The step from 1 to 2 is: multiply by $\gamma^0$ and trace over the result.
2->3: $Tr((\gamma^0)^2)=Tr(1)=1$ on the left hand side and $2Tr(\gamma^0\gamma^\mu)P_\mu=8\eta^{0\mu}P_\mu=8P_0=8P^0$ on the right hand side.
Moreover the anticommutator gives $2Q_r^2$ so you can just strip off a factor of 2 on both sides.
3->4: $P^0$ is the total energy by definition, as the right hand side is in fact mulitplied by an unit operator (as $Q$ are operators) one has $P^0 1 = H$ giving the final result.
For some magic involving the gamma matrices see the wikipedia article http://en.wikipedia.org/wiki/Gamma_matrices
This post imported from StackExchange Physics at 2014-04-15 16:42 (UCT), posted by SE-user A friendly helper