Lieb Lattice Flat Band at $E = 0$

Main result

For the nearest-neighbour tight-binding model on the Lieb lattice (three sublattices $A, B, C$ per unit cell, $A$ at square corners, $B$ and $C$ at the two edge midpoints) with hopping amplitude $t$, the $3\times 3$ Bloch Hamiltonian has one perfectly flat eigenvalue across the entire Brillouin zone,

\[\boxed{\; E_{\rm flat}(\mathbf{k}) \;=\; 0 \qquad\text{for every }\mathbf{k} \in \mathrm{BZ}, \;}\]

and two dispersive bands symmetric about $E = 0$,

\[\boxed{\; E_{\pm}(\mathbf{k}) \;=\; \pm\,2\,t\, \sqrt{\,\cos^{2}\!\tfrac{\theta_{1}}{2} + \cos^{2}\!\tfrac{\theta_{2}}{2}\,}, \qquad \theta_{j} \equiv \mathbf{k}\cdot\mathbf{a}_{j}. \;}\]

The dispersive bands touch the flat band at a single point — the M point $\mathbf{k} = (\pi/a, \pi/a)$ at the Brillouin-zone corner — where $\cos(\theta_{1}/2) = \cos(\theta_{2}/2) = 0$ and $E_{\pm}(M) = 0$. Expansion near $M$ yields a linear (Dirac- like) touching of the three bands, all meeting at $E = 0$.

The flat band's origin is the bipartite sublattice imbalance of the Lieb lattice: the $A$ sublattice has $1$ site per unit cell while $B\cup C$ has $2$. Lieb's theorem (1989) guarantees at least $|A| - |B\cup C|$ zero-energy eigenvectors in magnitude, which for $N_{c}$ unit cells is exactly $N_{c}$ — one per $\mathbf{k}$, i.e. a full flat band.

Setup

Lattice geometry

The Lieb lattice is a decoration of the square lattice: starting from a plain square of primitive vectors $\mathbf{a}_{1} = a\,(1, 0)$, $\mathbf{a}_{2} = a\,(0, 1)$, add one site at the midpoint of each horizontal edge and one at the midpoint of each vertical edge. Denote these

• \[A\]

  (corner / "copper"): position $\mathbf{r}_{A} = (0, 0)$,

• \[B\]

  (horizontal-edge midpoint / "oxygen $x$"): $\mathbf{r}_{B} = \mathbf{a}_{1}/2 = (a/2, 0)$,

• \[C\]

  (vertical-edge midpoint / "oxygen $y$"): $\mathbf{r}_{C} = \mathbf{a}_{2}/2 = (0, a/2)$.

(The naming anticipates the CuO$_{2}$ plane of cuprates, where $A$ carries the Cu orbital and $B, C$ carry the O$_{x}, \text{O}_{y}$ orbitals.) Set $a = 2$ so that the NN bond length |\mathbf{r}_{B}

• \mathbf{r}{A}| = 1$; then \mathbf{a}{1} = (2, 0)$,

\[\mathbf{a}_{2} = (0, 2)\]

.

Each $A$-site at $\mathbf{R}$ has four NN: two $B$-sites along $\pm \hat{x}$ and two $C$-sites along $\pm \hat{y}$. Each $B$-site has only two NN: the two $A$-sites flanking it along $\hat{x}$. Similarly each $C$-site has two NN (the two $A$-sites along $\hat{y}$). There is no direct $B$–$C$ hopping — $B$ and $C$ sites are not nearest neighbours (the $B$–$C$ distance is $\sqrt{(a/2)^{2} + (a/2)^{2}} = a/\sqrt{2}$, which is a second- neighbour distance).

Explicitly, from the $A$-site at $\mathbf{R}$:

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

The lattice is bipartite with partition $(A) \sqcup (B\cup C)$: every NN bond connects an $A$-site to a $B$- or $C$-site.

