2 Scattering and absorption cross sections of a metal nanoparticle

1 Surface Plasmon-Polariton Waves

(a) Excitation of SPP waves:

Figure 1: Dispersion relationship

Generating SPP requires energy and momentum conservation and this is done via coupling based on total internal reflection. So, as indicated below:

Figure 2: Evanescent waves between two media

θ > θ_c= arcsinn₂ n1

, we have TIR.

In mediumn1:

ω

cn1=q

k²_k+k²_z k_k = ωn1

c sinθ In mediumn₂:

kk is conserved:

kk= ωn1

c sinθ

A propagating wave inn₂ would have

ω

cn2 =k_k

ω

cn1sinθ >

ω

cn2=⇒n1sinθ > n2

θ >arcsinn2

n₁ =θc

Because the wave is evanescent, we must go close to the SPP wave.

Figure 3: Excitation mechanisms of SPP We need to create an evanescent wave in the dielectric.

Figure 4: Dispersion relationship for SPP excitation

β_o= ω_o

c n_prismsinθ=⇒θ

This means for each EM wave at frequencyω, there is an angle at which the energy and in-plane momentum are conserved, hence, the coupling occurs, as indicated in Fig. 5.

The reflectance drops at angleθ because of the energy transfer from EM wave to the surface plasmon.

(b) Propagation length of a SPP wave.

(ω) = 1− ω²_p

ω(ω+iγ) =_R+i_I

Figure 5: Reflectance as a function of angle For a complex refractive index:

β+i k_I = ω

c(n_R+i n_I ), wherek_I is related ton_I

e^i(β+ikI)z =e^−kIze^iβz I(z) =Ioe^−2kIz(Intensity)

Figure 6: Propagation length as a function of propagation axis z We assume that losses are small: γ ω, ωp

βR+βI ' ω c

r md

m+_d

If we notice from that equation thatβ has now also a complex value becausem has complex value. _d=1 for simplicity

β_R+β_I = ω c

s

(_R+i_I).1 R+iI+ 1

= ω c

s R+iI

_R+ 1(1 +i^I

R+1)

=βR

v u u u u t

1 +i^I

R

1 +i _{I R}+ 1

Now we can use approximation, e.g. we can consider_I <|_R| β_R+β_I 'β_R

r

(1 +i_I R

)(1−i _I R+ 1)

∼β_R r

1 +i_I( 1

_R − 1 _R+ 1)

∼βR

1 +iI

2 ( 1

_R − 1 _R+ 1)

β_I ' _I

2β_R 1 _R(_R+ 1) l= 1

2β_I = R(R+ 1) _Iβ_R We replaceβR with

ω c

r R

_R+ 1 l= _R(_R+ 1)

I

c ω

r_R+ 1 R

l'(c ω)1

I

√_R(_R+ 1)³²

(ω) = 1− ω²_p

ω(ω+iγ) =⇒R= 1− ω_p² ω²+γ² and_I = ω_p²

ω(ω²+γ²)γ l= c

ω ω³ ω_p²

s 1− ω²_p

ω²(2− ω²_p ω²)³²γ⁻¹

l=c ω

ωp

2s 1−ω_p²

ω² 2−ω²_p ω²

!³

2

γ⁻¹

1.1 Evanescent Wave

Figure 7: Evanescent waves

Under total internal reflection, the wavevector in medium n₂ reads:

ω²

c²n²₂ =k²_k+k²_z k²_k = ω²

c²n²₁sin²θ

=k_z²ω²

c²(n²₂−n²₁sin²θ)

kz= ωn2

c s

1−n²₁sin²θ n²₂ But:sin²θ >

n2

n₁ 2

=⇒1−n²₁

n²₂ sin²θ <1 k_z =iωn₂

c s

n²₁sin²θ n²₂ −1 E(x, z) =E_oe^ikkxe^ikzz

E(x, z) =E_o e

−^ω_cn2

s

n2 1 n2 2

sin²θ−1

e^ikkx

The enclosed part of the above equation is what gives evanescent wave. The figure below shows the behavior of evanescent waves.

