I've recently been watching this lecture series on Condensed Matter. The part I'm currently on covers band theory for the tight-binding model in a few different scenarios. We covered two different models in one-dimension. The first had a Hamiltonian given by $$H = -t \sum_i c_i^\dagger c_{i+1} + c_{i+1}^\dagger c_i,$$ where the sum runs over the $N$ sites in the one-dimensional chain with periodic boundary conditions and $c_i$ is a creation operator for an orbital at site $i$. We solved this model and got to the eigenenergies $$E(k) = - 2 t \cos(k a),$$ where $a$ is the lattice spacing and $k = \frac{2\pi n}{N a}$, with $n = 0, 1, \ldots, N-1$.

The second model had two sites per unit cell. We have $N$ unit cells and have the Hamiltonian $$H = - \sum_i t_1(c_{i,A}^\dagger c_{i,B} + c_{i,B}^\dagger c_{i,A}) + t_2(c_{i-1,B}^\dagger c_{i,A} + c_{i+1,A}^\dagger c_{i,B}),$$ where $A$ and $B$ label the two different sites per unit cell. The eigenenergies we found this time were $$E(k) = \pm \sqrt{t_1^2 + t_2^2 + 2 t_1 t_2 \cos(ka)},$$ where $a$ is the size of one unit cell and $k = \frac{2\pi n}{N a}$, $n = 0,1, \ldots, N_1$ (to the best of my knowledge).

What I find weird is that I would expect that the limit $t_1 \to t_2$ in the second model should reduce it to the first model with $t = t_1 = t_2$, albeit with $2N$ sites instead of $N$. However, that is not what happens, since the second model always has two energy bands, while the first has only one. My question is: why doesn't the second model reduce to the first one in the limit $t_1 \to t_2$?

I've recently been watching this lecture series on Condensed Matter. The part I'm currently on covers band theory for the tight-binding model in a few different scenarios. We covered two different models in one-dimension. The first had a Hamiltonian given by $$H = -t \sum_i c_i^\dagger c_{i+1} + c_{i+1}^\dagger c_i,$$ where the sum runs over the $N$ sites in the one-dimensional chain with periodic boundary conditions and $c_i$ is a creation operator for an orbital at site $i$. We solved this model and got to the single-particle eigenenergies $$E(k) = - 2 t \cos(k a),$$ where $a$ is the lattice spacing and $k = \frac{2\pi n}{N a}$, with $n = 0, 1, \ldots, N-1$.

The second model had two sites per unit cell. We have $N$ unit cells and have the Hamiltonian $$H = - \sum_i t_1(c_{i,A}^\dagger c_{i,B} + c_{i,B}^\dagger c_{i,A}) + t_2(c_{i-1,B}^\dagger c_{i,A} + c_{i+1,A}^\dagger c_{i,B}),$$ where $A$ and $B$ label the two different sites per unit cell. The single-particle eigenenergies we found this time were $$E(k) = \pm \sqrt{t_1^2 + t_2^2 + 2 t_1 t_2 \cos(ka)},$$ where $a$ is the size of one unit cell and $k = \frac{2\pi n}{N a}$, $n = 0,1, \ldots, N_1$ (to the best of my knowledge).

What I find weird is that I would expect that the limit $t_1 \to t_2$ in the second model should reduce it to the first model with $t = t_1 = t_2$, albeit with $2N$ sites instead of $N$. However, that is not what happens, since the second model always has two energy bands, while the first has only one. My question is: why doesn't the second model reduce to the first one in the limit $t_1 \to t_2$?