Plugging Majorana Spinor into Dirac Lagrangian does not give Majorana Lagrangian? This seems like it should be simple but somehow I do not see how.
The Majorana Lagrangian can be written in terms of a left handed Weyl spinor $\psi_L$ as
$$
\mathcal{L}_M= i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L - \tfrac{m}{2} \psi^T_L \epsilon \psi_L + \tfrac{m}{2} \psi^\dagger_L \epsilon \overline{\psi}_L \hspace{1cm}.
$$
Meanwhile the Dirac Lagrangian can be written in terms of a Left handed Weyl spinor $\psi_L$ and a right handed Weyl spinor $\psi_R$ as
$$
    \mathcal{L}_D = i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L + i  \psi_R^\dagger \sigma^\mu \partial_\mu \psi_R - m (\psi_L^\dagger \psi_R + \psi_R^\dagger \psi_L).
$$
Here I am using the convention $\sigma^\mu = (I, \sigma^i)$, $\bar\sigma^\mu = (I, -\sigma^i)$, and $\epsilon = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$.
The reality condition for the Majorana spinor is just
$$
\psi_R = - \epsilon \overline{\psi}_L.
$$
Plugging in the above $\psi_R$ into $\mathcal{L}_D$, and using the identity $- \epsilon \sigma^\mu \epsilon = (\bar\sigma^\mu)^*$, I get
$$
\mathcal{L}_D = i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L + i  \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L + m \psi_L^\dagger \epsilon \overline{\psi}_L - m \psi_L^T \epsilon \psi_L.
$$
It seems to me that I would have $\mathcal{L}_D = 2 \mathcal{L}_M$ if only I could prove
$$
i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L \stackrel{?}{=} i  \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L.
$$
However, I do not see why the above equation has to be true. An added layer of complication is that $\psi_L$ is really a vector of Grassmann variables that satisfy
$$
\{ \psi_L^a, \psi_L^b \}_+ = 0 \hspace{1 cm} \{ \overline{\psi^a}_L, \overline{\psi^b}_L \}_+ = 0 \hspace{1 cm} \{ \psi_L^a, \overline{\psi^b}_L \}_+ = \delta^{ab}
$$
for $a,b = 1,2$.
What are the correct manipulations to show that $\mathcal{L}_D = 2 \mathcal{L}_M$?
 A: Ah. Figured it out. I want to show that
$$
i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L \stackrel{?}{=} i  \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L.
$$
Let's manipulate the right hand side. Because it is a single number (in a linear algebra sense) it is equal to its own transpose. However, because it $\psi_L$ is really a 2 component column vector of anti commuting Grassmann numbers, when we take the transpose we also have to negate it when we implicitly flip the order of multiplication. So
\begin{align*}
i  \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L &=  \big(i  \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L \big)^T \\
&= -i \partial_\mu \overline{\psi}_L^\dagger (\bar{\sigma}^\mu)^\dagger \psi_L.
\end{align*}
Next, note that $(\bar{\sigma}^\mu)^\dagger = \bar{\sigma}^\mu$ because all the Pauli matrices are self adjoint.  Finally, integrate by parts, picking up an extra minus sign. This gives us our desired equation.
\begin{align*}
i  \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L 
&= -i \partial_\mu \psi_L^\dagger (\bar{\sigma}^\mu)^\dagger \psi_L \\
&= -i \partial_\mu \psi_L^\dagger \bar{\sigma}^\mu \psi_L \\
&= i \psi_L^\dagger \bar{\sigma}^\mu  \partial_\mu \psi_L \\
\end{align*}
Which is just what I wanted.
Note that I've been using $\overline{\psi}_L$ to mean what most people mean by $\psi_L^*$.
