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Using the Dirac equation with or $p=$ zero and the $\gamma^0$ matrix defined as $$\gamma^0=\begin{pmatrix}0 & \sigma_0 \\ \sigma_0 & 0\end{pmatrix} = \begin{pmatrix}0 & \bf{I} \\ \bf{I} & 0\end{pmatrix}$$

I get eigenvalues $\pm m=\pm E$ with $\hbar=c=1$.

Finding the eigenvectors I get the four possible $$\begin{pmatrix}1 \\ 0 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 1 \\ 0 \\ 1\end{pmatrix}, \begin{pmatrix}0 \\ 0 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix}1 \\ 1 \\ 1 \\ 1\end{pmatrix}.$$

I see in other texts they are using $$\gamma^0=\begin{pmatrix}\bf{I} & \\ & -\bf{I}\end{pmatrix}$$ and get possible eigenvectors $$\begin{pmatrix}1 \\ 0 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 1 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 0 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 0 \\ 0 \\ 1\end{pmatrix}.$$

Should I expect a different result from my definition of $\gamma^0$ for the eigenvectors?

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  • $\begingroup$ Hi! Your first set of eigenvalues is wrong: 0 is the null vector, and it is the trivial eigenvector of anything; 1 is the sum of the first two vectors. The right eigenvectors are something like: (1,0,1,0), (1,0,-1,0), (0,1,0,1) and (0,1,0,-1). $\endgroup$ – Dox Jan 14 at 8:38
  • $\begingroup$ thanks :) They are the nontrivial answers I got too but with two positive eigenvalues m=E and and two negative m=-E $\endgroup$ – manu fc Jan 14 at 12:15
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Gamma matrices are define as: $$ \{\gamma^\mu,\gamma^\nu \} =2\eta_{\mu\nu} $$

Matrices $S \gamma_\mu S^{-1}$ are also satisfying this equation. So we have different bases for gamma matrices. Now easy understand how your answers are related.

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