## Abstract

We show that Mach-Zehnder interferometers (MZIs) formed from waveguides in a perfectly reflecting cladding can display manifestly different transmission characteristics to conventional MZIs due to mode recirculation and resonant reflection. Understanding and exploiting this behavior, rather than avoiding it, may lead to improved performance of photonic crystal (PC) based MZIs, for which cladding radiation is forbidden for frequencies within a photonic bandgap. Mode recirculation in such devices can result in a significantly sharper switching response than in conventional interferometers. A simple and accurate analytic model is presented and we propose specific PC structures with both high and low refractive index backgrounds that display these properties.

©2004 Optical Society of America

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### Equations (9)

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(1)
$$|{\psi}_{\pm}\u3009=\frac{1}{\sqrt{2}}(|{\psi}_{1}\u3009\pm |{\psi}_{2}\u3009).$$
(2)
$${|\Psi (0)\u3009}^{(1)}=|{\psi}_{+}\u3009$$
(2)
$${\mid \Psi \left(L\right)\u3009}^{\left(1\right)}=\left({e}^{i\beta L}\mid {\psi}_{1}\u3009+{e}^{i\left(\beta L+\phi \right)}\mid {\psi}_{2}\u3009\right)/\sqrt{2}={e}^{i\chi}\left(\mathrm{cos}(\phi \u20442)\mid {\psi}_{+}\u3009-i\mathrm{sin}(\phi \u20442)\mid {\psi}_{-}\u3009\right),$$
(3)
$$T={\mid {e}^{i\chi}\mathrm{cos}(\phi \u20442)\mid}^{2}={\mathrm{cos}}^{2}(\phi \u20442)=\frac{1}{1+{\mathrm{tan}}^{2}(\phi \u20442)}.$$
(4)
$${\mid \Psi \left(0\right)\u3009}^{\left(2\right)}=-i{e}^{i\chi}\mathrm{sin}(\phi \u20442)\left({e}^{i\beta L}|{\psi}_{1}\u3009-{e}^{i\left(\beta L+\phi \right)}\mid {\psi}_{2}\u3009\right)\u2044\sqrt{2}$$
(4)
$$=-i{e}^{2i\chi}\mathrm{sin}(\phi \u20442)\left(\mathrm{cos}(\phi \u20442)\mid {\psi}_{-}\u3009-i\mathrm{sin}(\phi \u20442)\mid {\psi}_{+}\u3009\right).$$
(5)
$$T=\frac{4{\mathrm{sin}}^{2}\left(\chi \right){\mathrm{cos}}^{2}(\phi \u20442)\u2044{\mathrm{sin}}^{4}(\phi \u20442)}{1+4{\mathrm{sin}}^{2}\left(\chi \right){\mathrm{cos}}^{2}(\phi \u20442)\u2044{\mathrm{sin}}^{4}(\phi \u20442)}.$$
(6)
$$T=\frac{4{\mathrm{cot}}^{4}(\phi \u20442)}{1+4{\mathrm{cot}}^{4}(\phi \u20442)}=\frac{1}{1+{\mathrm{tan}}^{4}(\phi \u20442)\u20444}.$$
(7)
$$\U0001d4d5=\frac{\pi \mid \mathrm{cos}(\phi \u20442)\mid}{1-\mid \mathrm{cos}(\phi \u20442)\mid}.$$