费曼物理学（2） 第5次作业

Chasse_neige

1 势与场 (15-7)

Chern-Simons 理论是一类特殊的相互作用项，在量子反常，分数量子霍尔效应，任意子以及 Dyon（既带电荷又带磁荷的粒子）中有重要的应用。其特殊之处体现在其运动方程没有动力学项，即其运动方程就是一个恒等式，在下面的计算中我们可以看到这一点。4 维时空的 Chern-Simons 理论的拉氏量可以写为：

$$\mathcal{L}={ϵ}^{\mu \nu \rho \sigma }{F}_{\mu \nu }{F}_{\rho \sigma }$$

1. 请回忆第一次作业第三题，证明上述拉氏量可以写为：

   $$\mathcal{L}=-8\overrightarrow{E}\cdot \overrightarrow{B}$$

   证明：

   $$\begin{array}{c}\mathcal{L}={ϵ}^{\mu \nu \rho \sigma }{F}_{\mu \nu }{F}_{\rho \sigma }={ϵ}^{\mu \nu \rho \sigma }({\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu })({\partial }_{\rho }{A}_{\sigma }-{\partial }_{\sigma }{A}_{\rho })\\ =4{ϵ}^{\mu \nu \rho \sigma }{\partial }_{\mu }{A}_{\nu }{\partial }_{\rho }{A}_{\sigma }\\ =4({ϵ}^{ijk}{\partial }_0{A}_i{\partial }_j{A}_k-{ϵ}^{ijk}{\partial }_i{A}_0{\partial }_j{A}_k+{ϵ}^{ijk}{\partial }_i{A}_j{\partial }_0{A}_k-{ϵ}^{ijk}{\partial }_i{A}_j{\partial }_k{A}_0)\\ =8({\partial }_t\overrightarrow{A}\cdot (\mathrm{\nabla }\times \overrightarrow{A})+\mathrm{\nabla }\varphi \cdot (\mathrm{\nabla }\times \overrightarrow{A}))=-8\overrightarrow{E}\cdot \overrightarrow{B}\end{array}$$

2. 在低维情况下，Chern-Simons 理论有独特的性质，在 3 维时空当中，带有外源的 Chern-Simons 理论可以写为：

   $$\mathcal{L}=\frac{\kappa }{2}{ϵ}^{\mu \nu \rho }{A}_{\mu }{\partial }_{\nu }{A}_{\rho }-A_{\mu }{J}^{\mu }$$

   其中 $\mu ,\nu ,\rho =\{0,1,2\}$。请利用 Euler-Lagrange 方程证明 3 维时空 Chern-Simons 理论对应的运动方程为：

   $$\rho =\kappa B$$$$J^i=\kappa {ϵ}^{ij}{E}_j$$

   其中 $i,j=\{1,2\}$，第一个方程可以理解为每一个电荷都带着一个垂直纸面向外的磁场，第二个方程就是第一个方程的演化方程。

   利用 Euler-Lagrange 方程

   $$\frac{\partial }{\partial x_{\mu }}\frac{\partial }{\partial ({\partial }_{\mu }{A}_{\nu })}\mathcal{L}=\frac{\partial }{\partial A_{\nu }}\mathcal{L}$$$$\frac{\partial }{\partial ({\partial }_{\mu }{A}_{\nu })}\mathcal{L}=\frac{\partial }{\partial ({\partial }_{\mu }{A}_{\nu })}(\frac{\kappa }{2}{ϵ}^{\rho \mu \nu }{A}_{\rho }{\partial }_{\mu }{A}_{\nu }-A_{\mu }{J}^{\mu })=\frac{\kappa }{2}{ϵ}^{\rho \mu \nu }{A}_{\rho }$$$$\frac{\partial }{\partial x_{\mu }}\frac{\kappa }{2}{ϵ}^{\rho \mu \nu }{A}_{\rho }=\frac{\kappa }{2}{ϵ}^{\rho \mu \nu }{\partial }_{\mu }{A}_{\rho }$$$$\frac{\partial }{\partial A_{\nu }}(\frac{\kappa }{2}{ϵ}^{\nu \mu \rho }{A}_{\nu }{\partial }_{\mu }{A}_{\rho }-A_{\nu }{J}^{\nu })=\frac{\kappa }{2}{ϵ}^{\nu \mu \rho }{\partial }_{\mu }{A}_{\rho }-J^{\nu }$$$$\begin{array}{c}\frac{\kappa }{2}{ϵ}^{\rho \mu \nu }{\partial }_{\mu }{A}_{\rho }=\frac{\kappa }{2}{ϵ}^{\nu \mu \rho }{\partial }_{\mu }{A}_{\rho }-J^{\nu }\\ \kappa {ϵ}^{\mu \rho \nu }{\partial }_{\mu }{A}_{\rho }=J^{\nu }\end{array}$$

