CHEM 548: Materials Chemistry

Lecture 16: Reciprocal Space, 2D Band Structures, and the CuO₂²⁻ Layer

Wed 10/28/2026 · Meeting 18

Reciprocal lattices, the first Brillouin zone, 2D band structures, and the CuO₂²⁻ layer

Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 6 §§6.3–6.5

Band Structure: the Linear F Atom Chain

A linear chain of fluorine atoms above four candidate band structure diagrams labelled a through d, each plotting E of k from k equals zero to k equals pi over a with a Fermi level drawn in, asking which one is correct.

Working the F-Chain Band Structure

An empty E of k axis frame for the fluorine chain running from k equals zero to k equals pi over a, which fills in one band at a time: the 2s sigma band running uphill, then the 2p-z sigma band running downhill, then the narrower 2p-x and 2p-y pi bands, then the Fermi level cutting the 2p-z band, and finally the avoided crossing between the 2p-z sigma and 2s sigma-star bands at the zone boundary.

The frame starts empty, with k running from zero out to pi over a. We fill it in one band at a time.

Same approach as the H chain, but now with both s and p orbitals

[advance: draw the 2s sigma band]

The 2s orbital is bonding at k = 0 and anti-bonding at k = π/a — this band runs uphill

[advance: add the 2pz sigma band]

For 2p_z, along the internuclear axis: uniform phases at k = 0 give anti-bonding interactions, high energy
At k = π/a the phase alternation gives bonding interactions, so this band runs downhill, ending below the anti-bonding 2s
Note the large intrinsic 2s-2p energy separation on fluorine

[advance: add the narrower pi bands]

2p_x and 2p_y, perpendicular to the chain: bonding π at k = 0, anti-bonding π at k = π/a — uphill, like the 2s band
This band is narrower, because π overlap is weaker than σ overlap

[advance: place the Fermi level]

Fluorine has 7 valence electrons, which fills 3.5 bands at two electrons per band
The Fermi level cuts the middle of the 2p_z band, three bands full below it — this hypothetical chain would be a metal

[advance: show the avoided crossing]

More rigorous calculations mix σ 2s and σ 2p_z as their bands approach each other at k = π/a
The bands avoid crossing: the 2s gains p character and drops, the 2p_z gains s character and rises, giving sp-hybrid character
Analogous to s-p mixing in the O₂ molecular orbital diagram

Next we collect what the fluorine chain taught us.

Lessons from the Linear F Chain

There are now 4 orbitals in the unit cell (a single F atom with 1 2s + 3 2p orbitals) giving rise to 4 bands in the band structure.
The fact that the 2p wavefunction changes sign at the nucleus causes the 2p σ band to run downhill (opposite of the 2s σ band).
The reduced spatial overlap of the π interaction causes the π bands to be narrower than the σ bands.
The 2p orbitals start out at a higher energy than the 2s orbitals (because of the atomic orbital energies)

Key Concepts: 1D Band Structures

Band Formation

Bloch functions: $\psi = e^{ikx}u(x)$ determines phase relationships across unit cells
Zone boundaries: $k = 0$ (uniform phase) vs $k = \pi/a$ (alternating phase via $e^{i\pi} = -1$)
Orbital-band rule: Number of orbitals in unit cell = number of bands
Peierls distortion: Dimerization opens band gap by stabilizing filled states

Material Properties

Classification: Fermi level cuts band (metal) vs between bands (semiconductor/insulator)
Direct gap: VBM and CBM at same k → efficient optical transitions (LEDs)
Indirect gap: Different k-points → requires phonon for momentum (Si solar cells)
Bandwidth: σ orbitals wider than π orbitals due to stronger overlap

Let me pull the one-dimensional story together, starting with how bands form.

Bloch functions: psi equals e to the i k x times u of x, which sets the phase relationship from one unit cell to the next
Zone boundaries: uniform phase at k = 0, alternating phase at k = π/a, because e to the iπ is −1
Orbital-band rule: the number of orbitals in the unit cell is the number of bands
Peierls distortion: dimerizing the chain opens a band gap by stabilizing the filled states

Then the material properties.

