## Surface plasmon polaritons Last time we see the dispersion relation of bulk plasmon, and this time we will go through the derivation of surface modes solution and the dispersion relation of surface plasmon polaritons (SPPs), from the [[Maxwell equation recap|Maxwell equations]] and the boundary conditions at the interface. ### What is SPPs This is kind of a repetition (see [[Introduction to surface plasmon]]) but, surface plasmon polaritons is the combined excitation of electron motion (electron density wave) at the surface, and the associated electromagnetic waves. It is excited by photon and propagates at the surface as a guide wave. And it is not confined in a limited space like [[Localized surface plasmons|LSPs]]. SPPs exist at interface where the permittivity of two sides have different sign, like metal-dialectic interface. ### Derivation of surface modes and SPPs' dispersion As always, consider the case with no external current density, $J_{f} = 0$. Ignore nonlinear behavior and assumption we have non-magnetic media. we should have the following two equations directly from Maxwell. $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} = -\mu \frac{\partial \mathbf{H}}{\partial t} = -\mu_0 \frac{\partial \mathbf{H}}{\partial t}$ $\nabla \times \mathbf{H} = -\frac{\partial \mathbf{D}}{\partial t} = -\varepsilon \frac{\partial \mathbf{E}}{\partial t} = -\varepsilon_0 \varepsilon_r \frac{\partial \mathbf{E}}{\partial t}$ Use plane wave ansatz, $\mathbf{E}(\mathbf{r}, t) = \mathbf{E}(\mathbf{r})e^{-i\omega t},\ \mathbf{H}(\mathbf{r}, t) = \mathbf{H}(\mathbf{r})e^{-i\omega t}$, we may write the components in matrix form, $ \nabla \times \mathbf{E} = \begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ E_x & E_y & E_z \end{vmatrix} $ $ \nabla \times \mathbf{H} = \begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ H_x & H_y & H_z \end{vmatrix} $ Institute the plane wave ansatz, we have 6 sets of equations, From $\nabla \times \mathbf{E} = -\mu_0 \frac{\partial \mathbf{H}}{\partial t}$, we have $\begin{align*} \frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z} &= i\omega\mu_0 H_x \\ \frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x} &= i\omega\mu_0 H_y \\ \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} &= i\omega\mu_0 H_z \end{align*} $ From $\nabla \times \mathbf{H} = -\varepsilon_0 \varepsilon_r \frac{\partial \mathbf{E}}{\partial t}$, we have $\begin{align*} \frac{\partial H_z}{\partial y} - \frac{\partial H_y}{\partial z} &= -i\omega\varepsilon_0\varepsilon_r E_x \\ \frac{\partial H_x}{\partial z} - \frac{\partial H_z}{\partial x} &= -i\omega\varepsilon_0\varepsilon_r E_y \\ \frac{\partial H_y}{\partial x} - \frac{\partial H_x}{\partial y} &= -i\omega\varepsilon_0\varepsilon_r E_z \end{align*} $ ^5121a3 Now consider our problem set up, the metal/dielectric interface ![[Drawing 2024-03-31 23.59.40.excalidraw.svg]] The dielectric is assumed as air, and surface wave propagates in $x$ direction. Then we have no dependence on $y$, and the dielectric function $\varepsilon_{r}$ is a function of $z$ only (since in $xy$ plane the material has no difference). Therefore, the solution of the surface wave has the form $\mathbf{E}(\mathbf{r}) = \mathbf{E}(z)e^{i k_x x} = \mathbf{E}(z)e^{i \beta x}$ where $k_{x}$ is the wavevector