supplemental appendix derivation of equation 1 …

Supplemental Material: Annu. Rev. Phys. Chem. 2007. 58:267-97 doi: 10.1146/annurev.physchem.58.032806.104607 Localized Surface Plasmon Resonance Spectroscopy and Sensing Willets and Van Duyne

S-1

SUPPLEMENTAL APPENDIX In the supplemental material, we derive Equations 1 and 3 from the second section of the main text (Theory) in greater detail. DERIVATION OF EQUATION 1 Maxwell’s Equations To begin, we consider a single metal sphere irradiated by y-propagating and z-polarized laser light at wavelength λ. The sphere is embedded in an external medium (viz., vacuum, gas, or liquid) with dielectric constant, εo. The dielectric constant of the sphere is ε1 and is assumed to be independent of the sphere size. The sphere, laser E field, and Cartesian coordinate system are shown in Figure 1A for the case a/λ = 1.0.

The laser E field has the general form ˆ ˆ ˆ,x y zE E E= + +E x y z

r (S.1)

where zyx ˆ,ˆ,ˆ are unit vectors along the Cartesian axes x, y, z. The values of Ex, Ey, and Ez are controlled by polarization to be Ex = Ey = 0 and Ez = non-zero for an x-propagating, z-polarized wave:

0 cos 2 .z zxE E tπ υλ

⎡ ⎤⎛ ⎞= −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ (S.2)

Substituting 2 and 2 giveskπ πν ωλ

= =

0 cos( ).z zE E kx tω= − (S.3)

Figure 1. A sphere with radius a embedded in a dielectric medium (ε0) and irradiated by x-propagating, z-polarized light of wavelength λ. (a) a/λ = 1. (b) a/λ = 0.1 (the quasi-static field case).


S-2

The general solution to this problem involves solving Maxwell’s equations (Equations S.4) to find the time-dependent electric and magnetic fields in all regions of space:

0ερ

=⋅∇ E (S.4a) t∂

∂−=×∇

BE , (S.4b)

0

2

εjEB +

∂∂

=×∇t

c (S.4c) 0=⋅∇ B . (S.4d)

Substantial simplification in the mathematics of this situation, while retaining the essence of the physics, can be obtained by making the long wavelength approximation. If we require that λ > 400 nm, then kx << ωt so that Ez = 0 cos ,zE tω (S.5) and if we further require that a/λ < 0.1, the electric field of the laser will be Eo, a vector pointing along the z axis, that can be assumed to be spatially independent for distances on the order of the sphere size. These conditions are depicted in Figure 1B. The external applied field Eo is approximately constant in time so that

0 and 0 =∂∂

=∂∂

ttBE .

Thus Maxwell’s equations reduce to the equations for electrostatics (Equation S.6) and magnetostatics (Equations S.7). Electrostatics:

0ερ

=⋅∇ E (S.6a) 0=×∇ E (S.6b)

Magnetostatics:

20cεjB =×∇ (S.7a) 0=⋅∇ B (S.7b)

This means that for static charges and currents, electricity and magnetism are distinct phenomena. The problem we wish to focus on is the solution of the electrostatic equations for the electric field inside and outside the metal sphere. Thus we ignore the magnetostatic equations. The metal sphere is a conductor and has no net charge density (i.e. ρ = 0) so that 0=⋅∇ E (S.8a) 0=×∇ E . (S.8b)


S-3

In order to calculate the E(x,y,z) field, we need to know the scalar potential, Φ(x,y,z), so that the field can be obtained from .= −∇E Φ (S.9) The scalar potential, Φ(x,y,z), can be obtained from 0,∇⋅∇ =Φ (S.10)

which gives Laplace’s equation: 2 0.∇ =Φ (S.11)

Laplace’s Equation We start by converting Laplace’s equation in Cartesian coordinates to spherical coordinates using the relationships x = r sin θ cosφ y = r sin θ sinφ z = r cos θ

r = (x2 + y2 + z2)1/2 ⎟⎠⎞

⎜⎝⎛= −

rz1cosθ ⎟

⎠⎞

⎜⎝⎛= −

xy1tanφ

and the coordinate system defined in Figure 2:

Figure 2. Coordinate axes system for Cartesian and polar coordinates.


S-4

Thus we can rewrite Equation S.11 2 ( , , ) ( , , ) 0.r rθ φ θ φ∇ =Φ (S.12) Recall that in spherical coordinates,

2 2

22 2 2 2

1 1 1 1( , , ) sin ,sin sin

r rr r r

θ φ θθ θ θ θ φ

⎡ ⎤∂ ∂ ∂ ∂⎛ ⎞∇ = + +⎜ ⎟⎢ ⎥∂ ∂ ∂ ∂⎝ ⎠⎣ ⎦

(S.13)

or the equivalent form

2

2 22 2 2 2

1 1 1 1( , , ) sin .sin sin

r rr r r r

θ φ θθ θ θ θ φ

⎡ ⎤∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞∇ = + +⎜ ⎟ ⎜ ⎟⎢ ⎥∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦

(S.14)

We now can separate the variables into radial and angular components. Radial and Angular Separation of Variables Assume the solution is of the form ( , , ) ( ) ( , ),r F r Yθ φ θ φ=Φ (S.15)

and substitute Equation S.15 into S.12 to give [ ]2 ( , , ) ( ) ( , ) 0,r F r Y I II IIIθ φ θ φ∇ = + + = (S.16)

where (using Equation S.13)

[ ]),()(12

2

φθYrrFrr

I∂∂

= (S.17a)

⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=θ

φθθθθ

)],()([sinsin1

2

YrFr

II (S.17b)

2

2

22

)],()([sin1

φφθ

θ ∂∂

=YrF

rIII . (S.17c)

Because the r-dependent operators act only on F(r) and the φθ , operators act only on

),( φθY , we can separate terms and rewrite Laplace’s equation as


S-5

2

2

2

2 2 2 2

( ) 2 ( )( , )

( ) 1 ( , ) ( ) 1 ( , )sin 0.sin sin

F r F rI II III Yr r r

F r Y F r Yr r

θ φ

θ φ θ φθθ θ θ θ φ

⎡ ⎤∂ ∂+ + = + +⎢ ⎥∂ ∂⎣ ⎦

⎡ ⎤⎡ ∂ ∂ ⎤ ∂⎛ ⎞ + =⎜ ⎟ ⎢ ⎥⎢ ⎥∂ ∂ ∂⎝ ⎠⎣ ⎦ ⎣ ⎦

(S.18)

Now do the following to Equation S.18: (a) rearrange to get all the F(r) terms on one side of the equation and all the ),( φθY terms on the other side and (b) divide both sides first by ),( φθY followed by F(r)/r2 to get

2 2 2

2 2 2

( ) 2 ( ) 1 1 ( , ) 1 ( , )sin .( ) ( , ) sin sinr F r F r Y Y

F r r r r Yθ φ θ φθ

θ φ θ θ θ θ φ⎡ ⎤ ⎡ ⎤∂ ∂ ∂ ∂ ∂⎛ ⎞+ = − +⎜ ⎟⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂ ∂⎝ ⎠⎣ ⎦ ⎣ ⎦

(S.19) The only way to maintain this equality is for each side of this expression to be equal to a constant. We arbitrarily call this constant G. As a result Laplace’s equation is now separated into two equations, one dealing with the radial dependence (Equation S.20a) and one with the angular dependence (Equation S.20b):

GrrF

rrrF

rFr

=⎥⎦

⎤⎢⎣

⎡∂

∂+

∂∂ )(2)(

)( 2

22

(S.20a)

2

2 2

1 ( , ) 1 ( , )sin ( , ).sin sin

Y Y GYθ φ θ φθ θ φθ θ θ θ φ

⎧ ⎫∂ ∂ ∂⎛ ⎞ + = −⎨ ⎬⎜ ⎟∂ ∂ ∂⎝ ⎠⎩ ⎭ (S.20b)

Separating Polar and Azimuthal Angles A second separation of variables is required to isolate the - and -θ φ dependent parts of Laplace’s equation. For this, we assume ( , ) ( ) ( ).Y P Qθ φ θ φ= (S.21) Substituting Equation S.21 into Equation S.20b yields

[ ]2

2 2

1 [ ( ) ( )] 1 [ ( ) ( )]sin ( ) ( ) .sin sin

P Q P Q G P Qθ φ θ φθ θ φθ θ θ θ φ

⎧ ⎫∂ ∂ ∂⎛ ⎞ + = −⎨ ⎬⎜ ⎟∂ ∂ ∂⎝ ⎠⎩ ⎭ (S.22)

Because θ∂∂ only operates on P(θ) and 2

2

φ∂∂ operates only on ),(φQ we have


S-6

[ ] [ ] [ ]2

2 2

( ) ( )( ) ( )sin ( ) ( ) .sin sin

P QQ P G P Qθ φφ θθ θ φ

θ θ θ θ φ⎧ ⎫⎛ ⎞∂ ∂∂⎪ ⎪+ = −⎨ ⎬⎜ ⎟∂ ∂ ∂⎪ ⎪⎝ ⎠⎩ ⎭

(S.23)

Divide through both sides by )(φQ , then by P(θ), and multiply by sin2 θ to get

[ ] [ ]22 22

2 2

P( ) Q( )sin sinsin G sin .P( )sin Q( )sin

θ φθ θθ θθ θ θ θ φ θ φ

⎧ ⎫⎛ ⎞∂ ∂∂⎪ ⎪+ = −⎨ ⎬⎜ ⎟∂ ∂ ∂⎪ ⎪⎝ ⎠⎩ ⎭ (S.24)

Now rearrange to get all θ on the left-hand side and all φ on the right-hand side to get

[ ] [ ]22

2

( ) ( )sin 1sin sin .( ) ( )

P QG

P Qθ φθ θ θ

θ θ θ φ φ⎧ ⎫⎛ ⎞∂ ∂∂⎪ ⎪+ = −⎨ ⎬⎜ ⎟∂ ∂ ∂⎪ ⎪⎝ ⎠⎩ ⎭

(S.25)