   所以

   $$\rho =\kappa {ϵ}^{ij0}{\partial }_i{A}_j=\kappa (\mathrm{\nabla }\times A)=\kappa B$$$$J^i=\kappa ({ϵ}^{0ji}{\partial }_0{A}_j+{ϵ}^{j0i}{\partial }_j{A}_0)=\kappa {ϵ}^{ij}{E}_j$$

3. 考虑二维平面上存在两个电荷的情况，此时电荷密度与矢量势可以写为：

   $$\rho (\overrightarrow{x},t)=e\sum _{n=1}^2\delta (\overrightarrow{x}-{\overrightarrow{x}}_n(t))$$$$A_i({\overrightarrow{x}}_n,t)=\frac{e}{2\pi \kappa }\sum _{m\ne n}{ϵ}_{ij}\frac{x_n^j-x_m^j}{|{\overrightarrow{x}}_n-{\overrightarrow{x}}_m{|}^2}$$

   试计算 Aharonov-Bohm 相位：

   $$e{\oint }_C\overrightarrow{A}\cdot d\overrightarrow{x}$$

   注：一个粒子绕着另外一个粒子可被视为将两粒子交换两次。AB 相位的结果告诉我们两粒子交换两次多出一个相位。因此，如果相位不等于 1，则两粒子为任意子。（费米子和玻色子都要求交换两次的相位等于 1。）

   $$\begin{array}{c}e{\oint }_C\overrightarrow{A}\cdot d\overrightarrow{x}=e{\oint }_C\frac{e}{2\pi \kappa }\sum _{m\ne n}{ϵ}_{ij}\frac{x_n^j-x_m^j}{|{\overrightarrow{x}}_n-{\overrightarrow{x}}_m{|}^2}d{x}^i\\ =e{\iint }_S\mathrm{\nabla }\times \overrightarrow{A}\cdot d\overrightarrow{S}=e{\iint }_S\frac{\rho }{\kappa }d{x}_{1}d{x}_2\\ =e{\iint }_S\frac{e}{\kappa }\sum _{n=1}^2\delta (\overrightarrow{x}-{\overrightarrow{x}}_n(t))d{x}_{1}d{x}_2=\frac{e^2}{\kappa }\end{array}$$

2 最小作用量原理与磁单极 (19-1)

一般而言，考虑到磁单极子之后，麦克斯韦方程不能从一个作用量出发，利用最小作用量原理推导。特别地，如果存在磁单极子之后，为了满足环路定理，时空中会存在狄拉克弦，这也暗示着我们不能从一个简单的作用量出发，推导存在磁单极时候的麦克斯韦方程组。一个可行的方案是引入两个势能函数 $A^{\mu }$ 与 $B^{\mu }$。请利用最小作用量原理写出下面包含 $A^{\mu }$ 与 $B^{\mu }$ 的拉氏量对应的运动方程：

$$\mathcal{L}=-\frac{n^{\alpha }{n}^{\mu }}{8\pi n^2}{\eta }^{\beta \nu }(F_{\alpha \beta }^A{F}_{\mu \nu }^A+F_{\alpha \beta }^B{F}_{\mu \nu }^B)+\frac{n^{\alpha }{n}_{\mu }}{16\pi n^2}{ϵ}^{\mu \nu \gamma \delta }(F_{\alpha \nu }^B{F}_{\gamma \delta }^A-F_{\alpha \nu }^A{F}_{\gamma \delta }^B)-A_{\mu }{J}^{\mu }-\frac{4\pi }{e^2}{B}_{\mu }{K}^{\mu }$$

