An electron bound to a single carbon atom can have only discrete energies. What happens when many carbon atoms join together into a flat sheet? The electron waves (wavefunctions) spread onto neighbouring atoms, and the allowed energies spread over a range. This range of energies is called a band. The gaps between bands and the way they meet help determine whether a material conducts electricity. In graphene, the clue to calculating these bands lies in the honeycomb lattice. Each repeating unit, the unit cell, contains two atoms, A and B, and each atom is coupled to three neighbours. To find out what bands this structure produces, we will use the tight-binding model, which describes electronic states through atomic orbitals and their interactions with neighbouring atoms.
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.
Let us start by considering a single bond between neighbouring A and B atoms. Write the $p_z$ orbital wavefunctions localized near the two atoms as $\varphi_A$ and $\varphi_B$. Without coupling in this model, the two orbitals have the same energy. We take this common orbital energy as our reference zero.
When the atoms are coupled, an electron initially localized near A can later be found near B as well. To describe an electronic state of the coupled pair, we therefore need a combination of both orbitals. With the coefficient of each orbital still unknown, write the electron’s wavefunction as
$$\psi=c_A\varphi_A+c_B\varphi_B.$$
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 coefficient multiplying the original orbital is zero because we chose the orbital energy before coupling as our reference zero. The coefficient $-t$ multiplying the other orbital describes the coupling that allows an electron to move between A and B. This coefficient is the hopping matrix element, and it has 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 combination $\varphi_A+\varphi_B$, with matching phases, and $\varphi_A-\varphi_B$, 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 energies were zero. Coupling separates them by $2t$, so a larger $t$ gives a larger energy difference between the two states.
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, we assigned a coefficient to each atom and found . To avoid solving separately for the countless coefficients, let us use the regular repetition of the crystal, its translational symmetry.
In a periodic crystal, we can study electronic states as Bloch waves: shifting the position by a lattice translation changes only the phase. The wavevector $\mathbf k$ specifies how that phase changes. Once the coefficient on one A atom is known, multiplying by the phase factor for the displacement gives the coefficient on any other A atom. The B coefficients can also be found by multiplying by the corresponding phase factors. At a given wavevector, this leaves just two unknown coefficients: one for A and one for B.
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, we saw that coupling two orbitals through a single bond $-t$ gives different energies for different combinations of the orbitals. Now the A and B coefficients are linked by $-tf(\mathbf k)$, which brings together the contributions from all three bond directions. Let us find how this coupling separates 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 positive and negative energies correspond to different combinations of the A and B components. For example, at $\Gamma$ the phases in all three directions match, so $f=3$. Then 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 $0,2\pi/3,-2\pi/3$. In the complex plane, the three arrows have length 1 and point in directions 120° apart. Joining them head to tail forms an equilateral triangle and brings us back to the starting point. 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 contributions from the three neighbouring B atoms to the same A orbital cancel exactly because of their different phases. On the B side, $f^*(\mathbf K)=0$ as well. The energy splitting therefore vanishes at this wavevector, and the two bands meet at $E=0$.
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 plot energy as a height for every wavevector throughout the first Brillouin zone. With $k_x,k_y$ as the two horizontal axes and energy on the vertical axis, the lower $\pi$ band and the upper $\pi^*$ band each form a surface.
The two surfaces meet at pointed, cone-like tips at the six corners corresponding to $K$ and $K^{\prime}$. The three bond directions and the wavevector-dependent phases have given us the full shape of graphene’s bands.
Now that we know the shapes of the two bands, let us work out how graphene’s electrons fill them. Knowing which states are filled and which are empty will help us identify where an electron can go when it gains energy. Start at absolute zero, filling the lowest-energy states first.
Let us count the states using 100 unit cells as an example. First, count each spatial wavefunction as one state, before distinguishing 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.
Now include spin. Two electrons with different spin orientations can occupy the same spatial state. The lower $\pi$ band has 100 such states, so it can hold up to 200 electrons.
In neutral graphene, each carbon atom contributes one electron to the states derived from $p_z$ orbitals. For 100 unit cells, that gives exactly 200 electrons, just enough to fill the lower $\pi$ band.
At absolute zero, the states with $E<0$ are therefore filled and those with $E>0$ are empty. The energy at this boundary is called the Fermi energy, here $E_F=0$. In the 3D plot, the $E=0$ plane touches the bands only at the points corresponding to $K$ and $K^{\prime}$. Neutral graphene, whose filled and empty bands meet at points, is called a semimetal.
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 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.