This is only true for all φθ , if both sides equal a constant arbitrarily called D so that

[ ] DGPP

=+⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛

∂∂

∂∂ θ

θθθ

θθθ 2sin)(sin)(

sin (S.26a)

[ ]2

2

( )1 .( )

QD

Qφ

φ φ∂

− =∂

(S.26b)

[ ]

[ ] )()(

1:

sin)(sin)(

sin :

)(2)()( :

2

2

2

2

22

DQQ

EquationAzimuthal

DGPPEquationPolar

GrrF

rrrF

rFr

EquationRadial

VARIABLESOFSEPARATIONOFSUMMARY

=∂

∂−

=+⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=⎥⎦

⎤⎢⎣

⎡∂

∂+

∂∂

φφ

φ

θθθθ

θθθ


S-7

Solving the Azimuthal Equation Because the azimuthal equation (Equation S.26b) is a function of φ only, replace partial derivatives by ordinary derivatives and rearrange to get

2

2

( ) ( ) 0.d Q DQd

φ φφ

+ = (S.27)

Solutions to this equation have the form φφ imeQ =)( (S.28) only if m2 = D . (S.29)

This can be checked by direct substitution:

2

2 2 22

( ) d ( ) .im im imdQ Qime and i m e m ed d

φ φ φφ φφ φ

= = = − (S.30)

Therefore, 02 =+− φφ imim Deem if and only if m2 = D. (S.31)

The boundary condition on Q )(φ is )(φQ must be single-valued and continuous over

πφ 20 ≤≤ so that )Q( )( πφφ +=Q . Substituting φφ imeQ =)( into the boundary condition, we get )2( πφφ += imim ee . This reduces to 1 = eim2π = cos2πm + i sin 2πm, • for m = 0; cos0 = 1 and sin0 = 0 ∴ ei2π(0) = 1 • for m = ±1; cos ±2π = 1 and sin ± 2π = 0 ∴ ei2π(±1) = 1 • for m = 2±; cos ± 4π = 1 and sin ±π = 0 ∴ ei2π(±2) = 1. So the final result is φφ im

m eQ =)( (S.32)

where m2 = D and m = 0, ±1; ±2; ±3; ….


S-8

Solving the Polar Equation

Because the polar equation (Equation S.26a) is a function of θ only, replace partial derivatives by ordinary derivatives and substitute m2 = D to get

[ ] 2 2( )sin sin sin .( )

d Pd G mP d d

θθ θ θθ θ θ

⎧ ⎫⎛ ⎞⎪ ⎪+ =⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

(S.33)

Divide both sides by sin2 θ, multiply through by P(θ) and rearrange to get

[ ] 2

2

( )1 sin ( ) 0.sin sin

d Pd mG Pd d

θθ θ

θ θ θ θ⎛ ⎞ ⎛ ⎞

+ − =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

(S.34)

Carry out the indicated derivatives and then substitute x = cos θ and use the appropriate trigonometric relationships to get

2 2

22 2

( ) ( )(1 ) 2 ( ) 0.(1 )

d P dP mx x G Pdx dx x

θ θ θ⎛ ⎞

− − + − =⎜ ⎟−⎝ ⎠ (S.35)

Equation S.35 is the associated Legendre equation. The solution of this equation is summarized below:

G = l(l+1); where l = 0, 1, 2, 3, …

and

|m|2 1 ( | |)!( ) (cos ).2 ( | m |)! l

l l mP Pl

θ θ+ −=

+ (S.36)

Combining the Azimuthal and Axial Solutions Now we are in a position to combine the results for the polar and azimuthal equations to give the total angular dependence. By our original assumption, )()(),( φθφθ QPY = (from S.21) where: φφ im

m eQ =)( (from S.32) with m = 0, ±1, ±2, ±3, ….


S-9

and )(cos|)!m|( |)!|(

212)(P |m||m| θθ li P

lmll

+−+

= (from S.36)

with l = 0, 1, 2, 3, ….

so that: φθπ

φθ imlm eP

lmllY )(cos

|)!m|( |)!|(

412),( |m|

,1 +−+

= (S.37)

with l = 0, 1, 2, 3, …. and m = -l, –l + 1 , … 0, 1, 2, … l – 1, l. These are the NORMALIZED SPHERICAL HARMONICS. The table on the following page shows these relationships for varying values of l and m.


