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
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 zpolarized 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.
Figure 1. A sphere with radius a embedded in a dielectric medium (ε0) and irradiated by xpropagating, z-polarized light of wavelength λ. (a) a/λ = 1. (b) a/λ = 0.1 (the quasi-static field
case).
The laser E field has the general form
r
E = Ex xˆ + E y yˆ + Ez zˆ ,
(S.1)
where xˆ , yˆ , zˆ 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:
⎡ ⎛x
⎞⎤
Ez = Ez0 cos ⎢ 2π ⎜ − υ t ⎟ ⎥ .
(S.2)
⎠⎦
⎣ ⎝λ
2π
= k and 2πν = ω gives
Substituting
λ
Ez = Ez0 cos(kx − ωt ).
S-1
(S.3)
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
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:
∇⋅E =
ρ
ε0
c 2∇ × B =
∂E j
+
∂t ε 0
(S.4a)
∇×E = −
(S.4c)
∇⋅B = 0.
∂B
,
∂t
(S.4b)
(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 = Ez0 cos ωt ,
(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
∂E
∂B
= 0 and
= 0.
∂t
∂t
Thus Maxwell’s equations reduce to the equations for electrostatics (Equation S.6) and
magnetostatics (Equations S.7).
Electrostatics:
∇⋅E =
ρ
ε0
(S.6a)
∇×E = 0
(S.6b)
(S.7a)
∇⋅B = 0
(S.7b)
Magnetostatics:
∇×B =
j
ε 0c 2
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
∇⋅E = 0
(S.8a)
∇×E = 0.
(S.8b)
S-2
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
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
∇ ⋅∇Φ = 0,
(S.10)
∇ 2Φ = 0.
(S.11)
which gives Laplace’s equation:
Laplace’s Equation
We start by converting Laplace’s equation in Cartesian coordinates to spherical
coordinates using the relationships
x = r sin θ cos φ
y = r sin θ sin φ
r = (x2 + y2 + z2)1/2
θ = cos −1 ⎜ ⎟
⎛z⎞
⎝r⎠
and the coordinate system defined in Figure 2:
Figure 2. Coordinate axes system for Cartesian and polar coordinates.
S-3
z = r cos θ
⎛ y⎞
⎝x⎠
φ = tan −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
Thus we can rewrite Equation S.11
∇ 2 (r ,θ , φ )Φ(r , θ , φ ) = 0.
(S.12)
Recall that in spherical coordinates,
∇ 2 (r ,θ , φ ) =
1 ∂2
1
r+ 2
2
r ∂r
r
⎡ 1 ∂ ⎛
∂ ⎞
1 ∂2 ⎤
+
sin
θ
⎜
⎟
⎢
⎥,
∂θ ⎠ sin 2 θ ∂φ 2 ⎦
⎣ sin θ ∂θ ⎝
(S.13)
or the equivalent form
∇ 2 (r ,θ , φ ) =
1 ∂ ⎛ 2 ∂ ⎞ 1
⎜r
⎟+
r 2 ∂r ⎝ ∂r ⎠ r 2
⎡ 1 ∂ ⎛
1 ∂2 ⎤
∂ ⎞
sin
.
θ
+
⎟
⎢ sin θ ∂θ ⎜
∂θ ⎠ sin 2 θ ∂φ 2 ⎥⎦
⎝
⎣
(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
Φ(r ,θ , φ ) = F (r )Y (θ , φ ),
(S.15)
and substitute Equation S.15 into S.12 to give
∇ 2 (r , θ , φ ) [ F (r )Y (θ , φ ) ] = I + II + III = 0,
(S.16)
where (using Equation S.13)
1 ∂2
[rF (r )Y (θ , φ )]
r ∂r 2
(S.17a)
1
∂ ⎛
∂[ F (r )Y (θ , φ )] ⎞
⎜ sin θ
⎟
∂θ
r sin θ ∂θ ⎝
⎠
(S.17b)
I=
II =
2
III =
1
∂ 2 [ F (r )Y (θ , φ )]
.
