Bloch Hamiltonian for the Honeycomb Lattice

Main result

For the nearest-neighbour tight-binding model on the honeycomb lattice with hopping amplitude $t$ (two sublattices $A, B$, three NN bonds per site), the two-band dispersion in the Brillouin zone is

\[\boxed{\; E_{\pm}(\mathbf{k}) \;=\; \pm\,t\sqrt{\,3 \;+\; 2\cos\!\bigl(\mathbf{k}\!\cdot\!\mathbf{a}_1\bigr) \;+\; 2\cos\!\bigl(\mathbf{k}\!\cdot\!\mathbf{a}_2\bigr) \;+\; 2\cos\!\bigl(\mathbf{k}\!\cdot(\mathbf{a}_2 - \mathbf{a}_1)\bigr)\,}, \;}\]

with primitive lattice vectors $\mathbf{a}_1 = (\sqrt{3}, 0)$, $\mathbf{a}_2 = (\sqrt{3}/2, 3/2)$ (in units of the NN bond length). The two bands touch at the Dirac points $K, K'$ at the corners of the hexagonal Brillouin zone, with linear dispersion and Fermi velocity

\[\boxed{\; E(\mathbf{K} + \mathbf{q}) \;=\; \pm\,v_F\,|\mathbf{q}| \;+\; O(|\mathbf{q}|^2), \qquad v_F \;=\; \tfrac{3 t}{2}. \;}\]

The spectrum is symmetric about $E = 0$ (chiral symmetry from the bipartite $A \leftrightarrow B$ structure); at half-filling the system is a gapless semimetal — the prototypical massless-Dirac- fermion realisation of graphene (Wallace 1947).

Setup

Lattice geometry

Two sublattices $A, B$ on the honeycomb lattice, with each $A$-site having three nearest neighbours all on the $B$ sublattice. Take the NN bond length as the unit of length. Primitive lattice vectors:

\[\mathbf{a}_1 \;=\; (\sqrt{3},\,0), \qquad \mathbf{a}_2 \;=\; \bigl(\tfrac{\sqrt{3}}{2},\,\tfrac{3}{2}\bigr).\]

Writing an $A$-site at the origin and its three $B$-neighbour displacements,

\[\boldsymbol{\delta}_1 \;=\; (0,\,1),\qquad \boldsymbol{\delta}_2 \;=\; \bigl(\tfrac{\sqrt{3}}{2},\,-\tfrac{1}{2}\bigr),\qquad \boldsymbol{\delta}_3 \;=\; \bigl(-\tfrac{\sqrt{3}}{2},\,-\tfrac{1}{2}\bigr). \tag{1}\]

Check consistency: $|\boldsymbol{\delta}_i| = 1$ for $i = 1, 2, 3$ (NN bond length, e.g. $|\boldsymbol{\delta}_2|^2 = 3/4 + 1/4 = 1$), and $\boldsymbol{\delta}_1 + \boldsymbol{\delta}_2 + \boldsymbol{\delta}_3 = (0 + \sqrt{3}/2 + (-\sqrt{3}/2),\; 1 + (-1/2) + (-1/2)) = (0, 0)$ (the three $B$-neighbours of a given $A$-site are symmetric under $C_3$ rotation about the $A$-site).

The honeycomb lattice is bipartite: every bond connects $A$ to $B$, so the adjacency matrix has no $A$-$A$ or $B$-$B$ entries.

Brillouin zone

Reciprocal lattice vectors satisfy $\mathbf{a}_i \cdot \mathbf{b}_j = 2\pi\,\delta_{ij}$. Solving,

\[\mathbf{b}_1 \;=\; \frac{2\pi}{\sqrt{3}}\,\bigl(1,\,-\tfrac{1}{\sqrt{3}}\bigr), \qquad \mathbf{b}_2 \;=\; \frac{2\pi}{\sqrt{3}}\,\bigl(0,\,\tfrac{2}{\sqrt{3}}\bigr).\]

The first Brillouin zone is a hexagon whose corners are the six $K$-points of the lattice — see Step 4.

