# 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.

• Have you written out the terms (in terms of gamma matrices etc.) to see whether $\bar{\psi^c}\psi$ and $\bar{\psi}\psi^c$ are actually different? Jan 19, 2020 at 17:03
• See the added comment above. Jan 19, 2020 at 17:09

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.

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: $$\sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix}.$$

• Please note that $\psi$ is a 4-component spinor. A product like $\psi^T\sigma_2\psi$ is meaningless unless you write in terms of the Weyl fields in $\psi$. Also, $\bar{\psi}\psi$ is Lorentz invariant. That is the whole point of not writing a term $\psi^\dagger\psi$ but $\bar{\psi}\psi$. Jan 19, 2020 at 17:18
• Oh snap! Well, it is a 2-component spinor for Majorana fermions as far as I know. Let me update the answer to make it more specific. Jan 19, 2020 at 17:43

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 $$-\frac{m}{2}(\overline{\psi}_R(\psi_R)^c+\overline{(\psi_R)^c}\psi_R.$$