其中 $F_A^{\mu \nu }={\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu }$，$F_B^{\mu \nu }={\partial }^{\mu }{B}^{\nu }-{\partial }^{\nu }{B}^{\mu }$。此外 $J^{\mu }$ 与通常的电流，$K^{\mu }$ 为平常的磁流，$n^{\mu }$ 为狄拉克弦的方向矢量。

$$\begin{array}{c}\delta s=\delta \int -\frac{n^{\alpha }{n}^{\mu }}{8\pi n^2}{\eta }^{\beta \nu }(F_{\alpha \beta }^A{F}_{\mu \nu }^A+F_{\alpha \beta }^B{F}_{\mu \nu }^B)+\frac{n^{\alpha }{n}_{\mu }}{16\pi n^2}{ϵ}^{\mu \nu \gamma \delta }(F_{\alpha \nu }^B{F}_{\gamma \delta }^A-F_{\alpha \nu }^A{F}_{\gamma \delta }^B)-A_{\mu }{J}^{\mu }-\frac{4\pi }{e^2}{B}_{\mu }{K}^{\mu }d{x}^4\\ =\int -\frac{n^{\alpha }{n}^{\mu }}{8\pi n^2}{\eta }^{\beta \nu }(F_{\alpha \beta }^A\delta F_{\mu \nu }^A+F_{\mu \nu }^A\delta F_{\alpha \beta }^A+F_{\alpha \beta }^B\delta F_{\mu \nu }^B+F_{\mu \nu }^B\delta F_{\alpha \beta }^B)\\ +\frac{n^{\alpha }{n}_{\mu }}{16\pi n^2}{ϵ}^{\mu \nu \gamma \delta }(F_{\alpha \nu }^B\delta F_{\gamma \delta }^A+F_{\gamma \delta }^A\delta F_{\alpha \nu }^B-F_{\alpha \nu }^A\delta F_{\gamma \delta }^B-F_{\gamma \delta }^B\delta F_{\alpha \nu }^A)\\ -J^{\mu }\delta A_{\mu }-\frac{4\pi }{e^2}{K}^{\mu }\delta B_{\mu }d{x}^4\end{array}$$

现在对第一项的变分进行分析

$$\delta F_{\mu \nu }^A=\delta ({\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu })={\partial }_{\mu }\delta A_{\nu }-{\partial }_{\nu }\delta A_{\mu }$$$$\begin{array}{c}{n}^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }(F_{\alpha \beta }^A\delta F_{\mu \nu }^A+F_{\mu \nu }^A\delta F_{\alpha \beta }^A)\\ =n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }{F}_{\alpha \beta }^A{\partial }_{\mu }\delta A_{\nu }-n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }{F}_{\alpha \beta }^A{\partial }_{\nu }\delta A_{\mu }+n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }{F}_{\mu \nu }^A{\partial }_{\alpha }\delta A_{\beta }-n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }{F}_{\mu \nu }^A{\partial }_{\beta }\delta A_{\alpha }\\ =n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }{F}_{\alpha \beta }^A{\partial }_{\mu }\delta A_{\nu }-n^{\alpha }{n}^{\nu }{\eta }^{\beta \mu }{F}_{\alpha \beta }^A{\partial }_{\mu }\delta A_{\nu }+n^{\mu }{n}^{\alpha }{\eta }^{\nu \beta }{F}_{\alpha \beta }^A{\partial }_{\mu }\delta A_{\nu }-n^{\mu }{n}^{\alpha }{\eta }^{\nu \beta }{F}_{\alpha \beta }^A{\partial }_{\nu }\delta A_{\mu }\\ =2(n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }-n^{\alpha }{n}^{\nu }{\eta }^{\beta \mu })F_{\alpha \beta }^A{\partial }_{\mu }\delta A_{\nu }\end{array}$$

