Example of LiNbO3

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

doi:10.1107/97809553602060000633

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

pdf | chapter contents | chapter index | related articles

International Tables for Crystallography (2006). Vol. D. ch. 1.6, pp. 172-173

Section 1.6.6.2. Example of LiNbO₃

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

1.6.6.2. Example of LiNbO₃

| top | pdf |

In order to understand how tensors can be used in calculating the optical changes induced by an applied electric field, it is instructive to take a particular example and work out the change in refractive index for a given electric field. LiNbO₃ is the most widely used electro-optic material in industry and so this forms a useful example for calculation purposes. This material crystallizes in point group [3m] , for which the electro-optic tensor has the form (for the effect of symmetry, see Section 1.1.4.10 ) (with [x_1] perpendicular to m) $[ \pmatrix{ 0&-r_{22}& r_{13}\cr 0&r_{22}&r_{13}\cr 0&0&r_{33}\cr 0&r_{51}&0\cr r_{51}&0&0\cr -r_{22}&0&0}, \eqno (1.6.6.2)]$ with r₁₃ = 9.6, r₂₂ = 6.8, r₃₃ = 30.9 and r₅₁ = 32.5 × 10⁻¹² mV⁻¹, under the normal measuring conditions where the crystal is unclamped.

Calculation using dielectric impermeability tensor . Suppose, for example, a static electric field [E_3^0] is imposed along the [x_3] axis. One can then write $[\pmatrix{ \Delta\eta_1\cr \Delta\eta_2\cr \Delta\eta_3\cr \Delta\eta_4\cr \Delta\eta_5\cr \Delta\eta_6\cr } = \pmatrix{ 0&-r_{22}& r_{13}\cr 0&r_{22}&r_{13}\cr 0&0&r_{33}\cr 0&r_{51}&0\cr r_{51}&0&0\cr -r_{22}&0&0 }\pmatrix{ 0\cr 0\cr E_3^0} = \pmatrix{ r_{13}E_3^0\cr r_{13}E_3^0\cr r_{33}E_3^0\cr 0\cr 0\cr 0\cr }. \eqno (1.6.6.3)]$ Thus $[\eqalignno{\Delta\eta_1 & = r_{13}E^0 = \Delta\eta_2 &\cr \Delta\eta_3 & = r_{33}E^0 &\cr \Delta\eta_4 & = \Delta\eta_5 = \Delta\eta_6 = 0. & (1.6.6.4)}]$ Since the original indicatrix of LiNbO₃ before application of the field is uniaxial, $[\eqalignno{\eta_1 & = {1\over n_o^2} = \eta_2 &\cr \eta_3 & = {1\over n_e^2}, & (1.6.6.5)}]$ and so differentiating, the following are obtained: $[\eqalignno{\Delta\eta_1 & = \Delta\eta_2 = - {2\over n_o^3}\Delta n_o &\cr \Delta\eta_3 & = - {2\over n_e^3}\Delta n_e. & (1.6.6.6)}]$ Thus, the induced changes in refractive index are given by $[\eqalignno{\Delta n_1 & = \Delta n_2 = - {n_o^3\over 2} r_{13}E_3^0&\cr \Delta n_3 & = - {n_e^3\over 2 }r_{33}E_3^0. &(1.6.6.7)}]$ It can be seen from this that the effect is simply to change the refractive indices by deforming the indicatrix, but maintain the uniaxial symmetry of the crystal. Note that if light is now propagated along, say, [x_1] , the observed linear birefringence is given by $[ (n_e - n_o) - {\textstyle{1\over 2}} (n_e^3r_{33} - n_o^3r_{13}) E_3^0 .\eqno (1.6.6.8)]$

If, on the other hand, the electric field [E_2^0] is applied along [x_2] , i.e. within the mirror plane, one finds $[\eqalignno{\Delta\eta_1 & = - r_{22}E_2^0&\cr \Delta\eta_2 & = + r_{22}E_2^0&\cr \Delta\eta_4 & = + r_{51}E_2^0&\cr \Delta\eta_3 & = \Delta\eta_5 = \Delta\eta_6 = 0.& (1.6.6.9)}]$ Diagonalization of the matrix $[ \pmatrix{ \Delta\eta_1 & 0&0 \cr 0&\Delta\eta_2 & \Delta\eta_4 \cr 0&\Delta\eta_4 &\Delta\eta_3 }\eqno (1.6.6.10)]$ containing these terms gives three eigenvalue solutions for the changes in dielectric impermeabilities: $[\displaylines{\quad(1) \hfill -r_{22}E_2^0 \hfill\cr \quad(2) \hfill {r_{22}+(r_{22}^2+4r_{51}^2)^{1/2}\over 2} E_2^0 \hfill\cr \quad(3) \hfill {r_{22}-(r_{22}^2+4r_{51}^2)^{1/2}\over 2} E_2^0. \hfill \cr\hfill(1.6.6.11)}]$ On calculating the eigenfunctions, it is found that solution (1) lies along [x_1] , thus representing a change in the value of the indicatrix axis in this direction. Solutions (2) and (3) give the other two axes of the indicatrix: these are different in length, but mutually perpendicular, and lie in the [x_2 x_3] plane. Thus a biaxial indicatrix is formed with one refractive index fixed along [x_1] and the other two in the plane perpendicular. The effect of having the electric field imposed within the mirror plane is thus to remove the threefold axis in point group 3m and to form the point subgroup m (Fig. 1.6.6.1).