∂φ 2
r 2 sin 2 θ
(S.17c)
Because the r-dependent operators act only on F(r) and the θ , φ operators act only on
Y (θ , φ ) , we can separate terms and rewrite Laplace’s equation as
S-4
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
⎡ ∂ 2 F (r ) 2 ∂F (r ) ⎤
I + II + III = Y (θ , φ ) ⎢
+
+
2
r ∂r ⎥⎦
⎣ ∂r
F (r ) ⎡ 1 ∂ ⎛
∂Y (θ , φ ) ⎞ ⎤ F (r ) ⎡ 1 ∂ 2Y (θ , φ ) ⎤
= 0.
⎜ sin θ
⎟ + 2 ⎢ 2
r 2 ⎢⎣ sin θ ∂θ ⎝
r ⎣ sin θ ∂φ 2 ⎥⎦
∂θ ⎠ ⎥⎦
(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 Y (θ , φ ) terms on the other side and (b) divide both sides
first by Y (θ , φ ) followed by F(r)/r2 to get
r 2 ⎡ ∂ 2 F (r ) 2 ∂F (r ) ⎤
1 ⎡ 1 ∂ ⎛
1 ∂ 2Y (θ , φ ) ⎤
∂Y (θ , φ ) ⎞
sin
θ
+
=
−
+
⎜
⎟
⎢
⎥
⎢
⎥.
F (r ) ⎣ ∂r 2
r ∂r ⎦
Y (θ , φ ) ⎣ sin θ ∂θ ⎝
∂θ ⎠ sin 2 θ ∂φ 2 ⎦
(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):
r 2 ⎡ ∂ 2 F (r ) 2 ∂F (r ) ⎤
+
⎢
⎥=G
F (r ) ⎣ ∂r 2
r ∂r ⎦
(S.20a)
⎧ 1 ∂ ⎛
1 ∂ 2Y (θ , φ ) ⎫
∂Y (θ , φ ) ⎞
sin
+
θ
⎨
⎬ = −GY (θ , φ ).
⎜
⎟
∂θ ⎠ sin 2 θ ∂φ 2 ⎭
⎩ sin θ ∂θ ⎝
(S.20b)
Separating Polar and Azimuthal Angles
A second separation of variables is required to isolate the θ - and φ - dependent parts of
Laplace’s equation. For this, we assume
Y (θ , φ ) = P(θ )Q(φ ).
(S.21)
Substituting Equation S.21 into Equation S.20b yields
⎧ 1 ∂ ⎛
1 ∂ 2 [ P (θ )Q (φ )] ⎫
∂[ P (θ )Q (φ )] ⎞
sin
+
θ
⎨
⎬ = −G [ P (θ )Q (φ ) ] . (S.22)
⎜
⎟
2
∂θ
∂φ 2
⎠ sin θ
⎩ sin θ ∂θ ⎝
⎭
∂2
∂
Because
only operates on P(θ) and
operates only on Q(φ ), we have
∂θ
∂φ 2
S-5
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
⎪⎧ Q(φ ) ∂
⎨
⎩⎪ sin θ ∂θ
⎛
∂ [ P (θ ) ] ⎞ P(θ ) ∂ 2 [Q(φ ) ] ⎫⎪
θ
sin
⎬ = −G [ P(θ )Q(φ )] .
⎜
⎟+
∂θ ⎠ sin 2 θ ∂φ 2 ⎪⎭
⎝
(S.23)
Divide through both sides by Q(φ ) , then by P(θ), and multiply by sin2 θ to get
2
∂ [ P(θ ) ] ⎞
sin 2 θ ∂ [ Q(φ ) ] ⎪⎫
⎪⎧ sin 2 θ ∂ ⎛
2
+
sin
θ
⎨
⎬ = −G sin θ .
⎜
⎟
2
2
∂θ ⎠ Q(φ ) sin θ ∂φ
⎪⎩ P(θ ) sin θ ∂θ ⎝
⎪⎭
(S.24)
Now rearrange to get all θ on the left-hand side and all φ on the right-hand side to get
⎪⎧ sin θ ∂
⎨
⎪⎩ P(θ ) ∂θ
2
⎛
∂ [ P(θ ) ] ⎞ ⎪⎫
1 ∂ [Q(φ ) ]
2
sin
θ
sin
θ
.
