of the third order AC Stark shifts of the energy levels while they are driven by the fields with the properties we invoked. The AC Stark effect proposed here, will produce a pair of peaks that are shifted in equal and opposite directions from the positions of the peaks without the closed cycle. For example, if one is observing the frequency of the a↔d transition, then the third order Stark shift of state a can be observed by a shift in this resonance frequency when the field at ω is switched on. Note 2 that ω is far from resonance with any transition coming directly from state a. 2 The second order Stark terms can be expected to be larger than the third for the case where perturbation theory is reasonably accurate. However, by selecting the signs of the two detunings to the two virtual levels from a state of interest, one can, for this simple three state model, arrange to cancel the second order shifts. For example, if we selectdetuningsandfieldamplitudessuchthat|Ω |2∆ω =−|Ω |2∆ω ,thenetsecondorderACStarkshiftofstate ab c ac b a will be zero, leaving only the enantiometric distinguishing third order terms. As will be demonstrated below, this does not hold when the (2J +1) spatial degeneracy of states is considered. Going to 5’th order in perturbation theory, we can write the energy of dressed state a as: (cid:18)|Ω |2 |Ω |2(cid:19)(cid:18) |Ω |2 (cid:19) E /¯h = ω − ab + ac 1+ bc a a ∆ω ∆ω ∆ω ∆ω b c b c (cid:18)2(cid:60)(Ω Ω Ω )(cid:19)(cid:18) |Ω |2 (cid:18) 1 1 (cid:19)(cid:18)|Ω |2 |Ω |2(cid:19)(cid:19) + ac cb ba 1+ cb − + ab + ac (6) ∆ω ∆ω ∆ω ∆ω ∆ω ∆ω ∆ω ∆ω b c b c b c b c Thus,the4’thordertermsarethesameforthetwoenantiomersandthatselecting|Ω |2∆ω =−|Ω |2∆ω ,cancels ab c ac b both the 2’nd and 4’th order terms. The 5’th order terms are, like the 3’rd order terms, of opposite sign for a pair of enantiomers. If we select parameters to cancel out the 2’nd order term, then ∆ω ∆ω < 0 and the 5’th order term b c will cancel part of the 3’rd order contribution. This suggests that the optimal size of |Ω |2 ∼|∆ω ∆ω |. This I have bc b c confirmed with numerical calculations with the 3×3H . Figure 1 shows the AC Stark shifts of level a as a function ds of Ω for ∆ω = 1000,Ω = 100. The two curves are for ∆ω = −1000,Ω = 100 and ∆ω = −4000,Ω = 200, bc b ab c ac c ac selected such that the second order AC Stark shift is zero. In both cases, the range of AC Stark shift (as a function of Ω ) is ±10, which is the same as the individual contributions to the second order Stark shifts from levels b and c. bc (cid:112) In both cases, the maximum shifts occur when |Ω |= |∆ω ∆ω |. bc b c II. EFFECT OF SPATIAL DEGENERACY Transitionsingasphasemoleculeshavedegeneraciesthatarisefromtherotationalsymmetryofspace.10 Thesecan be distinguished by specifying the projection of the total angular moment, with M quantum number, on some axis, usually the Z axis. Like for the static Stark effect, this degeneracy will be at least partially lifted by the second and thirdorder,chiralACStarkeffects. Ingeneral,toaccountforthiseffect,wemustusedegenerateperturbationtheory and diagonalize the (2J+1)×(2J+1) matrix of M,M(cid:48) values for the level with total rotational angular momentum quantum number J. If we probe the spectrum of a transition J ↔ J where we can neglect the AC Stark effect in f stateJ ,theACStarkshiftandsplittingpatternofstateJ willbedirectlyobservedinthespectrum. Wecanexpand f the asymmetric top states in terms of symmetric top wavefunctions using:11 J (cid:88) |J,τ,M(cid:105)= |J,K,M(cid:105) A (7) K,τ K=−J where τ = 0...2J labels the different rotational energy levels for fixed J in increasing energy order. The more commonly used asymmetric rotor rotational state labels K and K can be determined using K =floor((τ +1)/2) a b a and K =floor((2J −τ)/2) where floor(x) is the largest integer ≤ x. The transition dipole matrix element between c symmetrictopstates|J(cid:48),K(cid:48),M(cid:48) >and|J,K,M >isgivenby(E µ /2)(cid:104)J,τ,M|φ |J(cid:48),τ(cid:48),M(cid:48)(cid:105),wheretheintegralis G α α,G knownasadirectioncosinematrixelementasitisthematrixelementofthedotproductofaunitvectoralignedwith the G = X,Y,Z axis of the laboratory axes system, and the unit vector along the α = A,B,C principle axes of the momentofinertiatensorofthemolecule,inorderofincreasingmomentofinertia. Townes&Schawlow11 giveexplicit expressions for these, using a notation (cid:104)J,K,M|φ |J(cid:48),K(cid:48),M(cid:48)(cid:105) = (cid:104)J|φ |J(cid:48)(cid:105)(cid:104)J,K|φ |J(cid:48),K(cid:48)(cid:105)(cid:104)J,M|φ |J(cid:48),M(cid:48)(cid:105). I α,G J α α,G (cid:112) will use a notation where (cid:104)J|φ |J(cid:48)(cid:105) is included in the other two factors as then each factor is ≤1. Transforming to J the asymmetric top eigenfunction basis gives matrix elements: (cid:104)J,τ,M|φ |J(cid:48),τ(cid:48),M(cid:48)(cid:105)=(cid:104)J,τ|φ |J(cid:48),τ(cid:48)(cid:105)(cid:104)J,M|φ |J(cid:48),M(cid:48)(cid:105) (8) α,G α G (cid:88) (cid:104)J,τ|φ |J(cid:48),τ(cid:48)(cid:105)= A A (cid:104)J,K|φ |J(cid:48),K(cid:48)(cid:105) (9) α K,τ K(cid:48),τ(cid:48) α K,K(cid:48) 4 10 A 5 el v e L f o t hif 0 S k r a t S C -5 A -10 -4000 -2000 0 2000 4000 Ω bc FIG. 1: Plot of AC Stark shift of level a as a function of Ω for Ω =100,∆ω =1000 and the cases (a) Ω =100,∆ω = bc ab b ac c −1000 and (b) Ω = 200,∆ω = −4000. Case (a) is represented by the red (more steeply rising) curve and case (b) by the ac c blue (more slowly rising) curve. An allowed rotational transition will have only one nonzero φ transition component11 and to have a closed cycle of α threetransitions,thethreeprincipleaxesdirectionsareeachrepresentedonce. ThetripleproductofRabifrequencies that contribute to the Chiral AC Stark effect can be written: E E E Ω Ω Ω = −e−i(φ1+φ2−φ3) 1 2 3µ µ µ ×(cid:104)J ,τ |φ |J ,τ (cid:105)(cid:104)J ,τ |φ |J ,τ (cid:105)(cid:104)J ,τ |φ |J ,τ (cid:105) (10) ab bc ca 8¯h3 A B C a a γ b b b b β c c c c α a a (cid:88) (cid:104)J ,M|φ |J ,M(cid:48)(cid:105)(cid:104)J ,M(cid:48)|φ |J ,M(cid:48)(cid:48)(cid:105)(cid:104)J ,M(cid:48)(cid:48)|φ |J ,M(cid:48)(cid:48)(cid:105) a X b b Y c c Z a M(cid:48)=M(cid:48)(cid:48)±1 To be nonzero, M = M(cid:48)(cid:48) or