Can Crystal Symmetry Explain the Lack of Optical Activity in Quartz?

[Paul Colby]

Hi,

I've been looking at the optics of $\alpha$-quartz which comes in two parities, left and right. Quartz is optically active which means that the plane of a linearly polarized beam propagating along the optic axis is rotated by an angle proportional to the distance traveled. I would like to express this in terms of tensor constitutive parameters, $\epsilon_{nm}$ and $\mu_{nm}$. Here is where this runs aground. Crystal symmetry limits the form of the permittivity tensor to the form,

$\epsilon_{nm} = \left(\begin{array}{ccc}\epsilon_{a}&0&0\cr 0&\epsilon_{a}&0\cr 0 & 0 & \epsilon_{b}\end{array}\right)$​

The very same symmetry arguments would require,

$\mu_{nm} = \left(\begin{array}{ccc}\mu_{a}&0&0\cr 0&\mu_{a}&0\cr 0 & 0 & \mu_{b}\end{array}\right)$​

Okay, for a beam propagating along the optic ($z$-axis) no optical activity can be generated from constitutive relations of this form? What gives?

 

[DrDu]

You have to take spatial dispersion into account. I.e., the electric polarisation does not only depend on the field at the same point in space but also on the field in nearby points. In Fourier space, this means that epsilon on the wavevector k. To obtain optical activity, the first term in a Taylor expansion of epsilon in powers of k has to be taken into account, i.e.,
$\epsilon_{ij}(k)=\epsilon^0_{ij}+\epsilon^{(1)}_{ijl} k_l+\ldots$. Magnetic effects can be taken into account with the quadratic terms in k, so no need for a separate tensor $\mu$.
However, if you really want so, you can alternatively introduce an additional tensor which describes the dependence of $P$ on $B$ to get optical activity.
Details can be found e.g. in Landau Lifshetz, Electrodynamics of continua.

 

[Paul Colby]

Thanks, I even have that book. Time to use it. Ah, $\epsilon_{ijl}$ clearly has different symmetry constrains.