I am not sure if this is more of a Chemistry or a Physics question, but in my Organic Chem class we discussed that chiral molecules will rotate plane polarized light. However, my professor did not discuss the mechanism at all. She also said that the only way to determine the angle of rotation was experimentally. I am really curious to know how and why this process occurs and if there is any way to calculate or even estimate the angle of rotation based on the structure of the compound.


3 Answers 3


You might start with understanding Rayleigh scattering, and then plane polarized light interacting with a simple anisotropic molecule before going onto chiral ones.

A plane polarized light wave is propagating in the direction given by the right hand rule, so let's say it's electric ($E$) field is in the $\hat{i}$ direction, the magnetic ($B$) field in the $\hat{j}$ direction so its wavevector is in the $\hat{k}$ direction.

Now let's say the light wave encounters a simple liquid crystal molecule--it's much smaller than the wavelength of the light. Forget about the chemical side-groups and other fine details, and just picture the molecule as a rod. When our light wave interacts with the rod, electrons of charge $q$ in the molecule will experience a force $Eq$ from the $E$ field of the light wave (see Lorentz force). But the electrons are bound to the molecule like a mass on a spring, so also experience a restoring force. Further, they would rather be displaced along the rod axis as opposed to away from it (the molecule's polarizability will be a tensor). This interaction of light with the electron is elastic scattering, and scattered light will have it's direction altered by the electron.

For instance, let's say that the rod is pointing at an angle $\theta$ from the $\hat{i}$ direction into the $\hat{j}$ direction. Assuming the displacement of the electron is purely along the rod's long axis, light scattered from that rod will have is polarization shifted by $\theta$.

So you don't need chiral molecules to rotate polarization of light. You could say, take some sheets of polarizing material (which is basically a sheet of rod-like molecules orientated in the same direction) and rotate them slightly with respect to each other, and these will rotate the polarization of the light passing through them.

A chiral molecular will act in a similar way--now just picture a helix of rods. Even if you have a solution of these randomly orientated, the net rotation of light will be in a given chirality. That is assuming the solution is not racemic.

If you knew the exact polarizability of your molecule of interest, you could calculate this angle of rotation. That can be a difficult thing to predict from first principles--but as one of the comments mentioned, DFT might get you that. You'd also have to know the number of your scatters, and their relative orientation, or average over all them if random. We have also assumed that the scattering is elastic--in reality there may be some inelastic scattering and absorption, and these can depend strongly on the wavelength of the light.

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    $\begingroup$ Since the anisotropic molecules are randomly oriented, the polarization shifts by each ($\theta$), will be different, making the light incident on the next molecule to have a different polarization. How then the final emergent light has a single polarization, though the individual shifts at different places (and times) were random? $\endgroup$ Dec 20, 2013 at 4:35
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    $\begingroup$ @SatwikPasani, in a polarizer, e.g. a polarizing sheet or a polarized sunglass lens, just about all the rod-like molecules have the same orientation. We wouldn't expect randomly oriented rods to polarize light. A solution of randomly oriented chiral molecules, all of the same chirality (not racemic), however, can rotate the light even through they are randomly oriented. I think you can understand this with the right hand rule too. The answer by valdo might also help with this point, particularly this quote: "Imagine now a coil spring. Note that it looks the same when you turn it." $\endgroup$ Sep 12, 2015 at 16:51

I generally agree with the above answer. But I can explain the same by simpler words IMHO.

Linear polarization may be expressed as a superposition of two opposite circular polarizations. And the polarization direction depends on the phase difference of the two circular polarizations. In simple words: imagine a situation where you add two vectors that rotate with the same angular velocity in different directions, both having the same amplitude. Their sum is the vector that oscillates in a constant direction.

Imagine now a coil spring. Note that it looks the same when you turn it. Hence If you take a "heap" of such springs, each pointing in a random direction, still they all exhibit a sort of an anisotropy. This is chirality.

If a circularly-polarized EM field propagates through the media consisting of such "springs" - it's likely that the interaction will be different for right/left handed polarizations.

Imagine that the interaction is elastic. Effectively this "slows down" the EM wave, however two circularly-polarized components are affected differently. Hence, after passing through the media, the phase difference of the two components will change. Hence the superposition of those two components will still be linearly-polarized, but the direction will change.

  • $\begingroup$ For what it's worth, Fresnel's memoir on this subject is now available in English. $\endgroup$ Jul 5, 2021 at 13:27
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    $\begingroup$ Congratulations @valdo! 3Blue1Brown used essentially your explanation to also describe this effect: youtu.be/aXRTczANuIs&t=991 $\endgroup$ Sep 3, 2023 at 2:24
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    $\begingroup$ @Polydynamical: wow, indeed. And this seems to be an interesting channel. Thanks! $\endgroup$
    – valdo
    Sep 4, 2023 at 10:35

A chiral molecule in terms of optics, is a circular birefringence. but before we get there, let's consider linear birefringence - a anisotropic material which has a different refractive index for the parallel component and perpendicular component of a polarised EM wave.

so a circular birefringence is pretty much the same but it has different refractive index for right circular and left circular components. and any linear polarisation is a superposition of right circular and left circular polarisations.

now why does it rotate the polarization state of a plane polarized light? because the two components now have a different phase difference and hence the superposition of these components now gives a linearly polarised light of different polarisation.


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