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Bloch's theorem states the following: suppose we have a Hamiltonian

$$ H = \frac{p^2}{2m} + V(x) $$

where $V(x + a) = V(x)$, then the wavefunctions take the form $\psi_k(x) = e^{ikx} u_k(x)$ and $u_k(x+a) = u_k(x)$.

If we have a lattice that has sublattices, such as graphene which has two sublattices $A$ and $B$, then I have read here in Eq. (2) and here in Eq. (2.5) that, as the translation between the two sublattices is not a symmetry of the Hamiltonian, we have to write the Bloch wavefunctions as

$$ \psi_k(x) = \psi_k^A(x) + \psi_k^B(x)$$

where $\psi_k^A$ and $\psi_k^B$ are Bloch wavefunctions for each sublattice. I do not see why this is the case. For graphene, I would expect the Hamiltonian can be written as

$$ H = \frac{p^2}{2m} + V_A(x) + V_B(x) $$

where $V_A(x)$ and $V_B(x)$ are the periodic potentials of each sublattice. However, as each sublattice has the same periodicity, then I could just write $V(x) = V_A(x) + V_B(x)$ and just apply Bloch's theorem as stated above with a single solution $\psi_k(x)$. How do I know that I can split the wavefunction up into the Bloch functions of the individual sublattices? How does Bloch's theorem "see" the sublattices?

I could rephrase my question in terms of group theory. The Hilbert space $\mathcal{H}$ of the theory forms a representation of the discrete translations $T = \{ T(a) : a \in \Lambda \}$, where $\Lambda$ is the Bravais lattice. If I added two atoms to the unit cell, such as in graphene, the Bravais lattice has not changed i.e. the periodicty of the lattice has not changed, so the symmetry group of the lattice is still $T$, however the above results imply that the representation space has split up as $\mathcal{H} = \mathcal{H}_A \oplus \mathcal{H}_B$, where $\mathcal{H}_{A/B}$ are irreps of $T$ for each sublattice. How do I show this? How does Bloch's theorem know about the sublattices?

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There's actually no contradiction between what you say and what the two linked documents say. In fact, the two documents describe the solution of the wave equation in the form of a tight-binding solution, which is indeed a Bloch wavefunction, but it's constructed as the sum of wavefunctions with terms centered in the two sublattices.

Recall that the idea of the tight-binding model is to construct the Bloch wavefunctions as a superposition of atomic orbitals associated to the atoms of the crystal lattice. Since the graphene lattice is a lattice with a two-atom basis, the solution considers the superposition of the atomic orbitals from each of the two atoms. About this, see also N. W. Ashcroft and N. D. Mermin, Solid state physics, chapter 10, section General Remarks on the Tight Binding Method, remark 4.

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    $\begingroup$ As an additional remark, the decomposition of $\psi$ into $\psi_A + \psi_B$ can be seen as arbitrary, since $\psi_A$ and $\psi_B$ are not even orthogonal in general, so one could redefine each term as one wishes (say, $\psi'_A(x) = \psi_A(x) + \epsilon(x)$ and $\psi'_B(x) = \psi_B(x) - \epsilon(x)$). $\endgroup$ – QuantumApple Jan 10 at 13:18
  • $\begingroup$ Does this argument still apply if we choose the Wannier basis instead? The Wannier basis is sometimes used for tight binding models instead of atomic orbitals as it forms an orthonormal basis. Wannier states are the Fourier transform of the Bloch states, and there is exactly one Wannier state for each Bravais lattice vector. For this reason, there would be half the number of Wannier states as atomic sites in graphene (because each Bravais lattice site contains two atomic sites per unit cell), so are we unable to use Wannier states as a basis for a tight binding model? $\endgroup$ – Hermitian_hermit Jan 12 at 1:08
  • $\begingroup$ @Hermitian_hermit Depending on how you decompose the Bloch wavefunctions you can construct Wannier functions centered on both atoms of the basis. Or you can consider the two-atom basis as a molecule and construct the Wannier functions from the lattice of molecules. In any case, recall that the point of the tight-binding method is to construct an approximate solution of the Schroedinger equation, and one usually tries, at least as a first approximation, to construct a simple solution. $\endgroup$ – Massimo Ortolano Jan 12 at 8:31
  • $\begingroup$ In any case, as far as I know, if you want to use the Wannier functions with the tight-binding method, you should have already found a solution of the Schroedinger equation with another method. So, it becomes more complicated. $\endgroup$ – Massimo Ortolano Jan 12 at 8:32

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