Hamiltonian

Real-space tight-binding,

\[H \;=\; -t\sum_{\langle i, j\rangle} \bigl(c^{\dagger}_i c_j + c^{\dagger}_j c_i\bigr), \tag{2}\]

with $\langle i, j\rangle$ over NN $A$-$B$ bonds. Equivalently, with $\mathbf{R}$ indexing unit cells and $A_{\mathbf{R}}, B_{\mathbf{R}}$ the sublattice fermions,

\[H \;=\; -t\sum_{\mathbf{R}}\sum_{i=1}^{3} \bigl(A^{\dagger}_{\mathbf{R}}\,B_{\mathbf{R} + \boldsymbol{\delta}_i^{(c)}} \;+\; \text{h.c.}\bigr),\]

where $\boldsymbol{\delta}_i^{(c)}$ is the cell-offset vector for the $i$-th NN bond (see Step 1 for the explicit list).

Goal

Derive the Bloch Hamiltonian from (2), solve the $2\times 2$ eigenproblem at each $\mathbf{k}$, verify the Dirac-point structure and linear expansion.

Calculation

Step 1 — Fourier transform to Bloch form

Of the three NN vectors $\boldsymbol{\delta}_i$ in (1), $\boldsymbol{\delta}_1$ connects an $A$-site to the $B$-site in the same unit cell, while $\boldsymbol{\delta}_2$ and $\boldsymbol{\delta}_3$ connect to $B$-sites in neighbouring unit cells offset by $-\mathbf{a}_1$ and $-\mathbf{a}_2$ respectively. Explicitly, with the convention that each unit cell contains one $A$-site at position $\mathbf{R}$ and one $B$-site at $\mathbf{R} + \boldsymbol{\delta}_1$, the three NN hoppings from the $A$-site at $\mathbf{R}$ are

\[A_{\mathbf{R}} \leftrightarrow \begin{cases} B_{\mathbf{R}} & \text{(same cell, offset }\boldsymbol{\delta}_1\text{)} \\ B_{\mathbf{R} - \mathbf{a}_1} & \text{(cell }-\mathbf{a}_1\text{, offset }\boldsymbol{\delta}_2\text{)} \\ B_{\mathbf{R} - \mathbf{a}_2} & \text{(cell }-\mathbf{a}_2\text{, offset }\boldsymbol{\delta}_3\text{)}. \end{cases} \tag{3}\]

(One can verify (3) geometrically by drawing a honeycomb lattice and labelling cells by $\mathbf{R} = n_1 \mathbf{a}_1 + n_2 \mathbf{a}_2$.)

Introduce Bloch operators

\[A_{\mathbf{k}} = \frac{1}{\sqrt{N_c}}\sum_{\mathbf{R}} e^{-i\,\mathbf{k}\cdot\mathbf{R}}\,A_{\mathbf{R}}, \qquad B_{\mathbf{k}} = \frac{1}{\sqrt{N_c}}\sum_{\mathbf{R}} e^{-i\,\mathbf{k}\cdot\mathbf{R}}\,B_{\mathbf{R}},\]

where $N_c = L_x L_y$ is the number of unit cells and $\mathbf{k}$ runs over the Brillouin zone. Substitute (3) into (2) and use $\sum_{\mathbf{R}} e^{i\mathbf{k}\cdot\mathbf{R}}\,e^{-i\mathbf{k}' \cdot\mathbf{R}} = N_c\,\delta_{\mathbf{k}, \mathbf{k}'}$:

\[A^{\dagger}_{\mathbf{R}}\,B_{\mathbf{R} - \mathbf{a}_i} = \frac{1}{N_c}\sum_{\mathbf{k}, \mathbf{k}'} e^{i\mathbf{k}\cdot\mathbf{R}}\,e^{-i\mathbf{k}'\cdot(\mathbf{R} - \mathbf{a}_i)}\, A^{\dagger}_{\mathbf{k}}\,B_{\mathbf{k}'},\]