同理

$$\begin{array}{c}{n}^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }(F_{\alpha \beta }^B\delta F_{\mu \nu }^B+F_{\mu \nu }^B\delta F_{\alpha \beta }^B)\\ =2(n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }-n^{\alpha }{n}^{\nu }{\eta }^{\beta \mu })F_{\alpha \beta }^B{\partial }_{\mu }\delta B_{\nu }\end{array}$$

再分析第二项

$$\begin{array}{c}{n}^{\alpha }{n}_{\mu }{ϵ}^{\mu \nu \gamma \delta }(F_{\alpha \nu }^B\delta F_{\gamma \delta }^A+F_{\gamma \delta }^A\delta F_{\alpha \nu }^B-F_{\alpha \nu }^A\delta F_{\gamma \delta }^B-F_{\gamma \delta }^B\delta F_{\alpha \nu }^A)\\ =(n^{\alpha }{n}_{\mu }{ϵ}^{\mu \nu \gamma \delta }-n^{\gamma }{n}_{\mu }{ϵ}^{\mu \delta \alpha \nu })(F_{\alpha \nu }^B\delta F_{\gamma \delta }^A+F_{\gamma \delta }^A\delta F_{\alpha \nu }^B)\\ =(2n^{\alpha }{n}_{\mu }{ϵ}^{\mu \nu \gamma \delta }-n^{\gamma }{n}_{\mu }{ϵ}^{\mu \delta \alpha \nu }+n^{\delta }{n}_{\mu }{ϵ}^{\mu \gamma \alpha \nu })F_{\alpha \nu }^B{\partial }_{\gamma }\delta A_{\delta }\\ +(n^{\alpha }{n}_{\mu }{ϵ}^{\mu \nu \gamma \delta }-n^{\nu }{n}_{\mu }{ϵ}^{\mu \alpha \gamma \delta }-2n^{\gamma }{n}_{\mu }{ϵ}^{\mu \delta \alpha \nu })F_{\gamma \delta }^A{\partial }_{\alpha }\delta B_{\nu }\end{array}$$

对这几项作分部积分并且舍去全微分项，得到

$$\begin{array}{c}\delta s=\int d\tau -\frac{1}{4\pi n^2}((n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }-n^{\alpha }{n}^{\nu }{\eta }^{\beta \mu })F_{\alpha \beta }^A{\partial }_{\mu }\delta A_{\nu }+(n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }-n^{\alpha }{n}^{\nu }{\eta }^{\beta \mu })F_{\alpha \beta }^B{\partial }_{\mu }\delta B_{\nu })\\ +\frac{1}{16\pi n^2}((2n^{\alpha }{n}_{\mu }{ϵ}^{\mu \nu \gamma \delta }-n^{\gamma }{n}_{\mu }{ϵ}^{\mu \delta \alpha \nu }+n^{\delta }{n}_{\mu }{ϵ}^{\mu \gamma \alpha \nu })F_{\alpha \nu }^B{\partial }_{\gamma }\delta A_{\delta }\\ +(n^{\alpha }{n}_{\mu }{ϵ}^{\mu \nu \gamma \delta }-n^{\nu }{n}_{\mu }{ϵ}^{\mu \alpha \gamma \delta }-2n^{\gamma }{n}_{\mu }{ϵ}^{\mu \delta \alpha \nu })F_{\gamma \delta }^A{\partial }_{\alpha }\delta B_{\nu })\\ -J^{\mu }\delta A_{\mu }-\frac{4\pi }{e^2}{K}^{\mu }\delta B_{\mu }d{x}^4=0\end{array}$$$$\begin{array}{c}\int d\tau \frac{1}{4\pi n^2}((n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }-n^{\alpha }{n}^{\nu }{\eta }^{\beta \mu })\delta A_{\nu }{\partial }_{\mu }{F}_{\alpha \beta }^A+(n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }-n^{\alpha }{n}^{\nu }{\eta }^{\beta \mu })\delta B_{\nu }{\partial }_{\mu }{F}_{\alpha \beta }^B)\\ -\frac{1}{16\pi n^2}((2n^{\alpha }{n}_{\mu }{ϵ}^{\mu \nu \gamma \delta }-n^{\gamma }{n}_{\mu }{ϵ}^{\mu \delta \alpha \nu }+n^{\delta }{n}_{\mu }{ϵ}^{\mu \gamma \alpha \nu })\delta A_{\delta }{\partial }_{\gamma }{F}_{\alpha \nu }^B\\ -(n^{\alpha }{n}_{\mu }{ϵ}^{\mu \nu \gamma \delta }-n^{\nu }{n}_{\mu }{ϵ}^{\mu \alpha \gamma \delta }-2n^{\gamma }{n}_{\mu }{ϵ}^{\mu \delta \alpha \nu })\delta B_{\nu }{\partial }_{\alpha }{F}_{\gamma \delta }^A)\\ -J^{\mu }\delta A_{\mu }-\frac{4\pi }{e^2}{K}^{\mu }\delta B_{\mu }d{x}^4=0\end{array}$$

