E = E 0 ŷe i(qx ωt), (15.1)

Transcription

1 15. ANISOTROPIC CRYSTALS 15.1 Optics of crystals The study of anisotropic materials, the subject of this chapter, is often called crystal optics, because many crystalline solids exhibit anisotropic properties. In fact, only one of the seven systems known to crystallography, the cubic crystal system, has isotropic optical properties. I ll define an anisotropic material as one having properties that depend on direction. Direction of what? is a fair question when faced with the definition I just gave. The answer is that it essentially always the direction of the electric field which controls wave propagation, charge transport, and optical absorption. Despite the large numbers of anisotropic crystals, anisotropic materials are not widely experienced in everyday life. Polarizing films are used as sunglasses and in photography; these devices are not single crystals but they are anisotropic. Polaroid sheet contains linear chains of iodine in a transparent polymer host. It absorbs light with linear polarization parallel to the chains and has maximum transmission for light with polarization rotated by 90. Research life differs from everyday life. Many of the solids that have attracted the most attention in condensed-matter physics over the past several decades are anisotropic. The list includes cuprate superconductors, Fe-based superconductors, certain topological insulators, organic linear-chain conductors, electroactive polymers, graphene, graphite, carbon nanotubes, and others Polarized light If you look around in textbooks or on the Internet, you can find statements like the following: A polarizer is an optical device that passes light of a specific polarization and blocks waves of other polarizations. This is just wrong: the linear polarizer will block (by absorption, by refraction out of the beam, or by reflection of the unwanted polarization) light with its electric field directed along one specific axis. In real polarizers the blockage is rarely 100%; however, the leakage can in the best cases be only a few parts per million. It will transmit more or less of the light in all other polarizations, with maximum transmission for light polarized perpendicular to the direction of blockage Linear polarization It should come as no surprise that I now assert that the electromagnetic waves I introduced in Chapter 2 are linearly polarized waves. The fields point in specific directions, unchanged as the waves propagate. I ll take the electric field direction as representing the plane of polarization. The plane-wave solutions are given by Eqs. 2.5 and 2.6; the direction of the electric field for waves in vacuum was defined right after Eq. 2.9 as ê for the electric field and ĥ for the magnetic field, with ê ĥ = 0. A cartoon of the fields may be seen in Fig The fields for waves in isotropic homogeneous linear media are the solutions to Eqs. 3.3a 3.3d, written as a polarized plane wave in Eq To be specific, let me orient the propagation vector q along the x axis and the electric field along the y axis. My linearly-polarized plane wave is then E = E 0 ŷe i(qx ωt), (15.1) 273

2 where E 0 is the (complex) amplitude of the field Unpolarized light Is is easy to write a formula for linearly-polarized light. It is not so easy to write one for unpolarized light. Suppose I try to add a second field, at right angles to the field in Eq. 15.1, thinking that I will then have the field along both y and z, which would seem to fill the bill of having the field along no single axis. The second field is If I add the fields in Eqs and 15.2 I get the sum to be E 2 = E 0 ẑe i(qx ωt), (15.2) E s = E 0 (ŷ + ẑ)e i(qx ωt) = 2E 0 êe i(qx ωt), (15.3) with ê =(ŷ + ẑ)/ 2. The sum is still linearly polarized; the direction of polarization (ê) is at 45 to the y and z axes. The superposition of the two linearly polarized beams has produced another linearly polarized beam. So what is unpolarized light and how to I write the fields for it? Let me start the answer by saying that the linearly polarized fields in Eqs are perfectly coherent. I can delay one by seconds or years or even the age of the Universe and still find coherent superposition with the prompt beam because the frequency and wavelength are unvarying. Unpolarized light consists of the superposition of uncorrelated fields oriented in random directions. There is enough variation in frequency that there is no coherent superposition of the constituent fields. The fields may also be partially polarized, in which case a polarizer finds power variations as it is rotated through 180 but not to zero at any orientation. I can represent partially polarized light as a superposition of unpolarized light and perfectly polarized light. The method of writing unpolarized and partially polarized light uses Stokes vectors and Mueller matrices, 2,338, matrices acting on 4-element vectors. This does the job, but is more than needed here. From the point of view of understanding and measuring the optical properties of solids, I can make use of the perfectly polarized light introduced above Circular polarization Photons have helicity; the two possible states are ± h and correspond to circularly polarized light. Circular polarization is the sum of two linearly polarized fields shifted in phase by 90.Iwrite E ± = E 0 (ŷ ± iẑ)e i(qx ωt), (15.4) where the minus sign is for right circular polarization and the plus is for left circular polarization. No actual source exists with coherence times of years, much less the age of the Universe. The initial LIGO laser interferometer had coherence times of s, about two days. 337 More typical values for very stable lasers are a few milliseconds. There are two conventions in use. These are generally described by saying that you put your right or left thumb along the direction of the wave the fingers then point in the direction of rotation of the corresponding circular polarization (right or left). The problem is that sometimes the thumb points towards the source of the radiation and sometimes away. I have used the second convention. 274

3 I ve said that the observable is the real part of the complex field. It is worth writing the real part of Eq. 15.4: E ± = E 0 [ŷ cos(qx ωt) ẑ sin(qx ωt)], The electric field vector of the circular polarized wave at fixed time follows a helical path along the direction of propagation. Fig Right circular polarized waves. The blue cosine wave is the ŷ component while the green sine wave is the ẑ component. The resultant is the red helix that rotates once per wavelength. Plane polarized light can be written as a sum of equal amplitudes of right and left circularly polarized light at various relative phases. If the relative phase is zero, I add them as 1/ 2 (ˆx + iŷ)+ 1 / 2 (ˆx iŷ) =ˆx, to get polarization along the x axis with unit amplitude. If the relative phase is π, 180, then I add them as 1/ 2 (ˆx + iŷ)+ 1 / 2 (ˆx iŷ)e iπ = 1 / 2 (ˆx + iŷ) 1 / 2 (ˆx iŷ) =iŷ. (To avoid shifting the phase of the plane polarized light, I would need to retard the phase of the left circular polarization by 90 and advance the right by the same amount.) In general, a phase difference of φ rotates the plane of polarization by θ = φ/ Elliptical polarization Looking up or down the beam, the E field of plane polarized light is a line along the polarization direction. The E field of circular polarized light is a circle. Elliptical polarized light is a linear combination of plane and circular polarizations. Looking along the beam, the E field of elliptical polarized light traces an ellipse. The ratio of major to minor axis of the ellipse is set by the field strengths of the plane and circular constituents. 275

