CHEM 548: Materials Chemistry

Lecture 15: 1D Band Structures and Bloch Functions

Mon 10/26/2026 · Meeting 17

Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 6, §§6.1–6.3

Learning Objectives: Intro to Band Theory

Learning Objectives — Intro to Band Theory (1D Hydrogen Chain)

Understand why crystals require band structures instead of molecular orbitals.
Define a crystal orbital as a Bloch function (periodic basis × lattice phase term).
Interpret k as both a wavevector and crystal momentum.
Identify bonding (k=0) and antibonding (k=π/a) limits in the 1st Brillouin zone.
Relate band width to orbital overlap (wide = delocalized, narrow = localized).
Recognize density of states (DOS) as a summary of orbital distribution vs. energy.

Electronic Band Structure

Molecular orbital diagram for a finite molecule beside the band structure diagram for an infinite crystal.

Solids and Surfaces (Hoffmann, 1989)

Cover of Roald Hoffmann's 1989 book Solids and Surfaces beside a photo of Hoffmann.

From Molecules to an Infinite Chain

Energy-level diagrams for H2 through H50 rings ending at the infinite hydrogen chain continuum.

To move toward understanding electronic structure in solids, we’ll start small and scale up.

Begin with molecules made only of hydrogen — each atom contributing one 1s orbital — and build larger systems until we reach an infinite, one-dimensional chain with translational symmetry
Smallest unit H₂: bonding and antibonding molecular orbitals
Rings of 6, 10, 14, or 50 H atoms show key trends: the number of orbitals equals the number of atoms, half are filled
The energy spacing between orbitals decreases as the system grows
Infinite limit: the levels form a continuum — half filled — with more closely spaced orbitals at the top and bottom of the energy range than in the middle

Next, we will recall how we built molecular orbitals from atomic orbitals.

Atomic and Molecular Orbitals

Hydrogen 1s atomic orbital wavefunction beside a molecular orbital built as a linear combination of atomic orbitals.

Crystal Orbitals: Bloch Functions

The Bloch function psi(x) = e^(ikx) u(x) with its wavefunction, periodicity, and basis-set parts labeled.

$\psi(x)=u(x)e^{ikx}$

First Brillouin Zone

Build asking what k is and what values it can take, developed on a ring of N hydrogen atoms.

Everything so far should feel familiar.

The u(x) term is just the hydrogen 1s wave function. The e^{ikx} term involves x, the position in the chain. But what about k? What does it represent physically, and what values can it take?

To answer that, let's start by determining the possible range of k values. Consider our one-dimensional chain, not infinite for now, but a ring of hydrogen atoms.

[advance: bring up the ring of N hydrogen atoms]

We can take the ring large enough that its curvature is negligible, say 10,000 hydrogen atoms.

Starting half an angstrom from the atom we call the origin, traveling once around the ring means moving through 10,000 lattice spacings, each of length A, the distance between atoms. After going 10,000A, we return to the same point, so the wavefunction must have the same value there.

[advance: show the periodic boundary condition]

The wavefunction must satisfy \psi(x) = \psi(x + N A), where N is the number of atoms and A is the lattice spacing. In the limit of an infinite chain, this becomes a periodic boundary condition: the wavefunction’s value at one end equals its value at the other.

We write \psi(x) = e^{ikx}u(x). At x + N A, \psi(x + N A) = e^{ik(x + N A)}u(x + N A).

Since u(x) is identical in each unit cell and periodic with A, u(x + N A) = u(x) and it cancels out.

That leaves e^{ikx} = e^{ik(x + N A)} = e^{ikx} e^{ik N A}.

Next, we will apply Euler's relationship to work out which k values are allowed.

First Brillouin Zone (cont)

Euler's relationship applied to e^(ikNa), leading to the allowed k values and the first Brillouin zone.

Let's use Euler's relationship now just for the second term.

Write e to the ikna as cosine of kna plus i sine of kna.

The term in brackets has to be one for this equality to hold. If it takes any other value, the equality would be violated.

[advance: show that the bracketed term must equal one]

In radians, the term in brackets must equal an integer multiple of 2 pi for the wavefunction to repeat. Cosine of 2 pi, 4 pi, and so on is 1, and sine of those values is 0.

So k, multiplied by the number of atoms in the chain (N) and the spacing between them (A), must equal an integer times 2 pi.

[advance: rearrange for the allowed values of k]

Rearranging gives k equal to an integer n times 2 pi divided by the number of atoms N and the lattice spacing a.

