LESSON NOTES · 02

A one-dimensional band, derived

Prerequisites: Localized-basis lesson; Fourier phases; delta functions

1. A periodic orthogonal chain

Take \(N\) orthonormal spinless sites on a ring, spacing \(a\), nearest-neighbor hopping \(-t\) with \(t>0\). Indices are periodic. The normalized Bloch state satisfies the ring boundary condition. There are \(N\) distinct wavevectors per orbital and spin channel in a Brillouin zone; its two endpoints are equivalent and must not both be counted.

\[ \begin{aligned}\hat H&=\varepsilon_0\sum_j|j\rangle\langle j|-t\sum_j(|j\rangle\langle j+1|+|j+1\rangle\langle j|),\\|k\rangle&=N^{-1/2}\sum_j e^{ikja}|j\rangle,\quad e^{ikNa}=1\Rightarrow k=\frac{2\pi m}{Na}.\end{aligned} \]

2. Act on the Bloch amplitudes

The neighbors of site \(j\) give factors \(e^{-ika}\) and \(e^{ika}\). Their sum is \(2\cos ka\). Negative hopping puts the minimum at \(k=0\); the width is \(4t\). Band folding does not change the physical states.

\[ \hat H|k\rangle=[\varepsilon_0-t(e^{ika}+e^{-ika})]|k\rangle,\quad E(k)=\varepsilon_0-2t\cos ka,\quad -\pi/a\leq k<\pi/a,\quad W=4t. \]

3. Change variables to obtain the DOS

Normalize DOS per site per spin channel. In the infinite-size limit use the delta-function identity to sum inverse band slopes at all roots. Generic interior energies have two roots. The edge divergences are integrable van Hove singularities, not infinitely many states. Finite chains have discrete delta peaks; artificial broadening is a plotting choice.

\[ \begin{aligned}g(E)&=\frac{a}{2\pi}\int_{-\pi/a}^{\pi/a}\delta(E-E(k))\,dk=\frac{a}{2\pi}\sum_{roots}\frac1{|dE/dk|},\\\frac{dE}{dk}&=2ta\sin ka,\\g(E)&=\frac1{\pi\sqrt{4t^2-(E-\varepsilon_0)^2}}\quad(|E-\varepsilon_0|<2t),\\g(E)&=0\quad\text{outside the band},\qquad\int g(E)dE=1.\end{aligned} \]

4. Velocity and curvature

Restore \(\hbar\) to show dimensions explicitly. Expand the cosine near the minimum and match its quadratic term to \(\hbar^2k^2/(2m^*)\). The effective mass is local to that extremum, not a universal mass throughout the band.

\[ v(k)=\frac1\hbar\frac{dE}{dk}=\frac{2ta}{\hbar}\sin ka,\quad E(k\simeq0)=\varepsilon_0-2t+ta^2k^2,\quad m^*=\frac{\hbar^2}{2ta^2}. \]

Teaching schematic drawn from the equations or algorithm steps in this lesson; not measured data.

Worked model and filling

Toy \(\varepsilon_0=0\), \(t=1\) eV, \(a=0.25\) nm gives \(W=4\) eV and \(g(0)=1/(2\pi)=0.1592\) eV\(^{-1}\) per site per spin. Using \(\hbar^2/(2m_e)=3.80998\) eV Å\(^2\) gives \(m^*=0.6096m_e\). At spinless half filling \(k_F=\pi/(2a)\) and \(E_F=0\). With spin degeneracy and no splitting, one electron per site is half filling; two is full filling.

Check yourself and limits

Why does a flat band segment give large DOS?

Answer and reasoning

A small \(|dE/dk|\) packs many wavevectors into a narrow energy interval.

Does this prove a real half-filled one-dimensional material is metallic?

Answer and reasoning

No: interactions, distortion, disorder and additional orbitals can alter the conclusion.

References and further reading

• MIT 6.730, Lecture 13: Tight-binding LCAO
• David Tong, Applications of Quantum Mechanics

Derivations and numerical examples here are original teaching constructions, not copied passages or reported research data.

---

← Previous lesson · Pathway overview · Next lesson →