| grid.py | ||
| README.md | ||
Lattice Momenta
A simple analytic model of momentum modes on a regular tight-binding lattice. The analytic derivation is summarised below.
Brillouin Zone
We work with Bravais lattices, defined by integer sums of some linearly independent basis vectors \vec{a}_i
\Lambda = \left\{n_1\vec{a}_1 + n_2\vec{a}_2\right\} \quad \mathrm{for}\ n_1, n_2 \in \mathbb{Z}\,.
To understand momenta on such a lattice, we consider the reciprocal or dual lattice \Lambda^* defined by the set of points
\Lambda^* = \left\{m_1\vec{b}_1 + m_2\vec{b}_2\right\} \quad \mathrm{for}\ m_1, m_2 \in \mathbb{Z}\,,
where the reciprocal basis vectors \vec{a}_i obey
\vec{a}_i \cdot \vec{b}_j = 2\pi \delta{ij}\,.
Such a dual lattice obeys the condition
\exp(i \vec{r} \cdot \vec{k}) = 1 \;\;\forall\;\; \vec{r} \in \Lambda\;,\;\vec{k}\in\Lambda^*\,,
this means that on our original lattice \Lambda, any momentum corresponding to a site on the dual lattice
is indistinguishable zero momentum.
The Brillouin zone is defined as the set of all momenta that are closer to the origin than to any other site on the dual lattice.
For lattices that are the union of multiple Bravais lattices, we require that every Bravais lattice has the same basis (but different offsets), and use the Brillouin zone of any one of the sublattices.
Hamiltonian
The energy levels are obtained from a diagonalisation of the tight-binding Hamiltonian in momentum space.
We begin with the simplest form for the tight-binding Hamiltonian for a Bravais lattice:
H = -t \sum_{\vec{r},j} (\ket{\vec{r}} \bra{\vec{r}+\vec{s_j}} + \ket{\vec{r}+\vec{s_j}} \bra{\vec{r}})\,,
where \bra{\vec{r}} are the position eigenstates at the lattice sites.
We take the Fourier transform
\ket{\vec{r}} = \sum_{\vec{k}} \exp(i \vec{k}\cdot\vec{r}) \ket{\psi(\vec{k})}\,,
Where \bra{\psi(\vec{k})} are the momentum eigenstates. We then obtain
H = -t \sum_{\vec{k},j} \left(\exp(i \vec{k}\cdot\vec{s}_j) \ket{\psi(\vec{k})}\bra{\psi(\vec{k})} + \exp(i \vec{k}\cdot\vec{s}_j) \ket{\psi(\vec{k})}\bra{\psi(\vec{k})}\right)\,,
By defining \Phi_{\vec{k}} = \sum_j \exp(i \vec{k}\cdot\vec{s}_j), we can write
H = -t \sum_{\vec{k},j} \mathbb{R}e(\Phi_{\vec{k}}) \ket{\psi(\vec{k})}\bra{\psi(\vec{k})}\,,
so we have energy eigenstates \ket{\psi(\vec{k})} with eigenvalues
E = -t \mathbb{R}e(\Phi_{\vec{k}})\,.
Similarly, for a hexagonal lattice, we have
H_{\mathrm{hex}} = -t \sum_{\vec{r},j} (\ket{a,\vec{r}} \bra{b,\vec{r}+\vec{s_j}} + \ket{b,\vec{r}+\vec{s_j}} a,\bra{\vec{r}})\,,
where \ket{a, \vec{r}} and \ket{b, \vec{r}} live on different sub-lattices.
In momentum space, this gives
H_{\mathrm{hex}} = -t \sum_{\vec{k},j} \left(\Phi_{\vec{k}} \ket{a, \psi(\vec{k})}\bra{b, \psi(\vec{k})} + \Phi_{\vec{k}}^* \ket{b, \psi(\vec{k})}\bra{a, \psi(\vec{k})}\right)\,.
Considering superpositions of the form \alpha \ket{a, \psi(\vec{k})} + \beta \ket{b, \psi(\vec{k})}, we find that the eigenstates have \beta=\pm \alpha, and the eigenvalues are
E_{\mathrm{hex}} = \mp t \left|\Phi_{\vec{k}}\right|\,,
So we have two sheets, symmetric about E=0.