Periodic dielectrics create optical stop bands
Use Bloch modes and a bilayer transfer relation to diagnose a directional band gap.
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The question for today
What you’ll understand
- State Bloch periodicity for a lossless periodic dielectric.
- Evaluate a normal-incidence bilayer dispersion relation.
- Distinguish a directional gap, finite-stack reflection, and a complete gap.
Repeated dielectric layers can reflect strongly even when each interface is only a moderate reflector. Their repeated scattering organizes the allowed optical modes into bands. A band gap means the ideal infinite periodic system has no propagating Bloch mode in a specified frequency and wavevector range; it is not synonymous with absorption. We will analyze a lossless one-dimensional stack and keep its normal-incidence result separate from stronger claims about all directions and polarizations.
Translation symmetry organizes the modes
Consider a nonmagnetic dielectric whose real positive refractive index repeats after distance a along z. At normal incidence, with scalar transverse field and no transverse variation, Maxwell propagation reduces to a second-order wave equation with periodic coefficient. Bloch theory allows a solution to be written as a plane-wave factor times an envelope periodic over one cell. Translating by one cell therefore multiplies the field by exp(iKa). The Bloch wavevector K describes the cell-to-cell phase, not the local wavenumber inside every layer.
For real K the Bloch multiplier has unit magnitude and a bulk mode can propagate through the ideal structure. In a stop band, the permitted multipliers for a real driving frequency are exponentially growing and decaying reciprocal partners rather than unit-modulus factors. The decaying branch describes penetration from a boundary into a semi-infinite crystal. Losslessness is compatible with this evanescence because energy can be reflected at the boundary. One must not interpret the decay as irreversible conversion of optical energy to heat.
One unit cell gives a testable dispersion relation
Build a cell from two layers with indices nH and nL and physical thicknesses dH and dL. Their optical phase thicknesses are δH and δL, each index times thickness times vacuum angular wavenumber. Continuity of tangential electric and magnetic fields at the interfaces determines a two-by-two transfer matrix. In this lossless normal-incidence model its determinant is one. Its trace is twice cos(Ka), giving a compact relation between real optical frequency and allowed Bloch wavevector.
The trace criterion makes the gap diagnosis elementary: a real K exists when the right-hand side lies between minus one and plus one. Outside that interval the mode is evanescent in the periodic direction. At a design vacuum wavelength λ₀, a quarter-wave stack has nH dH=nL dL=λ₀/4. Both phase thicknesses are then π/2, making the right-hand side minus one half the index ratio plus its reciprocal. Nonzero index contrast makes its magnitude exceed one at that design frequency.
Specify what kind of gap has been established
A finite stack has boundaries and a finite number of cells, so it generally allows some tunneling through a stop-band region. Its exact reflectance also depends on the incident and substrate media and on how the stack terminates. A defect can support a localized mode within the bulk gap and create a narrow transmission feature. Neither possibility contradicts the absence of propagating Bloch modes in the perfect infinite crystal. They involve different boundary or structural problems.
A gap for normal propagation in one dimension does not establish a complete three-dimensional gap. Oblique incidence changes phase thicknesses and polarization-dependent interface admittances, and some directions can remain open. In the waves lab, Reset and set both controls to phase zero degrees and ratio one, predicting reinforcement; Reset and set phase 180 degrees and ratio one, predicting cancellation. This illustrates interference of two contributions reaching an observer. Its same-direction fixed-frequency waves do not implement multilayer back-reflection, a Bloch eigenproblem, or a band diagram.
Work it through
Diagnose a quarter-wave design
A lossless normal-incidence bilayer has nH=2, nL=1 and design vacuum wavelength 600 nm. Choose quarter-wave layer thicknesses, calculate the period, and evaluate cos(Ka) at that wavelength to decide whether a propagating Bloch mode exists.
Set each optical thickness to 600/4=150 nm. The high-index physical thickness is 150/2=75 nm, and the low-index thickness is 150/1=150 nm.
Add the physical layer thicknesses to obtain period a=225 nm. Both layer phase thicknesses are π/2 at the design wavelength, so their cosine product vanishes and their sine product is one.
The trace relation gives cos(Ka)=−(2+1/2)/2=−1.25. No real K has this cosine, so the chosen frequency lies in the normal-incidence stop band of the infinite periodic model.
The design uses 75 nm and 150 nm layers with period 225 nm, and cos(Ka)=−1.25 establishes a normal-incidence gap for the ideal infinite stack.
The living notebook
Explore the model
Interactive lab · waves
Phase organizes repeated contributions
Reset and set BOTH controls to phase 0° and ratio 1; predict reinforcement. Reset and set phase 180° and ratio 1; predict cancellation. The model adds two same-direction waves and does not calculate reflected fields or optical bands.
- Sum at x = 0
- 0.866
- Resultant amplitude
- 1.732
- Phase difference
- 60°
Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). 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 forInterference is part of the mechanism, while a band-gap claim requires the periodic Maxwell or transfer-matrix calculation.
Pause & reason
Check your understanding
Your turn
Try it yourself
Replace both layers by the same refractive index 1.5 while retaining a quarter-wave design at 600 nm. Find each physical thickness, the period, and cos(Ka). Explain why this value does not demonstrate an open band gap.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
An index ratio of one makes the two layers optically identical; the trace relation reduces to the cosine of their total phase.
Compare with the solution
Each thickness is 600/(4×1.5)=100 nm, and the period is 200 nm. At the design frequency cos(Ka)=−(1+1)/2=−1. This lies on the allowed real-cosine boundary rather than outside it. The medium is uniform: labeling two identical slabs as a cell merely folds its dispersion into a chosen Brillouin zone and opens no gap.
Keep the idea
Take it with you
- Bloch modes organize fields in a periodic dielectric.
- A bilayer trace outside the real-cosine interval diagnoses a directional stop band.
- Lossless gaps cause evanescence and reflection; finite boundaries and complete gaps require further analysis.
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