Kagome Bloch Hamiltonian and the Flat Band

Main result

For the nearest-neighbour tight-binding model on the kagome lattice with hopping amplitude $t$, the $3\times 3$ Bloch Hamiltonian has one perfectly flat eigenvalue across the entire Brillouin zone,

\[\boxed{\; E_{\rm flat}(\mathbf{k}) \;=\; +2\,t \qquad\text{for every }\mathbf{k} \in \mathrm{BZ}, \;}\]

and two dispersive bands

\[\boxed{\; E_{\pm}(\mathbf{k}) \;=\; -\,t \pm t\,\sqrt{\,4 S(\mathbf{k}) - 3\,}, \qquad S(\mathbf{k}) \;\equiv\; \cos^{2}\!\tfrac{\theta_{1}}{2} + \cos^{2}\!\tfrac{\theta_{2}}{2} + \cos^{2}\!\tfrac{\theta_{2}-\theta_{1}}{2}, \;}\]

with $\theta_{j} \equiv \mathbf{k}\cdot\mathbf{a}_{j}$. The flat band and the upper dispersive band $E_{+}$ are degenerate at the $\Gamma$-point $\mathbf{k} = (0, 0)$ (both equal to $+2t$ there), forming a quadratic band-touching point; they separate for $\mathbf{k} \ne 0$.

The exact flatness holds for every finite kagome lattice with periodic boundary conditions: the adjacency matrix has an eigenvalue $-1$ with massive degeneracy, corresponding to real-space compact localised states supported on hexagonal plaquettes of the lattice (Sutherland 1986, Mielke 1991).

Setup

Lattice geometry

The kagome lattice has $3$ sites per primitive cell, labelled $A, B, C$, each at a corner of a triangle. With triangle side length $a$ (= twice the NN bond length $a/2$), the Bravais-lattice primitive vectors are

\[\mathbf{a}_{1} \;=\; a\,(1, 0), \qquad \mathbf{a}_{2} \;=\; a\,\bigl(\tfrac{1}{2},\,\tfrac{\sqrt{3}}{2}\bigr).\]

Set $a = 1$ henceforth. The three sublattice positions within a cell are

\[\mathbf{r}_{A} = (0, 0), \qquad \mathbf{r}_{B} = \bigl(\tfrac{1}{2}, 0\bigr) = \tfrac{\mathbf{a}_{1}}{2}, \qquad \mathbf{r}_{C} = \bigl(\tfrac{1}{4}, \tfrac{\sqrt{3}}{4}\bigr) = \tfrac{\mathbf{a}_{2}}{2}.\]

The NN bond length is $|\mathbf{r}_{B} - \mathbf{r}_{A}| = 1/2$; checking |\mathbf{r}_{C}

• \mathbf{r}{A}| = \sqrt{1/16 + 3/16} = 1/2$ and |\mathbf{r}{C}
• \mathbf{r}_{B}| = \sqrt{1/16 + 3/16} = 1/2$, so every in-cell pair

forms a NN bond (the three sublattices form an upward-pointing triangle in each cell, sharing edges with downward-pointing triangles in neighbouring cells).

Each site has four nearest neighbours: two in the same unit cell and two in neighbouring cells. Explicitly, from an $A$-site at $\mathbf{R}$ the four NN bonds go to

\[A_{\mathbf{R}} \leftrightarrow \begin{cases} B_{\mathbf{R}} & \text{(same-cell, via }\mathbf{r}_B - \mathbf{r}_A\text{)} \\ B_{\mathbf{R} - \mathbf{a}_{1}} & \text{(cell }-\mathbf{a}_{1}\text{)} \\ C_{\mathbf{R}} & \text{(same-cell)} \\ C_{\mathbf{R} - \mathbf{a}_{2}} & \text{(cell }-\mathbf{a}_{2}\text{)} \end{cases} \tag{1}\]

Similarly each $B$-site connects to two $A$ and two $C$ NN, and each $C$-site to two $A$ and two $B$ NN. Total bond count: on an $L_{1}L_{2}$-unit-cell PBC torus the kagome lattice has $3 L_{1}L_{2}$ sites and $6 L_{1}L_{2}$ bonds (4 bonds per site, divided by 2).

Hamiltonian

\[H \;=\; -\,t\sum_{\langle i, j\rangle} \bigl(c^{\dagger}_{i} c_{j} + c^{\dagger}_{j} c_{i}\bigr),\]

with $\langle i, j\rangle$ over NN bonds and $t > 0$.

