# Why does the free fermionic 2-point correlation matrix $C_{ij}=\langle c^{\dagger}_i c_j\rangle$ have eigenvalues equal to either $0$ or $1$?

Consider a system of free fermions with Hamiltonian $$H = \sum_{ij} t_{ij}c^{\dagger}_ic_j\quad \longrightarrow \quad H = \sum_k E_k d^{\dagger}_kd_k,$$ with $$t_{ij}$$ hermitian. An eigenstate $$|\psi \rangle$$ of $$H$$ is given by acting on $$|0\rangle$$ with $$d^{\dagger}$$s as usual: $$|\psi^{N_p}\rangle = \prod_{a \in N_p} d_a | 0 \rangle \quad \text{where} \quad d_a = \sum_i\phi^a_ic_i,$$ where the set $$\{\phi^a\}$$ are the eigenvectors of $$t_{ij}$$. Its 2-point correlator matrix $$C$$ given by $$C_{ij} = \langle \psi | c^{\dagger}_i c_j |\psi\rangle \equiv \langle c^{\dagger}_i c_j \rangle_{\psi}= \sum_k\overline{\phi_i^{k}}\phi^k_j\langle d^{\dagger}_kd_k\rangle_{\psi}= \sum_k\overline{\phi_i^{k}}\phi^k_j \delta_{k\in \psi}= \sum_{k\in \psi}\overline{\phi_i^{k}}\phi^k_j.$$ Why are the eigenvalues of $$C$$ all either $$0$$ or $$1$$?

When I try to find them I run into: $$\sum_j C_{ij} \xi^n_j = \Xi^n \xi^n_i \implies \sum_j \sum_{k\in \psi}\overline{\phi_i^{k}}\phi^k_j \xi^n_j = \Xi^n \xi^n_i,\tag{1}$$ and can’t go any further. I know that $$\{\phi^n\}$$ satisfy the usual properties of the eigenvectors of a hermitian matrix, namely $$\sum_{k}\overline{\phi_i^{k}}\phi^k_j =\delta_{ij} \quad \text{and} \quad \sum_{i}\overline{\phi_i^{k}}\phi^p_i = \delta^{kp},$$ but I cant use the first one in (1) as $$k$$ doesn’t run through all its values, i.e. it depends on $$\psi$$.

You are almost there. You just need to use that the $$\phi_i^k$$ form a unitary matrix $$U\equiv U_{ij}$$.
Then, your formula $$C_{ij} = \sum_k\overline{\phi_i^{k}}\phi^k_j\langle d^{\dagger}_kd_k\rangle_{\psi}$$ reads $$C = UD U^\dagger$$ (with $$D$$ a real diagonal matrix with entries $$0$$ and $$1$$), and thus $$C$$ and $$D$$ have the same spectrum (the spectrum is invariant under conjugation with $$U$$).
• Thanks again for your help. (For future learners: defining $U_{ij} = \phi_i^j$ would imply that if $U$ is unitary then $U^{\dagger}U=1$ and in terms of $\phi$’s, $(U^{\dagger})_{ij}U_{jk}=\overline{\phi^j_i}\phi^j_k = \delta_{ik}$ which indeed works out.) Apr 23 at 9:51