The graphite used in pencil leads consists of sheets of carbon atoms stacked in layers. A single layer, just one atom thick, is graphene. Inside this thin sheet taken from such a familiar material, low-energy electrons behave like particles with no mass. How does a single sheet of carbon give rise to such strange behaviour?
The shapes and energies of electron waves depend on how the atoms connect. In graphene, each carbon atom bonds to three neighbours, forming a honeycomb. Expressing these connections and their repetition in equations gives us the foundation for calculating the electron waves and their energies. Our story of graphene begins by working out this honeycomb geometry.
Sliding a copy of the pattern over the original reveals where it repeats: if every atom lines up, the displacement connects two occurrences of the same arrangement. First, pick one carbon atom and slide the whole honeycomb towards a neighbouring atom by exactly the carbon-carbon distance $a$.
Half of the atoms land on other atoms; the other half fall into the empty centres of hexagons. Look at the three bonds leaving each atom. One group has a bond pointing up, the other a bond pointing down. The slide carries an up-atom onto a down-atom, so the bonds no longer line up.
Now slide from an up-atom to another up-atom. This time the whole pattern lands exactly on itself. Look at the up-atoms alone: they form a mesh of equilateral triangles, a triangular lattice. The down-atoms alone form the same triangular lattice, shifted a little from the first.
Let us call the up-atoms the A sites and the down-atoms the B sites. Both are carbon; their surroundings within the lattice differ. The repeating lattice is triangular, and placing one A and one B at each of its points produces the honeycomb.
The honeycomb can be rebuilt by repeating an A and the B directly above it as a pair. Carve these two dots into a stamp. Each press will add another copy of the pair.
Put the origin on an A site. Call the arrow to the upper-right A $\mathbf{a}_1$ and the arrow to the upper-left A $\mathbf{a}_2$. These two arrows form the sides of a rhombic stamp. Align its A mark with each A site and press in the same orientation: the rhombi fit together and fill the honeycomb.
To locate a press, count how many times the stamp moves along each arrow from the origin. After $m$ moves along $\mathbf{a}_1$ and $n$ along $\mathbf{a}_2$, its A mark is at
$$\mathbf{R} = m\,\mathbf{a}_1 + n\,\mathbf{a}_2 \tag{1}$$
Here $m,n$ are integers, with negative values for moves in the opposite direction. For example, $m=1,n=-1$ gives the horizontal move $\mathbf{a}_1-\mathbf{a}_2$. Combining the two arrows reaches every A site.
The set of positions $\mathbf{R}$ where the A mark lands is the Bravais lattice. The two arrows $\mathbf{a}_1,\mathbf{a}_2$ are its primitive vectors. Translating the whole honeycomb by any $\mathbf{R}$ brings it onto itself; this is a lattice translation.
The pair carved into the stamp is the basis: the group of atoms placed at each lattice point. The rhombus covered by one press is a unit cell. Here, each cell contains one A and one B.
Let us now express the two sides of this stamp, $\mathbf{a}_1$ and $\mathbf{a}_2$, in terms of the carbon-carbon distance $a$.
The two sides of the stamp are the primitive vectors $\mathbf{a}_1,\mathbf{a}_2$: the arrows from the A at the origin to the upper-right and upper-left A sites. Each can be built by following two bonds through the B directly above the origin. Adding the two arrows along that A→B→A path gives one side of the stamp.
Take the $x$ axis to the right and the $y$ axis upward, and write vectors as ($x$ component, $y$ component). The carbon-carbon distance in graphene is $a=0.142\,\mathrm{nm}$.
