クライン-ゴルドン方程式の非相対論的極限(正のエネルギー解)

\begin{align*}E=+ c\sqrt{(\boldsymbol p-q\boldsymbol A)^2+m^2c^2}+qA^0\end{align*}

\begin{align*}\left\{\frac{1}{c^2}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)^2-\left(\boldsymbol\nabla-\frac{iq\boldsymbol A}{\hbar}\right)^2+\frac{m^2c^2}{\hbar^2}\right\}\phi=0\end{align*}

\begin{align*}i\hbar\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)\varPsi=-\frac{\hbar^2}{2m}\left(\boldsymbol \nabla-\frac{iq\boldsymbol A}{\hbar}\right)^2\varPsi\end{align*}

\begin{align*}\left\{\frac{1}{c^2}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)^2-\left(\boldsymbol\nabla-\frac{iq\boldsymbol A}{\hbar}\right)^2+\frac{m^2c^2}{\hbar^2}\right\}\phi&=0\end{align*}

\begin{align*}H=\pm c\sqrt{(\boldsymbol p-q\boldsymbol A)^2+m^2c^2}+qA^0\end{align*}

\begin{align*}H&=\frac{1}{2m}(\boldsymbol p-q\boldsymbol A)^2+qA^0\end{align*}

\begin{align*}E=\pm c\sqrt{(\boldsymbol p-q\boldsymbol A)^2+m^2c^2}+qA^0\tag{1}\end{align*}

\begin{align*}E=+ c\sqrt{(\boldsymbol p-q\boldsymbol A)^2+m^2c^2}+qA^0\tag{2}\end{align*}

\begin{align*}i\hbar\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)\varPsi=-\frac{\hbar^2}{2m}\left(\boldsymbol \nabla-\frac{iq\boldsymbol A}{\hbar}\right)^2\varPsi\tag{3}\end{align*}

\begin{align*}\varPsi&=e^{-\frac{i}{\hbar}(E’t-\boldsymbol p\cdot\boldsymbol x)}\tag{4}\\E’&=\frac{1}{2m}(\boldsymbol p-q\boldsymbol A)^2+qA^0\tag{5}\end{align*}

\begin{align*}\left\{\frac{1}{c^2}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)^2-\left(\boldsymbol\nabla-\frac{iq\boldsymbol A}{\hbar}\right)^2+\frac{m^2c^2}{\hbar^2}\right\}\phi=0\tag{6}\end{align*}

\begin{align*}\phi&=e^{-\frac{i}{\hbar}(Et-\boldsymbol p\cdot\boldsymbol x)}\tag{7}\\E&=+ c\sqrt{(\boldsymbol p-q\boldsymbol A)^2+m^2c^2}+qA^0\tag{2}\end{align*}

\begin{align*}E&=+c\sqrt{(\boldsymbol p-q\boldsymbol A)^2+m^2c^2}+qA^0\\&=mc^2\sqrt{1+\frac{1}{m^2c^2}(\boldsymbol p-q\boldsymbol A)^2}+qA^0\\&=mc^2\left\{1+\frac{1}{2}\frac{1}{m^2c^2}(\boldsymbol p-q\boldsymbol A)^2+\cdots\right\}+qA^0\\&\simeq mc^2+\frac{1}{2m}(\boldsymbol p-q\boldsymbol A)^2+qA^0\\&=mc^2+E’\tag{8}\end{align*}

\begin{align*}\phi&=e^{-\frac{i}{\hbar}(Et-\boldsymbol p\cdot\boldsymbol x)}\\&\simeq e^{-\frac{i}{\hbar}(mc^2t+E’t-\boldsymbol p\cdot\boldsymbol x)}\\&=e^{-\frac{imc^2}{\hbar}t}\varPsi\tag{9}\end{align*}

\begin{align*}\left\{\frac{1}{c^2}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)^2-\left(\boldsymbol\nabla-\frac{iq\boldsymbol A}{\hbar}\right)^2+\frac{m^2c^2}{\hbar^2}\right\}\phi=0\tag{6}\end{align*}