For a ring of 10,000 hydrogen atoms, there are 10,000 molecular orbitals. Allowing n to take both positive and negative values, plus or minus 1, 2, 3, and so on up to 5,000, produces one orbital for each atom.

As the ring grows larger, the allowed k values move closer together, but the endpoints stay fixed at plus and minus pi over a.

In the limit of an infinite chain, this range from minus pi over a to plus pi over a defines the first Brillouin zone. It plays the same role for electronic structure as the unit cell does for the crystal lattice; knowing the states within this zone gives the complete electronic structure.

Next, we will make this concrete with an infinite one-dimensional chain of hydrogen atoms.

1D Chain of H Atoms

An infinite one-dimensional chain of hydrogen atoms drawn as connected circles.

Visualizing Crystal Orbitals: k = 0

Build expanding the crystal orbital sum at k = 0, ending in the fully bonding hydrogen chain.

And let's write the mathematical functions for our different crystal orbitals.

At k equals zero, we're at the center of the first Brillouin zone. The crystal orbital is given by the same general formula, but with k equal to zero, the exponential term becomes one.

So each term in the sum, whether it's the 1s orbital at the origin, one unit cell away, two unit cells away, and so on, adds with the same phase.

[advance: reduce every coefficient to one and draw the bonding chain]

Since e to the zero equals one everywhere, each orbital has the same coefficient of one.

The total wavefunction is therefore just the sum of all the hydrogen 1s orbitals along the chain.

If we draw the first few orbitals, all overlap constructively, forming a fully bonding molecular orbital, the most bonding possible combination. This makes it the lowest-energy crystal orbital in the first Brillouin zone.

Next, we will draw that same fully bonding crystal orbital as a mathematical plot.

Crystal Orbital at k = 0

At k = 0 the basis set u(x) times the constant phase factor gives the summed 1s crystal orbital wavefunction.

Visualizing Crystal Orbitals: k = π/a

Build expanding the crystal orbital at k = pi over a, term by term, ending in the fully antibonding chain.

Now let’s move to the other extreme in the first Brillouin zone—the largest possible k value, k equals pi over a.

Here, a is the one-dimensional lattice constant, the spacing between hydrogen atoms.

[advance: write the expansion at k = pi over a]

We start with the same formula, but now k equals pi over a instead of zero. Substituting this value throughout, each orbital's coefficient depends on its position x.

For the atom at the origin, x equals zero, so its exponential term equals one.

[advance: step to the next orbital's coefficient]

For the orbital one unit cell from the origin, x equals one times the lattice spacing a. For the hydrogen two unit cells away, x equals two a, and so on for each successive atom along the chain.

[advance: reduce with Euler's relation and draw the antibonding chain]

Using Euler's relation, the first term equals e to the zero, or one. The next term, e to the i pi, gives cosine of pi equals negative one and sine of pi equals zero.

The following term, e to the i 2 pi, returns positive one, then e to the i 3 pi gives negative one again. The coefficients therefore alternate between +1 and –1 across the chain.

Graphically, each hydrogen 1s orbital alternates in phase, positive, negative, positive, negative, so neighboring orbitals interfere destructively. This produces a fully antibonding crystal orbital, the highest-energy and most antibonding state possible.

Next, we will draw that fully antibonding crystal orbital as a mathematical plot.

Crystal Orbital at k = π/a

At k = pi over a the basis set u(x) times a phase factor of wavelength 2a gives the alternating crystal orbital wavefunction.

Band Structure: Linear H-Atom Chain

Band structure of the linear hydrogen atom chain: energy versus k from minus pi over a to plus pi over a.

Putting this all together gives what’s called the band structure—the “spaghetti diagram.”

Only one band in this simple case, but we can still see key features
k equals zero, center of the first Brillouin zone: all orbital interactions bonding — the lowest-energy crystal orbital
k equals pi over a: all interactions antibonding — the highest-energy orbital in the zone
Negative k direction: the cosine and sine terms produce a mirror image
From what we've derived, we only know the energies at a few points — the actual curve between them could be straight or slightly curved
The shape shown here, calculated with extended Hückel theory, is roughly parabolic

Let’s consider what the crystal orbital looks like at intermediate points—such as k equals pi over 2a, halfway between the center and the zone boundary.

Intermediate Values of k

At k = pi over 2a the phase factor has wavelength 4a, giving a crystal orbital with nodes every two atoms.

At k equals pi over 2a, the function still varies sinusoidally, but because of the factor of 2 in the denominator, its wavelength doubles.

