Reduced Zone Scheme
11.4 Multiple Bands
Fig. 11.5Diatomic tight binding dis-persion in one dimension. Bottom:
Reduced zone scheme. Top: Extended zone scheme. Note that in obtaining the extended zone scheme from the re-duced zone scheme, one simply trans-lates pieces of the dispersion curve by appropriate reciprocal lattice vectors.
11.4 Multiple Bands
In the tight binding chain considered in this chapter, we considered only the case where there is a single atom in the unit cell and a single orbital per atom. However, more generally we might consider a case where we have multiple orbitals per unit cell.
One possibility is to consider one atom per unit cell, but several or-bitals per atom.10 Analogous to what we found for the tight binding
10Each atom actually has an infinite number of orbitals to be considered at higher and higher energy. But only a small number of them are filled, and within our level of approximation we can only consider very few of them.
model having only one orbital per atom, when the atoms are very far apart, one has only the atomic orbitals on each atom. However, as the atoms are moved closer together, the orbitals merge together and the energies spread to form bands.11Analogous to Fig. 11.3 we have shown
11This picture of atomic orbitals in the weak hopping limit merging together to form bands does not depend on the fact that the crystal of atoms is ordered.
Glasses and amorphous solids can have this sort of band structure as well!
how this occurs for the two band case in Fig. 11.6.
Fig. 11.6 Caricature of bands for a two-band model as a function of inter-atomic spacing. In the atomic limit, the orbitals have energies 1atomic and
2atomic. If the system has valence two per unit cell, then in the atomic limit the lower orbital is filled and the up-per orbital is empty. When the atoms are pushed together, the lower band remains filled, and the upper remains empty, until the bands start to over-lap, whereupon we have two bands both partially filled, which becomes a metal.
A very similar situation occurs when we have two atoms per unit cell but only one orbital per atom (see Exercises 11.2 and 11.4.) The general result will be quite analogous to what we found for vibrations of a diatomic chain in Chapter 10.
In Fig. 11.5 we show the spectrum of a tight-binding model with two different atoms per unit cell—each having a single orbital. We have shown results here in both the reduced and extended zone schemes.
As for the case of vibrations, we see that there are now two possible energy eigenstates at each value of k. In the language of electrons, we say that there are two bands (we do not use the words “acoustic” and
“optical” for electrons, but the idea is similar). Note that there is a gap between the two bands where there are simply no energy eigenstates.
Let us think for a second about what might result in this situation where there are two atoms per unit cell and one orbital per atom. If each atom (of either type) were divalent, then the two electrons donated per atom would completely fill the single orbital on each atom. In this case, both bands would be completely filled with both spin-up and spin-down electrons.
106 Tight Binding Chain (Interlude and Preview)
On the other hand, if each atom (of either type) is monovalent, then this means exactly half of the states of the system should be filled.
However, here, when one fills half of the states of the system, then all of the states of the lower band are completely filled (with both spins) but all of the states in the upper band are completely empty. In the extended zone scheme it appears that a gap has opened up precisely where the Fermi surface is (at the Brillouin zone boundary!).
In the situation where a lower band is completely filled but an upper band is completely empty, if we apply a weak electric field to the system can current flow? In this case, one cannot rearrange electrons within the lower band, but one can remove an electron from the lower band and put it in the upper band in order to change the overall (crystal) momentum of the system. However, moving an electron from the lower band requires a finite amount of energy—one must overcome the gap between the bands. As a result, for small enough electric fields (and at low temperature), this cannot happen. We conclude that a filled band is an insulator as long as there is a finite gap to any higher empty bands.
As with the single-band case, one can imagine the magnitude of hop-ping changing as one changes the distance between atoms. When the atoms are far apart, then one is in the atomic limit, but these atomic states spread into bands as the atoms get closer together, as shown in Fig. 11.6.
For the case where each of the two atoms is monovalent, in the atomic limit, half of the states are filled—that is, the lower-energy atomic orbital is filled with both spin-up and spin-down electrons, whereas the higher-energy orbital is completely empty (i.e., an electron is transferred from the higher-energy atom to the lower-energy atom and this completely fills the lower-energy band). As the atoms are brought closer together, the atomic orbitals spread into bands (the hopping t increases). However, at some point the bands get so wide that their energies overlap12—in
12See Exercise 11.4.
which case there is no gap to transfer electrons between bands, and the system becomes a metal as marked in Fig. 11.6. (If it is not clear how bands may overlap, consider, for example, the right side of Fig. 16.2. In fact band overlap of this type is very common in real materials!)
Chapter Summary
• Solving the tight-binding Schroedinger equation for electron waves is very similar to solving Newton’s equations for vibrational (phonon) waves. The structure of the reciprocal lattice and the Brillouin zone remains the same.
• We obtain energy bands where energy eigenstates exist, and gaps between bands.
• Zero hopping is the atomic limit. As hopping increases, atomic orbitals spread into bands.
Exercises 107
• Energies are parabolic in k near bottom of band—like free elec-trons, but with a modified effective mass.
