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

  1. Apr 11, 2016 #1
    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
     
  2. jcsd
  3. Apr 11, 2016 #2

    vanhees71

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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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