S-10

Table of Yl,m(θ,ф)

The first few normalized spherical harmonics, Yl,m(θ,ф)

Y00 = 2/1)(21π

Y10 = θπ

cos321 2/1

⎟⎠⎞

⎜⎝⎛

Y1±1= )exp(sin23

21 2/1

φθπ

i±⎟⎠⎞

⎜⎝⎛

m

Y20 = )1cos3(541 2

2/1

−⎟⎠⎞

⎜⎝⎛ θπ

Y2±1= )exp(cossin215

21 2/1

φθθπ

i±⎟⎠⎞

⎜⎝⎛

m

Y2±2= )2exp(sin215

41 2

2/1

φθπ

i±⎟⎠⎞

⎜⎝⎛

Y30 = )cos3cos5(741 3

2/1

θθπ

−⎟⎠⎞

⎜⎝⎛

Y3±1 = )1exp(sin)1cos5(2181 2

2/1

φθθπ

±−⎟⎠⎞

⎜⎝⎛

m

Y3±2 = )2exp(cossin2105

41 2

2/1

φθθπ

i±⎟⎠⎞

⎜⎝⎛

Y3±3 = )3exp(sin3581 3

2/1

φθπ

i±⎟⎠⎞

⎜⎝⎛

m

___________________________________________________________


S-11

Solving the Radial Equation Since the radial equation (Equation S.20a) is a function of r only, replace partial derivatives by ordinary derivatives and substitute G = l(l + 1) to get

( )( ) ( ) ( )12

2

22

+=⎥⎦

⎤⎢⎣

⎡+ ll

drrdF

rdrrFd

rFr ll

l

(S.38)

Rearrange to a more standard form:

2

2 2

( ) ( ) F ( )2 ( 1) 0.l l ld F r dF r rl ldr r dr r

+ − + = (S.39)

Assume a trial solution of the form nr(r) =lF and calculate the derivatives to get

1][)( −== nn

l nrdrrd

drrdF

(S.40a)

2

1 22

( ) [ ] ( 1) .n

n nld F r d d r d nr n n rdr dr dr dr

− −⎡ ⎤⎡ ⎤= = = −⎢ ⎥ ⎣ ⎦

⎣ ⎦ (S.40b)

Now substitute the derivatives (Equations S.40a and S.40b) and the trial function into the radial equation (Equation S.39) to get

n

2 12

2 ( 1)r( 1) 0.n n l ln n r nrr r

− − +− + − = (S.41)

Rearrange to get 2 2 n-2( 1) 2 ( 1)r 0.n nn n r nr l l− −− + − + = (S.42)

Divide through by rn-2 and simplify to get the quadratic: 2 ( 1) 0.n n l l+ − + = (S.43)

Solve the quadratic for n; this yields two solutions: ln = and ( 1).n l= − + (S.44)

Thus there are two solutions to the radial equation


S-12

( 1)

( ) ,

l

l l

rF r

r− +

⎧= ⎨⎩

(S.45)

which can be combined to a general solution with coefficients to be determined by the appropriate boundary conditions for the problem: ( 1)( ) .l l

lF r Ar Br− += + (S.46) Combining the Radial and Angular Solutions Combine the solutions for radial, polar, and azimuthal equations to give a general solution for Laplace’s Equation in spherical coordinates. So the general solution for ),,(),,(2 φθφθ rr ml,Φ∇ (from S.12)

is ),()(),,( m, φθφθ lll.m YrFr =Φ (S.47a) ( ) ( ) ( )1

, , , cos ,mll iml m l l lr A r B r P e φθ φ θ− +⎡ ⎤= +⎣ ⎦Φ (S.47b)

with l = 0, 1, 2, … and m = −l, −l + 1, . . . −2, −1, 0, 1, 2, . . . , l − 1, l .

Applying Boundary Conditions to the Solution At this point it is useful to go back to the physical picture of the metal sphere in a constant electric field (viz., the long wavelength approximation, Figure 1B) to see the origin of the geometrically derived boundary conditions. It is clear that the geometry of this problem is axially symmetric (Figure 1B/2). Thus there is no dependence on the azimuthal angle, φ. This corresponds to the m = 0 solutions of the azimuthal equation. In general the E field inside the sphere will be different from that outside the sphere. We know that the incident E field will not be affected by the sphere as you move far from the sphere (i.e. r → ∞). We also know that the field at the center of the sphere should have some finite value (i.e. the field will be nonzero). These three conditions can be expressed mathematically as follows: 1) outin EE ≠ 2) as r → ∞ zEout ˆ0E→ 3) as r → 0 finite→inE


S-13

Because we know that m = 0 by axial symmetry, let us examine some solutions of Equations S.46, S.37, and S.47 for specific values of l. ▪ For m = 0, l = 0:

⎥⎦⎤

⎢⎣⎡ +=

rBAF 0

00 001 ,4

Yπ

=

so that

000 0

1 .4

BArπ

⎡ ⎤= +⎢ ⎥⎣ ⎦Φ (S.48)

Evaluating Equation S.48 for r → 0 and r → ∞ we get

000 00( 0) and ( ) .

4Ar rπ

→ = ∞ →∞ =Φ Φ

Because E = −ΛΦ, these solutions correspond to

Ein(r → 0) = ∞ and Eout (r → ∞) = 0.

These solutions violate the three boundary conditions established above.

∴ A0 = B0 = 0

▪ For m = 0, l = 1:

⎥⎦⎤

⎢⎣⎡ += 2

111 r

BrAF 103 cos .

