advanced35 min studyLesson 276 in Physics

Dielectric tensor and birefringence

Resolve material principal axes, polarization eigenmodes and phase retardation in an anisotropic dielectric.

Before you begin

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The question for today

What you’ll understand

Interpret the dielectric tensor in a specified orthonormal basis.
Derive two transverse indices for propagation along a principal axis.
Compute retardation while distinguishing optical and propagation axes.

An isotropic dielectric polarizes in the direction of the electric field. A crystal can respond differently along different directions, so the induced displacement need not remain parallel to the field. The dielectric tensor records that directional response. Birefringence appears when a chosen propagation direction supports two polarization eigenmodes with different phase velocities. Finding those modes requires both the material tensor and the direction of propagation; three quoted refractive indices alone do not determine an arbitrary experiment.

Name the basis before interpreting components

Consider a reciprocal transparent dielectric in a frequency band where loss is negligible. Its relative permittivity can be represented by a real symmetric tensor. An orthogonal change of coordinates diagonalizes that tensor; the resulting directions are its principal dielectric axes. In those coordinates each displacement component equals its own permittivity times the corresponding electric-field component. The diagonal entries are dimensionless material properties at the specified frequency, not automatically the indices of every possible wave traveling through the crystal.

Rotating the coordinate system generally introduces off-diagonal entries even though the physical sample has not changed. Those entries can describe an ordinary rotated crystal rather than a nonreciprocal interaction. If two principal permittivities coincide, the material is uniaxial and the remaining distinct principal direction is its optic axis. Along that optic axis, the two propagating transverse polarizations are degenerate in the ideal uniaxial model. The word axis must therefore distinguish a material symmetry direction from whichever direction the beam happens to follow.

\mathbf D=\epsilon_0\boldsymbol\epsilon_r\mathbf E,\qquad \boldsymbol\epsilon_r^{\rm principal}=\begin{pmatrix}\epsilon_x&0&0\\0&\epsilon_y&0\\0&0&\epsilon_z\end{pmatrix},\qquad \boldsymbol\epsilon_r^{\rm lab}=R\boldsymbol\epsilon_r^{\rm principal}R^{T}

D is in C/m² and E in V/m. R is an orthogonal rotation whose columns are the principal axes expressed in laboratory coordinates. Real symmetric diagonalization assumes reciprocal negligible-loss response.↔ Scroll wide equations horizontally.

Use Maxwell transversality correctly

In a uniform source-free plane wave, Gauss's law makes displacement transverse to the wavevector. In an anisotropic dielectric the electric field itself can have a longitudinal component because it need not be parallel to displacement. This distinction is essential away from principal directions. It also helps explain why the direction of energy flow can differ from the wavevector for an extraordinary wave. A ray direction and a phase-normal direction should not be interchanged without checking the mode.

A particularly clean geometry sends the wave along the principal z direction with scalar relative permeability equal to one. The two electric eigenpolarizations are then x and y, both transverse, and their squared refractive indices are the respective transverse tensor entries. The longitudinal entry does not enter these two indices. If the transverse entries coincide, there is no splitting in that direction even if the longitudinal response is different. Rotating the beam generally changes the eigenmode problem rather than merely relabeling the same two numbers.

\mathbf k\cdot\mathbf D=0,\qquad \mathbf k\parallel\hat{\mathbf z}:\quad n_x^2=\epsilon_x,\quad n_y^2=\epsilon_y\quad(\mu_r=1)

The index formulas apply to propagation along the principal z axis with positive real transverse permittivities. Only D is guaranteed transverse in a general anisotropic source-free geometry.↔ Scroll wide equations horizontally.

Turn a phase difference into polarization change

At normal incidence along a principal axis, decompose an input field into the two eigenpolarizations. Neglect interface amplitude differences or compensate for them separately. The components accumulate different phases through a plate, creating a relative retardation proportional to thickness and index difference. An equal-amplitude input can emerge linear, elliptical or circular depending on that relative phase. A quarter-wave retardation gives circular polarization only when the two transmitted component amplitudes are equal.

