# Majorana Fermions: Lagrangean and equations of motion

• LCSphysicist
In summary, the given equations can be obtained from the given Lagrangian by taking partial derivatives and applying the charge conjugation property. The process is still being evaluated and further steps are needed to fully understand it.
LCSphysicist
Homework Statement
Show that the equations (below) can be obtained from the followong lagrangian
Relevant Equations
.
$$i \gamma^{\mu} \partial_{\mu} \psi = m \psi_c \\ i \gamma^{\mu} \partial_{\mu} \psi_c = m \psi$$

Where ##\psi_c = C \gamma^0 \psi^*##

Show that the above equations can be obtained from the followong lagrangian

$$L = \overline{\psi} i \gamma^{\mu} \partial_{\mu} \psi - \frac{1}{2} m \left ( \psi^T C \psi + \overline{\psi} C \overline{\psi}^T \right )$$

Where ##C## is charge conjugation

\begin{align*} L = \overline{\psi} i \gamma^{\mu} \partial_{\mu} \psi - \frac{1}{2} m \left ( \psi^T C \psi + \overline{\psi} C \overline{\psi}^T \right ) = \overline{\psi}_a i \gamma^{\mu} \partial_{\mu} \psi^a - \frac{1}{2} m \left ( \psi^a C_{ab} \psi^b + \overline{\psi}^a C_{ab} \overline{\psi}^b \right ) \end{align*}

\begin{align*}
\frac{\partial L}{\partial \psi^r} = -\frac{1}{2} m \left ( C_{ra} \psi^a + \psi^a C_{ar} \right ) = - \frac{1}{2} m \left ( C_{ra} \psi^a - \psi^a C_{ra} \right )
\end{align*}

\begin{align*}
\frac{\partial L}{\partial \overline{\psi}^r} = i \gamma^{\mu} \partial_{\mu} \psi_r -\frac{1}{2} m \left ( C_{ra} \overline{\psi}^a + \overline{\psi}^a C_{ar} \right ) = i \gamma^{\mu} \partial_{\mu} \psi_r -\frac{1}{2} m \left ( C_{ra} \overline{\psi}^a - \overline{\psi}^a C_{ra} \right )
\end{align*}

\begin{align*}
\frac{\partial}{\partial x^{\mu}} \frac{\partial L}{\partial \partial_{\mu} \psi^r} = \frac{\partial}{\partial x^{\mu}} \left ( \overline{\psi_r} i \gamma^{\mu}\right) = \partial_{\mu} \overline{\psi}_r i \gamma^{\mu}
\end{align*}

\begin{align*}
\frac{\partial}{\partial x^{\mu}} \frac{\partial L}{\partial \partial_{\mu} \overline{\psi}^r} = 0
\end{align*}

\begin{align*}
-\frac{1}{2} m \left ( C_{ra} \psi^a - \psi^a C_{ra} \right ) - i \partial_{\mu} \overline{\psi_r} \gamma^{\mu} = 0 \\
i \gamma^{\mu} \partial_{\mu} \psi_r -\frac{1}{2} m \left ( C_{ra} \overline{\psi}^a - \overline{\psi}^a C_{ra} \right ) = 0
\end{align*}

But i am not sure how to proceed!

Can someone give me a tip? I am still trying to evaluate it, but i can't found out what i have to do.

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