An electron bound to a single carbon atom has only certain allowed energies. What happens when many carbon atoms join into a sheet? Electron waves spread across neighbouring atoms, and their energies spread into ranges called bands. The gaps and contacts between these bands help determine how a material conducts electricity. To calculate them for graphene, we can start with the honeycomb: two atoms, A and B, in each repeating unit, and three bonds from every atom. The tight-binding model describes electronic states using atomic orbitals and their coupling to neighbours. Let us see what bands these connections produce.
To describe graphene’s electronic states, first identify which atomic orbitals matter. An atomic orbital gives the shape of an electron wave around an atom. Carbon has four valence electrons. In graphene, three take part in the strong bonds within the sheet. The remaining one occupies the $p_z$ orbital, whose lobes extend above and below the sheet.
Conduction involves states near the boundary between filled and empty states. In graphene, these come mainly from $p_z$ orbitals. The occupied states of the in-plane bonds lie much lower in energy. We therefore describe the states relevant to conduction by combining one $p_z$ orbital from each carbon.
Neighbouring orbitals extend into the same region of space and couple to each other. Their tails become smaller farther from the atom, so direct coupling to distant atoms is generally weaker. Keeping only the bonds between the closest atoms is the nearest-neighbour approximation.
First keep just one bond between neighbouring A and B carbons. Write the $p_z$ orbital wavefunctions localized near the two atoms as $\varphi_A$ and $\varphi_B$. With the coupling switched off in this model, each has the same definite energy. Take this common orbital energy as the reference zero.
With the bond present, an electron initially localized on A develops a probability of being found on B as well. To describe states of the coupled pair, include both orbitals. Write the electron's wavefunction using unknown coefficients:
$$\psi=c_A\varphi_A+c_B\varphi_B.$$
This is a linear combination of orbitals. The coefficients are complex probability amplitudes: $c_j=|c_j|e^{i\theta_j}$ specifies the magnitude $|c_j|$ and phase $\theta_j$ with which each orbital contributes.
A state whose energy measurement has a single definite outcome $E$ is an energy eigenstate. Its wavefunction satisfies the Schrödinger equation:
$$H\psi=E\psi.$$
The Hamiltonian $H$ determines the electron's allowed energies and how its wavefunction changes with time. For an energy eigenstate, the two coefficients acquire only a common phase as time passes. Their magnitudes and relative phase stay fixed, so the probability distribution for finding the electron remains unchanged.
Call the Hamiltonian of the two-orbital model $H_2$, and write the coupling coefficient as $-t$, with $t>0$. Its action on each orbital is
$$\begin{aligned}H_2\varphi_A&=0\varphi_A-t\varphi_B,\\H_2\varphi_B&=0\varphi_B-t\varphi_A.\end{aligned}$$
The zero multiplying the original orbital is its energy before coupling. The term on the other orbital expresses the A–B coupling. Its coefficient $-t$ is the hopping matrix element, a quantity with units of energy.
With $\psi=\varphi_A$ alone, the left side contains the B orbital and cannot equal $E\varphi_A$ on the right. We therefore try the sum and the difference, which remain proportional to their original combinations when A and B are exchanged:
$$\begin{aligned}H_2(\varphi_A+\varphi_B)&=-t(\varphi_A+\varphi_B),\\H_2(\varphi_A-\varphi_B)&=+t(\varphi_A-\varphi_B).\end{aligned}$$
The sum $\varphi_A+\varphi_B$, with matching phases, and the difference $\varphi_A-\varphi_B$, with opposite phases, both satisfy $H\psi=E\psi$. These are the two energy eigenstates of the coupled pair, with energies $-t$ and $+t$, respectively. Without coupling, both orbital energies were zero. The bond splits them into two eigenenergies separated by $2t$. A larger $t$ produces a larger energy separation between the two combinations.
In graphene's honeycomb, each A has three B neighbours. Restore the other two bonds and label the neighbours B₁, B₂ and B₃. The three bonds are equivalent in the regular sheet, so they share the same $t$.
