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Noether current (2.16) の別導出法

場の量子論

physicstan.hatenablog.com

前回，保存された Noether current が (2.16) で与えられることを示した．そこでは \phi と \phi^{\ast} を独立な場として扱ったのだった．今回は \phi を実部と虚部に分けて計算していく．つまり， \phi = \phi_{1} + i\phi_{2} とする． 無限小変換の式： \phi \rightarrow e^{i\alpha}\phi ， \phi \rightarrow \phi + \alpha\Delta\phi より，

\begin{aligned} \alpha\Delta\phi &= e^{i\alpha}\phi - \phi \\\ &\sim (1+i\alpha)\phi - \phi \\\ &= i\alpha(\phi_{1} + i\phi_{2}) \\\ &= -\alpha\phi_{2} + i\alpha\phi_{1} \end{aligned}


なので，

\displaystyle{ \alpha\Delta\phi_{1} = -\alpha\phi_{2} ,\hspace{0.7cm} \alpha\Delta\phi_{2} = \alpha\phi_{1}}


\begin{aligned} \mathcal{L} &= |{\partial}_{\mu}\phi|^2 - m^2|\phi|^2 \\\ &= {\partial}_{\mu}\phi\cdot{\partial}^{\mu}\phi^{\ast} - m^2\phi\phi^{\ast} \\\ &= {\partial}_{\mu}(\phi_{1}+i\phi_{2})\cdot{\partial}^{\mu}(\phi_{1}-i\phi_{2}) - m^2\left({\phi_{1}}^2+{\phi_{2}}^2\right) \\\ &= {\partial}_{\mu}\phi_{1}\cdot{\partial}^{\mu}\phi_{1} + {\partial}_{\mu}\phi_{2}\cdot{\partial}^{\mu}\phi_{2} - m^2\left({\phi_{1}}^2+{\phi_{2}}^2\right) \end{aligned}


を用いると，

\displaystyle{ \frac{\partial\mathcal{L}}{\partial({\partial}_{\mu}\phi_{1})} = 2({\partial}_{0}\phi_{1}) - 2({\partial}_{1}\phi_{1}) -2({\partial}_{2}\phi_{1}) - 2({\partial}_{3}\phi_{1}) = 2{\partial}^{\mu}\phi_{1}}


なので，

\begin{aligned} j^{\mu} &= \frac{\partial\mathcal{L}}{\partial({\partial}_{\mu}\phi_{1})}\Delta\phi_{1} + \frac{\partial\mathcal{L}}{\partial({\partial}_{\mu}\phi_{2})}\Delta\phi_{2} \\\ &= 2{\partial}^{\mu}\phi_{1}\cdot(-\phi_{2}) + 2{\partial}^{\mu}\phi_{2}\cdot\phi_{1} \\\ &= 2\left[\phi_{1}{\partial}^{\mu}\phi_{2} - \phi_{2}{\partial}^{\mu}\phi_{1}\right] \end{aligned}


ここで，

\begin{aligned} i\left[({\partial}^{\mu}\phi^{\ast})\phi - \phi^{\ast}({\partial}^{\mu}\phi)\right] &= i\left[({\partial}^{\mu}\phi_{1} -i{\partial}^{\mu}\phi_{2})(\phi_{1}+i\phi_{2}) - (\phi_{1}-i\phi_{2})({\partial}^{\mu}\phi_{1}+i{\partial}^{\mu}\phi_{2})\right] \\\ &= 2\left[\phi_{1}{\partial}^{\mu}\phi_{2} - \phi_{2}{\partial}^{\mu}\phi_{1}\right] \end{aligned}


したがって，

\displaystyle{ j^{\mu} = i\left[({\partial}^{\mu}\phi^{\ast})\phi - \phi^{\ast}({\partial}^{\mu}\phi)\right]}


を得る．