Energy spectrum of a tight-binding model

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?

-
sorry, how did you do the calculation with 'each unit cell containing two sites'? – nervxxx Jan 17 '13 at 2:13
@nervxxx To do so, we split the system into the two sublattices consisting of the even and odd sites. Then, expressing the operators in each of the sublattices in momentum space, we can write the Hamiltonian as a $2\times 2$ matrix coupling the two sublattices. Finally, diagonalizing the matrix yields the energy spectrum. – leongz Jan 17 '13 at 21:54

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 $k$.
What would experiments say about the excitation spectrum then? Would the energy at zero momentum be $2t$ or $\pm 2t$? – leongz Jan 17 '13 at 19:31
It depends what you mean by momentum. If it means that the phase changes by $\exp ik$ every time you shift to the next atom, $x\to x+1$, then it's just $2t$. If you mean by momentum $k$ the number $k$ from $\exp 2ik$ that one gets by moving by two lattice sites, $x\to x+2$, then there are two possible states obeying this condition and their energies are $\pm 2t$. But I have already told you so. – Luboš Motl Jan 18 '13 at 16:53