4 15.3 Crystal symmetry Symmetry is a fundamental concept in physics. The basic idea is that any physical system which is unchanged the correct word is invariant under some operation will have observables that must respect the symmetry. Mathematically, the symmetry operations define a mathematical group; group theory 28,29 is the mathematical theory of these groups and their manipulation. Most books on solid-state physics cover crystal symmetry. I particularly like the discussion by Burns 23 but Kittel 24 and Ashcroft and Mermin 19 are also good Translational symmetry Translational symmetry is the basic symmetry that defines a crystal. Each crystal structure can be constructed by starting at some point in space and moving by the translation vector (actually a set of vectors) that defines the crystal to generate all the points of the crystal lattice. For example, I can define the NaCl lattice by starting at the center of an Na atom and then moving by a translation vector to the center of a neighboring Na atom, then by a translation vector to the center of another Na atom, and so on. Now, if I move to the nearest Cl atom, the same set of translation vectors will trace out all the Cl atoms in the crystal. The translation vector is usually written as T = na + mb + pc where n, m, andp are (positive and negative) integers and the vectors a, b, andc define the unit cell of the crystal. The set of points defined by T as the integers run over the range from minus to plus infinity is known as a Bravais lattice. The name is that of French crystallographer Auguste Bravais ( ), who proved that in three-dimensional space only fourteen distinct lattices may be constructed. To each point of the Bravais lattice one attaches a basis of atoms. (The attachments is done in an identical fashion for each point in order to preserve translational symmetry. The basis in the example above is one Na and one Cl atom. Note that there are usually two ways to set up the vectors a, b, andc. In one, they are the primitive vectors, the smallest vectors possible; therefore, they reach every point in the Bravais lattice. The second way uses the conventional vectors. These define a unit cell which is larger than the primitive cell; however, the conventional cell generally clearly exhibits the symmetry of the lattice. The difference is perhaps best illustrated with an example. Figure 15.2 shows a face centered cubic (fcc) lattice. The primitive vectors go from he origin at the front bottom corner of the cube across the three adjacent faces to the atoms nearest neighbors all in the centers of the faces. The conventional vectors go from the origin along the cube edges to the second-neighbor atoms at the adjacent corners. If the primitive vectors are If the primitive vectors are used, all the points of the Bravais lattice will be reached. If the conventional vectors are used, only a fraction are reached by movements by T; the other points are assigned to the basis of the structure. Although the need to define a basis even in monatomic crystals seems like an added complication, the conventional cell is much easier to visualize and to use to illustrate the crystal symmetry; consequently, I ll use the conventional cell. 276

5 Fig A sketch of a face centered cubic (fcc) lattice, showing the primitive translation vectors (red) and the conventional translation vectors (blue). The vectors a, b, andc define a rectangular prism with edges of length a, b, andc; there are angles α, β, andγ defined respectively in the following way. α is the angle between b and c. β is the angle between c and a. γ is the angle between a and b Seven crystal systems Every crystal belongs to one of seven crystal systems. Each system is defined in terms of the relative lengths of the unit vectors a, b, andc of the lattice unit cell, as well as the angles between these vectors. The seven crystal systems are cubic, tetragonal, orthorhombic, monoclinic, triclinic, hexagonal, and trigonal. They are illustrated in Fig Fig The seven crystal systems. Cubic. I start with the system with the highest symmetry: cubic. The cubic crystal has all axes equal (a = b = c) and all angles 90. The rectangular prism of the cubic system is, not surprisingly, a cube. Tetragonal. If I take the cubic crystal and expand (or compress) it along one axis, I get a tetragonal crystal. Here, a = b c; all angles are 90. Orthorhombic. If I take the tetragonal crystal and expand (or compress) it along a second axis, I get an orthorhombic crystal. Here, a b c; all angles are

6 Monoclinic. If now I take the orthorhombic crystal and tip one axis away from the normal to the plane defined by the other two in a way that preserves two right angles, I get a monoclinic crystal. For the monoclinic crystal, a b c; the angle β 90 while the other two are 90. Triclinic. To make triclinic, I take monoclinic and tip the other axes so that none are perpendicular to any other, making α β γ 90. It remains true that a b c also. I ve now made the lattice as low in symmetry as I can. Hexagonal. I return to the tetragonal system and open the angle γ to 120.Theformerly square ab plane is now a hexagonal figure. The hexagonal crystal has a = b c and α = β =90 ; γ = 120. Trigonal. Finally, I return to the cube and pull it along one body diagonal, rendering all angles different from 90 but all still equal. The axes also remain equal, a = b = c. Trigonal sometimes called rhombohedral. The seven crystal systems contain the fourteen Bravais lattices. The criterion for defining a Bravais lattice is that the surroundings of the lattice points be identical. The Bravais lattices are shown in many books on solid-state physics and crystallography. 19,23,24 All of the Bravais lattices have lattice points located at the corners of the prism and some have other lattice points located at symmetric locations, in the middle of a face or at the body center. For example, the cubic system has simple cubic (SC), body-centered cubic (BCC), and face-centered cubic (FCC) Bravais lattices. It is worth noting that the hexagonal Bravais lattice (the only member of that crystal system) requires that there be a lattice point at the center of the hexagon, so that the lattice points on the ab plane form a triangular mesh The dielectric tensor Crystal symmetry has a dominant effect on the dielectric function. In a general crystal, the relation between D and E is a 3 3 tensor as I wrote already in Eq. 2.11: D = ɛ E where (still in general) ɛ = ɛ xx ɛ xy ɛ xz ɛ yx ɛ yy ɛ yz ɛ zx ɛ zy ɛ zz (15.5) The off-diagonal terms allow an electric field along ˆx to produce a response (polarization or current) along ŷ or ẑ as well as ˆx. When I take into account reflection, rotation, inversion, and improper rotation symmetry operations, I will find that there are thirty-two different crystal classes. Furthermore, the combination of all possible symmetry operations results in a total of 230 different space groups. Fortunately, for considering optical anisotropy, I only need to think about the seven crystal systems. If I want to consider selection rules, such a which vibrational modes appear in an infrared spectrum. then I would need to dig deeper. It is a 3 3 tensor because there are three directions in space. 278

