I am trying to show that

\[\bar{\psi}\gamma^{\mu}\psi =\bar{\psi}_L\gamma^{\mu}\psi_L + \bar{\psi}_R\gamma^{\mu}\psi_R\]

For $\mu = 0$, this is using $\gamma^0 \gamma^0 = 1$ rather straightforward.

However, for the case $\mu \neq 0$ I obtain an annoying sign error (using the Weyl representation for the gamma matrices:

\[\bar{\psi}\gamma^1\psi = \left( \begin{array}{cc} \psi_L^*,\psi_R^*\\ \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right) \left( \begin{array}{cc} 0 & -\sigma_1 \\ \sigma_1 & 0 \\ \end{array} \right) \left( \begin{array}{c} \psi_L \\ \psi_R \\ \end{array} \right)\]

which when multiplying out stuff gives me the wrong result

\[\bar{\psi}\gamma^{1}\psi =\bar{\psi}_L\gamma^{1}\psi_L - \bar{\psi}_R\gamma^{1}\psi_R\]

For the other two components, the same would happen. How can I get rid of this annoying minus sign in front of the second term? Or is my way to show this identity not a good one in the first place?