Bloch Theorem
Theorem: let a quantum mechanical system be described by the Hamiltonian \hat{H}, where the potential energy V(\mathbf{r}) possesses the periodicity of a Bravais lattice. This means for any position vector \mathbf{r} and for every lattice vector \mathbf{R} of the Bravais lattice, the potential satisfies V(\mathbf{r} + \mathbf{R}) = V(\mathbf{r}). The Hamiltonian is given by:
\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r})
The eigenfunctions \psi(\mathbf{r}) of this Hamiltonian can be chosen to have the form of a plane wave modulated by a function u_{\mathbf{k}}(\mathbf{r}) which has the same periodicity as the lattice. Specifically,
\psi_{\mathbf{k}}(\mathbf{r}) = e^{i \mathbf{k} \cdot \mathbf{r}} u_{\mathbf{k}}(\mathbf{r})
where u_{\mathbf{k}}(\mathbf{r} + \mathbf{R}) = u_{\mathbf{k}}(\mathbf{r}) for all lattice vectors \mathbf{R}. The vector \mathbf{k} is known as the crystal momentum vector. An equivalent statement is that the eigenfunctions are simultaneous eigenfunctions of the Hamiltonian and the lattice translation operators, satisfying
\psi(\mathbf{r} + \mathbf{R}) = e^{i \mathbf{k} \cdot \mathbf{R}} \psi(\mathbf{r})
Proof: we define the set of translation operators \hat{T}_{\mathbf{R}} which act on any function f(\mathbf{r}) as:
\hat{T}_{\mathbf{R}} f(\mathbf{r}) = f(\mathbf{r} + \mathbf{R})
where \mathbf{R} is a Bravais lattice vector. The Hamiltonian of the system is:
\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r})
We examine the action of the translation operator on the Hamiltonian acting on an arbitrary wavefunction \psi(\mathbf{r}). The kinetic energy operator, being a function of the Laplacian \nabla^2, is invariant under translation since \nabla involves derivatives with respect to components of \mathbf{r} and is unaffected by a constant shift \mathbf{R}. The potential energy operator is periodic by hypothesis, V(\mathbf{r} + \mathbf{R}) = V(\mathbf{r}).
Let us compute the commutator of the Hamiltonian and a translation operator:
\begin{aligned} [\hat{H}, \hat{T}_{\mathbf{R}}]\psi(\mathbf{r}) &= (\hat{H}\hat{T}_{\mathbf{R}} - \hat{T}_{\mathbf{R}}\hat{H})\psi(\mathbf{r}) \\ &= \hat{H}\psi(\mathbf{r}+\mathbf{R}) - \hat{T}_{\mathbf{R}}\hat{H}\psi(\mathbf{r}) \\ &= \left(-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r})\right)\psi(\mathbf{r}+\mathbf{R}) - \left(-\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r}) + V(\mathbf{r})\psi(\mathbf{r})\right)_{\mathbf{r} \to \mathbf{r}+\mathbf{R}} \end{aligned}
Since V(\mathbf{r}+\mathbf{R})=V(\mathbf{r}), the term V(\mathbf{r})\psi(\mathbf{r}+\mathbf{R}) in the first part is equivalent to V(\mathbf{r}+\mathbf{R})\psi(\mathbf{r}+\mathbf{R}), which is the translated form of V(\mathbf{r})\psi(\mathbf{r}). The kinetic operator is also translationally invariant. Therefore, the application of \hat{H} at position \mathbf{r} to a function \psi(\mathbf{r}+\mathbf{R}) is identical to translating the result of \hat{H}\psi(\mathbf{r}):
\hat{H}(\mathbf{r})\psi(\mathbf{r}+\mathbf{R}) = (\hat{H}\psi)(\mathbf{r}+\mathbf{R}) = \hat{T}_{\mathbf{R}} (\hat{H}\psi)(\mathbf{r})
This gives:
[\hat{H}, \hat{T}_{\mathbf{R}}] = 0
The operators \hat{T}_{\mathbf{R}} also commute with themselves, as translations can be performed in any order:
\hat{T}_{\mathbf{R}_1}\hat{T}_{\mathbf{R}_2} = \hat{T}_{\mathbf{R}_2}\hat{T}_{\mathbf{R}_1} = \hat{T}_{\mathbf{R}_1+\mathbf{R}_2}
Because the Hamiltonian \hat{H} and the set of all translation operators \{\hat{T}_{\mathbf{R}}\} form a set of mutually commuting operators, they must possess a common set of eigenfunctions. Let \psi(\mathbf{r}) be such an eigenfunction. It must satisfy the eigenvalue equations for both operators simultaneously:
\begin{aligned} \hat{H} \psi(\mathbf{r}) &= E \psi(\mathbf{r}) \\ \hat{T}_{\mathbf{R}} \psi(\mathbf{r}) &= c(\mathbf{R}) \psi(\mathbf{r}) \end{aligned}
Here, E is the energy eigenvalue and c(\mathbf{R}) are the eigenvalues of the translation operators. Let us examine the properties of these eigenvalues c(\mathbf{R}). Consider the action of two successive translations:
\hat{T}_{\mathbf{R}_1} \hat{T}_{\mathbf{R}_2} \psi(\mathbf{r}) = \hat{T}_{\mathbf{R}_1} (c(\mathbf{R}_2) \psi(\mathbf{r})) = c(\mathbf{R}_2) \hat{T}_{\mathbf{R}_1} \psi(\mathbf{r}) = c(\mathbf{R}_1) c(\mathbf{R}_2) \psi(\mathbf{r})
Also, from the composition property of the translation operators themselves,
\hat{T}_{\mathbf{R}_1} \hat{T}_{\mathbf{R}_2} \psi(\mathbf{r}) = \hat{T}_{\mathbf{R}_1 + \mathbf{R}_2} \psi(\mathbf{r}) = c(\mathbf{R}_1 + \mathbf{R}_2) \psi(\mathbf{r})
Comparing these two results, we find a condition on the eigenvalues:
c(\mathbf{R}_1 + \mathbf{R}_2) = c(\mathbf{R}_1) c(\mathbf{R}_2)
The imposition of periodic boundary conditions on the macroscopic crystal requires that |c(\mathbf{R})|^2 = 1, which constrains c(\mathbf{R}) to be a complex number of unit modulus. Let the primitive lattice vectors be \mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3.