Bipartite structure and sublattice imbalance

On an $L_{1}\times L_{2}$ unit-cell torus ($N_{c} = L_{1}L_{2}$),

\[|A| = N_{c},\qquad |B| = N_{c},\qquad |C| = N_{c}, \qquad |B\cup C| = 2 N_{c}.\]

The imbalance $|B\cup C| - |A| = N_{c}$ is the key input for Lieb's theorem (Step 5), which predicts at least $N_{c}$ zero-energy eigenvectors — the entire flat band.

Hamiltonian

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

with the sum over NN $A$–$B$ and $A$–$C$ bonds only (no $B$–$C$).

Goal

Derive the $3\times 3$ Bloch Hamiltonian, find the exact flat-band eigenvector, and present the bipartite-imbalance argument that explains its origin.

Calculation

Step 1 — Fourier transform to the Bloch Hamiltonian

Use the site-centered Fourier convention (as in the kagome note, Step 2):

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

Each NN bond between $\alpha$ at $\mathbf{r}_{\alpha} + \mathbf{R}$ and $\beta$ at $\mathbf{r}_{\beta} + \mathbf{R}'$ contributes $-t\,e^{-i\mathbf{k}\cdot\mathbf{d}}$ in the Bloch Hamiltonian $(\alpha, \beta)$ block, with \mathbf{d} \equiv (\mathbf{r}_{\beta}

• \mathbf{R}') - (\mathbf{r}_{\alpha} + \mathbf{R})$ the bond

vector. A Hermitian-conjugate pair sums to $-2t\cos(\mathbf{k}\cdot \mathbf{d})$.

For the two $A$–$B$ bonds of (1): same-cell $\mathbf{d} = +\mathbf{a}_{1}/2$, cross-cell $\mathbf{d} = -\mathbf{a}_{1}/2$. Summing,

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

Similarly $\widetilde{H}_{AC}(\mathbf{k}) = -2t\cos(\theta_{2}/2)$. The $(B, C)$ entry is zero (no direct $B$–$C$ hopping). Assembling,

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

The matrix is real symmetric and has the generic bipartite block structure

\[\widetilde{H}(\mathbf{k}) \;=\; \begin{pmatrix} 0 & \mathbf{q}^{T}(\mathbf{k}) \\ \mathbf{q}(\mathbf{k}) & \mathbf{0}_{2\times 2}\end{pmatrix}, \qquad \mathbf{q}(\mathbf{k}) \;=\; -2t\,\begin{pmatrix} \cos(\theta_{1}/2) \\ \cos(\theta_{2}/2)\end{pmatrix} \in \mathbb{R}^{2}. \tag{3}\]

The $2\times 2$ lower-right zero block reflects the absence of direct $B$–$C$ hopping.

Step 2 — Secular equation

Set $\lambda = -2t\,\mu$ so that $\mu$ is an eigenvalue of $M\equiv\widetilde{H}/(-2t)$. By cofactor expansion along the first row of $M - \mu\mathbb{I}$ with $s_{1} = \cos(\theta_{1}/2)$, $s_{2} = \cos(\theta_{2}/2)$,

\[\det(M - \mu\mathbb{I}) = -\mu\,\det\!\begin{pmatrix}-\mu & 0 \\ 0 & -\mu\end{pmatrix} - s_{1}\,\det\!\begin{pmatrix}s_{1} & 0 \\ s_{2} & -\mu\end{pmatrix} + s_{2}\,\det\!\begin{pmatrix}s_{1} & -\mu \\ s_{2} & 0\end{pmatrix}\]

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

Setting the determinant to zero,

\[\boxed{\; \mu\,\bigl[\,\mu^{2} - (s_{1}^{2} + s_{2}^{2})\,\bigr] \;=\; 0. \;} \tag{4}\]

The three roots are

\[\mu_{0} = 0, \qquad \mu_{\pm} = \pm\,\sqrt{s_{1}^{2} + s_{2}^{2}}.\]

