Suppose we are in a $2d$ QFT for a Majorana fermion with action given by $$ S = \frac{1}{2}\int dx^0dx^1\; \overline{\Psi} \gamma^\mu\partial_\mu \Psi.$$ We are in Euclidean spacetime and the Clifford algebra is given by \begin{equation} \gamma^0 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad \qquad \gamma^1 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}. \end{equation}
It is straightforward to write the action under the coordinate transformation $z = x^0 + i x^1$ and $\overline{z} = x^0 - i x^1$ as $$S= \int dz d\overline{z} \;\bigg[\psi \overline{\partial} \psi + \overline{\psi} \partial \overline{\psi}\bigg]$$ where $\Psi = \begin{pmatrix} \psi \\ \overline{\psi} \end{pmatrix}$ and $\partial \equiv \partial_z$ and $\overline{\partial} \equiv \partial_{\overline{z}}$.
The EOM of this system is $\partial \overline{\psi} = 0 = \overline{\partial} \psi$.
However it seems like we cannot obtain this immediately from the second version of the action and applying the Euler-Lagrange equations, since we would obtain e.g. $$\partial \overline{\psi} - \partial \overline{\psi} = 0.$$ I think the explanation boils down to the fact that $\psi$ and $\overline{\psi}$ are not our fundamental fields, but $\Psi$ is.
Moreover, a second question question relates to the action in the first form: because we are looking at a Majorana fermion, we have that $\Psi^* = \Psi$. But if we insert this immediately into the action there, it seems that it would mess up the Euler-Lagrange equations there as well?
