# Bloch's theorem for a lattice with sublattices

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?

• 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 $\psi'_B(x) = \psi_B(x) - \epsilon(x)$). – QuantumApple Jan 10 at 13:18