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
Today we start on the electronic structure of solids, with a one-dimensional chain of hydrogen atoms.
Learning Objectives: Intro to Band Theory
Band structures why crystals require them instead of molecular orbitals.
Crystal orbital a Bloch function, periodic basis times a lattice phase term.
k a wavevector and a crystal momentum.
Bonding limits k = 0 bonding and k = π/a antibonding in the first Brillouin zone.
Band width tracks orbital overlap: wide means delocalized, narrow means localized.
Density of states a summary of orbital distribution against energy.
For several lectures we have worked out the electronic structure of atoms and then of molecules, and now in chapter 6 we move on to an infinite crystal. By the end you should be able to explain why a crystal needs a band structure rather than a molecular orbital diagram, you should understand what a bloch function is. You should be able to say what k, the wavevector, is. You should be able to show that k equals zero gives the most bonding crystal orbital and k equals pi over a the most antibonding, and that those are the limits of the first Brillouin zone. You should be able to relate the width of a band to orbital overlap, and to read a density of states.
Electronic Band Structure
MO diagram for finite sized molecule
Band structure diagram for infinite crystal
This slide shows where we are going. On the left is the molecular orbital diagram for an octahedral transition-metal complex, where we combine the orbitals on the metal with the ligand SALCs. A real crystal has an enormous number of atoms, so it does not make sense to treat it as one huge cluster and calculate its molecular orbitals. Instead we describe the electronic structure of a crystal with a band structure. On the right is the band structure of rhenium trioxide, which is also built from transition-metal-centered octahedra. Chemists sometimes call these spaghetti diagrams, and the goal of this chapter is to understand how a diagram like this relates to the atomic orbitals and the bonding.
Solids and Surfaces (Hoffmann, 1989)
Published in 1989
Roald Hoffmann (Cornell)
We will follow chapter 6 of Solid State Materials Chemistry closely. A lot of the early thinking about how to describe the electronic structure of crystals in the language of chemistry was done by Roald Hoffmann at Cornell. His book Solids and Surfaces, published in 1989, is only about 140 pages and was very influential. If you want to go deeper than chapter 6 it is an excellent resource.
From Molecules to an Infinite Chain
energy highestenergy MO lowestenergy MO H2 H6 H10 H14 H50 H∞
Per Solid State Materials Chemistry , Figure 6.1, §6.1.
To get to an infinite solid, we start with molecules and make them bigger and bigger. Let's start with hydrogen, so each atom has one orbital, in a one-dimensional chain. The smallest piece of the chain is the H₂ molecule, with a bonding and an antibonding orbital. A ring of six hydrogens has the same MO diagram as the pi system of benzene, and then come rings of 10, 14 and 50. The number of molecular orbitals equals the number of atoms, and half of them are filled, because each orbital holds two electrons and each hydrogen has one. The spacing gets smaller as the ring gets bigger, and by a million atoms the levels are very close together. The infinite chain gives a continuous band that is half filled. Notice that the levels at the bottom and top are closer together than in the middle, something we'll come back to later.
Atomic and Molecular Orbitals
To describe the structures of molecules we approximated MOs as linear combinations of atomic orbitals.
-8 -6 -4 -2 0 2 4 6 8 1.5 1.0 0.5 0.0 1s wavefunction, ψ(r) distance from nucleus, r (Å) Atomic orbital -4 -3 -2 -1 0 1 2 3 4 ψ1 ψ2 wavefunction ψ z (Å) Molecular orbital
$\psi_+ = 0.56(\psi_1 + \psi_2)$
On the last slide I shaded each orbital light or dark until I got to the infinite chain, and then I just filled the band in. So how do we describe the orbitals in an infinite solid? For a molecule, we took each molecular orbital to be a linear combination of atomic orbitals, like the H₂ bonding orbital built from the two hydrogen 1s orbitals, and we are going to do the same thing for a crystal.
