SU(2)_V, SU(2)_A transformations

• A
Within my project thesis I stumbled over the term SU(2)_V, SU(2)_A transformations. Although I know U(1)_V, U(1)_A transformations from the left and right handed quarks( U(1)_V transformations transform left and right handed quarks the same way, while U(1)_A transformations transform them with a phase which just differs by a minus sign) I have no idea what vectorial /axial stands for in the SU(2) context.
Thank you again so much in advance,
Bob

Answers and Replies

vanhees71
Science Advisor
Gold Member
It's easier to explain with the corresponding Lie algebras, i.e., for "infinitesimal transformations". You start with the ##\mathrm{SU}(2)_L \times \mathrm{SU(2)_R## transformations, which act independently on the left and right-handed parts of the Dirac spinor
$$\psi_R=\frac{1+\gamma^5}{2}, \quad \psi_L=\frac{1-\gamma_5}{2}.$$

An arbitrary infinitesimal transformation acts like
$$\delta \psi_R = \delta \vec{\alpha} \cdot \vec{\tau} \psi_R, \quad \delta \psi_L=\delta \vec{\beta} \cdot \vec{\tau} \psi_L,$$
where ##\vec{\tau}=\vec{\sigma}/2## with the Pauli matrices acting in spinor space.
Now you can decompose these infinitesimal transformation in a different, i.e., in one that acts in the same way on both the left and the right handed component and one that acts differently. To that end we write
$$\delta \psi = \mathrm{i} \delta \vec{\alpha} \cdot \vec{\tau} \frac{1+\gamma^5}{2} \psi + \mathrm{i} \delta \vec{\beta}| \cdot \vec{\tau}_L \frac{1-\gamma^5}{2} \psi = \frac{\mathrm{i}}{2} (\delta \vec{\alpha}+\delta \vec{\beta}) \cdot \vec{\tau} \psi + \frac{\mathrm{i}}{2} (\delta \vec{\alpha}-\delta \vec{\beta}) \gamma^5 \psi.$$
Now define
$$\delta \vec{\alpha}_V=\frac{1}{2} (\delta \vec{\alpha}+\delta \vec{\beta}), \quad \delta \vec{\alpha}_A=\frac{1}{2} (\delta \vec{\alpha}+\delta \vec{\beta})$$
Then the transformation reads
$$\delta \psi = \mathrm{i} (\delta \vec{\alpha}_V \cdot \vec{\tau}_V + \delta \vec{\alpha}_A \cdot \vec{\tau}_A) \psi, \quad \vec{\tau}_V=\vec{\tau}, \quad \vec{\tau}_A=\vec{\tau} \gamma^5.$$
Note that the ##\vec{\tau}_V## build an su(2) subalgebra (and it's exponential thus an SU(2) subgroup) of the chiral group, while the ##\vec{\tau}_A## don't form a sub algebra/group but a rest class of the group.

If you have a chirally symmetric theory (of then necessarily massless fermions), the corresponding Noether currents are
$$\vec{j}_V^{\mu} = \bar{\psi} \gamma^{\mu} \vec{\tau} \psi, \quad j_A^{\mu} = \bar{\psi} \gamma^{\mu} \vec{\tau} \gamma^5 \psi,$$
which are vectors and axial vectors under space reflections, respectively and both are isovectors.