4Y θ

π=

If we subsume π43 into the constants A1 and B1 to be determined by the boundary

conditions, we get

110 1 2

cos( , ) cos .Br A rr

θθ θ= +Φ (S.49)

Convert to Cartesian coordinates using

θcosrz = and ( ) 2/1222 zyxr ++=

to get


S-14

( )

110 1 3/ 22 2 2

( , , ) .B zx y z A zx y z

= ++ +

Φ (S.50)

Now calculate the E field from E(x,y,z) = −ΛΦ10(x,y,z) recalling that

zyx ˆˆˆ zyx ∇+∇+∇=∇ and

, , .x y zx y z∂ ∂ ∂

∇ = ∇ = ∇ =∂ ∂ ∂

Thus,

( ) ( )zyxzyx

zyx ,,ˆˆˆ,, 10ΦzyxE ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

−=

( )( )

11 3

2 2 2 2ˆ ˆ ˆ, , .B zx y z A z

x y z x y z

⎡ ⎤⎛ ⎞∂ ∂ ∂ ⎢ ⎥= − + + +⎜ ⎟ ⎢ ⎥∂ ∂ ∂⎝ ⎠ + +⎢ ⎥⎣ ⎦

E x y z (S.51)

Compute E(x,y,z) by taking the partial derivatives for each component of the field separately and applying the chain rule as needed to get

( )( )

( )( )

1 1 3 52 2 2 2 2 22 2

ˆ ˆ ˆ3ˆˆ, , .z x y z

x y z A Bx y z x y z

⎡ ⎤+ +⎢ ⎥= − − −⎢ ⎥+ + + +⎢ ⎥⎣ ⎦

x y zzE z (S.52)

Now we must solve for A1 and B1 using the boundary conditions at the surface of the sphere (viz., r = a). Remember that

( )1/ 22 2 2 .r x y z= + +

Summary of boundary conditions

Inside vs. outside of sphere Sphere surface (r = a)

Condition 1A:

( ) zE ˆ,, 0Ezyxout → as r → ∞

Condition 2A:

outroutinrin ,, EE εε = (radial BC)

Condition 1B:

( ) finitezyxin →,,E as r → 0

Condition 2B:

Eθ and Eφ are continuous (tangential BC)


S-15

▪ Applying condition 1A to Equation S.52: ( ) zE ˆ,, 0Ezyxout → as r → ∞

( ) ( )⎥⎦⎤

⎢⎣⎡ ++

−−−= 53,1,1ˆˆˆ3ˆˆ,,

rzyxz

rBAzyx outoutout

zyxzzE . (S.53)

Term in [ ] goes to 0 as r → ∞. Thus, 1, 0.outA E− = (S.54) ▪ Applying condition 1B: ( ) finitezyxin →,,E as r → 0

( ) ( )⎥⎦⎤

⎢⎣⎡ ++

−−−= 53,1,1ˆˆˆ3ˆˆ,,

rzyxz

rBAzyx ininin

zyxzzE (S.55)

Term in [ ] goes to ∞ as r → 0. Thus to keep the field finite, 1, 0.inB = (S.56) ▪ Applying condition 2A: outroutinrin ,, EE εε = (radial BC) First we must compute the radial components of E. The relationship between the spherical coordinate components of E and the Cartesian coordinate components of E is given by (see D. Fitts, Vector Analysis in Chemistry, McGraw-Hill, NewYork, 1974, pp. 124--25) ( ) zyxE ˆˆˆ,, zyx EEEzyx ++= (S.57a) and ( ) ˆˆ ˆ, , .rr E E Eθ φθ φ = + +E r θ φ (S.57b)

To get Er, Eθ, and Eф from Ex, Ey, and Ez, use the following:

Er = sinθcosφEx + sinθsinφEy + cosθEz

Eθ = cosθcosφEx + cosθsinφEy – sinθEz

Eφ = −sinφEx + cosφEy,

as well as the following relationships between the unit vectors:

φθ cossinˆˆ =⋅ xr φθ coscosˆˆ =⋅ xθ φsinˆˆ −=⋅ xφ φθ sinsinˆˆ =⋅ yr φθ sincosˆˆ =⋅ yθ φcosˆˆ =⋅ yφ

θcosˆˆ =⋅ zr θsinˆˆ −=⋅ zθ 0ˆˆ =⋅ zφ .


S-16

So we see that these can be combined to give zyxr EEEE zryrxr ˆˆˆˆˆˆ ⋅+⋅+⋅=

zyx EEEE zθyθxθ ˆˆˆˆˆˆ ⋅+⋅+⋅=θ

zyx EEEE zφyφxφ ˆˆˆˆˆˆ ⋅+⋅+⋅=φ . Now rearrange Ein(x,y,z) into the form ( ) , , ,ˆ ˆ ˆ, , .in x in y in z inx y z E E E= + +E x y z (S.58) And rearrange Equation S.55 to match the form of Equation S.58:

( )2

1, 1, 1, 1,1,5 5 5 3

3 3 3ˆ ˆ ˆ, , .in in in in

in in

B zx B zy B z Bx y z A

r r r r⎡ ⎤⎡ ⎤ ⎡ ⎤

= + + − −⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

E x y z (S.59)