7 The principal axes This dielectric tensor is a symmetric tensor; hence, it is possible to find a Cartesian coordinate system in which it can be rendered diagonal; in this coordinate system, I have ɛ = ɛ x ɛ y 0. (15.6) 0 0 ɛ z The axes of the coordinate system in which ɛ is diagonal are known as the principal axes. The diagonal components are the principal components. I will write diagonal component with a single subscript, as ɛ x rather than ɛ xx. An alert reader might ask as this point: Does the fact that the dielectric function is diagonal in its principal axes mean that I have removed the situation where polarization or current are not in the same direction as the electric field? The answer is yes in the special case where the field is along one of the principal axes; in this case the polarization and current are also along the principal axis. However, if the field is in an arbitrary direction, then the answer is definitely no! Polarization and current will be in directions different from the electric field. As an example, let me take E =(E 0 / 2)(ŷ + ẑ), a field of amplitude E 0 at 45 to the y and z principal axes. The displacement vector is the dot product of ɛ and E, i.e., D = ɛ E = E 0 2 (ɛ y ŷ + ɛ z ẑ), making D not parallel to E if ɛ y ɛ z, i. e., if the y-z plane is not isotropic Complex dielectric tensor That the dielectric function is a tensor does not stop it from being a complex function, and I ll write it the same way I typically did when it was a scalar: ɛ = ɛ 1 + 4πi ω σ 1, (15.7) where both ɛ 1 and σ1 are 3 3 diagonalizable tensors with real elements. For example the conductivity (in its principal axes) is in σ 1 = σ 1x σ 1y 0. (15.8) 0 0 σ 1z The symmetry may be proved in a number of ways: from thermodynamics, using Onsager s reciprocity relation, by energy conservation. 23 The process for multiplying the dielectric tensor ɛ and the (column vector) electric field E is as follows: ɛ E = ɛ xx ɛ xy ɛ xz ɛ yx ɛ yy ɛ yz ɛ zx ɛ zy ɛ zz E x E y E z = ˆx(ɛ xx E x + ɛ xy E y + ɛ xz E z )+ŷ(ɛ yx E x + ɛ yy E y + ɛ yz E z )+ẑ(ɛ zx E x + ɛ zy E y + ɛ zz E z ) 279

8 Crystal symmetry restricts the dielectric tensor The dielectric tensor must respect the symmetry of the crystal. If the crystal is indistinguishable along ˆx and ŷ, then a field in those two directions will induce identical polarizations and currents; the corresponding elements in ɛ must be identical. There are a number of ways I could understand why for example the dielectric tensor of an orthorhombic crystal has 3 different elements and where the principal axes point. I can use group theory. I can consider the properties of tensors. What I will do here is make a (perhaps somewhat hand-waving) argument based on things I can do to the crystal that leave it unchanged. The operations I need are rotation and reflection. I ll state a sort of Murphy s law for the optics of crystals: if the dielectric tensor can possibly display anisotropy, it will do so. Murphy affects the principal-axis values: if they can be different, they will be different. Moreover, if the symmetry does not fix the directions of the principal axes, the axes will change direction as the wavelength varies. Cubic. I start again with the cubic crystal system. The cubic system has 3 orthogonal 4-fold (90 ) rotation axes along the a, b, andc crystal axes. (There are also three orthogonal mirror planes, all normal to the crystal axes, four 120 axes along the body diagonals, a center of inversion, and others, making a total of 36 cubic space groups. 23 The amount of detail can become quite large.) The three four-fold axes mean that the crystal properties are unchanged for 90 rotations; the diagonal components of the dielectric tensor must be the same. The tensor can then be written as ɛ = ɛ = ɛ1, Cubic. (15.9) where ɛ is a complex scalar and 1 is the identity tensor. But the dot product of the identity tensor and a vector is just the vector, so I have D = ɛ E = ɛ1 E = ɛe. The dielectric function of a cubic crystal is a scalar; the material is isotropic. Tetragonal. The tetragonal lattice has mirror planes in the a-b, a-c, andb-c planes. A reflection leaves unchanged a vector lying in the plane of the mirror but reverses one perpendicular to the mirror. To be unchanged under reflection, the axes must lie in the mirror or be perpendicular to it. Hence the principal axes are along the crystallographic axes. The structure has one fourfold rotation axis along the c axis. The components of the dielectric tensor for a and b are therefore identical; the c axis component is different. I then have ɛ = ɛ a ɛ a 0. Tetragonal. (15.10) 0 0 ɛ c Thedielectrictensorissaidtobeuniaxial: there is an optic axis along c and if the electric field is along c, thend = ĉɛ c E. For electric field direction ê in the a-b plane, D = êɛ a E. There are of course counterexamples; Murphy s laws are not theorems. For example, the components of the dielectric function vary with frequency and it is possible that if I graph them over a wide frequency range they will cross one or more times. At the frequencies where they cross, the anisotropy does not exist. However, as the mathematicians say, this is a set of measure zero. 280

