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The matrices present in the Dirac equation must have the following properties:

  1. $\{a^j,a^k\}_{ab} = 2\delta^{jk}\delta_{ab}$
  2. $\{a^j,\beta\} = 0$
  3. $(\beta^2)_{ab} = \delta_{ab}$

How can one show from this that the trace of $\beta$ must be $0$?

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up vote 3 down vote accepted

Yup. For example, because $(\alpha^1)^2=1$ from the first equation, $${\rm Tr}(\beta) = {\rm Tr}(\beta\alpha^1\alpha^1) = {\rm Tr}(\alpha^1\beta\alpha^1)=-{\rm Tr}(\beta\alpha^1\alpha^1) = -{\rm Tr}(\beta)$$ In the first step, I inserted a unit matrix. In the second step, I used cyclicity of the trace. In the third step, I used the equation 2 (that $\alpha^i$ and $\beta$ anticommute). In the final step, I recombined the square of $\alpha^1$ into the unit matrix again. Because the trace of $\beta$ is equal to minus itself, it must be zero.

I didn't use the identity 3.

Well, I didn't fully need the identity 1, either. It's enough that $\alpha^1$ anticommutes with $\beta$ and is invertible. If we use such an anticommuting $\alpha^1$ to conjugate $\beta$, we revert all the eigenvalues of $\beta$ and because of the cyclicity of the trace, we may erase $\alpha^1$ times its inverse again.

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Thanks, well answered. –  Mew Sep 20 '12 at 14:39
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