Nanoscale antennas

Said R. K. Rodriguez24/04/2018

The problem with nanoscale optics

Ε𝑖𝑖(𝑘𝑘𝑘𝑘−𝜔𝜔𝜔𝜔)

~1-10 nm

~400-800 nm

How to interface light emitters & receiverswith plane waves?

Antenna

Tx/Rx

An antenna is a device that converts free-space radiation into localized energy, and vice versa

Radiation

Novotny & van Hulst, Nat.Photon. 5, 83 (2011).

What is an antenna?

Antenna

Tx/Rx

An antenna is a device that converts free-space radiation into localized energy, and vice versa

Radiation

What is an antenna?

At radio frequencies, E = 0 inside the metal – perfect metal

Novotny & van Hulst, Nat.Photon. 5, 83 (2011).

Antenna

Tx/Rx

An antenna is a device that converts free-space radiation into localized energy, and vice versa

Radiation

What is an antenna?

At optical frequencies, E ≠ 0 inside the metal.

Consequence: radio-freq. antenna designs cannot be directly scaled

Novotny & van Hulst, Nat.Photon. 5, 83 (2011).

200 400 600 800 1000 1200 1400 1600 1800-150

-100

-50

0

50

Measured data: ε' ε"

Drude model: ε' ε"

Modified Drude model: ε'

ε"

ε

Wavelength (nm)

ε'

Dielectric constant for Ag

Finite ε’ leads to field penetration

Plasmons in the bulk oscillate at ωp determined by the free electron density and effective mass

Plasmons confined to surfaces that can interact with light to form propagating “surface plasmon polaritons (SPP)”

Localized surface plasmons in nanoparticles

+ + +

- - -

+ - +

k 0

2

εω

mNedrude

p =

From plasmons to plasmonics

2/1

"'

+

=+=dm

dmxxx c

ikkkεε

εεω

optical resonance frequency depends on shape & size; k is irrelevant

transmissionreflection

Colors of gold nanoparticles Stained glass @ Notre Dame de Paris

Lycargus cup, 4thC AD 1260

Observables

Extinction cross section [m2]

Power removed from beamIncident intensity

Extinction = scattering + absorption

removed from the beam re-radiated into all angles lost as heat in the scatterer

Linear response to applied field

Small object kd <<1 - incident field is approximately constant

Volume polarization (weak index so E=Ein)

Total dipole moment

Larger particles & ε : larger dipole moments

Electrostatic sphere

Consider a sphere in a static field E0 εmε

z

rθ

a( )( )ar

ar>=∆Φ<=∆Φ

00

2

1

Laplace equation:

( ) ( ) zEarrr

arr

m 0221

21 −=Φ=∂Φ∂

=∂Φ∂

=Φ=Φ∞→

lim,, εε

Boundary conditions set by 0)()( =Φ∇⋅−∇=⋅∇=⋅∇ εε ED

Solution

1 0 0 0

32 0 0 02 2

0

3cos cos cos2 2

cos coscos cos2 4

m m

m m

m

m m

E r E r E r

pE r a E E rr r

ε ε εθ θ θε ε ε ε

ε ε θ θθ θε ε πε ε

−Φ = − + = − + +

−Φ = − + = − + +

E0 εm

εz

rθ

a[ see J. D. Jackson, Classical Electrodynamics, Ch. 4]

30 0 with 4

2m

SI SI mm

p E a ε εα α πε εε ε

−= = +

rr

Inside sphere: homogeneous fieldOutside sphere: background field plus field of a dipole with

In the ball:

Outside:

Metal sphere

Drude model for a metal: Lorentzian `plasmon resonance’

• Resonance at ε(ω0) = -2 εm• Response scales with the volume V• α exceeds V by factor 5 to 10• Shape shifts condition ε = -2 εm• γ still needs to include radiation damping

𝑝𝑝 = 4𝜋𝜋𝜀𝜀0𝛼𝛼𝐸𝐸0 𝛼𝛼 = 𝑎𝑎3 𝜀𝜀 − 𝜀𝜀𝑚𝑚

𝜀𝜀 + 2𝜀𝜀𝑚𝑚 0

𝛼𝛼 = 𝑎𝑎3 𝜔𝜔02

𝜔𝜔02 − 𝜔𝜔2 + 𝑖𝑖𝜔𝜔𝑖𝑖 0

𝜀𝜀 = 1 −𝜔𝜔𝑝𝑝

2

𝜔𝜔(𝜔𝜔 + 𝑖𝑖𝑖𝑖𝜔𝜔)means

Revisiting polarizabilityClassical model of harmonically bound electron describes atom, and scatterer alike, as an oscillating dipole

20 0

2 20

3 ( ) ( )i t i tSI

Vt e ei

ω ωε ω α ωω ω ωγ

= =− −

p E E

Lorentzian resonance

Extinction: how much power is taken from the beam ?