9 Orthorhombic. The orthorhombic lattice has mirror planes in the a-b, a-c, andb-c planes. Hence the principal axes are along the crystallographic axes. There are only twofold axes, which take an axis into itself. So the three components are different, ɛ = ɛ a ɛ b ɛ c. Orthorhombic. (15.11) The orthorhombic system is biaxial. Monoclinic. Monoclinic crystals are also biaxial. The b axis is at right angles to both a and c, making one of the principal axes parallel to b. The other two axes are not fixed relative to the crystal axes and will change directions as the frequency of the electromagnetic field changes. I must re-diagonalize the tensor for each frequency of the light that encounters the crystal. ɛ = ɛ x ɛ y 0. Monoclinic. (15.12) 0 0 ɛ z with ɛ y = ɛ b. Triclinic. The only symmetry operations are identity and inversion. The crystal is biaxial with axes all changing direction with frequency. ɛ = ɛ x ɛ y ɛ z. Triclinic. (15.13) Hexagonal. The hexagonal class has a 6-fold axis perpendicular to the hexagonal or basal plane. This is as good as a 4-fold axis for forcing the response of the plane to be independent of the direction of the in-plane field. Hexagonal crystals are uniaxial with the optic axis perpendicular to the basal plane ɛ = ɛ a ɛ a ɛ c. Hexagonal. (15.14) Trigonal. The optic axis is fixed to the direction the cube was pulled or pushed. Trigonal (rhombohedral) crystals are uniaxial. ɛ = ɛ x ɛ x ɛ z. Trigonal. (15.15) 281

10 15.5 Plane-wave propagation in anisotropic materials Coordinates It may seem that I have a choice in how to align the axes of my coordinate system. On the one hand, I could, as I did on p. 11 for vacuum, align the wave vector q and the fields D and H along the Cartesian axes and let the dielectric tensor be in non-diagonal form. On the other hand, I can choose as my Cartesian system the principal axes and let the fields fall on the crystal as they may, with the wave vector and field vectors of the electromagnetic wave possibly having components along two or three axes. In practice, the choice is not there. I will have to use the principal axes as my coordinate axes Maxwell s equations revisited I will continue to write the electric field E and the magnetic field H of the light as complex plane waves, as in Eqs. 2.5 and 2.6. I will also write D(r,t)=D 0 e i(q r ωt). (15.16) I can start with Maxwell s equations given in Eqs. 2.4a 2.4d and then make use of the simplifications of local, nonmagnetic, and linear materials. I will plan to use D = ɛ E with ɛ = ɛ 1 +(4πi/ω)σ 1, so that, after substituting the plane-wave fields into Maxwell s equations, I have q D = 0 q H = 0 q E = ω c H q H = ω c D. (15.17a) (15.17b) (15.17c) (15.17d) These Maxwell s equations for anisotropic materials tell me that it is q, D, andh that determine my right-handed coordinate system. The electric field E is not in general parallel to D, so the electric field will not in general be perpendicular to q. Equation 15.17c tells me that the field E is perpendicular to H Field directions From Maxwell s equations, Eqs a 15.17d I can construct the picture of the field directions shown in Fig The coordinates here are those of the fields, q, D, andh, so that in general the dielectric tensor is not diagonal. The E field is inclined to these axes and has a component along either ±q as shown in the left or right sides of the figure. The direction of energy flow is determined by the Poynting vector S, given(asihave said before) by S =(c/8π)ree H. In vacuum and in isotropic materials the wave vector and Poynting vector point in the same direction. This is not the case in anisotropic crystals. Thewavevectorq is related to the complex refractive index in the usual way: q = N ω ˆq. (15.18) c 282

11 Fig Fields for a wave propagating along x in an anisotropic medium with principal axes at an arbitrary angle with respect to the fields. The vectors q, D, andh form a right handed set and determine the primed coordinate system. I combine Eqs c and to write H = N ˆq E. With this, and using the identity for the triple cross product, the Poynting vector is S =(ReN) c 8π E 2 [ˆq ê(ˆq ê)] (15.19) Thus S, E and H form a right handed set and S points away from q by the same angle that E points away from D. Energy flow and the ray direction directions differ, as shown in Fig The refractive index Now, I find a formal solution for the refractive index (and, hence, for the wave vector) of plane-polarized light in anisotropic crystals. The easiest way to continue is to take the cross product of q with Eq c and use the vector identity for the triple cross product on the left side and Eq d on the right side: q (q E) = ω c q H q(q E) q 2 E = ω2 c 2 D I then obtain (after using Eq , canceling ω 2 /c 2,usingD = ɛ E, changing signs, and swapping left and right sides): ɛ E = N 2 [E ˆq(ˆq E)]. (15.20) As a comment, the amplitude and the e i(q r ωt) terms in the electric field E will divide out; I could replace E with ê in Eq if I wanted to do so. The refractive index N is determined by the dielectric tensor, by the wave propagation direction ˆq, and by the electric field direction ê. Before outlining the solution for the most general anisotropic crystal, I ll look at the important case of the uniaxial crystal. An equally valid approach would be to take the curl of Eqs. 2.4c and 2.4d, use some vector identities, and obtain a wave equation for E. The use of plane waves form for the fields leads to the same equation for the refractive index that I get here. 283

