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Local lorentz tranformations of fermion action

  1. Aug 18, 2010 #1
    The action for a fermion in curved spacetime is

    [itex]S = -\int d^4 x \sqrt{- \det(\eta^{ab} e_{a\mu}e_{b\nu})} \left[ i\overline{\psi} e^\mu_a \gamma^a D_\mu \psi + i m \overline{\psi}\psi \right][/itex]

    where [itex]g_{\mu\nu} = \eta^{ab} e_{a\mu} e_{b\nu}[/itex] and the derivative operator acting on fermions is given by

    [itex]D_\mu \psi = \partial_\mu \psi - \frac{i}{2} \omega_{ab\mu} S^{ab} \psi[/itex]

    where [itex]S^{ab} = -(i/4)[\gamma^a,\gamma^b][/itex].

    I am failing to show that [itex]D_\mu \psi[/itex] transforms covariantly under local Lorentz transformations.

    As far as I understand, the relation [itex]\omega_{ab\mu} = e_a^\nu \nabla_\mu e_{b\nu}[/itex] implies that the following transformation rules

    [itex]\psi(x) \mapsto \Lambda_{1/2}(x) \psi(x)[/itex]
    [itex]e_{a\mu}(x) \mapsto {\Lambda^a}_b(x) e_{b\mu}(x)[/itex]
    [itex]\omega_{ab\mu}(x) \mapsto {\Lambda^{\alpha}}_a(x) \omega_{\alpha\beta\mu}(x) {\Lambda^\beta}_b(x) + \eta_{\alpha\beta}{\Lambda^\alpha}_a(x)\partial_\mu {\Lambda^\beta}_b(x)[/itex].

    Plugging these transformations into [itex]D_\mu \psi[/itex] for an infinitesimal Lorentz transformation, for which
    [itex]{\Lambda^a}_b = \delta^a_b + \theta^a_b[/itex]
    [itex]\Lambda_{1/2} = 1 + (i/2)\theta_{ab} S^{ab}[/itex]
    I obtain

    [itex]D_\mu \psi \mapsto D_\mu' \psi' = (\partial_\mu \Lambda_{1/2})\psi + \Lambda_{1/2} \partial_\mu \psi - \frac{i}{2}\left[{\Lambda^{\alpha}}_a \omega_{\alpha\beta\mu} {\Lambda^\beta}_b + \eta_{\alpha\beta}{\Lambda^\alpha}_a \partial_\mu {\Lambda^\beta}_b \right]S^{ab}\Lambda_{1/2}\psi[/itex]

    In infinitesimal form:

    [itex]D_\mu' \psi' = \frac{i}{2} \partial_\mu \theta_{ab} S^{ab}\psi + (1 + i/2\, \theta_{ab}S^{ab})\partial_\mu \psi -\frac{i}{2}\left[(\delta^\alpha_a + \theta^\alpha_a)\omega_{\alpha\beta\mu}(\delta^\beta_b + \theta^\beta_b) + \eta_{\alpha\beta} (\delta^\alpha_a + \theta^\alpha_a ) \partial_\mu \theta^\beta_b \right] S^{ab}(1 + i/2\, \theta_{cd}S^{cd})\psi[/itex]

    Ignorning quadratic terms and using the anti-symmetry of [itex]\theta_{ab}[/itex]:

    [itex]D_\mu' \psi' = (1 + i/2\, \theta_{ab}S^{ab})\partial_\mu \psi -\frac{i}{2} \left( \omega_{ab\mu} S^{ab} + \frac{i}{2} \omega_{ab\mu} \theta_{cd} S^{ab} S^{cd}\right)\psi - \frac{i}{2}\left( \omega_{a\beta\mu} \theta^\beta_b + \omega_{\alpha b \mu} \theta^\alpha_a\right) S^{ab} \psi[/itex]

    The first two terms combine to give [itex](1+ i/2 \, \theta_{cd}S^{cd})(\partial_\mu - i/2\, \omega_{ab\mu}S^{ab})\psi = \Lambda_{1/2} D_\mu \psi[/itex] as required, but the last term does not have anything to cancel with. Does anyone have any suggestions about where I might be going wrong?
    Last edited: Aug 18, 2010
  2. jcsd
  3. Aug 18, 2010 #2
    I think the error lies in the fact that I should be considering the transformation of [itex]{\omega^a}_{b}[/itex] rather than [itex]\omega_{ab}[/itex]. The transformation rule for [itex]{\omega^a}_{b}[/itex] is

    [itex]{\omega^a}_{b\mu}(x) \mapsto {\Lambda^\beta}_b(x) {\omega^\alpha}_{\beta\mu}(x) {\Lambda_{\alpha}}^a(x) + [\partial_\mu {\Lambda^\alpha}_b(x)]{\Lambda_\alpha}^a(x)[/itex]

    expanding this into infinitesimal form gives extra minus signs because of [itex]{\Lambda^a}_b = \delta^a_b + \theta^a_b[/itex] and [itex]{\Lambda_a}^b = \delta^b_a - \theta^b_a[/itex]. These cause the last term in the expression for [itex]D_\mu' \psi'[/itex] to cancel.
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