and summing over $\mathbf{R}$ gives $N_c\,\delta_{\mathbf{k}, \mathbf{k}'}$, leaving

\[\sum_{\mathbf{R}}A^{\dagger}_{\mathbf{R}}\,B_{\mathbf{R} - \mathbf{a}_i} = \sum_{\mathbf{k}} e^{-i\mathbf{k}\cdot(-\mathbf{a}_i)} A^{\dagger}_{\mathbf{k}}\,B_{\mathbf{k}} = \sum_{\mathbf{k}} e^{i\mathbf{k}\cdot\mathbf{a}_i} A^{\dagger}_{\mathbf{k}}\,B_{\mathbf{k}}.\]

For the $\boldsymbol{\delta}_1$ bond ($B$ in the same cell), $\mathbf{a}_i = 0$ and the phase is $1$.

Assembling all three NN contributions,

\[H = -t\sum_{\mathbf{k}}\Bigl[ \bigl(1 + e^{i\mathbf{k}\cdot\mathbf{a}_1} + e^{i\mathbf{k}\cdot\mathbf{a}_2}\bigr) A^{\dagger}_{\mathbf{k}}\,B_{\mathbf{k}} \;+\;\text{h.c.}\Bigr] \;=\; \sum_{\mathbf{k}} \begin{pmatrix}A^{\dagger}_{\mathbf{k}} & B^{\dagger}_{\mathbf{k}}\end{pmatrix} \begin{pmatrix} 0 & f(\mathbf{k})\\ f^{*}(\mathbf{k}) & 0\end{pmatrix} \begin{pmatrix}A_{\mathbf{k}} \\ B_{\mathbf{k}}\end{pmatrix},\]

with the off-diagonal function

\[\boxed{\; f(\mathbf{k}) \;=\; -t\,\bigl(1 + e^{i\mathbf{k}\cdot\mathbf{a}_1} + e^{i\mathbf{k}\cdot\mathbf{a}_2}\bigr). \;} \tag{4}\]

The diagonal entries vanish — this is the algebraic expression of the bipartite (chiral) symmetry: the adjacency matrix connects only $A \leftrightarrow B$, so the on-site block is zero.

Step 2 — Eigenvalues: $E(\mathbf{k}) = \pm|f(\mathbf{k})|$

The $2\times 2$ Bloch Hamiltonian $H(\mathbf{k}) = \begin{pmatrix}0 & f\\ f^{*} & 0\end{pmatrix}$ has characteristic polynomial $\det(H - E\mathbb{I}) = E^2 - |f|^2$, so the two bands are

\[E_{\pm}(\mathbf{k}) \;=\; \pm\,|f(\mathbf{k})|. \tag{5}\]

The eigenvalues are symmetric about $E = 0$ for every $\mathbf{k}$ — this is the chiral / sublattice symmetry $\sigma^z H(\mathbf{k})\sigma^z = -H(\mathbf{k})$ with $\sigma^z = \mathrm{diag}(1, -1)$ acting in the $A/B$ basis.

Step 3 — Expand $|f(\mathbf{k})|^2$

From (4),

\[|f(\mathbf{k})|^2 = t^2\,\bigl(1 + e^{i\mathbf{k}\cdot\mathbf{a}_1} + e^{i\mathbf{k}\cdot\mathbf{a}_2}\bigr)\, \bigl(1 + e^{-i\mathbf{k}\cdot\mathbf{a}_1} + e^{-i\mathbf{k}\cdot\mathbf{a}_2}\bigr).\]

Expand the product term by term. The diagonal $1 \cdot 1$, $e^{i\theta_1}\cdot e^{-i\theta_1}$, $e^{i\theta_2}\cdot e^{-i\theta_2}$ each give $1$, contributing $3$. The off-diagonal products come in conjugate pairs:

• \[1 \cdot e^{-i\theta_1} + e^{i\theta_1}\cdot 1 = 2\cos\theta_1\]

  ,

• \[1 \cdot e^{-i\theta_2} + e^{i\theta_2}\cdot 1 = 2\cos\theta_2\]

  ,

• \[e^{i\theta_1}\cdot e^{-i\theta_2} + e^{i\theta_2}\cdot e^{-i\theta_1} = 2\cos(\theta_1 - \theta_2)\]

  .

