I am working through the exercises in the ‘Spinors’ section of Stewart’s book ‘Advanced General Relativity’. Problem 2.5.7 asks to derive the spinor formula $$ \square \Psi_{ABCD} = 6\Psi_{EF(AB} \Psi_{CD)}{}^{EF} $$ by differentiating (using $\nabla_{AA’}$) the expression $$ \nabla_{F}{}^{A’} \Psi^F{}_{BCD}=0. $$ Here, $\Psi_{ABCD}$ is the totally symmetric spinor known as the gravitational spinor or the Weyl conformal spinor, and $\square = \nabla_{AA’} \nabla^{AA’}$ is the wave operator.
It is far from obvious to me the strategy for deriving this formula. I fear it may be necessary to use the definition given for the Weyl spinor: $\Psi_{ABCD} = \epsilon_{\hat B A} \square_{C(D} \epsilon^{\hat B}{}_{B)}$ (where indices with a $\hat{}$ represent components rather than abstract indices, $\epsilon_{\hat A}{}^A = (o^A, \iota^A)$, and $\square_{CD} = \nabla_{C’(C}\nabla_{D)}{}^{C’}$). However, I find I pretty quickly get lost in the calculations trying to do this, and I can’t see why it should work out.
I suppose I am really just looking for some sort of indication or hint as to how this problem should be done, and whether it is supposed to be difficult and tedious or if there is some trick I am missing.