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Taking your second equation an setting $t_1,t_2\to t$, we get $$E(k)=\pm t \sqrt2 \sqrt{1+\cos(ka)}=\pm 2 t \cos\left(\frac{ka}2\right)$$

where I used $1+\cos x = 2 \cos x/2$. The prefactor $2t$ comes because your Hamiltonian is degenerate (try to put $t_1,t_2\to t$ in the second Hamiltonian). The $k/2$ comes because $N$ is different by a factor of 2.

However notice that there is still a $\pm$. This happens because the equation used is not right, it was diagonalized as if I was handling a two-species unit cell, so we get a duplication of the bands.

Taking your second equation an setting $t_1,t_2\to t$, we get $$E(k)=\pm t \sqrt2 \sqrt{1+\cos(ka)}=\pm 2 t \cos\left(\frac{ka}2\right)$$

where I used $1+\cos x = 2 \cos x/2$. The $k/2$ comes because $N$ is different by a factor of 2.

However notice that there is still a $\pm$. This happens because the equation used is not right, it was diagonalized as if I was handling a two-species unit cell, so we get a duplication of the bands.

Taking your second equation an setting $t_1,t_2\to t$, we get $$E(k)=\pm t \sqrt2 \sqrt{1+\cos(ka)}=\pm 2 t \cos\left(\frac{ka}2\right)$$

where I used $1+\cos x = 2 \cos x/2$. The prefactor $2t$ comes because your Hamiltonian is degenerate (try to put $t_1,t_2\to t$ in the second Hamiltonian). The $k/2$ comes because $N$ is different by a factor of 2.

Taking your second equation an setting $t_1,t_2\to t$, we get $$E(k)=\pm t \sqrt2 \sqrt{1+\cos(ka)}=\pm 2 t \cos\left(\frac{ka}2\right)$$

where I used $1+\cos x = 2 \cos x/2$. The prefactor $2t$ comes because your Hamiltonian is degenerate (try to put $t_1,t_2\to t$ in the second Hamiltonian). The $k/2$ comes because $N$ is different by a factor of 2.

However notice that there is still a $\pm$. This happens because the equation used is not right, it was diagonalized as if I was handling a two-species unit cell, so we get a duplication of the bands.