Goal

Derive the $3\times 3$ Bloch Hamiltonian, diagonalise, and prove that one eigenvalue is constant (independent of $\mathbf{k}$) — yielding the flat band.

Calculation

Step 1 — Fourier transform to the Bloch Hamiltonian

Define Bloch operators without sublattice offsets,

\[c_{\alpha, \mathbf{k}} \;=\; \frac{1}{\sqrt{N_{c}}} \sum_{\mathbf{R}} e^{-i\mathbf{k}\cdot\mathbf{R}}\,c_{\alpha, \mathbf{R}}, \qquad \alpha \in \{A, B, C\},\]

for $N_{c} = L_{1}L_{2}$ unit cells. The two $A$–$B$ bonds from (1) give the same-cell contribution (phase $1$) and the cross-cell contribution ($\mathbf{R}$ to $\mathbf{R} - \mathbf{a}_{1}$, phase $e^{i\mathbf{k}\cdot\mathbf{a}_{1}}$ by the same-site-collapse argument used in the honeycomb note Step 1):

\[\sum_{\mathbf{R}}\bigl(A^{\dagger}_{\mathbf{R}} B_{\mathbf{R}} + A^{\dagger}_{\mathbf{R}} B_{\mathbf{R} - \mathbf{a}_{1}}\bigr) \;=\; \sum_{\mathbf{k}} \bigl(1 + e^{i\mathbf{k}\cdot\mathbf{a}_{1}}\bigr)\, A^{\dagger}_{\mathbf{k}} B_{\mathbf{k}}.\]

Similarly for the two $A$–$C$ bonds (cross-cell shift $-\mathbf{a}_{2}$) and the two $B$–$C$ bonds. The geometry for the latter: $B$ at $\mathbf{r}_{B}$ connects to $C$ at $\mathbf{r}_{C}$ in the same cell, and to $C$ at $\mathbf{r}_{C} + \mathbf{a}_{1} - \mathbf{a}_{2}$ in the cell shifted by $\mathbf{a}_{1} - \mathbf{a}_{2}$ (check: $|\mathbf{r}_{C} + \mathbf{a}_{1} - \mathbf{a}_{2} - \mathbf{r}_{B}| = |(1/4, \sqrt{3}/4) + (1/2, -\sqrt{3}/2) - (1/2, 0)| = |(1/4, -\sqrt{3}/4)| = 1/2$, NN ✓). Fourier gives the factor $1 + e^{i\mathbf{k}\cdot(\mathbf{a}_{1} - \mathbf{a}_{2})}$.

Collecting contributions (and using $-t\,(1 + e^{i\phi})\,A^{\dagger}_{\mathbf{k}}B_{\mathbf{k}} + \text{h.c.}$ for each bond type),

\[H = \sum_{\mathbf{k}}\,\Psi^{\dagger}_{\mathbf{k}}\, \widetilde{H}(\mathbf{k})\,\Psi_{\mathbf{k}},\qquad \Psi_{\mathbf{k}} = (A_{\mathbf{k}}, B_{\mathbf{k}}, C_{\mathbf{k}})^{T},\]

\[\widetilde{H}(\mathbf{k}) = -t\,\begin{pmatrix} 0 & 1 + e^{i\theta_{1}} & 1 + e^{i\theta_{2}} \\ 1 + e^{-i\theta_{1}} & 0 & 1 + e^{-i(\theta_{2} - \theta_{1})} \\ 1 + e^{-i\theta_{2}} & 1 + e^{i(\theta_{2} - \theta_{1})} & 0 \end{pmatrix}. \tag{2}\]

with $\theta_{j} = \mathbf{k}\cdot\mathbf{a}_{j}$. This is the complex-gauge Bloch Hamiltonian.