The wavefunction now repeats every four unit cells instead of two
In general, the wavelength equals 2 pi divided by k; substituting k equals pi over 2a gives a wavelength of four times the lattice spacing a
When this oscillating term multiplies the 1s basis set: the crystal orbital alternates in phase every two atoms — positive, node, negative, node
Nodal planes two atoms apart: less antibonding, and therefore lower in energy, than the one at k equals pi over a, but higher in energy than the fully bonding orbital at k equals zero

Next, we will step back and ask directly: what is k?

What Is k?

k is a quantum number that “labels” wave functions (analogous to SALCs) generated by the translational symmetry of the crystal (it determines the coefficients of the basis set)
Reciprocal space wavevector. It tells us how rapidly the crystal orbital changes phase

$\lambda = 2π/k$

It determines the crystal momentum of an electron

$p = h/\lambda$$\;\longrightarrow\; p = hk/2π = \hbar k$

So what exactly is this variable k? We can think of it in several ways.

Each crystal orbital has its own k value — 10,000 distinct k values in a finite ring of 10,000 atoms; in an infinite crystal, k can take infinitely many values, each describing a different crystal orbital with a unique phase pattern
k serves as a label that defines the phase relationships, or coefficients, of the atomic orbitals in the crystal
It also tells us how quickly the coefficients alternate in sign, which relates to how rapidly the orbital oscillates within the lattice
From that oscillation we can define a wavelength; by de Broglie's relation, the momentum of the electron is Planck's constant divided by that wavelength

[advance: plug the wavelength into de Broglie's relation]

If we use this value of lambda from up here and plug it into Broglie's relationship, then we get this relationship.

The momentum of the electron is equal to h bar times the value of k.

And so we see that k also tells us something about the momentum of the electron, and in that sense it's called a crystal momentum.

And we'll see for things like optical transitions that has important consequences.

Next, we will relate the band structure to orbital overlap in real space.

Bandwidth and Orbital Overlap

Build comparing the hydrogen-chain band at H-H distances of 1.0 and 2.0 angstroms to define bandwidth.

To relate the band structure to the real-space crystal, consider this one-dimensional elemental solid.

Plotted band structure assumes hydrogen atoms spaced one angstrom apart: highest-energy crystal orbital about +20 electron volts, lowest about –20
Increase the spacing to two angstroms: the parabolic shape remains, but the bonding interactions at k equals zero weaken due to reduced orbital overlap
As a result, the lowest-energy, fully bonding orbital rises in energy compared to when the atoms were closer together

[advance: mark the weakened antibonding band edges]

At the edges of the Brillouin zone, the antibonding interactions also weaken significantly, lowering the energy there as well
Bandwidth: the energy difference between the top and bottom of a band
Strong overlap between neighboring unit cells: large bandwidth, wide or highly dispersed band, more delocalized electrons; small overlap: narrower bandwidth, more localized electrons

Next, we will use the density of states to summarize the band.

Density of States

The final concept is the density of states plot, a simplified way to represent the full band structure or “spaghetti diagram.”

DOS: how many crystal orbitals exist within an infinitesimal energy range between E and E plus delta E
The band flattens near the top and bottom: many states within a small energy range — DOS peaks at both ends of the band
In the middle the band is steeper: a small energy window includes fewer orbitals — a minimum in the DOS
Matches the 50-atom hydrogen ring we saw earlier: orbitals closely spaced at the bottom and top, farther apart in the middle
Omits detailed features of the full band structure, but gives essential information about the electronic structure of a crystal

Next, we will pull the whole lecture together in a summary.

Lecture Summary

Goal: Bridge from molecular orbitals → crystal band structure.
Approach: Extend MO theory to infinite periodic solids using Bloch functions:
$\psi(x) = u(x)\cdot e^{ikx}$
- u(x): periodic basis (hydrogen 1s) e^ikx: imposes lattice periodicity
k (“crystal momentum”):
- Labels distinct crystal orbitals
- Determines wavelength $\lambda = 2π/k$ → momentum $p = \hbar k$
- Allowed range: $−π/a \le k \le +π/a$ (1st Brillouin zone)
Extremes:
- $k = 0$ → all bonding (ψ in-phase) lowest energy
- $k = π/a$ → all antibonding (ψ alternating ±) highest energy
Bandwidth: $\Delta E \approx E(π/a) − E(0) \propto$ orbital overlap
- Large overlap → wide band → delocalized electrons
- Small overlap → narrow band → localized electrons
Density of States (DOS): number of states per energy interval dE
- Peaks where bands are flat (bottom/top)
- Minima where bands steep (midband)
Essence: Infinite chain → continuous bands; k maps real-space bonding to energy-momentum space.

