1

Solid State Physics Semiclassical motion in a magnetic field 16 Lecture notes by Quantization of the cyclotron orbit: Landau levels 16 Michael Hilke Magneto-oscillations 17 McGill University (v. 10/25/2006) Phonons: lattice vibrations 17 Mono-atomic phonon dispersion in 1D 17 Optical branch 18 Experimental determination of the phonon Contents dispersion 18 Origin of the elastic constant 19 Introduction 2 Quantum case 19

The Theory of Everything 3 Transport (Boltzmann theory) 21 Relaxation time approximation 21 H2O - An example 3 Case 1: F~ = −eE~ 21 Diffusion model of transport (Drude) 22 Binding 3 Case 2: Thermal inequilibrium 22 Van der Waals attraction 3 Physical quantities 23 Derivation of Van der Waals 3 Repulsion 3 Semiconductors 24 Crystals 3 Band Structure 24 Ionic crystals 4 Electron and hole densities in intrinsic (undoped) Quantum mechanics as a bonder 4 semiconductors 25 Hydrogen-like bonding 4 Doped Semiconductors 26 Covalent bonding 5 Carrier Densities in Doped semiconductor 27 Metals 5 Metal-Insulator transition 27 Binding summary 5 In practice 28 p-n junction 28 Structure 6 Illustrations 6 One dimensional conductance 29 Summary 6 More than one channel, the quantum point Scattering 6 contact 29 Scattering theory of everything 7 1D scattering pattern 7 Quantum Hall effect 30 Point-like scatterers on a Bravais lattice in 3D 7 General case of a Bravais lattice with basis 8 superconductivity 30 Example: the structure factor of a BCC lattice 8 BCS theory 31 Bragg’s law 9 Summary of scattering 9

Properties of Solids and liquids 10 single electron approximation 10 Properties of the free electron model 10 Periodic potentials 11 Kronig-Penney model 11 Tight binging approximation 12 Combining Bloch’s theorem with the tight binding approximation 13 Weak potential approximation 14 Localization 14 Electronic properties due to periodic potential 15 Density of states 15 Average velocity 15 Response to an external field and existence of holes and electrons 15 Bloch oscillations 16 2

INTRODUCTION derived based on a periodic lattice. However it appeared that most of the properties are very similar independent What is Solid State Physics? of the presence or not of a lattice. But there are many Typically properties related to crystals, i.e., period- exceptions, for example, localization due to disorder or icity. the disappearance of the periodicity. What is Condensed Matter Physics? Properties related to solids and liquids including crys- Even in crystals, liquid-like properties can arise, such tals. For example: liquids, polymers, carbon nanotubes, as a Fermi liquid, which is an interacting electron system. rubber,... The same is also true the other way around since liquids Historically, SSP was considers as he basis for the un- can form liquid crystals. derstanding of solids since many of the properties were Hence, the study of SSP and CMP are strongly interre- lated and can not be separated. These inter-correlations are illustrated below.

The Big Bang of Condensed Matter Physics

Crystal Crystal

Supraconductivity

Opto-effects

Magnetism Metals Insulators Semiconductors properties 3

THE THEORY OF EVERYTHING

See ppt notes

H2O - AN EXAMPLE

See ppt notes

BINDING

What holds atoms together and also keeps them from collapsing? FIG. 1: Graphical representation of a H2 molecule We will start with the simplest molecule: H2 to ask what holds it together. The total Hamiltonian can then be written as the sum between the non interacting H atoms and the cross terms Van der Waals attraction due to Coulomb interactions. Hence,

The Van der Waals force arises simply from the change H = (H0) + (Hint) (1) in energy due to the cross Coulomb interactions between atom a and b, which is simply a dipole-dipole interaction. with

µ 2 2 2 ¶ µ ¶ ~ ¡ 2 2¢ e e 2 1 1 1 1 H = − ∇1 + ∇2 − − + e + − − (2) 2m r1a r2b rab r12 r1b r2a

Using standard perturbation theory it is then possible which leads to a total potential of the form to evaluate the gain in energy due to Hint. This is left µ ¶ σ6 σ12 as an exercise. The result is φ(r) = −4² − , (4) r6 r12 −αaαb ∆E ' , (3) which is usually referred to as the Lennard-Jones po- r6 ab tential. The choice of the repulsive term is somewhat arbitrary but it reflects the short range nature of the where α are the atomic polarizabilities. x interaction and represents a good approximation to the full problem. The parameters ² and σ depend on the molecule. Derivation of Van der Waals

Crystals Problem 1 What about crystals? Let’s first think about what kind of energy scales are involved in the problem. If we assume Repulsion that the typical distance between atoms is of the order of 1A˚ we have We just saw that there is an attractive potential of e2/1A˚ ' 14.4eV for the Coulomb energy and (5) the form −1/r6. If there were only a Coulomb repul- sion of strength 1/r this would lead to the collapse of our µ ¶2 molecule. In fact there is a very strong repulsion, which 1 ~2 ' 3.8eV for the potential in a 1A˚ quantum box comes from the Pauli principle. For the general pur- 1A˚ pose this repulsive potential is often taken to be ∼ 1/r12, (6) 4

