I'm considering the following free Lagrangian (density):

$$\mathcal{L}= \bar{\Psi} \left(i \displaystyle{\not}{\partial}\mathbb{1}_{2}\right)\Psi$$

Where $\Psi$ is a doublet of two fermion fields $\Psi = \begin{pmatrix}\psi_{1}\\\psi_{2}\end{pmatrix}$.

This Lagrangian is obviously invariant under $U(2)$ transformations:

$\Psi \rightarrow U \Psi$, such that $U^{\dagger} U=\mathbb{1}_{2}$,

Using the Pauli matrices as the generators of the $U(2)=SU(2)\times U(1)$ group we can write:

$$U=e^{-i \left(\theta^{0}\mathbb{1}_2+\theta^{1}\sigma_{1}+\theta^{2}\sigma_{2}+\theta^{3}\sigma_{3}\right)}$$

In all the book I have read they stops here with the symmetries of the lagrangian, the problem is that in my opinion there is an additional symmetry:

$$\Psi \rightarrow e^{-i \gamma^{5} (q_{1}\mathbb{1}_{5}+q_{2}\sigma_{1})} \Psi = \begin{pmatrix}e^{-i \gamma^{5} (q_{1}+q_{2})}\psi_{1}\\e^{-i \gamma^{5} (q_{1}-q_{2})}\psi_{2}\end{pmatrix}$$

That corresponds to the chiral (axial) symmetries for each $\psi_{k}$, and so the symmetry group should be $[U(2)]\otimes [U_A(1)]_{q_1+q_2}\otimes [U_A(1)]_{q_1-q_2}$.

This symmetry can't be obtained from the one above because the coefficients $\theta^{i}$ are not numbers anymore but rather 4x4 matrices $(\gamma^{5})$.

Is this right or am I missing something?


1 Answer 1


The full symmetry group of the massless Lagrangian is actually: $$ SU(2)_L\times U_L(1)\times SU(2)_R\times U_R(1) $$ which covers the additional symmetries you noticed.

It can be rearranged as: $$ SU(2)_V\times U_V(1)\times SU(2)_A\times U_A(1) $$ where $$ U_A(1) $$ (related to your $q_1$ portion) is usually broken by the quantum anomaly, and $$ SU_A(2) $$ (related to your $q_2$ portion) might be dynamically violated by chiral symmetry breaking, depending on the circumstances (e.g. via QCD or NJL interactions).


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