Classification: the Fermi level cutting a band gives a metal; falling between bands gives a semiconductor or an insulator
Direct gap: valence band maximum and conduction band minimum at the same k — efficient optical transitions, what LEDs are built on
Indirect gap: different k-points, so the transition needs a phonon for momentum — silicon solar cells
Bandwidth: σ bands are wider than π bands, because σ overlap is stronger

That is the whole one-dimensional picture; next we take these same ideas into two dimensions.

Learning Objectives: 2D Band Structures

By the end of this lecture, you will be able to:

Construct reciprocal lattice vectors from real space lattice vectors using perpendicularity and length requirements
Identify the first Brillouin zone as the Wigner-Seitz cell of reciprocal space
Locate and interpret special k-points (Γ, X, M) in 2D square lattices
Predict crystal orbital phase relationships at different k-points in two dimensions
Read and construct band structure diagrams for 2D materials using cuts through reciprocal space

Real and Reciprocal Lattices in 2D

A real-space rectangular lattice generated by vectors a and b next to its reciprocal-space lattice generated by a-star and b-star.

Defining the Reciprocal Lattice Vectors

The reciprocal lattice with its defining relations: a-star perpendicular to b with a-star dot a equal to two pi, then the resulting formulas a-star equals two pi over a and b-star equals two pi over b.

Let's consider a little bit more carefully how the reciprocal space lattice vectors are related to the real space lattice vectors.

Perpendicularity: A star must be perpendicular to the real space B lattice vector, and B star must be perpendicular to the real space A lattice vector
For this rectangular lattice, right-angle vectors mean A star is parallel to A, and B star is parallel to B
Length requirement: the dot product of A star times A is equal to 2 pi, and the dot product of B star times B is equal to 2 pi

[advance: put the two requirements together]

Putting both together gives the length of A star = 2 pi divided by the length of the real space A lattice vector, and the analogous equation for B star
So the longer the lattice vector in real space, the shorter the reciprocal space lattice vector

That inverse relationship is exactly what we need to connect reciprocal space to band structures.

Reciprocal Space in 1D

A one-dimensional real-space lattice and its reciprocal lattice, then the k values minus pi over a, zero, and plus pi over a, then the first Brillouin zone boundaries at one half a-star.

How does reciprocal space relate to band structure? Let's return to one dimension.

We previously skipped reciprocal lattices in 1D because they're trivial — only one direction, so real and reciprocal space differ only in unit cell length
Reciprocal vector length = 2π over real space vector length — the inverse relationship we saw earlier

[advance: label the k values]

But remember our k vector also had a 1 over distance component
First B1 zone: to know everything about the electronic structure, examine k between minus pi over A and plus pi over A
That interval is exactly the reciprocal space unit cell — so the first B1 zone is the unit cell of the reciprocal space lattice, partly why it's so important

[advance: mark the Brillouin zone boundaries]

To find the B1 zone: start at a lattice point, take one half of the reciprocal space lattice vector in either direction

Next we do exactly the same construction in two dimensions.

The First Brillouin Zone in 2D

The first Brillouin zone of a 2D reciprocal lattice drawn by bisecting the reciprocal lattice vectors, then its formal definition as a Wigner-Seitz cell.

So let's take that idea now and bring it into two dimensions.

Our 2D rectangular reciprocal space lattice
Define a rectangle that bisects both unit cell vectors halfway to the next reciprocal space lattice point — that would be our first B1 zone

[advance: add the formal definition]

Formal definition: the region of k-points closer to the origin than to any other lattice point
A unit cell formed by bisecting lattice vectors: a Wigner-Seitz cell
Unlike crystallographic unit cells, the first BZ doesn't need to be a parallelogram (2D) or parallelepiped (3D) for non-orthogonal lattices like hexagonal or oblique
Representing 2D band structure needs energy at every point in the first BZ — a 3D plot with kx, ky, and energy axes; we typically take cuts through it

Those cuts run between a few special high-symmetry points.

Special Points in a 2D Square Lattice

The gamma, X, and M special points of a 2D square lattice Brillouin zone, then their coordinates in reciprocal lattice vectors, then the same points written as Cartesian wave vectors.