所以可以整理得到包含 $A^{\mu }$ 与 $B^{\mu }$ 的拉氏量对应的运动方程为

$$\frac{1}{4\pi n^2}(n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }-n^{\alpha }{n}^{\nu }{\eta }^{\beta \mu }){\partial }_{\mu }{F}_{\alpha \beta }^A-\frac{1}{16\pi n^2}(2n^{\alpha }{n}_{\mu }{ϵ}^{\mu \delta \gamma \nu }-n^{\gamma }{n}_{\mu }{ϵ}^{\mu \nu \alpha \delta }+n^{\nu }{n}_{\mu }{ϵ}^{\mu \gamma \alpha \delta }){\partial }_{\gamma }{F}_{\alpha \delta }^B=J^{\nu }$$$$\frac{1}{4\pi n^2}(n^{\alpha }{n}^{\mu }{\eta }^{\beta \nu }-n^{\alpha }{n}^{\nu }{\eta }^{\beta \mu }){\partial }_{\mu }{F}_{\alpha \beta }^B-\frac{1}{16\pi n^2}(n^{\nu }{n}_{\mu }{ϵ}^{\mu \alpha \gamma \delta }+2n^{\gamma }{n}_{\mu }{ϵ}^{\mu \delta \alpha \nu }-n^{\alpha }{n}_{\mu }{ϵ}^{\mu \nu \gamma \delta }){\partial }_{\alpha }{F}_{\gamma \delta }^A=\frac{4\pi }{e^2}{K}^{\mu }$$

3 各向异性介质 (蒋济帆)

对于一般的晶体，由于各向异性，应该用介电张量 ${ϵ}_{ij}$ 来表示其电学性质。取右手系 Oxyz，记电位移矢量 $\overrightarrow{D}=(D_x,D_y,D_z)$，电场强度矢量 $\overrightarrow{E}=(E_x,E_y,E_z)$，介电张量 $ϵ$ 为：

$$ϵ=\left(\begin{array}{ccc}{ϵ}_{11}& {ϵ}_{12}& {ϵ}_{13}\\ {ϵ}_{21}& {ϵ}_{22}& {ϵ}_{23}\\ {ϵ}_{31}& {ϵ}_{32}& {ϵ}_{33}\end{array}\right)$$

则：

$$\overrightarrow{D}=ϵ\cdot \overrightarrow{E}\text{i.e.}$$$$\left(\begin{array}{c}{D}_x\\ D_y\\ D_z\end{array}\right)=\left(\begin{array}{ccc}{ϵ}_{11}& {ϵ}_{12}& {ϵ}_{13}\\ {ϵ}_{21}& {ϵ}_{22}& {ϵ}_{23}\\ {ϵ}_{31}& {ϵ}_{32}& {ϵ}_{33}\end{array}\right)\left(\begin{array}{c}{E}_x\\ E_y\\ E_z\end{array}\right)$$