Call the three vectors from an A site to its neighbouring B sites $\boldsymbol{\delta}_1,\boldsymbol{\delta}_2,\boldsymbol{\delta}_3$. Each has length $a$. Since the interior angle of a regular hexagon is $120°$, the vectors are separated by $120°$. Take the first one straight up:
$$\boldsymbol{\delta}_1 = a\,(0,\ 1) \tag{2}$$
Turning this vector $120°$ to the right gives components $a(\sin120°,\cos120°)$; turning it to the left changes the sign of the $x$ component. Since $\sin120°=\tfrac{\sqrt3}{2}$ and $\cos120°=-\tfrac12$,
$$\begin{aligned} \boldsymbol{\delta}_2 &= a\!\left(\tfrac{\sqrt{3}}{2},\ -\tfrac{1}{2}\right) \\ \boldsymbol{\delta}_3 &= a\!\left(-\tfrac{\sqrt{3}}{2},\ -\tfrac{1}{2}\right) \end{aligned} \tag{3}$$
The vectors $\boldsymbol{\delta}_j$ point from A to B, so the vectors from B back to neighbouring A sites are $-\boldsymbol{\delta}_j$. Go from A to the B directly above it by $\boldsymbol{\delta}_1$, then from that B to the upper-right A by $-\boldsymbol{\delta}_3$. The total displacement is $\mathbf{a}_1$. Using $-\boldsymbol{\delta}_2$ for the second step gives $\mathbf{a}_2$.
$$\begin{aligned} \mathbf{a}_1 &= \boldsymbol{\delta}_1 - \boldsymbol{\delta}_3 = a\!\left(\tfrac{\sqrt{3}}{2},\ \tfrac{3}{2}\right) \\ \mathbf{a}_2 &= \boldsymbol{\delta}_1 - \boldsymbol{\delta}_2 = a\!\left(-\tfrac{\sqrt{3}}{2},\ \tfrac{3}{2}\right) \end{aligned} \tag{4}$$
From , $|\mathbf{a}_1|=a\sqrt{\tfrac34+\tfrac94}=\sqrt3\,a$. This is the lattice constant of the triangular lattice, the distance between neighbouring lattice points. With $a=0.142\,\mathrm{nm}$, it is $0.246\,\mathrm{nm}$.
On this page $a$ denotes the carbon-carbon distance. Some books instead use $a$ for the longer lattice constant $\sqrt3\,a$, so this difference in notation needs care when comparing formulas.
A lattice translation $\mathbf{R}=m\mathbf{a}_1+n\mathbf{a}_2$ brings the atoms back onto the same pattern. How far do the crests and troughs of a wave shift over that same displacement?
Place two small floats on water and send straight wave crests towards them. In some positions the floats rise and fall together; in others, one rises while the other falls. This difference in where two points lie in the wave's cycle is expressed as a phase difference.
A wave whose crests form parallel straight lines is a plane wave. At a fixed time, write it as $\cos(\mathbf{k}\cdot\mathbf{r})$. Here $\mathbf{r}=(x,y)$ is position, and the argument of the cosine, $\mathbf{k}\cdot\mathbf{r}$, is the phase. The wavevector $\mathbf{k}$ is perpendicular to the crests and has magnitude $2\pi/\lambda$, where $\lambda$ is the wavelength.
Moving from $\mathbf{r}$ to $\mathbf{r}+\mathbf{R}$ changes the phase from $\mathbf{k}\cdot\mathbf{r}$ to $\mathbf{k}\cdot(\mathbf{r}+\mathbf{R})$: the phase difference is $\mathbf{k}\cdot\mathbf{R}$. If it is an integer multiple of $2\pi$, the wave's crests and troughs coincide before and after the translation. Otherwise the wave shifts, even though the atoms line up.
The complex number $e^{i\mathbf{k}\cdot\mathbf{R}}$ represents the phase difference $\mathbf{k}\cdot\mathbf{R}$ and is called a phase factor. Phase differences separated by an integer multiple of $2\pi$ give the same phase factor.