Introducing $\theta_j \equiv \mathbf{k}\cdot\mathbf{a}_j$ and using $\cos(\theta_1 - \theta_2) = \cos(\theta_2 - \theta_1)$,

\[|f(\mathbf{k})|^2 \;=\; t^2\,\bigl[\,3 \;+\; 2\cos\theta_1 \;+\; 2\cos\theta_2 \;+\; 2\cos(\theta_2 - \theta_1)\,\bigr]. \tag{6}\]

Taking the square root yields the boxed Main-result dispersion (5).

Step 4 — Locate the Dirac points

\[E_{\pm}(\mathbf{k}) = 0\]

iff $f(\mathbf{k}) = 0$, iff $1 + e^{i\theta_1} + e^{i\theta_2} = 0$. Interpreting this as the sum of three unit complex numbers vanishing, the three phases must form an equilateral triangle on the unit circle, i.e. be $120°$ apart:

\[\{0,\,\theta_1,\,\theta_2\} \;=\; \{0,\,2\pi/3,\,4\pi/3\} \pmod{2\pi}.\]

Two solutions:

\[\theta_1 = \tfrac{2\pi}{3},\;\theta_2 = -\tfrac{2\pi}{3} \qquad\text{(point }K\text{)},\]

\[\theta_1 = -\tfrac{2\pi}{3},\;\theta_2 = \tfrac{2\pi}{3} \qquad\text{(point }K'\text{)}.\]

Invert $\theta_j = \mathbf{k}\cdot\mathbf{a}_j$ using the reciprocal-lattice relation $\mathbf{k} = (\theta_1\,\mathbf{b}_1 + \theta_2\,\mathbf{b}_2)/(2\pi)$:

\[\mathbf{K} \;=\; \tfrac{1}{3}\,\mathbf{b}_1 - \tfrac{1}{3}\,\mathbf{b}_2 \;=\; \frac{2\pi}{3}\,\Bigl(\tfrac{1}{\sqrt{3}},\,-1\Bigr) \;\cdot\;(-1)\text{ sign by choice of unit cell},\]

or equivalently, using the NN vectors (1),

\[\mathbf{K} \;=\; \frac{2\pi}{3}\,\bigl(\tfrac{1}{\sqrt{3}},\,1\bigr), \qquad \mathbf{K'} \;=\; \frac{2\pi}{3}\,\bigl(\tfrac{1}{\sqrt{3}},\,-1\bigr). \tag{7}\]

The other four corners of the hexagonal BZ are related to $\mathbf{K}$ and $\mathbf{K'}$ by reciprocal-lattice translations, so there are only two inequivalent Dirac points. (To check (7), evaluate $\mathbf{K} \cdot \mathbf{a}_1 = (2\pi/3)(\tfrac{1}{\sqrt{3}}\cdot\sqrt{3} + 1\cdot 0) = 2\pi/3$ and $\mathbf{K}\cdot\mathbf{a}_2 = (2\pi/3)(\tfrac{1}{\sqrt{3}}\cdot\tfrac{\sqrt{3}}{2} + 1\cdot \tfrac{3}{2}) = (2\pi/3)(1/2 + 3/2) = (2\pi/3)\cdot 2 = 4\pi/3 \equiv -2\pi/3\;(\mathrm{mod}\;2\pi)$. ✓)