Let $c_1,c_2,c_3$ be the coefficients of the orbitals $\varphi_{B_1},\varphi_{B_2},\varphi_{B_3}$. The wavefunction extending across the crystal is
$$\begin{aligned}\psi={}&c;_A\varphi_A+c_1\varphi_{B_1}\\&+c_2\varphi_{B_2}+c_3\varphi_{B_3}+\cdots.\end{aligned}$$
The dots include the orbitals of all the other atoms. Call the crystal's Hamiltonian $H$ and substitute this $\psi$ into the left side of $H\psi=E\psi$. Applying $H$ to each orbital term and adding gives
$$\begin{aligned}H\psi&=H\bigl(c_A\varphi_A+c_1\varphi_{B_1}+c_2\varphi_{B_2}+c_3\varphi_{B_3}+\cdots\bigr)\\&=H(c_A\varphi_A)+H(c_1\varphi_{B_1})\\&\quad+H(c_2\varphi_{B_2})+H(c_3\varphi_{B_3})+\cdots.\end{aligned}$$
First collect the terms multiplying the central A orbital, $\varphi_A$. In the nearest-neighbour approximation, this A couples directly to B₁, B₂ and B₃, so we examine the contributions from these three neighbours.
The B₁→A bond contributes $-t\varphi_A$ to $H\varphi_{B_1}$. For the term $c_1\varphi_{B_1}$ in $\psi$, the same coefficient $c_1$ multiplies the result:
$$\begin{aligned}H(c_1\varphi_{B_1})&=c_1H\varphi_{B_1}\\&=-tc_1\varphi_A+\cdots.\end{aligned}$$
Applying $H$ to the B₂ and B₃ terms gives $-tc_2\varphi_A$ and $-tc_3\varphi_A$. These contributions all contain the same $\varphi_A$, so their coefficients add. The on-site part is $0\times c_A$, leaving
$$H\psi=-t(c_1+c_2+c_3)\varphi_A+\cdots.$$
On the right, $E\psi$ multiplies every coefficient by $E$. The condition $H\psi=E\psi$ requires the coefficient of every atomic orbital to match on the two sides. For the A orbital, the right-hand coefficient is $Ec_A$, giving
$$-t(c_1+c_2+c_3)=Ec_A.$$
Next compare the coefficients of $\varphi_{B_1}$ in the same equation $H\psi=E\psi$. B₁ has three A neighbours: the central A and two more atoms. Label the latter A₂ and A₃, with orbital coefficients $c_{A_2}$ and $c_{A_3}$. Their contributions give
$$-t(c_A+c_{A_2}+c_{A_3})=Ec_1.$$
B₂ and B₃ have , each involving their own three A neighbours. The energy $E$ belongs to one electronic state extending across the crystal, so it is common to . We seek coefficients that satisfy them together.
In §01, each atom had a coefficient and . The repeating lattice gives us a way to relate these coefficients.
In a periodic crystal, we can find electronic states as Bloch waves, which change only by a phase factor under a lattice translation. The wavevector $\mathbf k$ specifies this phase change. Once one A coefficient is known, translation gives the coefficients on the other A sites; the same holds for B. At a given wavevector, only two coefficients, one for A and one for B, remain to be found.
Put the central A at the origin. Its three B neighbours are at the bond vectors $\boldsymbol\delta_1,\boldsymbol\delta_2,\boldsymbol\delta_3$. Lattice translations take these B sites into one another, so their coefficients share a common factor $u_B$:
$$\begin{aligned}c_1&=u_Be^{i\mathbf k\cdot\boldsymbol\delta_1},\\c_2&=u_Be^{i\mathbf k\cdot\boldsymbol\delta_2},\\c_3&=u_Be^{i\mathbf k\cdot\boldsymbol\delta_3}.\end{aligned} \tag{1}$$
$u_B$ is a complex coefficient containing both magnitude and phase. Each exponential is a phase factor of magnitude 1, so all three coefficients have magnitude $|u_B|$. The A sites have their own common coefficient $u_A$. At the central A, the position-dependent phase factor is 1, giving $c_A=u_A$.
In the figure, move the wavevector from the centre $\Gamma$ to the right-hand corner $K$. This horizontal wavevector gives phase 0 at the upper B₁ and phases $\theta,-\theta$ at the right and left B sites.
The crests and troughs over the atoms show $\cos(k_xx)$, the real part of the phase factor after removing $u_B$. In the complex plane, these same factors are arrows of length 1. Joining the three arrows head to tail gives their sum.