Crystal Orbitals: Bloch Functions
$\psi(x) = e^{ikx}\,u(x)$
Wavefunction(x is the position along the chain)
Imposes the periodicity of the lattice$e^{ikx} = \cos(kx) + i\sin(kx)$
Basis setRepresents the molecular orbitals inside the unit cell (repeats exactly from one unit cell to the next)
For a crystal we build crystal orbitals as linear combinations of atomic orbitals using the Bloch function. Psi of x is the wavefunction, and x is the position along the chain. U of x is the basis set. Even a one-dimensional crystal has a unit cell, and u of x is exactly the same in every unit cell. It is made from the orbitals of the atoms or the molecule inside the cell, so for the hydrogen chain it is just the hydrogen 1s orbital. The other term, e to the i k x, imposes the periodicity of the lattice. It sets the coefficient in front of the orbitals in each unit cell, in the same way that symmetry set the phases of the orbitals in a ligand SALC. Using Euler's relation, e to the i k x can also be written as cosine k x plus i sine k x.
First Brillouin Zone
What is k? What is its physical significance? What are its allowed values?
Consider a ring of N hydrogen atoms, with N large enough that we can neglect curvature. If we start at any point on the chain, x, and go all the way around the chain to the same point, x + Na, the wavefunction must repeat (periodic boundary conditions).
$\psi(x) = \psi(x + Na)$
$e^{ikx}u(x) = e^{ik(x+Na)}u(x+Na)$
But u(x) is periodic. That is it repeats identically from one unit cell to the next. So, u(x) = u(x + Na) and
$e^{ikx} = e^{ik(x+Na)} = e^{ikx}e^{ikNa}$
Most of this should be familiar. U of x is the hydrogen 1s wavefunction, and x is the position in the chain. But k is new, so we don't yet know what it means physically and what values it can take, so let's start by working out the range of k.
[advance: bring up the ring of N hydrogen atoms]
It is easier for now to think of a ring of hydrogen atoms than an infinite chain, with enough atoms that we can neglect the curvature, say 10,000. If we start at some point, say half an angstrom from the atom we call the origin, and go all the way around the ring, we travel 10,000 times the lattice constant a, the spacing between the hydrogens. We are back at the same point, so the wavefunction has to have the same value there.
[advance: show the periodic boundary condition]
So psi of x has to equal psi of x plus N a, where capital N is the number of atoms in the ring and a is their spacing. We can write both sides as Bloch functions. u of x equals u of x plus N a and that term drops out. That leaves e to the i k x equal to e to the i k x times e to the i k N a.
First Brillouin Zone (cont)
Applying Euler’s relationship
$e^{ikx} = e^{ikx}e^{ikNa} = e^{ikx}$ $\left[\cos(kNa) + i\sin(kNa)\right]$
Obviously, the term in brackets must be equal to 1. That will only happen when kNa = n(2π) (where n is an integer) because cos (n2π) = 1 and sin(n2π ) = 0. So
$kNa = n(2\pi)$
Rearranging we get
$k = n\frac{(2\pi)}{Na} = \pm\frac{2\pi}{Na}, \pm\frac{4\pi}{Na}, \pm\frac{6\pi}{Na}, \ldots \pm (N/2)\frac{2\pi}{Na}$
Notice that even as the length of the chain (N) goes to infinity the relevant values of k are limited to the range −π/a to +π/a. This range is called the First Brillouin Zone .
Using Euler's relation on the second term, e to the i k N a becomes cosine k N a plus i sine k N a. For the equality to hold, the term in brackets has to be one, because only then is e to the i k x equal to e to the i k x. That happens when k N a is an integer times 2 pi, since the cosine of 2 pi is 1 and the sine is 0. So k, times the number of atoms N, times the spacing a, has to be an integer times 2 pi.
[advance: rearrange for the allowed values of k]
Solving for k, k equals a small integer n, times 2 pi, divided by capital N times a. A ring of 10,000 atoms has 10,000 molecular orbitals, and if n runs over plus and minus 1, 2, 3, up to plus and minus 5,000, we have one orbital for every atom. If we make the ring bigger, a hundred thousand or a million atoms, the values of k get closer together, but the end points do not change. They stay at minus pi over a and plus pi over a, and that range of k is called the first Brillouin zone. The first Brillouin zone is to the electronic structure what the unit cell is to the crystal, so once we know the states in this range of k we know everything we need about the electronic structure.
1D Chain of H Atoms
To make this less abstract, let's go back to the one-dimensional chain of hydrogen atoms, and this time let it be infinite.
Visualizing Crystal Orbitals: k = 0
What does the crystal orbital look like when k = 0 (at the center of the Brillouin zone)?
