It is said that in 1+1 dimension, if we take $\gamma^0=i\sigma^2$ and $\gamma^1=\sigma^1$, then the two components of dirac spinor $\psi_L$(upper component) and $\psi_R$(lower component) decouple in the Dirac equation $(i\gamma^u\partial_u-m)\psi=0$ But, the Pauli matrices $\sigma^1$ and $\sigma^2$ are off-diagonal matrices. So the components of the spinor $\psi$ clearly mixes together in the equation. So what does it mean by the components decouple?

Also I think that it is wrong to choose $\gamma^0$ and $\gamma^1$ like above. In order to satisfy the Clifford algebra, I think it must be that $\gamma^0=\sigma^2$ and $\gamma^1=i\sigma^1$. Is this correct?


1 Answer 1


I think it's a fine basis, but there is no decoupling like you say. It's $\sigma^3$ whose eigenvalues distinguish the left and right movers in this basis, and while $\sigma^3$ commutes with $m$ it anticommutes with $ \gamma^\mu \partial_\mu$, so the evolution mixes them. If the mass was zero, they really would decouple, because you could multiply the equation by $\gamma^0$ and then the whole thing would commute with $\sigma^3$.

  • $\begingroup$ You mean that $i\sigma^2$ and $\sigma^1$ are fine basis? But they do not satisfy $\{\gamma^u, \gamma^v\}=-2\eta^{uv}$ where $\eta^{uv}=diag(-1,1,1,1)$ $\endgroup$
    – Keith
    May 2, 2018 at 18:06
  • 1
    $\begingroup$ They anticommute, and $(i \sigma_2)^2 = -1$ and $\sigma_1^2 = 1$. Looks like Cl(1,1) to me! This is the 1+1D Majorana basis. $\endgroup$ May 2, 2018 at 18:27

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