One-band 1D tight-binding model: how to find the two-particle eigenstates?

Consider the simple hopping model in second quantization,

$\hat{H} = -J \sum_{i,j=1}^\infty \left(\hat{c}_i \hat{c}^\dagger_j + h.c.\right)$

where $J$ is real and $\hat{c}_i$ are annihilation fermionic operators for an infinite 1D periodic lattice.

As well known, a single-excitation eigenstate of this Hamiltonian can be found in the form

$|\psi^{(1)}\rangle = \hat{d}^\dagger_k |0\rangle$

where $\hat{d}^\dagger_k \propto \sum_{i=1}^\infty \hat{c}^\dagger_i e^{i k x_i}$, $x_i$ is the position of the i-th site and $k$ is a wavevector in the first Brillouin zone.

Is there a standard recipe to find the two-excitation eigenstates? That is, for states of the form

$|\psi^{(2)}\rangle = \sum_{i,j<i}^\infty A_{i,j} \hat{c}^\dagger_i \hat{c}^\dagger_j |0\rangle$

is there a known Ansatz to find analytically the matrix $A_{i,j}$ such that $|\psi^{(2)}\rangle$ is an eigenstate of $\hat{H}$?

• Hint: write the Hamiltonian in terms of the $\hat d_k$ modes – Ruben Verresen May 2 '18 at 22:48
• Thanks for the hint! It's indeed much easier than what I was expecting! – Michele Cotrufo May 2 '18 at 22:58