in the propagation direction, and by convention it is often written as $\beta$, aka propagation constant. Now write $\mathbf{E}(z)$ on $x,y,z$ directions, one should have $\mathbf{E}(\mathbf{z}) = E'_x(z) \mathbf{\hat{i}} + E'_y(z) \mathbf{\hat{j}} + E'_z(z) \mathbf{\hat{k}}$ $\begin{aligned} E_x &= E'_x(z)e^{i\beta x} \\ E_y &= E'_y(z)e^{i\beta x} \\ E_z &= E'_z(z)e^{i\beta x} \end{aligned}$ Similarly, for magnetic field we have $\mathbf{H}(\mathbf{r}) = \mathbf{H}(z) e^{i\beta x}$ $\mathbf{H}(z) = H'_x(z) \mathbf{\hat{i}} + H'_y(z) \mathbf{\hat{j}} + H'_z(z) \mathbf{\hat{k}}$ $\begin{aligned} H_x &= H'_x(z) e^{i\beta x} \\ H_y &= H'_y(z) e^{i\beta x} \\ H_z &= H'_z(z) e^{i\beta x} \end{aligned}$ >[!Note] >1. Here $\mathbf{E}(z)$ is not $E_{z}$, $E_{z}$ is the $z$ component of the electric field, while $\mathbf{E}(z)$ means the electric field of this propagating wave is a function (and only a function) of $z$, cause we remove the $x$ dependence outside to the exponential term, and it's independent on $y$, so we only have the dependence on $z$ due to the permittivity change. >2. The $\mathbf{E}(r)$ and $\mathbf{E}(z)$ are a vector field, and $E_{x}$... are the components of this vector field, the field itself is propagating along x direction. See the sketch below. **This is just an illustration, does not reflect the actual field!** >![[Drawing 2024-04-01 00.35.48.excalidraw.svg]] >3. For EM wave in free space, due to the transverse property, the component on the propagating direction should be zero, but here we have a guided wave, and it does not have to be transverse, so the $x$ component $E_{x}$ could be non-zero. And if consider the frequency domain, we have $\frac{\partial}{\partial x} = i\beta$ and $\frac{\partial}{\partial y} = 0$, the [[#^5121a3|6 sets of equations]] can be simplified as $\begin{aligned} \frac{\partial E_y}{\partial z} &= -i\omega\mu_0 H_x, \\ \frac{\partial E_x}{\partial z} - i\beta E_z &= i\omega\mu_0 H_y, \\ i\beta E_y &= i\omega\mu_0 H_z. \end{aligned}$ and $\begin{aligned} \frac{\partial H_y}{\partial z} &= i\omega\varepsilon_0 \varepsilon_r E_x, \\ \frac{\partial H_x}{\partial z} - i\beta H_z &= -i\omega\varepsilon_0 \varepsilon_r E_y, \\ i\beta H_y &= -i\omega\varepsilon_0 \varepsilon_r E_z. \end{aligned}$ From Chapter 8 in [Introduction to Electrodynamics](https://books.google.ch/books/about/Introduction_to_Electrodynamics.html?id=J9ygBwAAQBAJ&redir_esc=y)by Griffiths, these equations gives 2 solutions as TM and TE. - Transverse magnetic, TM, or p polarized: $H_{x} =0 \quad H_{z} =0 \quad E_{y}=0$ $E_{z} \neq 0 \quad H_{y}\neq 0 \quad E_{x}\neq 0$ - Transverse electric, TE, or s polarized: $E_{x} =0 \quad E_{z} =0 \quad H_{y}=0$ $H_{z} \neq 0 \quad E_{y}\neq 0 \quad H_{x}\neq 0$ ![[Drawing 2024-04-01 01.16.09.excalidraw.svg]] Those zero components gets diminished, we now have three equations for each case, here $k_{0}=\frac{\omega}{c}$ ($k_{0}$ is our defined variable in dispersion diagram). For TM, we have $\begin{aligned} E_x &= -\frac{i}{\omega \varepsilon_0 \varepsilon_r} \frac{\partial H_y}{\partial z}, \\ E_z &= -\frac{\beta}{\omega \varepsilon_0 \varepsilon_r} H_y, \\ \frac{\partial^2 H_y}{\partial z^2} + (k_0^2 \varepsilon_r - \beta^2) H_y &= 0 \end{aligned}$ And for TE, we have $\begin{aligned} H_x &= \frac{i}{\omega \mu_0} \frac{\partial E_y}{\partial z}, \\ H_z &= \frac{\beta}{\omega \mu_0} E_y, \\ \frac{\partial^2 E_y}{\partial z^2} + (k_0^2 \varepsilon_r - \beta^2) E_y &= 0. \end{aligned}$ And with the [[Maxwell equation recap#Boundary conditions for waves across an interface|boundary conditions for maxwell equation]], we can already solve these sets of equations. Assume surface mode exist and consider TM case, **the electromagnetic wave should bound to the surface and decay in $\pm z$ direction**. ($\leftarrow$ this is an very important assumption!) ![[Drawing 2024-04-01 01.37.47.excalidraw.svg]] | TM | $z>0$ | $z<0$ | | :-------------------------: | ------------------------------------------------------------------------------------ | ------------------------------------------------------------------------------------ | | Assumption:<br>decay in $z$ | $H_y = A_d e^{i\beta x - k_d z}$ | $H_y = A_m e^{i\beta x + k_m z}$ | | Solve from above equation | $E_x = \frac{i A_d k_d}{\omega \varepsilon_0 \varepsilon_d} e^{i\beta x - k_d z}$ | $E_x = -\frac{i A_m k_m}{\omega \varepsilon_0 \varepsilon_m} e^{i\beta x + k_m z}$ | | Solve from above equation | $E_z = -\frac{A_d \beta}{\omega \varepsilon_0 \varepsilon_d} e^{i\beta x - k_d z}$ | $E_z = -\frac{A_m \beta}{\omega \varepsilon_0 \varepsilon_m} e^{i\beta x + k_m z}$ | Here $A_{m},\ A_{d}$ are field amplitude, $k_{m},\ k_{d}$ are decay constants in metal and dielectric, respectively. And apply the boundary condition. $\begin{aligned} H_y|_{z=+\delta} &= H_y|_{z=-\delta} \\ A_d e^{i\beta x} &= A_m e^{i\beta x} \end{aligned}$ This gives $A_d = A_m = A$, and $\begin{aligned} E_x|_{z=+\delta} &= E_x|_{z=-\delta} \\ \frac{i A_k k_d}{\omega\varepsilon_0 \varepsilon_d} e^{i\beta x} &= -\frac{i A_k k_m}{\omega\varepsilon_0 \varepsilon_m} e^{i\beta x} \end{aligned} $ This gives $\frac{k_d}{\varepsilon_d} = -\frac{k_m}{\varepsilon_m}$. If using the other two boundary conditions we get exactly the same result. And if we institute our assumptions for $H_{y}$ into the previously obtained equation $\frac{\partial^2 H_y}{\partial z^2} + (k_0^2 \varepsilon_r - \beta^2) H_y = 0$ we can have one more relation between $k$ and $\varepsilon$, which is $\begin{aligned} k_m^2 &= \beta^2 - k_0^2 \varepsilon_m \\ k_d^2 &= \beta^2 - k_0^2 \varepsilon_d \end{aligned}$ Combining this with $\frac{k_d}{\varepsilon_d} = -\frac{k_m}{\varepsilon_m}$ we get $\beta = k_0 \sqrt{\frac{\varepsilon_m \varepsilon_d}{\varepsilon_m + \varepsilon_d}}$, or $k_x = \frac{\omega}{c} \sqrt{\frac{\varepsilon_m \varepsilon_d}{\varepsilon_m + \varepsilon_d}}$. This is the dispersion relation for surface plasmon polaritons. And if we do the same procedures for TE case, we will have no solution. >[!Note] >This $\beta$, or $k_{x}$ could be imaginary, $\beta =k_{x}=k_{x}'+ik_{x}''$. The imaginary part indicates the loss behavior along the propagation direction. This will be discussed later. ### Dispersion of SPPs From the derivation, we have the relation $\beta=k_{x} = \frac{\omega}{c} \sqrt{\frac{\varepsilon_m \varepsilon_d}{\varepsilon_m + \varepsilon_d}}$ plot the dispersion diagram for lossless metal, we have the following sketch. ![[Drawing 2024-04-01 02.05.11.excalidraw.svg]] When $\varepsilon_{m}+ \varepsilon_{d}=0$, $k_{x} \rightarrow +\infty$, the corresponding