Figure 1.6.6.1 | top | pdf |

(a) Symmetry elements of point group 3m. (b) Symmetry elements after field applied along [x_2] . (c) Effect on circular section of uniaxial indicatrix.

Relationship between linear electro-optic coefficients $[r_{ijk}]$ and the susceptibility tensor $[\chi_{ijk}^{(2)}]$ . It is instructive to repeat the above calculation using the normal susceptibility tensor and equation (1.6.3.14). Consider, again, a static electric field along [x_3] and light propagating along [x_1] . As before, the only coefficients that need to be considered with the static field along [x_3] are $[\chi_{113} = \chi_{223}]$ and $[\chi_{333}]$ . Equation (1.6.3.14) can then be written as $[\displaylines{ \pmatrix{\varepsilon_1+\varepsilon_o\chi_{13}E_3^0+n^2&0&0\cr 0&\varepsilon_1+\varepsilon_o\chi_{13}E_3^0&0\cr 0&0&\varepsilon_3+\varepsilon_o\chi_{33}E_3^0} \pmatrix{ E_1\cr E_2\cr E_3}\cr\hfill\quad = \pmatrix{ n^2&0&0\cr 0&n^2&0\cr 0&0&n^2 }\pmatrix{ E_1\cr E_2\cr E_3},\hfill (1.6.6.12)}]$ where for simplicity the Voigt notation has been used. The first line of the matrix equation gives $[(\varepsilon_1+\varepsilon_o\chi_{13}E_3^0+n^2) E_1 = n^2E_1. \eqno (1.6.6.13)]$ Since only a transverse electric field is relevant for an optical wave (plasma waves are not considered here), it can be assumed that the longitudinal field [ E_1 = 0] . The remaining two equations can be solved by forming the determinantal equation $[ \left| \matrix{\varepsilon_1+\varepsilon_o\chi_{13}E_3^0-n^2&0\cr 0&\varepsilon_3+\varepsilon_o\chi_{33}E_3^0-n^2 }\right| = 0, \eqno (1.6.6.14)]$ which leads to the results $[ n_1^2 = \varepsilon_1+\varepsilon_o\chi_{13}E_3^0 \;\hbox{ and } \; n_2^2 = \varepsilon_3+\varepsilon_o\chi_{33}E_3^0. \eqno (1.6.6.15)]$ Thus $[ n_1^2 = n_o^2+\varepsilon_o\chi_{13}E_3^0 \;\hbox{ and } n_2^2 = n_e^2+\varepsilon_o\chi_{33}E_3^0, \eqno (1.6.6.16)]$ and so $[(n_1 - n_o)(n_1 + n_o) = \varepsilon_o\chi_{13}E_3^0 \;\hbox{ and }\; (n_2- n_e)(n_2 + n_e) = \varepsilon_o\chi_{33}E_3^0, \eqno(1.6.6.17)]$ and since $[n_1 \simeq n_o]$ and $[n_2 \simeq n_e]$ , $[n_1 - n_o = {\varepsilon_o\chi_{13}E_3^0\over 2n_o} \;\hbox{ and }\; n_2- n_e = {\varepsilon_o\chi_{33}E_3^0\over 2n_e}. \eqno(1.6.6.18)]$ Subtracting these two results, the induced birefringence is found: $[(n_e - n_o) - {\textstyle{1\over 2}}\left({\varepsilon_o\chi_{33}\over n_e} - {\varepsilon_o\chi_{13}\over n_o}\right) E_3^0 .\eqno(1.6.6.19)]$ Comparing with the equation (1.6.6.8) calculated for the linear electro-optic coefficients, $[(n_e - n_o) - {\textstyle{1\over 2}}\left(n_e^3r_{33} - n_o^3r_{13}\right) E_3^0, \eqno (1.6.6.20)]$ one finds the following relationships between the linear electro-optic coefficients and the susceptibilities $[\chi^{(2)}]$ : $[ r_{13} = {\varepsilon_o\chi_{13}\over n_o^4} \;\hbox{ and } \; r_{33} = {\varepsilon_o\chi_{33}\over n_e^4}. \eqno(1.6.6.21)]$