Chiral symmetry in massless QCD

Question

The QCD Lagrangian for two flavors is:

$-\frac{1}{4} G\tilde{G}+i\bar{u}\displaystyle{\not} D u+i\bar{d} \displaystyle{\not} D d-m_u\bar{u} u-m_d\bar{d}d$

or alternaively

$-\frac{1}{4} G\tilde{G}+i\bar{u_L}\displaystyle{\not} D u_L+i\bar{d_R} \displaystyle{\not} D d_R-(m_u\bar{u_R} u_L+m_d\bar{d_R}d_L+ h.c.)$

where the bar means bar($q_{L,R}$), where $q=u,d$

In the massless case the Lagrangian should be invariant under $SU(2)_L\times SU(2)_R$. The invariance would mean $q_{L,R}\rightarrow U_{L,R}q_{L,R},$ where $U_{L,R}$ are $U(2)$ matrices.

I want to see this explicitly for the $\bar{u_L}\displaystyle{\not} D u_L$ term.

I have

$u_L\rightarrow Uu_L$

and

$\overline{u_L}\rightarrow \overline{U u_L}=\overline{\begin{pmatrix}U_{11}&U_{12}\\U_{21}&U_{22}\end{pmatrix}\begin{pmatrix}u_L\\d_L\end{pmatrix}}=\overline{\begin{pmatrix}U_{11}u_L+U_{12}d_L\\U_{21}u_L+U_{22}d_L\end{pmatrix}}=\begin{pmatrix}U^*_{11}u^\dagger_L+U^*_{12}d^\dagger_L&U^*_{21}u^\dagger_L+U^*_{22}d^\dagger_L\end{pmatrix}\gamma_0=\begin{pmatrix}u_L^\dagger&d_L^\dagger\end{pmatrix}\gamma_0\begin{pmatrix}U_{12}^*&U_{22}^*\\U_{11}^*&U_{21}^*\end{pmatrix}$

where I have used $\gamma_0=\begin{pmatrix}0&1\\1&0\end{pmatrix}$

Summarized this means

$\bar{u_L}\displaystyle{\not} D u_L\rightarrow \bar{u_L}\begin{pmatrix}U_{12}^*&U_{22}^*\\U_{11}^*&U_{21}^*\end{pmatrix}\displaystyle{\not} D \begin{pmatrix}U_{11}&U_{12}\\U_{21}&U_{22}\end{pmatrix} u_L$

I did a mistake, since the matrix is unitary and $\begin{pmatrix}U_{12}^*&U_{22}^*\\U_{11}^*&U_{21}^*\end{pmatrix}\begin{pmatrix}U_{11}&U_{12}\\U_{21}&U_{22}\end{pmatrix}\neq 1$

Can you spot my mistake?

Answer 1

You just have to be careful with the Dirac adjoint (the bar operation) $$q \rightarrow q'=Uq$$ By definition $\bar{q} = q^\dagger \gamma^0$, therefore the transformation for $\bar{q}$ is: $$\bar{q} \rightarrow (q')^\dagger\gamma^0 = q^\dagger U^\dagger \gamma^0,$$ since matrices from $SU(2)$ will be unitary, they satisfy $U^\dagger U = 1$. Recalling that these matrices do not act on the spinor indices, we can just move them through the covariant derivate: $$\bar{q}'\displaystyle{\not} Dq' = q^\dagger U^\dagger\gamma^0\displaystyle{\not}D\, U q = \bar{q}\displaystyle{\not}D\,q$$

Your $\bar{Uq}$ should look like: $$(u_L^* \;\; d_L^*)\gamma^0\begin{pmatrix} U_{11}^* & U_{21}^*\\ U_{12}^* & U_{22}^* \end{pmatrix}$$