frequency would be $\omega_{sp}$. If we assume the metal is lossless and apply lossless Drude model, we have $\varepsilon_{m}(\omega)=1-\frac{\omega_{p}^{2}}{\omega^{2}}$ $\varepsilon_{m}+\varepsilon_{d}=1-\frac{\omega_{p}^{2}}{\omega^{2}}+\varepsilon_{d}=0$ $\omega^{2}=\frac{\omega_{p}^{2}}{1+\varepsilon_{d}}$ $\omega_{sp}=\omega=\frac{\omega_{p}}{\sqrt{1+\varepsilon_{d}}}$ And for air or vacuum, $\varepsilon_{d} = 1$, so $\omega_{sp}=\frac{\omega_{p}}{\sqrt{2}}$. >[!Note] >$\omega_{sp}=\frac{\omega_{p}}{\sqrt{1+\varepsilon_{d}}}$ is an important relation. One should remember this as well as the dispersion relation. Some critical points in the (lossless metal) dispersion diagram: 1. At here, surface wave is essentially the light wave in dielectric, but slightly slower (than $\frac{c}{n_{d}}$); 2. At here, we have SPP, namely partly photon and partly surface plasmon; 3. At here, surface plasmon has $\beta=k_{x}=+\infty$, this means $v_{g}=\frac{\mathrm{d}\omega}{\mathrm{d}k}=0$ at $\omega_{sp}$; 4. No solution between $\omega_{sp}$ and $\omega_{p}$, and; 5. above $\omega_{p}$, the wave would propagate through metal, as shown in [[Introduction to surface plasmon#What is plasmon?]] At point 3, we have a strong confinement of the field at z direction, cause $E_{z}$ decay exponentially at both directions. The decay constants are $k_{d}$ and $k_{m}$ in dielectric and metal, this parameter is called confinement length. And the confinement can be quantified by $\hat{z}=\frac{1}{|k_{d}|}$or $\frac{1}{|k_{m}|}$, depending in dielectric or metal. Recall the relation we had during our derivation, $\begin{aligned} k_m^2 &= \beta^2 - k_0^2 \varepsilon_m \\ k_d^2 &= \beta^2 - k_0^2 \varepsilon_d \end{aligned}$ this means $k_{m}$ and $k_{d}$ will be larger if $\beta=k_{x}$ goes big, and we have smaller $\hat{z}$, indicating we're having stronger confinement at 3, but only slightly confinement at 1. If we add loss, i.e., let the $\beta=k_{x}$ becomes complex, we have $\beta=k_{x}=k_{x}'+ik_{x}''$, and at the propagation direction the field will have an exponentially decay as $e^{-k_{x}'' x}$, and the intensity $I$ of SPP will decay as $e^{-2k_{x}'' x}$. We may define SPP propagation length as $L_{spp}=\frac{1}{2\Im(\beta)}=\frac{1}{2k_{x}''}$ This is defined in a common way, i.e., the intensity dropped to $\frac{1}{e}$ of its initial amplitude. >[!Note] > For flat metal interface, the $L_{spp}$ would typically be 10-100 microns, depending on specific metal and $\omega$. One example given in lecture from *Kress et al., Nano Letters 14, 5827 (2014)* has a propagation length of only $7.8\ \mu m$. > The loss can be considered as *Ohmic loss* due to dragging electrons along the electromagnetic field. And if the system is lossy, we will have a different form of dispersion. The $\beta$ would have a finite value, the confinement is now limited and not goes to 0, and $\omega$ between $\omega_{sp}$ and $\omega_p$ is now allowed. ![[Drawing 2024-04-01 02.51.00.excalidraw.svg]] >[!Info] >See more on: >https://en.wikipedia.org/wiki/Surface_plasmon_polariton >https://en.wikipedia.org/wiki/Polarization_(waves)#s_and_p_designations >https://en.wikipedia.org/wiki/Transverse_mode >https://en.wikipedia.org/wiki/Transverse_wave >[[Introduction to surface plasmon]] >[[Localized surface plasmons]]