## Localized surface plasmons In this chapter, we adapt the plasmon behavior from the a surface to some confined metal particle. Under such case, the plasmons are called localized surface plasmons, or LSPs. The behavior is studied in an exactly same way, i.e., write the Maxwell equation and define the boundary conditions. Then have the differential equations solved. ### Laplace equation for sphere in static electric field #### Derivation ![[Drawing 2024-07-20 15.38.33.excalidraw.svg]] The problem setup is shown in the figure. We have a sphere in a static electric field $E_{0}$. The radius is $a$, and dielectric constant inside and outside the sphere is $\varepsilon_{sph}(\omega)$ and $\varepsilon_{d}(\omega)$. The geometry makes this problem easier in spherical system, so we define $\phi$, $\theta$ and $r$. Under the given condition, the standard electrostatic problem, $\mathbf{E}=-\nabla V$ under no charge density condition, is $-\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_{0}}=0$ $\nabla \cdot (-\nabla V)=0=\nabla^{2} V$ becomes $\nabla^2 V = \frac{\partial}{\partial r} \left( r^2 \frac{\partial V}{\partial r} \right) + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial V}{\partial \theta} \right) = 0$ If we assume the solution $V(r,\theta)$ is separatable, namely $V(r, \theta) = R(r) \Theta(\theta)$ we would have a (solved) general solution $V(r, \theta) = \sum_{\ell=0}^{\infty} \left( A_{\ell} r^{\ell} + \frac{B_{\ell}}{r^{\ell+1}} \right) P_{\ell} (\cos \theta)$ Here $P_{\ell}$ is $\ell^{th}$-order [Legendre polynomial](https://en.wikipedia.org/wiki/Legendre_polynomials). Here we list the first several terms, $\begin{array}{c|c} n & P_n(x) \\ \hline 0 & 1 \\ 1 & x \\ 2 & \frac{1}{2} (3x^2 - 1) \\ 3 & \frac{1}{2} (5x^3 - 3x) \\ 4 & \frac{1}{8} (35x^4 - 30x^2 + 3) \\ 5 & \frac{1}{8} (63x^5 - 70x^3 + 15x) \\ \end{array}$ Now, consider our model. We have different dielectric constant inside and outside the sphere, so we should separate the inside and outside as $V_{in}(r, \theta) = \sum_{\ell=0}^{\infty} \left( A_{\ell} r^{\ell} + \frac{B_{\ell}}{r^{\ell+1}} \right) P_{\ell} (\cos \theta)$ $V_{out}(r, \theta) = \sum_{\ell=0}^{\infty} \left( C_{\ell} r^{\ell} + \frac{D_{\ell}}{r^{\ell+1}} \right) P_{\ell} (\cos \theta)$ For $V_{in}$, if $r \rightarrow 0$, the $V$ should have a finite value, so $B_{\ell}=0$ for all $\ell$. Similarly, the sphere could only affect finite region. Therefore, for $r \rightarrow \infty$, $V_{\text{out}} \rightarrow -E_0 z = -E_0 r \cos \theta$. This requires $C_{1}=-E_{0}$, and $C_{\ell}=0$ for $\ell \neq 1$. Now consider the boundary condition, exactly same as what we did in [[Surface plasmon polaritons]], we consider the interface following [[Maxwell equation recap#Boundary conditions for waves across an interface]]. $E_{\parallel,\text{in}} = E_{\parallel,\text{out}} \qquad \varepsilon_{r,1} E_{\perp,1} = \varepsilon_{r,2} E_{\perp,2}$ Matching the field, we have $E_{\parallel}: \quad -\frac{1}{a} \left( \frac{\partial V_{\text{in}}}{\partial \theta} \right) \bigg|_{r=a} = -\frac{1}{a} \left( \frac{\partial V_{\text{out}}}{\partial \theta} \right) \bigg|_{r=a}$ $E_{\perp}: \quad -\varepsilon_{\text{sph}} \left( \frac{\partial V_{\text{in}}}{\partial r} \right) \bigg|_{r=a} = -\varepsilon_{\text{d}} \left( \frac{\partial V_{\text{out}}}{\partial r} \right) \bigg|_{r=a}$ This