# Driving a solution of optical isomer molecules with the resonant frequency

What happens when we drive a solution of optical isomer molecules (enantiomers) with a microwave radiation in resonance with the tunneling frequency of the molecules (the frequency of the transition between the eigenstates of the Hamiltonian)? I expect it will become a racemic mixture. Is that correct?

Update: Any reference for an experiment that does that is appreciated.

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Usually, optical isomers are defined when tunneling can be neglected (that is, when the torsional barrier between the isomers is very large that the isomers can be isolated and are considered stable). Then the ground state of the Hamiltonian is degenerate. From Quantum Mechanics it is a question of taste weather you choose the localized basis or the symmetric basis to define your states $\psi_\pm = \psi_R \pm \psi_L$, where $\psi_{R/L}$ are the ground state eigenfunctions for the right/left isomers and $\psi_\pm$ are the symmetric (antisymmetric) eigenfunctions.

However, due to decoherence (and the process called einselection by Zurek) in fact the molecules in the ground states are described by the localized states (the isomer states). You can use a laser to drive the population from the ground state to an excited state with energy above the torsional barrier (or with large tunneling) and then wait for torsion to occur and then dump again the population to the ground state of the other isomer. Usually when you do this you move $\psi_R \leftrightarrow \psi_L$ so that if you start in a racemic mixture you end up in a racemic mixture, but if you start in a single isomer, you can convert to the other isomer. However, there are also procedures to "break" this symmetry. For references, see Shapiro et al, Phys. Rev. Lett. 84, 1669 (2000) and Phys. Rev. Lett. 90, 033001 (2003).

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Aren't $\psi_{L/R}$ the chiral states which are not eigenstates of the Hamiltonian, (and hence the paradox of Hund), while $\psi_\pm$ are the true eigenstates (which are not degenerate)? –  Tarek Sep 12 '12 at 11:47
If the barrier is much higher than the energy of the ground state of the isomers (and its width is also large) you can neglect tunneling. Then $E(\psi_+) = E(\psi_-)$ to the best resolution available. Since the two states are degenerate to all practical purposes, whatever superposition of them is also an eigenstate of the Hamiltonian. However, because of interactions with the environment already at very low temperature, it is the localized states the only superposition states that survive. –  perplexity Sep 12 '12 at 12:39
Sorry folks for communicating the wrong result: the system simply oscillates between the two states, there is no 50%-50% separation. This came because I forgot an $\sqrt{-1}$ in the time evolution. To model the situation you need to address the full Master Equation with decoherence included (Lindblad Eq.). –  Lupercus Sep 13 '12 at 8:33