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Let $\psi(x)$ be solution of Dirac equation $$ (\gamma^\mu\Pi_\mu-mc) \psi(x)=0 $$ where $\Pi_\mu=i\partial_\mu-eA_\mu$ is momentum operator in present electromagnetic field . We consider tow operators $$ P_1=\frac{1}{2}\begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix} \quad ,\quad P_2=\frac{1}{2}$\begin{pmatrix} 1 & -1\\ -1 & 1 \end{pmatrix} $$ where verify $P_1+P_2=1$ and $P_2\gamma^\mu=\gamma^\mu P_1$ . If we put $\psi_2=P_1\psi$ and $\psi_1=P_1\psi$, and we write $\psi(x)$ as form $\psi=\begin{pmatrix} \chi\\ \phi \end{pmatrix}$, then we observe that $\psi_1=\frac{1}{2}\begin{pmatrix} \chi+\phi\\ \chi+\phi \end{pmatrix}$

My question, what is Explanation of this result?

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    $\begingroup$ What is so surprising about $ \frac{1}{2}\begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix} \begin{pmatrix} \chi\\ \phi \end{pmatrix} = \frac{1}{2}\begin{pmatrix} \chi+\phi\\ \chi+\phi \end{pmatrix}$ ? $\endgroup$ – Thomas Fritsch Jul 17 at 20:09
  • $\begingroup$ My guess is that the OP wants to know what is the significance of these operators. $\endgroup$ – G. Smith Jul 17 at 20:35
  • $\begingroup$ Thomas Fritsch, Note that two component of $\psi_1$ are equal, what is physical mean of this? $\endgroup$ – Mohamed ELarbi Gadja Jul 18 at 19:27

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