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

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SUMMARY

The discussion focuses on SU(2)_V and SU(2)_A transformations in the context of quantum field theory, specifically regarding their application to Dirac spinors. It explains that SU(2)_V transformations act identically on both left and right-handed components, while SU(2)_A transformations differ by a phase factor. The infinitesimal transformations are expressed using the Pauli matrices and the decomposition into vectorial and axial components is clarified. The Noether currents associated with these transformations are also defined, highlighting their roles as vectors and axial vectors under space reflections.

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  • Understanding of SU(2) and U(1) groups in particle physics
  • Familiarity with Dirac spinors and chirality
  • Knowledge of Lie algebras and infinitesimal transformations
  • Basic concepts of Noether's theorem and current conservation
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bob2
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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
 
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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.
 

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