Let's read back through what we did today.

Goal: bridge from molecular orbitals to the crystal band structure
Approach: extend molecular orbital theory to infinite periodic solids using Bloch functions — psi of x equals u of x times e to the i k x; u of x is the periodic basis, the hydrogen 1s orbital, and e to the i k x imposes the lattice periodicity
k, the crystal momentum: labels the distinct crystal orbitals; determines the wavelength, lambda equals 2 pi over k, and so the momentum, p equals h-bar k; its allowed range, minus pi over a to plus pi over a, is the first Brillouin zone
Extremes: k equals zero all bonding, psi in phase, lowest energy; k equals pi over a all antibonding, psi alternating in sign, highest energy
Bandwidth, the energy difference between those two, tracks orbital overlap: large overlap gives a wide band and delocalized electrons, small overlap a narrow band and localized electrons
Density of states counts the number of states per energy interval: peaks where the band is flat, at the bottom and top; minimum where the band is steep, in the middle
Essence: an infinite chain gives continuous bands, and k maps real-space bonding into energy and momentum space

Next, we will test these ideas with two review questions, starting with the basis set.

Check Your Understanding: The Basis Set

Question asking what the basis set is in the context of Bloch functions and how it is chosen for the 1D hydrogen chain, then its answer.

Check Your Understanding: k and Bonding

Question asking how the value of k determines whether a crystal orbital is bonding or antibonding, then its bulleted answer.

Second question. How does the value of k determine whether a crystal orbital is bonding or antibonding? Think back to the two extremes we drew.

[advance: reveal the answer]

k determines how the phases of the atomic orbitals vary along the crystal
k equals zero: all orbitals in phase, all overlaps constructive — maximum bonding, lowest energy
k equals pi over a: adjacent orbitals alternate in sign, plus minus plus minus — complete destructive overlap, maximum antibonding, highest energy
Intermediate k values: partial phase alternation — mixed bonding and antibonding character, intermediate energies
Thus, k controls the bonding character and energy of each crystal orbital across the band

That leaves the Lecture 15 practice questions.

Learning Objectives: Molecular Chains and Band Gaps

Learning-objectives page for electronic band structures in 1D systems: build band structures for molecular chains and see how dimerization creates band gaps, predict the number of bands from the orbitals in the unit cell, distinguish direct from indirect band gaps, classify metals, semiconductors and insulators by Fermi-level position, and analyze how s, p-sigma and p-pi orbitals contribute to band formation.

Linear Chain of H₂ Molecules

A chain of hydrogen atoms dimerized into H2 molecules: paired circles with a short intramolecular H-H distance and a longer intermolecular distance, with the repeat distance a spanning one pair.

Two Orbitals per Unit Cell, Two Bands

Molecular orbital diagram of one H2 molecule: two hydrogen 1s orbitals combine into a filled bonding MO psi-plus and an empty antibonding MO psi-minus, so the two-atom unit cell carries two molecular orbitals and the chain has two bands.

Bonding MO Band: k = 0 and k = π/a

Build developing the bonding MO band from the Bloch function psi of x equals e to the i k x times u of x, adding k = 0 with u = psi-plus and then the chain of H2 dimers with every orbital in phase, and ending at k = pi over a where the phase alternates from one dimer to the next.

Let's start with the bonding molecular orbital.

The Bloch function for this band is psi equals e to the i k x times u of x
u of x is the basis set, the molecular orbitals inside the unit cell — here the bonding MO, psi-plus

[advance: write k = 0 with u = psi-plus under the Bloch function]

[advance: draw the chain of H2 molecules with every orbital in phase]

At k = 0 the e to the i k x term is one, so every unit cell takes the same coefficient
Phases stay constant along the chain: bonding within each molecule and between molecules

[advance: flip the chain to the zone edge, k = pi over a]

At k = pi over a the term alternates +1 and -1, so the phase flips from one unit cell to the next
Intramolecular stays bonding, intermolecular turns antibonding
The band spans these two extremes, so the endpoint energies are enough to sketch it

Next, we will run the same two k points with the antibonding molecular orbital as the basis.

Antibonding MO Band

Build running the antibonding molecular orbital through both k points, from the k = 0 chain in which each antibonding H2 dimer repeats identically, ending at k = pi over a where the antibonding dimer flips sign from one unit cell to the next.

