- #1

LCSphysicist

- 646

- 161

- 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!