亥姆霍兹方程（Helmholtz equation)
考虑真空中波动方程，( ∇ 2 − 1 c 2 ∂ 2 ∂ t 2 ) u ( r , t ) = 0 \left(\nabla^{2}-\frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}}\right) u(\mathbf{r}, t)=0 ( ∇ 2 − c 2 1 ∂ t 2 ∂ 2 ) u ( r , t ) = 0 u ( r , t ) = A ( r ) T ( t ) u(\mathbf{r},t)=A(\mathbf{r})T(t) u ( r , t ) = A ( r ) T ( t ) A ( r ) A(\mathbf{r}) A ( r ) T ( t ) T(t) T ( t ) 波动方程的解u ( r , t ) = A ( r ) T ( t ) u(\mathbf{r},t)=A(\mathbf{r})T(t) u ( r , t ) = A ( r ) T ( t ) A ( r ) A(\mathbf{r}) A ( r ) ( ∇ 2 + k 2 ) A = 0 \left(\nabla^{2}+k^{2}\right) A=0 ( ∇ 2 + k 2 ) A = 0
Helmholtz方程推导
通过分离变量法可得方程：∇ 2 A A = 1 c 2 T d 2 T d t 2 \frac{\nabla^{2} A}{A}=\frac{1}{c^{2} T} \frac{d^{2} T}{d t^{2}} A ∇ 2 A = c 2 T 1 d t 2 d 2 T A ( r ) A(\mathbf{r}) A ( r ) T ( t ) T(t) T ( t ) u ( r , t ) = u ~ 0 e i ( k ⋅ r − w t ) u(\mathbf{r},t)=\widetilde{u}_0e^{i(\mathbf{k} \cdot \mathbf{r}-wt)} u ( r , t ) = u 0 e i ( k ⋅ r − w t ) ∇ 2 A A = − k 2 \frac{\nabla^{2} A}{A}=-k^{2} A ∇ 2 A = − k 2 1 c 2 T d 2 T d t 2 = − k 2 \frac{1}{c^{2} T} \frac{d^{2} T}{d t^{2}}=-k^{2} c 2 T 1 d t 2 d 2 T = − k 2 A ( r ) A(\mathbf{r}) A ( r ) ∇ 2 A + k 2 A = ( ∇ 2 + k 2 ) A = 0 . \nabla^{2} A+k^{2} A= \boxed{\left(\nabla^{2}+k^{2}\right) A=0}. ∇ 2 A + k 2 A = ( ∇ 2 + k 2 ) A = 0 . T ( t ) T(t) T ( t ) d 2 T d t 2 + ω 2 T = ( d 2 d t 2 + ω 2 ) T = 0 \frac{d^{2} T}{d t^{2}}+\omega^{2} T=\left(\frac{d^{2}}{d t^{2}}+\omega^{2}\right) T=0 d t 2 d 2 T + ω 2 T = ( d t 2 d 2 + ω 2 ) T = 0
wikipedia-Helmholtz equation
表面等离极化激元（surface palsmon polarirons）
自由电子气模型表明，体自由电子集体共振（即等离激元，plasmon）只存在纵波模式 ，因此三维等离激元无法与电磁横波 耦合形成表面等离极化激元。在电介质（介电常数大于零）-金属（介电函数实部小于零）界面处可以形成等离激元与电磁场的耦合，即表面等离激元（surface palsmon polarirons, spps）。
单界面spps色散关系推导
单界面spps的推导是利用无自由电荷、无自由电流、无磁化的Maxwell方程组，代入边界条件，得到色散关系。束缚电荷的信息包含在了介电函数ε ( r ) \varepsilon(r) ε ( r )
在介质中的波动方程为，∇ × ∇ × E = − μ 0 ∂ 2 D ∂ t 2 \nabla \times \nabla \times \mathbf{E}=-\mu_{0} \frac{\partial^{2} \mathbf{D}}{\partial t^{2}} ∇ × ∇ × E = − μ 0 ∂ t 2 ∂ 2 D ∇ × ∇ × E ≡ ∇ ( ∇ ⋅ E ) − ∇ 2 E \nabla \times \nabla \times \mathbf{E} \equiv \nabla(\nabla \cdot \mathbf{E})-\nabla^{2} \mathbf{E} ∇ × ∇ × E ≡ ∇ ( ∇ ⋅ E ) − ∇ 2 E ∇ ⋅ ( ε E ) ≡ E ⋅ ∇ ε + ε ∇ ⋅ E = ∇ ⋅ D = 0 \nabla \cdot(\varepsilon \mathbf{E}) \equiv\mathbf{E} \cdot \nabla \varepsilon+\varepsilon \nabla \cdot \mathbf{E}=\nabla \cdot \mathbf{D}=0 ∇ ⋅ ( ε E ) ≡ E ⋅ ∇ ε + ε ∇ ⋅ E = ∇ ⋅ D = 0 ∇ ε ( r ) = 0 \nabla \varepsilon(\mathbf{r})=0 ∇ ε ( r ) = 0 ∇ 2 E − ε c 2 ∂ 2 E ∂ t 2 = 0 \nabla^{2} \mathbf{E}-\frac{\varepsilon}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}=0 ∇ 2 E − c 2 ε ∂ t 2 ∂ 2 E = 0 ∇ E = 0 \nabla \mathbf{E}=0 ∇ E = 0 ∇ ( ∇ ⋅ E ) = K ( K ⋅ E ) = 0 \nabla (\nabla \cdot \mathbf{E})=\mathbf{K} (\mathbf{K} \cdot \mathbf{E})=0 ∇ ( ∇ ⋅ E ) = K ( K ⋅ E ) = 0 K \mathbf{K} K E \mathbf{E} E ∇ ε ( r ) = 0 \nabla \varepsilon(\mathbf{r})=0 ∇ ε ( r ) = 0
定义k 0 = ω c k_{0}=\frac{\omega}{c} k 0 = c ω k 0 k_0 k 0 ∇ 2 E + k 0 2 ε E = 0 \nabla^{2} \mathbf{E}+k_{0}^{2} \varepsilon \mathbf{E}=0 ∇ 2 E + k 0 2 ε E = 0
考虑如下结构，x x x y y y ε ( r ) = ε ( z ) \varepsilon(\mathbf{r})=\varepsilon(z) ε ( r ) = ε ( z ) E ( x , y , z ) = E ( z ) e i β x \mathbf{E}(x, y, z)=\mathbf{E}(z) \mathrm{e}^{i \beta x} E ( x , y , z ) = E ( z ) e i β x β = k x \beta=k_{x} β = k x ∂ 2 E ( y , z ) ∂ y 2 + ∂ 2 E ( y , z ) ∂ z 2 + ( k 0 2 ε − β 2 ) E = 0 \frac{\partial^{2} \mathbf{E}(y,z)}{\partial y^{2}}+\frac{\partial^{2} \mathbf{E}(y,z)}{\partial z^{2}}+\left(k_{0}^{2} \varepsilon-\beta^{2}\right) \mathbf{E}=0 ∂ y 2 ∂ 2 E ( y , z ) + ∂ z 2 ∂ 2 E ( y , z ) + ( k 0 2 ε − β 2 ) E = 0 ∂ ∂ y = 0 \frac{\partial}{\partial y}=0 ∂ y ∂ = 0 ∂ 2 E ( z ) ∂ z 2 + ( k 0 2 ε − β 2 ) E = 0 \frac{\partial^{2} \mathbf{E}(z)}{\partial z^{2}}+\left(k_{0}^{2} \varepsilon-\beta^{2}\right) \mathbf{E}=0 ∂ z 2 ∂ 2 E ( z ) + ( k 0 2 ε − β 2 ) E = 0 在spps中只存在TM模式 ，即只有H y H_y H y E x E_x E x E z E_z E z E x = − i 1 ω ε 0 ε ∂ H y ∂ z E_{x}=-i \frac{1}{\omega \varepsilon_{0} \varepsilon} \frac{\partial H_{y}}{\partial z} E x = − i ω ε 0 ε 1 ∂ z ∂ H y E z = − β ω ε 0 ε H y E_{z}=-\frac{\beta}{\omega \varepsilon_{0} \varepsilon} H_{y} E z = − ω ε 0 ε β H y H y H_y H y ∂ 2 H y ∂ z 2 + ( k 0 2 ε − β 2 ) H y = 0 \frac{\partial^{2} H_{y}}{\partial z^{2}}+\left(k_{0}^{2} \varepsilon-\beta^{2}\right) H_{y}=0 ∂ z 2 ∂ 2 H y + ( k 0 2 ε − β 2 ) H y = 0 z z z z > 0 z>0 z > 0 H y ( z ) = A 2 e i β x e − k 2 z E x ( z ) = i A 2 1 ω ε 0 ε 2 k 2 e i β x e − k 2 z E z ( z ) = − A 2 β ω ε 0 ε 2 e i β x e − k 2 z \begin{aligned} H_{y}(z) &=A_{2} \mathrm{e}^{i \beta x} \mathrm{e}^{-k_{2} z} \\ E_{x}(z) &=i A_{2} \frac{1}{\omega \varepsilon_{0} \varepsilon_{2}} k_{2} e^{i \beta x} \mathrm{e}^{-k_{2} z} \\ E_{z}(z) &=-A_{2} \frac{\beta}{\omega \varepsilon_{0} \varepsilon_{2}} e^{i \beta x} \mathrm{e}^{-k_{2} z} \end{aligned} H y ( z ) E x ( z ) E z ( z ) = A 2 e i β