G
+
=
−
⎜
⎟⎬
Q(φ ) ∂φ 2
∂θ ⎠ ⎪⎭
⎝
(S.25)
This is only true for all θ , φ if both sides equal a constant arbitrarily called D so that
⎧ sin θ ∂ ⎛
∂[P(θ )] ⎞⎫
2
⎜ sin θ
⎟⎬ + G sin θ = D
⎨
∂θ ⎠⎭
⎩ P(θ ) ∂θ ⎝
2
1 ∂ [Q(φ ) ]
−
= D.
Q(φ ) ∂φ 2
(S.26b)
SUMMARY OF SEPARATION OF VARIABLES
Radial Equation :
Polar Equation :
Azimuthal Equation :
(S.26a)
r 2 ⎡ ∂ 2 F (r ) 2 ∂F (r ) ⎤
+
⎢
⎥=G
F (r ) ⎣ ∂r 2
r ∂r ⎦
⎧ sin θ ∂ ⎛
∂[P(θ )] ⎞⎫
2
⎜ sin θ
⎟⎬ + G sin θ = D
⎨
∂θ ⎠⎭
⎩ P(θ ) ∂θ ⎝
1 ∂ 2 [Q(φ )]
−
=D
Q(φ ) ∂φ 2
S-6
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
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
d 2Q(φ )
+ DQ(φ ) = 0.
dφ 2
(S.27)
Solutions to this equation have the form
Q(φ ) = e imφ
only if
(S.28)
m2 = D .
(S.29)
This can be checked by direct substitution:
dQ(φ )
= imeimφ and
dφ
d 2Q(φ ) 2 2 imφ
= i m e = − m 2 eimφ .
dφ 2
(S.30)
− m 2 e imφ + De imφ = 0
if and only if m2 = D.
(S.31)
Therefore,
The boundary condition on Q (φ ) is Q(φ ) must be single-valued and continuous over
0 ≤ φ ≤ 2π so that Q(φ ) = Q(φ + π ) .
Substituting Q(φ ) = e imφ into the boundary condition, we get e imφ = e im (φ + 2π ) . 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
Qm (φ ) = e imφ
where m2 = D and m = 0, ±1; ±2; ±3; ….
S-7
(S.32)
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
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
⎪⎧ sin θ d
⎨
⎪⎩ P(θ ) dθ
⎛
d [ P(θ ) ] ⎞ ⎪⎫
2
2
⎜ sin θ
⎟ ⎬ + G sin θ = m .
θ
d
⎝
⎠ ⎪⎭
(S.33)
Divide both sides by sin2 θ, multiply through by P(θ) and rearrange to get
d [ P(θ ) ] ⎞ ⎛
1 d ⎛
m2
θ
+
−
sin
G
⎜
⎟ ⎜
sin θ dθ ⎝
sin 2 θ
dθ ⎠ ⎝
⎞
⎟ P(θ ) = 0.
⎠
(S.34)
Carry out the indicated derivatives and then substitute x = cos θ and use the appropriate
trigonometric relationships to get
(1 − x 2 )
d 2 P (θ )
dP (θ ) ⎛
m2 ⎞
−
+
−
2
x
G
P(θ ) = 0.
⎜
2 ⎟
−
(1
)
dx 2
dx
x
⎝
⎠
(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
P (θ ) =
2l + 1 (l − | m |)! |m|
Pl (cos θ ).
2 (l + | m |)!
(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,
Y (θ , φ ) = P(θ )Q(φ )
(from S.21)
where:
Qm (φ ) = e imφ
(from S.32)
with
m = 0, ±1, ±2, ±3, ….
S-8
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
2l + 1 (l − | m |)! |m|
Pl (cos θ )
2 (l + | m |)!
and
Pi|m| (θ ) =
with
l = 0, 1, 2, 3, ….
so that:
Y1,m (θ , φ ) =
with
l = 0, 1, 2, 3, …. and m = -l, –l + 1 , … 0, 1, 2, … l – 1, l.
2l + 1 (l − | m |)! |m|
Pl (cos θ )e imφ
4π (l + | m |)!
(from S.36)
(S.37)
These are the NORMALIZED SPHERICAL HARMONICS. The table on the following
page shows these relationships for varying values of l and m.
