# 狄拉克方程式

$i \hbar \frac{\partial\psi (\mathbf{x},t)}{\partial t} = \left( \frac{\hbar c}{i}\boldsymbol{\alpha \cdot \nabla} +\beta m c^2 \right) \psi (\mathbf{x},t)$

## 狄拉克的最初推导

$i \hbar \frac{\partial\psi (\mathbf{x},t)}{\partial t} = H \psi (\mathbf{x},t) \equiv -\frac{\hbar^2}{2m}\nabla^2\psi (\mathbf{x},t)$

$i \hbar \frac{\partial\psi(\mathbf{x},t) }{\partial t} = H \psi (\mathbf{x},t) \equiv \left(c(\alpha_1 p_1 + \alpha_2 p_2 + \alpha_3 p_3) + \beta mc^2 \right) \psi (\mathbf{x},t)$

$p_i=\frac{\hbar}{i} \frac{\partial}{\partial x_i}, i=1,2,3$

 狄拉克方程式（原始版本） $i \hbar \frac{\partial\psi (\mathbf{x},t)}{\partial t} = \left[ \frac{\hbar c}{i} \left( \alpha_1\frac{\partial}{\partial x_1} + \alpha_2\frac{\partial}{\partial x_2} + \alpha_3\frac{\partial}{\partial x_3}\right) + \beta m c^2\right] \psi (\mathbf{x},t) \equiv H \psi (\mathbf{x},t)$

$i \hbar \frac{\partial\psi (\mathbf{x},t)}{\partial t} = \left( \frac{\hbar c}{i}\boldsymbol{\alpha \cdot \nabla} +\beta m c^2 \right) \psi (\mathbf{x},t)$

$\psi (\mathbf{x},t)= \begin{pmatrix} \psi_1 (\mathbf{x},t)\\ \psi_2 (\mathbf{x},t)\\ \psi_3 (\mathbf{x},t)\\ \vdots \\ \psi_N (\mathbf{x},t)\\ \end{pmatrix}$

$\rho(x) = \psi^\dagger \psi = \sum_{k=1}^N \psi_i^* \psi_i$

$\alpha_i\alpha_j + \alpha_j\alpha_i = 2 \delta_{ij}I$
$\alpha_i\beta + \beta\alpha_i = 0$
$\alpha_i^2 = \beta^2 = I$

$\beta = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix} \quad \alpha_i = \begin{pmatrix} 0 & \sigma_i \\ \sigma_i & 0 \end{pmatrix}$

$\sigma_1 = \begin{pmatrix}0&1\\1&0\end{pmatrix} \quad \sigma_2 = \begin{pmatrix}0&-i\\i&0\end{pmatrix} \quad\sigma_3 = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$

$\beta = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}, \quad \alpha_1 = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix},$
$\alpha_2 = \begin{pmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & -i& 0 & 0 \\ i & 0 & 0 & 0 \end{pmatrix}, \quad \alpha_3 = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}$

$i\frac{\partial\psi (\mathbf{x},t)}{\partial t} = \left( \frac{1}{i}\boldsymbol{\alpha \cdot \nabla} +\beta m \right) \psi (\mathbf{x},t)$

## 狄拉克方程的洛伦兹协变形式

$\left\{ \gamma^\mu, \gamma^\nu \right\} = -2\eta^{\mu\nu}$，其中ημν是平直时空的度规
$\eta = \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$

$\left( \gamma^\mu \partial _\mu \right)^2 = \frac{1}{2}\left\{ \gamma^\mu, \gamma^\nu \right\}\partial _\mu \partial _\nu = -\partial _\nu \partial ^\nu = \frac{\partial^2}{\partial t^2} - \nabla^2$

 狄拉克方程式（協變形式） $i \hbar \gamma^\mu \partial_\mu \psi - m c \psi = 0$

$i \gamma^\mu \partial_\mu \psi - m\psi = 0$

$\gamma^\mu=(\gamma^0,\boldsymbol\gamma) \equiv (\gamma^0,\gamma^1,\gamma^2,\gamma^3)$
$\gamma^0=\beta$
$\boldsymbol\gamma=\beta\boldsymbol\alpha$，或寫成$\gamma^i=\beta \alpha^i, i=1,2,3$

${\partial\!\!\!\big /} \equiv \gamma^\mu \partial_\mu$

$i \hbar {\partial\!\!\!\big /} \psi - m c \psi = 0$

 狄拉克方程式（自然單位制） $(i{\partial\!\!\!\big /} - m) \psi = 0\,$