Step 2 — Gauge transformation to real-symmetric form

Switch to the site-centered Fourier convention

\[c_{\alpha, \mathbf{k}} \;=\; \frac{1}{\sqrt{N_{c}}} \sum_{\mathbf{R}} e^{-i\mathbf{k}\cdot(\mathbf{R} + \mathbf{r}_{\alpha})} c_{\alpha, \mathbf{R}}, \qquad \alpha \in \{A, B, C\},\]

which replaces the cell-index phase $e^{-i\mathbf{k}\cdot\mathbf{R}}$ of Step 1 by a full-position phase $e^{-i\mathbf{k}\cdot(\mathbf{R} + \mathbf{r}_{\alpha})}$ including the sublattice offset. Under this convention every bond $\alpha_{\mathbf{R}} \leftrightarrow \beta_{\mathbf{R}'}$ contributes, in the Bloch Hamiltonian,

\[-t\,e^{-i\mathbf{k}\cdot(\mathbf{r}_{\beta} + \mathbf{R}' - \mathbf{r}_{\alpha} - \mathbf{R})} \;=\; -t\,e^{-i\mathbf{k}\cdot\mathbf{d}},\]

with $\mathbf{d} = (\mathbf{r}_{\beta} + \mathbf{R}') - (\mathbf{r}_{\alpha} + \mathbf{R})$ the bond vector from $\alpha$ to $\beta$. Summing a Hermitian-conjugate pair gives $-2t\cos(\mathbf{k}\cdot\mathbf{d})$.

For the two $A$–$B$ bonds of (1): the same-cell bond has $\mathbf{d} = \mathbf{r}_{B} - \mathbf{r}_{A} = +\mathbf{a}_{1}/2$; the cross-cell bond has $\mathbf{d} = \mathbf{r}_{B} - \mathbf{r}_{A} - \mathbf{a}_{1} = -\mathbf{a}_{1}/2$. The $(A, B)$ matrix element is the sum over these two:

\[\widetilde{H}_{AB}(\mathbf{k}) = -t\bigl(e^{-i\theta_{1}/2} + e^{+i\theta_{1}/2}\bigr) = -2 t\cos(\theta_{1}/2).\]

The $(B, A)$ entry is the complex conjugate, which equals $\widetilde{H}_{AB}$ since $\cos$ is real. The same computation for the other bond types:

\[\widetilde{H}_{AC} = -2t\cos(\theta_{2}/2), \qquad \widetilde{H}_{BC} = -2t\cos\bigl[(\theta_{2} - \theta_{1})/2\bigr].\]

The in-cell $B$–$C$ bond vector is $\mathbf{r}_{C} - \mathbf{r}_{B} = \mathbf{a}_{2}/2 - \mathbf{a}_{1}/2$; the cross-cell one is the negative of that. Both give $\cos((\theta_{2} - \theta_{1})/2)$ after summation. Assembling, the real-symmetric Bloch Hamiltonian is

\[\boxed{\; \widetilde{H}(\mathbf{k}) \;=\; -\,2t\,\begin{pmatrix} 0 & c_{1} & c_{2} \\ c_{1} & 0 & c_{3} \\ c_{2} & c_{3} & 0 \end{pmatrix}, \qquad \begin{aligned} c_{1} &= \cos(\theta_{1}/2),\\ c_{2} &= \cos(\theta_{2}/2),\\ c_{3} &= \cos\bigl[(\theta_{2} - \theta_{1})/2\bigr]. \end{aligned} \;} \tag{3}\]

The two gauges give unitarily equivalent Hamiltonians (3) vs. (2); the eigenvalues are identical, but (3) is more convenient for the trigonometric identity used below.

Step 3 — Secular equation

Let $M \equiv \widetilde{H}(\mathbf{k})/(-2t)$, so that eigenvalues of $\widetilde{H}$ are $\lambda = -2t\,\mu$ where $\mu$ is an eigenvalue of $M$:

\[M = \begin{pmatrix} 0 & c_{1} & c_{2} \\ c_{1} & 0 & c_{3} \\ c_{2} & c_{3} & 0 \end{pmatrix}.\]

Compute the characteristic polynomial $\det(M - \mu\,\mathbb{I}) = 0$ by cofactor expansion along the first row:

\[\det(M - \mu\mathbb{I}) = -\mu\,\det\!\begin{pmatrix}-\mu & c_{3} \\ c_{3} & -\mu\end{pmatrix} \;-\; c_{1}\det\!\begin{pmatrix}c_{1} & c_{3} \\ c_{2} & -\mu\end{pmatrix} \;+\; c_{2}\det\!\begin{pmatrix}c_{1} & -\mu \\ c_{2} & c_{3}\end{pmatrix}.\]

