Multiple clusters on the kagome lattice

TODO

Single-site case. Three-site cluster case.

In this tutorial you will use :py:class:LatticeSolver to solve the single-orbital Hubbard model on the kagome lattice. We will do this in two ways.

First, as three inequivalent correlated subspaces \(\mathcal{C}\) for \(i \in \{A, B, C\}\). This will ignore spatial correlations within a triangle in a unit cell. Doing it this way requires constructing projectors onto the different correlated subspaces.

The second way is take a single three-site cluster and have one correlated subspace \(\mathcal{C}\). This will include spatial correlations within a triangle in a unit cell.

Tip

In examples/kagome_hubbard.py we provide an example if you are stuck. But you will learn a lot more if you write it yourself.

The model

The tight-binding model on the kagome lattice can be written as

\[ \hat{H}^0 = \sum_{k\sigma} \vec{c}_{k\sigma}^{\dagger} \mathbf{H}_{\sigma}(k) \vec{c}_{k\sigma}, \]

where the second-quantized operators are written as vectors as

\[ \vec{c}^{\dagger}_{k\sigma} = \left( \hat{c}^{\dagger}_{kA\sigma}, \hat{c}^{\dagger}_{kB\sigma}, \hat{c}^{\dagger}_{kC\sigma} \right), \]

where \(\hat{c}^{\dagger}_{k\alpha\sigma}\) creates an electron on site \(\alpha \in \{A,B,C\}\) with spin \(\sigma\) within a unit cell, and the dispersion matrix is given by

\[\begin{split} \mathbf{H}_{\sigma}(k) = \begin{pmatrix} 0 & - 2 t \cos(k_1/2) & -2 t \cos(k_2/2) \\ -2 t \cos(k_1/2) & 0 & -2 t \cos(k_3/2) \\ -2 t \cos(k_2/2) & -2 t \cos(k_3/2) & 0, \end{pmatrix} \end{split}\]

where \(t\) is the hopping amplitude. For lattice vectors in Cartesian coordinates given by \(\vec{a}_1 = (1, 0)\) and \(\vec{a}_2 = (1/2, \sqrt{3}/2)\), the variables within the cosine functions are given by \(k_1 = k_x\), \(k_2 = k_x/2 + \sqrt{3} k_y / 2\) and \(k_3 = - k_x / 2 + \sqrt{3} k_y / 2\).

Construct a matrix of the dispersion relations