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

\[E_{\rm flat} \;=\; 0 \qquad\text{(all }\mathbf{k}\text{)}, \qquad E_{\pm}(\mathbf{k}) \;=\; \mp\,2 t\,\sqrt{s_{1}^{2} + s_{2}^{2}} \;=\; \pm\,2 t\,\sqrt{\cos^{2}\!\tfrac{\theta_{1}}{2} + \cos^{2}\!\tfrac{\theta_{2}}{2}}. \tag{5}\]

The flat band sits at $E = 0$ throughout the BZ, independent of $\mathbf{k}$.

Step 3 — Explicit flat-band eigenvector

Solve $\widetilde{H}(\mathbf{k})\,\mathbf{v} = 0$. In components $\mathbf{v} = (v_{A}, v_{B}, v_{C})^{T}$, (2) gives

\[\text{row }A: \quad -2t\bigl[s_{1} v_{B} + s_{2} v_{C}\bigr] = 0,\]

\[\text{row }B: \quad -2t\,s_{1} v_{A} = 0,\]

\[\text{row }C: \quad -2t\,s_{2} v_{A} = 0.\]

For generic $\mathbf{k}$ with $s_{1}, s_{2} \ne 0$, rows $B$ and $C$ force $v_{A} = 0$. Row $A$ then requires $s_{1} v_{B} + s_{2} v_{C} = 0$, a single linear constraint on $(v_{B}, v_{C})$. A convenient normalisation picks

\[\boxed{\; |\psi_{\rm flat}(\mathbf{k})\rangle \;=\; \frac{1}{\sqrt{s_{1}^{2} + s_{2}^{2}}}\, \bigl(\,0,\;\; -s_{2},\;\; +s_{1}\,\bigr)^{T}, \;} \tag{6}\]

so that $\langle\psi_{\rm flat}|\psi_{\rm flat}\rangle = 1$ for all $\mathbf{k}$ except the single $M$-point $\theta_{1} = \theta_{2} = \pi$ where both $s_{j} = 0$ (see Step 6). The flat-band wavefunction has zero weight on the $A$ sublattice for every $\mathbf{k}$ — this is the algebraic expression of the bipartite imbalance argument (Step 5).

Step 4 — Dispersive bands and the $E_{\pm}$ eigenvectors

The $E_{\pm}$ eigenvectors satisfy $\widetilde{H}\,\mathbf{v}_{\pm} = \pm E \mathbf{v}_{\pm}$ with $E = 2t\sqrt{s_{1}^{2} + s_{2}^{2}}$. From rows $B$ and $C$,

\[-2t\,s_{1}\,v_{A} = \pm E\,v_{B}, \qquad -2t\,s_{2}\,v_{A} = \pm E\,v_{C},\]

giving $v_{B} = -2t s_{1} v_{A}/(\pm E)$, $v_{C} = -2t s_{2} v_{A}/(\pm E)$. Normalising,

\[|\psi_{\pm}(\mathbf{k})\rangle \;=\; \frac{1}{\sqrt{2(s_{1}^{2} + s_{2}^{2})}}\, \begin{pmatrix}\pm\sqrt{s_{1}^{2} + s_{2}^{2}} \\ -s_{1} \\ -s_{2}\end{pmatrix}. \tag{7}\]

Both $\psi_{\pm}$ have non-zero weight on the $A$ sublattice; only the flat-band state (6) is $A$-null. The three bands are orthogonal for every $\mathbf{k}$ where $E(\mathbf{k}) > 0$.

