Generalized tight-binding model - how to solve it? Consider a generalized 1D tight-binding model (without spin) with the following Hamiltonian
\begin{equation}
\mathcal{H}\left(\{\chi_{r,r+1}\}\right)=\sum_{r}(\chi_{r,r+1}c^\dagger_rc_{r+1}+h.c.)~,
\end{equation}
and suppose that the complex number $\chi_{r,r+1}$ is a variational parameter. The $r$ index runs over a lattice sites of a 1D chain (you can think of this as a periodic boundary condition problem with $N$ lattice sites). 
Q: Given an arbitrary set of $\chi_{r,r+1}$, is there a systematic procedure to find the ground-state of the model (numerically or analytically) ? I'm not looking for a detailed answer with the calculation here. Perhaps just a good reference or starting point. For instance, in the case where $\chi_{r,r+1}=\chi$ $\forall r$, the system can be diagonalized by a simple lattice Fourier transform.
 A: OK, it is probably a bad idea to exchange in comments. Let me expand what I said in the comments.
If my understanding is correct, the OP wants to know, as the first step toward solving the whole problem, the ground state energy of the many-body Hamiltonian $\mathcal{H}$ defined by
$$
\mathcal{H} = \sum_{r,s}H_{rs}c^\dagger_r c_{s},
$$
for a given set of parameters $\{ H_{rs}\}$. Here $c^\dagger_{r}$ and $c_{r}$ are standard fermion creation and annihilation operators. The subscripts $r,s$ run over all lattice sites from 1 to $N$. The Hermiticity requires that
$$
H_{rs} = H^\ast_{sr}.
$$
In other words, the $N\times N$ matrix $H$, whose $(r,s)$ entry is defined to be $H_{rs}$, must be Hermitian. In some literature, $H$ is known as the "first-quantized Hamiltonian". Note that the above $\mathcal{H}$ takes a slightly more general form than the one described by OP.
The first step is to diagonalize $\mathcal{H}$. To this end, we introduce a new set of fermion operators:
$$
c_{r} = \sum_{m}V_{rm}f_{m};\quad{}c^\dagger_{r}=\sum_{m}V^\ast_{rm}f^\dagger_{m}.
$$
We demand that the new fermion operators obey the standard fermion algebra. It can be seen that this is amount to demand
$$
\sum_{m}V_{rm}V^\ast_{sm}=\delta_{rs},
$$
or equivalently $VV^\dagger=1_N$, i.e. $V$ is a unitary $N\times N$ matrix.
Substituting the above in, we find $\mathcal{H}$ written in terms of new fermion operators,
$$
\mathcal{H} = \sum_{r,s,m,n}V^\ast_{rm}V_{sn}H_{rs}f^\dagger_m f_n = \sum_{m,n}(V^\dagger HV)_{mn}f^\dagger_m f_n.
$$
Since $H$ is Hermitian, we can always find a unitary $V$ so that $H$ is diagonalized:
$$
V^\dagger HV = \Lambda.
$$
Here $\Lambda = \textrm{diag}(\lambda_1,\lambda_2\cdots,\lambda_N)$. $\lambda_i\in\mathbb{R}$ are eigenvalues of $H$. Thus,
$$
\mathcal{H} = \sum_{m}\lambda_m f^\dagger_m f_m.
$$
This is the desired diagonalized form of $\mathcal{H}$.
The second step is to find the ground state energy of $\mathcal{H}$. We see that all eigenstates of $\mathcal{H}$ are labeld by the occupation numbers $f^\dagger_mf_m$. It is easy to see that the ground state of $\mathcal{H}$ is constructed by filling up all modes with negative energy. In other words, in the ground state,
$$
f^\dagger_m f_m=\left\{\begin{array}{cc}
1 & \lambda_m<0\\
0 & \lambda_m>0
\end{array}
\right. .
$$
There will be degeneracy if some $\lambda_m = 0$. Then, the ground state energy is
$$
E_{G}=\sum_{m,\lambda_m<0}\lambda_m.
$$