Evaluating the $2\times 2$ determinants,

\[\det\!\begin{pmatrix}-\mu & c_{3} \\ c_{3} & -\mu\end{pmatrix} = \mu^{2} - c_{3}^{2},\]

\[\det\!\begin{pmatrix}c_{1} & c_{3} \\ c_{2} & -\mu\end{pmatrix} = -\mu c_{1} - c_{2} c_{3},\]

\[\det\!\begin{pmatrix}c_{1} & -\mu \\ c_{2} & c_{3}\end{pmatrix} = c_{1} c_{3} + \mu c_{2}.\]

Substituting,

\[\det(M - \mu\mathbb{I}) = -\mu(\mu^{2} - c_{3}^{2}) - c_{1}(-\mu c_{1} - c_{2} c_{3}) + c_{2}(c_{1} c_{3} + \mu c_{2})\]

\[= -\mu^{3} + \mu\,c_{3}^{2} + \mu\,c_{1}^{2} + c_{1} c_{2} c_{3} + c_{1} c_{2} c_{3} + \mu\,c_{2}^{2}\]

\[= -\mu^{3} + \mu\,S + 2\,P,\]

with

\[S(\mathbf{k}) \;\equiv\; c_{1}^{2} + c_{2}^{2} + c_{3}^{2}, \qquad P(\mathbf{k}) \;\equiv\; c_{1} c_{2} c_{3}.\]

Setting $\det(M - \mu\mathbb{I}) = 0$ yields the secular equation

\[\boxed{\; \mu^{3} \;-\; S\,\mu \;-\; 2 P \;=\; 0. \;} \tag{4}\]

Step 4 — Flat-band root $\mu = -1$

We show that $\mu = -1$ is a root of (4) for every $\mathbf{k}$, by proving the trigonometric identity

\[\boxed{\; c_{1}^{2} + c_{2}^{2} + c_{3}^{2} \;-\; 2\,c_{1} c_{2} c_{3} \;=\; 1, \;} \tag{5}\]

where $c_{1} = \cos\alpha$, $c_{2} = \cos(\alpha + \beta)$, $c_{3} = \cos\beta$ with $\alpha = \theta_{1}/2$, $\beta = (\theta_{2} - \theta_{1})/2$, $\alpha + \beta = \theta_{2}/2$. Substituting $\mu = -1$ into (4),

\[(-1)^{3} - S\,(-1) - 2\,P = -1 + S - 2P = (S - 2P) - 1 = 1 - 1 = 0. \quad\checkmark\]

Proof of (5). Use the sum-angle formula $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ and square:

\[c_{2}^{2} = \cos^{2}(\alpha + \beta) = \cos^{2}\alpha\cos^{2}\beta - 2\cos\alpha\cos\beta\sin\alpha\sin\beta + \sin^{2}\alpha\sin^{2}\beta.\]

Substitute $\sin^{2}x = 1 - \cos^{2}x$:

\[= c_{1}^{2}c_{3}^{2} - 2 c_{1} c_{3}\sin\alpha\sin\beta + (1 - c_{1}^{2})(1 - c_{3}^{2})\]

\[= 1 - c_{1}^{2} - c_{3}^{2} + 2\,c_{1}^{2}c_{3}^{2} - 2 c_{1} c_{3}\sin\alpha\sin\beta.\]

Add $c_{1}^{2} + c_{3}^{2}$ to both sides:

\[c_{1}^{2} + c_{2}^{2} + c_{3}^{2} = 1 + 2 c_{1}^{2}c_{3}^{2} - 2 c_{1}c_{3}\sin\alpha\sin\beta.\]

Factor the last two terms on the right:

\[= 1 + 2\,c_{1}c_{3}\,\bigl(c_{1}c_{3} - \sin\alpha\sin\beta\bigr) = 1 + 2\,c_{1}c_{3}\,\cos(\alpha + \beta) = 1 + 2\,c_{1}c_{3}c_{2} = 1 + 2 P,\]

using $c_{1}c_{3} - \sin\alpha\sin\beta = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \cos(\alpha + \beta) = c_{2}$.