Ordering the Four Crystal Orbitals

Build ordering the crystal orbitals of the H2 chain by energy, from the two psi-plus orbitals at k = 0 and k = pi over a labelled by their intramolecular and intermolecular bonding character, ending in all four stacked from the doubly antibonding k = 0 psi-minus orbital down to the doubly bonding k = 0 psi-plus orbital.

Two Bands and the Fermi Level

Build of the band structure of the linear H2 chain, energy against k from 0 to pi over a with the psi-plus band rising and the psi-minus band falling either side of a gap at pi over a, ending with the Fermi level drawn as a dashed line inside that gap between the filled and the empty band.

We plot the bonding MO at k = 0 (lowest energy) and k = π/a (higher), with k on the x-axis.

The antibonding MO at k = pi over a sits just above the bonding MO at that same k, and the antibonding MO at k = 0 is highest
Connect the four points and we have two bands, one from each molecular orbital

[advance: draw in the Fermi level]

The Fermi energy separates filled states below from empty states above at T = 0
Here the bonding band is filled and the antibonding band is empty
Without dimerization the two bands would be degenerate at k = pi over a
Dimerization stabilizes the filled band and destabilizes the empty one, so only the filled band gains and the distortion is favorable
This is a Peierls distortion, the solid-state analog of a Jahn-Teller distortion

Next, we will name the two bands the way a semiconductor chemist would, and read the same picture as a density of states.

Valence Band, Conduction Band, and the DOS

Build of the extended-Hückel band structure and density of states of the H2 chain, from the filled psi-plus valence band and empty psi-minus conduction band either side of the Fermi level with the molecular orbital levels of one H2 molecule at the left and N of E at the right, ending with a red arrow tracing the antibonding psi-minus molecular orbital into the conduction-band peak of the density of states.

Using extended Hückel theory for a more rigorous calculation shows the anti-bonding band is wider than the bonding band—anti-bonding orbitals destabilize more than bonding orbitals stabilize (though extended Hückel may exaggerate this asymmetry).

The basic features hold: the lower band rises from k = 0 to pi over a, the upper band falls, and the Fermi level lies in the gap
Semiconductor names: filled bands are valence bands, empty bands are conduction bands, and the separation is the band gap

[advance: trace the antibonding molecular orbital into the peak in the density of states]

Set the MO diagram of one H2 molecule beside the plot: the density of states is each orbital broadened into a band

Next, we will ask what happens when the top of one band and the bottom of the next sit at different values of k.

Direct vs. Indirect Band Gaps

Direct and indirect band gaps compared: on the left the conduction-band minimum sits directly above the valence-band maximum and a photon alone bridges the gap; on the right the minimum is displaced in k, so the transition also needs a lattice vibration and the material absorbs less efficiently.

When the Fermi level cuts through a band (like in the H-atom chain), infinitesimal thermal energy excites electrons between states—these are metals.

Fermi level between bands: a real HOMO-LUMO separation — semiconductors, or insulators once the gap is large
Direct gap: valence band maximum and conduction band minimum at the same k, so a photon alone excites the electron vertically — efficient absorption and emission
Indirect gap: the two extrema sit at different k, and k is momentum while a photon carries almost none
The electron then needs a photon for the energy and a phonon for the momentum; that three-body process is inefficient, like a forbidden molecular transition
For LEDs and other optical uses, direct band gap semiconductors are strongly preferred

Next, we will finish with the Lecture 15 practice questions.

Practice questions

Work the Lecture 15 practice questions before the next class. They cover Bloch functions and the crystal orbital, the allowed values of k and the first Brillouin zone, the bonding and antibonding limits of the hydrogen-chain band, bandwidth and orbital overlap, the density of states, the two-band structure of a chain of H₂ molecules, the Peierls distortion, and direct versus indirect band gaps.

Open the Lecture 15 practice questions

Every question carries a worked explanation, so you can check your reasoning as you go.

The homework for this lecture now lives in the online practice bank.

Work through it before the next class
Covers the Bloch function and how a crystal orbital is built from a periodic basis and a phase factor
The periodic boundary condition that fixes the allowed values of k and the first Brillouin zone
The fully bonding orbital at k equals zero and the fully antibonding one at k equals pi over a
How bandwidth tracks orbital overlap, and how the density of states summarizes the band
The two-band structure of a chain of H2 molecules and the Peierls distortion that opens the gap between those two bands
The difference between a direct and an indirect band gap

Every question has a worked explanation attached, so you can check your reasoning as you go.