S-9
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
Table of Yl,m(θ,ф)
The first few normalized spherical harmonics, Yl,m(θ,ф)
Y00 =
1
2(π )1 / 2
Y10 =
1⎛ 3⎞
⎜ ⎟
2⎝π ⎠
1/ 2
cos θ
1⎛ 3 ⎞
Y1±1= m ⎜
⎟
2 ⎝ 2π ⎠
Y20 =
1⎛ 5⎞
⎜ ⎟
4⎝π ⎠
1/ 2
sin θ exp(±iφ )
1/ 2
(3 cos 2 θ − 1)
1 ⎛ 15 ⎞
Y2±1= m ⎜
⎟
2 ⎝ 2π ⎠
Y2±2=
1 ⎛ 15 ⎞
⎟
⎜
4 ⎝ 2π ⎠
Y30 =
1⎛ 7⎞
⎜ ⎟
4⎝π ⎠
1/ 2
sin θ cos θ exp(±iφ )
1/ 2
sin 2 θ exp(±2iφ )
1/ 2
(5 cos 3 θ − 3 cos θ )
1 ⎛ 21 ⎞
Y3±1 = m ⎜ ⎟
8⎝ π ⎠
1 ⎛ 105 ⎞
Y3±2 =
⎟
⎜
4 ⎝ 2π ⎠
1/ 2
(5 cos 2 θ − 1) sin θ exp(±1φ )
1/ 2
sin 2 θ cos θ exp(±2iφ )
1/ 2
1 ⎛ 35 ⎞
Y3±3 = m ⎜ ⎟ sin 3 θ exp(±3iφ )
8⎝ π ⎠
___________________________________________________________
S-10
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
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
r 2 ⎡ d 2 Fl (r ) 2 dFl (r ) ⎤
+
⎥ = l (l + 1)
⎢
Fl (r ) ⎣ dr 2
r dr ⎦
(S.38)
Rearrange to a more standard form:
d 2 Fl (r ) 2 dFl (r )
F (r )
+
− l (l + 1) l 2 = 0.
2
dr
r dr
r
(S.39)
Assume a trial solution of the form Fl (r) = r n and calculate the derivatives to get
dFl (r ) d [r n ]
=
= nr n −1
dr
dr
(S.40a)
d 2 Fl (r ) d ⎡ d [r n ] ⎤ d
= ⎢
= ⎡⎣ nr n −1 ⎤⎦ = n(n − 1)r n − 2 .
⎥
2
dr
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
2
l (l + 1)r n
= 0.
n(n − 1)r n − 2 + nr n −1 −
r
r2
(S.41)
n(n − 1)r n − 2 + 2nr n − 2 − l (l + 1)r n-2 = 0.
(S.42)
Rearrange to get
Divide through by rn-2 and simplify to get the quadratic:
n 2 + n − l (l + 1) = 0.
(S.43)
Solve the quadratic for n; this yields two solutions:
n = l and n = −(l + 1).
Thus there are two solutions to the radial equation
S-11
(S.44)
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
⎧r l
Fl (r ) = ⎨ − (l +1) ,
⎩r
(S.45)
which can be combined to a general solution with coefficients to be determined by the
appropriate boundary conditions for the problem:
Fl (r ) = Ar l + Br − (l +1) .
(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 (r , θ , φ )Φ l,m (r , θ , φ )
is
(from S.12)
Φ l.m (r ,θ , φ ) = Fl (r )Yl ,m (θ , φ )
(S.47a)
Φl ,m ( r ,θ , φ ) = ⎡⎣ Al r l + Bl r −( l +1) ⎤⎦ Pl m ( cos θ ) eimφ ,
(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) E in ≠ E out
2) as r → ∞ E out → E 0 zˆ
E in → finite
3) as r → 0
S-12
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
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:
B ⎤
⎡
F0 = ⎢ A0 + 0 ⎥
r ⎦
⎣
Y00 =
1
,
4π
so that
Φ 00 =
1
4π
B0 ⎤
⎡
⎢ A0 + r ⎥ .
⎣
⎦
(S.48)
Evaluating Equation S.48 for r → 0 and r → ∞ we get
A0
.
4π
Φ 00 (r → 0) = ∞ and Φ 00 (r → ∞) =
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:
B ⎤
⎡
F1 = ⎢ A1r + 21 ⎥
r ⎦
⎣
Y10 =
3
cos θ .