$\psi_k = e^{ikx}u(x)$
$\psi_{k=0} = e^{0}\psi_{1s(0)} + e^{0}\psi_{1s(1)} + e^{0}\psi_{1s(2)} + e^{0}\psi_{1s(3)} + e^{0}\psi_{1s(4)} + e^{0}\psi_{1s(5)} + \ldots$
$\psi_{k=0} = \psi_{1s(0)} + \psi_{1s(1)} + \psi_{1s(2)} + \psi_{1s(3)} + \psi_{1s(4)} + \psi_{1s(5)} + \ldots$
n =
0
1
2
3
4
5
All H-H interactions are bonding, this is the lowest energy crystal orbital
Per Solid State Materials Chemistry , Figure 6.3, §6.1.
We start at k equals zero, the center of the first Brillouin zone. The crystal orbital is e to the zero times the 1s orbital at the origin, plus e to the zero times the 1s orbital one unit cell away, then two, three and four unit cells away, and so on. E to the zero is one wherever we are on the chain, so the coefficient in front of every orbital is one.
[advance: reduce every coefficient to one and draw the bonding chain]
So the crystal orbital is just the sum of the hydrogen 1s orbitals on every atom. If we draw the first six, every orbital interacts constructively with its neighbors, so every interaction is bonding. There is no way to combine these 1s orbitals that is more bonding than this, so this is the lowest energy crystal orbital.
Crystal Orbital at k = 0
k = 0
u(x) eikx ψ(x) ×
→
Basis set
Crystal orbital
wavefunction
λ = ∞
a
Per Solid State Materials Chemistry , Figure 6.4, §6.1.
We can also plot this. The dots are the positions of the hydrogen nuclei, and the curve on the left is the sum of a 1s wavefunction on each atom. When k is zero, e to the i k x is just one, a constant with an infinite wavelength. So when we multiply the two, the crystal orbital on the right looks just like the sum of the hydrogen 1s orbitals.
Visualizing Crystal Orbitals: $k = \pi/a$
What does the crystal orbital look like when k = π/a (at the edge of the Brillouin zone)?
$\psi_k = e^{ikx}u(x)$
$\psi_{k=\pi/a} =$ $e^{i(\pi/a)0a}\psi_{1s(0)}$ $+$ $e^{i(\pi/a)1a}\psi_{1s(1)}$ $+ e^{i(\pi/a)2a}\psi_{1s(2)} + e^{i(\pi/a)3a}\psi_{1s(3)}$
$\psi_{k=\pi/a} = e^{0}\psi_{1s(0)} + e^{i\pi}\psi_{1s(1)} + e^{i2\pi}\psi_{1s(2)} + e^{i3\pi}\psi_{1s(3)} +$
$\psi_{k=\pi/a} = \psi_{1s(0)} - \psi_{1s(1)} + \psi_{1s(2)} - \psi_{1s(3)} + \psi_{1s(4)} - \psi_{1s(5)} + \cdots$
n =
0
1
2
3
4
5
All H-H interactions are antibonding, this is the highest energy crystal orbital
Now we go to the largest value of k we can have and still be in the first Brillouin zone, k equals pi over a, the edge of the zone. Remember that a is the unit cell constant, the spacing between hydrogen atoms.
[advance: write the expansion at k = pi over a]
We use the same formula, but now k is pi over a in every term and only x changes. For the atom at the origin, x is zero.
[advance: step to the next orbital's coefficient]
For the atom one unit cell from the origin, x is a, and for the one two unit cells away it is 2a, and so on. The a's cancel, so the exponents are zero, i pi, 2 i pi and 3 i pi. E to the zero is one. Pi radians is 180 degrees, where the cosine is minus one and the sine is zero, so the next coefficient is minus one. At 2 pi it is plus one again, and at 3 pi minus one. So the coefficients alternate, plus one, minus one, plus one, minus one.
[advance: reduce with Euler's relation and draw the antibonding chain]
So the 1s orbitals alternate in phase along the chain. Every interaction between nearest neighbors is destructive, so every one is antibonding, and there is no crystal orbital more antibonding than this one. This is the highest energy crystal orbital.
Crystal Orbital at $k = \pi/a$
k = π/a
u(x) eikx ψ(x) ×
→
Basis set
Crystal orbital
wavefunction
λ = 2a
Per Solid State Materials Chemistry , Figure 6.4, §6.1.