$\mathbf{k}$ is the wave vector; $\mathbf{a}^{*}$, $\mathbf{b}^{*}$ are the reciprocal lattice vectors. Per Solid State Materials Chemistry, §6.4.

Special high-symmetry points are convenient for representation because band extrema (maxima and minima) typically occur there.

2D square lattice, three special points: Γ (gamma) at the BZ center, X at the zone boundary in the A* direction, M at the zone corner
A point at the B* boundary would be equivalent to X by the square lattice's fourfold symmetry
Points defined by coordinates in reciprocal lattice vectors

[advance: list the special-point coordinates]

Gamma: zero a-star plus zero b-star; X: one-half a-star plus zero b-star; M: one-half a-star plus one-half b-star
Recall for this square lattice: a-star is two pi over a along x, b-star is two pi over b along y

[advance: rewrite them as wave vectors]

k is now a vector, not a scalar — the wave vector, or crystal momentum vector, written in terms of A* and B* (per Solid State Materials Chemistry, section 6.4)
With A* = 2π/A and B* = 2π/B, and A = B for a square lattice, both equal 2π/A
Rewritten: at Γ, kx = 0 and ky = 0; X is (1/2)A*, so π/A times the kx unit vector plus 0 times ky; M: π/A in both kx and ky
We already know how to treat these: k = 0 → orbital phase doesn't change between unit cells; k = π/A → reverse orbital phase between neighboring unit cells

Now let's put these k vectors to work.

Band Structure of a 2D H-Atom Sheet

A square lattice of hydrogen atoms, then its crystal orbitals at the gamma point, the X point, and the M point, then the resulting band structure diagram from gamma to X to M and back to gamma.

So let's illustrate these k vectors with a simple example.

2D hydrogen atom sheet, hydrogens forming a square lattice — here's the positions of our atoms

[advance: show the gamma-point crystal orbital]

At gamma, all hydrogen 1s orbitals have the same phase, because kx is 0 and ky is 0 — moving to a neighboring unit cell, phases stay the same; e to the ikx is just 1, a constant

[advance: add the X-point crystal orbital]

At X, kx becomes π/A — phase changes every unit cell in x (positive, negative, positive), but ky = 0, so no phase change in y
Essentially non-bonding: each atom anti-bonding with its two x-neighbors, bonding with its two y-neighbors

[advance: add the M-point crystal orbital]

At M, phases alternate in both x and y — each hydrogen has four anti-bonding interactions: the most anti-bonding crystal orbital
Energy ordering: gamma to X to M

[advance: draw the band structure]

y-axis energy, as in 1D; x-axis: special k-points — labeled positions, not a continuous coordinate
Γ all-bonding, minimum energy; X anti-bonding one direction, bonding the other — approximately non-bonding, intermediate; M anti-bonding both directions, maximum
Path Γ to X to M, often continuing M back to Γ — somewhat arbitrary; points between are a line through the first Brillouin zone
Like the H atom chain: half-filled band, Fermi level cuts through around X — a metal

Before we leave the hydrogen sheet, let's check your understanding with a few quick questions.

Check Your Understanding: Reciprocal Vectors

Question 1 asking for the defining relationship of the reciprocal lattice vector A-star, then its answer: perpendicular to B with A-star dot A equal to two pi.

Question 1, on reciprocal lattice vectors. In a 2D real space lattice defined by vectors A and B, what is the fundamental defining relationship for the reciprocal space lattice vector A star?

A: A star parallel to A, length of A star is two pi over the length of A
B: A star parallel to B, A star dot B equals two pi
C: A star perpendicular to A, A star dot A equals one
D: A star perpendicular to B, A star dot A equals two pi
Take a moment and think about it

[advance: reveal the answer]

The answer is D. This is the correct general definition. A star must be perpendicular to the other real space vector, B, and its dot product with its corresponding real vector, A, is two pi.

Question 2 asks about the first Brillouin zone.

Check Your Understanding: The Brillouin Zone

Question 2 asking what the first Brillouin zone is, then its answer: the Wigner-Seitz cell of the reciprocal space lattice.