12 15.6 The uniaxial crystal The tetragonal, hexagonal, or trigonal systems are uniaxial: there is a single optic axis, typically taken as being in the z direction. The dielectric tensor is given by one of Eqs , 15.14, or I ll use the notation of the first two of these. Two of the diagonal components are the the same (ɛ a ) and one is different (ɛ c ). My goal is to obtain an equation for N in terms of the components of ɛ for a given ray direction ˆq. My second goal is to find the electric field (to within an arbitrary constant). My third goal is to calculate the Poynting vector and to determine its direction relative to ˆq. The equation for N will be the equivalent for anisotropic materials of Eq. 3.10, which reads N = ɛ. It is equivalent but in no way as simple. The a-b plane of the uniaxial crystal is isotropic; the properties of the crystal do not change as it is rotated around the optic (z) axis. So I can set ˆq in the x-z plane and at an angle θ to the optic axis, so that ˆq ŷ = 0 while ˆq ẑ =cosθ and ˆq ˆx =sinθ, making ˆq =sinθˆx +cosθẑ (15.21) If I think now about the electric field direction, there seem to be two possibly distinct cases: (1) The field is perpendicular to the x-z direction (along ŷ) and (2) the field lies in the x-z plane and hence has no ŷ component. In case (1) the waves are transverse whereas in case (2) they may have an E-field component parallel to ˆq The refractive index Case 1: E perpendicular to the optic axis If E is perpendicular to the optic axis (along ŷ) thenitisperpendiculartoq also. Equation becomes ɛ a E y = N 2 E y. so that N = ɛ a. The magnetic field is given by Eq c; ˆq, ê, andĥ formaright-handed set just as in the isotropic case, Eq Moreover D(= ɛ a E) points in the same direction as E. This polarization sees no hint of the anisotropy. It is known as the ordinary ray or ordinary wave propagation: N o = ɛ a. (15.22) Case 2: E in the plane containing the optic axis and q For the extraordinary wave, q and E both have ˆx and ẑ components; neither has ŷ components. Equation thus becomes two equations: ɛ a E x = N 2 E x N 2 sin θ(ˆq E) ɛ c E z = N 2 E z N 2 cos θ(ˆq E) It is not the optical axis. The optical axis is the axis about which an optical system, such as a lens, has a certain degree of symmetry. It is complicated enough for uniaxial materials; for biaxial crystals, coming later, it is more so. 284

13 I can rewrite these in the following way (where I ve multiplied the first by sin θ and the second by cos θ): E x sin θ = N 2 sin 2 θ N 2 (ˆq E) ɛ a E z cos θ = N 2 cos 2 θ N 2 (ˆq E) ɛ c I can add these, and then ˆq E = E x sin θ + E z cos θ cancels out, leaving 1 N 2 = sin2 θ N 2 ɛ a + cos2 θ N 2 ɛ c This equation appears to be quadratic in N 2, but, when fractions are cleared, the N 4 terms cancel, leaving only the quadratic term. After a minute amount of algebra I get N(θ) = ɛa ɛ c ɛc cos 2 θ + ɛ a sin 2 θ, (15.23) where I have written explicitly that the refractive index is a function of the angle the ray makes to the optic axis. The refractive index moves smoothly between two extremes, N(θ) = { ɛa N o θ =0 ɛc N e θ =90 where N e is known as the extraordinary refractive index. Note that in fact the extraordinary wave has a refractive index N(θ) and a propagation velocity that depends on the direction of q. N(θ) equals what one sees in the literature and in tables of materials properties for N e when θ =90. This is the propagation direction for which N(θ) differsthemost from N o Ray surfaces As I have just shown, the refractive index of a uniaxial crystal has two values; the electric field directions (or polarizations) that correspond to these two values are orthogonal. One polarization, the extraordinary ray, has a refractive index that varies with the direction of q. Consequently, light traveling in the crystal moves with a speed that depends on polarization and on direction of travel. Imagine that there is a point source of light at the center of the crystal capable of emitting in all directions a short flash of unpolarized light. The flash expands out from the point with a speed set by c/n where n is the real part of the refractive index N. Inanisotropicmedium, the light will be located in an expanding spherical shell. In an anisotropic material, there are two shells of light. One, generated by ordinary rays, travels in a spherical shell at c/n o and with electric field direction perpendicular to the optic axis. The other set of rays, the extraordinary ones, give an ellipsoid of revolution about the optical axis. The distance from the origin of this ellipsoid is c/n(θ), where θ is the angle of the ray to the optic axis. The 285

14 sphere and ellipsoid touch along the optical axis (θ =0)wheren(θ) =n o = ɛ a. They have a maximum separation for propagation in the ab plane, where n o still equals ɛ a and n(θ = n e = ɛ c. There are two possibilities for the ellipsoidal shell. If n e > n o (a positive uniaxial crystal), the extraordinary waves travel more slowly than the ordinary, and the extraordinary surface is a prolate ellipsoid inside the ordinary sphere. If, in contrast, n e <n o (a negative uniaxial crystal), the extraordinary waves travel more quickly than the ordinary, and the extraordinary surface is an oblate ellipsoid surrounding the ordinary sphere. These surfaces are illustrated in Fig for both negative and positive uniaxial materials. The surfaces are shown looking down the optical axis (left side) and with the optical axis lying in the plane of the figure (right side). Field directions are also shown. Fig Ray surfaces for negative and positive uniaxial materials. The polarization vectors are indicated on the surfaces The fields Now I ll address the electric field, magnetic field, and Poynting vector in the uniaxial crystal. I used Faraday s law, Eq c, and Ampère s law, Eq d, to obtaining the equation for N(θ) and will use one of these again to find the relationship between magnetic and electric fields. What I need now is Gauss law, Eq a. I will write D =0=q D or ˆq ɛ E =0 where I have used that D = ɛ E and divided out either n o ω/c or N(θ)ω/c depending on whether the E direction is for the ordinary or extraordinary direction. The ray direction is ˆq =sinθˆx +cosθẑ. 286