M(cid:48)(cid:48) ± 2. Both the products of matrix elements (cid:104)J ,τ |φ |J ,τ (cid:105)(cid:104)J ,τ |φ |J ,τ (cid:105)(cid:104)J ,τ |φ |J ,τ (cid:105) and (cid:104)J ,M|φ |J ,M(cid:48)(cid:105)(cid:104)J ,M(cid:48)|φ |J ,M(cid:48)(cid:48)(cid:105)(cid:104)J ,M(cid:48)(cid:48)|φ |J ,M(cid:48)(cid:48)(cid:105) a a γ b b b b β c c c c α a a a X b b Y c c Z a are imaginary, giving real values for the products of direction cosine matrix elements. The effective Stark Hamiltonian for the state J can be written with nonzero matrix elements: a (cid:104)J ,M ±2|H |J ,M(cid:105) = B(cid:104)J ,M ±2|φ |J ,M ±1(cid:105)(cid:104)J ,M ±1|φ |J ,M(cid:105) (11) a eff a a X b b X a +C(cid:104)J ,M ±2|φ |J ,M ±1(cid:105)(cid:104)J ,M ±1|φ |J ,M(cid:105)(cid:104)J ,M|φ |J ,M(cid:105) a X b b Y c c Z a −C(cid:104)J ,M ±2|φ |J ,M ±2(cid:105)(cid:104)J ,M ±2|φ |J ,M ±1(cid:105)(cid:104)J ,M ±1|φ |J ,M(cid:105) a Z c c Y b b X a (cid:104)J ,M|H |J ,M(cid:105) = A|(cid:104)J ,M|φ |J ,M(cid:105)|2+(cid:88)(cid:104)B|(cid:104)J ,M|φ |J ,M ±1(cid:105)|2 (12) a eff a a Z c a X b ± +C(cid:104)J ,M|φ |J ,M ±1(cid:105)(cid:104)J ,M ±1|φ |J ,M(cid:105)(cid:104)J ,M|φ |J ,M(cid:105) a X b b Y c c Z a −C(cid:104)J ,M|φ |J ,M(cid:105)(cid:104)J ,M|φ |J ,M ±1(cid:105)(cid:104)J ,M ±1|φ |J ,M(cid:105)] a Z c c Y b b X a 5 with (cid:18)µ E (cid:19)2 |(cid:104)J ,τ |φ |J ,τ (cid:105)|2 A = − α 3 a a α c c (13) 2¯h ∆ω c (cid:18)µ E (cid:19)2 |(cid:104)J ,τ |φ |J ,τ (cid:105)|2 B = − β 1 a a β b b (14) 2h¯ ∆ω b E E E C = −cos(φ +φ −φ ) 1 2 3µ µ µ ×(cid:104)J ,τ |φ |J ,τ (cid:105)(cid:104)J ,τ |φ |J ,τ (cid:105)(cid:104)J ,τ |φ |J ,τ (cid:105) (15) 1 2 3 8¯h3 A B C a a γ b b b b β c c c c α a a IfwetakeB,C =0,wehavethetraditionalACStarkEffectwithM alongtheaxisofthefieldagoodquantumnumber √ and Stark Shifts of A times ((J+1)2−M2)/(J+1)(cid:112)(2J +1)(2J +3), M2/J(J+1), or (cid:0)J2−M2(cid:1)/J 4J2−1 for J = J +1,J , and J −1 respectively. If we take A,C = 0, we get the same energy levels after substituting C for c a a a A and J for J . However, in this case the states are eigenstates of J , not J . If we take C = 0, we have a double b c X Z second order AC Stark effect. Unfortunately, we can only partially cancel the two Stark Effects! Figures 2,3, and 4 show plots of the AC Stark shifts of the M components of a J = 5 state coupled by near resonant fields to states a with (J =5,J =4), (J =4,J =4), and (J =5,J =5) respectively. In each case, calculations are for B =1 and b c b c b c A is varied from -2 to +2. For the case where J (cid:54)= J the smallest variance of the eigenvalues for fixed B is when b c A=−0.302B for which the variance of the energy levels is reduced by a factor of 0.75 from the A=0 case. For both the J = J cases, the smallest variance for fixed B is for A = 0.5B with, again, the variance 0.75 times as much as b c for B =0. Thus, we can realize only a limited degree of cancelation of the second order Stark Effects, regardless the relative strengths or detunings of the two near resonant excitations. However, note that in the J = J cases, the b c optimal second order shift occurs when B and A have the same signs, which implies we can take ∆ω and ∆ω to a c have the same signs, including setting them equal to one another. If we do this, we will drive the b ↔ c transition exactly on resonance, for which we can expect a much larger effect. 