\begin{align*}\frac{1}{c^2}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)^2e^{-\frac{imc^2}{\hbar}t}\varPsi-e^{-\frac{imc^2}{\hbar}t}\left(\boldsymbol\nabla-\frac{iq\boldsymbol A}{\hbar}\right)^2\varPsi+\frac{m^2c^2}{\hbar^2}e^{-\frac{imc^2}{\hbar}t}\varPsi=0\tag{10}\end{align*}

\begin{align*}\frac{1}{c^2}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)^2e^{-\frac{imc^2}{\hbar}t}\varPsi&=\frac{1}{c^2}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)e^{-\frac{imc^2}{\hbar}t}\varPsi\\&=\frac{1}{c^2}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)e^{-\frac{imc^2}{\hbar}t}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}-\frac{imc^2}{\hbar}\right)\varPsi\\&=\frac{1}{c^2}e^{-\frac{imc^2}{\hbar}t}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}-\frac{imc^2}{\hbar}\right)\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}-\frac{imc^2}{\hbar}\right)\varPsi\\&=\frac{1}{c^2}e^{-\frac{imc^2}{\hbar}t}\left\{\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}-\frac{imc^2}{\hbar}\right)\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)-\frac{imc^2}{\hbar}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}-\frac{imc^2}{\hbar}\right)\right\}\varPsi\\&=\frac{1}{c^2}e^{-\frac{imc^2}{\hbar}t}\left\{\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}-2\frac{imc^2}{\hbar}\right)\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)-\frac{m^2c^4}{\hbar^2}\right\}\varPsi\tag{11}\end{align*}

\begin{align*}\vert mc^2\varPsi\vert\gg \left| i\hbar\frac{\partial \varPsi}{\partial t}\right|,\vert qA^0\varPsi\vert\tag{12}\end{align*}

\begin{align*}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}-2\frac{imc^2}{\hbar}\right)\end{align*}

\begin{align*}\frac{1}{c^2}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)^2e^{-\frac{imc^2}{\hbar}t}\varPsi&\simeq\frac{1}{c^2}e^{-\frac{imc^2}{\hbar}t}\left\{-2\frac{imc^2}{\hbar}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)-\frac{m^2c^4}{\hbar^2}\right\}\varPsi\\&=e^{-\frac{imc^2}{\hbar}t}\left\{-2\frac{im}{\hbar}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)-\frac{m^2c^2}{\hbar^2}\right\}\varPsi\tag{13}\end{align*}

\begin{align*}&e^{-\frac{imc^2}{\hbar}t}\left\{-2\frac{im}{\hbar}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)\right\}-e^{-\frac{imc^2}{\hbar}t}\left(\boldsymbol\nabla-\frac{iq\boldsymbol A}{\hbar}\right)^2\varPsi=0\\\rightarrow& i\hbar\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)\varPsi=-\frac{\hbar^2}{2m}\left(\boldsymbol \nabla-\frac{iq\boldsymbol A}{\hbar}\right)^2\varPsi\tag{3}\end{align*}

\begin{align*}E=- c\sqrt{(\boldsymbol p-q\boldsymbol A)^2+m^2c^2}+qA^0\end{align*}

\begin{align*}\left\{\frac{1}{c^2}\left(\frac{\partial}{\partial t}+\frac{iqA^0}{\hbar}\right)^2-\left(\boldsymbol\nabla-\frac{iq\boldsymbol A}{\hbar}\right)^2+\frac{m^2c^2}{\hbar^2}\right\}\phi=0\end{align*}

\begin{align*} i\hbar\bigg(\frac{\partial}{\partial t}&+\frac{iqA^0}{\hbar}\bigg)\chi=\frac{\hbar^2}{2m}\left(\boldsymbol \nabla-\frac{iq\boldsymbol A}{\hbar}\right)^2\chi\end{align*}