Two waves can therefore have different wavelengths or directions yet give the same phase factor for every lattice translation. How much can their wavevectors differ for this to happen? Write that difference as $\mathbf{G}=\mathbf{k}'-\mathbf{k}$:
$$e^{i\mathbf{k}'\cdot\mathbf{R}} = e^{i(\mathbf{k}+\mathbf{G})\cdot\mathbf{R}} = e^{i\mathbf{k}\cdot\mathbf{R}}\,e^{i\mathbf{G}\cdot\mathbf{R}} \tag{5}$$
The two phase factors agree for every lattice translation if the extra factor $e^{i\mathbf{G}\cdot\mathbf{R}}$ is always 1:
$$e^{i\mathbf{G}\cdot\mathbf{R}} = 1 \quad\text{for every } \mathbf{R} \tag{6}$$
The difference $\mathbf{G}$ gives us a way to picture : draw $\cos(\mathbf{G}\cdot\mathbf{r})$ and look for crests that line up with every A site. Try changing their spacing in the figure.
Set $\lambda=1.50a$ and align the crests with the rows along $\mathbf{a}_2$. There is now one crest on each row. Denote this upper-right-pointing wavevector difference $\mathbf{G}$ by $\mathbf{b}_1$.
Substituting $\mathbf{G}=\mathbf{b}_1$ into the plotted wave $\cos(\mathbf{G}\cdot\mathbf{r})$ gives the phase $\mathbf{b}_1\cdot\mathbf{r}$. The phase difference is the dot product of the wavevector and the displacement, so moving by $\mathbf{a}_1$ changes the phase by $\mathbf{b}_1\cdot\mathbf{a}_1$. In the figure, that move reaches the next crest: one full cycle, or $2\pi$. Thus $\mathbf{b}_1\cdot\mathbf{a}_1=2\pi$. A displacement $\mathbf{a}_2$ stays on the same crest, so $\mathbf{b}_1\cdot\mathbf{a}_2=0$.
Keep the same wavelength and turn the crests to lie along $\mathbf{a}_1$. Denote the resulting upper-left-pointing $\mathbf{G}$ by $\mathbf{b}_2$. Now a displacement $\mathbf{a}_1$ stays on the same crest, giving phase difference 0, while $\mathbf{a}_2$ reaches the next crest, giving $2\pi$. Writing both waves' phase differences as dot products gives
$$\begin{aligned} \mathbf{b}_1\cdot\mathbf{a}_1 &= 2\pi, & \mathbf{b}_1\cdot\mathbf{a}_2 &= 0, \\ \mathbf{b}_2\cdot\mathbf{a}_1 &= 0, & \mathbf{b}_2\cdot\mathbf{a}_2 &= 2\pi \end{aligned} \tag{7}$$
Halve the crest spacing of the $\mathbf{b}_1$ wave. The original crests remain, with an extra crest between each pair, so every A site still lies on a crest. Since wavevector magnitude is inversely proportional to wavelength, this wave has $\mathbf{G}=2\mathbf{b}_1$. The vectors satisfying extend beyond $\mathbf{b}_1$ and $\mathbf{b}_2$.
To see how these vectors are arranged, use the components $G_x$ and $G_y$ as the two axes of a coordinate plane. This plane is reciprocal space, also called wavevector space, and the tip of each $\mathbf{G}$ is plotted as a point. The points $\mathbf{b}_1$ and $2\mathbf{b}_1$ lie in the same direction from the origin, with $2\mathbf{b}_1$ twice as far away. Points farther from the origin correspond to waves with more closely spaced crests.
On this plane, mark the positions of the $\mathbf{G}$ satisfying . These points form a lattice: the reciprocal lattice. A vector from the origin to one of these points is a reciprocal lattice vector. The two vectors $\mathbf{b}_1,\mathbf{b}_2$ are its primitive vectors.
Each marked point on the wavevector map represents a wave whose crests line up with all the A sites. Move away from the marked points and the crests no longer line up at every A site.