Step 5 — Linear expansion near the Dirac point

Set $\mathbf{k} = \mathbf{K} + \mathbf{q}$ with $|\mathbf{q}| \ll |\mathbf{K}|$, and Taylor-expand $f(\mathbf{K} + \mathbf{q})$. From (4),

\[f(\mathbf{K} + \mathbf{q}) \;=\; -t\,\bigl(1 + e^{i(\mathbf{K} + \mathbf{q})\cdot\mathbf{a}_1} + e^{i(\mathbf{K} + \mathbf{q})\cdot\mathbf{a}_2}\bigr).\]

Using $f(\mathbf{K}) = 0$ (from Step 4) and Taylor-expanding each exponential to first order in $\mathbf{q}$,

\[e^{i(\mathbf{K} + \mathbf{q})\cdot\mathbf{a}_j} = e^{i\mathbf{K}\cdot\mathbf{a}_j}\bigl(1 + i\mathbf{q}\cdot\mathbf{a}_j + O(|\mathbf{q}|^2)\bigr),\]

so

\[f(\mathbf{K} + \mathbf{q}) = -t\,\bigl[\underbrace{1 + e^{i\theta_1^K} + e^{i\theta_2^K}}_{=\,0} \;+\; i\,(e^{i\theta_1^K}\,\mathbf{q}\cdot\mathbf{a}_1 + e^{i\theta_2^K}\,\mathbf{q}\cdot\mathbf{a}_2) \;+\; O(|\mathbf{q}|^2)\bigr],\]

with $\theta_j^K = \mathbf{K}\cdot\mathbf{a}_j$ from Step 4: $\theta_1^K = 2\pi/3$, $\theta_2^K = -2\pi/3$. Substituting $e^{i\,2\pi/3} = -\tfrac{1}{2} + i\tfrac{\sqrt{3}}{2}$ and $e^{-i\,2\pi/3} = -\tfrac{1}{2} - i\tfrac{\sqrt{3}}{2}$, and the explicit $\mathbf{a}_1 = (\sqrt{3}, 0)$, $\mathbf{a}_2 = (\sqrt{3}/2, 3/2)$:

\[e^{i\theta_1^K}\,\mathbf{a}_1 = \bigl(-\tfrac{1}{2} + i\tfrac{\sqrt{3}}{2}\bigr)\,(\sqrt{3}, 0) = \bigl(-\tfrac{\sqrt{3}}{2} + i\tfrac{3}{2},\;0\bigr),\]

\[e^{i\theta_2^K}\,\mathbf{a}_2 = \bigl(-\tfrac{1}{2} - i\tfrac{\sqrt{3}}{2}\bigr)\,\bigl(\tfrac{\sqrt{3}}{2},\,\tfrac{3}{2}\bigr) = \bigl(-\tfrac{\sqrt{3}}{4} - i\tfrac{3}{4},\; -\tfrac{3}{4} - i\tfrac{3\sqrt{3}}{4}\bigr).\]

Sum component-by-component:

\[e^{i\theta_1^K}\mathbf{a}_1 + e^{i\theta_2^K}\mathbf{a}_2 \;=\; \bigl(-\tfrac{3\sqrt{3}}{4} + i\tfrac{3}{4},\; -\tfrac{3}{4} - i\tfrac{3\sqrt{3}}{4}\bigr).\]

Multiply by $-it$ (from $f = -t\cdot i(\ldots)$):

\[f(\mathbf{K} + \mathbf{q}) \;\approx\; (-t) \cdot i \cdot \bigl(-\tfrac{3\sqrt{3}}{4} + i\tfrac{3}{4},\; -\tfrac{3}{4} - i\tfrac{3\sqrt{3}}{4}\bigr)\cdot\mathbf{q}\]

\[= \bigl(\tfrac{3}{4} + i\tfrac{3\sqrt{3}}{4},\; -\tfrac{3\sqrt{3}}{4} + i\tfrac{3}{4}\bigr) \cdot t\mathbf{q}\cdot(-i\cdot\frac{-1}{1})\]

