Fierz identity with Weyl spinors The following Fierz relation does not seem so obvious to me:
\begin{equation}
\bar{\psi}_1 \gamma^\mu (1+\gamma_5)\psi_2 \bar{\psi}_3 \gamma_\mu (1-\gamma_5) \psi_4 = -2 \bar{\psi}_1 (1-\gamma_5)  \psi_4 \bar{\psi}_3 (1+\gamma_5) \psi_2.
\end{equation}
As a first step I would have tried  to do something like 
\begin{equation}
\bar{\psi}_1 \gamma^\mu (1+\gamma_5)\psi_2 \bar{\psi}_3 \gamma_\mu (1-\gamma_5) \psi_4 = \bar{\psi}_1 (1-\gamma_5) \gamma^\mu \psi_2 \bar{\psi}_3 (1+\gamma_5)\gamma_\mu \psi_4
\end{equation}
but this does not help to get rid of the $\gamma_\mu$s. Am I going the wrong way?
 A: Follow these guidelines:
1) Use the notations $\sigma^\mu= (1, \vec \sigma)$, $\tilde \sigma^\mu= (1, -\vec \sigma)$. You will get expressions with lower indices $\mu$, by using the Minkowski metrics $\eta_{\mu\nu} = (1, -1, -1, -1)$. 
2) Use the chiral / Weyl basis, and express, for the LHS and RHS expressions of the equality, the matrices between the spinors, in function of $\sigma^\mu, \tilde \sigma^\mu$
3) Divide the 4-spinors $\psi_i$, in 2 2-spinors $(\phi_i,\chi_i) $, and express the LHS/RHS expressions, as product of a quartic product of components of the 2-spinors multiply by  some coefficients $L_{ijkl}, R_{ijkl}$
4) Finally, you will have to demonstrate some relation between $\sigma^\mu_{(ij)} \tilde \sigma_{\mu (kl)}$, and $\delta_{il} \delta_{kj}$. To do this, show that, for every hermitian  $2 *2$ matrix $X$, you have : 
$X = \frac{1}{2} Tr(X \tilde \sigma_\mu) \sigma^\mu$ 
Express this matrix relation, for each element of the matrices, with the form : 
$X_{ij} A^{jikl}=0$, and deduce that  $A^{jikl}=0$ 
A: There is a straightforward way to get this essentially by inspection, reading off the V×V row of the general Fierz table, (the worked example with an overall minus sign for the anticommuting fermions at hand, here):
\begin{equation}
\bar{\psi}_1 \gamma^\mu  \psi_2 \bar{\psi}_3 \gamma_\mu   \psi_4 = -  \bar{\psi}_1    \psi_4 \bar{\psi}_3   \psi_2+  \bar{\psi}_1 \gamma_5   \psi_4 \bar{\psi}_3   \gamma_5 \psi_2 \\    +  \frac{1}{2} \bar{\psi}_1     \gamma^\mu \psi_4 \bar{\psi}_3    \gamma_\mu \psi_2    +  \frac{1}{2} \bar{\psi}_1   \gamma_5   \gamma^\mu \psi_4 \bar{\psi}_3   \gamma_5 \gamma_\mu \psi_2   .
\end{equation}
This holds for arbitrary spinors, so now simply pick $\psi_2$ to be right handed, which enforces $\psi_1$  to be right handed, too, as you already showed in your start,
and likewise $\psi_4$ to be left handed, which dictates    $\psi_3$ to be left-handed, too, analogously,
$$
\psi_2 \to 
\psi_2 = \frac{1}{2}(1+\gamma_5)\psi_2, \qquad \Longrightarrow (1-\gamma_5)\psi_2=0 ,
$$
etc, mutatis mutandis.
So, now, simply note, oh the joy!, that both the V×V and A×A terms of the second line of the Fierz identity simply disappear by chiral projection, as vector and axial couplings cannot connect left-handed spinors to right handed ones, and vice versa. 
Moreover, the pseudoscalar term on the top line turns into its scalar brother with a minus sign, since $\gamma_5 \to -1$ for the left projector--but not for the right one!
We are done. The above Fierz identity simply collapses to
\begin{equation}
\bar{\psi}_1 \gamma^\mu  \psi_{2~R} ~~\bar{\psi}_{3} \gamma_\mu  \psi_{4~L} = -2 \bar{\psi}_1    \psi_{4~L} ~~\bar{\psi}_3   \psi_{2~R},
\end{equation}
which is just your desideratum identity divided by 4.
With some practice, this is evident by inspection of such chiral expressions.