Check Your Understanding: The Γ Point

Question 3 asking what the special k-point gamma represents, then its answer: the center of the first Brillouin zone where k x and k y are zero.

Check Your Understanding: The M Point

Question 4 asking for the nature of the hydrogen-sheet crystal orbital at the M point, then its answer: phases alternate in both directions, all-antibonding.

Question 4. For the 2D square lattice of hydrogen atoms, what is the nature of the crystal orbital at the M point, where k x and k y are both pi over A?

A: all orbitals in phase, all-bonding
B: orbitals alternate phase in both the x and y directions, all-antibonding
C: orbitals randomly phased, zero net overlap
D: orbitals alternate phase in the x direction but are in phase in the y direction
Take a moment and think about it

[advance: reveal the answer]

The answer is B. At the M point, the phase reverses in both directions, creating a checkerboard pattern where every orbital is out of phase with all its neighbors, resulting in the highest energy, all-antibonding orbital.

One last question, on lengths in reciprocal space.

Check Your Understanding: Vector Lengths

Question 5 asking what happens to the reciprocal vector length when the real-space vector is made shorter, then its answer: it becomes longer.

Learning Objectives: The CuO₂²⁻ Layer

By the end of this lecture, you will be able to:

Construct band structures for multi-atom unit cells from molecular orbital diagrams
Interpret density of states (DOS) and partial DOS diagrams to identify orbital contributions
Predict orbital overlap patterns and symmetry-allowed interactions at different k-points
Relate band width to orbital overlap strength and electron delocalization
Extract key properties from band structure diagrams: metallic vs semiconducting behavior, localized vs delocalized electrons

La₂CuO₄: Home of the CuO₂²⁻ Layer

The layered perovskite structure of lanthanum copper oxide next to the square CuO2 sheet viewed down the stacking axis.

Today's topic is the CuO₂²⁻ layer.

Chosen for two reasons: our first structure with two atom types, and more than a textbook case — this copper oxide layer is the key building block in cuprate superconductors
Example: lanthanum copper oxide, La₂CuO₄ — a layered perovskite structure
Replacing some lanthanum with divalent cations like strontium or barium oxidizes copper from +2 to +3, triggering superconductivity — this late-1980s discovery sparked tremendous excitement
Drawing shows octahedral coordination around copper
But Cu²⁺ has a pronounced Jahn-Teller distortion — very long axial bonds; ignoring them, we can treat the copper as square planar

Let me pull that sheet out of the structure.

The CuO₂²⁻ Sheet

The CuO2 sheet with its square unit cell, copper at the corners and oxygen at the edge midpoints, next to the d-orbital energies of square planar Cu2+ with dx2-y2 highest.

MO Diagram → DOS Plot

The extended Hueckel MO diagram of the CuO2 sheet broadened into the total density of states with regions one, two, and three, then a highlight on the copper 3d partial DOS, then the highlight cleared.

Let's start with the band structure results, then work backwards to rationalize them.

Left: calculated MO diagram for square planar Cu²⁺, extended Hückel theory — four d orbitals (dxz, dyz, dxy, dz²) clustered around -14 eV, dx²-y² several eV higher as the sigma antibonding MO; lower down, predominantly O 2p orbitals with some Cu mixing forming bonding MOs
Broadening those MOs gives the density of states plot in the middle, three regions
Region I: O 2p nonbonding and Cu-O bonding bands — most bonding at the bottom, least at the top
Region II: the four d orbitals not pointing at ligands — dxy, dxz, dyz form pi antibonding bands, dz² weakly antibonding
Region III: the sigma star band from dx²-y², half-filled since this orbital holds one electron

[advance: highlight the partial density of states]

PDOS (right): where specific orbitals contribute
dx²-y²: contributions at the bottom (bonding with oxygen) and top (antibonding) — the large splitting reflects its strongest interaction with oxygen
dxy, in the xy plane: next highest splitting via π and π* interactions — bonding character in the oxygen bands below, antibonding in region II; dxz and dyz behave similarly
dz²: very narrow energy range — weak interactions with oxygen, essentially nonbonding

[advance: clear the highlight]

Relative areas reflect orbital counting: regions II and III are both Cu d, but II has four orbitals and III only one — so region III's area is one-fourth of region II
Region I: dominated by O 2p — two oxygens per copper, three p orbitals each, six bands total; region I's area should be about 50% larger than region II, which contains only four bands
The DOS is as far as the MO diagram takes us — for band widths and the actual band structure, the spaghetti diagram, we need to analyze orbital overlap

Let's do that overlap analysis now, starting with our basis set.