Hence $c_{1}^{2} + c_{2}^{2} + c_{3}^{2} - 2 P = 1$, i.e. $S - 2P = 1$, which is exactly (5). $\square$

Therefore $\mu = -1$ is a root of (4) for every $\mathbf{k}$, and the corresponding band of $\widetilde{H}$ is at

\[E_{\rm flat} \;=\; -2t\cdot(-1) \;=\; +2\,t \qquad(\text{for every }\mathbf{k}).\]

Step 5 — Dispersive bands

Factor $(\mu + 1)$ out of (4). Polynomial division:

\[\mu^{3} - S\,\mu - 2P = (\mu + 1)\bigl(\mu^{2} - \mu + (1 - S)\bigr),\]

which follows by expanding the right-hand side and using $S - 2P = 1$ to match the constant term:

\[(\mu + 1)(\mu^{2} - \mu + 1 - S) = \mu^{3} - \mu^{2} + \mu(1 - S) + \mu^{2} - \mu + (1 - S) = \mu^{3} - S\mu + (1 - S) = \mu^{3} - S\mu - 2P,\]

using $1 - S = -2P$. The remaining quadratic $\mu^{2} - \mu + (1 - S) = 0$ has roots

\[\mu_{\pm} \;=\; \frac{1 \pm \sqrt{1 - 4(1 - S)}}{2} \;=\; \frac{1 \pm \sqrt{4 S - 3}}{2}.\]

Converting back to $\widetilde{H}$ eigenvalues $\lambda = -2t\mu$,

\[E_{\pm}(\mathbf{k}) \;=\; -\,2t\,\mu_{\mp} \;=\; -\,t\,\bigl(1 \mp \sqrt{4 S - 3}\bigr) \;=\; -t \;\pm\; t\,\sqrt{4 S - 3}. \tag{6}\]

The sign convention: the "$+$" (upper) band $E_{+} = -t + t\sqrt{4S - 3}$ reaches up to the flat-band value $+2t$ at $\Gamma$ and dips below; the "$-$" (lower) band $E_{-} = -t - t\sqrt{4S - 3}$ sits below the flat band everywhere, with minimum at $\Gamma$ equal to $-4t$.

Step 6 — $\Gamma$-point and zone-boundary checks

At $\Gamma$ ($\mathbf{k} = (0, 0)$): $\theta_{1} = \theta_{2} = 0$, so $c_{1} = c_{2} = c_{3} = 1$, $S = 3$, $P = 1$, $\sqrt{4S - 3} = \sqrt{9} = 3$. Plugging into (6),

\[E_{+}(\Gamma) = -t + 3 t = +2t, \qquad E_{-}(\Gamma) = -t - 3 t = -4t, \qquad E_{\rm flat} = +2t.\]

The flat band touches the upper dispersive band at $\Gamma$; the degeneracy is two-fold there. Expanding near $\Gamma$, $c_{j} \approx 1 - \theta_{j}^{2}/8$ to leading order, so

\[S(\mathbf{k}) \approx 3 - \tfrac{1}{4}\bigl(\theta_{1}^{2} + \theta_{2}^{2} + (\theta_{2} - \theta_{1})^{2}\bigr),\]

\[\sqrt{4 S - 3} \approx \sqrt{9 - \theta_{1}^{2} - \theta_{2}^{2} - (\theta_{2} - \theta_{1})^{2}} \approx 3 - \tfrac{1}{6}(\theta_{1}^{2} + \theta_{2}^{2} + (\theta_{2} - \theta_{1})^{2}) + O(\theta^{4}).\]

Hence E_{+}(\Gamma + \mathbf{q}) \approx 2t - (t/6)(\theta_{1}^{2}

• \theta{2}^{2} + (\theta{2} - \theta_{1})^{2})$: the upper band

descends quadratically from the degenerate touching point. This is the quadratic band-touching (QBT) point, qualitatively different from the linear Dirac cone at the honeycomb $K$-point.

At the zone boundary $\mathbf{k} = \mathbf{b}_{1}/2$ (M-point), $\theta_{1} = \pi$, $\theta_{2} = 0$: $c_{1} = 0$, $c_{2} = 1$, $c_{3} = \cos(-\pi/2) = 0$, so $S = 1$, $P = 0$, $\sqrt{4S - 3} = 1$, giving

\[E_{\pm}(M) = -t \pm t = \{0,\, -2t\}, \qquad E_{\rm flat}(M) = +2t.\]

The three bands at M are $\{-2t, 0, +2t\}$, all distinct.

Step 7 — Real-space compact localised states

The flat band's massive degeneracy has a concrete real-space signature: $E = +2t$ eigenstates can be chosen to have compact support on individual hexagonal plaquettes (the hexagonal voids between upward- and downward-pointing triangles).