15 Case 1: E perpendicular to the optic axis: Electric and magnetic fields The E field is the ŷ direction by construction. The refractive index is N o = ɛ a as given on p Then, I can find the H field using Eq c, which I write as H = N o ˆq E. The cross product is ˆx ŷ ẑ H = N o ˆq E = N o sin θ 0 cosθ 0 E y 0 = N oe y cos θˆx + N o E y sin θẑ. The magnitude of this field is H = N o E y cos 2 θ +sin 2 θ = N o E y. So the fields are in every way ordinary. The wave is a transverse electromagnetic wave with the same relation between q, E, andh as in an isotropic materials The refractive index is N = N o. The Poynting vector is parallel to the wave vector, so the waves and energy flow in the same direction Case 2: E in the plane containing the optic axis and q: electric field Light with this polarization is called the extraordinary wave. I ll start by writing a general electric field vector in the x-z plane: E = E xˆx + E z ẑ. (15.24) What I want to know is the direction of E as a function of θ; to know this field requires me to find E x and E z in terms of the overall magnitude of the electric field E. Now the displacement vector D is D = ɛ E = ɛ a E xˆx + ɛ c E z ẑ, andthengauss lawis ˆq D =0=ɛ a E x sin θ + ɛ c E z cos θ. After a couple of lines of algebra I get E z = E x ɛ a ɛ c tan θ. (15.25) Although Eq describes the ratio of the two components of the electric field, it is not very illuminating. So I ll write E = E xˆx E x ɛ a ɛ c tan θẑ and write the magnitude of the electric field E = E E. Several lines of algebra later I arrive at ɛ c cos θ E x = E. (15.26) ɛ 2 a sin 2 θ + ɛ 2 c cos 2 θ There is no constraint on the magnitude of the field. I can always make the overall electric field larger or smaller by turning up or down my lamp, or my laser, or my RF generator. 287

16 I now put this into Eq and find ɛ a sin θ E z = E. (15.27) ɛ 2 a sin 2 θ + ɛ 2 c cos 2 θ I said that the electric field is not perpendicular to the wave vector. Let me check that by calculating the dot product of ˆq and E, which I ll call E : E = ˆq E = E x sin θ + E z cos θ. I substitute Eqs and to find E = (ɛ c ɛ a )sinθcos θ E. (15.28) ɛ 2 a sin 2 θ + ɛ 2 c cos 2 θ I like Eq very much indeed. It shows that the ˆq-parallel (extraordinary) part of the electric field in uniaxial crystals is zero if (1) ɛ c = ɛ a (the material is not anisotropic) or (2) θ = 0 (the wave is propagating along the optic axis) or (3) θ =90 (the wave is propagating perpendicular to the optic axis). In all other cases, there is a component parallel to ˆq Case 2: E in the plane containing the optic axis and q: magnetic field Now that I have an equation for E, I can calculate H from Ampère s law: H = N(θ)ˆq E. I do this calculation, using Eqs (for ˆq), and (to start) (for E): H = N(θ)(E x cos θ E z sin θ)ŷ. Now, E z is negative for θ in the range (See Eq ) Thus, H is definitely in the plus ŷ direction. I substitute Eqs , (for fields), and (for N(θ)) to obtain H = ɛ a ɛ c (ɛ a sin 2 θ + ɛ c cos 2 θ) ɛ 2 a sin 2 θ + ɛ 2 c cos 2 θ Eŷ. (15.29) I do not like Eq very much but see no way to clean it up. It has the advantage of being dimensionally correct and of giving H = { No Eŷ θ =0 N e Eŷ θ =90 (15.30) where N o = ɛ a is the ordinary refractive index and N e = ɛ c is the extraordinary refractive index. As ever, it is the E-field direction that determines which refractive index is employed. 288

17 Case 2: E in the plane containing the optic axis and q: Poynting vector The next (and last) step in the treatment of wave propagation in uniaxial materials is to calculate the time-average Poynting vector, S =(c/8π)e H. I ll use E = E xˆx + E y ẑ (Eq ) and H = Hŷ. Then S = c 8π ( E zh ˆx + E x H )ẑ and, using Eqs , 15.27, S = c ɛ a sin θ ˆx + ɛ c cos θ ẑ EH (15.31) 8π ɛ 2 a sin 2 θ + ɛ 2 c cos 2 θ The magnitude of the Poynting vector is S =(c/8π)eh, (which I get either by calculating S S or by noting that the angle between the electric and magnetic fields is 90 ). The direction of the Poynting vector is more interesting than the magnitude; the unit vector in the direction of S is ŝ = ɛ a sin θ ˆx + ɛ c cos θ ẑ ɛ 2 a sin 2 θ + ɛ 2 c cos 2 θ (15.32) and ˆq ŝ cos φ = ɛ a sin 2 θ + ɛ c cos 2 θ ɛ 2 a sin 2 θ + ɛ 2 c cos 2 θ where φ is the angle between ˆq and ŝ. The angle is zero for θ =0and90 and not zero for other values. As sketched in Fig. 15.4, the Poynting vector can lie either between q and D or on the other side of q from D, depending on whether E = ˆq E is negative or positive. (See Eq ) So I ll also specify the angle α between S and the optic axis, given by cos α = ẑ ŝ. The result is cos θ cos α = (ɛa ) 2 ɛc sin 2 θ +cos 2 θ Some uniaxial materials Many materials (quartz, sapphire, ice) are uniaxial crystals. The refractive indices of a number of these materials are in Table 15.1, from Ref The refractive index is for a frequency in the yellow (590 nm, 16,950 cm 1, 2.10 ev), close to the wavelength of lowpressure sodium lamps, like those used in some street lamps. Most of these materials are reasonably transparent at this wavelength. The table gives the name, chemical formula, crystal system, n o, n e,andδn = n e n o. All three of the uniaxial crystal systems are represented. Calcite has the largest δn of the materials listed, a reason why calcite is used in many polarizers supplied by commercial vendors. 289

