In Mussardo's Statistical field theory Chapter 12, section 12.3 about the conformal field theory of a free fermion field he talks about the complex fermion field (Dirac field) $$ \Psi(z,\bar{z}) = \left( \begin{array}{c} \chi(z, \bar{z})\\ \bar{\chi}(z, \bar{z})\\ \end{array} \right) = \frac{1}{\sqrt{2}} \left( \begin{array}{c} \psi_1 +i\, \psi_2\\ \bar{\psi_1} +i\, \bar{\psi_2}\\ \end{array} \right) \tag{1} $$

Where he calls $\psi_1$ and $\psi_2$ real Majorana fermions. $z$ and $\bar{z}$ are complex conjugates of each other, while the bar over the $\chi$ denote that it is anti-analytic, that is it depends only on $\bar{z}$ and viceversa for the $\chi$ without the bar, as one can see from the equation of motion derived from this euclidean action

$$ S= \frac{\lambda}{2\pi} \int d^2x \, \bar{\Psi}\, \gamma^\mu \partial_\mu \Psi \tag{2} $$ where $\bar{\Psi}= {\Psi}^\dagger \gamma^0$. The euclidean gamma matrices satisfy $\{\gamma^\mu,\gamma^\nu \} = \delta^{\mu \nu}$ and we use this representation: $\gamma^0=\sigma_1$ and $\gamma^1=\sigma_2$ where the $\sigma_i$ are the usual Pauli matrices, in such a way that the dirac operator is: $$ \gamma^\mu \partial_\mu = \gamma^0 \partial_0 +\gamma^1 \partial_1 = \begin{pmatrix} 0 & \partial_0 -i \, \partial_1 \\ \partial_0 +i \, \partial_1 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 2\frac{\partial}{\partial{z}} \\ 2\frac{\partial}{\partial \bar{z}} & 0 \end{pmatrix}= \begin{pmatrix} 0 & 2{\partial} \\ 2\bar{\partial} & 0 \end{pmatrix} $$

Then he says he will denote with ${\psi}$ and $\bar{\psi}$ the analytic and anti-analytic components of the Majorana fermion, and taking $\lambda=1$ in $(2)$ their action is

$$ S= \frac{1}{2\pi} \int d^2x \,\, \Big( \psi \, \bar{\partial}\, \psi \, + \bar{\psi} \, {\partial} \, \bar{\psi} \Big)\tag{3} $$

Now, since so far he only used the name Majorana fermion for the $\psi\,$, I was a bit confused in this passage: I was unsure if the $\psi$ in $(3)$ where the $\chi$ and $\bar{\chi}$ in $(1)$ or, if the were, for example, $\psi_1$ and $\bar{\psi_1}$in $(1)$. So I tried to derive $(3)$ from $(2)$.

After boring computations I obtained this lagrangian: $$ \mathcal{L}=2(\, \chi^\dagger \, \bar{\partial} \, \chi \, + \, \bar{\chi}^\dagger \, \partial \, \bar{\chi} \, ) \tag{4} $$

which, beside a factor of two, if $\chi$ and $\bar{\chi}$ are real, reproduces $(3)$

Still a factor of two is missing and i thought I could get rid of it by considering the second equality in $(1)$, putting it into $(3)$ to obtain this: $$ \mathcal{L}=\, \psi_1 \, \bar{\partial} \, \psi_1 + \, i \, \psi_1 \, \bar{\partial} \, \psi_2 \, - \, i \, \psi_2 \, \bar{\partial} \, \psi_1 \, + \, \psi_2 \, \bar{\partial} \, \psi_2 \, + \, \bar{\psi_1} \, \partial \, \bar{\psi_1} \, + \, i \, \bar{\psi_1} \, \partial \, \bar{\psi_2} \, - \, i \, \bar{\psi_2} \, \partial \, \bar{\psi_1} \, + \, \bar{\psi_2} \, \partial \, \bar{\psi_2} \, \tag{5} $$

Now the factor of two is gone, and if we consider only one fermion, setting $\psi_2 = \bar{\psi_2} = 0$ we exactly obtain the action $(3)$.

I have a few doubts about all of this:

  1. Is this the correct way to obtain this action?
  2. If it is, why do we treat (in Mussardo's book) just one Majorana fermion on its own and not both $\psi_1$ and $\psi_2$? I don't think we are allowed to do that if we want to talk about the complex fermion $(1)$ and the dynamics given by $(2)$ since the equivalent lagrangian is the one given in $(5)$, which contain mixed terms, and not the one in $(3)$
  3. Could it be that we're only interested in the stress-energy tensor, and since that is given by the sum of the stress energy tensors for $\psi_1$ and $\psi_2$ considered on their own, that is having the action $(3)$?

I'm very confused so I hope it's an understandable question.


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