Real versus complex Hamiltonian While a Hamiltonian must be a Hermitian matrix, it can either be real or complex. Is there a  significance for having a real Hamiltonian? Does it have any additional physical symmetries? 
For example, a two-band tight-binding model has Hamiltonian of the form 
$$\sum_k \left( \epsilon_1 c_{k1}^\dagger c_{k1}+\epsilon_2 c_{k2}^\dagger c_{k2}+t c_{k1}^\dagger c_{k2} + t^* c_{k2}^\dagger c_{k1}\right).$$
Does a real $t$ tell us anything physical about the system?
 A: The realness of the interorbital coupling $t$ can indeed tell us something about the symmetries of the system. Here, it entails at least one symmetry: spinless time-reversal symmetry.
Consider the Hamiltonian of non-interacting spinless particles that live on a periodic chain with two orbitals in each unit cell as in OP's example. In this case, the Hamiltonian can be block-diagonalized by using the periodicity of our system:
\begin{equation}
H = \sum_k \begin{pmatrix} c^\dagger_{k1} & c^\dagger_{k2} \end{pmatrix} H(k) \begin{pmatrix} c_{k1} \\ c_{k2} \end{pmatrix} 
\end{equation}
where $H(k)$ is a $2\times 2$ block, labelled by the momentum $k$, called the Bloch Hamiltonian. Now assume that the system has time-reversal symmetry. This means that $[H,T]=0$ with $T$ the time-reversal operator. For spinless particles it is given by $T=K$ with $K$ complex conjugation with respect to the position representation (see e.g. Sakurai). Since $T c_{k,i} = c_{-k,i}$, we find
\begin{equation}
H = THT^{-1} \Rightarrow H(-k) = TH(k)T^{-1} = H(k)^*.
\end{equation}
In OP's example, a periodic chain of cells with no tunnelling between them is considered. Essentially an infinite amount of isolated atoms. The Bloch Hamiltonian is
\begin{equation}
H(k) = \begin{pmatrix} \epsilon_1 & t \\ t^* & \epsilon_2 \end{pmatrix}
\end{equation}
so that the total Hamiltonian $H$ is not time-reversal symmetric if $t$ is not real.
Another example which illustrates the effect of spinless time-reversal symmetry in a periodic system is given by graphene. Here, spinless $T$ and inversion protect the nodes in the band spectrum (Dirac points) locally since combined they forbid any $\sigma_z$ term which would open the energy gap. The two Dirac points can only be destroyed if they merge in a time-reversal-invariant point of the Brillouin zone.
Generally, if time-reversal symmetry is present, the energy spectrum plotted as a function of $k$ (the so-called energy bands) is symmetric with respect to $k$: $E_i(-k) = E_j(k)$. However, the converse is not true since spatial symmetries such as inversion can also enforce a symmetric spectrum.
Spinful time reversal
In the case of a half-integer spin system, it can be shown that the time-reversed state $T\left| \psi \right>$ is orthogonal to $\left| \psi \right>$ as a result of $T^2 = -1$ (Kramers' theorem). This means that the energy bands have to become degenerate in time-reversal-invariant points $k = -k + G$ where $G$ is a reciprocal lattice vector. A system with two bands is therefore always a metal. The simplest model for an insulator requires four bands.
