Majorana and Weyl fermions

Last updated Dec 29, 2024

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We found the Dirac mass term $$\mathcal{L}_\text{dirac mass}=m(\psi^\dagger_L\psi_R+\psi^\dagger_R\psi_L),$$ to be Lorentz invariant, while terms like $\psi^\dagger_R\psi$ are not.

It turns out there is another Lorentz invariant mass term

Majorana mass term

The Majorana mass term for Spin $\frac{1}{2}$ particles is $$\mathcal{L}_\text{Maj}=m\psi^T_R\sigma_2\psi_R.$$

Lorentz invariance of Majorana mass term

The transformation of the Spinors $\psi_R$ and $\psi^\dagger_R$ was given as $$\delta\psi_R = \frac{1}{2}(i\Theta_j+\beta_j)\sigma_j\psi_R,\quad \delta\psi^\dagger_R=\frac{1}{2}(-i\Theta_j+\beta_j)\psi^\dagger_R\sigma_j$$ We can now calculate the transformation property of the majorana mass term $$\delta(\psi^T_R\sigma_2\psi_R) = \delta(\psi^T_R\sigma_2)\psi_R + \psi^T_R\sigma_2\delta\psi_R$$ using $$\sigma^T_j\sigma_2 = -\sigma_2\sigma_j$$ to be $$\delta(\psi^T_R\sigma_2\psi_R)=-\frac{1}{2}(i\Theta_j+\beta_j)\psi^T_R\sigma_2\sigma_j\psi_R + \frac{1}{2}(i\Theta_j+\beta_j)\psi^T_R\sigma_2\sigma_j\psi_R=0$$

Explicitly plugging in $\sigma_2$ one can write $$\psi^T_R\sigma_2\psi_R=-i(\psi_1\psi_2-\psi_2\psi_1).$$

However this term is only non-trivial if $$\psi_1\psi_2\neq\psi_2\psi_1,$$ meaning the fermion components must not commute!

One finds that the fermion components must anti-commute, meaning they are Grassmann numbers.

We can also write the mass term in terms of a four component spinor $$\psi=\begin{pmatrix}\psi_L\\ i\sigma_2\psi^*_L\end{pmatrix}$$ as $$\frac{m}{2}\bar{\psi}\psi=i\frac{m}{2}(\psi^\dagger_L\sigma_2\psi^*_L- \psi^T_L\sigma_2\psi_L)$$

These Majorana spinors are invariant under Charge conjugation

One writes $$\psi=\psi_L, \quad \tilde{\psi}=\psi_R,$$ And the indices are included as $\psi^\alpha$ and $\psi_{\dot{\beta}}$ so that a Dirac Spinor can be written as $$\psi=\begin{pmatrix}\psi^{\alpha}\\ \tilde{\psi}_{\dot{\beta}}\end{pmatrix}$$ Using this notation the Majorana mass term can be written as $$\mathcal{L}_\text{maj}=\psi^T_R\sigma_2\psi_R=i\psi_\alpha\epsilon^{\alpha\beta}\psi_\beta,$$ with the $\epsilon$ tensor $\epsilon^{\alpha\beta}=-i\sigma^{\alpha\beta}_2$.

The $\epsilon$ tensor can raise and lower indices for Weyl spinors, like $g_{\mu\nu}$ can do for four vectors. $$\psi\chi = \psi_\alpha\chi^\alpha = \psi_\alpha\epsilon^{\alpha\beta}\chi_\beta=-\chi_\beta\epsilon^{\alpha\beta}\psi_\alpha=\chi_\beta\epsilon^{\beta\alpha}\psi_\alpha=\chi_\psi$$ \tilde{\psi}\tilde{\chi}=\tilde{\psi}^{\dot{\alpha}}\tilde{\chi}_\dot{\alpha}=-\tilde{\chi}_\dot{\alpha}\tilde{\psi}^{\dot{\alpha}}=\tilde{\chi}^\dot{\alpha}\tilde{\psi}_\dot{\alpha}=\tilde{\chi}\tilde{\psi}