# Equivalence between Dirac and Majorana action in CFT

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.