A half-wave plate gives a relative phase of pi and can rotate the direction of a suitable linear input. An input already aligned with one eigenaxis, however, has no second component to interfere with and remains linearly polarized. This diagnostic guards against describing every birefringent plate as a polarization rotator regardless of its orientation. The vector lab supports component decomposition only; it contains no complex phase, tensor multiplication or propagating optical eigenmode and must be interpreted at that limited level.

\Delta\phi=\frac{2\pi d}{\lambda_0}(n_y-n_x),\qquad \begin{pmatrix}E_x(d)\\E_y(d)\end{pmatrix}=\begin{pmatrix}e^{i k_0n_xd}&0\\0&e^{i k_0n_yd}\end{pmatrix}\begin{pmatrix}E_x(0)\\E_y(0)\end{pmatrix}

d and vacuum wavelength λ0 are lengths, k0 = 2π/λ0, and Δφ is the y-minus-x phasor phase in radians. This single-pass Jones description neglects loss, reflections and mode coupling.↔ Scroll wide equations horizontally.

Work it through

A quarter-wave crystal plate

A transparent illustrative crystal has principal relative permittivities $(4,9,16)$ and $\mu_r=1$ . A beam propagates along z at vacuum wavelength $600\,\mathrm{nm}$ . Find the smallest positive thickness giving a quarter-wave retardation and the output for equal in-phase x and y amplitudes, neglecting reflections.

For propagation along the principal z axis, the two transverse indices are $n_x=\sqrt4=2$ and $n_y=\sqrt9=3$ . The z entry sixteen is not the index for either of these transverse modes.
Set $\Delta\phi=\pi/2$ . The smallest positive thickness is $d=\lambda_0/[4(n_y-n_x)]=600/4=150\,\mathrm{nm}$ . The result has length units and is less than one vacuum wavelength because the illustrative birefringence is large.
The components retain equal magnitudes while their relative phasor phase becomes $+\pi/2$ . Their real electric-field components are in quadrature, yielding a circle in the transverse plane. Handedness requires an explicitly stated viewing convention.

The result

A 150 nm plate produces quarter-wave retardation and circular output under the stated equal-amplitude, reflection-free assumptions.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Resolve the eigenaxis components

Reset and set BOTH Angle = 45° and Magnitude = 1; predict equal x and y projections. Reset and set Angle = 0° and Magnitude = 1; predict zero y projection. Interpret x and y as two dielectric eigenaxes. This real vector instrument has no relative-phase control, dielectric tensor or waveplate propagation, so it illustrates the input decomposition only.

Vectorx / y projectionsTrace marker

x component: 1.393
y component: 0.975
Magnitude: 1.7

Angle35°Magnitude1.7

t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA plate can create relative retardation only between components that are actually present.

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Check your understanding

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Read the checkpoint explanation

Along the symmetry axis the two transverse principal responses coincide, so the polarization splitting vanishes. Anisotropy can remain for other propagation directions.

Your turn

Try it yourself

A plate has transverse indices $n_x=1.50$ and $n_y=1.52$ at $\lambda_0=500\,\mathrm{nm}$ . Find its smallest positive half-wave thickness. What happens to linear light entering exactly along the x eigenpolarization?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Set the relative retardation to π. Then distinguish a relative phase between two nonzero components from an overall phase of one component.

Compare with the solution

The birefringence is $0.02$ , so $d=\lambda_0/[2(0.02)]=12500\,\mathrm{nm}=12.5\,\mu\mathrm m$ . An x-polarized input has zero y amplitude throughout this ideal uncoupled plate, and therefore remains x-polarized. Its x component accumulates an overall phase, but there is no second component to make that phase into a polarization change. The same plate can rotate a linear input that has both components, so the output depends on the input orientation as well as the material and thickness.

Keep the idea

Take it with you

Principal dielectric axes belong to a stated tensor and coordinate convention.
In anisotropic source-free propagation D is transverse; E need not be.
Retardation changes polarization only through the populated eigencomponents.

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