Step 5 — Bipartite-imbalance argument (Lieb 1989)

The flat band of (5) is no accident. It follows from a general theorem on bipartite lattices: if a tight-binding Hamiltonian has the block structure $M = \begin{pmatrix}0 & Q\\ Q^{T} & 0\end{pmatrix}$ with $Q \in \mathbb{R}^{N_{1}\times N_{2}}$ ($N_{1}$ sites in one sublattice, $N_{2}$ in the other), then the at-least-$|N_{1} - N_{2}|$-degeneracy at $E = 0$ follows from

\[M\,\mathbf{v} = 0 \quad\Longleftrightarrow\quad Q\,\mathbf{v}_{2} = 0\quad\text{and}\quad Q^{T}\,\mathbf{v}_{1} = 0.\]

By rank-nullity, $\dim\ker Q = N_{2} - \mathrm{rank}(Q) \ge N_{2} - \min(N_{1}, N_{2}) = \max(0, N_{2} - N_{1})$, so there are at least $\max(0, N_{2} - N_{1})$ zero-energy eigenvectors supported on sublattice 2. Similarly at least $\max(0, N_{1} - N_{2})$ such vectors on sublattice 1. In total, at least $|N_{1} - N_{2}|$ zero-energy eigenvectors regardless of the detailed structure of $Q$.

For the Lieb lattice, sublattice 1 is $A$ ($N_{1} = N_{c}$) and sublattice 2 is $B\cup C$ ($N_{2} = 2 N_{c}$). The imbalance is $|N_{2} - N_{1}| = N_{c}$, giving $\ge N_{c}$ zero-energy states on $B\cup C$ (consistent with our real-space expectation "$v_{A} = 0$"). Distributing these over the $N_{c}$ values of $\mathbf{k}$ gives one zero mode per momentum, i.e. a complete flat band. This is Lieb's theorem (Phys. Rev. Lett. 62, 1201 (1989)) specialised to the Lieb-lattice adjacency matrix.

The argument generalises: any bipartite lattice with unequal sublattice sizes has at least $|N_{1} - N_{2}|$ zero-energy modes, and if the lattice is connected (every site reachable via NN bonds) the degeneracy saturates the bound. For the Lieb lattice at half-filling this macroscopic degeneracy produces flat-band ferromagnetism in the Hubbard model (Lieb 1989, Mielke 1991, Tasaki 1998).

Step 6 — Band-touching at the $M$-point

The flat band and the dispersive bands (5) coincide at $E = 0$ wherever $s_{1}^{2} + s_{2}^{2} = 0$, i.e. $\cos(\theta_{1}/2) = \cos(\theta_{2}/2) = 0$, i.e. $\theta_{1} = \theta_{2} = \pi$. This is the BZ corner (the M-point) at $\mathbf{k} = (\pi/a, \pi/a)$.

Expand (5) near $M$. With our normalisation $\mathbf{a}_{j} = 2 \hat{e}_{j}$, we have $\theta_{j} = \mathbf{k}\cdot\mathbf{a}_{j} = 2 k_{j}$; the $M$-point in the $\mathbf{k}$-variable is $\mathbf{M} = (\pi/2, \pi/2)$. Set $\mathbf{q} = \mathbf{k} - \mathbf{M}$, so $\theta_{j} = \pi + 2 q_{j}$ and

\[\cos(\theta_{j}/2) \;=\; \cos(\pi/2 + q_{j}) \;=\; -\sin q_{j} \;\approx\; -q_{j} + O(q_{j}^{3}).\]

Therefore

\[s_{1}^{2} + s_{2}^{2} \;\approx\; q_{1}^{2} + q_{2}^{2} \;=\; |\mathbf{q}|^{2},\]

and

\[E_{\pm}(\mathbf{M} + \mathbf{q}) \;=\; \pm\,2t\,\sqrt{q_{1}^{2} + q_{2}^{2}} \;=\; \pm\,2t\,|\mathbf{q}| \;+\; O(|\mathbf{q}|^{3}), \tag{8}\]

so the dispersive bands touch the flat band linearly — a three-way Dirac-like touching at $M$. The two dispersive bands plus the flat band all meet at $E = 0$, with the dispersive cones meeting the plane tangentially along a cone and a flat pancake respectively. The Fermi velocity is $v_{F} = 2t$.

