Answer by Adam for Wick's theorem and transverse field Ising model

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 find $$I=\frac1N\sum_{q,q'}e^{-i q R_i +i q' R_j}(u_q+i v_q)(u_{q'}-iv_{q'})\langle \psi_0 | \eta_q \eta_{q'}^\dagger|\psi_0\rangle ,$$which is simplified, using $u_q^2+v_q^2=1$, into $$ I = \frac1N \sum_q e^{-i q (R_i-R_j)}=\delta_{ij}.$$

This is indeed the identity we wanted to prove.