已知真空磁导率为 ${\mu }_0$，真空介电常数为 ${ϵ}_0$。

1.证明：介电张量为对称张量。 Hint：利用电磁场的能量。

证明：由于电场能量密度可以写成

$$w_E=\frac{1}{2}{ϵ}_{ij}{E}_i{E}_j$$

由于能量密度关于电场强度是连续的，所以其偏导数满足可交换性

$$\frac{{\partial }^2}{\partial E_i\partial E_j}{w}_E=\frac{{\partial }^2}{\partial E_j\partial E_i}{w}_E$$

所以 ${ϵ}_{ij}={ϵ}_{ji}$

2.对于普通的晶体，介质是非磁性的、透明的，此时 $\overrightarrow{B}={\mu }_0\overrightarrow{H}$，其中 $\overrightarrow{B}$ 为磁感应强度，$\overrightarrow{H}$ 为磁场强度。选取合适的坐标系 Oxyz，可使得介电张量对角化，即 $ϵ=\text{diag}\{{ϵ}_x,{ϵ}_y,{ϵ}_z\}$（${ϵ}_x,{ϵ}_y,{ϵ}_z>0$)，设晶体中某平面单色电磁波波矢 $\overrightarrow{k}$ 与 x 轴和 y 轴的夹角分别为 $\alpha$ 与 $\beta$（$0⩽\alpha ,\beta ⩽{90}^{\circ }⩽\alpha +\beta$)

(i) 求 $|\overrightarrow{k}|$ 所有可能的值。

$$\overrightarrow{k}=(k\cos \alpha ,k\cos \beta ,k\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }{)}^T$$$$\begin{array}{c}\mathrm{\nabla }\times \overrightarrow{E}=-\frac{\partial }{\partial t}\overrightarrow{B}\\ \overrightarrow{k}\times \overrightarrow{E}=\omega \overrightarrow{B}\end{array}$$$$\begin{array}{c}\mathrm{\nabla }\times \overrightarrow{B}={\mu }_0\frac{\partial }{\partial t}\overrightarrow{D}\\ \overrightarrow{k}\times \overrightarrow{B}=-{\mu }_0\omega \overrightarrow{D}\end{array}$$

所以

$$\begin{array}{c}\overrightarrow{k}\times (\overrightarrow{k}\times \overrightarrow{E})=-{\mu }_0{\omega }^2\overrightarrow{D}\\ (\overrightarrow{k}\cdot \overrightarrow{E})\overrightarrow{k}-k^2\overrightarrow{E}=-{\mu }_0{\omega }^2\overrightarrow{D}\\ k_i{E}_i{k}_j-k^2{E}_j=-{\mu }_0{\omega }^2{ϵ}_{ij}{E}_i\end{array}$$$$\begin{array}{c}(k\cos \alpha E_x+k\cos \beta E_y+k\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }{E}_z)k\cos \alpha -k^2{E}_x=-{\mu }_0{\omega }^2{ϵ}_x{E}_x\\ (k\cos \alpha E_x+k\cos \beta E_y+k\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }{E}_z)k\cos \beta -k^2{E}_y=-{\mu }_0{\omega }^2{ϵ}_y{E}_y\\ (k\cos \alpha E_x+k\cos \beta E_y+k\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }{E}_z)k\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }-k^2{E}_z=-{\mu }_0{\omega }^2{ϵ}_z{E}_z\end{array}$$$$\left(\begin{array}{ccc}{\mu }_0{\omega }^2{ϵ}_x-k^2{\sin }^2\alpha & k^2\cos \beta \cos \alpha & k^2\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }\cos \alpha \\ k^2\cos \alpha \cos \beta & {\mu }_0{\omega }^2{ϵ}_y-k^2{\sin }^2\beta & k^2\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }\cos \beta \\ k^2\cos \alpha \sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }& k^2\cos \beta \sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }& {\mu }_0{\omega }^2{ϵ}_z-k^2({\cos }^2\alpha +{\cos }^2\beta )\end{array}\right)\overrightarrow{E}=\overrightarrow{0}$$