x e − k 2 z = i A 2 ω ε 0 ε 2 1 k 2 e i β x e − k 2 z = − A 2 ω ε 0 ε 2 β e i β x e − k 2 z z < 0 z<0 z < 0 H y ( z ) = A 1 e i β x e k 1 z E x ( z ) = − i A 1 1 ω ε 0 ε 1 k 1 e i β x e k 1 z E z ( z ) = − A 1 β ω ε 0 ε 1 e i β x e k 1 z \begin{aligned} H_{y}(z) &=A_{1} \mathrm{e}^{i \beta x} \mathrm{e}^{k_{1} z} \\ E_{x}(z) &=-i A_{1} \frac{1}{\omega \varepsilon_{0} \varepsilon_{1}} k_{1} e^{i \beta x} \mathrm{e}^{k_{1} z} \\ E_{z}(z) &=-A_{1} \frac{\beta}{\omega \varepsilon_{0} \varepsilon_{1}} e^{i \beta x} \mathrm{e}^{k_{1} z} \end{aligned} H y ( z ) E x ( z ) E z ( z ) = A 1 e i β x e k 1 z = − i A 1 ω ε 0 ε 1 1 k 1 e i β x e k 1 z = − A 1 ω ε 0 ε 1 β e i β x e k 1 z k i k_i k i H y H_{y} H y ε i E z \varepsilon_{i} E_{z} ε i E z E x E_x E x k 2 k 1 = − ε 2 ε 1 \boxed {\frac{k_{2}}{k_{1}}=-\frac{\varepsilon_{2}}{\varepsilon_{1}}} k 1 k 2 = − ε 1 ε 2 H y H_{y} H y k 1 2 = β 2 − k 0 2 ε 1 k_{1}^{2}=\beta^{2}-k_{0}^{2} \varepsilon_{1} k 1 2 = β 2 − k 0 2 ε 1 k 2 2 = β 2 − k 0 2 ε 2 k_{2}^{2}=\beta^{2}-k_{0}^{2} \varepsilon_{2} k 2 2 = β 2 − k 0 2 ε 2 β = k 0 ε 1 ε 2 ε 1 + ε 2 \boxed {\beta=k_{0} \sqrt{\frac{\varepsilon_{1} \varepsilon_{2}}{\varepsilon_{1}+\varepsilon_{2}}}} β = k 0 ε 1 + ε 2 ε 1 ε 2 k = w c ε 1 ε 2 ε 1 + ε 2 \boxed {k=\frac{w}{c} \sqrt{\frac{\varepsilon_{1} \varepsilon_{2}}{\varepsilon_{1}+\varepsilon_{2}}}} k = c w ε 1 + ε 2 ε 1 ε 2
单界面spps色散关系讨论
在推导spps色散关系时曾通过连续性条件得到k 2 k 1 = − ε 2 ε 1 \frac{k_{2}}{k_{1}}=-\frac{\varepsilon_{2}}{\varepsilon_{1}} k 1 k 2 = − ε 1 ε 2 ℜ ( k i ) > 0 \Re (k_i)>0 ℜ ( k i ) > 0 ℜ ( k i ) > 0 \Re (k_i)>0 ℜ ( k i ) > 0 ℜ ( ε 1 ) \Re (\varepsilon_1) ℜ ( ε 1 ) ℜ ( ε 2 ) \Re (\varepsilon_2) ℜ ( ε 2 )
单界面spps满足色散关系，k = w c ε 1 ε 2 ε 1 + ε 2 k=\frac{w}{c} \sqrt{\frac{\varepsilon_{1} \varepsilon_{2}}{\varepsilon_{1}+\varepsilon_{2}}} k = c w ε 1 + ε 2 ε 1 ε 2 ε 1 = 1 \varepsilon_{1}=1 ε 1 = 1 ε ( ω ) = 1 − ω p 2 ω 2 \varepsilon(\omega)=1-\frac{\omega_{\mathrm{p}}^{2}}{\omega^{2}} ε ( ω ) = 1 − ω 2 ω p 2 ε a i r = 1 , ε s i l i c a = 2.25 \varepsilon_{air}=1, \varepsilon_{silica}=2.25 ε a i r = 1 , ε s i l i c a = 2 . 2 5 β \beta β β \beta β ω s p = ω p 1 + ε 2 \omega_{\mathrm{sp}}=\frac{\omega_{\mathrm{p}}}{\sqrt{1+\varepsilon_{2}}} ω s p = 1 + ε 2 ω p v g = d w ( k ) d k = 0 v_g=\frac{dw(k)}{dk}=0 v g = d k d w ( k ) = 0
对于相同角频率的波，在spps中波矢k x k_x k x k 0 k_0 k 0
surface palsmon polarirons wikipedia-surface palsmon polarirons