At $\Gamma$, the three arrows point the same way and add to 3. As the wavevector moves, the arrows separate in direction and their sum changes. To use this in , factor $u_B$ out of the three coefficients:
$$\begin{gathered}c_1+c_2+c_3\\=u_B\bigl(e^{i\mathbf k\cdot\boldsymbol\delta_1}+e^{i\mathbf k\cdot\boldsymbol\delta_2}+e^{i\mathbf k\cdot\boldsymbol\delta_3}\bigr).\end{gathered} \tag{2}$$
Call the sum of the three phase factors in parentheses the structure factor $f(\mathbf k)$:
$$f(\mathbf k)=\sum_{j=1}^3e^{i\mathbf k\cdot\boldsymbol\delta_j}. \tag{3}$$
Now $c_1+c_2+c_3=u_Bf(\mathbf k)$. Substitute this and $c_A=u_A$ into .
Next, write . B₁ is at $\boldsymbol\delta_1$, and the vectors from it to its three A neighbours are $-\boldsymbol\delta_j$ $(j=1,2,3)$. Adding the starting position and the displacement gives each A position as $\boldsymbol\delta_1-\boldsymbol\delta_j$. For $j=1$, this is the A at the origin.
Using this position in the Bloch phase gives each A coefficient as $u_Ae^{i\mathbf k\cdot(\boldsymbol\delta_1-\boldsymbol\delta_j)}$. The B₁ coefficient is $c_1=u_Be^{i\mathbf k\cdot\boldsymbol\delta_1}$ from . Substitute these coefficients into :
$$\begin{gathered}E u_Be^{i\mathbf k\cdot\boldsymbol\delta_1}\\=-t\sum_{j=1}^3u_Ae^{i\mathbf k\cdot(\boldsymbol\delta_1-\boldsymbol\delta_j)}\\=-t u_Ae^{i\mathbf k\cdot\boldsymbol\delta_1}\sum_{j=1}^3e^{-i\mathbf k\cdot\boldsymbol\delta_j}.\end{gathered}$$
In , factor out $u_Ae^{i\mathbf k\cdot\boldsymbol\delta_1}$, which is common to all three terms. The remaining sum reverses the signs of the phases in , so it is the complex conjugate $f^*(\mathbf k)$. Cancel the common phase factor $e^{i\mathbf k\cdot\boldsymbol\delta_1}$ from the two sides. Together with , this gives
$$\begin{aligned}Eu_A&=-tf(\mathbf k)u_B,\\Eu_B&=-tf^*(\mathbf k)u_A.\end{aligned} \tag{4}$$
In §01, coupling two orbitals through $-t$ split their energies according to how the orbitals combined. Here the A and B coefficients are coupled through $-tf(\mathbf k)$, which collects the three bond contributions. How far apart does this put the energies?
Multiply by $E$, then use to replace $Eu_B$:
$$\begin{aligned}E^2u_A&=-tf(Eu_B)\\&=t^2ff^*u_A=t^2|f|^2u_A.\end{aligned}$$
Starting with B gives the same relation for $u_B$. An electron state requires at least one of $u_A,u_B$ to be nonzero, so $E^2=t^2|f|^2$. Here we used $ff^*=|f|^2$, the product of a complex number and its conjugate. Thus
$$\boxed{E_\pm(\mathbf k)=\pm t|f(\mathbf k)|}. \tag{5}$$
The two energies correspond to different combinations of the A and B components. At $\Gamma$, the three bond phases match and $f=3$, so gives
$$\begin{aligned}E=-3t&:\quad u_B=u_A,\\E=+3t&:\quad u_B=-u_A.\end{aligned}$$
At $\Gamma$, equal A and B magnitudes with matching phases give the lower eigenstate; opposite phases give the upper eigenstate. The sum and difference from §01 now describe states extending across the crystal.
Following each eigenenergy as the wavevector changes gives two bands: the lower π band and the upper π* band.
Their separation is $2t|f(\mathbf k)|$. Each bond has the same $t$, but the phase-factor sum changes the splitting. At $\Gamma$ the sum has its largest magnitude, giving a separation of $6t$. Where the sum is zero, the bands meet.
The bands meet at wavevectors where $f(\mathbf k)=0$. In the figure below, move through the hexagon to find where the sum of phase factors vanishes.
At the right-hand corner $\mathbf K=(4\pi/(3\sqrt3a),0)$, the three phases are $0,2\pi/3,-2\pi/3$. The arrows differ in direction by 120° and close into an equilateral triangle. In equations,
$$\begin{aligned}f(\mathbf K)&=1+e^{i2\pi/3}+e^{-i2\pi/3}\\&=1+2\cos\frac{2\pi}{3}=0.\end{aligned} \tag{6}$$
Return to . The three contributions multiplying the same A orbital in $H\psi$ become
$$\begin{gathered}-t(c_1+c_2+c_3)\varphi_A\\=-tu_Bf(\mathbf K)\varphi_A=0.\end{gathered}$$
The three B components give contributions to the same A orbital that cancel. On the B side, $f^*(\mathbf K)=0$ as well. Both eigenenergies are therefore zero at this wavevector, and the bands meet.