Figure 8: Exponential decay of Evanescent waves

2 Scattering and absorption cross sections of a metal nanoparticle

Here we have a nanoparticle, with radius ’a’ being hit by an EM plane waves as depicted below.

The generated polarizabilityp=oαEo, whereα= ^αo

1−i^k3

6παo

and αo= 4πa^3m−1

m+2

Incident Field: E_i=E_o

Scattered Field (from the dipole): Es= _4π^P

o(...) Total Field: Et=Ei+Es

Figure 9: EM plane waves impinged onto a nanoparticle Power through the integration sphere reads:

P = Z

4π

S~tot.·~nr²dΩ S~_tot = 1

2Re{E_tot×H_tot^∗ } S~tot =S~inc+S~sca+S~int

where

S~inc= 1

2Re{E_inc×H_inc^∗ }, S~_sca= 1

2Re{E_sca×H_sca^∗ }, S~_int= 1

2Re{E_sca×H_int^∗ }+1

2Re{E_int×H_sca^∗ } Pinc=

Z

4π

S~inc·~nr²dΩ = 0.

This is because the EM field coming from outside to the nanoparticle goes out. In other words, no field remains in the nanoparticle.

P_sca = Z

4π

S~_sca·~nr²dΩ =scattered power P_int=

Z

4π

S~_int·~nr²dΩ6= 0

P =Psca+Pint=−d

dtEsphere ,

whereE_sphereis the energy in the integration sphere. The variation of the energy in the volume is related to the power absorbed in the sphere, so−_dt^dE_sphere =P_abs where P_abs is the power absorbed by the particle−P_abs =P_sca+P_int with rearrangement of the terms, we get

−P_int=P_sca+P_abs P_sca+P_abs =P_tot ,

this is the total power dissipated by the induced dipole. If we consider the nanoparticle as a dipole, then

P_tot=ωImn

~ p ~E_i^∗o

=ωoIm nE~iE~_i^∗

o

=ω_o |E~_i|² Im{α}

Incident intensity: I_i = _Z¹ |E~_i|², whereZ =q_µ

o

o and it is the vacuum impedance Ptot =ωoZIiIm{α}

Ptot=IikIm{α}

σ_tot =kIm{α}

cross-section, which has the dimensions of an area, and it expresses the effective area of the nanoparticle when it interacts with light. Psca=Power scattered by~p= Iiσsca , whereasσsca

is the scattering cross-section.

This result can be obtained by calculated the power radiated by the dipole.

σsca= k⁴ 6π |α|² P_abs=Ptot−Psca= (kIm{α} − k⁴

6π |α|²)Ii

σ_abs is the absorption cross-section=kIm{α} − k⁴ 6π |αo |² Im{α}=Im

( αo

1−i_6π^k3α_o )

= Im{α_o}

1 +_(6π)^k62 |αo |² + k³ 6π

|αo|² 1 +_(6π)^k62 |αo|² where

σtot =kIm{α} and σsca = k⁴ 6π |α|² σ_abs = kIm{α_o}

1 +_(6π)^k62 |α_o|²

σtot=σsca+σ_abs

−P_int=Psca+Pabs

Ptot=−P_int

From the figure above, the EM plane waves propagate toward the detector, therefore if we

Figure 10: Measurement of radiated power by the nanoparticle

integrate across the whole integration sphere; the only non-zero part will be the part at the detector. So, without nanoparticle the detector would measure the incident power. But with nanoparticle the detector would measureP_inc+P_int+

P_scabecause the detector is small P =P_inc+P_int=P_inc−P_sca−P_abs

P =P_inc−P_tot

If we make an experiment and measure the curve will look like the one shown below.

Figure 11: Exctinction of power at the detector position

The dip is called Extinction, whereas P_tot = P_ext =⇒ The Extinction cross-section σ_ext = σsca+σabs. From that we see in a simple experiment, we can measure the power dissipated by a nanoparticle through this extinction dip, and this is also called optical theorem. That theorem states that the dip measured in transmission by an infinitesimal detector is equal to the total power dissipated (scattering + absorption) by the nanoparticle.