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I'm having a little difficulty with understanding the normalisation process of the $\gamma$-matrices.

In Thomson Modern Particle Physics 2013, the normalisation of the $\gamma$-matrices are quoted as:

$$ (\gamma^{\mu})^{\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0} $$ Where $\mu=0,1,2,3$ or sometimes just $\mu=0, k$ where, obviously $k=1,2,3$. I have attempted to start this given example, but I'm not sure on the next steps. Thus far, I have:

$$ (\gamma^{0})^{\dagger}=\gamma^{0} $$ $$ (\gamma^{k})^{\dagger}=-\gamma^{k} $$ I also know that $$ (\gamma^{0})^{2}=I\,\,\mathrm{and}\,\,(\gamma^{k})^{2}=-I $$ I'm just not sure how to put this together. If anyone could give a quick run through or some prods in the right direction that would be excellent.

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    $\begingroup$ Hint: $\{\gamma^\mu,\gamma^\nu\}=??\quad$ (take $\mu=k$ and $\nu=0$) $\endgroup$ – AccidentalFourierTransform Feb 1 '16 at 14:31
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While studying the $\gamma$-matrices, I also faced the same question. Here maybe an solution about this.

First, the convention is the same: $$(\gamma^{0})^{\dagger}=\gamma^{0}$$ $$(\gamma^{k})^{\dagger}=-\gamma^{k}$$ $$(\gamma^{0})^{2}=I\,\,\mathrm{and}\,\,(\gamma^{k})^{2}=-I$$ Additionally, here used the eqution, $$\{\gamma^{\mu},\gamma^{\nu}\}=2g^{\mu\nu}I$$ Especially, for the $\gamma^{0}$ and $\gamma^{k}$, $$\{\gamma^{0},\gamma^{k}\}=0$$ Then we can see, $$(\gamma^{0} \gamma^{k})^{\dagger}=(\gamma^{k})^{\dagger} (\gamma^{0})^{\dagger}$$ and $$\gamma^{0} \gamma^{k}=-\gamma^{k} \gamma^{0}$$ so $$(-\gamma^{k} \gamma^{0})\dagger=(\gamma^{k})^{\dagger} (\gamma^{0})^{\dagger}$$ $$-\gamma^{0} (\gamma^{k})^{\dagger}=-\gamma^{k} \gamma^{0}$$ $$(\gamma^{k})^{\dagger}=\gamma^{0}\gamma^{k} \gamma^{0}$$ also we known that $$(\gamma^{0})^{\dagger}=\gamma^{0}\gamma^{0} \gamma^{0}$$ we can combined these together to see $$(\gamma^{\mu})^{\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0}$$ Noticing that here we have already selected a specific representation.

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