solves the remaining coefficients, from $E_{\parallel}$, we have $\begin{array}{ll} A_1 = -E_0 + \frac{D_1}{a^3}&(1) \\ A_{\ell} = \frac{D_{\ell}}{a^{2\ell+1}} \quad \text{for } \ell \neq 1 &(2)\\ \end{array}$ and $E_{\perp}$, we have $\begin{array}{ll} \varepsilon_{\text{sph}} A_1 = -\varepsilon_{\text{d}} \left[ E_0 + \frac{2D_1}{a^3} \right] &(3) \\ \varepsilon_{\text{sph}} \ell A_{\ell} = -\varepsilon_{\text{d}} (\ell + 1) \frac{D_{\ell}}{a^{2\ell+1}} \quad \text{for } \ell \neq 1 &(4) \\ \end{array}$ From (2) and (4), we have $A_{\ell} = D_{\ell} = 0$ and (1) and (3), we have $A_1 = -\frac{3 \varepsilon_{\text{d}}}{\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}}} E_0 \qquad D_1 = \frac{\varepsilon_{\text{sph}} - \varepsilon_{\text{d}}}{\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}}} a^3 E_0$ So we solve the static field problem, the potential is given as $V_{\text{in}}(r, \theta) = -\frac{3 \varepsilon_{\text{d}}}{\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}}} E_0 r \cos \theta$ $V_{\text{out}}(r, \theta) = -E_0 r \cos \theta + \frac{\varepsilon_{\text{sph}} - \varepsilon_{\text{d}}}{\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}}} E_0 a^3 \frac{\cos \theta}{r^2}$ Note this is just the electrostatic potential, instead of the field $\mathbf{E}$. And we did not define the material properties. #### Physical intuitions Check the expression of $V_{\text{out}}$, which has two parts: $V_{\text{out}}(r, \theta) = -E_0 r \cos \theta + \frac{\varepsilon_{\text{sph}} - \varepsilon_{\text{d}}}{\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}}} E_0 a^3 \frac{\cos \theta}{r^2}$ The first part is the external field, while the second part is a $r^{2}$ dependent term. Recall the expression of an ideal point dipole ([[Maxwell equation recap]]): $V_{\text{ideal dipole}} (r, \theta) = \frac{\mathbf{p_d} \cdot \mathbf{r}}{4 \pi \varepsilon_0 \varepsilon_{\text{d}} r^3} = \frac{p_d \cos \theta}{4 \pi \varepsilon_0 \varepsilon_{\text{d}} r^2}$ So we can rewrite the latter part as a applied field plus an ideal dipole, namely $V_{\text{out}}(r, \theta) = -E_0 r \cos \theta + \frac{p_d \cos \theta}{4 \pi \varepsilon_0 \varepsilon_{\text{d}} r^2}$ with $\mathbf{p_d} = 4 \pi \varepsilon_0 \varepsilon_{\text{d}} a^3 \frac{\varepsilon_{\text{sph}} - \varepsilon_{\text{d}}}{\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}}} \mathbf{E_0}$. If consider under the polarization picture, we have $\mathbf{p_d} = \varepsilon_0 \varepsilon_{\text{d}} \alpha \mathbf{E_0}$ So we may write the effect of the sphere as a dipole, with the polarizability being $\alpha = 4 \pi a^3 \frac{\varepsilon_{\text{sph}} - \varepsilon_{\text{d}}}{\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}}}$ and this $\alpha$ depends on the geometry and dielectric constant of the sphere also the surroundings. #### Fröhlich condition As the polarizability suggested, when the denominator becomes zero, the value of $\alpha$ becomes infinitely large. This is call Fröhlich condition, namely $\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}} \rightarrow 0$ This happens if one of these dielectric constant is negative and the other is positive, and the summation is zero. In this case, the polarizability blows up. Notice we are talking about the result in a static field, but $\varepsilon$ (especially $\varepsilon_{\text{sph}}$, cause $\varepsilon_{sph}(\omega) = 1-\frac{\omega_{p}^{2}}{\omega^{2}}$ for lossless Drude model) is a function of frequency, so Fröhlich condition only occurs at specific $\omega$. And the resonance is localized surface plasmon. >[!Note] >In actual case, the LSP does not necessarily happen for spherical particles/structures. This could also occur for pyramid tips, or alternative structures. But the expression would be different. A special case is $\varepsilon''_{sph}$ is small or almost constant, namely lossless condition. We can write $\operatorname{Re} \left\{ \varepsilon_{\text{sph}}(\omega) \right\} = -2 \varepsilon_{\text{d}}$ And insert this result to our lossless Drude model, $\varepsilon_{sph}(\omega) = 1-\frac{\omega_{p}^{2}}{\omega^{2}}$ we have $\omega_{\text{LSP}} = \frac{\omega_p}{\sqrt{3 \varepsilon_{\text{d}}}}$ For air and vacuum, the result is $\omega_{\text{LSP}} = \frac{\omega_p}{\sqrt{3}}$ This is a good reference. Recall for SPP, under vacuum condition we also have $\omega_{sp}=\frac{\omega_{p}}{\sqrt{2}}$ The $\varepsilon_{d}$ dependence of behavior of LSP frequency makes it a tool for environment detection, like refractive index detection (since it shift to lower frequency with increasing $\omega_{d}$). ### Oscillatory field and quasi-static approximation From previous static field derivation, we can have the electric field inside and outside the given sphere as $\mathbf{E_{\text{in}}} = \frac{3 \varepsilon_{\text{d}}}{\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}}} \mathbf{E_0}$ $\mathbf{E_{\text{out}}} = \mathbf{E_0} + \frac{3 (\mathbf{p_d} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \mathbf{p_d}}{4 \pi \varepsilon_0 \varepsilon_{\text{d}} r^3}$ The field inside the sphere is a constant and outside is $\mathbf{r}$ dependent (applied field + depolarization term). Both field is enhanced at $\omega_\text{LSP}$ (Cause $\mathbf{p_d} \propto \frac{\varepsilon_{\text{sph}} - \varepsilon_{\text{d}}}{\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}}}$). But this is the result from the static field (for sure, it cannot be enhanced all the time, otherwise energy conservation would be broken). To deal with the oscillating field, consider the quasi-static approximation, namely, $\mathbf{E}(\mathbf{r}, t) = \mathbf{E_0} \exp[-i \omega t]$ and if the wavelength is far larger than the particle size ($\lambda \gg a$, or Rayleigh scattering condition), then field is constant over sphere at any $t$, and the polarization is $\mathbf{p_d} = \varepsilon_0 \varepsilon_{\text{d}} \alpha \mathbf{E_0} \exp[-i \omega t]$ with the $\alpha$ is the polarizability given by the electrostatic model above. >[!Note] >It is worth noting that the actual case could be more complicated. Electric field of oscillating dipole could have alternative expression due to retardation since the electromagnetic wave needs time to travel. If we write the expression of (complicated) field of oscillating dipole under $\mathbf{E}(t) = \mathbf{E} \exp[-i \omega t] \qquad \mathbf{H}(t) = \mathbf{H} \exp[-i \omega t]$ we will get $\mathbf{E} = \frac{1}{4 \pi \varepsilon_0 \varepsilon_{\text{d}}} \left[ k^2 (\hat{\mathbf{r}} \times \mathbf{p_d}) \times \hat{\mathbf{r}} \frac{e^{ikr}}{r} + (3 (\mathbf{p_d} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \mathbf{p_d}) \left( \frac{1}{r^3} - \frac{ik}{r^2} \right) e^{ikr} \right]$ $\mathbf{H} = \frac{ck^2}{4\pi} (\hat{\mathbf{r}} \times \mathbf{p_d}) \frac{e^{ikr}}{r} \left( 1 - \frac{1}{ikr} \right)$ (Jackson, Classical Electrodynamics, pp. 394--395) And in "near zone", or say $kr \ll 1$ and dipole size much smaller than $\lambda$, the field would be $\mathbf{E} = \frac{3 (\mathbf{p_d} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \mathbf{p_d}}{4 \pi \varepsilon_0 \varepsilon_{\text{d}} r^3}$ This is the same result as the electrostatic limit, and $\mathbf{H} = \frac{i \omega}{4 \pi r^2} (\hat{\mathbf{r}} \times \mathbf{p_d})$ this field vanishes for $\omega \rightarrow 0$. In "radiation zone", or say $kr\gg 1$ and dipole size is much smaller than $\lambda$, $\mathbf{E} = \sqrt{\frac{\mu_0}{\varepsilon_0 \varepsilon_{\text{d}}}} (\mathbf{H} \times \hat{\mathbf{r}})$ $\mathbf{E}$ and $\mathbf{H}$ are in phase, mutually perpendicular, and transverse, while $\mathbf{H} = \frac{ck^2}{4\pi} (\hat{\mathbf{r}} \times \mathbf{p_d}) \frac{e^{ikr}}{r}$ which describes the spherical waves. ### Applications and more What LSPs can do? The typical metallic nanoparticle model could: #### Enhance absorption and scattering This is done by changing the cross-section. Using electrostatic solution and quasi-static approximation, we have for LSPs, $\sigma_{\text{abs}} \equiv 4 \pi k a^3 \, \Im \left[ \frac{\varepsilon_{\text{sph}} - \varepsilon_{\text{d}}}{\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}}} \right]$ $\sigma_{\text{sca}} \equiv \frac{8 \pi}{3} k^4 a^6 \left| \frac{\varepsilon_{\text{sph}} - \varepsilon_{\text{d}}}{\varepsilon_{\text{sph}} + 2 \varepsilon_{\text{d}}} \right|^2$ This is valid for dielectric or metallic spheres in Rayleigh limit. The radius is $a$ and the wavevector is $k$. It is worth noting that $\sigma_{\text{abs}} \propto a^3$ and $\sigma_{\text{sca}} \propto a^6$, so $\frac{\sigma_{\text{sca}}}{\sigma_{\text{abs}}} = a^3$ This indicates that scattering increases for larger particles. For gold nanoparticles, the intersection of scattering and absorption cross-section lies at 90 nm diameter. #### LSPs for different geometries The geometry of the nanoparticles could be different from sphere. For spheroids, there would be two axes (major and minor) so there would be two LSP resonances. The longer axis would have a lower $\omega_{\text{LSP}}$. The spheres could also have core-shell structure. This would make the polarizability have a more complicated expression. The core-shell structure provides wider turnability in near infrared. #### Beyond Rayleigh limit As mentioned above (and presented in publish data), the $\omega_{\text{LSP}}$ gets smaller with increasing size. This is inconsistence with our size-independent model at [[#Laplace equation for sphere in static electric field]]. This is because the Rayleigh limit no longer holds for larger spheres, which can not be considered as a point. The depolarization field would affect by retardation. And the modified Fröhlich condition is $\varepsilon_{\text{sph}} = -\left[ 2 + \frac{12}{5} (ka)^2 \right] \varepsilon_{\text{d}}$ which is wavevector and size-dependent. We will cover this in [[Mie resonances]]. An intuitive way to understand this is the LSP resonance shifts to lower frequency because the spheres are larger, and opposing charges feel weaker restoring force, so the restoration energy, or say the resonance energy gest lower.