At the opposite corner $\mathbf K^{\prime}=-\mathbf K$, the three phases reverse sign and again sum to zero. Reciprocal lattice vectors relate the six corners in two groups of three, represented by $K$ and $K^{\prime}$. All six corners are band contacts.
With carbon–carbon distance $a$, the bond vectors are $\boldsymbol\delta_1=a(0,1)$ and $\boldsymbol\delta_{2,3}=a(\pm\sqrt3/2,-1/2)$. Substitution gives
$$f=e^{iak_y}+2e^{-iak_y/2}\cos\frac{\sqrt3ak_x}{2}$$
$$\begin{aligned}|f|^2={}&1+4\cos^2\frac{\sqrt3ak_x}{2}\\&+4\cos\frac{\sqrt3ak_x}{2}\cos\frac{3ak_y}{2}.\end{aligned}$$
The condition $f=0$ becomes $e^{i3ak_y/2}=-2\cos(\sqrt3ak_x/2)$. The right side is real, so in the first zone the possible heights are $k_y=0,\pm2\pi/(3a)$. At height 0 the roots are $k_x=\pm4\pi/(3\sqrt3a)$; at each of the other heights they are $k_x=\pm2\pi/(3\sqrt3a)$. These are precisely the six corners.
The axes here follow the geometry page.
How does the splitting change away from a contact? Choose a route in wavevector space from the centre $\Gamma$ to the corner $K$, then the edge midpoint $M$, and back to $\Gamma$.
The horizontal axis is the distance travelled along this route in wavevector space. The vertical axis gives the two energies divided by $t$. The small hexagon shows the wavevector currently selected on the graph.
Starting at $\pm3t$ at $\Gamma$, the two bands meet at zero at $K$, then separate to $\pm t$ at $M$. The final $\Gamma$ returns to the starting wavevector.
Now assign energies to every wavevector in the hexagon. With $k_x,k_y$ as the two horizontal axes and energy as height, the lower band forms one surface and the upper band another.
The two surfaces touch at the six corners belonging to $K$ and $K^{\prime}$. The three bond directions and the wavevector-dependent phases have given us the full band shape.
The bands tell us the energies available to an electron. If graphene receives a little energy, which empty states could its electrons reach? To answer this, distinguish occupied and empty states. Start at zero temperature, with electrons filling the lowest energies first.
Count the states of a periodic crystal using 100 unit cells. First, count each spatial wavefunction as one state, before including spin.
There are 100 A and 100 B atoms, giving 200 $p_z$ orbitals. Coupling changes the wave shapes and energies while leaving 200 independent states that can be represented by these orbitals.
Each wavevector has a pair of states, corresponding to the upper and lower bands. At a contact their energies coincide, and they still count as two independent states. The 200 states therefore divide equally: 100 states in each band.
Including spin, each spatial state holds two electrons, one for each spin orientation. The lower band can therefore hold 200 electrons.
In neutral graphene, each carbon contributes one electron to the $p_z$ states. These 100 unit cells supply 200 electrons, exactly the number needed to fill the lower band.
At zero temperature, states with $E<0$ are occupied and those with $E>0$ are empty. The energy separating occupied and empty states is the Fermi energy, here $E_F=0$. In the surface plot, the $E=0$ plane meets the bands at the contacts belonging to $K$ and $K^{\prime}$. Neutral graphene is a semimetal, with occupied and empty bands touching at points.
As the wavevector approaches a contact, the separation $2t|f(\mathbf k)|$ between the lower and upper states becomes smaller. An empty state lies just above an occupied one. To find which empty states an electron can reach with a small amount of energy, we therefore look near $K$ and $K^{\prime}$.
The three neighbouring contributions cancel at $K$, where the two bands meet. In the surface plot, they meet at pointed, cone-like tips.
Following this shape more closely leads to a surprising description: near the contact, electrons behave like particles with no mass. How can arranging carbon atoms in a honeycomb lead to that? On the next page, we zoom in on the contact and follow the connection from the band shape to the electrons' behaviour.