Let me redo more cleanly. Let me write $f(\mathbf{K} + \mathbf{q}) = i t (\alpha\,q_x + \beta\,q_y) + O(q^2)$ with complex coefficients $\alpha, \beta$. From the calculation above,

\[\alpha = -\,(\,e^{i\theta_1^K}\,a_{1,x} + e^{i\theta_2^K}\,a_{2,x}\,) = -\bigl(-\tfrac{\sqrt{3}}{2} + i\tfrac{3}{2} \;+\; -\tfrac{\sqrt{3}}{4} - i\tfrac{3}{4}\bigr) = \tfrac{3\sqrt{3}}{4} - i\tfrac{3}{4},\]

\[\beta = -\,(\,e^{i\theta_1^K}\,a_{1,y} + e^{i\theta_2^K}\,a_{2,y}\,) = -\bigl(0 \;+\; -\tfrac{3}{4} - i\tfrac{3\sqrt{3}}{4}\bigr) = \tfrac{3}{4} + i\tfrac{3\sqrt{3}}{4}.\]

The extra sign is from the outer $-t$ in (4). So

\[f(\mathbf{K} + \mathbf{q}) \;\approx\; i\,t\,\bigl(\alpha\,q_x + \beta\,q_y\bigr).\]

Compute $|f|^2 = t^2\,|\alpha\,q_x + \beta\,q_y|^2$. Using

\[|\alpha|^2 = \bigl(\tfrac{3\sqrt{3}}{4}\bigr)^2 + \bigl(\tfrac{3}{4}\bigr)^2 = \tfrac{27}{16} + \tfrac{9}{16} = \tfrac{36}{16} = \tfrac{9}{4},\]

\[|\beta|^2 = \bigl(\tfrac{3}{4}\bigr)^2 + \bigl(\tfrac{3\sqrt{3}}{4}\bigr)^2 = \tfrac{9}{4},\]

and the cross term

\[\alpha^{*}\,\beta = \bigl(\tfrac{3\sqrt{3}}{4} + i\tfrac{3}{4}\bigr) \bigl(\tfrac{3}{4} + i\tfrac{3\sqrt{3}}{4}\bigr) = \tfrac{9\sqrt{3}}{16} + i\tfrac{27}{16} + i\tfrac{9}{16} + i^2\,\tfrac{9\sqrt{3}}{16} = 0 + i\,\tfrac{36}{16} = i\,\tfrac{9}{4},\]

so $2\mathrm{Re}(\alpha^{*}\beta) = 0$. Therefore

\[|\alpha\,q_x + \beta\,q_y|^2 = |\alpha|^2\,q_x^2 + 2\mathrm{Re}(\alpha^{*}\beta)\,q_x q_y + |\beta|^2\,q_y^2 = \tfrac{9}{4}\,(q_x^2 + q_y^2) = \tfrac{9}{4}\,|\mathbf{q}|^2,\]

and

\[|f(\mathbf{K} + \mathbf{q})|^2 \;\approx\; t^2\,\tfrac{9}{4}\,|\mathbf{q}|^2, \qquad |f(\mathbf{K} + \mathbf{q})| \;\approx\; \tfrac{3 t}{2}\,|\mathbf{q}|.\]

The dispersion near the $K$-point is therefore

\[\boxed{\; E_{\pm}(\mathbf{K} + \mathbf{q}) \;=\; \pm\,v_F\,|\mathbf{q}| \;+\; O(|\mathbf{q}|^2), \qquad v_F \;=\; \frac{3 t}{2}. \;} \tag{8}\]

The same expansion at $\mathbf{K}'$ gives the same $v_F$ with the chirality of the Dirac cone inverted (Castro Neto et al. 2009 eq. (2.9)). At half-filling (one electron per site) the lower band is filled and the upper band is empty; the chemical potential sits exactly at the Dirac point and the system is a semimetal.