The basis is still the sum of the hydrogen 1s wavefunctions, but e to the i k x is different now. I am plotting its real part, which oscillates from plus one at one nucleus to minus one at the next, so its wavelength is 2a. When we multiply the two, the wavefunction is positive on one hydrogen and negative on the next, and in between it has to go through zero. That node is the characteristic feature of an antibonding orbital.
Band Structure: Linear H-Atom Chain
20 10 0 -10 -20 −π/a 0 π/a energy (eV) k
Per Solid State Materials Chemistry , Figure 6.5, §6.1.5.
Putting this together gives us a band structure, a spaghetti diagram with only one piece of spaghetti. At k equals zero all the interactions are bonding, so that is the lowest energy crystal orbital, and at pi over a they are all antibonding, so that is the highest. Going to negative k gives a mirror image because of the cosine and sine terms. So far we have only found the energies of three points, so we do not know whether they should be joined by straight lines or a curve. The curve shown here was calculated with extended Hückel theory, and it is almost parabolic. To see what the crystal orbitals look like in between, we can take k equals pi over 2a, halfway between the center of the first Brillouin zone and its edge.
Intermediate Values of k
k = π/2a
u(x) eikx ψ(x) ×
=
Basis set
Crystal orbital
wavefunction
λ = 4a
$\lambda = \dfrac{2\pi}{k}$
Per Solid State Materials Chemistry , Figure 6.4, §6.1.
At pi over 2a there is a 2 in the denominator, so e to the i k x still oscillates, but now it takes four unit cells to repeat, a wavelength of 4a. In general the wavelength is 2 pi over k, and with k equal to pi over 2a that gives 4a. Multiplying by the basis gives this crystal orbital, with a negative orbital, then a node where there is no orbital at all, then a positive orbital, another node, and a negative one. The nodes are now two atoms apart, where at pi over a there was a node between every pair of atoms. So this orbital is less antibonding and lower in energy than the one at pi over a, but higher in energy than the one at k equals zero.
What Is k?
Quantum number labels the wave functions (analogous to SALCs) generated by the translational symmetry of the crystal; it determines the coefficients of the basis set.
Reciprocal-space wavevector tells us how rapidly the crystal orbital changes phase, $\lambda = 2\pi/k$.
Crystal momentum $p = h/\lambda$$\;\longrightarrow\; p = hk/2\pi = \hbar k$
Now we can come back to the question of what k is, and there are a few ways to think about it. A ring of 10,000 atoms has 10,000 values of k, each a different molecular orbital, and an infinite chain has an infinite number, each a crystal orbital with a different phasing of the hydrogen 1s orbitals. K determines the coefficients in front of the atomic orbitals, so in that sense it is a label. K also tells us how rapidly those coefficients oscillate from positive to negative, which gives the crystal orbital a wavelength. De Broglie's relation says the momentum of a matter wave is Planck's constant divided by its wavelength.
[advance: plug the wavelength into de Broglie's relation]
If we put lambda equals 2 pi over k into de Broglie's relation, the momentum of the electron is h bar times k. So k also tells us the momentum of the electron, and in that sense it is called the crystal momentum. That has important consequences for things like optical transitions.
Bandwidth and Orbital Overlap
20 10 0 -10 -20 −π/a 0 π/a energy (eV) k H-H distance = 1.0 Å H-H distance = 2.0 Å
Bandwidth the energy difference between the top and bottom of a band; large where neighbouring unit cells overlap strongly (wide, dispersed band, delocalized electrons), small where they overlap weakly (narrow band, localized electrons).
Per Solid State Materials Chemistry , Figure 6.5, §6.1.5.
For an elemental chain in one dimension, about the only thing we can change in the structure is the spacing of the atoms. This band structure is for hydrogen atoms 1 angstrom apart, and the highest energy crystal orbital is at about plus 20 electron volts and the lowest at about minus 20. If we move the atoms 2 angstroms apart, the band still has the same roughly parabolic shape. But at k equals zero there is less overlap of the orbitals, so it is less bonding and its energy is not as deep. At the edges of the Brillouin zone the antibonding interactions are also much weaker, so the energy there comes down a lot.
[advance: mark the weakened antibonding band edges]
The energy between the bottom and the top of the band is called the bandwidth. When there is a lot of overlap from one unit cell to the next, the band is wide, or dispersed, and wide bands can mean delocalized electrons. When the orbitals barely overlap from one unit cell to the next, the band is very narrow, and narrow bands go with localized electrons.
