Optics 099 · Ring-resonator lattice

Topological Photonic Edge State

Why does light on a topological ring lattice flow around a defect instead of bouncing back? At that frequency the lattice has states only along its edges, and on each edge they all run one way. Light that meets a detuned or missing ring cannot turn back and cannot enter the gapped interior, so it flows around the defect. In a plain chain of rings the same defect sends most of the light back. The protection lasts while each ring's clockwise and counter-clockwise light stay uncoupled.

Send a pulse into a ring-resonator lattice with synthetic flux and into a plain ring chain. Edge light flows around a defect that the chain reflects.

Ring lattice: pulse energy reaching the output $\text{—}$

Ring lattice: energy sent back to the input waveguide $\text{—}$

Plain chain: pulse energy reaching the output $\text{—}$

Plain chain: energy sent back to the input waveguide $\text{—}$

Lattice light in the middle, two or more rings from every edge $\text{—}$

Lattice output delay after the input peak $\text{—}$

Edge light leaving the input runs $\text{—}$

Ring lattice: energy lost in the rings $\text{—}$

Plain chain: energy lost in the rings $\text{—}$

Ring lattice, 24 random chips with spread one coupling rate: steady delivered fraction $\text{—}$

Plain chain, the same 24 random chips: steady delivered fraction $\text{—}$

Replay frames recorded $\text{—}$

Optics 099

Why does light on a topological ring lattice flow around a defect instead of bouncing back?

Take the defect ring outRemoves the ring and its couplings from the lattice and from the chain.Drive frequency

\omega

-1.8\,J

Synthetic flux per square of rings

\alpha

\alpha=0.25

Defect ring detuning

U

5\,J

Random spread of ring resonances

\sigma

0\,J

Defect position on the edge

\text{Ring 3 past the input}

Loss rate of every ring

\kappa_{\mathrm{in}}

0.02\,J

Random seed

1

Each change sends a new pulse through both chips. The replay shows the light stored in each ring as it happens.

Coupled-mode model of the ring-resonator lattice proposed by Hafezi, Demler, Lukin and Taylor (2011): a 10 by 10 square lattice of identical ring resonators, each holding one circulating mode and coupled to its four neighbours with the same strength, the coupling rate. Hopping one ring along a row adds a phase that changes by the same step from one row to the next, so every square of four rings encloses the same synthetic flux, set as a fraction of a full turn of phase. Frequencies are measured from the ring resonance and, like rates, given in units of the coupling rate; times are in units of its inverse. Every ring loses light at 0.02 times the coupling rate, the value Hafezi et al. (2011) used; the rates measured on the 2013 chip give about 0.06. Waveguides couple at six times the coupling rate to the second and ninth rings of the edge nearest the viewer. The plain chain has 20 rings in a line, with waveguides at twice the coupling rate on its end rings, and is always driven at its band centre. The defect ring has its resonance shifted by the set amount, or is removed with all its couplings. Disorder shifts every ring resonance by an independent Gaussian amount. The input is a Gaussian pulse whose half-width at 1/e of its peak is 4 time units, integrated with fourth-order Runge–Kutta in steps of 0.05 for 90 time units. Delivered and returned fractions are the energies leaving by the output waveguide and by the input waveguide’s through port, divided by the pulse energy. The steady spectra and the 24 random chips at each disorder level use the exact continuous-wave solution. The bands come from a strip 24 rings tall and infinitely long. Not included: coupling between the clockwise and counter-clockwise light in each ring, which would let light turn back; fabrication spread in the couplings; nonlinearity and noise. Ring size and spacing in the scene are schematic.

Physics tutorial

Send light around a defect

BackgroundIn an ordinary chain of coupled ring resonators, light hops from ring to ring. A ring tuned off resonance acts like a wall: most of the light turns back and leaves by the input waveguide.

Why it mattersHafezi and colleagues proposed coupling the rings through links that add a phase, so that light going round any square of four rings picks up the same extra phase, as an electron in a magnetic field does. The lattice then guides light along its edge in one direction only. The same pulse goes into such a lattice and into a plain chain with the same defect.

Start with the essentials

Focus question: Why does light on the lattice edge flow around the defect while the chain sends it back?
One-sentence intuition: With the synthetic flux $\alpha=1/4$ the lattice has band gaps between $|\omega|\approx1.06\,J$ and $2.62\,J$ . Inside a gap the only states are edge states, and on a given edge they all move the same way. Light that meets the defect cannot turn back, because no state on that edge runs backward at that frequency, and it cannot enter the bulk. It goes round the defect and carries on.