Thus, using the unit vector relationships, we get

1, 1,, 5 5

21, 1,

1,5 3

3 3(sin cos ) (sin sin )

3(cos ) ,

in inr in

in inin

B zx B zyr r

B z BA

r r

θ φ θ φ

θ

⎡ ⎤ ⎡ ⎤= + +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤

− −⎢ ⎥⎣ ⎦

E

(S.60)

and similarly from Equation S.53,

1, 1,, 5 5

21, 1,

1,5 3


3(cos ) .

out outr out

out outout

B zx B zyr r

B z BA

r r

θ φ θ φ

θ

⎡ ⎤ ⎡ ⎤= + +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤

− −⎢ ⎥⎣ ⎦

E

(S.61)

Evaluate Equation S.60 using B1,in = 0 determined from the boundary condition 1B: , 1, cos .r in inA θ= −E (S.62) Now evaluate Equation S.61 using A1,out = −E0:

1, 1,, 5 5

21, 1,

05 3


3(cos ) .

out outr out

out out

B zx B zyr r

B z BE

r r

θ φ θ φ

θ

⎡ ⎤ ⎡ ⎤= + +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤

− +⎢ ⎥⎣ ⎦

E

(S.63)

Rearrange Equation S.63 and substitute the following:


S-17

,cossinrx

=φθ ,sinsinry

=φθ and yz

=θcos

to give

2 2

2, 0 1, 3 5

cos 3cos cos .r out outzy zxE B z

r r r rθθ θ

⎡ ⎤⎛ ⎞−= + + + +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦E (S.64)

We are now ready to use boundary condition 2A: outroutinrin ,, EE εε = ; inserting Equations S.62 and S.64 into the boundary condition:

1,1, 0 3

2 21, 2

5

coscos cos

3cos .

out outin in out

out out

BA E

rB zy zxz

r r r

ε θε θ ε θ

εθ

− = − +

⎛ ⎞+ +⎜ ⎟

⎝ ⎠

(S.65)

Divide through by cosθ and substitute z = rcosθ to give

2 2

1, 1, 21, 0 3 5

3.out out out out

in in out

B B zy zxA E zr r z z

ε εε ε

⎛ ⎞− = − + + +⎜ ⎟

⎝ ⎠ (S.66)

Recall that r2 = (x2 + y2 + z2) so Equation S.66 can be simplified to

1, 1, 21, 0 3 5

3( ).out out out out

in in out

B BA E r

r rε ε

ε ε− = − + (S.67)

Canceling r2, combining terms, and factoring out εout gives

1,1, 0 3

2.out

in in out

BA E

rε ε

⎡ ⎤− = +⎢ ⎥

⎣ ⎦ (S.68)

Letting r = a as we are interested in the surface of the sphere,

1,1, 0 3

2.out

in in out

BA E

aε ε

⎡ ⎤− = +⎢ ⎥

⎣ ⎦ (S.69)

We cannot simplify this expression any further, so we now turn to our next boundary condition. ▪ Applying condition 2B: Eθ and Eφ are continuous (tangential BC)


S-18

Consider the axial case where outin EE ,, θθ = Recall ˆ ˆ ˆˆ ˆ ˆ .x y zE E E Eθ = ⋅ + ⋅ + ⋅θ x θ y θ z (S.70) Using Equation S.59 and the unit vector relationships, we get

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−−

+⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡=

ininin

ininin

Ar

Br

zB

rzyB

rzxB

E

,13,1

5

2,1

5,1

5,1

,

3)sin(

3)sin(cos

3)cos(cos

θ

φθφθθ

(S.71)

and similarly from Equation S.53 in analogy to Equation S.71 above:

1, 1,, 5 5

21, 1,

1,5 3

3 3(cos cos ) (cos sin )

3( sin ) .

out outout

out outout

B zx B zyE

r r

B z BA

r r

θ θ φ θ φ

θ

⎡ ⎤ ⎡ ⎤= + +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤

− − −⎢ ⎥⎣ ⎦

(S.72)

Evaluate Equation S.71 using B1,in = 0 (Equation S.56) determined from boundary condition 1B: , 1, sin .in inE Aθ θ= (S.73) Now evaluate Equation S.72 using A1,out = -E0 (Equation S.54) determined from boundary condition 1A:

1, 1,, 5 5

21, 1,

05 3

3 3(cos cos ) (cos sin )

3( sin ) .

out outout

out out

B zx B zyE

r r

B z BE

r r

θ θ φ θ φ

θ

⎡ ⎤ ⎡ ⎤= + +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤

− − −⎢ ⎥⎣ ⎦

(S.74)

Rearrange Equation S.74 to get

21, 1,

, 0 3 5

1, 1,5 5

sin 3 sinsin

3 cos sin 3 cos cos.

out outout

out out

B B zE E

r rB zy B zx

r r

θ

θ θθ

θ φ θ φ

= − + − +

+

(S.75)


