I am trying to understand calculation of correlation function in the ground state of the Transverse Field Ising model, from the following book, which is freely available: http://link.springer.com/book/10.1007/978-3-642-33039-1
The calculations can be found in Chapter 2 of the book. I shall follow the notation from this book and try to describe most of the steps.
Set up:
Consider a spin chain with $N$ sites. The hamiltonian for transverse field Ising model is (Page $17$ of the book)$$H= -\sum_i S^z_i - \lambda\sum_i S^x_i\otimes S^x_{i+1}.$$Now, the book follows the well known process of using Jordan Wigner transformation to map Pauli operators ($S^x_i,S^y_i,S^z_i$) to fermionic operators $c_i, c^{\dagger}_i$. After this, a fourier transform is performed (equation $2.2.7$), defining new operators $c_q, c^{\dagger}_q$, which are the Fourier transforms of original $c_i,c^{\dagger}_i$ and now the hamiltonian looks like (equation $2.2.8$):$$H=N-2\sum_q(1+\lambda \cos(q))c^{\dagger}_qc_q - \lambda\sum_q(e^{-iq}c^{\dagger}_qc^{\dagger}_{-q}-e^{iq}c_qc_{-q}).$$
Then a Bogoliubov transformation is performed, which is the source of my confusion. They define operators $\eta_q,\eta^{\dagger}_q$ in the following way (equation $2.2.11$):$$\eta_q = u_qc_q + iv_qc^{\dagger}_{-q} , \quad \eta^{\dagger}_q = iv_qc_q + u_qc^{\dagger}_{-q}.$$
This transformation diagonalizes the hamiltonian $H$, with appropriate choice of $u_q,v_q$ and one infers that the ground state $|\psi_0\rangle$ is the state annihilated by all $\eta_q$: $\eta_q|\psi_0\rangle = 0.$
Main question:
Now in appendix $2.A.3$ (Page $42$), correlation function $\langle \psi_0|S^x_iS^x_{i+n}|\psi_0\rangle$ is computed. This is a complicated expression when written in terms of operators $c_i, c^{\dagger}_i$ and for this Wick's theorem is used. But, as can be seen in equation $2.A.30$, calculation is done as if $|\psi_0\rangle$ is annihilated by $c_i$ themselves. Whereas, we saw that $|\psi_0\rangle$ is actually annihilated by $\eta_q$, which is a mixture of both $c_i$ and $c^{\dagger}_i$.
In fact, all the further calculations appear to be done in same manner, assuming that $|\psi_0\rangle$ is annihilated by $c_i$. I traced equation $2.A.32$ to the following reference: http://pcteserver.mi.infn.it/~molinari/NOTES/Wick.pdf
In this reference, wick's theorem has been stated as Theorem $IV.4$ (Page $4$). Equation $2.A.32$ (of the book) looks very similar to corollary $IV.6$ (of the reference). But the corollary is true only if $|\psi_0\rangle$ has $0$ expectation value with all normal-ordered operators.
So how can $|\psi_0 \rangle$ have $0$ expectation value with normal-ordered form of a product of $c_i,c^{\dagger}_i$? Shouldn't this be true only with $\eta_q,\eta^{\dagger}_q$? Is there a underlying principle here, that expectation values do not change under Bogoliubov transformation?