Denoting by $\gamma^a$ the Minkowski space gamma matrices with respect to the Lorentz tetrad $\{e^a\}$, and covariant derivative $D_a$, then the gammas are covariantly constant.

Start with the massless Dirac equation $$ \gamma^{b}D_{b}\Psi = 0$$

Act again with the Dirac operator $$\gamma^{a}D_{a}\gamma^{b}D_{b}\Psi=0 $$ So, since $D$ annihilates $\gamma$ $$\gamma^{a}\gamma^{b}D_{a}D_{b}\Psi = 0 $$ so $$\frac{1}{2}\{\gamma^{a},\gamma^{b}\}D_{a}D_{b}\Psi + \frac{1}{2}\gamma^{a}\gamma^{b}[D_a,D_b]\Psi = 0 \ \ (1) $$ But $$\{\gamma^{a},\gamma^{b}\}=2\eta^{ab} $$ and $$ [D_a,D_b]\Psi = {\mathcal{R}_{ab}}\Psi $$ Where ${\mathcal{R}}_{ab}$ is the spin-curvature (antisymmetric in a and b). ${\mathcal{R}}_{ab}$ satisfies the identity $$ -\gamma^b{\mathcal{R}}_{ab} = {\mathcal{R}}_{ab}\gamma^b = \frac{1}{2}\gamma^b R_{ab}$$ where $R_{ab}$ is the Ricci tensor (in the Lorentz tetrad). so (1) becomes $$ [D^aD_a+\frac{1}{4}\gamma^a\gamma^bR_{ab}]\Psi = 0 $$ i.e. $$ [D^aD_a-\frac{1}{4}R]\Psi = 0 $$

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user twistor59