Density of States
-10 -12 -14 -16 0 π/a energy (eV) k EF -10 -12 -14 -16 EF empty states filled states DOS, N(E)
Density of states, N (E ) which is the number of allowed energy levels per unit volume of the solid over the energy range E to E + dE , as dE goes to zero.
Shape Flat bands give sharp peaks in the DOS, wide bands give shallow features in the DOS.
Per Solid State Materials Chemistry , Figure 6.6, §6.1.6.
Full band structures can get complicated, so we often summarize them with a density of states plot, the number of crystal orbitals between E and E plus delta E as delta E goes to zero. The band for the hydrogen chain flattens out at the bottom and the top, so a narrow energy window there, say a tenth of an electron volt, contains a lot of states, and the density of states has peaks at the bottom and the top. In the middle the band is steeper, so the same window contains fewer states and the density of states goes through a minimum. That is what we saw with the 50-atom ring, where the levels were closer together at the top and bottom than in the middle. A density of states does not show everything in the band structure, but it tells us a lot.
Lecture Summary
Concept What it says
Bloch function $\psi(x) = u(x)\,e^{ikx}$; u(x) the periodic basis (hydrogen 1s), $e^{ikx}$ the lattice periodicity. k, the crystal momentum labels the crystal orbitals; $\lambda = 2\pi/k$, $p = \hbar k$; allowed range $-\pi/a \le k \le +\pi/a$, the first Brillouin zone. Extremes k = 0 all bonding, in phase, lowest energy; k = π/a all antibonding, alternating sign, highest energy. Bandwidth $\Delta E \approx E(\pi/a) - E(0)$, proportional to orbital overlap: large overlap, wide band, delocalized; small overlap, narrow band, localized. Density of states states per energy interval dE: peaks where the band is flat (top and bottom), minimum where it is steep (midband).
In summary, we learned the Bloch function builds each crystal orbital from a basis, here the hydrogen 1s orbital repeated in every unit cell, times e to the i k x, which imposes the periodicity of the lattice. K labels the crystal orbitals, it sets their wavelength, 2 pi over k, and their crystal momentum, h bar k, and its allowed range from minus pi over a to plus pi over a is the first Brillouin zone. At k equals zero the crystal orbital is all bonding and lowest in energy, and at pi over a it is all antibonding and highest. The bandwidth is proportional to the orbital overlap, so a wide band means delocalized electrons and a narrow band localized ones. And the density of states peaks where the band is flat and has a minimum where it is steep.
Check Your Understanding: k and Bonding
How does the value of k determine whether a crystal orbital is bonding or antibonding?
Role of k The value of k determines how the phases of the atomic orbitals vary along the crystal.
k = 0 When k = 0 , all orbitals are in phase , so all overlaps are constructive → maximum bonding → lowest energy .
k = π/a When k = π/a , adjacent orbitals alternate in sign (+ − + − ) → complete destructive overlap → maximum antibonding → highest energy .
Intermediate k Intermediate k values produce partial phase alternation, giving orbitals with mixed bonding/antibonding character and intermediate energies .
Thus k controls the bonding character and energy of each crystal orbital across the band.
Question: how does the value of k determines whether a crystal orbital is bonding or antibonding.
[advance: reveal the answer]
K determines how the phases of the orbitals change along the chain. At k equals zero all the orbitals are in phase, so every overlap is constructive, and that is the most bonding and lowest energy crystal orbital. At pi over a neighboring orbitals alternate in sign, so every overlap is destructive, and that is the most antibonding and highest in energy. In between, the phases only partly alternate, so those crystal orbitals are partly bonding and partly antibonding, with energies in between.
Learning Objectives: Molecular Chains and Band Gaps
By the end of this lecture, you will be able to:
Molecular chains Construct band structures for one-dimensional molecular chains and understand how dimerization creates band gaps
Counting bands Predict the number of bands from the number of atomic orbitals in the unit cell
Direct vs indirect Distinguish between direct and indirect band gaps and explain their impact on optical properties
Fermi level Classify materials as metals, semiconductors, or insulators based on band structure and Fermi level position
Orbital contributions Analyze how different atomic orbitals (s, p-sigma, p-pi) contribute to band formation and dispersion
In this part we stay in one dimension, but with slightly more complicated chains. By the end you should be able to build the band structure of a chain of molecules, and to see how pairing the atoms into molecules opens a band gap. You should be able to predict the number of bands from the number of atomic orbitals in the unit cell, and to use the position of the Fermi level to classify a material as a metal, a semiconductor or an insulator. You should be able to tell a direct band gap from an indirect one, and explain why that matters for absorbing and emitting light.
