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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 $\psi$.

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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 $U$).

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  • $\begingroup$ 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.) $\endgroup$ Commented Apr 23, 2021 at 9:51
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    $\begingroup$ @FriendlyLagrangian Comment on your edit: The point is not really that U diagonalized C - the point is that any two matrices related by conjugation with U have the same spectrum. $\endgroup$ Commented Apr 23, 2021 at 10:14

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