4π
3
into the constants A1 and B1 to be determined by the boundary
4π
conditions, we get
If we subsume
Φ10 (r , θ ) = A1r cos θ +
B1 cos θ
.
r2
(S.49)
Convert to Cartesian coordinates using
z = r cos θ
and
(
r = x2 + y2 + z 2
to get
S-13
)
1/ 2
Supplemental Material: Annu. Rev. Phys. Chem. 2007. 58:267-97
doi: 10.1146/annurev.physchem.58.032806.104607
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Willets and Van Duyne
Φ10 ( x, y, z ) = A1 z +
B1 z
( x2 + y2 + z 2 )
3/ 2
.
(S.50)
Now calculate the E field from E(x,y,z) = −ΛΦ10(x,y,z) recalling that
∇ = ∇ x xˆ + ∇ y yˆ + ∇ z zˆ and
∇x =
∂
,
∂x
∇y =
∂
,
∂y
∇z =
∂
.
∂z
Thus,
⎛∂
∂
∂ ⎞
E( x, y, z ) = −⎜⎜ xˆ + yˆ + zˆ ⎟⎟Φ10 ( x, y, z )
∂y
∂z ⎠
⎝ ∂x
⎡
⎤
⎛ ∂
B1 z
∂
∂ ⎞⎢
⎥.
E ( x, y, z ) = − ⎜ xˆ + yˆ + zˆ ⎟ A1 z +
3 ⎥
⎢
2
2
2
2
∂
∂
∂
x
y
z
⎝
⎠⎢
( 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
⎡
⎤
3z ( xxˆ + yyˆ + zzˆ ) ⎥
zˆ
⎢
ˆ
.
−
E ( x, y, z ) = − A1z − B1
3
5 ⎥
⎢ 2
2
2
2
2
2
2
2
x
y
z
x
y
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
r = ( x2 + y 2 + z 2 ) .
1/ 2
Summary of boundary conditions
Inside vs. outside of sphere
Condition 1A:
Sphere surface (r = a)
Condition 2A:
E out ( x, y, z ) → E 0 zˆ as r → ∞
Condition 1B:
ε in E r ,in = ε out E r ,out (radial BC)
Condition 2B:
E in ( x, y, z ) → finite as r → 0
Eθ and Eφ are continuous (tangential BC)
S-14
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
▪ Applying condition 1A to Equation S.52: E out ( x, y, z ) → E 0 zˆ as r → ∞
⎡ zˆ 3z ( xxˆ + yyˆ + zzˆ ) ⎤
E out ( x, y, z ) = − A1,out zˆ − B1,out ⎢ 3 −
⎥⎦ .
r5
⎣r
(S.53)
Term in [ ] goes to 0 as r → ∞. Thus,
− A1,out = E0 .
(S.54)
▪ Applying condition 1B: E in ( x, y, z ) → finite as r → 0
⎡ zˆ 3z ( xxˆ + yyˆ + zzˆ ) ⎤
E in ( x, y, z ) = − A1,in zˆ − B1,in ⎢ 3 −
⎥⎦
r5
⎣r
(S.55)
Term in [ ] goes to ∞ as r → 0. Thus to keep the field finite,
B1,in = 0.
(S.56)
▪ Applying condition 2A: ε in E r ,in = ε out E r ,out (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)
E( x, y, z ) = E x xˆ + E y yˆ + E z zˆ
(S.57a)
and
E ( r ,θ , φ ) = Er rˆ + Eθ θˆ + Eφ φˆ .
(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:
φˆ ⋅ xˆ = − sin φ
rˆ ⋅ yˆ = sin θ sin φ
θˆ ⋅ xˆ = cos θ cos φ
θˆ ⋅ yˆ = cos θ sin φ
rˆ ⋅ zˆ = cos θ
θˆ ⋅ zˆ = − sin θ
φˆ ⋅ zˆ = 0 .
rˆ ⋅ xˆ = sin θ cos φ
S-15
φˆ ⋅ yˆ = cos φ
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
So we see that these can be combined to give
E r = rˆ ⋅ xˆ E x + rˆ ⋅ yˆ E y + rˆ ⋅ zˆ E z
E = θˆ ⋅ xˆ E + θˆ ⋅ yˆ E + θˆ ⋅ zˆ E
θ
x
y
z
Eφ = φˆ ⋅ xˆ E x + φˆ ⋅ yˆ E y + φˆ ⋅ zˆ E z .