Any lattice vector \mathbf{R} can be expressed as a linear combination of these with integer coefficients n_j: \mathbf{R} = n_1 \mathbf{a}_1 + n_2 \mathbf{a}_2 + n_3 \mathbf{a}_3. The functional equation for c(\mathbf{R}) implies that c(\mathbf{R}) = [c(\mathbf{a}_1)]^{n_1} [c(\mathbf{a}_2)]^{n_2} [c(\mathbf{a}_3)]^{n_3}. We can represent the eigenvalues c(\mathbf{a}_j) as:
c(\mathbf{a}_j) = e^{i \theta_j}
We define a vector \mathbf{k} in the reciprocal space such that its components are determined by these phases, for example through the relation \mathbf{k} \cdot \mathbf{a}_j = \theta_j. This allows us to write the eigenvalue c(\mathbf{R}) in a general form:
c(\mathbf{R}) = e^{i \mathbf{k} \cdot \mathbf{R}}
Substituting this result back into the eigenvalue equation for the translation operator yields:
\psi(\mathbf{r} + \mathbf{R}) = \hat{T}_{\mathbf{R}} \psi(\mathbf{r}) = e^{i \mathbf{k} \cdot \mathbf{R}} \psi(\mathbf{r})
This is one form of the theorem. To arrive at the other standard form, we define a function u_{\mathbf{k}}(\mathbf{r}) as:
u_{\mathbf{k}}(\mathbf{r}) = e^{-i \mathbf{k} \cdot \mathbf{r}} \psi_{\mathbf{k}}(\mathbf{r})
This implies \psi_{\mathbf{k}}(\mathbf{r}) = e^{i \mathbf{k} \cdot \mathbf{r}} u_{\mathbf{k}}(\mathbf{r}). We now must show that u_{\mathbf{k}}(\mathbf{r}) has the periodicity of the lattice. We evaluate this function at a translated coordinate \mathbf{r} + \mathbf{R}:
\begin{aligned} u_{\mathbf{k}}(\mathbf{r} + \mathbf{R}) &= e^{-i \mathbf{k} \cdot (\mathbf{r} + \mathbf{R})} \psi_{\mathbf{k}}(\mathbf{r} + \mathbf{R}) \\ &= e^{-i \mathbf{k} \cdot \mathbf{r}} e^{-i \mathbf{k} \cdot \mathbf{R}} \psi_{\mathbf{k}}(\mathbf{r} + \mathbf{R}) \end{aligned}
Using the property \psi_{\mathbf{k}}(\mathbf{r} + \mathbf{R}) = e^{i \mathbf{k} \cdot \mathbf{R}} \psi_{\mathbf{k}}(\mathbf{r}), we substitute for the translated wavefunction:
\begin{aligned} u_{\mathbf{k}}(\mathbf{r} + \mathbf{R}) &= e^{-i \mathbf{k} \cdot \mathbf{r}} e^{-i \mathbf{k} \cdot \mathbf{R}} (e^{i \mathbf{k} \cdot \mathbf{R}} \psi_{\mathbf{k}}(\mathbf{r})) \\ &= e^{-i \mathbf{k} \cdot \mathbf{r}} \psi_{\mathbf{k}}(\mathbf{r}) \\ &= u_{\mathbf{k}}(\mathbf{r}) \end{aligned}
This demonstrates that the function u_{\mathbf{k}}(\mathbf{r}) is periodic with the Bravais lattice, u_{\mathbf{k}}(\mathbf{r} + \mathbf{R}) = u_{\mathbf{k}}(\mathbf{r}), completing the proof.