Consider the one-dimensional tight-binding Hamiltonian $$\mathcal{H}=t\sum_m\left(a^\dagger_m a_{m+1}+a^\dagger_{m+1} a_{m}\right).$$ With the lattice constant set to 1, the energy spectrum is given by $$\epsilon(k)=2t\cos{k},$$ for $k\in[-\pi,\pi]$ in the first Brillouin zone. However, if we repeat the calculation with each unit cell containing two sites, we would obtain an energy spectrum of $$\epsilon(k)=\pm2t\cos{k}$$ instead, with the size of the first Brillouin zone reduced by half. Since the energy spectrum of a system can be determined unambiguously by experiments such as neutron scattering, it seems that we have a contradiction. What am I missing here?
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What you're missing is that the two treatments are the same. In the second one, the Brillouin zone is 2 times smaller but $$ \cos(k+\pi) =-\cos(k)$$ so choosing the opposite sign in front of the energy is equivalent to shifting to the now-forbidden half of the Brillouin zone. When you identify the states correctly, there are of course the same states in both cases. Note that changing $k$ to $k+\pi$ is equivalent to the alternating sign flips at the lattice sites, $$ a_m\to (-1)^m a_m $$ in the position representation. Most typically, the original large Brillouin zone may be divided to two – one for which $a_{m+1}$ is closer to $+a_m$ and one-half for which it is closer to $-a_m$ and it's sensible to redefine the sites by the alternating signs. At any rate, the doubling of the states coming from the $\pm$ sign exactly compensates the halving of the Brillouin zone for $k$. |
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