Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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).

share|improve this answer
    
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
1  
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
add comment

Your intuition is right, as far as your solution of optically active (chiral) molecules can be assimilated to an ensemble of harmonically driven two-level systems. For the two level system (left- and right handed molecules being the respective states) which starts out with only left handed molecules, the radiation drives the system to a state where left- and right handed molecules are present in proportion of 50%.

However, the fine tuning of the frequency is not important in reaching the final 50% distribution. A larger difference between the driving- and the eigenfrequency of the left-right transition leads only to a smoother, more prolonged transition to the final 50% state.

I cannot understand this by pure theory, but numerical simulation shows this .

share|improve this answer
    
The left and right-handed states are in fact degenerate. The two level system should be thought of as composed of the eigenstates of the hamiltonian/parity operator. Each of these eigenstates is either a sysmmetric or anti-symmetric superposition of the chirality states (left and right handed). –  Tarek Sep 12 '12 at 8:33
    
This are the same suppositions on which my simulation was based. If you are interested I can send you the details. –  Lupercus Sep 12 '12 at 8:45
    
I am indeed interested in the simulation details of this. If you wish, you can send to [my email]. In order to understand the theory behind this, we should first know how the chiral states are stabilized in the first place. One famous stabilization mechanism is [decoherence] (prl.aps.org/abstract/PRL/v103/i2/e023202). It seems to me that the predictions of this paper can be tested based on the setup in the question. –  Tarek 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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.