泡利方程：
\[ i \hbar \frac{\partial}{\partial t} \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} = \left[ \frac{1}{2m} \left( \mathbf{p} - q \mathbf{A} \right)^2 + q \phi + \mu_B \mathbf{B} \cdot \boldsymbol{\sigma} \right] \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \]
即：
\[ \boxed{ \begin{aligned} i \hbar \frac{\partial \psi_1}{\partial t} &= \left[ \frac{1}{2m} \left( -i \hbar \nabla - q \mathbf{A} \right)^2 + q \phi \right] \psi_1 + \mu_B \left( B_z \psi_1 + (B_x - i B_y) \psi_2 \right), \\ i \hbar \frac{\partial \psi_2}{\partial t} &= \left[ \frac{1}{2m} \left( -i \hbar \nabla - q \mathbf{A} \right)^2 + q \phi \right] \psi_2 + \mu_B \left( (B_x + i B_y) \psi_1 - B_z \psi_2 \right). \end{aligned} } \]
狄拉克方程：
\[ (i \gamma^\mu \partial_\mu - m) \psi = 0 \]
即：
\[ \boxed{ \begin{aligned} \partial_t \psi_1 &= -\partial_z \psi_3 - (\partial_x - i \partial_y) \psi_4 - i m \psi_1, \\ \partial_t \psi_2 &= -(\partial_x + i \partial_y) \psi_3 + \partial_z \psi_4 - i m \psi_2, \\ \partial_t \psi_3 &= \partial_z \psi_1 + (\partial_x - i \partial_y) \psi_2 + i m \psi_3, \\ \partial_t \psi_4 &= (\partial_x + i \partial_y) \psi_1 - \partial_z \psi_2 + i m \psi_4. \end{aligned} } \]
对平面波正能解 \(\psi_j(x,t) = u_j e^{-i (p \cdot x - E t)} = u_j e^{-i (p_x x + p_y y + p_z z - E t)}\),
\(\partial_t \psi_j = i E \psi_j =i E u_j e^{-i (p \cdot x - E t)}\)
\(\partial_x \psi_j = -i p_x \psi_j =-i p_x u_j e^{-i (p \cdot x - E t)}\)
\(\partial_y \psi_j = -i p_y \psi_j =-i p_y u_j e^{-i (p \cdot x - E t)}\)
\(\partial_z \psi_j = -i p_z \psi_j =-i p_z u_j e^{-i (p \cdot x - E t)}\)
有：
\( (E + m) u_1 = p_z u_3 + (p_x - i p_y) u_4 \)
\( (E + m) u_2 = (p_x + i p_y) u_3 - p_z u_4 \)
\( (E - m) u_3 =p_z u_1 + (p_x - i p_y) u_2 \)
\( (E - m) u_4 =p_z u_2 - (p_x + i p_y) u_1 \)
其中 \( E^2 =p^2 + m^2 \)，方程1和2等价于3和4,
方程的解为：
\(u_1 = \frac{p_z u_3 + (p_x + i p_y) u_4}{E + m}\),
\( u_2 = \frac{(p_x - i p_y) u_3 - p_z u_4}{E + m}\).
\( u_3 \) 和 \( u_4 \) 对应粒子的 自旋自由度（如电子自旋向上/向下），
对上旋，\(u_3=1,u_4=0\)，有：
\(u_1 = \frac{p_z}{E + m}\),对应动量的\(p_z\)分量
\(u_2 = \frac{(p_x - i p_y)}{E + m}\)，对应动量的\(p_x\)和\(p_y\)自旋分量.
对平面波负能解 \(\psi_j(x,t) = u_j e^{i (p \cdot x - E t)} = u_j e^{i (p_x x + p_y y + p_z z - E t)}\),
\(\partial_t \psi_j = -i E \psi_j =-i E u_j e^{i (p \cdot x - E t)}\)
\(\partial_x \psi_j = i p_x \psi_j =i p_x u_j e^{i (p \cdot x - E t)}\)
\(\partial_y \psi_j = i p_y \psi_j =i p_y u_j e^{i (p \cdot x - E t)}\)
\(\partial_z \psi_j = i p_z \psi_j =i p_z u_j e^{i (p \cdot x - E t)}\)
有：
\( (E - m) u_1 = p_z u_3 + (p_x - i p_y) u_4 \)
\( (E - m) u_2 = (p_x + i p_y) u_3 - p_z u_4 \)
\( (E + m) u_3 = p_z u_1 + (p_x - i p_y) u_2 \)
\( (E + m) u_4 =p_z u_2 - (p_x + i p_y) u_1 \)
其中 \( E^2 =p^2 + m^2 \)，方程1和2等价于3和4,
方程的解为：
\( u_1 = \frac{p_z u_3 + (p_x - i p_y) u_4}{E - m}\),
\(u_2 = \frac{(p_x + i p_y) u_3 - p_z u_4}{E - m} \).