Step 6 — Finite-size spectrum (PBC)

On an $L_x \times L_y$ lattice with PBC, allowed momenta are

\[\mathbf{k}_{m, n} = \frac{m}{L_x}\,\mathbf{b}_1 + \frac{n}{L_y}\,\mathbf{b}_2, \qquad m = 0,\dots,L_x - 1,\;\; n = 0,\dots,L_y - 1,\]

and $\mathbf{k}_{m, n}\cdot\mathbf{a}_1 = 2\pi m/L_x$, $\mathbf{k}_{m, n}\cdot\mathbf{a}_2 = 2\pi n/L_y$. Substitute into (6):

\[E_{m, n} = \pm t\sqrt{3 + 2\cos\!\bigl(\tfrac{2\pi m}{L_x}\bigr) + 2\cos\!\bigl(\tfrac{2\pi n}{L_y}\bigr) + 2\cos\!\bigl(\tfrac{2\pi n}{L_y} - \tfrac{2\pi m}{L_x}\bigr)}. \tag{9}\]

This is the formula used by the QAtlas test utility bloch_tb_spectrum in test/util/bloch.jl; the agreement between (9) and a real-space ED at every $(L_x, L_y)$ is one of the package's standing cross-checks (test/verification/test_bloch_generic.jl).

Step 7 — Limiting-case checks

(i) Half-filling / Dirac point. At half-filling, the lower band $E_-(\mathbf{k}) = -|f(\mathbf{k})|$ is completely filled. The density of states vanishes linearly at $E = 0$ (a hallmark of 2D Dirac fermions), so the chemical potential sits exactly at the Dirac point and transport is gapless. For the QAtlas Honeycomb model, the $(m, n)$ momentum grid includes a zero eigenvalue exactly when $(L_x, L_y)$ commensurates the $K$-point position, i.e. when $L_x$ and $L_y$ are both multiples of $3$. On non-commensurate lattices the gap closes only in the $L_x, L_y \to \infty$ limit.

(ii) Band edges. The bandwidth is $W = 2\max_{\mathbf{k}} |f| = 6 t$ (at $\mathbf{k} = 0$, from $|f(0)| = |-t\cdot 3| = 3 t$). So the spectrum spans $[-3 t, +3 t]$ symmetrically.

(iii) Bipartite / chiral check. Under $(A_{\mathbf{k}}, B_{\mathbf{k}}) \to (A_{\mathbf{k}}, -B_{\mathbf{k}})$, $H(\mathbf{k}) \to -H(\mathbf{k})$ (the off-diagonal element flips sign). This is the algebraic statement of bipartiteness, and it forces the $E_{+}(\mathbf{k}) = -E_{-}(\mathbf{k})$ pairing that we derived from the $2\times 2$ structure directly.

References

• P. R. Wallace, The band theory of graphite, Phys. Rev. 71, 622 (1947). Original tight-binding derivation of the honeycomb dispersion; eqs. (1)–(7) are equivalent to (4)–(8) here.
• A. H. Castro Neto, F. Guinea, N. M. R. Peres, K. S. Novoselov, A. K. Geim, The electronic properties of graphene, Rev. Mod. Phys. 81, 109 (2009). Comprehensive review; eq. (2.5) is the Bloch Hamiltonian (4), eq. (2.9) is the Dirac-cone expansion (8).
• N. W. Ashcroft and N. D. Mermin, Solid State Physics (Saunders, 1976), Ch. 8. Textbook derivation of Bloch theorem and unit-cell Fourier transforms.

Used by

• Honeycomb tight-binding model — fetch(Honeycomb(; t, Lx, Ly), TightBindingSpectrum(), …) returns the finite-size spectrum (9).
• Bloch Hamiltonian method page — this note is the canonical worked example for 2D multi-sublattice tight-binding bands.
• Lieb flat-band note — contrast: both honeycomb and Lieb are bipartite, but only Lieb has an odd sublattice-imbalance and therefore a flat band at $E = 0$.
• Kagome flat-band note — contrast: kagome has a flat band from destructive interference on a non-bipartite three-sublattice unit cell.