Just as $\mathbf{a}_1,\mathbf{a}_2$ locate the stamp positions, integer combinations of $\mathbf{b}_1,\mathbf{b}_2$ locate the reciprocal lattice points. With integers $m',n'$:
$$\mathbf{G} = m'\mathbf{b}_1 + n'\mathbf{b}_2 \tag{8}$$
For graphene, substitute and $\mathbf{b}_1=(x,y)$ into the first row of :
$$\begin{aligned} a\!\left(\tfrac{\sqrt{3}}{2}\,x + \tfrac{3}{2}\,y\right) &= 2\pi \\ a\!\left(-\tfrac{\sqrt{3}}{2}\,x + \tfrac{3}{2}\,y\right) &= 0 \end{aligned} \tag{9}$$
Adding gives $3ay=2\pi$, so $y=\tfrac{2\pi}{3a}$; substituting into gives $x=\tfrac{2\pi}{\sqrt3\,a}$. Solving for $\mathbf{b}_2$ in the same way yields
$$\begin{aligned} \mathbf{b}_1 &= \left(\frac{2\pi}{\sqrt{3}\,a},\ \frac{2\pi}{3a}\right) \\ \mathbf{b}_2 &= \left(-\frac{2\pi}{\sqrt{3}\,a},\ \frac{2\pi}{3a}\right) \end{aligned} \tag{10}$$
Both vectors have length $\tfrac{4\pi}{3a}$ and generate a triangular reciprocal lattice.
Wavevectors separated by one of these reciprocal lattice vectors give the same phase factor for every lattice translation. We group $\mathbf{k}$ and $\mathbf{k}+\mathbf{G}$ together by this shared pattern of phase differences.
Wavevector space extends without end, but wavevectors separated by a reciprocal lattice vector belong to the same group: they give the same phase factor for every lattice translation. To avoid counting that pattern over and over, we can keep one wavevector from each group.
Take a point $\mathbf{k}$ near $\mathbf{b}_1$. Subtracting $\mathbf{b}_1$ brings it close to the origin, while keeping it in the same group. If we always choose the point closest to the origin, how large a region will all these points fill?
Compare keeping $\mathbf{k}$ where it is with subtracting $\mathbf{b}_1$. The new point's distance from the origin equals the old point's distance from $\mathbf{b}_1$. So subtracting helps whenever $\mathbf{k}$ is closer to $\mathbf{b}_1$ than to the origin. The dividing line passes through the midpoint of the segment joining the origin and $\mathbf{b}_1$, at right angles to it: the perpendicular bisector.
There are six nearest reciprocal lattice points around the origin. Drawing a boundary with each one encloses a regular hexagon. Inside it, every point is closer to the origin than to any other reciprocal lattice point, so no reciprocal lattice translation can bring it closer. This region is the first Brillouin zone. Wavevectors outside it can be brought into it by subtracting a suitable $\mathbf{G}$.
On an edge, however, two reciprocal lattice points are equally close; at a corner, three are. The boundary therefore still contains multiple wavevectors from the same group. The six corners look alike. Do they all belong to one group?
Call the rightmost corner $K$ and translate it by reciprocal lattice vectors. Adding $\mathbf{b}_2$ reaches the upper-left corner; subtracting $\mathbf{b}_1$ reaches the lower-left one. Three alternating corners are connected.
Starting at the opposite corner $-K$ instead, adding $\mathbf{b}_1$ reaches the upper-right corner and subtracting $\mathbf{b}_2$ reaches the lower-right one. These are the other three corners. No integer combination of $\mathbf{b}_1$ and $\mathbf{b}_2$ takes $K$ to $-K$, so the two groups remain separate. The first three are called $K$ points, and the other three $K'$ points.
We now have two kinds of corners on one hexagon. Its centre $\Gamma$ and edge midpoints $M$ also give us landmarks for comparing waves. What differences appear when we lay the waves at these points over the honeycomb? In particular, how does the distinction between $K$ and $K'$ show up at the atoms?
