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I'm studying the Dirac equation using Walter Greiner's textbook, my question is about the following passage

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Firstly I don't understand what the author means with "the rest energy being separeted by (2.78)". Secondly I don't see how (2.78) implies in (2.79).

Qmechanic
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Bruno Piveta
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  • Apply product rule on $\tilde{X}=X e^{-i(mc^2/\hbar)t}$ with $X=(\varphi,\chi)^T$ one gets an additional term on the left side $mc^2 X e^\theta$ with $e^\theta = e^{-imc^2/\hbar t}$which brought on the RHS together with $mc^2(\varphi,-\chi)^T e^\theta $ yields $-mc^2 (\varphi,\chi)^T e^\theta+mc^2(\varphi,-\chi)^T e^\theta =-2mc^2 (0,\chi)^T e^\theta $. Of course at the end one cancels out the phase factor $e^\theta$ on both sides. – Frederic Thomas Feb 15 '23 at 16:10
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    Related/worth checking: https://physics.stackexchange.com/q/489990/226902, https://physics.stackexchange.com/q/694598/226902, https://physics.stackexchange.com/q/468605/226902 – Quillo Feb 15 '23 at 16:22

1 Answers1

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About the rest energy: That's just a definition. They define $\varphi$ as $\frac{\widetilde{\varphi}}{e^{-i (m c^2/\hbar) t}}$ and analogously $\chi$. Since in the case of H not being explicitely time-dependent the general solution of the Dirac equation is a linear combination of eigenstates multiplied with $e^{-i E t/\hbar}$, $\varphi$ only changes in time with E - mc^2 (as opposed to $\widetilde{\varphi}$ changing in time with all of E, the same is true for $\chi$), they have "seperated" the rest energy. About the last equation: If you use $\varphi = \phi \cdot e^{...}$ and the product rule, the left hand site of (2.77) gets an extra term of $m c^2 \cdot (\begin{array}{c} \varphi \\ \chi \end{array}) \cdot e^{...}$ (the other term is just $i \hbar (\frac {\partial}{\partial t} (\begin{array}{c} \varphi \\ \chi \end{array})) \cdot e^{...}$), so if you subtract that second term on both sides and after that divide everything by $e^{...}$ (which you are allowed to since that's never zero), you get exactly equation (2.79).

Tarik
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