2
$\begingroup$

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 $\mathcal{L}_D = 2 \mathcal{L}_M$?

$\endgroup$
  • $\begingroup$ Good question. I'd like to know too... $\endgroup$ – Mozibur Ullah Sep 16 '18 at 7:40
  • 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 Sep 17 '18 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 Sep 17 '18 at 14:49
0
$\begingroup$

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 left 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 $\psi_L^*$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.