2.5 2 1.5 hift 1 S k r 0.5 a St C 0 A -0.5 -1 -1.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 A FIG. 2: Plot of second order AC Stark Effect of the 11 components of a J = 5 state with two near resonant fields to states a withJ =5andJ =4andwithB=1(couplingbetweenstatesaandb)andvariationinAfrom-2to+2. (couplingbetween b c states a and c) 6 1.5 1 hift 0.5 S k r a St C 0 A -0.5 -1 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 A FIG. 3: Plot of second order AC Stark Effect of the 11 components of a J = 5 state with two near resonant fields to states a withJ =4andJ =4andwithB=1(couplingbetweenstatesaandb)andvariationinAfrom-2to+2. (couplingbetween b c states a and c) 2 1.5 1 hift 0.5 S k ar 0 t S C A -0.5 -1 -1.5 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 A FIG. 4: Plot of second order AC Stark Effect of the 11 components of a J = 5 state with two near resonant fields to states a withJ =5andJ =4andwithB=1(couplingbetweenstatesaandb)andvariationinAfrom-2to+2. (couplingbetween b c states a and c) 7 If we drive the b ↔ c transition exactly on resonance, we cannot expect to be able to use perturbation theory, but it straightforward to simply diagonalize the [2(J +J +J )+3]×[2(J +J +J )+3] matrix that includes all a b c a b c the M components of all three rotational levels. This is a generalization of the time independent H above. I use ds the notation Ω = Ω∗ = µ (cid:104)J ,τ |φ |J ,τ (cid:105)E exp(−iφ )/2h¯ and corresponding expressions for Ω and Ω , and ba ab α a a α b b 1 1 ac bc now the off-diagonal matrix elements between state |i = a,b,c,M > and |j,M > is given by Ω (cid:104)i,M |φ |j,M (cid:105), i j ij i G J where G corresponds to the polarization of radiation nearly resonate with the i ↔ j transition. One of three Ω, corresponding to the B polarized rotational transition, will be purely imaginary. It makes no difference to the results if we permute the factor of i between Ω ,Ω , or Ω . The eigenstates are made up of functions with M even (odd), ab bc ac a M even (odd), and M odd (even). c b As an example, consider the case J = 2,J = J =1,∆ω = ∆ω =1000,Ω = Ω = 100. Figure 5 gives a plot a c b b c ab ac of the five eigenvalues of states that adiabatically evolve from the J = J eigenstates as a function of Ω (which is a bc assumedtoimaginary). ItisevidentthatthreeofthefiveJ =2stateshaveACStarkshiftsthatareinsensitivetothe value of Ω but the lowest energy states are shifted down in frequency by an amount that depends strongly on the bc sign of (cid:61)(Ω ) and thus will differentially shift the transitions of a pair of enantiomers. As |Ω | becomes comparable bc bc to ∆ω = ∆ω , the chiral sensitive AC Stark shifts become large compare to the shifts when Ω = 0. This can be b c bc understood to be a consequence of the largest resonant (and thus first order) Stark Shift of