Now rearrange Ein(x,y,z) into the form
Ein ( x, y, z ) = Ex ,in xˆ + E y ,in yˆ + Ez ,in zˆ .
(S.58)
And rearrange Equation S.55 to match the form of Equation S.58:
⎡ 3B z 2 B
⎤
⎡ 3B zx ⎤
⎡ 3B zy ⎤
Ein ( x, y, z ) = ⎢ 1,in5 ⎥ xˆ + ⎢ 1,in5 ⎥ yˆ + ⎢ 1,in5 − 1,3in − A1,in ⎥ zˆ . (S.59)
r
⎣ r
⎦
⎣ r
⎦
⎣ r
⎦
Thus, using the unit vector relationships, we get
⎡ 3B zx ⎤
⎡ 3B zy ⎤
Er ,in = (sin θ cos φ ) ⎢ 1,in5 ⎥ + (sin θ sin φ ) ⎢ 1,in5 ⎥ +
⎣ r
⎦
⎣ r
⎦
2
⎡ 3B z
⎤
B
(cos θ ) ⎢ 1,in5 − 1,3in − A1,in ⎥ ,
r
⎣ r
⎦
(S.60)
and similarly from Equation S.53,
⎡ 3B zx ⎤
⎡ 3B zy ⎤
+ (sin θ sin φ ) ⎢ 1,out
Er ,out = (sin θ cos φ ) ⎢ 1,out
⎥
⎥+
5
5
⎣ r
⎦
⎣ r
⎦
2
⎡ 3B z
⎤
B1,out
A
−
−
(cos θ ) ⎢ 1,out
1, out ⎥ .
5
r3
⎣ r
⎦
(S.61)
Evaluate Equation S.60 using B1,in = 0 determined from the boundary condition 1B:
Er ,in = − A1,in cos θ .
(S.62)
Now evaluate Equation S.61 using A1,out = −E0:
⎡ 3B zx ⎤
⎡ 3B zy ⎤
+ (sin θ sin φ ) ⎢ 1,out
Er ,out = (sin θ cos φ ) ⎢ 1,out
⎥
⎥+
5
5
⎣ r
⎦
⎣ r
⎦
2
⎡ 3B z
⎤
B1,out
E
−
+
(cos θ ) ⎢ 1,out
0 ⎥.
5
r3
⎣ r
⎦
Rearrange Equation S.63 and substitute the following:
S-16
(S.63)
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
sin θ cos φ =
x
,
r
sin θ sin φ =
z
y
, and cos θ =
r
y
to give
Er ,out
⎡ − cos θ 3 ⎛ 2
zy 2 zx 2 ⎞ ⎤
= E0 cos θ + B1,out ⎢
+ 5 ⎜ z cos θ +
+
⎟⎥ .
3
r ⎝
r
r ⎠⎦
⎣ r
(S.64)
We are now ready to use boundary condition 2A: ε in E r ,in = ε out E r ,out ; inserting Equations
S.62 and S.64 into the boundary condition:
−ε in A1,in cos θ = ε out E0 cos θ −
ε out B1,out cos θ
r3
+
(S.65)
3ε out B1,out ⎛ 2
zy 2 zx 2 ⎞
+
⎜ z cos θ +
⎟.
r5
r
r ⎠
⎝
Divide through by cosθ and substitute z = rcosθ to give
−ε in A1,in = ε out E0 −
ε out B1,out
r3
+
3ε out B1,out ⎛ 2 zy 2 zx 2 ⎞
+
⎜z +
⎟.
r5
z
z ⎠
⎝
(S.66)
Recall that r2 = (x2 + y2 + z2) so Equation S.66 can be simplified to
−ε in A1,in = ε out E0 −
ε out B1,out
r
3
+
3ε out B1,out
r
5
(r 2 ).
