1 Optical Interactions in the Context of Nano Optics : Near-field optics

1 Optical Interactions in the Context of Nano Optics : Near-field optics

• We consider a dipole in front of a perfectly conductive interface. The boundary value problem can be solve using image charges. In other words, the charges of the dipole induce opposite charges in the perfect conductor as shown in the figure below.

Figure 1: Dipoles near a perfect mirror with orientation perpendicular and parallel to the in- terface. The dashed arrows represent the image dipoles.

The modification of the emission rate can be easily obtained by calculating the power radiated by the real and image dipolesP, normalized by the power emitted by the real dipole in free spaceP0, using the fact that Γ/Γ0 =P/P0 (see lecture notes). The results depends on the dipole orientation and it reads

Γ⊥

Γ0

= 1−3Im

1

(2kz)³ − i (2kz)²

exp(2ikz)

, (1)

Γ_k

Γ₀ = 1 +3 2Im

1

(2kz)³ − i

(2kz)² + 1 (2kz)

exp(2ikz)

, (2)

where⊥ and k refer to the dipole orientations with respect to the interface, k =ω/c is the wave-vector andz is the distance of the real dipole from the perfect conductor.

Whenz→0,Γ⊥/Γ₀→2, because the two dipoles oscillate in phase, whereasΓ_k/Γ₀ →0, because the two dipoles oscillate in opposite phase. The termexp(2ikz)is the interference between the two dipoles, which depends on2z, i.e. the distance between real and image charges. When z → ∞, the decay rates approach the value Γ₀, but the perpendicular dipole approaches it faster, since it does not contain the far-field term1/(2kz).

• In free-space, the partial local density of statesρµis identical to the total density of states ρ. To show this, prove that

nµ.Im

_↔ G0

.nµ

= 1 3Im

Tr

_↔ G0

, (3)

where

↔

G₀ is the free-space dyadic.

In this problem we would like to prove

The free space dyadic Green function is given by

↔

G(R) =

(

↔

I −RˆR) +ˆ i kR(

↔

I −3 ˆRR)ˆ − 1 k²R²(

↔

I −3 ˆRR)ˆ

G₀,

with

G₀ = e^−ikr 4πr . By using

Tr(

↔

I) = 3 , Tr( ˆRR) = 1ˆ

n_µ

↔

In_µ= 1 , n_µ( ˆRR)nˆ _µ= 1/3 Now we have

Tr(

↔

G₀) =

3−1 + i

kr(3−3)− 1

k²r²(3−3)

G₀ = 2G₀,

n_µ·^↔G₀·n_µ=

(1−1 3) + i

kr(1−1)− 1

k²r²(1−1)

G₀ = 2 3G₀.

Im

Tr(

↔

G0)

= 2Im(G0) Im(nµ·^↔G0·nµ) = (2/3)G0







n_µ·Im{^↔G₀} ·n_µ

= 1 3Im{Tr

_↔ G₀

}.

• Two molecules, fluorescein (donor) and alexa green 532 (acceptor), are located in a plane centered between two perfectly conducting surfaces separated by the distance d. The emission spectrum of the donor(f_D) and the absorption spectrum of the acceptor (σ_A) are approximated by a superposition of two Gaussian distribution functions. Use the fit parameters from Section 8.6.2 in the text book Principles of Nano-Optics (Second edition) by Lukas Novotny.

1. Determine the Green’s function for this configuration.

2. Calculate the decay rate γ₀ of the donor in the absence of the acceptor.

3. Determine the transfer rate γD→Aas a function of the separation R between donor and acceptor. Assume random dipole orientations.

4. Plot the Förster radiusR0 as a function of the separationd.

Two molecules, fluorescein (donor) and alexa green 532 (acceptor), are located in a plane centered between two perfectly conducting surfaces separated by the distanced(see the figure).

1. To find the Green’s function one can use angular spectrum representation for a dipole.

e^ikr r = i

2π

¨∞

−∞

e^{ikxx+ikyy+ikzz} kz

dkx dky,

G0(r, r⁰) = e^ik|r−r0| 4π|r−r⁰|,

↔

G(r, r⁰) =

1+ 1 k²∇∇

G0.

Then

G0(r, r⁰) = i 8π²

ˆ e^{ikxx+ikyy+ikzz}

k_z dkxdky, wherek_zd=πnand as a resultk_z=πn/d. We also have

ω =kc= q

k²_x+k_y²+k_z²c= r

k²_k+π²n² d² ,

G0(r, r⁰) =

∞

X

n=1

1 8π²

ˆ e^ikxx+ikyy

πn d

h 2isin

πn d z

i

dkxdky,

wheree^ikzz+e^−ikzz= 2isinkzz.

2. The decay rateγ0 of the donor in the absence of the acceptor is related to γ ∝Im[Tr

↔

G] = Im[Gxx+Gyy+Gzz] = Im[3G0+ 1

k²∇²G0] = Im[2G0], where

∇²G₀=− 1 4π²

∞

X

n=1

ˆ d πn

−(k_x²+k_y²+π²n² d² )

e^ikxx+ikyysinπnz d

dk_xdk_y =−ω² c²G₀,

0

∞

X 1 ˆ

cos(kxx+kyy) +isin(kxx+kyy)h πn i

where

k_k²=k_x²+k²_y, dk_x dk_y =k_k dΦ dk_k, i(kxx+kyy) =ik_kρ.

Thus we have

G₀(r, r⁰) = 2π 4π²

∞

X

n=1

ˆ

k_kcos k_kρ

+isin k_kρ

πn d

h

sinπn d zi

dk_k,

Im[2G0] =−1 π

∞

X

n=1



 ˆ ^qω2

c2−^n2π2

d2

0

kk

cos k_kρ

sin ^nπz_d

nπ d

dkk+ ˆ ∞

qω2 c2−^n2π2

d2

dkk· · ·





Sincesin(πnz/d) for z= 0, it makes the second integral zero. At the end we have γ ∝ 1

π X

ˆ ^qω2

c2−^n2π2

d2

0

k_k

nπ d

sin²(nπz d )dkk,

which contains only the terms above cutoff (the decay rate is determined by the cavity modes).

3. The transfer rateγD→Aas a function of the separation R between donor and acceptor is

γD→A

γ₀ = 1 R⁶

ˆ ∞ 0

f0(ω)σ(ω)

n⁴(ω)ω⁴ T(ω)dω, where

T(ω) = 16π²k⁴R⁶

nA·^↔G·nD

2

.

Since the absolute value for random orientation is equal toTr[

↔

G], as a result we have

T(ω) =k⁴R⁶

X

n

ˆ kk

e^ikkρsin ^nπz_d

nπ d

dkk

2

.

We remark that here the integration includes both near (evanescent) and far field contri- butions ofk_k.

4. The Förster radiusR0 is related to transfer rate γD→A

γ₀ = R₀

R 6

,

R⁶₀ ≈R⁶

X

n

ˆ

k_ke^ikkρsin ^nπz_d

nπ d

dk_k

2

.

References

Principles of Nano-Optics (Second edition) by Lukas Novotny and Bert Hecht

Molecular scattering and fluorescence in strongly confined optical fields by Mario Agio.