- #1
LCSphysicist
- 644
- 162
- 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!
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!