S-19

Equate Eθ,in (Equation S.73) and Eθ,out (Equation S.75) according to boundary condition 2B and divide by sinθ to get

2

1, 1, 1, 1,1, 0 3 5 5 5

3 3 cos sin 3 cos cos.

sin sinout out out out

in

B B z B zy B zxA E

r r r rθ φ θ φθ θ

= − + − + + (S.76)

Substitute φθ

cossin

=r

x and φθ

sinsin

=r

y into Equation S.76 and simplify to

get

2 2 2

1, 1, 1, 1,1, 0 3 5 6 2 6 2

3 3 cos 3 cos.

sin sinout out out out

in

B B z B zy B zxA E

r r r rθ θ

θ θ= − + − + + (S.77)

Substitute z = rcosθ into Equation S.77 and simplify to get

2 2

1, 1, 1, 2 21, 0 3 5 5 2

3 3 cos( ).

sinout out out

in

B B z BA E x y

r r rθ

θ= − + − + + (S.78)

Substitute (x2 + y2) = r2 sin2θ and simplify to get

2 2 2

1, 1, 1,1, 0 3 5 5

3 3 cos.out out out

in

B B z B rA E

r r rθ

= − + − + (S.79)

Substitute z2 = r2 cos2 θ to get

2 2

1, 1, 1,1, 0 3 5 5

3 3.out out out

in

B B z B zA E

r r r= − + − + (S.80)

And this finally simplifies to

1,1, 0 3 .out

in

BA E

r= − + (S.81)

Let r = a to yield the final result:

1,1, 0 3 .out

in

BA E

a= − + (S.82)

Solving for the Final E Field Expressions (Equation 1) We now have two equations in two unknowns to solve for A1,in and B1,out, i.e. Equations S.69 and S.82 reproduced below:


S-20

⎥⎦

⎤⎢⎣

⎡+=− 3

,10,1

2aB

EA outoutinin εε (S.69)

1,1, 0 3 .out

in

BA E

a= − + (S.82)

Solve for A1,in first by eliminating B1,out. Multiply both sides of Equation S.82 by −2εout and add to Equation S.69 to get

1, 03 .

(2 )out

inout in

A Eεε ε

⎡ ⎤− = ⎢ ⎥+⎣ ⎦

(S.83)

Now solve for B1,out to get

31, 0 .

( 2 )in out

outin out

B a E ε εε ε

⎡ ⎤−= ⎢ ⎥+⎣ ⎦

(S.84)

Recall our previous expressions for Ein(x,y,z) (Equation S.55) and Eout(x,y,z) (Equation S.53). Substituting Equations S.56 and S.83 into Equation S.55, we obtain

( ) 03 ˆ, , .

2out

inout in

x y z Eεε ε

⎛ ⎞= ⎜ ⎟+⎝ ⎠

E z (S.85)

Substituting Equations S.54 and S.84 into Equation S.53, we obtain

( ) ( ) ( )30 0 3 5

ˆ 3ˆ ˆ ˆ ˆ, , .2

in outout

in out

zx y z E a E x y zr r

ε εε ε

⎡ ⎤− ⎡ ⎤= − − + +⎢ ⎥ ⎢ ⎥+ ⎣ ⎦⎣ ⎦

zE z x y z (S.86)

This is Equation 1 from the main text. DERIVATION OF EQUATION 3 This derivation originates from References 25 and 40 in the main text. We begin by considering the shift in the localized surface plasmon resonance (LSPR) originating from the adsorption of a layer of thickness d and refractive index nA onto the nanoparticle surface. At all distances greater than d from the nanoparticle surface, we assume that there is an external environment with refractive index nE. Thus the distance-dependent refractive index can be written as

0

( ) .A

E

n z dn z

n d z≤ ≤⎧

= ⎨ < < ∞⎩ (S.87)

We are interested in measuring the wavelength-shift in the LSPR spectrum due to the introduction of the adsorption layer; this can be expressed as follows:


S-21

( )initialfinal nnm −⋅=Δλ , (S.88) where m is the bulk refractive index sensitivity of the nanoparticles (in units of energy RIU-1 where RIU is a refractive index unit) and ninitial and nfinal are the refractive indices of the initial state and the final state, respectively. We can define ninitial as the external environment in the absence of the adsorption layer; i.e. ninitial = nE. However, the refractive index of the final state is more complex because it involves some combination of the refractive index of both the adsorbed layer as well as the external environment, as shown in Equation S.87 above. Moreover, the refractive index should be weighted by the distant-dependent intensity of the electromagnetic (EM) field since the fields that probe the local refractive index are more intense closer to the nanoparticle surface (see, for example, Figure 5 in the text). We can approximate the EM field as decaying exponentially with some characteristic length given by ld:

( ) ⎟⎠⎞⎜

⎝⎛−=

dlzEzE exp0 . (S.89)

Thus, to normalize the refractive index by the relative field intensity (e.g. I/I0), we need to use the EM field squared and our normalization factor becomes

2

2exp exp .d d

z zl l

⎡ ⎤⎛ ⎞ ⎛ ⎞− −=⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ (S.90)