18 Table Uniaxial crystals, at 590 nm Material Formula Crystal n o n e Δn system barium borate BaB 2 O 4 Trigonal beryl Be 3 Al 2 (SiO 3 ) 6 Hexagonal calcite CaCO 3 Trigonal calomel Hg 2 Cl 2 Tetragonal cinnabar HgS Trigonal hematite Fe 2 O 3 Trigonal ice H 2 O Hexagonal lithium niobate LiNbO 3 Trigonal magnesium fluoride MgF 2 Tetragonal quartz SiO 2 Trigonal rutile TiO 2 Tetragonal sapphire Al 2 O 3 Trigonal sodium nitrate NaNO 3 Trigonal silicon carbide SiC Hexagonal tourmaline (a silicate) Trigonal zircon, high ZrSiO 4 Tetragonal zircon, low ZrSiO 4 Tetragonal The biaxial crystal My goal now is to obtain a general equation for N in terms of the components of ɛ without restricting the symmetry. The algebra is similar to what I did for the uniaxial crystal. First, I write Eq in components, using the form in Eq for the dielectric tensor and dotting from the left with the appropriate unit vector: ɛ x E x = N 2 E x N 2 (ˆx ˆq)(ˆq E) ɛ y E y = N 2 E y N 2 (ŷ ˆq)(ˆq E) ɛ z E z = N 2 E z N 2 (ẑ ˆq)(ˆq E) (15.33) where (ˆx ˆq) isaclumsywaytowrite thex component of the unit vector ˆq. (q x is already taken as the x component of the wave vector q.) In order to be slightly less clumsy I will define ˆx ˆq α, ŷ ˆq β, andẑ ˆq γ, noting that the dot product of two unit vectors is just the cosine of the angle between them. But maybe a bit more obscure. I m using the conventional notation for direction cosines and hoping that the use of α, β, andγ earlier for the angles in the unit cell prism does not lead to too much confusion. In my discussion of uniaxial crystals, cos θ = γ. 290

19 These three equations can be reorganized (solve for E i and then multiply by α, β, and γ respectively) to give αe x = N 2 α 2 N 2 (ˆq E) ɛ x βe y = N 2 β 2 N 2 (ˆq E) ɛ y γe z = N 2 γ 2 N 2 ɛ z (ˆq E) Now, I sum these three equations. I recognize the sum of the terms on the left as ˆq E = αe x + βe y + γe z, so that the sum is ˆq E = N 2 [ α 2 N 2 ɛ x + β2 N 2 + γ2 ɛ y N 2 ˆq E ɛ z If I was working with isotropic systems, I would be in trouble because ˆq E would be zero and N 2 would equal ɛ. But this trouble does not arise for anisotropic materials. I can cancel ˆq E from both sides and obtain the desired equation for N 2 in terms of ɛ : 1 N 2 = α 2 N 2 + β2 ɛ x N 2 + γ2 ɛ y N 2 (15.34) ɛ z Eq appears to be a cubic equation for N 2. Fortunately the N 6 term cancels on both sides because α 2 + β 2 + γ 2 = 1. The quadratic equation for N 2 is N 4 [ ɛ x α 2 + ɛ y β 2 + ɛ z γ 2] N 2 [ { ɛ x ɛ y α 2 + β 2} { + ɛ y ɛ z β 2 + γ 2} { + ɛ z ɛ x γ 2 + α 2}] (15.35) + ɛ x ɛ y ɛ z =0. Equation is of the form a(n 2 ) 2 + b(n 2 )+c = 0 so I know the solution because I learned the quadratic equation some years ago. The three terms that go into the quadratic formula are a = ɛ x α 2 + ɛ y β 2 + ɛ z γ 2 b = ɛ x ɛ y { α 2 + β 2} ɛ y ɛ z { β 2 + γ 2} ɛ z ɛ x { γ 2 + α 2} c = ɛ x ɛ y ɛ z (15.36) If ever I need it, I can write several pages of algebra and have a result. There will in fact be two results and, when they are plugged back into Eq , I will find that the two results correspond to orthogonal directions of E. The solutions depend also on direction of propagation, so that there are two ellipsoidal ray surfaces. These surfaces intersect at four points in pairs along a line joining the intersection points and the origin. At these points the two refractive indices are equal. The two lines are the two optical axes of the biaxial crystal. Or a sixth-order equation for N. But it is N 2 Iwant. There is a nice discussion at the Wolfran website on the method to preserve numerical accuracy when b 2 4ac so one of the roots involves the difference between two almost identical numbers. See mathworld.wolfram.com/quadraticequation.html. 291 ]

20 Rays and fields along principal axes Optical effects in the biaxial crystal are considerably simplified if q and E are along principal axes. Let me set E ˆx but allow (for the moment) q to be arbitrary. With this direction for the electric field, the dot product ˆq E = αe x where α is the cosine of the angle between q and ˆx. (See p. 290.) I can then rewrite Eq as ɛ x E x = N 2 E x N 2 α 2 E x 0=N 2 βαe x 0=N 2 γαe x If I suppose that α 0, then the second and third of these require β = γ =0. Butthen α = 1 and the first equation leads to E x = 0, contradicting my starting with E = E xˆx. Thus, I have to take α = 0, putting q in the y-z plane, perpendicular to E. Then, Eq become N 2 a = ɛ x (15.37) Similar arguments give N 2 b = ɛ y when the field is in the ŷ direction and N 2 c = ɛ z for field in the ẑ direction. The anisotropy is evident in the different values of N affecting wave propagation for the three field directions but the wave propagation is purely ordinary. Moreover, the arguments that led to Eq tell me that the magnitudes of the corresponding magnetic field are in each case H = N i E, just as in the ordinary case Anisotropic material with boundaries: R and T Now let me turn to the case of light incident on the surface of an anisotropic crystal. The general case is quite complicated. The surface may be cut at an arbitrary angle to the principal axes; the angle of incidence may vary, as may the polarization. A nice discussion of the general case can be found in Born and Wolf. 8 The transmission and reflection of light at an interface is as ever determined by the boundary conditions. As I ve already shown, the electric field in the anisotropic medium generally consists of two orthogonal contributions: the ordinary wave and the extraordinary wave. Each has an associated magnetic field and q vector. So I will need to decompose the incident fields along these two directions and apply the boundary conditions to both Normal incidence with field along a principal axis I will start with a simple case: The field is normally incident on a crystal that was grown (or cut and polished) with two principal axes lying in the surface and the third parallel to the normal. For definiteness, let me consider an orthorhombic crystal (biaxial) with the a axis along the normal, parallel to ˆx, andtheb and c axes in the surface, parallel to ŷ and ẑ, respectively. I ll then have q parallel to ˆx and can polarize the incoming light in the y-z plane. Suppose that the incident field is polarized E ŷ. Continuity of the electric field means that the field inside is also along ŷ and that it is traveling in a medium characterized In the absence of surface charges or currents the boundary conditions are the continuity of the tangential component of E, the normal component of D, the normal component of B, and the tangential component of H. 292

