# Tight Binding Hamiltonian for graphene

The TB Hamiltonian for the tetragonal lattice is $$\hat H_0 = -J\sum_{m,n} (\hat a_{m+1,n}^\dagger \hat a_{m,n}+\hat a_{m,n}^\dagger \hat a_{m,n+1}+h.c.)$$

How can this be derived for the hexagonal lattice?

• Your title talks about a tight-binding Hamiltonian but the body of your question talks about a Hubbard Hamiltonian, which has an extra on site term which does not appear in Hamiltonian you have written down. What are you actually after? – By Symmetry Jul 12 at 8:54
• TB Hamiltonian. – John Typas Jul 13 at 1:52

## 1 Answer

Graphene has two atoms per unit cell belonging to sublattices A and B. Nearest neighbor hopping takes place only between atoms of different sublattices:

$$\hat{H} = -t \sum_j a^\dagger\left(\mathbf{r}_j\right) b\left(\mathbf{r}_j\right) + a^\dagger\left(\mathbf{r}_j\right) b\left(\mathbf{r}_j + \mathbf{d}_1\right) + a^\dagger\left(\mathbf{r}_j\right) b\left(\mathbf{r}_j + \mathbf{d}_2\right) + \mathrm{h.c.}\,.$$

The Hamiltonian tells you that for an A-sublattice atom inside the unit cell at $$\mathbf{r}_j$$, there are three nearest neighbors: one in the same unit cell and two more in the unit cells shifted by the lattice vectors $$\mathbf{d}_1$$ and $$\mathbf{d}_2$$. Write the Hamiltonian in the momentum space:

$$\hat{H} = -t \sum_j \left[\frac{1}{\sqrt{N}}\sum_\mathbf{q} a^\dagger_\mathbf{q}e^{-i\mathbf{r}_j \cdot\mathbf{q}}\right] \left[\frac{1}{\sqrt{N}}\sum_{\mathbf{q}'} b_{\mathbf{q}'}e^{i\mathbf{r}_j \cdot\mathbf{q}'}\right] \\ -t \sum_j \left[\frac{1}{\sqrt{N}}\sum_\mathbf{q} a^\dagger_\mathbf{q}e^{-i\mathbf{r}_j \cdot\mathbf{q}}\right] \left[\frac{1}{\sqrt{N}}\sum_{\mathbf{q}'} b_{\mathbf{q}'}e^{i\left(\mathbf{r}_j+\mathbf{d}_1\right) \cdot\mathbf{q}'}\right] \\ -t \sum_j \left[\frac{1}{\sqrt{N}}\sum_\mathbf{q} a^\dagger_\mathbf{q}e^{-i\mathbf{r}_j \cdot\mathbf{q}}\right] \left[\frac{1}{\sqrt{N}}\sum_{\mathbf{q}'} b_{\mathbf{q}'}e^{i\left(\mathbf{r}_j+\mathbf{d}_2\right) \cdot\mathbf{q}'}\right] + \mathrm{h.c.}\,.$$

Here, $$N$$ is the number of unit cells in the system. Performing the summation over $$j$$ gives $$N\delta_{\mathbf{qq}'}$$ so that

$$\hat{H} = -t \sum_\mathbf{q} a^\dagger_\mathbf{q} b_\mathbf{q} -t \sum_\mathbf{q} a^\dagger_\mathbf{q} b_\mathbf{q} e^{i \mathbf{d}_1 \cdot \mathbf{q}} -t \sum_\mathbf{q} a^\dagger_\mathbf{q}b_\mathbf{q}e^{i \mathbf{d}_2\cdot \mathbf{q}} + \mathrm{h.c.} \\ =-t\sum_\mathbf{q}a^\dagger_\mathbf{q}b_\mathbf{q} \left( 1 + e^{i\mathbf{d}_1\cdot\mathbf{q}} + e^{i\mathbf{d}_2\cdot\mathbf{q}} \right) + \mathrm{h.c.} \\ =-t\sum_\mathbf{q} \begin{pmatrix} a^\dagger_\mathbf{q} & b^\dagger_\mathbf{q} \end{pmatrix} \begin{pmatrix} 0&1 + e^{i\mathbf{d}_1\cdot\mathbf{q}} + e^{i\mathbf{d}_2\cdot\mathbf{q}} \\ 1 + e^{-i\mathbf{d}_1\cdot\mathbf{q}} + e^{-i\mathbf{d}_2\cdot\mathbf{q}} &0 \end{pmatrix} \begin{pmatrix} a_\mathbf{q} \\ b_\mathbf{q} \end{pmatrix}\,.$$

There you go!

• thank you al lot ! – John Typas Jul 30 at 13:51