Linear Chain of H2 Molecules
Let’s see what happens to the band structure of our H atom chain if we let the atoms dimerize into H2 molecules.
a
Now there are two H-H distances, a short intramolecular distance and a longer intermolecular distance.
The chain of hydrogen atoms gave one band, half filled, and that would make it a metal. We do not usually think of hydrogen as a metal, although at very high pressures, like in the interior of Jupiter or Saturn, people do expect it to become metallic. Under normal conditions hydrogen atoms pair up into H₂ molecules, so let's look at the band structure of a linear chain of H₂ molecules. Now there are two hydrogen-hydrogen distances, a short one within each molecule and a longer one between molecules. The unit cell is about twice as long, and it contains two atoms, a whole H₂ molecule.
Two Orbitals per Unit Cell, Two Bands
The unit cell now contains two atoms (each with one orbital), so there are two molecular orbitals per unit cell and there will be two bands in the band structure diagram.
antibonding MO (ψ− ) bonding MO (ψ+ )
The most important consequence is that there are now two atoms in the unit cell, each with one orbital, so there will be two bands. The number of bands always equals the number of atomic orbitals in the unit cell. The easiest way to think about it is with the molecular orbitals of H₂, a bonding orbital and an antibonding orbital, and each of those gives us a band.
Bonding MO Band: $k = 0$ and $k = \pi/a$
$\psi(x) = e^{ikx}\,u(x)$
$k = 0,\; u = \psi_+$
a
$k = \pi/a,\; u = \psi_+$
a
Per Solid State Materials Chemistry , Figure 6.7, §6.2.
Let's start with the bonding orbital. The crystal orbitals in this band are e to the i k x times u of x, and the basis u of x is the molecular orbital inside the unit cell. For this band, u of x is the bonding molecular orbital of H₂, psi plus.
[advance: write k = 0 with u = psi-plus under the Bloch function]
We look at this band at two k points, starting at k equals zero. There e to the i k x is one, so the coefficient in front of the molecular orbital is one in every unit cell, and the phase does not change from one unit cell to the next.
[advance: draw the chain of H2 molecules with every orbital in phase]
These are the first four unit cells, and they all look the same going to infinity. Because all the 1s orbitals have the same phase, the interaction within each molecule is bonding, as it must be since that is our basis, and the interaction between molecules is bonding too.
[advance: flip the chain to the zone edge, k = pi over a]
At the edge of the first Brillouin zone, k equals pi over a, we start with the same basis in the first unit cell. But now e to the i k x alternates between plus one and minus one from one unit cell to the next, so neighboring molecules have opposite phases. The interaction within each molecule is still bonding, but the interaction between molecules is now antibonding. Every crystal orbital in this band lies between these two, so knowing the energies of the end points is enough to sketch the band.
Antibonding MO Band
$\psi(x) = e^{ikx}\,u(x)$
$k = 0,\; u = \psi_-$
a
$k = \pi/a,\; u = \psi_-$
a
For the antibonding band, the only difference is that the basis we repeat in each unit cell is the antibonding molecular orbital, psi minus. At k equals zero it has the same phase in every unit cell, so the interaction within each molecule is antibonding, as expected, and the interaction between molecules is antibonding too. This crystal orbital is doubly antibonding, so it is the highest energy crystal orbital in the whole band structure.
[advance: switch the chain to k = pi over a]
At k equals pi over a, e to the i k x reverses the phase of the orbital in every unit cell. It is still antibonding within each molecule, but the interaction between molecules is now bonding. So this crystal orbital is lower in energy than the one at k equals zero.
Ordering the Four Crystal Orbitals
$k = 0,\; u = \psi_-$ → intramolecular = antibonding; intermolecular = antibonding
$k = \pi/a,\; u = \psi_-$ → intramolecular = antibonding ; intermolecular = bonding
$k = \pi/a,\; u = \psi_+$ → intramolecular = bonding ; intermolecular = antibonding
$k = 0,\; u = \psi_+$ → intramolecular = bonding; intermolecular = bonding
Per Solid State Materials Chemistry , Figure 6.7, §6.2.
