狄拉克方程式

${\displaystyle (i{\partial \!\!\!{\big /}}-m)\psi =0\,}$

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

狄拉克的最初推导

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

${\displaystyle i\hbar {\frac {\partial \psi (\mathbf {x} ,t)}{\partial t}}=H\psi (\mathbf {x} ,t)\equiv \left(c(\alpha _{1}p_{1}+\alpha _{2}p_{2}+\alpha _{3}p_{3})+\beta mc^{2}\right)\psi (\mathbf {x} ,t)}$

${\displaystyle p_{i}={\frac {\hbar }{i}}{\frac {\partial }{\partial x_{i}}},i=1,2,3}$

${\displaystyle i\hbar {\frac {\partial \psi (\mathbf {x} ,t)}{\partial t}}=\left[{\frac {\hbar c}{i}}\left(\alpha _{1}{\frac {\partial }{\partial x_{1}}}+\alpha _{2}{\frac {\partial }{\partial x_{2}}}+\alpha _{3}{\frac {\partial }{\partial x_{3}}}\right)+\beta mc^{2}\right]\psi (\mathbf {x} ,t)\equiv H\psi (\mathbf {x} ,t)}$

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

${\displaystyle \psi (\mathbf {x} ,t)={\begin{pmatrix}\psi _{1}(\mathbf {x} ,t)\\\psi _{2}(\mathbf {x} ,t)\\\psi _{3}(\mathbf {x} ,t)\\\vdots \\\psi _{N}(\mathbf {x} ,t)\\\end{pmatrix}}}$

${\displaystyle \rho (x)=\psi ^{\dagger }\psi =\sum _{i=1}^{N}\psi _{i}^{*}\psi _{i}}$

${\displaystyle \alpha _{i}\alpha _{j}+\alpha _{j}\alpha _{i}=2\delta _{ij}I}$
${\displaystyle \alpha _{i}\beta +\beta \alpha _{i}=0}$
${\displaystyle \alpha _{i}^{2}=\beta ^{2}=I}$

${\displaystyle \beta ={\begin{pmatrix}I&0\\0&-I\end{pmatrix}}\quad \alpha _{i}={\begin{pmatrix}0&\sigma _{i}\\\sigma _{i}&0\end{pmatrix}}}$

${\displaystyle \sigma _{1}={\begin{pmatrix}0&1\\1&0\end{pmatrix}}\quad \sigma _{2}={\begin{pmatrix}0&-i\\i&0\end{pmatrix}}\quad \sigma _{3}={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}}$

${\displaystyle \beta ={\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}},\quad \alpha _{1}={\begin{pmatrix}0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}},}$
${\displaystyle \alpha _{2}={\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix}},\quad \alpha _{3}={\begin{pmatrix}0&0&1&0\\0&0&0&-1\\1&0&0&0\\0&-1&0&0\end{pmatrix}}}$

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

狄拉克方程的洛伦兹协变形式

${\displaystyle \left\{\gamma ^{\mu },\gamma ^{\nu }\right\}=2\eta ^{\mu \nu }}$ ，其中ημν是闵可夫斯基时空的度规
${\displaystyle \eta ={\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}}}$

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

${\displaystyle i\hbar \gamma ^{\mu }\partial _{\mu }\psi -mc\psi =0}$

${\displaystyle i\gamma ^{\mu }\partial _{\mu }\psi -m\psi =0}$

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

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

${\displaystyle i\hbar {\partial \!\!\!{\big /}}\psi -mc\psi =0}$

${\displaystyle (i{\partial \!\!\!{\big /}}-m)\psi =0\,}$

參考資料

1. ^ Dirac Equation and Hydrogen Atom (PDF). [2014-09-10]. （原始内容存档 (PDF)于2016-03-05）.
2. ^ see for example Brian Pendleton: Quantum Theory 2012/2013, section 4.3 The Dirac Equation页面存档备份，存于互联网档案馆