Step 7 — Limiting-case and finite-size checks

(i) Half-filling. With $3 N_{c}$ sites, half-filling is $3 N_{c}/2$ electrons. If $N_{c}$ is odd the count is half-integer so one assumes a grand-canonical ensemble. At $T = 0$ the $N_{c}$ states of the lower dispersive band ($E_{-} < 0$) are filled; the flat band at $E = 0$ is half-filled (in the spinless case) or fully-filled per spin (in the spinful case, with additional flat spin degeneracy). This macroscopic degeneracy is the input for flat-band ferromagnetism (Lieb 1989).

(ii) $\Gamma$-point. At $\mathbf{k} = (0, 0)$: $s_{1} = s_{2} = 1$, $s_{1}^{2} + s_{2}^{2} = 2$, so $E_{\pm}(\Gamma) = \pm 2t \sqrt{2}$. Three-band spectrum at $\Gamma$: $\{-2t\sqrt{2},\, 0,\, +2t\sqrt{2}\}$, all distinct (bandwidth $= 4t\sqrt{2}$).

(iii) Bipartite chiral symmetry. The block form (3) makes the spectrum symmetric about $E = 0$: for every dispersive eigenvalue $+E(\mathbf{k})$ there is $-E(\mathbf{k})$. The sublattice operator $\Sigma = \mathrm{diag}(+1, -1, -1)$ (positive on $A$, negative on $B\cup C$) satisfies $\Sigma \widetilde{H}(\mathbf{k}) \Sigma = -\widetilde{H}(\mathbf{k})$, the algebraic expression of bipartiteness.

(iv) Honeycomb vs Lieb comparison. The honeycomb lattice (see bloch-honeycomb-dispersion) is also bipartite but with equal sublattice sizes $|A| = |B|$, so Lieb's theorem gives zero zero-mode lower bound — and indeed the honeycomb spectrum has Dirac touching but no flat band. The Lieb lattice differs precisely by its $1 : 2$ sublattice imbalance.

(v) Kagome vs Lieb comparison. The kagome flat band sits at $+2t$, not at $E = 0$, and arises from destructive interference on hexagonal plaquettes — not from a bipartite sublattice imbalance (the kagome lattice is not bipartite). The two flat-band mechanisms are genuinely distinct paradigms.

References

• E. H. Lieb, Two theorems on the Hubbard model, Phys. Rev. Lett. 62, 1201 (1989). Bipartite sublattice imbalance → ground-state spin $S = |N_{A} - N_{B}|/2$; flat-band ferromagnetism in the half-filled Hubbard model.
• A. Mielke, Ferromagnetism in the Hubbard model on line graphs and further considerations, J. Phys. A 24, 3311 (1991). Flat bands on line graphs; Lieb-lattice case as a decorated bipartite lattice.
• H. Tasaki, From Nagaoka's ferromagnetism to flat-band ferromagnetism and beyond, Prog. Theor. Phys. 99, 489 (1998). Rigorous review of flat-band magnetism mechanisms including the Lieb / Mielke constructions.
• Y.-F. Wang, C.-D. Gong, Frustration induced non-trivial flat bands in Lieb lattice, Phys. Rev. B 85, 214428 (2012). Modern treatment including perturbations.

Used by

• Lieb tight-binding model — fetch(QAtlas.Lieb(; t, Lx, Ly), TightBindingSpectrum(), …) returns the finite-size three-band spectrum, with exact zeros at $N_{c}$ momenta (the full flat band) and the dispersive $\pm E(\mathbf{k})$.
• Honeycomb dispersion note — bipartite lattice with equal sublattice sizes gives a linear Dirac cone but no flat band; the Lieb $1:2$ imbalance produces the flat band.
• Kagome flat-band note — a second, fundamentally different flat-band paradigm: destructive interference on a non-bipartite three-sublattice lattice.