# Optical Activity and Quantum Mechanics

Is the problem of molecular chirality been addressed completely? i.e, why one enantiomer rotates plane polarized light in one direction and the other one in the opposite direction? I would like to know whether quantum mechanics has furnished a proper explanation for this? What is the inherent property in the molecule that changes its behavior towards plane-polarized light?

• The inherent property of the molecule is the chirality - since the molecule has a rotation (with handed-ness) built in to the structure, it will interact with left vs right circularly polarized beams differently. So, it interacts with plane-polarized light by rotating the polarization. – Jon Custer Aug 3 '17 at 13:48
• Do you know the classical case of dipole scattering a la Jackson? Calculate the expected dipole moment in the ground states of each molecule. – AHusain Aug 3 '17 at 14:01

The subject was already mainstream enough in 1937 that a review appeared in Rev. Mod. Phys.  Here is a quick summary. The phase $\psi_r$ (resp. $\psi_l$) of a right (resp. left) circularly polarised wave after traversing a depth $d$ of the material is

\begin{aligned} \psi_r &= 2\pi\nu\left(t-n_r \frac{d}{c}\right)\\ \psi_l &= 2\pi\nu\left(t-n_l \frac{d}{c}\right) \end{aligned}

where the refractive indices, $n_r$ and $n_l$, are therefore different for the two polarisations. Then, one introduces the mean difference of phase $\delta = \frac{1}{2}(\psi_r - \psi_l)$: the right polarised wave is advanced by $\delta$ whereas the left polarised wave is retarded by $\delta$. The result of their superposition is then a linearly polarised wave whose plane of polarisation is turned by an angle $\delta$. The rotation per unit of path length, $\varphi=\delta/d$, is also called the rotatory power. This is the quantity to compute using Quantum Mechanics.

Assuming all molecules are in the same quantum state $a$ to simplify the exposition,

$$\varphi \propto \sum_b \frac{R_{ba}}{\nu_{ba}^2-\nu^2}$$

where $b$ is another quantum state, $\nu_{ba}$ is the frequency of the light absorbed in the transition $b\to a$, $\nu$ is the frequency of the incident light, and $R_{ba}$ is a constant. Condon shows that

$$R_{ba} = \mathop{\mathrm{Im}}\left\{\langle a|p|b\rangle\langle b|m|a\rangle\right\}$$

where $\mathrm{Im}$ stands for the imaginary part, and where $p$ and $m$ are respectively the electric and magnetic dipole moments of the molecule. Since for a molecule with an inversion center, $p=0$, only chiral molecules can lead to any effect. Since the inversion transfors $p$ into $-p$ and leaves $m$ invariant, this shows that a molecule and its enantiomer have nearly exactly opposite values for all $R_{ab}$'s, and therefore for $\varphi$. I wrote "almost" because there is theoretically a very slight difference between the quantum states of $|a\rangle$ and $|b\rangle$ for the molecule and its enantiomer, but this is not significant for the level of precision of polarisation experiments and applications.

 E. U. Condon. Theories of optical rotatory power. Rev. Mod. Phys., 9:432–457, Oct 1937.

• I don't understand why this (correct) answer got downvoted. – Emilio Pisanty Aug 4 '17 at 22:41

Yes. It boils down to the orientation-averaged polarizability of the molecule, which is a perfectly calculable property of the molecule and it has been accessible to quantum-chemistry methods for many decades.

In essence, you take the molecule in its ground state (or in some thermal ensemble, as appropriate) $|\Psi_\mathcal O\rangle$ corresponding to some orientation $\mathcal O$, you add a linearly polarized light field, you calculate the resulting time-dependent dipole moment $$\mathbf d_\mathcal O(t)=\langle\Psi_\mathcal O(t)|\hat{\mathbf d}|\Psi_\mathcal O(t)\rangle$$ using first-order time-dependent perturbation theory, and then you calculate the orientation average of $\mathbf d_\mathcal O(t)$, which ─ for chiral molecules ─ can point in some other direction than the original polarization even after the orientation averaging. This can sometimes entail some heavy calculation, depending on how big your molecule is, how many electrons you want to include in the calculation, how precise you want to be, and how OK you are with taking computational shortcuts, but it is all perfectly accessible to quantum mechanics as formulated in the late 1920s and early 1930s and there is no mystery about it.