• A filled band with a gap to the next band is an insulator (a band insulator), a partially filled band has a Fermi surface and is a metal.
• Whether a band is filled depends on the valence of the atoms.
• As we found for phonons, gaps open at Brillouin zone boundaries.
Group velocities are also zero at zone boundaries.
References
No book has an approach to tight binding that is exactly like what we have here. The books that come closest do essentially the same thing, but in three dimensions (which complicates life a bit). These books are:
• Ibach and Luth, section 7.3
• Kittel, chapter 9, section on tight-binding
• Burns, sections 10.9–10.10
• Singleton, chapter 4
Possibly the nicest (albeit short) description is given by
• Dove, section 5.5.5.
Also a nice short description of the physics (without detail) is given by
• Rosenberg, section 8.19.
Finally, an alternative approach to tight binding is given by
• Hook and Hall, section 4.3.
This is a good discussion, but they insist on using time-dependent Schroedinger equation, which is annoying.
Exercises
(11.1) Monatomic Tight Binding Chain
Consider a one-dimensional tight binding model of electrons hopping between atoms. Let the distance between atoms be called a, and here let us label the atomic orbital on atom n as|n for n = 1 . . . N (you may assume periodic boundary conditions, and you may assume orthonormality of orbitals, i.e., n|m = δnm). Suppose there is an on-site energy and a hopping matrix element −t. In other words, suppose n|H|m = for n = m and
n|H|m = −t for n = m ± 1.
Derive and sketch the dispersion curve for elec-trons. (Hint: Use the effective Schroedinger equa-tions of Exercise 6.2a. The resulting equation
should look very similar to that of Exercise 9.2.)
How many different eigenstates are there in this system?
What is the effective mass of the electron near the bottom of this band?
What is the density of states?
If each atom is monovalent (it donates a single electron) what is the density of states at the Fermi surface?
Give an approximation of the heat capacity of the system (see Exercise 4.3).
What is the heat capacity if each atom is diva-lent?
108 Exercises
(11.2) Diatomic Tight Binding Chain
We now generalize the calculation of the previous exercise to a one-dimensional diatomic solid which might look as follows:
−A − B − A − B − A − B−
Suppose that the onsite energy of type A is differ-ent from the onsite energy of type B. I.e,n|H|n
is A for n being on a site of type A and is B for n being on a site of type B. (All hopping matrix elements−t are still identical to each other.)
Calculate the new dispersion relation. (This is extremely similar to Exercise 10.1. If you are stuck, try studying that exercise again.)
Sketch this dispersion relation in both the re-duced and extended zone schemes.
What happens if A= B?
What happens in the “atomic” limit when t be-comes very small.
What is the effective mass of an electron near the bottom of the lower band?
If each atom (of either type) is monovalent, is the system a metal or an insulator?
*Given the results of this exercise, explain why LiF (which has very ionic bonds) is an extremely good insulator.
(11.3) Tight Binding Chain Done Right
Let us reconsider the one-dimensional tight bind-ing model as in Exercise 11.1. Again we assume an on-site energy and a hopping matrix element−t.
In other words, suppose n|H|m = for n = m andn|H|m = −t for n = m ± 1. However, now, let us no longer assume that orbitals are orthonor-mal. Instead, let us assumen|m = A for n = m andn|m = B for n = m + 1 with n|m = 0 for
|n − m| > 1.
Why is this last assumption (the |n − m| > 1 case) reasonable?
Treating the possible non-orthogonality of orbitals here is very similar to what we did in Exercise 6.5.
Go back and look at that exercise.
Use the effective Schroedinger equation from Exercise 6.5 to derive the dispersion relation for this one-dimensional tight binding chain.
(11.4) Two Orbitals per Atom
(a) Consider an atom with two orbitals, A and B having eigenenergies Aatomic and Batomic. Now suppose we make a one-dimensional chain of such atoms and let us assume that these orbitals remain
orthogonal. We imagine hopping amplitudes tAA
which allows an electron on orbital A of a given atom to hop to orbital A on the neighboring atom.
Similarly we imagine a hopping amplitude tBBthat allows an electron on orbital B of a given atom to hop to orbital B on the neighboring atom. (We as-sume that V0, the energy shift of the atomic orbital due to neighboring atoms, is zero).
Calculate and sketch the dispersion of the two resulting bands.
If the atom is diatomic, derive a condition on the quantities Aatomic − Batomic, as well as tAA
and tBB which determines whether the system is a metal or an insulator.
(b)* Now suppose that there is in addition a hop-ping term tABwhich allows an electron on one atom in orbital A to hop to orbital B on the neighbor-ing atom (and vice versa). What is the dispersion relation now?
(11.5) Electronic Impurity State*
Consider the one-dimensional tight binding Hamil-tonian given in Eq. 11.4. Now consider the situ-ation where one of the atoms in the chain (atom n = 0) is an impurity such that it has an atomic orbital energy which differs by Δ from all the other atomic orbital energies. In this case the Hamilto-nian becomes
Hn,m= 0δn,m− t(δn+1,m+ δn−1,m) + Δδn,mδn,0. (a) Using an ansatz
φn= Ae−qa|n|
with q real, and a the lattice constant, show that there is a localized eigenstate for any negative Δ, and find the eigenstate energy. This exercise is very similar to Exercise 9.6.