We can now define an effective refractive index, neff, which normalizes the z-dependent refractive index by the intensity factor in Equation S.90:

( )0

2 2exp .effdd

zn n z dzll

∞⎛ ⎞−= ⎜ ⎟⎝ ⎠∫ (S.91)

The factor of

dl2 normalizes the integral such that neff = nE in the absence of an

adsorption layer. Inserting Equation S.87 into Equation S.91 and expanding the integral, we get

0

2 2 2exp exp .d

eff A Ed dd d

z zn n dz n dzl ll

∞⎡ ⎤⎛ ⎞ ⎛ ⎞− −= +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∫ ∫ (S.92)

Evaluating this integral yields the following expression:

2 21 exp exp .eff A Ed d

d dn n nl l⎡ ⎤⎛ ⎞ ⎛ ⎞− −= − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

(S.93)


S-22

If we now substitute neff for nfinal and nE for ninitial in Equation S.88 above, we get the following result:

( ) 2 21 exp exp .eff E A E Ed d

d dm n n m n n nl lλ ⎧ ⎫⎡ ⎤⎛ ⎞ ⎛ ⎞− −Δ = ⋅ − = ⋅ − + −⎜ ⎟ ⎜ ⎟⎨ ⎬⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎩ ⎭ (S.94)

Grouping the nE terms and rearranging, we get

( ) 21 exp .A Ed

dm n n lλ ⎡ ⎤⎛ ⎞−Δ = ⋅ − − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ (S.95)

This is Equation 3 from the main text. It is important to note that this equation reduces simply to ( )EA nnm −⋅=Δλ in bulk solvents (where d goes to infinity) where nA is the refractive index of the new solvent and nE is the refractive index of the original bulk medium. Moreover, this approach also works for multilayer systems, as shown in References 25 and 40, following the same derivation. Because this equation is valid for both SPR and LSPR wavelength-shift measurements, we can use it to directly compare the performance of the two types of sensors:

,

,

21 exp.

21 exp

d LSPRLSPR LSPR

SPR SPR

d SPR

dlm

m dl

λλ

⎡ ⎤⎛ ⎞−− ⎜ ⎟⎢ ⎥Δ ⎝ ⎠⎣ ⎦=Δ ⎡ ⎤⎛ ⎞−− ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

(S.96)

Here the (nA – nE) terms cancel out because we are comparing sensor performance under identical experimental conditions. From the literature, we know the following relationships (References 25 and 40 from main text): mSPR = 9000 nm RIU-1 and mLSPR = 200 nm RIU-1. This indicates that the SPR outperforms the LSPR sensor for sensing changes in bulk refractive index by a factor of 45. However, as previously stated, the response of the two techniques becomes comparable when measuring short range changes in refractive index due to a molecular adsorption layer. This is due to the difference in the EM-field decay length, i.e. ld,SPR = 200 nm and ld,LSPR = 5 nm. Inserting the values for ld and m into Equation S.96, we can now directly compare the wavelength-shift performance of the two sensors as a function of layer thickness as shown in Figure 3.


S-23

As the layer thickness approaches zero, the response of the two sensors becomes nearly identical (i.e, the ratio approaches one). Thus, despite the large difference in the bulk refractive index responses, the two sensors are comparable for thin adsorption layers, which is important for most sensing applications (such as sensing antigen-antibody binding). At this point, we have only considered the bulk refractive index response (m) and the EM-field decay length (ld) for comparing the LSPR and surface plasmon resonance (SPR) sensor response. However, one should also account for the different surface areas of the two substrates by normalizing against the number of adsorbed molecules to get the wavelength-shift response per molecule. We can approximate this by comparing the area occupied by the metal for an SPR thin film with the area occupied by a nanoparticle array. Consider a 10 × 10-μm region of an NSL-fabricated sample. In the case when 500-nm diameter nanospheres are used as a deposition mask, approximately 460 spheres can occupy that space (owing to the fact that it is a hexagonally close-packed array rather than a square array). Thus, one can subtract the area occupied by the spheres (460*π*0.252 = 90.3 μm2) from the area of the region of interest (100 μm2) to get an approximate value of the area occupied by the nanoparticles (9.7 μm2). Compared to a continuous thin film which will occupy all 100 μm2 of the region of interest, the nanoparticles occupy only 9.7 μm2, or 90.3% less area. If we now consider a full-coverage adsorbed monolayer, the LSPR sensor response originates from 90.3% fewer molecules than the SPR sensor. Thus, in terms of sensor response per adsorbed molecule, the LSPR sensor actually outperforms the SPR sensor. As future experiments push toward sensing smaller concentrations (even single molecules), this feature of LSPR sensors will become increasingly important.

Figure 3. Comparison of the LSPR and SPR wavelength-shift responses using Equation S.96 and values mSPR = 9000 nm RIU-1, mLSPR = 200 nm RIU-1, ld,SPR = 200 nm, and ld,LSPR = 5 nm.

supplemental appendix derivation of equation 1 …

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