记 $\vec{k}=(k_1,k_2,k_3)$, 设 $\vec{r}=(x,y,z)$. 于是
\[
\vec{k}\cdot\vec{r}=k_1 x+k_2 y+k_3 z.
\]
\[
\nabla=(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})=(\partial_x, \partial_y, \partial_z).
\]
\[
\frac{\partial}{\partial x}e^{i(\vec{k}\cdot\vec{r}-\omega t)}=e^{i(\vec{k}\cdot\vec{r}-\omega t)}\cdot i \frac{\partial}{\partial x}(k_1 x+k_2 y+k_3 z)=ik_1 e^{i(\vec{k}\cdot\vec{r}-\omega t)}.
\]
因此,
\[
\nabla e^{i(\vec{k}\cdot\vec{r}-\omega t)}=ie^{i(\vec{k}\cdot\vec{r}-\omega t)}(k_1,k_2,k_3)=i\vec{k}e^{i(\vec{k}\cdot\vec{r}-\omega t)}.
\]
\[
\begin{split}
\nabla^2 e^{i(\vec{k}\cdot\vec{r}-\omega t)}&=\nabla(\nabla e^{i(\vec{k}\cdot\vec{r}-\omega t)})\\
&=\nabla(i\vec{k}e^{i(\vec{k}\cdot\vec{r}-\omega t)})\\
&=i\vec{k}^t\nabla(e^{i(\vec{k}\cdot\vec{r}-\omega t)})\\
&=i\vec{k}^t\cdot i\vec{k}e^{i(\vec{k}\cdot\vec{r}-\omega t)}\\
&=(-1)\vec{k}^t\cdot\vec{k}e^{i(\vec{k}\cdot\vec{r}-\omega t)}\\
\end{split}
\]
因此,
\[
\mathrm{tr}\nabla^2(e^{i(\vec{k}\cdot\vec{r}-\omega t)})=(-1)|\vec{k}|^2\cdot e^{i(\vec{k}\cdot\vec{r}-\omega t)}.
\]
另一方面,
\[
\begin{aligned}
\frac{\partial}{\partial t}A&=A_0 e^{i(\vec{k}\cdot\vec{r}-\omega t)}\cdot(-i\omega)\\
\frac{\partial^2}{\partial t^2}A&=A_0 e^{i(\vec{k}\cdot\vec{r}-\omega t)}\cdot(-i\omega)^2\\
\end{aligned}
\]
代入方程, 得
\[
\frac{1}{c^2}\cdot(-1)A_0\cdot\omega^2\cdot e^{i(\vec{k}\cdot\vec{r}-\omega t)}-(-1)A_0|\vec{k}|^2\cdot e^{i(\vec{k}\cdot\vec{r}-\omega t)}=0
\]
这推出
\[
-\frac{\omega^2}{c^2}+|\vec{k}|^2=0.
\]
即
\[
\frac{\omega^2}{c^2}=k^2.
\]
