# Dirac equation plane with solution

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?

• 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).
– Dox
Jan 14 '20 at 8:38
• thanks :) They are the nontrivial answers I got too but with two positive eigenvalues m=E and and two negative m=-E Jan 14 '20 at 12:15

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.