Optical rotation along the optic axis of a uniaxial crystal

Glazer, A. M.; Cox, K. G.

doi:10.1107/97809553602060000633

International
Tables for
Crystallography
Volume D
Physical properties of crystals
Edited by A. Authier

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International Tables for Crystallography (2006). Vol. D. ch. 1.6, pp. 168-170

Section 1.6.5.5. Optical rotation along the optic axis of a uniaxial crystal

A. M. Glazer^a ^* and K. G. Cox^b ^‡

^a Department of Physics, University of Oxford, Parks Roads, Oxford OX1 3PU, England, and ^bDepartment of Earth Sciences, University of Oxford, Parks Roads, Oxford OX1 3PR, England
Correspondence e-mail: glazer@physics.ox.ac.uk

1.6.5.5. Optical rotation along the optic axis of a uniaxial crystal

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Consider a uniaxial crystal such as quartz, crystallizing in point group 32. In this case, the only dielectric tensor terms (for the effect of symmetry, see Section 1.1.4.10 ) are $[\varepsilon_{11} = \varepsilon_{22} \ne \varepsilon_{33}]$ , with the off-diagonal terms equal to zero. The equations for the dielectric displacements along the three coordinate axes [x_1] , [x_2] and [x_3] are then given, according to equations (1.6.5.14) and (1.6.5.15), by $[\eqalignno{D_1& = \varepsilon_{o}\varepsilon_{11}E_1 - i\varepsilon_{o}\left[G_{12}E_2 + G_{13}E_3\right]&\cr D_2& = \varepsilon_{o}\varepsilon_{22}E_2 - i\varepsilon_{o}\left[G_{23}E_3 - G_{21}E_1\right]&\cr D_3 &= \varepsilon_{o}\varepsilon_{33}E_3 - i\varepsilon_{o}\left[G_{31}E_1 - G_{32}E_2\right]. & (1.6.5.20)}]$ If the light is taken to propagate along [x_3] , the optic axis, the fundamental optics equation (1.6.3.14) is expressed as $[\displaylines{ \pmatrix{ \varepsilon_{11}& - iG_{12}& -iG_{13}\cr iG_{12}&\varepsilon_{11}& -iG_{23}\cr iG_{13}&iG_{23}&\varepsilon_{33}+n^2} \pmatrix{E_1\cr E_2\cr E_3 } \cr\hfill\quad= \pmatrix{ n^2&0&0\cr 0&n^2&0\cr 0&0&n^2} \pmatrix{ E_1\cr E_2\cr E_3} .\hfill(1.6.5.21)}]$ As usual, the longitudinal solution given by $[iG_{13}E_1 + iG_{23}E_2 + \varepsilon_{33}E_3 = 0 \eqno(1.6.5.22)]$ can be ignored, as one normally deals with a transverse electric field in the normal case of propagating light. For a non-trivial solution, then, $[ \left| \matrix{ \varepsilon_{11}-n^2&-iG_{12}\cr iG_{12}&\varepsilon_{11}-n^2} \right| = 0, \eqno (1.6.5.23)]$ which gives $[ n^2 = \varepsilon_{11} \pm G_{12}. \eqno (1.6.5.24)]$ This results in two eigenvalue solutions, [n_1] and [n_2] , from which one has $[ (n_1 - n_2)(n_1 + n_2) = 2G_{12} = -2G_3 \eqno (1.6.5.25)]$ and thus $[ n_1 - n_2 = {G_{12}\over {\bar n}} = - {G_{3}\over {\bar n}}. \eqno (1.6.5.26)]$ The optical rotatory power (1.6.5.19) is then given by $[\rho = {\pi G_{3}\over \lambda {\bar n}} = {\pi (n_2 - n_1)\over \lambda}. \eqno(1.6.5.27)]$ Note that in order to be consistent with the definition of rotatory power used here, $[n_2\,\gt\, n_1]$ for a dextrorotatory solution. This implies that [n_2] should be identified with [n_L] and [n_1] with [n_R] . To check this, find the eigenvectors corresponding to the two solutions (1.6.5.24).

For $[n_1^2 = \varepsilon_{11} - G_{3} ]$ , the following matrix is found from (1.6.5.21): $[ \pmatrix{G_3&iG_{3}\cr-iG_{3}&G_3} = 0, \eqno (1.6.5.28)]$ giving the Jones matrix $[(1/2^{1/2})\pmatrix{ 1\cr -i} = 0. \eqno (1.6.5.29)]$ This corresponds to a right-circularly polarized wave. It should be noted that there is confusion in the optics textbooks over the Jones matrices for circular polarizations . Jones (1948) writes a right-circular wave as $[\pmatrix{ 1\cr i},]$ but this is for a definition of right-circularly polarized light as that for which an instantaneous picture of the space distribution of its electric vector describes a right spiral. The modern usage is to define the sense of circular polarization through the time variation of the electric vector in a given plane as seen by an observer looking towards the source of the light. This reverses the definition given by Jones.

For $[n_2^2 = \varepsilon_{11} + G_{3} ]$ , the following matrix is found: $[ \pmatrix{-G_3&iG_{3}\cr -iG_{3}&-G_3} = 0, \eqno (1.6.5.30)]$ giving $[ (1/{2}^{1/2})\pmatrix{ 1\cr i} = 0. \eqno (1.6.5.31)]$

This corresponds to a left-circularly polarized wave . Therefore it is proved that the optical rotation arises from a competition between two circularly polarized waves and that in equation (1.6.5.26) [n_1 = n_R] and [n_2 = n_L] , the refractive indices for right- and left-circularly polarized light , respectively. Note that Fresnel's original idea of counter-rotating circular polarizations fits nicely with the eigenvectors rigorously determined in (1.6.5.29) and (1.6.5.31). Thus $[\rho = {\pi (n_L - n_R)\over \lambda}. \eqno(1.6.5.32)]$ Finally, for light propagating along [x_3] in quartz, one can write the direction of the wave normal as $[ {\hat s}_1 = 0; \quad {\hat s}_2 = 0;\quad {\hat s}_3 = 1 \eqno(1.6.5.33)]$ and then the gyration vector is given by $[ G = g_{i3} {\hat s}_i {\hat s}_3 = g_{13} + g_{23} + g_{33} = g_{33} \eqno(1.6.5.34)]$ as $[g_{12}= g_{23} = 0]$ in point group 32. Thus from (1.6.5.26) it is seen that $[ g_{33} = G_3 = -G_{12}, \eqno(1.6.5.35)]$ thus linking one of the components of the gyration tensor $[g_{m\ell}]$ with a component of the gyration vector $[\bf G]$ and a tensor component of [[G]] .

References

Jones, R. C. (1948). A new calculus for the treatment of optical systems. VII. Properties of N-matrices. J. Opt. Soc. Am. 38, 671–685.Google Scholar

International Tables for Crystallography (2006). Vol. D. ch. 1.6, pp. 168-170