Bloch theorem

Now consider the hamiltonian $\hat{H}$ of a periodic system with lattice vectors ${\bf a}_1,{\bf a}_2,{\bf a}_3$ and let the translation operators $\hat{T}_{\bf\tau}$ be associated to ${\bf\tau} = n {\bf a}_1 + m {\bf a}_2 + l {\bf a}_3$, with $n,m,l$ any three integers (positive or negative). If we neglect the electron-electron interaction (which is not periodic), or we rather include it with a (periodic) mean field approach, then $\hat{H}$ has the same periodicity of the lattice. This means that any shift of the reference frame by $\tau$ leaves the hamiltonian invariant:

$$\langle {\bf r^\prime} \vert \hat{H} \vert {\bf r} \rangle = \lan... ... r^\prime} \vert \hat{T}^\dag _\tau \hat{H} \hat{T}_\tau \vert {\bf r} \rangle.$$ (12.12)

Since this has to be valid for every ${\bf r}, {\bf r^\prime}$, and considering that $\hat{T}^\dag _\tau = \hat{T}^{-1}_\tau$, we have $\hat{H} = \hat{T}^{-1}_\tau \hat{H} \hat{T}_\tau$, which implies that the two operators commute and we can find a common set of eigenstates. The eigenvalues of $\hat{T}_\tau$ are of the form $e^{i \theta_\tau}$, and the angle $\theta_\tau$ can be written as $\theta_\tau = {\bf k} \cdot \tau$, with ${\bf k}$ a vector of reciprocal space. Such a vector can be written as the sum of a vector in the BZ plus a reciprocal lattice vector of the type ${\bf g} = n^\prime {\bf b}_1 + m^\prime {\bf b}_2 + l^\prime {\bf b}_3$, and since $\tau$ is a lattice vector, ${\bf g} \cdot \tau$ is a multiple of $2\pi$, which gives $e^{i {\bf (k + g)} \cdot \tau} = e^{i {\bf k} \cdot \tau}$. Therefore, the vector ${\bf k}+{\bf g}$, with ${\bf k}$ restricted to the BZ, cannot be used to label an eigenvector of $\hat{T}_\tau$ which is independent from the one labelled by ${\bf k}$. Therefore, we can write:

$$\hat{T}_{\bf\tau} \psi({\bf r}) = \psi({\bf r}-{\bf\tau}) = e^{-i {\bf k} \cdot \tau} \psi({\bf r}),$$ (12.13)

where the first equality applies to any function $\psi$, and the second to the eigenvectors of $\hat{T}_{\bf\tau}$. Let us now define:

$$u_{\bf k} ({\bf r}) = e^{-i {\bf k} \cdot {\bf r} } \psi({\bf r}).$$ (12.14)

If we apply a translational operator to it we obtain:

$$\hat{T}_{\bf\tau} u_{\bf k}({\bf r}) = u_{\bf k} ({\bf r}-{\bf\ta... ...{\bf k} \cdot {\bf r}} e^{i {\bf k} \cdot {\bf\tau}} \psi ({\bf r}-{\bf\tau}) =$$

$$e^{-i {\bf k} \cdot {\bf r}} e^{i {\bf k} \cdot {\bf\tau}} e^{-i ... ... ({\bf r}) = e^{-i {\bf k} \cdot {\bf r}} \psi ({\bf r}) = u_{\bf k} ({\bf r}),$$ (12.15)

which shows that $u_{\bf k}$ is periodic with period equal to $\tau$. We therefore found that the eigenstates of the operator $\hat{T}_\tau$ are of the type:

$$\psi_{\bf k}({\bf r}) = e^{i {\bf k}\cdot {\bf r}} u_{\bf k} ({\bf r}),$$ (12.16)

that is, for any function that has the periodicity of the crystal, and for any k vector in the BZ, Eq.  gives an eigenvector of $\hat{T}_\tau$. This applies always, but in the case where $\hat{T}_\tau$ commutes with $\hat{H}$, the eigenstates of $\hat{H}$ must also have the form . Since there is (at least) one eigenstate of $\hat{T}_\tau$ for each vector ${\bf k}$, then the hamiltonian $\hat{H}$ also has at least one eigenstate for each vector ${\bf k}$ of the BZ.

Coming back to –, since eigenstates of the hamiltonian for different ${\bf k}$ vectors are orthogonal, it is easy to verify that, replacing $i$ with ${\bf k'}$, $j$ with ${\bf k}$, and the sum over $i$ with an integral over ${\bf k'}$,  is equal to .

Note that since $\tau$ can be any arbitrary lattice vector, and that in particular we can have $\tau = {\bf a}_j; j=1,3$, the functions $u_{\bf k}$ must have the periodicity of the primitive cell, $u_{\bf k}({\bf r}+{\bf a}_j) = u_{\bf k}({\bf r})$ . Periodicity w.r.t any lattice vector is then automatic.