(S.67)
Canceling r2, combining terms, and factoring out εout gives
2B ⎤
⎡
−ε in A1,in = ε out ⎢ E0 + 1,3out ⎥ .
r ⎦
⎣
(S.68)
Letting r = a as we are interested in the surface of the sphere,
2B ⎤
⎡
−ε in A1,in = ε out ⎢ E0 + 1,3out ⎥ .
a ⎦
⎣
(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-17
Supplemental Material: Annu. Rev. Phys. Chem. 2007. 58:267-97
doi: 10.1146/annurev.physchem.58.032806.104607
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Willets and Van Duyne
Consider the axial case where Eθ ,in = Eθ ,out
Recall
Eθ = θˆ ⋅ xˆ Ex + θˆ ⋅ yˆ E y + θˆ ⋅ zˆ Ez .
(S.70)
Using Equation S.59 and the unit vector relationships, we get
⎡ 3B1,in zy ⎤
⎡ 3B1,in zx ⎤
Eθ ,in = (cos θ cos φ ) ⎢
θ
φ
(cos
sin
)
+
⎢
⎥+
⎥
5
5
⎣ r
⎦
⎣ r
⎦
⎤
⎡ 3B1,in z 2 B1,in
A
(− sin θ ) ⎢
−
−
1,in ⎥
5
r3
⎦⎥
⎣⎢ r
and similarly from Equation S.53 in analogy to Equation S.71 above:
⎡ 3B zx ⎤
⎡ 3B zy ⎤
Eθ ,out = (cos θ cos φ ) ⎢ 1,out
+ (cos θ sin φ ) ⎢ 1,out
⎥
⎥+
5
5
⎣ r
⎦
⎣ r
⎦
⎡ 3B1,out z 2 B1,out
⎤
− 3 − A1,out ⎥ .
( − sin θ ) ⎢
5
r
⎣ r
⎦
(S.71)
(S.72)
Evaluate Equation S.71 using B1,in = 0 (Equation S.56) determined from boundary
condition 1B:
Eθ ,in = A1,in sin θ .
(S.73)
Now evaluate Equation S.72 using A1,out = -E0 (Equation S.54) determined from
boundary condition 1A:
⎡ 3B zx ⎤
⎡ 3B zy ⎤
Eθ ,out = (cos θ cos φ ) ⎢ 1,out
+ (cos θ sin φ ) ⎢ 1,out
⎥
⎥+
5
5
⎣ r
⎦
⎣ r
⎦
⎡ 3B z 2 B1,out
⎤
− 3 − E0 ⎥ .
( − sin θ ) ⎢ 1,out
5
r
⎣ r
⎦
Rearrange Equation S.74 to get
Eθ ,out = − E0 sin θ +
3B1,out zy cos θ sin φ
r5
B1,out sin θ
+
3B1,out z 2 sin θ
−
r3
r5
3B1,out zx cos θ cos φ
r5
S-18
(S.74)
+
(S.75)
.
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
Equate Eθ,in (Equation S.73) and Eθ,out (Equation S.75) according to boundary condition
2B and divide by sinθ to get
A1,in = − E0 +
Substitute
B1,out
r
3
−
3B1,out z 2
r
5
x
= cos φ
r sin θ
+
3B1,out zy cos θ sin φ
r sin θ
5
+
3B1,out zx cos θ cos φ
r 5 sin θ
.
(S.76)
y
= sin φ into Equation S.76 and simplify to
r sin θ
and
get
A1,in = − E0 +
B1,out
r3
−
3B1,out z 2
r5
+
3B1,out zy 2 cos θ
r 6 sin 2 θ
+
3B1,out zx 2 cos θ
r 6 sin 2 θ
.
(S.77)
Substitute z = rcos θ into Equation S.77 and simplify to get
A1,in = − E0 +
B1,out
r3
−
3B1,out z 2
r5
+
3B1,out cos 2 θ
r 5 sin 2 θ
( x 2 + y 2 ).
(S.78)
Substitute (x2 + y2) = r2 sin2 θ and simplify to get
A1,in = − E0 +
B1,out
r3
−
3B1,out z 2
r5
+
3B1,out r 2 cos 2 θ
r5
.
(S.79)
Substitute z2 = r2 cos2 θ to get
A1,in = − E0 +
B1,out
r3
−
3B1,out z 2
r5
+
3B1,out z 2
r5
.
(S.80)
And this finally simplifies to
A1,in = − E0 +
B1,out
r3
.
(S.81)
.