Cycle average work done by E on p

in ImdpW Edt

α∝ ⋅ ∝

Revisiting polarizabilityExtinction: how much power is taken from the beam (in SI units) ?

0 0

1 1Re[ ] Re[ ] Re[ ] Re[ ]T Ti t

i t i t i td eW e dt e i e dtT dt T

ωω ω ωωα= ⋅ = ⋅∫ ∫

pE E E

* * *

0

1 ( ) ( )4

Ti t i t i t i tW e e i e i e dt

Tω ω ω ωωα ωα− −= + ⋅ −∫ E E E E

* 2 2

0

1 ( | | | | ) oscill.terms ( 2 )4

T

W i i dtT

ωα ωα ω= − + + ±∫ E E

2Im | |2

W ω α= E

Revisiting polarizabilityClassical model of harmonically bound electron describes atom, and scatterer alike as an oscillating dipole

20 0

2 20

3 ( ) ( )i t i tSI

Vt e ei

ω ωε ω α ωω ω ωγ

= =− −

p E E

Lorentzian resonance

Scattering: how much power does p radiate ?

22

0

2

0 0

22dipole

2

0 0

2 ||4

sinsin||sin W απε

θθϕθϕπ ππ π

∝=∝⋅∝ ∫ ∫∫ ∫∫ rprdErddA

sphere

nS

Equate extinction to scattering (energy conservation)

Scattering

Rayleigh / Larmor

Extinction

Work done to drive p

≥

Optical theorem

1. Very small particles scatter like r6/λ4 (Rayleigh)2. For very small particles absorption wins ~ r3/λ3. Big |α|2 implies large Im α

4𝜋𝜋𝑘𝑘 Im 𝛼𝛼 [𝑚𝑚2] 8𝜋𝜋3

𝑘𝑘4 𝛼𝛼 2 [𝑚𝑚2]

ScatteringExtinction

≥

Optical theorem

Since

Upper bound on the strongest possible dipole scatterer

Rayleigh / LarmorWork done to drive p

Equate extinction to scattering (energy conservation)

4𝜋𝜋𝑘𝑘 Im 𝛼𝛼 [𝑚𝑚2] 8𝜋𝜋3

𝑘𝑘4 𝛼𝛼 2 [𝑚𝑚2]

𝛼𝛼 ≤32

𝜆𝜆2𝜋𝜋

3

Im 𝛼𝛼 < 𝛼𝛼

Extinction – Interference effect𝑃𝑃𝑒𝑒𝑘𝑘𝜔𝜔 = ∬𝐷𝐷 Sext��erdA = ∬𝐷𝐷

12

Re Ei×Hs∗ +Es ×Hi∗ ��erdA

𝑄𝑄𝑒𝑒𝑘𝑘𝜔𝜔 =𝜎𝜎𝑒𝑒𝑘𝑘𝜔𝜔

𝐴𝐴𝑝𝑝𝑝𝑝𝑝𝑝𝜔𝜔Out of resonanceOn resonance

2𝜋𝜋𝜋𝜋𝜆𝜆

= 0.3

Extinction – Interference effect𝑃𝑃𝑒𝑒𝑘𝑘𝜔𝜔 = ∬𝐷𝐷 Sext��erdA = ∬𝐷𝐷

12

Re Ei×Hs∗ +Es ×Hi∗ ��erdA

𝑄𝑄𝑒𝑒𝑘𝑘𝜔𝜔 =𝜎𝜎𝑒𝑒𝑘𝑘𝜔𝜔

𝐴𝐴𝑝𝑝𝑝𝑝𝑝𝑝𝜔𝜔

σext = Apart

r=20 nm Ag particle, in n=1.5 (glass)

