It is useful to try to understand how the nearly-free electron model results in band structure in two dimensions. Let us consider a square lattice of monovalent atoms. The Brillouin zone is correspondingly square, and since there is one electron per atom, there should be enough electrons to half fill a single Brillouin zone. In absence of a periodic potential, the Fermi sea forms a circle as shown in the left of Fig. 15.3. The area of this circle is precisely half the area of the zone. Now when a periodic potential is added, gaps open up at the zone boundaries. This means that states close to the zone boundary get moved down in energy — and the closer they are to the boundary, the more they get moved down. As a result, states close to the boundary get filled up preferentially at the expense of states further from the boundary. This deforms the Fermi surface1 roughly as shown in the right of Fig. 15.3. In either case, there are low energy excitations possible and therefore the system is a metal.
Figure 15.4: Fermi Surfaces that Touch Brillouin Zone Boundaries. Left: Fermi Sea of a square lattice of monovalent atoms in two dimensions with strong periodic potential. The Fermi surface touches the Brillouin zone boundary. Right: The Fermi surface of copper, which is monovalent (the lattice structure is fcc, which determines the shape of the Brillouin zone, see Fig. 12.6).
If the periodic potential is strong enough the Fermi surface may even touch2 the Brillouin
zone boundary as shown in the left of Fig. 15.4. This is not uncommon in real materials. On the right of Fig. 15.4 the Fermi surface of copper is shown, which similarly touches the zone boundary.
1Recall that the Fermi surface is the locus of points at the Fermi energy (so all states at the Fermi surface have
the same energy), separating the filled from unfilled states. Keep in mind that the area inside the Fermi surface is fixed by the total number of electrons in the system.
2Note that whenever a Fermi surface touches the Brillouin zone boundary, it must do so perpendicularly. This is
due to the fact that the group velocity is zero at the zone boundary — i.e., the energy is quadratic as one approaches normal to the zone boundary. Since the energy is essentially not changing in the direction perpendicular to the zone boundary, the Fermi surface must intersect the zone boundary normally.
15.2. ENERGY BANDS IN TWO (OR MORE) DIMENSIONS 167
Figure 15.5: Fermi Sea of a Square Lattice of Divalent Atoms in Two Dimensions. Left: In the absence of a periodic potential, the Fermi sea forms a circle whose area is precisely that of the Brillouin zone (the black square). Right: when a sufficiently strong periodic potential is added, states inside the zone boundary are pushed down in energy so that all of these states are filled and no states outside of the first Brillouin zone are filled. Since there is a gap at the zone boundary, this situation is an insulator. (Note that the area of the Fermi sea remains fixed).
Let us now consider the case of a two-dimensional square lattice of divalent atoms. In this case the number of electrons is precisely enough to fill a single zone. In the absence of a periodic potential, the Fermi surface is still circular, although it now crosses into the second Brillouin zone, as shown in the left of Fig. 15.5. Again, when a periodic potential is added a gap opens at the zone boundary — this gap opening pushes down the energy of all states within the first zone and pushes up energy of all states in the second zone. If the periodic potential is sufficiently strong3,
then the states in the first zone are all lower in energy than states in the second zone. As a result, the Fermi sea will look like the right of Fig. 15.5. I..e, the entire lower band is filled, and the upper band is empty. Since there is a gap at the zone boundary, there are no low energy excitations possible, and this system is an insulator.
It is worth considering what happens for intermediate strength of the periodic potential. Again, states outside of the first Brillouin zone are raised in energy and states inside the first Brillouin zone are lowered in energy. Therefore fewer states will be occupied in the second zone and more states occupied in the first zone. However, for intermediate strength of potential, there will remain some states occupied in the second zone and some states empty within the first zone. This is precisely analogous to what happens in the right half of Fig. 15.2. Analogously, there will
3We can estimate how strong the potential needs to be. We need to have the highest energy state in the first
Brillouin zone be lower energy than the lowest energy state in the second zone. The highest energy state in the first zone, in the absence of periodic potential, is in the zone corner and therefore has energy corner= 2(π/2)2/(2m).
The lowest energy state in the second zone is it the middle of the zone boundary edge and in the absence of periodic potential has energy edge= (π/2)2/(2m). Thus we need to open up a gap at the zone boundary which is sufficiently
168 CHAPTER 15. INSULATOR, SEMICONDUCTOR, OR METAL
Figure 15.6: Fermi Sea of a Square Lattice of Divalent Atoms in Two Dimensions. Left: For intermediately strong periodic potential, there are still some states filled in the second zone, and some states empty in the first zone, thus the system is still a metal. Right: The states in the second zone can be moved into the first zone by translation by a reciprocal lattice vector. This is the reduced zone scheme representation of the occupancy of the second Brillouin zone.
still be some low energy excitations available, and the system remains a metal.
We emphasize that in the case where there are many atoms per unit cell, we should count the total valence of all of the atoms in the unit cell put together to determine if it is possible to obtain a filled-band insulator. If the total valence in of all the atoms in the unit cell is even, then for strong enough periodic potential, it is possible that some set of low energy bands will be completely filled, there will be a gap, and the remaining bands will be empty – i.e., it will be a band insulator.