the components of the mixed b and c state shifting into near resonance with the dressed state energies of the components of state a. For Ω =−1500i, the lowest states that arise from the two J =1 states are at 104.5 and 106.5, i.e., have AC Stark shifts bc of 90% of the splittings without Stark shifts. If we keep everything else the same and change ∆ω = −∆ω we get the variation of dress state eigenenergies b c shown in Figure 6. The resonance enhancement of the mixing of states J = J and J in evident in the much larger b c AC Stark Shifts in the ∆ω =∆ω case. If we take Ω to be real, the calculated energy values are symmetric under b c bc Ω ↔−Ω , i.e. therewouldbenodifferenceinthespectraofenantiomers. Figures7and8displaytheenergylevels bc bc asfunctionsofΩ forthecasewithJ =J =2,J =1and∆ω =±∆ω =1000. Again,theresonanceenhancement bc a c b b c is evident. In this case, three of the five states are strongly shifted, which matches the number of states of the J ,J b c resonance that are strongly shifted down towards the energy of J . a The AC Stark shifts for a spectrum of a J = 2 → 3 transition is displayed in Figure 9 for J = J = 1,∆ω = a b c b ∆ω =1000,Ω =100,Ω =200 and for Ω =−1200i,0,+1200i. The plot is for an X polarized probe. All Stark c ab ac bc components appear in the spectrum regardless of polarization of the probe field. Figure 10 displays the AC Stark shifts for the same transitions when ∆ω =∆ω =0, Ω =Ω =100 and Ω =−100i,0,+100i. In the cases with b c ab ac bc zero detunings of the 3-cycle transitions, the ∆ω spectrum is symmetric for Ω =0. This is lifted for Ω (cid:54)=0 but ad bc bc it is symmetric for simultaneous change in the signs of ∆ω and Ω . ad bc 8 0 -50 hift S k r -100 a St C A -150 -200 -1500 -1000 -500 0 500 1000 1500 Im(Ω ) 23 FIG.5: PlotofACStarkShiftsoftheJ =2stateasfunctionofΩ withJ =J =1,Ω =100,Ω =200,∆ω =∆ω =1000 1 bc 2 3 ab ac b c 25 20 15 hift S 10 k r a St 5 C A 0 -5 -10 -1500 -1000 -500 0 500 1000 1500 Im(Ω ) 23 FIG. 6: Plot of AC Stark Shifts of the J = 2 state as function of Ω with J = J = 1,Ω = 100,Ω = 200,∆ω = 1 bc 2 3 ab ac b 1000,∆ω =−1000 c 9 0 -50 hift S k r -100 a St C A -150 -200 -1500 -1000 -500 0 500 1000 1500 Im(Ω ) 23 FIG. 7: Plot of AC Stark Shifts of the J =2 state as function of Ω with J =1,J =2,Ω =100,Ω =200,∆ω =∆ω = 1 bc 2 3 ab ac b c 1000 35 30 25 hift 20 S k r 15 a St C 10 A 5 0 -5 -1500 -1000 -500 0 500 1000 1500 Im(Ω ) 23 FIG. 8: Plot of AC Stark Shifts of the J = 1 state as function of Ω with J = 1,J = 2,Ω = 100,Ω = 200,∆ω = 1 bc 2 3 ab ac b 1000,∆ω =−1000 c 10 4 3.5 n o siti 3 n a r T 1 2.5 + a J 2 > - a J 1.5 r o y f sit 1 n e nt 0.5 I 0 -40 -30 -20 -10 0 AC Stark Shift FIG.9: SimulatedJ =2↔3spectrumdisplayingACStarkshiftsforJ =J =1,Ω =100,Ω =200,∆ω =∆ω =1000. a 2 3 ab ac b c The spectra are offset for clarity, with (from bottom to top) Ω =−1200i,0,+1200i bc 2 1.5 1 0.5 0 -150 -100 -50 0 50 100 150 Δω ad FIG. 10: Simulated J =2↔3 spectrum displaying AC Stark shifts for J =J =1,Ω =100,Ω =100,∆ω =∆ω =0. a 2 3 ab ac b c The spectra are offset for clarity, with (from bottom to top) Ω =−100i,0,+100i bc