方程有解要求左边系数矩阵的行列式为 $0$

$$\begin{array}{c}({\mu }_0{\omega }^2{ϵ}_x-k^2{\sin }^2\alpha )({\mu }_0{\omega }^2{ϵ}_y-k^2{\sin }^2\beta )({\mu }_0{\omega }^2{ϵ}_z-k^2({\cos }^2\alpha +{\cos }^2\beta ))\\ +k^2\cos \beta \cos \alpha k^2\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }\cos \beta k^2\cos \alpha \sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }\\ +k^2\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }\cos \alpha k^2\cos \alpha \cos \beta k^2\cos \beta \sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }\\ -k^2\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }\cos \alpha ({\mu }_0{\omega }^2{ϵ}_y-k^2{\sin }^2\beta )k^2\cos \alpha \sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }\\ -({\mu }_0{\omega }^2{ϵ}_x-k^2{\sin }^2\alpha )k^2\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }\cos \beta k^2\cos \beta \sqrt{1-{\cos }^2\alpha -{\cos }^2\beta }\\ -k^2\cos \beta \cos \alpha k^2\cos \alpha \cos \beta ({\mu }_0{\omega }^2{ϵ}_z-k^2({\cos }^2\alpha +{\cos }^2\beta ))=0\end{array}$$

解得

行列式展开后，通过对称性化简，发现$k^6$项抵消，最终得到一个关于$k^2$的二次方程：

$$\begin{array}{c}{\mu }_0^3{\omega }^6{ϵ}_x{ϵ}_y{ϵ}_z-{\mu }_0^2{\omega }^4{k}^2[{ϵ}_x{ϵ}_y({\cos }^2\alpha +{\cos }^2\beta )+{ϵ}_x{ϵ}_z{\sin }^2\beta +{ϵ}_y{ϵ}_z{\sin }^2\alpha ]\\ +{\mu }_0{\omega }^2{k}^4[{ϵ}_x{\cos }^2\alpha +{ϵ}_y{\cos }^2\beta +{ϵ}_z({\sin }^2\alpha {\sin }^2\beta -{\cos }^2\alpha {\cos }^2\beta )]=0\end{array}$$

将方程除以${\mu }_0{\omega }^2$并设$\lambda =k^2$，得到二次方程：

$$\begin{array}{c}{\mu }_0^2{\omega }^4{ϵ}_x{ϵ}_y{ϵ}_z-{\mu }_0{\omega }^2\lambda [{ϵ}_x{ϵ}_y({\cos }^2\alpha +{\cos }^2\beta )+{ϵ}_x{ϵ}_z{\sin }^2\beta +{ϵ}_y{ϵ}_z{\sin }^2\alpha ]\\ +{\lambda }^2[{ϵ}_x{\cos }^2\alpha +{ϵ}_y{\cos }^2\beta +{ϵ}_z(1-{\cos }^2\alpha -{\cos }^2\beta )]=0\end{array}$$

解得$\lambda$即$k^2$为：

$$\lambda =\frac{{\mu }_0{\omega }^2[{ϵ}_x{ϵ}_y({\cos }^2\alpha +{\cos }^2\beta )+{ϵ}_x{ϵ}_z{\sin }^2\beta +{ϵ}_y{ϵ}_z{\sin }^2\alpha \pm \sqrt{D}]}{2({ϵ}_x{\cos }^2\alpha +{ϵ}_y{\cos }^2\beta +{ϵ}_z(1-{\cos }^2\alpha -{\cos }^2\beta ))}$$

