Lecture 9 – Light-Matter Interaction Part 3

Structure Dependence

EECS 598-002 Winter 2006Nanophotonics and Nano-scale Fabrication

P.C.Ku

2EECS 598-002 Nanophotonics and Nanoscale Fabrication by P.C.Ku

Schedule for the rest of the semesterIntroduction to light-matter interaction (1/26):

How to determine ε(r)? The relationship to basic excitations.

Basic excitations and measurement of ε(r). (1/31)Structure dependence of ε(r) overview (2/2)Surface effects (2/7 & 2/9):

Surface EM waveSurface polaritonsSize dependence

Case studies (2/14 – 2/21):Quantum wells, wires, and dotsNanophotonics in microscopyNanophotonics in plasmonics

Dispersion engineering (2/23 – 3/9):Material dispersionWaveguide dispersion (photonic crystals)

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Last time…

PolaritonsMacroscopic Maxwell’s equations describe the interaction of EM wave with a macroscopic medium (e.g. propagation, diffraction, scattering, etc.) through constitutive relations. In dielectric materials, constitutive relations are reduced to a dielectric function ε(r).Polaritons are quasi-particles resulting from the quantization of an EM wave propagating through a macroscopic media.In QM, “propagation” in a medium is described by the coupling of two SHO’s.

0

ckωε

=( )

ckωε ω

=

free-space polariton

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Overview

SizeWhen the linear dimension of a structure is smaller than the wavelength of the light, the light sees the composite of the structure and its surroundings.When the size of the structure is comparable to the wavelength of the basic excitations, dispersion relations for polaritons are greatly modified. (Discussed last time)

SurfaceWhen the structure is small enough that its surface area is comparable to its volume, effects that are pertaining to the surface start to dominate. Examples are electron scattering from the surface (which increases the decay rates of plasmons) and surface EM waves.

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Plasmons in mesoscopic structures2

2

0

Plasmon frequency: pne

mω

ε=

Ref: J. B. Pendry et al., Phys. Rev. Lett. 76 (1996) 4773.

If the structure consists of thin metal wires (r~1um) with air inthe surroundings, we have (assuming wires are arrangedinto a cubic lattice of a period a):

2

2

2 20 ln

2

eff

rn na

m m

r e n ar

π

µ ππ

→

→

→

Due to inductanceMuch lower ωp

e.g. aluminum wireswith r=1um, a=5mm:ωp 3622 THz->8.2 GHz.

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Negative refractive index

Negative ε and negative µ negative n

Ref: D. R. Smith et al., Phys. Rev. Lett. 84 (2000) 4184.J. B. Pendry et al., IEEE Trans. Micro. Theory & Tech., 47 (1999) 2075.

2

2 20

1effF

iωµ

ω ω ω= −

− + Γ

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Subwavelength grating broadband mirror

Ref: C. Mateus et al., Photon. Technol. Lett. 16 (2004) 1676.

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Surface EM wave motivation

Interface phenomena revisited:

1 0ε,μ 2 0ε,μ

θi

θr θt

1 1

1 1 2 2

1 1 2 2

1 1 2 2

2 // // // /

ix tx rx

t z

i z z

r z z

i z z

k k kH kH k kH k kH k k

εε εε εε ε

= =

=+−

=+

1 1 2 2/ /z zk kε ε= −When ,reflection and refractioncan be finite even withoutan incident wave.

x

z

ε1 ε2

E

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Surface EM wave (TM polarized)

Surface EM wave is an EM “mode” propagating along the surface (i.e. with its energy confined near the surface).

z

x

ε1

ε2

1

2

1

1

0:

0:

x

x

ik x z

ik x z

z E e e

z E e e

α

α

−⎧ >⎪⎨

<⎪⎩

1 1 2 2

1 1 2 2

1 1

2 2

/ // /

z zk ki iε εα ε α εα εα ε

= −⇒ = −

⇒ = −

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TE polarized surface EM wave?

If the electric field is TE polarized, we can show that the surface EM mode can not be supported.

