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Eq. 2.A.30 is a somewhat non-trivial identity for the ground-state $|\psi_0\rangle$ which only uses the ground-state property that $\eta_q|\psi_0\rangle =0$. (Of course, as the OP has noted, $c_i|\psi_0\rangle \neq 0$.) What we need to show is that $$I=\langle \psi_0 | (c_j+c_j^\dagger)(c_i+c_i^\dagger)|\psi_0\rangle =\delta_{ij}.$$ Using Eq. 2.A.37a, we ...



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