Extinction – Interference effect𝑃𝑃𝑒𝑒𝑘𝑘𝜔𝜔 = ∬𝐷𝐷 Sext��erdA = ∬𝐷𝐷

12

Re Ei×Hs∗ +Es ×Hi∗ ��erdA

𝑄𝑄𝑒𝑒𝑘𝑘𝜔𝜔 =𝜎𝜎𝑒𝑒𝑘𝑘𝜔𝜔

𝐴𝐴𝑝𝑝𝑝𝑝𝑝𝑝𝜔𝜔

2𝜋𝜋𝜋𝜋𝜆𝜆

= 0.3

Question: what does the above expression tells us aboutthe detector needed to measure the full extinction?

Summary

• Antennas convert free-space radiation into localized energy & viceversa

• At optical frequencies, E-field penetrates into the metal. This leads to surface plasmon resonances

• Extinction:– Work done by E on p– ∝ Im(α)– Interference of incident & scattered field

• Subwavelength particles can absorb and scatter much more light than is geometrically incident on them. In general, Qext >1 on resonance and Qext <1 off resonance

10 min. break

Approaches to controlling lightResonant nanoparticles Photonic crystals

Surface Plasmon Polaritons

Dipole radiation

𝑬𝑬(𝒓𝒓) = 𝜇𝜇0𝜔𝜔2𝐺𝐺(𝒓𝒓) � 𝒑𝒑1 dipole (vector):

1 dipole (scalar): 𝜓𝜓1 𝜋𝜋 =𝑒𝑒𝑖𝑖𝑘𝑘𝑝𝑝

𝜋𝜋� 𝑝𝑝

Dipole arrays

𝑬𝑬(𝒓𝒓) = 𝜇𝜇0𝜔𝜔2𝐺𝐺(𝒓𝒓) � 𝒑𝒑1 dipole (vector):

1 dipole (scalar): 𝜓𝜓1 𝜋𝜋 =𝑒𝑒𝑖𝑖𝑘𝑘𝑝𝑝

𝜋𝜋� 𝑝𝑝

Dipole array (scalar): 𝜓𝜓𝑡𝑡𝑡𝑡𝑡𝑡 𝜋𝜋 = 𝜓𝜓1(𝜋𝜋) � 𝐴𝐴𝐴𝐴

Depends on positions & complex amp. of scatterersFourier transform of geometry (more ahead)

a ≈ λ

Far-field of 2 dipoles

𝜓𝜓𝑡𝑡 = 𝜓𝜓1 + 𝜓𝜓2 = 𝑝𝑝𝑒𝑒−𝑖𝑖(𝑘𝑘𝑝𝑝1−𝛽𝛽/2)

𝜋𝜋1cos 𝜃𝜃1 +

𝑒𝑒−𝑖𝑖(𝑘𝑘𝑝𝑝2+𝛽𝛽/2)

𝜋𝜋2cos 𝜃𝜃2

●r1

r2

θ1

θ2

β = phase difference between dipoles

𝜓𝜓𝑡𝑡 𝜋𝜋 = 𝑝𝑝𝑒𝑒𝑖𝑖𝑘𝑘𝑝𝑝

𝜋𝜋� 𝐴𝐴𝐴𝐴

●r1

r2

θ1

θ2 𝐴𝐴𝐴𝐴 = cos12

(𝑘𝑘𝑘𝑘𝑘𝑘𝑡𝑡𝑘𝑘 𝜃𝜃 + 𝛽𝛽)

β = phase difference between dipoles

Exercise:

Far-field of 2 dipoles

𝜓𝜓𝑡𝑡 = 𝜓𝜓1 + 𝜓𝜓2 = 𝑝𝑝𝑒𝑒−𝑖𝑖(𝑘𝑘𝑝𝑝1−𝛽𝛽/2)

𝜋𝜋1cos 𝜃𝜃1 +

𝑒𝑒−𝑖𝑖(𝑘𝑘𝑝𝑝2+𝛽𝛽/2)