21 by N b. (See Eq and what immediately follows that.) The field inside is smaller on account of the fact that there is also a reflected wave moving out along the normal. The situation is in fact exactly the same as the one I worked out starting on p. 24. See also Fig Using the notation of the current chapter, and letting the incident field be in vacuum, I can just write r b = 1 N b. (15.38) 1+N b What I measure is the reflectance, R b = 1 N b 1+N b = (n b 1) 2 + κ 2 b (n b +1) 2 + κ 2, b 2 (15.39) with n b and κ b the real and imaginary parts of N b = ɛ b. Now I can rotate the polarizer 90 and measure R c. I can find a surface (or cut and polish one) containing the a axis, polarize the light appropriately, and measure R a These three measurements are done for the cases where the wave propagation is the simplest. But they are exactly what I want if I am interested in the dielectric function of the anisotropic material. Causality still applies, and so I can use the Kramers-Kronig method to estimate the principal components of the dielectric tensor in the same way as for isotropic solids. Transmittance is equally straightforward. The transmittance discussion in Chapter 7 apply to the polarized transmittance. The light polarized along the a axis travels at speeds and with absorption governed by N a. I say that it is simple or straightforward but there are subtleties that should be mentioned. For instance, measurements are typically made at near normal incidence not exactly normal incidence. Normally the reflectance from the Fresnel equations does not vary much with angle out to 10 or so. (See Appendix E.) In some cases, however, the reflectance for anisotropic media can show anomalies, especially with p-polarized light (light where the electric field is in the plane of incidence and thus light at non-normal incidence has a vertical component). If the dielectric function for the principal axis parallel to the surface normal direction has a zero at some frequency, anomalies may appear in the p-polarized spectrum. A good example has been given by Duarte, Sanjurjo, and Katiyar. 341 The fix is to use s-polarized light as much as possible. On p. 24 a and b represented the two media. I can turn the polarizer again and measure one of the other two. In principal, neglecting edge states, this fourth measurement is redundant. 293

22 Normal incidence with field in a plane containing an optical axis I ve just seen that reflectance and transmittance is pretty simple if the field is along a principal axis. What about the case if the field direction is between these two directions? Let me suppose that surface contains both the b and c axes, as above. The incident field has E i at an angle φ to the c axis, so that E i = E i sin(φ)ŷ + E i cos(φ)ẑ with E i the incident electric field strength. The reflected field has E r = E i r b sin(φ)ŷ + E i r c cos(φ)ẑ, makingthe reflected intensity be R (φ) =R b sin 2 (φ)+r c cos 2 (φ). One of R b and R c is the larger than the other. The reflectance interpolates smoothly between these two values as φ is varied. The principal axes correspond the the direction where the reflectance is a maximum and the (orthogonal) direction where it is a minimum. Note that the outgoing waves will not be polarized at φ. (Consider the case where R b = 0. The reflected field will be polarized always along ẑ, increasing in intensity as cos 2 (φ) as the incident polarization direction is varied.) I could place an analyzing polarizer in the optical train behind the sample also at angle φ. I would then need to take the dot product ê E r,wheree =sin(φ)ŷ +cos(φ)ẑ before squaring to get the intensity Normal incidence reflectance with an inclined optical axis I now turn to the case of a uniaxial crystal cut so that the optical axis dips below the surface. I ve simplified to a uniaxial crystal (tetragonal, hexagonal, or trigonal) and will make use of the fields that I calculated starting on p I ll continue to use a coordinate system in which the optical axis is along ẑ; consequently the incoming wave vector and fields will be inclined relative to the coordinates. A diagram of the geometry is in Fig Fig Light is normally incident on a uniaxial crystal with optic axis at an angle θ to the surface normal. The electric field of the incident light is oriented so as to excite the extraordinary wave in the crystal. 294

23 The optical axis is along the ẑ direction and the normal to the surface lies in the x-z plane, at an angle θ to the optic axis. The normal vector is then ˆn = ẑ cos(θ) +ˆx sin(θ). The wave vector direction ˆq will equal ˆn for normal incidence The ordinary wave If I orient my electric field along ŷ, not as shown in Fig but instead perpendicular to the plane of the diagram, I will have ordinary waves inside the crystal, with wave propagation and reflectance governed by N o = ɛ a according to Eq The reflected field and reflectance follow as usual The extraordinary wave If I orient my electric field in the x-z plane, as shown in Fig. 15.6, I will launch the extraordinary wave in the crystal. For simplicity (and without loss of generality in this linear medium) let me set the amplitude of the incident field to unity. Then I have The reflected field is written E i = ẑ sin(θ) ˆx cos(θ). E r = ẑr e sin(θ) ˆxr e cos(θ), where r e is the complex amplitude reflectivity. For the field inside, I can write E t = ẑe tz + ˆxE tx, where E tz is the strength of the field component along ẑ and E tx the strength along ˆx. I know these fields. I worked out the components of the extraordinary wave earlier in this chapter; they are given in Eqs and I also know the magnetic field (along ŷ in this geometry); it is in Eq The electric field and magnetic fields are continuous across the boundary, so of course all I now need to do is to equate the fields on the two sides of the boundary. I After a moderate amount of algebra I obtain r e = 1 N(θ) (15.40) 1+N(θ) where N(θ) = ɛ a ɛ c / ɛ c cos 2 θ + ɛ a sin 2 θ, as given in Eq I think that Eq is both surprising and not surprising. It is surprising that things worked out so nicely to give a solution formally like the isotropic case. It is not surprising in that I d expect a wave in the medium characterized by a particular value N(θ) of the refractive index to have a reflectance R e = 1 N(θ) 2 1+N(θ) = [n(θ) 1]2 + κ(θ) 2 (15.41) [n(θ)+1] 2 + κ(θ) 2, with n(θ) andκ(θ) the real and imaginary parts of N(θ). 295