Core mathematical model

Rings coupled with a synthetic flux

\begin{gathered}H=-J\sum_{x,y}\big(e^{-2\pi i\alpha y}\,a^{\dagger}_{x+1,y}a_{x,y}\\+\,a^{\dagger}_{x,y+1}a_{x,y}+\mathrm{h.c.}\big)\end{gathered}

Here $a_{x,y}$ is the light amplitude in the ring at column $x$ and row $y$ , and $J$ is the coupling rate. Going once round a square of four rings adds the phase $2\pi\alpha$ (Hafezi et al. 2011, eq. 1).

Bands of a long strip

\begin{gathered}\omega\,\phi_y=-2J\cos(k+2\pi\alpha y)\,\phi_y\\-J(\phi_{y+1}+\phi_{y-1})\end{gathered}

A wave $e^{ikx}\phi_y$ along a strip that is infinite in $x$ obeys this equation. States confined to one edge of the strip cross the gaps; their slope $v=d\omega/dk$ tells which way they carry light.

Light delivered to the output

\begin{gathered}T(\omega)=\nu^2\left|\langle\mathrm{out}|G(\omega)|\mathrm{in}\rangle\right|^2\\G=\left(\omega-H+i\kappa_{\mathrm{in}}+i\tfrac{\nu}{2}P\right)^{-1}\end{gathered}

The waveguides couple at the rate $\nu$ to the input and output rings, $P$ projects onto those two rings, and $\kappa_{\mathrm{in}}$ is the loss of every ring (Hafezi et al. 2011, eq. 6).

Where the pulse energy goes

1=\frac{E_{\mathrm{out}}+E_{\mathrm{back}}+2\kappa_{\mathrm{in}}\int\sum_j|a_j|^2\,dt}{E_{\mathrm{pulse}}}

Every bit of the pulse either reaches the output, leaves again by the input waveguide or is absorbed in the rings. The three parts of each bar in the energy chart add up to one.

A defect in a plain chain

T_{\mathrm{chain}}=\frac{4J^2\sin^2k}{4J^2\sin^2k+U^2}

A single ring detuned by $U$ in a chain lets this fraction through at wavenumber $k$ . At the band centre, $k=\pi/2$ , a defect with $U=5J$ passes only 14 % and reflects the rest.

Common difficulties

The edge is not a waveguide wall

Typical misconceptionLight stays on the edge because the edge rings form a separate guide.

Better mental modelEvery ring is identical. The edge state exists because the bulk has a gap with a topological band structure, and it appears on any edge, including the new edge round a missing ring.

Protection is not perfect

Typical misconceptionA topological edge state is immune to every defect.

Better mental modelHere it is protected against anything that keeps the light circulating one way in each ring. Coupling between clockwise and counter-clockwise light, which real rings have through sidewall roughness, would let it turn back.

The flux is not a magnetic field

Typical misconceptionPhotons feel a magnetic field in this chip.

Better mental modelPhotons carry no charge. The link ring between two site rings is shifted off centre, so light hopping one way travels a longer path than light hopping back. That difference in phase plays the role of the magnetic flux.

Run the experiment

01
Watch the edge carry light
Keep the default preset, Edge light flows around a defect, and watch the replay.
What to observe: Light leaves the input ring along the bottom edge, climbs round the amber defect through the next row and drops back to the edge. About 63 % reaches the output. The plain chain delivers 9 % and sends 61 % back to its input.
02
Move the defect
Drag the amber ring along the edge, or move the defect position slider.
What to observe: The lattice still delivers 54 to 63 % wherever the defect sits. The chain stays near 9 %.
03
Take the ring out
Load A ring taken out.
What to observe: The edge light simply follows the new edge round the hole and 62 % arrives. The chain is cut in two: nothing arrives and 71 % comes back.
04
Take the long way round
Load Opposite edge band: the long way round.
What to observe: At a positive drive frequency the edge state on this side runs the other way. The pulse leaves the input heading away from the output, travels round the other three edges and arrives about 24 time units after the input peak, against 8 for the default. More of it is absorbed on the longer path, so 32 % arrives.
05
Leave the gap
Load Drive in a bulk band, then No synthetic flux.
What to observe: In a bulk band about 30 % of the light spreads into the middle of the lattice and 18 % arrives. Without the flux there is no gap at all: 35 % ends up in the middle and 23 % arrives.
06
Make every ring a little wrong
Load Random rings: lattice against chain, then open Arrival time and random chips below the data.
What to observe: With every ring resonance spread by 0.8 coupling rates, the lattice still delivers 52 % on average over 24 random chips. The chain averages 9 %.

Topological Photonic Edge State

Send light around a defect

Core mathematical model

Rings coupled with a synthetic flux

Bands of a long strip

Light delivered to the output

Where the pulse energy goes

A defect in a plain chain

Common difficulties

The edge is not a waveguide wall

Protection is not perfect

The flux is not a magnetic field

Run the experiment

Watch the edge carry light

Move the defect

Take the ring out

Take the long way round

Leave the gap

Make every ring a little wrong

Related experiments and articles

Magnetism, disorder and topological response