The edge midpoint $\mathbf{M}$ lies halfway from the origin $\Gamma$ to $\mathbf{b}_1$, so $\mathbf{M}=\tfrac12\mathbf{b}_1=\left(\tfrac{\pi}{\sqrt3\,a},\tfrac{\pi}{3a}\right)$. The rightmost corner $\mathbf{K}$ is equally far from $\Gamma$, $\mathbf{b}_1$ and $-\mathbf{b}_2$. These three points form an equilateral triangle, whose centre gives
$$\mathbf{K} = \frac{\mathbf{0} + \mathbf{b}_1 - \mathbf{b}_2}{3} = \left(\frac{4\pi}{3\sqrt{3}\,a},\ 0\right) \tag{11}$$
The opposite corner has both coordinates reversed:
$$\mathbf{K}' = -\mathbf{K} = \left(-\frac{4\pi}{3\sqrt{3}\,a},\ 0\right) \tag{12}$$
To check whether opposite corners belong to the same group, write their displacement in terms of the reciprocal basis:
$$\mathbf{K}'-\mathbf{K}=-2\mathbf{K}=-\frac23\mathbf{b}_1+\frac23\mathbf{b}_2 \tag{13}$$
The coefficients $-\tfrac23$ and $\tfrac23$ are not integers, so this displacement is not a reciprocal lattice vector. The six corners fall into two distinct groups, $K$ and $K'$.
Choose a segment to highlight it on the hexagon. The distances below use the carbon–carbon bond length $a=0.142\,\mathrm{nm}=1.42\,\text{Å}$.
| Segment | Exact value | nm⁻¹ | Å⁻¹ |
|---|---|---|---|
| Γ–M | $\frac{2\pi}{3a}$ | 14.75 | 1.475 |
| Γ–K | $\frac{4\pi}{3\sqrt3\,a}$ | 17.03 | 1.703 |
| M–K | $\frac{2\pi}{3\sqrt3\,a}$ | 8.52 | 0.852 |
| Adjacent K–K′ (one edge) | $\frac{4\pi}{3\sqrt3\,a}$ | 17.03 | 1.703 |
| K–K (same family) | $\frac{4\pi}{3a}$ | 29.50 | 2.950 |
| Opposite K–K′ | $\frac{8\pi}{3\sqrt3\,a}$ | 34.06 | 3.406 |
| Nearest reciprocal lattice points | $\frac{4\pi}{3a}$ | 29.50 | 2.950 |
| Between the tips of b₁ and b₂ | $\frac{4\pi}{\sqrt3\,a}$ | 51.09 | 5.109 |
| Opposite edges | $\frac{4\pi}{3a}$ | 29.50 | 2.950 |
The hexagon has area $\dfrac{8\pi^2}{3\sqrt3\,a^2}$, approximately $753.6\,\mathrm{nm}^{-2}=7.536\,\text{Å}^{-2}$.
| Point / vector | $(k_x,k_y)$ |
|---|---|
| Γ | $(0,0)$ |
| M | $\left(\frac{\pi}{\sqrt3\,a},\frac{\pi}{3a}\right)$ |
| K · 0° | $\left(\frac{4\pi}{3\sqrt3\,a},0\right)$ |
| K · 120°, 240° | $\left(-\frac{2\pi}{3\sqrt3\,a},\pm\frac{2\pi}{3a}\right)$ |
| K′ · 180° | $\left(-\frac{4\pi}{3\sqrt3\,a},0\right)$ |
| K′ · 60°, 300° | $\left(\frac{2\pi}{3\sqrt3\,a},\pm\frac{2\pi}{3a}\right)$ |
| $\mathbf b_1,\mathbf b_2$ | $\left(\pm\frac{2\pi}{\sqrt3\,a},\frac{2\pi}{3a}\right)$ |
Distances and coordinates follow from the basis in .
Move from the centre $\Gamma$ to an edge midpoint $M$ or a corner $K$. Moving farther from the origin shortens the wavelength; moving around it turns the wave. Laid over the honeycomb, these waves also give different phases at the atoms.
Switching from $K$ to the opposite $K'$ keeps the same stripe pattern, but reverses how the phase arrows turn along the row.
The atomic positions have shown us how the honeycomb's repetition relates to the phases of a wave. What energies can an electron moving through this honeycomb have? To understand the opening puzzle—why electrons behave like particles with no mass—we need to find how their energy changes with wavevector.
The tight-binding (TB) method starts from atomic orbitals and includes the electron's hopping between atoms. Applied to the honeycomb, it gives the energies at each wavevector: the band structure. The shape of these bands near $K$ and $K'$ will lead us to the electrons' unusual behaviour.