Now we put these four crystal orbitals in order from lowest to highest energy. The bonding basis at k equals zero is bonding both within and between molecules, so no crystal orbital can be lower. The bonding basis at pi over a keeps the bonding interaction within the molecule but is antibonding between molecules, so it is higher.
[advance: stack all four crystal orbitals from lowest to highest]
The antibonding basis at pi over a is the opposite, antibonding within the molecule and bonding between molecules. The intramolecular antibonding orbital is higher in energy than the intermolecular antibonding orbital, because the hydrogen-hydrogen distance within the molecule is shorter than the distance between molecules. The highest energy crystal orbital is the antibonding basis at k equals zero, which is antibonding both within and between molecules.
Two Bands and the Fermi Level
0 π/a k energy ψ− band ψ+ band … … … … … … … … Fermi level, E F
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.
Now we can sketch the band structure, with k on the x axis. The bonding orbital at k equals zero is the lowest, and the bonding orbital at pi over a is above it. The antibonding orbital at pi over a is a little higher than that, and the highest energy crystal orbital is at k equals zero. Connecting the points gives the band from the bonding molecular orbital and the band from the antibonding molecular orbital, and that is our band structure.
[advance: draw in the Fermi level]
One more thing a band structure diagram tells us is which bands are filled. At T equals zero, every crystal orbital below the Fermi level is occupied and every one above it is empty. Here the bonding band is completely filled and the antibonding band is completely empty, and that tells us why the chain dimerizes. If the distance within the molecule were the same as the distance between molecules, the two bands would have the same energy at pi over a. The dimerization stabilizes the psi plus band and destabilizes the psi minus band, and with a half-filled band like this that kind of distortion is clearly favorable. A distortion of a half-filled band into pairs like this is called a Peierls distortion, the extended-solid version of a Jahn-Teller distortion.
Valence Band, Conduction Band, and the DOS
10 0 -10 -20 energy (eV) 0 π/a k ψ− ψ+ … … … … … … … … EF EF conductionband valenceband DOS, N(E)
Each molecular orbital of one H2 molecule broadens into a band: the density of states is the MO diagram smeared out.
Per Solid State Materials Chemistry , Figure 6.8, §6.2.
If we do the calculation more rigorously, with extended Hückel theory, the details change a little. The antibonding band is quite a bit wider than the bonding band, because antibonding orbitals are destabilized more than bonding orbitals are stabilized. The basic features are the same, two bands, the lower one running uphill from k equals zero to pi over a and the upper one running downhill, with the Fermi level in the gap between them. For semiconductors, the filled band is called the valence band, the empty band the conduction band, and the energy between them the band gap.
[advance: trace the antibonding molecular orbital into the peak in the density of states]
If we compare the density of states with the MO diagram of an H₂ molecule, we can think of the density of states as each molecular orbital broadened out into a band.
Direct vs. Indirect Band Gaps
direct bandgap energy k conductionband valenceband h ν Eg indirect bandgap energy k conductionband valenceband h ν Eg
For an indirect transition electron momentum (k ) must change, this requires a lattice vibration (phonon). Hence less efficient absorber.
Per Solid State Materials Chemistry , Figure 6.11, §6.3.1.
When the Fermi level cuts through a band, as in the hydrogen atom chain, thermal energy alone can excite electrons into empty states, so those solids are metals. When it falls between bands there is a gap, like a HOMO-LUMO gap, and those are semiconductors, or insulators if the gap is large. In a direct band gap the valence band maximum and the conduction band minimum are at the same k, so a photon alone can excite the electron. In an indirect band gap they are at different k, so the electron also has to change its momentum, and a photon does not carry enough momentum to do this. That momentum has to come from a lattice vibration, a phonon, so indirect gap semiconductors absorb light much less efficiently. So for a light-emitting diode, a direct band gap is generally better.
Band Structure: the Linear F Atom Chain
Which of the following is the correct band structure for a linear chain of F atoms?
Let's finish one dimension with another example, a linear chain of fluorine atoms. I want you to figure out what its band structure looks like. I have given you four possibilities, A through D, and one of them is correct. Analyze the chain and decide which one it is, and after you have done that we will go over the answer.
