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I'm having a problem following a derivation of Bloch's theorem, looking at a one dimensional lattice with $N$ nodes and spacing a, we impose periodic boundary conditions, meaning that the wave-function of a single electron $\psi(x)$ can be decomposed in the following modes:



$$q = \frac{2\pi m}{N a}\; m\in \mathbb{Z}.$$

Otherwise it would be multivalued.

Schrodinger's equation gives:$$(\epsilon - \frac{\hbar^2q^2}{2m})\phi(q)=\sum_{k \in B}\tilde{V}(k)\phi(q+k)$$ where $\epsilon$ is the energy eigenvalue of $\psi$, $B$ is the reciprocal Bravais lattice ($k=\frac{2\pi n}{a}$ with $n\in \mathbb{Z}$), $\tilde{V}$ is the Fourier transform of the potential. So the $\phi$ modes separated by a vector belonging to the reciprocal Bravais lattice are coupled but I don't see how it implies that only $q$ vectors that are in this class of equivalence (whose difference belongs to $B$) should appear in the first sum (the expansion of $\psi$ in modes).

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up vote 2 down vote accepted

This is not true--- you can make superpositions of degenerate energy eigenstates and the are still eigenstates. For example, moving at wavenumber k and -k have the same energy, but the difference is not in the reciprocal lattice.

The proper thing to say is that the Hamiltonian is translationally invariant with translation operator the lattice translations, and this means that H commutes with lattice translations, so that the eigenvalues of the lattice translations (the equivalence class of k under the reciprocal lattice) are each separately sectors where you find eigenvectors of H.

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