Problem: What is the state space \(\mathcal H_{\text{TB}}\) of the tight-binding model on a lattice \(\Lambda\) in \(\mathbf R^d\), ignoring spin degrees of freedom and assuming a single state/”orbital” per lattice point in \(\Lambda\).
Solution: The tight-binding model assumes a single electron is moving in a \(\Lambda\)-periodic potential \(\sum_{\mathbf x’\in\Lambda}V_V(\mathbf x-\mathbf x’)\). If the “lattice” \(\Lambda\) had consisted of a single lattice point \(\mathbf x’\in\mathbf R^d\), then the electron would be bound in some state \((|\mathbf p|^2/2m+V_V(\mathbf x-\mathbf x’))|\psi_{\mathbf x’}\rangle=E_0|\psi_{\mathbf x’}\rangle\). The point of tight-binding is to assume that upon modifying the Hamiltonian from \(|\mathbf p|^2/2m+V_V(\mathbf x-\mathbf x’)\mapsto H:=|\mathbf p|^2/2m+\sum_{\mathbf x’\in\Lambda}V_V(\mathbf x-\mathbf x’)\), the general state of the electron in this full \(\Lambda\)-periodic landscape \(\sum_{\mathbf x’\in\Lambda}V_V(\mathbf x-\mathbf x’)\) may be approximated by a Bloch superposition \(|\psi_{\mathbf k}\rangle=N^{-1/2}_{\Lambda}\sum_{\mathbf x\in\Lambda}e^{i\mathbf k\cdot\mathbf x}|\psi_{\mathbf x}\rangle\) of the localized bound states \(|\psi_{\mathbf x}\rangle\) (assuming \(\langle\psi_{\mathbf x’}|\psi_{\mathbf x}\rangle=\delta_{\mathbf x,\mathbf x’}\Leftrightarrow \langle\psi_{\mathbf k}|\psi_{\mathbf k}\rangle=1\)). In particular the tight-binding state space is \(\mathcal H_{\text{TB}}:=\text{span}_{\mathbf C}\{|\mathbf \psi_{\mathbf x}\rangle:\mathbf x\in\Lambda\}\) (the true state space \(\mathcal H\cong L^2(\mathbf R^d)\otimes\mathbf C^2\) of the electron is of course infinite-dimensional, so tight-binding represents a form of dimensionality reduction from \(\infty\to N_{\Lambda}\)). Armed with the eigenstates \(|\psi_{\mathbf k}\rangle\in\mathcal H_{\text{TB}}\), the eigenvalues may be immediately computed:
\[E(\mathbf k)=\langle\psi_{\mathbf k}|H|\psi_{\mathbf k}\rangle=N^{-1}_{\Lambda}\sum_{\mathbf x’\in\Lambda}\sum_{\mathbf x\in\Lambda}e^{i\mathbf k\cdot(\mathbf x-\mathbf x’)}\langle\psi_{\mathbf x’}|H|\psi_{\mathbf x}\rangle\]
\[=N^{-1}_{\Lambda}\sum_{\Delta\mathbf x\in\Lambda}\sum_{\mathbf x\in\Lambda}e^{-i\mathbf k\cdot\Delta\mathbf x}\langle\psi_{\mathbf x+\Delta\mathbf x}|H|\psi_{\mathbf x}\rangle\]
By translational symmetry of the lattice \(\Lambda\), the matrix element \(E_t(\Delta\mathbf x):=\langle\psi_{\mathbf x+\Delta\mathbf x}|H|\psi_{\mathbf x}\rangle\) is \(\mathbf x\)-independent. The two series thus decouple, with \(\sum_{\mathbf x\in\Lambda}=N_{\Lambda}\) cancelling the factor of \(N^{-1}_{\Lambda}\) in front. The upshot is that the spectrum \(E(\mathbf k)\) is the discrete Fourier transform of the tunneling energy \(E_t(\mathbf x)\):
\[E(\mathbf k)=\sum_{\mathbf x\in\Lambda}e^{-i\mathbf k\cdot\mathbf x}E_t(\mathbf x)\]
where \(\mathbf k\odot\Delta\mathbf x^*\in 2\pi\mathbf Z\) is quantized by BVK periodic boundary conditions and \(E(\mathbf k)\in\mathbf R\Leftrightarrow E_t^{\dagger}(\mathbf x)=E_t(-\mathbf x)\). Often, one makes the nearest-neighbours approximation in which \(E_t(\mathbf x)\approx 0\) for \(|\mathbf x|_1\geq 2\) and \(E_t(\mathbf 0)\approx E_0\). Finally, note that often \(E_t(\mathbf x)\leq 0\), so one sometimes sees the dispersion relation written in terms of \(|E_t(\mathbf x)|=-E_t(\mathbf x)\).
Problem: Compute the group velocity and effective mass tensor fields in \(\mathbf k\)-space for a tight-binding electron on a lattice \(\Lambda\).
Solution:
\[\mathbf v(\mathbf k):=\frac{\partial E}{\partial\hbar\mathbf k}=-\frac{i}{\hbar}\sum_{\mathbf x\in\Lambda}e^{-i\mathbf k\cdot\mathbf x}\mathbf xE_t(\mathbf x)\]
\[m^*(\mathbf k):=\left(\frac{\partial^2E}{\partial(\hbar\mathbf k)^{\otimes 2}}\right)^{-1}=-\hbar^2\left(\sum_{\mathbf x\in\Lambda}e^{-i\mathbf k\cdot\mathbf x}\mathbf x^{\otimes 2}E_t(\mathbf x)\right)^{-1}\]
Problem: Write down the nearest-neighbours tight-binding electron dispersion relations for a \(d=3\) tetragonal lattice \(\Lambda_{a,a,c}\), a \(d=3\) body-centered cubic lattice \(\Lambda_{\text{BCC}}\), a \(d=3\) face-centered cubic lattice \(\Lambda_{\text{FCC}}\), and a \(d=2\) sheet of graphene \(\Lambda_A+\Lambda_B\).
Solution: For \(\Lambda_{a,a,c}\):
\[E_{a,a,c}(k_x,k_y,k_z)=E_0+2E_{t,xy}(\cos k_xa+\cos k_ya)+2E_{t,z}\cos(k_zc)\]
For \(\Lambda_{\text{BCC}}\) with \(8\) nearest neighbours:
\[E_{\text{BCC}}(k_x,k_y,k_z)=E_0+E_t\sum_{\pm,\pm,\pm}e^{i(\pm k_x\pm k_y\pm k_z)a/2}=E_0+E_t\sum_{\pm}e^{\pm ik_xa/2}\sum_{\pm}e^{\pm ik_ya/2}\sum_{\pm}e^{\pm ik_za/2}\]
\[=E_0+8E_t\cos\frac{k_xa}{2}\cos\frac{k_ya}{2}\cos\frac{k_za}{2}\]
For \(\Lambda_{\text{FCC}}\) with \(12\) nearest neighbours:
\[E_{\text{FCC}}(k_x,k_y,k_z)=E_0+4E_t\left(\cos\frac{k_xa}{2}\cos\frac{k_ya}{2}+\cos\frac{k_xa}{2}\cos\frac{k_za}{2}+\cos\frac{k_ya}{2}\cos\frac{k_za}{2}\right)\]
For \(\Lambda_A+\Lambda_B\):