Working the F-Chain Band Structure
Avoided crossing
A more accurate treatment of the band structure would show an avoided
crossing between the 2pz σ and 2s σ* interactions at
k=π/a. There would be mixing between these two bands (creating sp-hybrid like
crystal orbitals).
We build up the band structure of the fluorine chain exactly the way we did for the hydrogen atom chain, except now we have to think about both the s and the p orbitals. Fluorine has one 2s and three 2p orbitals in the unit cell, so there are four bands.
[advance: draw the 2s sigma band]
We do the 2s orbital first. It is bonding at k equals zero and antibonding at k equals pi over a, just like the hydrogen 1s, so this band runs uphill.
[advance: add the 2pz sigma band]
Now the p orbital that points along the chain, the 2p z. At k equals zero the phase of the p orbital is the same in every unit cell, and two p orbitals pointing at each other with the same phase give an antibonding interaction, so this band starts at a high energy. At pi over a we invert the phase every time we move from one unit cell to the next, so the interaction turns bonding, and the energy of this band drops as we go from zero to pi over a. At pi over a we would expect it to end up below the antibonding 2s, with the caveat that there is a pretty large energy difference between the 2s and the 2p orbitals on a fluorine atom.
[advance: add the narrower pi bands]
Finally the 2p x and the 2p y, which are perpendicular to the chain. At k equals zero they give a bonding pi interaction and at pi over a an antibonding pi interaction, so this band runs uphill like the 2s band. But the pi overlap is smaller than the sigma overlap, so the energy changes less from zero to pi over a, and this band is narrower than the others.
[advance: place the Fermi level]
The last thing we need is the Fermi level. A fluorine atom has seven valence electrons, the 2s is full and the 2p orbitals are five-sixths full. Each band holds two electrons, so seven electrons fill three and a half bands. The Fermi level cuts through about the middle of the 2p z band and the other three bands are below it. So if we could make a linear chain of fluorine atoms, it would be a metallic conductor.
[advance: show the avoided crossing]
There is one subtlety I have glossed over. If we do the calculation more rigorously, the sigma 2s and the sigma 2p z orbitals can mix, and that mixing becomes important where the two bands approach each other at pi over a. The bands avoid crossing. The 2p z band picks up some 2s character, which raises its energy, the 2s band picks up some p character, which lowers its energy, and we get something that looks a little like sp hybrids. We saw something very similar with s p mixing in the molecular orbital diagram of O₂.
Lessons from the Linear F Chain
Four orbitals per cell one F atom with 1 × 2s + 3 × 2p gives four bands
2p σ runs downhill the 2p wavefunction changes sign at the nucleus, so this band runs opposite to 2s σ
π bands narrower reduced spatial overlap of the π interaction
2s below 2p the atomic orbital energies set the starting heights
So here is what the fluorine chain shows us. There are four orbitals in the unit cell, a 2s and three 2p orbitals, so there are four bands. Sketching the orbitals at k equals zero and pi over a shows that the 2p z band runs downhill in energy and the other three run uphill. The pi bands from the 2p x and the 2p y are narrower than the sigma bands, because the pi overlap is smaller. And the 2s atomic orbital is quite a bit lower in energy than the 2p.
Key Concepts: 1D Band Structures
Concept What it says
Band Bloch functions $\psi = e^{ikx}u(x)$ sets the phase relationship across unit cells Zone boundaries $k = 0$ uniform phase, $k = \pi/a$ alternating phase ($e^{i\pi} = -1$) Orbital-band rule orbitals in the unit cell = bands Peierls distortion dimerization opens a gap by stabilizing filled states Material Classification Fermi level cuts a band (metal) or falls between bands (semiconductor, insulator) Direct gap VBM and CBM at the same k, efficient optical transitions (LEDs) Indirect gap different k-points, a phonon supplies the momentum (Si solar cells) Bandwidth σ bands wider than π bands from stronger overlap
In summary: The Bloch function, e to the i k x times u of x, sets how the phase changes from one unit cell to the next. At k equals zero the phase is the same in every cell, and at pi over a it alternates. The number of bands equals the number of orbitals in the unit cell, and a Peierls distortion opens a gap by stabilizing the filled states. If the Fermi level cuts a band we have a metal, and if it falls between bands, a semiconductor or an insulator. A direct gap gives efficient optical transitions, as in LEDs, and an indirect gap needs a phonon, as in silicon solar cells. lastly, we saw sigma bands are wider than pi bands because their overlap is larger.
