狄拉克旋量

${\displaystyle \psi =\omega _{\vec {p}}\;e^{-ipx}\;}$

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

${\displaystyle \scriptstyle \psi }$相對論性自旋½
${\displaystyle \scriptstyle \omega _{\vec {p}}}$是狄拉克旋量，與波向量${\displaystyle \scriptstyle {\vec {p}}}$的平面波有關，
${\displaystyle \scriptstyle px\;\equiv \;p_{\mu }x^{\mu }}$,
${\displaystyle \scriptstyle p^{\mu }\;=\;\{\pm {\sqrt {m^{2}+{\vec {p}}^{2}}},\,{\vec {p}}\}}$為平面波的四維波向量，而${\displaystyle \scriptstyle {\vec {p}}}$為任意的，
${\displaystyle \scriptstyle x^{\mu }}$為一給定慣性系中的四維空間座標

${\displaystyle \omega _{\vec {p}}={\begin{bmatrix}\phi \\{\frac {{\vec {\sigma }}{\vec {p}}}{E_{\vec {p}}+m}}\phi \end{bmatrix}}\;,}$

${\displaystyle \scriptstyle \phi }$為任意的雙旋量，
${\displaystyle \scriptstyle {\vec {\sigma }}}$包立矩陣
${\displaystyle \scriptstyle E_{\vec {p}}}$為正根號${\displaystyle \scriptstyle E_{\vec {p}}\;=\;+{\sqrt {m^{2}+{\vec {p}}^{2}}}}$

源自狄拉克方程式的推導

${\displaystyle \left(-i{\vec {\alpha }}\cdot {\vec {\nabla }}+\beta m\right)\psi =i{\frac {\partial \psi }{\partial t}}\,}$

${\displaystyle {\vec {\alpha }}={\begin{bmatrix}\mathbf {0} &{\vec {\sigma }}\\{\vec {\sigma }}&\mathbf {0} \end{bmatrix}}\quad \quad \beta ={\begin{bmatrix}\mathbf {I} &\mathbf {0} \\\mathbf {0} &-\mathbf {I} \end{bmatrix}}\,}$

${\displaystyle \psi =\omega e^{-ip\cdot x}}$,

${\displaystyle \omega ={\begin{bmatrix}\phi \\\chi \end{bmatrix}}\,}$.

結果

${\displaystyle E{\begin{bmatrix}\phi \\\chi \end{bmatrix}}={\begin{bmatrix}m\mathbf {I} &{\vec {\sigma }}{\vec {p}}\\{\vec {\sigma }}{\vec {p}}&-m\mathbf {I} \end{bmatrix}}{\begin{bmatrix}\phi \\\chi \end{bmatrix}}\,}$.

${\displaystyle \left(E-m\right)\phi =\left({\vec {\sigma }}{\vec {p}}\right)\chi \,}$
${\displaystyle \left(E+m\right)\chi =\left({\vec {\sigma }}{\vec {p}}\right)\phi \,}$

${\displaystyle \omega ={\begin{bmatrix}\phi \\\chi \end{bmatrix}}={\begin{bmatrix}\phi \\{\frac {{\vec {\sigma }}{\vec {p}}}{E+m}}\phi \end{bmatrix}}\,}$.

${\displaystyle \omega ={\begin{bmatrix}\phi \\\chi \end{bmatrix}}={\begin{bmatrix}-{\frac {{\vec {\sigma }}{\vec {p}}}{-E+m}}\chi \\\chi \end{bmatrix}}\,}$.

細節

2-旋量

2-旋量最常見的定義為：

${\displaystyle \phi ^{1}={\begin{bmatrix}1\\0\end{bmatrix}}\quad \quad \phi ^{2}={\begin{bmatrix}0\\1\end{bmatrix}}\,}$

${\displaystyle \chi ^{1}={\begin{bmatrix}0\\1\end{bmatrix}}\quad \quad \chi ^{2}={\begin{bmatrix}1\\0\end{bmatrix}}\,}$

包立矩陣

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

${\displaystyle {\vec {\sigma }}\cdot {\vec {p}}=\sigma _{1}p_{1}+\sigma _{2}p_{2}+\sigma _{3}p_{3}={\begin{bmatrix}p_{3}&p_{1}-ip_{2}\\p_{1}+ip_{2}&-p_{3}\end{bmatrix}}}$

4-旋量

粒子

${\displaystyle u({\vec {p}},s)={\sqrt {E+m}}{\begin{bmatrix}\phi ^{(s)}\\{\frac {{\vec {\sigma }}\cdot {\vec {p}}}{E+m}}\phi ^{(s)}\end{bmatrix}}\,}$

${\displaystyle u({\vec {p}},1)={\sqrt {E+m}}{\begin{bmatrix}1\\0\\{\frac {p_{3}}{E+m}}\\{\frac {p_{1}+ip_{2}}{E+m}}\end{bmatrix}}\quad \mathrm {and} \quad u({\vec {p}},2)={\sqrt {E+m}}{\begin{bmatrix}0\\1\\{\frac {p_{1}-ip_{2}}{E+m}}\\{\frac {-p_{3}}{E+m}}\end{bmatrix}}}$

反粒子

${\displaystyle v({\vec {p}},s)={\sqrt {E+m}}{\begin{bmatrix}{\frac {{\vec {\sigma }}\cdot {\vec {p}}}{E+m}}\chi ^{(s)}\\\chi ^{(s)}\end{bmatrix}}\,}$

${\displaystyle v({\vec {p}},1)={\sqrt {E+m}}{\begin{bmatrix}{\frac {p_{1}-ip_{2}}{E+m}}\\{\frac {-p_{3}}{E+m}}\\0\\1\end{bmatrix}}\quad \mathrm {and} \quad v({\vec {p}},2)={\sqrt {E+m}}{\begin{bmatrix}{\frac {p_{3}}{E+m}}\\{\frac {p_{1}+ip_{2}}{E+m}}\\1\\0\\\end{bmatrix}}}$