This seems like it should be simple but somehow I do not see how.

The Majorana Lagrangian can be written in terms of a left handed Weyl spinor $\psi_L$ as

$$ \mathcal{L}_M= i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L - \tfrac{m}{2} \psi^T_L \epsilon \psi_L + \tfrac{m}{2} \psi^\dagger_L \epsilon \overline{\psi}_L \hspace{1cm}. $$ Meanwhile the Dirac Lagrangian can be written in terms of a Left handed Weyl spinor $\psi_L$ and a right handed Weyl spinor $\psi_R$ as $$ \mathcal{L}_D = i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L + i \psi_R^\dagger \sigma^\mu \partial_\mu \psi_R - m (\psi_L^\dagger \psi_R + \psi_R^\dagger \psi_L). $$

Here I am using the convention $\sigma^\mu = (I, \sigma^i)$, $\bar\sigma^\mu = (I, -\sigma^i)$, and $\epsilon = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$.

The reality condition for the Majorana spinor is just $$ \psi_R = - \epsilon \overline{\psi}_L. $$ Plugging in the above $\psi_R$ into $\mathcal{L}_D$, and using the identity $- \epsilon \sigma^\mu \epsilon = (\bar\sigma^\mu)^*$, I get

$$ \mathcal{L}_D = i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L + i \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L + m \psi_L^\dagger \epsilon \overline{\psi}_L - m \psi_L^T \epsilon \psi_L. $$

It seems to me that I would have $\mathcal{L}_D = 2 \mathcal{L}_M$ if only I could prove $$ i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L \stackrel{?}{=} i \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L. $$

However, I do not see why the above equation has to be true. An added layer of complication is that $\psi_L$ is really a vector of Grassmann variables that satisfy $$ \{ \psi_L^a, \psi_L^b \}_+ = 0 \hspace{1 cm} \{ \overline{\psi^a}_L, \overline{\psi^b}_L \}_+ = 0 \hspace{1 cm} \{ \psi_L^a, \overline{\psi^b}_L \}_+ = \delta^{ab} $$ for $a,b = 1,2$.

What are the correct manipulations to show that $\mathcal{L}_D = 2 \mathcal{L}_M$?

  • 1
    $\begingroup$ Hey pal, your Dirac mass $- m (\psi_L^\dagger \psi_R + \psi_R^\dagger \psi_L)$ does not sound right. It should read $- m (\bar{\psi}_L \psi_R + \bar{\psi}_R\psi_L)$. $\endgroup$
    – MadMax
    Commented Sep 17, 2018 at 14:35
  • $\begingroup$ In a chiral basis the missing $\gamma^0$ in $\bar{\psi} = \psi^\dagger \gamma^0 $ toggles right/left spinors. $\endgroup$
    – MadMax
    Commented Sep 17, 2018 at 14:49

1 Answer 1


Ah. Figured it out. I want to show that

$$ i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L \stackrel{?}{=} i \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L. $$

Let's manipulate the right hand side. Because it is a single number (in a linear algebra sense) it is equal to its own transpose. However, because it $\psi_L$ is really a 2 component column vector of anti commuting Grassmann numbers, when we take the transpose we also have to negate it when we implicitly flip the order of multiplication. So

\begin{align*} i \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L &= \big(i \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L \big)^T \\ &= -i \partial_\mu \overline{\psi}_L^\dagger (\bar{\sigma}^\mu)^\dagger \psi_L. \end{align*} Next, note that $(\bar{\sigma}^\mu)^\dagger = \bar{\sigma}^\mu$ because all the Pauli matrices are self adjoint. Finally, integrate by parts, picking up an extra minus sign. This gives us our desired equation. \begin{align*} i \psi_L^T (\bar{\sigma}^\mu)^* \partial_\mu \overline{\psi}_L &= -i \partial_\mu \psi_L^\dagger (\bar{\sigma}^\mu)^\dagger \psi_L \\ &= -i \partial_\mu \psi_L^\dagger \bar{\sigma}^\mu \psi_L \\ &= i \psi_L^\dagger \bar{\sigma}^\mu \partial_\mu \psi_L \\ \end{align*}

Which is just what I wanted.

Note that I've been using $\overline{\psi}_L$ to mean what most people mean by $\psi_L^*$.


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