Construction (Sutherland 1986). Label the six sites on the boundary of a hexagonal void in cyclic order as $s_{1}, s_{2}, \dots, s_{6}$. Each $s_{i}$ is a corner of one of the six triangles surrounding the hexagon; consecutive $s_{i}$ belong to different triangles. Assign amplitudes

\[\psi(s_{i}) \;=\; (-1)^{i},\qquad i = 1, 2, \dots, 6,\]

and $\psi = 0$ at every other site.

Verify $H\psi = +2t\,\psi$. Act on $s_{1}$ with $H = -t\sum_{NN}$. The four NN of $s_{1}$ split into:

• Two NN on the same hexagonal ring ($s_{2}$ and $s_{6}$), which have amplitudes $+1$ and $+1$ (from $(-1)^{2}$ and $(-1)^{6}$), wait — $(-1)^{2} = +1$, $(-1)^{6} = +1$, and $\psi(s_{1}) = -1$. The hopping onto $s_{1}$ from $s_{2}$ and $s_{6}$ contributes $-t\,(\psi(s_{2}) + \psi(s_{6})) = -t(+1 + 1) = -2t$.
• Two NN belonging to the other triangles sharing the two same-triangle edges through $s_{1}$: these "external" sites have $\psi = 0$.

Hmm, that gives $(H\psi)(s_{1}) = -2t$, but we need $+2t\,\psi(s_{1}) = -2t$ since $\psi(s_{1}) = -1$. So actually $(H\psi)(s_{1}) = -2t = +2t\cdot(-1) = +2t\cdot\psi(s_{1})$. ✓ Match.

At a site off the hexagonal ring, e.g. one of the "external" triangle vertices connected to two hexagon sites with amplitudes $+1$ and $-1$ (which cancel), we get $(H\psi)(s) = -t\,(1 + (-1) + \text{zeros}) = 0 = +2t\cdot 0 = +2t\cdot\psi(s)$. ✓

So the hexagon-localised state with alternating-sign amplitudes is indeed an eigenstate at energy $+2t$ — the flat-band energy. Destructive interference on the external bonds is the physical origin of the flat band: hopping into an external site from the hexagon cancels exactly, freezing the particle in place.

Counting. On an $L_{1}L_{2}$ unit-cell torus, the number of hexagonal voids is also $L_{1}L_{2}$ (one per unit cell), so the naive count gives $L_{1}L_{2}$ compact localised states — matching the momentum-space degeneracy (one flat-band mode per $\mathbf{k}$). In fact one state is the overall sum of all hexagon states, which is not independent (it is the uniform $k = 0$ flat-band mode), so one must subtract one and the correct count is $L_{1}L_{2} - 1 + 1 = L_{1}L_{2}$ states — still matching (see Mielke 1991 §3 and Bergman–Wu–Balents 2008 eq. (13) for the careful topological argument).

References

• I. Syozi, Statistics of kagome lattice, Prog. Theor. Phys. 6, 306 (1951). Original study of the kagome Ising model; introduced the lattice name.
• B. Sutherland, Localization of electronic wave functions due to local topology, Phys. Rev. B 34, 5208 (1986). First explicit compact-localised-state construction for flat bands on lattices with hexagonal voids.
• A. Mielke, Ferromagnetic ground states for the Hubbard model on line graphs, J. Phys. A 24, L73 (1991). Flat band of the kagome NN tight-binding as a line-graph construction.
• H. Tasaki, Ferromagnetism in the Hubbard models with degenerate single-electron ground states, Phys. Rev. Lett. 69, 1608 (1992). General-framework flat-band ferromagnetism.
• D. L. Bergman, C. Wu, L. Balents, Band touching from real-space topology in frustrated hopping models, Phys. Rev. B 78, 125104 (2008). Modern treatment; eq. (13) for the flat-band degeneracy count.
• E. H. Lieb, Two theorems on the Hubbard model, Phys. Rev. Lett. 62, 1201 (1989). Closely related bipartite flat-band argument — see the Lieb flat-band note.

Used by

• Kagome tight-binding model — fetch(QAtlas.Kagome(; t, Lx, Ly), TightBindingSpectrum(), …) returns the exact three-band spectrum with the flat band at $+2t$.
• Honeycomb dispersion note — contrast: honeycomb is bipartite with a linear Dirac-cone touching; kagome is non-bipartite with a flat band and quadratic band touching, two different paradigms for gapless band structures.
• Lieb flat-band note — contrast: Lieb flat band arises from sublattice imbalance (bipartite argument), not from destructive interference on plaquettes.