Help me in my attempt of finding the Lagrangian for the Majorana field The Majorana field $\psi$ can be thought of as a reality condition $\psi=\psi^c$ (and $\overline{\psi}=\overline{\psi^c}$) on the Dirac field. So how does one write the Lagrangian for the Majorana field? 
The way I am going about it consist of first writing down the Dirac field Lagrangian $$\mathcal{L}=i\overline{\psi}\gamma^\mu\partial_\mu\psi-m\overline{\psi}\psi$$ and substitute $\psi=\psi^c$ and $\overline{\psi}=\overline{\psi^c}$ into it. This apparently changes the Lagrangian to $$\mathcal{L}^\prime=i\overline{\psi^c}\gamma^\mu\partial_\mu\psi^c-m\overline{\psi^c}\psi^c.$$ But since $\overline{\psi^c}\psi^c\sim \overline{\psi}\psi$ and $\overline{\psi^c}\gamma^\mu\psi^c\sim \overline{\psi}\gamma^\mu\psi$ (the notation '$\sim$' means equality apart from a sign), $\mathcal{L}^\prime=\mathcal{L}$. This means that Lagrangian of the Dirac field $\mathcal{L}$ and the Majorana field $\mathcal{L}^\prime$ are same. I think this is wrong. 
Next, I can try the following. Maybe changing both $\psi$ and $\overline{\psi}$ simultaneously to $\psi^c$ and $\overline{\psi^c}$ respectively was wrong. Only $\overline{\psi}$ to $\overline{\psi^c}$ in which case the correct Lagrangian is either $$\mathcal{L}^{\prime\prime}=i\overline{\psi^c}\gamma^\mu\partial_\mu\psi-m\overline{\psi^c}\psi$$ or $$\mathcal{L}^{\prime\prime\prime}=i\overline{\psi}\gamma^\mu\partial_\mu\psi^c-m\overline{\psi}\psi^c.$$
Response to the comment I have checked that $\overline{\psi^c}\psi=\overline{(\psi_L)^c}\psi_L+\overline{(\psi_R)^c}\psi_R$ and $\psi\overline{\psi^c}=\overline{\psi_L}(\psi_L)^c+\overline{\psi_R}(\psi_R)^c$ which means that they are different. In fact, the terms $\overline{\psi^c}\psi$ and $\psi\overline{\psi^c}$ are hermitian conjugates of each other. I think there is a problem of lack of hermiticity of here which came from the original Dirac Hamiltonian which was non-hermitian. See this.
Please help! Tell me which one is correct and which are wrong and why. 
 A: You can use
$$
S[\psi]= \frac 12 \int d^dx \,\psi^T {\mathcal C}({D\!\! /  }+m)\psi.
$$
where ${\mathcal C}$ is the charge conjugation matrix that gives
$$
{\mathcal C}\gamma^\mu {\mathcal C}^{-1}=- (\gamma^\mu)^T
$$
and in terms of which
$$
\psi^c= {\mathcal C}^{-1} \bar\psi^T={\mathcal C}^{-1}\gamma^0 \psi^*.
$$
The ${\mathcal C}$ matrix must be   antisymmetric for this action to be non-zero, and so we must be in $d=$ 2, 3, 4 (mod 8) dimensions. These are the same  dimensions in which $(\psi^c)^c=\psi$ making  $\psi^c=\psi$  consistent, and allowing    Majorana fermions to be possible.  
A: The mass term has to stay invariant under general Lorentz transformations. Since Majorana fermions are their own antiparticles the usual 4-component spinor $\psi = \left( \psi_R, \psi_L \right)$ reduces to $\psi_R$.  It is easy to check that $\psi_R^\dagger \psi_R$ is not invariant, since the infinitesimal transformation is:
$$
\delta \psi_R = \frac{1}{2} \left( i \theta_j + \beta_j \right) \sigma_j \psi_R.
$$

The usual mass term is: $\psi_R^T \sigma_2 \psi_R$, where $\sigma_2$ is a Pauli matrix:
$$
\sigma_2 = \begin{pmatrix}
0 & -i \\
i & 0 \\
\end{pmatrix}.
$$
A: The standard Lagrangian for a Majorana field is $$\mathcal{L}=\frac{1}{2}\overline{\psi}(i\gamma^\mu\partial_\mu-m)\psi.$$ Since $\psi^c=\psi$ for a Majorana field, we first re-express the Dirac Lagrangian  $$\mathcal{L}=\overline{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$ in terms of $\psi^c$. Remembering, $\overline{\psi^c}\psi^c=\overline{\psi}\psi$ and $\overline{\psi^c}\gamma^\mu\psi^c=\overline{\psi}\gamma^\mu\psi$ the Dirac Lagrangian is transformed into, $$\mathcal{L}=\overline{\psi^c}(i\gamma^\mu\partial_\mu-m)\psi^c.$$ The final task is to make the replacement $\psi^c=\psi$, to get $$\mathcal{L}=\frac{1}{2}\overline{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$ where the factor $1/2$ in front is written by hand in accordance with the answer here. The is the required Majorana Lagragian. If $\psi=\psi_L+(\psi_L)^c$, the mass term becomes $$-\frac{m}{2}(\overline{\psi}_L(\psi_L)^c+\overline{(\psi_L)^c}\psi_L,$$ and if $\psi=\psi_R+(\psi_R)^c$, the mass term becomes $$-\frac{m}{2}(\overline{\psi}_R(\psi_R)^c+\overline{(\psi_R)^c}\psi_R.$$