(b) Consider instead a continuum one-dimensional Hamiltonian with a delta-function potential
H =− 2
2m∗∂x2+ (aΔ)δ(x).
Similarly show that there is a localized eigenstate for any negative Δ and find its energy. Compare your result to that of part (a).
(11.6) Reflection from an Impurity*
Consider the tight binding Hamiltonian from the previous exercise representing a single impurity in a chain. Here the intent is to see how this impu-rity scatters a plane wave incoming from the left
Exercises 109
with unit amplitude (this is somewhat similar to Exercise 9.5). Use an ansatz wavefunction
φn=
T e−ikna n 0
e−ikna+ Re+ikna n < 0
to determine the transmission T and reflection R as a function of k.
(11.7) Transport in One Dimension*
(a) Consider the one-dimensional tight binding chain discussed in this chapter at (or near) zero temperature. Suppose the right end of this chain is attached to a reservoir at chemical potential μR
and the left end of the chain is attached to a reser-voir at chemical potential μL and let us assume μL > μR. The particles moving towards the left will be filled up to chemical potential μR, whereas the particles moving towards the right will be filled up to chemical potential μL, as shown in the bot-tom of Fig. 11.4, and also diagrammed schemati-cally in the following figure
μL μR
μR
μL
(i) Argue that the total current of all the particles moving to the right is
jR=
∞
0
dk
π v(k)nF(β(E(k)− μL)) with v(k) = (1/)d(k)/dk the group velocity and nF the Fermi occupation factor; and an analogous equation holds for left moving current.
(ii) Calculate the conductance G of this wire, de-fined as
Jtotal= GV
where Jtotal = jL− jR and eV = μL− μR, and show G = 2e2/h with h Planck’s constant. This
“quantum” of conductance is routinely measured in disorder free one-dimensional electronic systems.
(iii) In the context of Exercise 11.6, imagine that an impurity is placed in this chain between the two reservoirs to create some backscattering. Argue that the conductance is reduced to G = 2e2|T |2/h.
This is known as the Landauer formula and is a pillar of nano-scale electronics.
(b) Now suppose that the chemical potentials at both reservoirs are the same, but the temperatures are TLand TR respectively.
(i) Argue that the heat current jq of all the parti-cles moving to the right is
jqR=
∞
0
dk
πv(k) (E(k)− μ) nF(βL(E(k)− μ)) and a similar equation holds for left-moving heat current.
(ii) Define the thermal conductance K to be Jq= K(TL− TR)
where Jq = jLq− jqRand TL− TR is assumed to be small. Derive that the thermal conductance can be rewritten as
Evaluating this expression, confirm the Wiedemann–Franz ratio for clean one-dimensional systems
K
T G =π2k2B 3e2
(Note that this is a relationship between conduc-tance and thermal conducconduc-tance rather than be-tween conductivity and thermal conductivity.) In evaluating the above integral you will want to use
∞
If you are very adventurous, you can prove this nasty identity using the techniques analogous to those mentioned in footnote 20 of Chapter 2, as well as the fact that the Riemann zeta function takes the value ζ(2) = π2/6 which you can prove analogous to the appendix of that chapter.
(11.8) Peierls Distortion*
Consider a chain made up of all the same type of atom, but in such a way that the spacing between atoms alternates as long-short-long-short as follows
−A = A − A = A − A = A−
In a tight binding model, the shorter bonds (marked with =) will have hopping matrix ele-ment tshort = t(1 + ) whereas the longer bonds (marked with −) have hopping matrix element tlong = t(1− ). Calculate the tight-binding en-ergy spectrum of this chain. (The onsite enen-ergy is the same on every atom). Expand your result to linear order in . Suppose the lower band is filled and the upper band is empty (what is the valence of each atom in this case?). Calculate the total ground-state energy of the filled lower band, and show it decreases linearly with increasing .
110 Exercises
Now consider a chain of equally spaced identical A atoms connected together with identical springs with spring constant κ. Show that making a distor-tion whereby alternating springs are shorter/longer by δx costs energy proportional to (δx)2. Conclude that for a chain with the valence as in the first part of this problem, a distortion of this sort will occur spontaneously. This is known as a Peierls distor-tion.
(11.9) Tight Binding in 2d*
Consider a rectangular lattice in two dimensions as shown in the figure. Now imagine a tight bind-ing model where there is one orbital at each lat-tice site, and where the hopping matrix element is
n|H|m = t1 if sites n and m are neighbors in the horizontal direction and is = t2 if n and m are neighbors in the vertical direction. Calculate the dispersion relation for this tight binding model.
What does the dispersion relation look like near the bottom of the band? (The two-dimensional
structure is more difficult to handle than the one-dimensional examples given in this chapter. In Chapters 12 and 13 we study crystals in two and three dimensions, and it might be useful to read those chapters first and then return to try this ex-ercise again.)
t1 t2