# Relativistic quantum field theory

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?

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