Exact envelope solitons in topological Floquet insulators

Subimal Deb; Subimal Deb; Dheerendra Singh; Munmun Chakraborty; Prasanta K. Panigrahi; Avinash Khare

doi:10.1364/OL.495156

Optics Letters
Vol. 48,
Issue 19,
pp. 4997-5000
(2023)
•https://doi.org/10.1364/OL.495156

Exact envelope solitons in topological Floquet insulators

Subimal Deb, Dheerendra Singh, Munmun Chakraborty, Prasanta K. Panigrahi, and Avinash Khare

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Subimal Deb, Dheerendra Singh, Munmun Chakraborty, Prasanta K. Panigrahi, and Avinash Khare, "Exact envelope solitons in topological Floquet insulators," Opt. Lett. 48, 4997-5000 (2023)

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Abstract

The existence of new types of four-wave mixing Floquet solitons were recently realized numerically through a resonant phase matching in a photonic lattice of type-I Dirac cones; specifically, a honeycomb lattice of helical array waveguides imprinted on a weakly birefringent medium. We present a wide class of exact solutions in this system for the envelope solitons in dark–bright pairs and a “molecular” form of bright–dark combinations. Some of the solutions, red or blue detuned, are mode-locked in their momenta, while the others offer a spectrum of allowed momenta subject to constraints amongst the system and solution parameters. We show that the characteristically different solutions exist at and away from the band edge, with the exact band edge possessing a periodic pair of sinusoidal excitations akin to that of two-level systems apart from localized solitons. These could have possible applications for designing quantum devices.

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	$ϕ_{1}$	$ϕ_{2}$	Constraints
H1.	$A s e c h (β Y)$	$B \tanh (β Y)$	$β^{2} = \frac{δ}{b_{0}^{″}}$ , $B^{2} = \frac{Ω}{χ}$ , $A^{2} = \frac{Ω - δ}{χ}$
H2.	$A \tanh (β Y)$	$B s e c h (β Y)$	$β^{2} = - \frac{δ}{b_{0}^{″}}$ , $B^{2} = \frac{Ω + δ / 2}{χ}$ , $A^{2} = \frac{Ω - δ / 2}{χ}$
H3.	$A {s e c h}^{2} (β Y)$ $+ B$	$D s e c h (β Y) \tanh (β Y)$	$β^{2} = - δ / b_{0}^{″}$ , $B^{2} = - 4 δ / χ$ , $D^{2} = A^{2}$ , $Ω = - 7 δ / 2$ , $A^{2} = - 9 δ / χ$ , $A B = 6 δ / χ$
H4.	$D s e c h (β Y) \tanh (β Y)$	$A {s e c h}^{2} (β Y)$ $+ B$	$β^{2} = δ / b_{0}^{″}$ , $D^{2} = A^{2}$ , $B^{2} = 4 δ / χ$ , $Ω = 4 δ$ , $A^{2} = 9 δ / χ$ , $A B = - 6 δ / χ$
H5.	$A {s e c h}^{2} (β Y)$	$D s e c h (β Y) \tanh (β Y)$	$D^{2} = A^{2} = - δ / χ$ , $β^{2} = δ / (3 b_{0}^{″})$ , $Ω = - δ / 6$
H6.	$A s e c h (β Y) \tanh (β Y)$	$D {s e c h}^{2} (β Y)$	$D^{2} = A^{2} = δ / χ$ , $β^{2} = - δ / (3 b_{0}^{″})$ , $Ω = 2 δ / 3$

	$ϕ_{1}$	$ϕ_{2}$	Constraints
D1.	$A \tanh (β Y)$	$α ϕ_{1}$	$(1 + α^{2}) A^{2} = b_{0}^{''} β^{2} / χ$ , $Ω = b_{0}^{''} β^{2}$
D2.	$A s e c h (β Y)$	$α ϕ_{1}$	$(1 + α^{2}) A^{2} = - b_{0}^{''} β^{2} / χ$ , $Ω = - b_{0}^{''} β^{2} / 2$
D3.	$A \cos (β Y)$	$A \sin (β Y)$	$χ A^{2} + b_{0}^{''} β^{2} / 2 = Ω$

	$ϕ_{1}$	$ϕ_{2}$	Constraints
H1.	$A s e c h (β Y)$	$B \tanh (β Y)$	$β^{2} = \frac{δ}{b_{0}^{″}}$ , $B^{2} = \frac{Ω}{χ}$ , $A^{2} = \frac{Ω - δ}{χ}$
H2.	$A \tanh (β Y)$	$B s e c h (β Y)$	$β^{2} = - \frac{δ}{b_{0}^{″}}$ , $B^{2} = \frac{Ω + δ / 2}{χ}$ , $A^{2} = \frac{Ω - δ / 2}{χ}$
H3.	$A {s e c h}^{2} (β Y)$ $+ B$	$D s e c h (β Y) \tanh (β Y)$	$β^{2} = - δ / b_{0}^{″}$ , $B^{2} = - 4 δ / χ$ , $D^{2} = A^{2}$ , $Ω = - 7 δ / 2$ , $A^{2} = - 9 δ / χ$ , $A B = 6 δ / χ$
H4.	$D s e c h (β Y) \tanh (β Y)$	$A {s e c h}^{2} (β Y)$ $+ B$	$β^{2} = δ / b_{0}^{″}$ , $D^{2} = A^{2}$ , $B^{2} = 4 δ / χ$ , $Ω = 4 δ$ , $A^{2} = 9 δ / χ$ , $A B = - 6 δ / χ$
H5.	$A {s e c h}^{2} (β Y)$	$D s e c h (β Y) \tanh (β Y)$	$D^{2} = A^{2} = - δ / χ$ , $β^{2} = δ / (3 b_{0}^{″})$ , $Ω = - δ / 6$
H6.	$A s e c h (β Y) \tanh (β Y)$	$D {s e c h}^{2} (β Y)$	$D^{2} = A^{2} = δ / χ$ , $β^{2} = - δ / (3 b_{0}^{″})$ , $Ω = 2 δ / 3$

	$ϕ_{1}$	$ϕ_{2}$	Constraints
D1.	$A \tanh (β Y)$	$α ϕ_{1}$	$(1 + α^{2}) A^{2} = b_{0}^{''} β^{2} / χ$ , $Ω = b_{0}^{''} β^{2}$
D2.	$A s e c h (β Y)$	$α ϕ_{1}$	$(1 + α^{2}) A^{2} = - b_{0}^{''} β^{2} / χ$ , $Ω = - b_{0}^{''} β^{2} / 2$
D3.	$A \cos (β Y)$	$A \sin (β Y)$	$χ A^{2} + b_{0}^{''} β^{2} / 2 = Ω$

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Optics Letters