Homework 2 (Hint: Consider the boundary condition at the interface and show that the boundary condition can not be satisfied if the TE wave decays exponentially on both sides of the interface.)

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Electric field polarization

The electric field has components both in parallel and in perpendicular to the x direction.

Both transverse and longitudinal wave components exist in a surface EM wave.

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Dispersion relation of surface EM waves

Dispersion relation = ω versus kx2

2 2 11 2

22 2 2

2 2

1 2

1 2

( )

( )

x

x

x

k ic

k ic

ck

ω εα

ω εα

ε εωε ε

⎧+ =⎪⎪

⎨⎪ + =⎪⎩

+⇒ =

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Surface polaritons

Surface polaritons:

e.g. surface plasmon polaritons b/w air and metal

1 2

1 2

( ) ( )( ) ( )xck ε ω ε ωω

ε ω ε ω+

=

2

1 2 2

22 2 22 4

1 ; 1

2 2

p

p pk kc

ωε ε

ω

ω ωω

= = −

⎛ ⎞⎛ ⎞⇒ = + ± + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

We need to pick the “-” solutionas ε2<0.

ω

k

Surface plasmon

Photonin air

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Short wavelength limit

As kx infinity

But since ω is finite, we need to demand:

Since in this limit, the wave sees only a very small region around the interface, the waves see a composite of these two different materials with a effective dielectric constant of . The wave is a longitudinal wave in this limit.

1 2

1 2

( ) ( )( ) ( )xck ε ω ε ωω

ε ω ε ω+

= →∞

1 2ε ε= −

1 2 0ε ε+ =

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Attenuation

Attenuation length L = the propagation distance at which the field intensity drops to 1/e.

( )

221 2

2 21 2 2 1

1 2 2 2 2 2 1

2 21 2 2 1

121 2 2 2

11

2

23/ 2

1

Re( )| | Im | | Im

As a special case when real, ' '' and '' ' and

( ' '') 1' '' ''

12 ''

''

x

x

x

cL k

c k i

k iccL

ε εω ε ε ε ε

ε ε ε ε ε ε ε

ε ε ε εεω ε ε ε ε

εω εε

εω ε

+⎛ ⎞= ⎜ ⎟ +⎝ ⎠= = +

⎛ ⎞+⇒ = ≈ +⎜ ⎟+ + ⎝ ⎠

⎛ ⎞⇒ ≈ +⎜ ⎟

⎝ ⎠

⇒ = is very large

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Optical excitation of surface polaritons

ω

k

Surface plasmon

Photonin air

Need to increase the k vector in order to radiatively excitethe surface polaritons.

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Total internal reflection

With light coming into the interface and experiencing total internal reflection, the kx component in the second media is larger than that in the free space.

20

sinsin

ik θθ

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Thin slabx

z

( )( )

( ) ( )

( ) ( )3

ˆ 1 3

ˆ ; 0

ix iz ix iz

ix iz ix iz

i k x k z i k x k zi i i

i k x k z i k x k zi i i i

E y E e E e i

H ik x E e E e E

+ −+ −

+ −+ − −

= + =

= − − =(TE polarized)

Applying boundary conditions at each interface, we have:

z=0 z=d

1 0ε,μ 2 0ε,μ 1 0ε,μ

1 2 3

222 2

1 2 1 2 22 2

x x x

x z x

k k k

n nn n k k kπ πλ λ

= =

⎛ ⎞> > = −⎜ ⎟⎝ ⎠

In the case and when , is imaginary.

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Thin slab (cont.)

0 20 40 60 80 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Incident Angle

d = wavelength/4

TransmissionReflection

0 20 40 60 80 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Incident Angle

d = wavelength

TransmissionReflection

When the incident angle meets the total internal reflection condition at the 1-2 interface, there is no wave propagated in region 2 but there is still a finite transmission through the thin slab. The wave seems to “tunnel”through the slab!

For example, in the following plots, n1=1.5 and n2=1.

Tunneling region

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Prism coupling technique

glass

metal

air

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Reading

Prasad, chapters 2, 5 and 6.