(S.82)
Let r = a to yield the final result:
A1,in = − E0 +
B1,out
a3
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-19
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
2B ⎤
⎡
− ε in A1,in = ε out ⎢ E0 + 13,out ⎥
a ⎦
⎣
A1,in = − E0 +
B1,out
a3
(S.69)
(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
⎡ 3ε out
⎤
(S.83)
− A1,in = ⎢
⎥ E0 .
⎣ (2ε out + ε in ) ⎦
Now solve for B1,out to get
⎡ ε −ε
⎤
B1,out = a 3 E0 ⎢ in out ⎥ .
(S.84)
⎣ (ε in + 2ε out ) ⎦
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
⎛ 3ε out ⎞
Ein ( x, y, z ) = ⎜
⎟ E0 zˆ .
ε
ε
+
2
in ⎠
⎝ out
(S.85)
Substituting Equations S.54 and S.84 into Equation S.53, we obtain
⎡ ε −ε
⎤
⎡ zˆ 3z
⎤
Eout ( x, y, z ) = E0 zˆ − ⎢ in out ⎥ a 3 E0 ⎢ 3 − 5 ( xxˆ + yyˆ + zzˆ ) ⎥ .
r
⎣r
⎦
⎣ ( ε in + 2ε out ) ⎦
(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
⎧n
n( z ) = ⎨ A
⎩ nE
0≤ z≤d
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-20
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
Δλ = m ⋅ (n final − ninitial ) ,
(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:
E ( z ) = E 0 exp⎛⎜ − z ⎞⎟ .
⎝ ld ⎠
(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
⎡
⎛ − z ⎞⎤
⎛ −2 z ⎞ .
⎢⎣ exp ⎜⎝ ld ⎟⎠ ⎥⎦ = exp ⎜⎝
ld ⎟⎠
(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:
∞
neff =
2
n ( z ) exp ⎛⎜ −2 z ⎞⎟ dz.
∫
ld ⎠
⎝
ld 0
(S.91)
The factor of 2
normalizes the integral such that neff = nE in the absence of an
ld
adsorption layer. Inserting Equation S.87 into Equation S.91 and expanding the integral,
we get
neff =
d
∞
2⎡
⎛ −2 z ⎞ dz + n exp ⎛ −2 z ⎞ dz ⎤ .
exp
n
⎜
⎢ A
∫d E ⎜⎝ ld ⎟⎠ ⎥⎦
ld ⎟⎠
⎝
ld ⎣ ∫0
(S.92)
Evaluating this integral yields the following expression:
⎡
⎤
neff = nA ⎢1 − exp ⎛⎜ −2d ⎞⎟ ⎥ + nE exp ⎛⎜ −2d ⎞⎟ .
l
ld ⎠
d ⎠⎦
⎝
⎝
⎣
S-21
(S.93)
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
If we now substitute neff for nfinal and nE for ninitial in Equation S.88 above, we get the
following result:
⎧ ⎡
⎫
⎤
Δλ = m ⋅ ( neff − nE ) = m ⋅ ⎨nA ⎢1 − exp ⎛⎜ −2d ⎞⎟ ⎥ + nE exp ⎛⎜ −2d ⎞⎟ − nE ⎬ .
ld ⎠ ⎦
ld ⎠
⎝
⎝
⎩ ⎣
⎭
(S.94)
Grouping the nE terms and rearranging, we get
⎡
⎤
Δλ = m ⋅ ( nA − nE ) ⎢1 − exp ⎛⎜ −2d ⎞⎟ ⎥ .
l
d ⎠⎦
⎝
⎣
(S.95)
This is Equation 3 from the main text.
It is important to note that this equation reduces simply to Δλ = m ⋅ (n A − n E ) 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:
ΔλLSPR mLSPR
=
mSPR
ΔλSPR
⎡
⎛ −2d
⎞⎤
⎢1 − exp ⎜
ld , LSPR ⎠⎟ ⎥⎦
⎝
⎣
.
⎡
⎛ −2 d
⎞⎤
⎢1 − exp ⎜
ld , SPR ⎠⎟ ⎥⎦
⎝
⎣
(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 wavelengthshift performance of the two sensors as a function of layer thickness as shown in Figure 3.
S-22
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
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.
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.
S-23