\(\sigma^\mu \partial_\mu = \sum_{\mu=0}^3 \sigma^\mu \partial_\mu = \sigma^0 \partial_0 + \sigma^1 \partial_1 + \sigma^2 \partial_2 + \sigma^3 \partial_3\)
即： \[ \sigma^\mu \partial_\mu = \begin{pmatrix} \partial_t + \partial_z & \partial_x - i \partial_y \\ \partial_x + i \partial_y & \partial_t - \partial_z \end{pmatrix} \]
右手系：
满足的方程 \( (E I + \vec{\sigma} \cdot \vec{p}) u(p) = 0 \)
\( u(p) =(u_1, u_2) \)
即：
\(\begin{pmatrix} E + p_z & p_x - i p_y \\ p_x + i p_y & E - p_z \end{pmatrix} u(p) = 0\)
得两个方程：
$$ \begin{cases} (E + p_z) u_1 + (p_x - i p_y) u_2 = 0 \quad \text{(a)}\\ (p_x + i p_y) u_1 + (E - p_z) u_2 = 0 \quad \text{(b)} \end{cases} \tag{2} $$
\( \frac{u1}{u2}=-\frac{p_x-i p_y}{E+p_z} =-\frac{E-p_z}{p_x+i p_y} \)
左手系：
满足的方程：\( (E I - \vec{\sigma} \cdot \vec{p}) v(p) = 0 \)
\(\begin{pmatrix} E - p_z & -p_x + i p_y \\ -p_x - i p_y & E + p_z \end{pmatrix} v(p) = 0 \)
得：
$$ \begin{cases} (E - p_z) v_1 + (-p_x + i p_y) v_2 = 0 \quad \text{(a)}\\ (-p_x - i p_y) v_1 + (E + p_z) v_2 = 0 \quad \text{(b)} \end{cases} \tag{2} $$
\( \frac{v1}{v2}=-\frac{p_x-i p_y}{E-p_z} =-\frac{E+p_z}{p_x+i p_y} \)
都满足 \( E^2 =p_x^2+p_y^2+p_z^2 \)，
即： \( (E+p_z)(E-p_z) = (p_x+i p_y)(p_x-i p_y) \)
或： \( (E+p_z)(E-p_z)u_1 u_2 = (p_x+i p_y)(p_x-i p_y)u_1 u_2 \)
\( D^2 u(p) =\Box I u(p) \)