photonic forces group - amolfoptical theorem. 1. very small particles scatter like r. 6 /Ξ» ......
TRANSCRIPT
Nanoscale antennas
Said R. K. Rodriguez24/04/2018
The problem with nanoscale optics
Ξππ(ππππβππππ)
~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 β 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 β 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: Ξ΅' Ξ΅"
Drude model: Ξ΅' Ξ΅"
Modified Drude model: Ξ΅'
Ξ΅"
Ξ΅
Wavelength (nm)
Ξ΅'
Dielectric constant for Ag
Finite Ξ΅β leads to field penetration
Plasmons in the bulk oscillate at Οp determined by the free electron density and effective mass
Plasmons confined to surfaces that can interact with light to form propagating βsurface plasmon polaritons (SPP)β
Localized surface plasmons in nanoparticles
+ + +
- - -
+ - +
k 0
2
Ξ΅Ο
mNedrude
p =
From plasmons to plasmonics
2/1
"'
+
=+=dm
dmxxx c
ikkkΡΡ
ΡΡΟ
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 & Ξ΅ : larger dipole moments
Electrostatic sphere
Consider a sphere in a static field E0 Ξ΅mΞ΅
z
rΞΈ
a( )( )ar
ar>=βΞ¦<=βΞ¦
00
2
1
Laplace equation:
( ) ( ) zEarrr
arr
m 0221
21 β=Ξ¦=βΞ¦β
=βΞ¦β
=Ξ¦=Ξ¦ββ
lim,, ΡΡ
Boundary conditions set by 0)()( =Ξ¦ββ ββ=β β=β β ΡΡ 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
Ρ Ρ Ρθ θ θΡ Ρ Ρ Ρ
Ξ΅ Ξ΅ ΞΈ ΞΈΞΈ ΞΈΞ΅ Ξ΅ ΟΞ΅ Ξ΅
βΞ¦ = β + = β + +
βΞ¦ = β + = β + +
E0 Ξ΅m
Ξ΅z
rΞΈ
a[ see J. D. Jackson, Classical Electrodynamics, Ch. 4]
30 0 with 4
2m
SI SI mm
p E a Ξ΅ Ρα Ξ± ΟΞ΅ ΡΡ Ξ΅
β= = +
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 resonanceβ
β’ Resonance at Ξ΅(Ο0) = -2 Ξ΅mβ’ Response scales with the volume Vβ’ Ξ± exceeds V by factor 5 to 10β’ Shape shifts condition Ξ΅ = -2 Ξ΅mβ’ Ξ³ still needs to include radiation damping
ππ = 4ππππ0πΌπΌπΈπΈ0 πΌπΌ = ππ3 ππ β ππππ
ππ + 2ππππ 0
πΌπΌ = ππ3 ππ02
ππ02 β ππ2 + ππππππ 0
ππ = 1 βππππ
2
ππ(ππ + ππππππ)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
Ο ΟΞ΅ Ο Ξ± ΟΟ Ο ΟΞ³
= =β β
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
Ξ±β β β
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
ΟΟ Ο ΟΟΞ±= β = β β« β«
pE E E
* * *
0
1 ( ) ( )4
Ti t i t i t i tW e e i e i e dt
TΟ Ο Ο ΟΟΞ± ΟΞ±β β= + β ββ« E E E E
* 2 2
0
1 ( | | | | ) oscill.terms ( 2 )4
T
W i i dtT
ΟΞ± ΟΞ± Ο= β + + Β±β« E E
2Im | |2
W Ο Ξ±= 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
Ο ΟΞ΅ Ο Ξ± ΟΟ Ο ΟΞ³
= =β β
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 Ξ±ΟΞ΅
ΞΈΞΈΟΞΈΟΟ ΟΟ Ο
β=ββ β β« β«β« β«β« rprdErddA
sphere
nS
Equate extinction to scattering (energy conservation)
Scattering
Rayleigh / Larmor
Extinction
Work done to drive p
β₯
Optical theorem
1. Very small particles scatter like r6/Ξ»4 (Rayleigh)2. For very small particles absorption wins ~ r3/Ξ»3. Big |Ξ±|2 implies large Im Ξ±
4ππππ Im πΌπΌ [ππ2] 8ππ3
ππ4 πΌπΌ 2 [ππ2]
ScatteringExtinction
β₯
Optical theorem
Since
Upper bound on the strongest possible dipole scatterer
Rayleigh / LarmorWork done to drive p
Equate extinction to scattering (energy conservation)
4ππππ Im πΌπΌ [ππ2] 8ππ3
ππ4 πΌπΌ 2 [ππ2]
πΌπΌ β€32
ππ2ππ
3
Im πΌπΌ < πΌπΌ
Extinction β Interference effectππππππππ = β¬π·π· SextοΏ½οΏ½erdA = β¬π·π·
12
Re EiΓHsβ +Es ΓHiβ οΏ½οΏ½erdA
ππππππππ =ππππππππ
π΄π΄ππππππππOut of resonanceOn resonance
2ππππππ
= 0.3
Extinction β Interference effectππππππππ = β¬π·π· SextοΏ½οΏ½erdA = β¬π·π·
12
Re EiΓHsβ +Es ΓHiβ οΏ½οΏ½erdA
ππππππππ =ππππππππ
π΄π΄ππππππππ
Οext = Apart
r=20 nm Ag particle, in n=1.5 (glass)
Extinction β Interference effectππππππππ = β¬π·π· SextοΏ½οΏ½erdA = β¬π·π·
12
Re EiΓHsβ +Es ΓHiβ οΏ½οΏ½erdA
ππππππππ =ππππππππ
π΄π΄ππππππππ
2ππππππ
= 0.3
Question: what does the above expression tells us aboutthe detector needed to measure the full extinction?
