Consider the Dirac Lagrangian,
$$L=\overline{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi$$
and take the Dirac spinor chiral decomposition with $\psi_{L}=\frac{1}{2}\left(1-\gamma^{5}\right)\psi$ and $\psi_{R}=\frac{1}{2}\left(1+\gamma^{5}\right)\psi$. How can I obtain the Lagrangian in the form,
$$ L=\overline{\psi}_{L}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{L}+\overline{\psi}_{R}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{R}-m\left(\psi_{L}\psi_{R}+\psi_{R}\psi_{L}\right)\;?
$$
Plugging $\psi=\psi_{L}+\psi_{R}$ in the first expression doesn't seem to work, as I get crossed terms (I guess). Could you advise?
Attempt:
I mean,
$$
\begin{array}{cl}
L & =\overline{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi\\
& =\left(\overline{\psi}_{L}+\overline{\psi}_{R}\right)\left(i\gamma^{\mu}\partial_{\mu}-m\right)\left(\psi_{L}+\psi_{R}\right)\\
& =\overline{\psi}_{L}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{L}+\overline{\psi}_{R}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{R}+\overline{\psi}_{L}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{R}+\overline{\psi}_{R}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{L}-m\left(\overline{\psi}_{L}\psi_{R}+\overline{\psi}_{R}\psi_{L}+\overline{\psi}_{L}\psi_{L}+\overline{\psi}_{R}\psi_{R}\right)\\
& =???\\
& =\overline{\psi}_{L}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{L}+\overline{\psi}_{R}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{R}-m\left(\overline{\psi}_{L}\psi_{R}+\overline{\psi}_{R}\psi_{L}\right)
\end{array}
$$
But why do we have $\overline{\psi}_{L}\psi_{L}+\overline{\psi}_{R}\psi_{R}=0$
and $\overline{\psi}_{L}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{R}+\overline{\psi}_{R}\left(i\gamma^{\mu}\partial_{\mu}\right)\psi_{L}=0$?
For example, let's just look at the mass term for now. We have
$$
\overline{\psi}_{L}=\psi_{L}^{\dagger}\gamma^{0}=\frac{1}{2}\psi^{\dagger}\left(1-\left(\gamma^{5}\right)^{\dagger}\right)\gamma^{0}
$$
so,
$$
\begin{array}{cl}
\overline{\psi}_{L}\psi_{L} & =\frac{1}{2}\psi^{\dagger}\left(1-\left(\gamma^{5}\right)^{\dagger}\right)\gamma^{0}\frac{1}{2}\left(1-\gamma^{5}\right)\psi\\
& =\frac{1}{4}\psi^{\dagger}\left(1-\gamma^{0}\gamma^{5}\gamma^{0}\right)\gamma^{0}\left(1-\gamma^{5}\right)\psi\\
& =\frac{1}{4}\psi^{\dagger}\left(\gamma^{0}-\gamma^{0}\gamma^{5}\gamma^{0}\gamma^{0}\right)\left(1-\gamma^{5}\right)\psi\\
& =\frac{1}{4}\psi^{\dagger}\left(\gamma^{0}-\gamma^{0}\gamma^{5}1\right)\left(1-\gamma^{5}\right)\psi\\
& =\frac{1}{4}\psi^{\dagger}\gamma^{0}\left(1-\gamma^{5}\right)\left(1-\gamma^{5}\right)\psi\\
& =\frac{1}{4}\overline{\psi}\left(1-\gamma^{5}-\gamma^{5}-\gamma^{5}\left(-\gamma^{5}\right)\right)\psi\\
& =\frac{1}{4}\overline{\psi}\left(1-2\gamma^{5}+1\right)\psi\\
& =\frac{1}{2}\overline{\psi}\left(1-\gamma^{5}\right)\psi
\end{array}
$$
and similarly,
$$
\overline{\psi}_{R}\psi_{R}=\frac{1}{2}\overline{\psi}\left(1+\gamma^{5}\right)\psi
$$
hence,
$$
\begin{array}{cl}
\overline{\psi}_{L}\psi_{L}+\overline{\psi}_{R}\psi_{R} & =\frac{1}{2}\overline{\psi}\left(1-\gamma^{5}\right)\psi+\frac{1}{2}\overline{\psi}\left(1+\gamma^{5}\right)\psi\\
& =\overline{\psi}-\frac{1}{2}\overline{\psi}\gamma^{5}\psi+\frac{1}{2}\overline{\psi}\gamma^{5}\psi\\
& =\overline{\psi}\psi.
\end{array}
$$
I didn't get zero. What am I missing?
Best Answer
The $\psi_L$ spinor transforms in the $(\frac12,0)$ representation of $SL(2,\mathbb C)$, and $\psi_R$ under the conjugate representation $(0,\frac12)$, so we can construct a Dirac spinor as the column,
$$\psi = \begin{pmatrix} \psi_L\\ \psi_R \end{pmatrix}$$
and it suffices to treat $\psi$ as a column vector of two elements, with $\bar \psi = \psi^\dagger \gamma^0$. You simply need to apply basic linear algebra to work out the Lagrangian now. Recall that,
$$\gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu\\ \tilde \sigma^\mu & 0 \end{pmatrix}$$
where $\sigma^\mu = (\mathbb I_2, \vec \sigma)$ and $\tilde \sigma^\mu = (\mathbb I_2,-\vec \sigma)$ where $\vec \sigma$ is the vector whose components are the Pauli matrices. Carrying out this calculation is a matter of multiplying columns, matrices and rows.