Correlation Functions: How can I prove this simple equation The correlation functions of the Transverse Ising Model is beautifully explained in "Quantum Ising Phases and Transitions in Transverse Ising Models" Quantum Ising Phases and Transitions in Transverse Ising Models book.
Second Edition
"https://link.springer.com/book/10.1007/978-3-642-33039-1"
But I cannot deduce equation (2.A.30), page 42, even though it seems straightforward, where:
$$
{\left\langle {\psi_0 } \right|A_iA_j \left| {\psi_0 } \right\rangle }={\left\langle {\psi_0 } \right|(\delta_{ij}-c_j^{\dagger}c_i+c_i^{\dagger}c_j)\left| {\psi_0 } \right\rangle }=\delta_{ij},
$$
where $A_l=c_l^{\dagger}+c_l$ and $c_i^{\dagger}/c$ are the fermion creation/annihilation operators.
Does anyone know where this equation came from?
 A: It's quite straightforward. You need the anti-commutator $[c_i,c_j^\dagger]_+ = \delta_{ij}$ (anti because they're fermionic) and then
$$A_iA_j = (c_i^\dagger + c_i)(c_j^\dagger + c_j) = c_i^\dagger c_j^\dagger + c_i^\dagger c_j + c_i c_j^\dagger + c_i c_j$$
We want to calculate the matrix element with $|\psi_0\rangle$. We have the relation that the annihilation operator annihilates the ground state ket while the creation operator annihilates the ground state bra, that is $c_i |\psi_0\rangle = 0$ and $\langle\psi_0| c_i^\dagger = 0$. Therefore, we drop the terms that have two creation and annihilation operators. What remains is
$$\langle \psi_0 | A_i A_j | \psi_0 \rangle
= \langle \psi_0 | c_i^\dagger c_j  + c_i c_j^\dagger  | \psi_0 \rangle
= \langle \psi_0 | c_i^\dagger c_j  - c_j^\dagger c_i + [c_i,c_j^\dagger]_+  | \psi_0 \rangle.$$
We can insert the anti-commutator and realize that the other terms act with $c_i$ or $c_j$ on the ground state ket, so they vanish. That is
$$\langle \psi_0 | A_i A_j | \psi_0 \rangle
= \langle \psi_0 | \delta_{ij} + c_i^\dagger c_j  - c_j^\dagger c_i | \psi_0 \rangle
= \delta_{ij}\langle \psi_0 | \psi_0 \rangle
= \delta_{ij}.$$
We used the identity
$$AB = AB + BA - BA = [A,B]_+ - BA$$
along the line.