其中判别式$\mathrm{\Delta }$为：

$$\begin{array}{c}\mathrm{\Delta }={({ϵ}_x{ϵ}_y({\cos }^2\alpha +{\cos }^2\beta )+{ϵ}_x{ϵ}_z{\sin }^2\beta +{ϵ}_y{ϵ}_z{\sin }^2\alpha )}^2\\ -4{ϵ}_x{ϵ}_y{ϵ}_z({ϵ}_x{\cos }^2\alpha +{ϵ}_y{\cos }^2\beta +{ϵ}_z(1-{\cos }^2\alpha -{\cos }^2\beta ))\end{array}$$

最终，$|\overrightarrow{k}|$的值为：

$$|\overrightarrow{k}|=\sqrt{\frac{{\mu }_0{\omega }^2[{ϵ}_x{ϵ}_y({\cos }^2\alpha +{\cos }^2\beta )+{ϵ}_x{ϵ}_z{\sin }^2\beta +{ϵ}_y{ϵ}_z{\sin }^2\alpha \pm \sqrt{\mathrm{\Delta }}]}{2({ϵ}_x{\cos }^2\alpha +{ϵ}_y{\cos }^2\beta +{ϵ}_z(1-{\cos }^2\alpha -{\cos }^2\beta ))}}$$

(ii) 若 $|\overrightarrow{k}|$ 只有一个可能的取值，求此时的 $\alpha$ 与 $\beta$

$$\mathrm{\Delta }=0$$$$\begin{array}{c}{({ϵ}_x{ϵ}_y({\cos }^2\alpha +{\cos }^2\beta )+{ϵ}_x{ϵ}_z{\sin }^2\beta +{ϵ}_y{ϵ}_z{\sin }^2\alpha )}^2\\ =4{ϵ}_x{ϵ}_y{ϵ}_z({ϵ}_x{\cos }^2\alpha +{ϵ}_y{\cos }^2\beta +{ϵ}_z(1-{\cos }^2\alpha -{\cos }^2\beta ))\end{array}$$

所以

$$\begin{array}{c}({ϵ}_x{ϵ}_y({\cos }^2\alpha +{\cos }^2\beta ){)}^2+({ϵ}_x{ϵ}_z{\sin }^2\beta {)}^2+({ϵ}_y{ϵ}_z{\sin }^2\alpha {)}^2\\ =2{ϵ}_x{ϵ}_y{ϵ}_z({ϵ}_x(2{\cos }^2\alpha -({\cos }^2\alpha +{\cos }^2\beta ){\sin }^2\beta )+{ϵ}_y(2{\cos }^2\beta -({\cos }^2\alpha \\ +{\cos }^2\beta ){\sin }^2\alpha )+{ϵ}_z(2-2{\cos }^2\alpha -2{\cos }^2\beta -{\sin }^2\alpha {\sin }^2\beta ))\end{array}$$

解得，当

$$\alpha =\beta =0,{ϵ}_x={ϵ}_y$$

或

$$\alpha =\frac{\pi }{2},\beta =0,{ϵ}_y={ϵ}_z$$

或

$$\alpha =0,\beta =\frac{\pi }{2},{ϵ}_x={ϵ}_z$$

时波矢只有一个可能的取值

3.对于单轴晶体，设 $\sqrt{\frac{{ϵ}_x}{{ϵ}_0}}=\sqrt{\frac{{ϵ}_y}{{ϵ}_0}}=n_o$，$\sqrt{\frac{{ϵ}_z}{{ϵ}_0}}=n_e$，求 $|\overrightarrow{k}|$，并据此论述它的双折射的性质。

此时

$$|\overrightarrow{k}|=n_0\frac{\omega }{c}$$

或

$$|\overrightarrow{k}|=\frac{\omega }{c}\sqrt{\frac{n_o^2{n}_e^2}{n_o^2({\cos }^2\alpha +{\cos }^2\beta )+n_e^2(1-{\cos }^2\alpha -{\cos }^2\beta )}}$$

即单轴晶体中存在两种传播模式，分别对应 $o$ 光和 $e$ 光