Basis Set Orbitals: the d_xy Band

The basis set for the dxy band: one copper dxy orbital and the two oxygen 2p pi orbitals of the unit cell.

The d_xy Band at Γ

The copper dxy and oxygen 2p orbital lattices at the gamma point, filled in with identical phases, then the bonding lobe, then the antibonding lobe, then the conclusion that mixing is symmetry forbidden and the band is nonbonding.

So let's start at the gamma point where KX and KY are both zero. Here I've separated the copper orbitals and the oxygen orbitals on two different diagrams just for clarity.

[advance: fill in all the orbital phases]

So because we're at gamma, it means that if we move to a neighboring unit cell, the phase of the orbital doesn't change.

And so we can see here that all of the copper DXY orbitals, they all have exactly the same phase.

Now the same must be true for the oxygen, so if we apply that rule, we would get these oxygen two p orbitals.

Notice that for the vertical ones, the shaded lobe is pointing up in all cases, and for the horizontal ones, the shaded lobe is pointing to the left in all cases.

The next thing is to say, all right, how are these going to interact with each other?

And the answer is the interaction is going to be zero.

To explain why that is, let's take one of these copper DXY orbitals, the one right in the middle, plop it in over here on the oxygen and look at the net interaction between copper and oxygen.

[advance: highlight the bonding lobe]

The lobe that goes off this way, that would have a bonding interaction with the oxygen two p orbitals.

[advance: highlight the antibonding lobe]

But the lobe that goes in the opposite direction, also shaded dark, is going to have an anti-bonding interaction with the oxygen two p orbitals.

[advance: show that mixing is symmetry forbidden]

If we were to look at the other two lobes, the one that comes upper right and lower left, they're going to encounter oxygen two p orbitals of opposite phase, and so that's also going to be equal bonding and anti-bonding overlap.

And so the net interaction is equal parts bonding and anti-bonding.

And when we see that, what it means is that these two sets of orbitals have different symmetries, and they're not permitted to mix by symmetry, so they're going to remain non-bonding at the gamma point.

And this copper DXY band that we're interested in, we can just call it non-bonding copper DXY at the gamma point.

Next, the M point.

The d_xy Band at M

The dxy band at the M point: the starting orbitals, then the copper phases alternating in x and y, then the oxygen phases alternating, then the resulting pi star crystal orbital.

Well, how does this picture change if we go to the M point where the KX and the KY are both equal to pi over A?

M means: translate by one unit cell vector in either the X or the Y direction, and the orbital phase has to change

[advance: flip the copper phases]

Every time I move one unit cell in either the X or the Y direction, the phases of the copper orbitals change.

[advance: flip the oxygen phases]

We're going to do exactly the same thing for the oxygens.

So here, notice I've translated by one unit cell in the X, I change the phase.

Translate by one unit cell in the X and I change the phase.

[advance: overlay the two sets]

Overlay the two orbital sets — they're all part of the same crystal
We see antibonding π interactions everywhere — the most antibonding π interaction possible between Cu dxy and O 2p orbitals

Now for the band at the top of the diagram.

Basis Set Orbitals: the d_x²−y² Band

The basis set for the dx2-y2 band: the copper dx2-y2 orbital and the oxygen 2p sigma orbitals pointing along the bonds.

The d_x²−y² Band at Γ

The dx2-y2 and oxygen 2p sigma lattices at gamma where mixing is symmetry forbidden, then the lower-lying oxygen 2s orbitals mixing weakly to give a sigma star crystal orbital.

We can apply the same analysis for the dx²-y² band at gamma.