Summary
β’ Antennas convert free-space radiation into localized energy & viceversa
β’ At optical frequencies, E-field penetrates into the metal. This leads to surface plasmon resonances
β’ Extinction:β Work done by E on pβ β Im(Ξ±)β Interference of incident & scattered field
β’ 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
π¬π¬(ππ) = ππ0ππ2πΊπΊ(ππ) οΏ½ ππ1 dipole (vector):
1 dipole (scalar): ππ1 ππ =ππππππππ
πποΏ½ ππ
Dipole arrays
π¬π¬(ππ) = ππ0ππ2πΊπΊ(ππ) οΏ½ ππ1 dipole (vector):
1 dipole (scalar): ππ1 ππ =ππππππππ
πποΏ½ ππ
Dipole array (scalar): πππ‘π‘π‘π‘π‘π‘ ππ = ππ1(ππ) οΏ½ π΄π΄π΄π΄
Depends on positions & complex amp. of scatterersFourier transform of geometry (more ahead)
a β Ξ»
Far-field of 2 dipoles
πππ‘π‘ = ππ1 + ππ2 = ππππβππ(ππππ1βπ½π½/2)
ππ1cos ππ1 +
ππβππ(ππππ2+π½π½/2)
ππ2cos ππ2
βr1
r2
ΞΈ1
ΞΈ2
Ξ² = phase difference between dipoles
πππ‘π‘ ππ = ππππππππππ
πποΏ½ π΄π΄π΄π΄
βr1
r2
ΞΈ1
ΞΈ2 π΄π΄π΄π΄ = cos12
(πππππππ‘π‘ππ ππ + π½π½)
Ξ² = phase difference between dipoles
Exercise:
Far-field of 2 dipoles
πππ‘π‘ = ππ1 + ππ2 = ππππβππ(ππππ1βπ½π½/2)
ππ1cos ππ1 +
ππβππ(ππππ2+π½π½/2)
ππ2cos ππ2
πππ‘π‘ ππ = ππππππππππ
πποΏ½ π΄π΄π΄π΄
βr1
r2
ΞΈ1
ΞΈ2 π΄π΄π΄π΄ = cos12
(πππππππ‘π‘ππ ππ + π½π½)
Ξ² = phase difference between dipoles
Exercise:
Note: Ξ²=Ο & ΞΈ=Ο/2 β AF = 0 β kd
Far-field of 2 dipoles
πππ‘π‘ = ππ1 + ππ2 = ππππβππ(ππππ1βπ½π½/2)
ππ1cos ππ1 +
ππβππ(ππππ2+π½π½/2)
ππ2cos ππ2
Dimer in static approximationDimer in a static approximation
Linear problem
β’ Symmetric, but not real matrixβ’ 1/polarizability on the diagonalβ’ 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 β 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
π΄π΄πΈπΈ =ππ(ππ0, ππππππ)ππ0(ππ0, ππππππ)
π·π·(ππ0, ππππππ)π·π·0(ππ0, ππππππ)
πΈπΈ (ππ0, ππππππππ) 2
πΈπΈ0(ππ0, ππππππππ) 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
Ξ»= 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
β’ Small particles of size < Ξ»/10 scatter like dipolesβ’ Arrays of dipoles can be described by effective polarizabilityβ’ Nanoantennas can be used to enhance:
β local fieldsβ absorption & spontaneous emissionβ Sensingβ Biological imagingβ Nonlinearities