𝜋𝜋2cos 𝜃𝜃2

𝜓𝜓𝑡𝑡 𝜋𝜋 = 𝑝𝑝𝑒𝑒𝑖𝑖𝑘𝑘𝑝𝑝

𝜋𝜋� 𝐴𝐴𝐴𝐴

●r1

r2

θ1

θ2 𝐴𝐴𝐴𝐴 = cos12

(𝑘𝑘𝑘𝑘𝑘𝑘𝑡𝑡𝑘𝑘 𝜃𝜃 + 𝛽𝛽)

β = phase difference between dipoles

Exercise:

Note: β=π & θ=π/2 → AF = 0 ∀ kd

Far-field of 2 dipoles

𝜓𝜓𝑡𝑡 = 𝜓𝜓1 + 𝜓𝜓2 = 𝑝𝑝𝑒𝑒−𝑖𝑖(𝑘𝑘𝑝𝑝1−𝛽𝛽/2)

𝜋𝜋1cos 𝜃𝜃1 +

𝑒𝑒−𝑖𝑖(𝑘𝑘𝑝𝑝2+𝛽𝛽/2)

𝜋𝜋2cos 𝜃𝜃2

Dimer in static approximationDimer in a static approximation

Linear problem

• Symmetric, but not real matrix• 1/polarizability on the diagonal• Interaction on the off-diagonal - this will shift resonances

Hybrid modes

Hybridization (exercise)

Arrays of coupled dipoles

Arrays of coupled dipoles

d= 100 nmax = ay = 450 nmn = 1.5

ax n

1 dipole

array of dipoles

Light cone & diffraction

Light emission from plasmonic arrayNA of objective

kx

ky

S.R.K. Rodriguez et al., Phys. Rev. X 1, 021019 (2011).

LSPR

Extinction of Au nanorod arrays

Bright – even / Dark - odd

Diffraction / Bloch theorem determines mode dispersionMode symmetry + illumination determines what you excite

S.R.K. Rodriguez et al., Physica B 407, 4081 (2012).

Coupled dipole calculations

Measurements

Shaper resonances by adding nanoparticles

Collective resonances

Uses of resonant nanostructures

Enhanced local fieldsOn resonance, ~ 104 enhanced intensity

Au spheres 5,8,20 nm, gaps of 1-3 nm

100 102 104

|E|2/|Ein|2

Single molecule Fluorescence Enhancement

100 nm

𝐴𝐴𝐸𝐸 =𝜂𝜂(𝜋𝜋0, 𝜔𝜔𝑒𝑒𝑚𝑚)𝜂𝜂0(𝜋𝜋0, 𝜔𝜔𝑒𝑒𝑚𝑚)

𝐷𝐷(𝜋𝜋0, 𝜔𝜔𝑒𝑒𝑚𝑚)𝐷𝐷0(𝜋𝜋0, 𝜔𝜔𝑒𝑒𝑚𝑚)

𝐸𝐸 (𝜋𝜋0, 𝜔𝜔𝑝𝑝𝑎𝑎𝑎𝑎) 2

𝐸𝐸0(𝜋𝜋0, 𝜔𝜔𝑝𝑝𝑎𝑎𝑎𝑎) 2

A. Kinkhabwala et al., Nat. Phot. 3, 654 (2009)

Yagi-Uda nanoantenna

A. F. Koenderink, Nano Lett. 9, 4228 (2009)

Directional emission from localized sources

100 nm

A. Curto et al., Science 329, 930 (2010)

100 nm

Directional emission from extended sources

emitting layer

G. Lozano et al., Light Sci. Appl. e66 (2013)

100 nm

Directional emission from extended sources

emitting layer emitting layer

SensingSingle protein binding/unbinding Refractive index sensing

n

P. Offermans et al., ACS Nano 5, 5151 (2011)

Biological imaging

Novotny & van Hulst, Nat.Photon. 5, 83 (2011).

Diffraction unlimited resolution

λ= 633 nm

Nonlinear effectsGenerating new frequenciesEnhanced mode mapping

P. Ghenuche et. al. Phys. Rev. Lett. 10, 116805 (2008) H. Harutyunyan et. al. Phys. Rev. Lett. 108, 217403 (2012)

Summary

• Small particles of size < λ/10 scatter like dipoles• Arrays of dipoles can be described by effective polarizability• Nanoantennas can be used to enhance:

– local fields– absorption & spontaneous emission– Sensing– Biological imaging– Nonlinearities