All copper orbitals identical phases, as are the oxygen 2p sigma orbitals — horizontal ones with shaded lobes pointing left, vertical ones pointing down
Just as with dxy at gamma, symmetry prevents mixing: everywhere a bonding overlap exists, there's an equal and opposite antibonding overlap — these orbitals don't mix at gamma

[advance: bring in the oxygen 2s orbitals]

But Cu dx²-y² isn't completely nonbonding at gamma
The lower-lying O 2s orbitals, previously neglected, have the correct symmetry to mix in an antibonding way with dx²-y²
The interaction is weak — poor energetic matching between O 2s and Cu 3d — but it's not zero

Now let's take this sigma band to the M point.

The d_x²−y² Band at M

The dx2-y2 band at the M point: the starting copper and oxygen sigma orbitals, then the phases alternating in both directions, then the resulting most antibonding sigma star crystal orbital.

What about the d x squared minus y squared band at the M point, where k x and k y are both equal to pi over a?

Just as before: the orbital phase has to change every time we translate by one unit cell in x or y
Starting point: the copper d x squared minus y squared orbital and the two oxygen 2p sigma orbitals of the unit cell, on separate panels

[advance: alternate the phases across both lattices]

Every time I translate by one unit cell in the x or the y direction, I flip the phase of the copper orbital, and I apply exactly the same rule to the oxygen 2p sigma orbitals.

[advance: combine them into the crystal orbital]

Overlay the two sets: every copper lobe meets an oxygen lobe of the opposite phase — every copper-oxygen contact is antibonding
This is the sigma star crystal orbital at M — the most antibonding sigma interaction possible between the copper d x squared minus y squared orbitals and the oxygen 2p orbitals

Let's collect the two bands at all three special points.

Two Bands at Γ, X, and M

A grid comparing the sigma star dx2-y2 band and the pi star dxy band at gamma, X, and M: weakly antibonding or nonbonding at gamma, antibonding in x at X, antibonding in x and y at M.

Band Structure of the CuO₂²⁻ Sheet

The full band structure of the CuO2 sheet from gamma to X to M to gamma with the Fermi level cutting the copper dx2-y2 sigma star band, next to the total density of states.

Key Concepts: Band Structure

What is being plotted? Energy vs. k, where k is the wavevector that gives the phase of the MO’s on moving from one unit cell to the next (as well as the crystal momentum of the electron).

How many lines are there in a band structure diagram? As many as there are atomic orbitals in the unit cell.

How is the center of gravity energy level of each band determined? It usually follows from the MO diagram.

How do we determine whether a band runs uphill or downhill? By comparing the orbital overlap at k=0 and k=π/a.

How do we distinguish metals from semiconductors and insulators? The Fermi level cuts a band in a metal, whereas there is a gap between the filled and empty states in a semiconductor.

Why are some bands flat and others steep? This depends on the degree of orbital overlap between building units.

Wide bands → Large intermolecular overlap → delocalized e⁻
Narrow bands → Weak intermolecular overlap → localized e⁻

What's plotted: Energy (Y-axis) vs K vector (X-axis).

K vector: how orbital phases change between unit cells; relates to electron crystal momentum
Number of bands: one band per atomic orbital in the unit cell
Band energy position: the band's center of gravity derives from the orbital energy in the building unit's MO diagram
Band slope: determined by orbital overlap — predicts whether bands rise or fall from Γ to M
Property predictions: Fermi level cutting through a band indicates metallic conductivity; gaps between filled and empty bands indicate semiconducting or insulating behavior
Band width: wide bands — strong overlap between unit cells, delocalized electrons (if partially filled); narrow or flat bands, like DZ² — minimal overlap, localized electrons

That completes the copper oxide story; your practice questions for both halves of this lecture are on the next slide.

Practice questions

Work the Lecture 16 practice questions before the next class. They cover the band structure of a linear fluorine atom chain, reciprocal lattice vectors and the first Brillouin zone, the special k-points of a square lattice, the band structure of a 2D hydrogen sheet, and the CuO₂²⁻ layer from its density of states through the d_xy and d_x²−y² bands.

Open the Lecture 16 practice questions

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