In comparison to room temperature (300K' 25meV) the ground state energy is then given by E0=-13.6 eV. these energies are huge. Hence ionic Crystals like NaCl What happens if we add one proton or H+ to the system are extremely stable, with binding energies of the order which is R away. The potential energy for the electron is of 1eV. then e2 e2 U(r) = − − (10) Ionic crystals r |r − R| The lowest eigenfunction with eigenvalue -13.6eV is Some solids or crystals are mainly held together by the µ ¶ 1 1 3/2 electrostatic potential and they include the alkali-halides ξ(r) = √ e−r/a0 , (11) + − like (NaCl −→ Na Cl ). π a0

where a0 is the Bohr radius. But now we have two protons. If the protons were infinitely apart then the d general solution to potential 10 is a linear superposi- tion ,i.e., ψ(r) = αξ(r) + βξ(r − R) with a degener- ate lowest eigenvalue of E0 = E0=-13.6 eV. When R is not infinite, the two eigenfunctions corresponding to the lowest eigenvalues Eb and Ea can be approximated by ψb = ξ(r) + ξ(r − R) and ψa = ξ(r) − ξ(r − R). See figure 3. - +

Ψ (r) b

FIG. 2: A simple ionic crystal such as NaCl Ψ (r) V(r) a 0 The energy per ion pair is Energy e2 C 14.4eV C = −α + n = −α + n with 6 < n ≤ 12, R Nionpair d d [d/A˚] d (7) + + where α is the Madelung constant and can be calcu- 0 r lated from the crystal structure. C can be extracted ex- perimentally from the minimum in the potential energy and typically n = 12 is often used to model the effect of the Pauli principle. Hence, from the derivative of the FIG. 3: The potential for two protons with the bonding and potential we obtain: anti-binding wave function of the electron µ ¶ 12C 1/11 d = (8) 0 e2α The average energy (or expectation value of Eb) is ∗ ∗ so that Eb = hψb Hψbi/hψb ψbi (12) 0 A + B Energy 11αe2 = E − with (13) = (9) Z 1 + ∆ Nionpair 12d0 A = e2 drξ2(r)/|r − R| (14) How good is this model? See table below: Z B = e2 drξ(r)ξ(|r − R|)/r (15) Quantum mechanics as a bonder Z ∆ = drξ(r)ξ(|r − R|), (16) Hydrogen-like bonding R where h·i ≡ ·dr and similarly, Let’s start with one H atom. We fix the proton at A − B E = E0 − (17) r = o then we know form basic quantum mechanics that a 1 − ∆ 5

Hence, the total energy for state ψb is now This figure illustrates why they are called bonding and total 2 Eb = Eb + e /R (18) anti-bonding, since in the bonding case the energy is low- ered when the distance between protons is reduced as When plugging in the numbers Etotal has a minimum at b long as R > R . 1.5A˚. See figure 4 0

-8 Covalent bonding

Covalent bonds are very similar to Hydrogen bonds, E tot a only that we have to extend the problem to a linear com- Antibonding -12 bination of atomic orbitals for every atom. N atoms E would lead to N levels, in which the ground state has a a

E [eV] E bonding wave-function. tot -13.6 eV E b

-16 E b R Bonding 0 Metals R [distance between protons] In metals bonding is a combination of the effects dis- FIG. 4: The energies as a function of the distance between cussed above. The idea is to consider a cloud of electrons them for the bonding and anti-binding wave functions of the only weakly bound to the atomic lattice. The total elec- electron trostatic energy can then be written as

Z X 2 X 2 Z 2 e e e n(r1)n(r2) Eel = − drn(r) + 0 + 1/2 · dr1dr2 (19) r − R R − R |r1 − r2| R R>R0

which corresponds to an ionic contribution of the form This last expressions leads to a minimum at rs/a0 = 1.6. We can now compare this with experimental values and 2 µ ¶1/3 αe 3 the result is off by a factor between 2 and 6. What went Eel = − where rs = (20) 2rs 4πn wrong. Well, we treated the problem on a semiclassi- cal level, without incorporating all the electron-electron and α is the Madelung constant. Deriving this requires interactions in a quantum theory. This is very very diffi- quite a bit of effort. On top of this one has to add the cult, but in the large density case this can be estimated kinetic energy of the electrons, which is of the form: and a better agreement with experiments is obtained. µ ¶ 9π 2/3 3~2 E = (21) kin 4 10mr2 s Binding summary And finally we have to add the exchange energy, with is a consequence of the Pauli principle. The expression for There are essentially three effects which contribute to this term is given by the binding of solids: µ ¶ 9π 1/3 3 • Van der Waals (a dipole-dipole like interaction) E = − (22) ex 4 4πr s • ionic (Coulomb attraction between ions) Putting all this together we obtain in units of the Bohr radius: • Quantum mechanics (overlap of the wave-function) µ 2 ¶ 24.35a0 30.1a0 12.5a0 In addition we have two effects which prevent the collapse E = − + 2 − eV/atom (23) rS rS rs of the solids: 6

• Coulomb Miller indices can be obtained through the construc- tion illustrated in the figure below: • Quantum mechanics (Pauli)

STRUCTURE

FIG. 6: Plane intercepts the axes at (3a ˆ1, 2a ˆ2, 2a ˆ3). The in- verse of these numbers are (1/3, 1/2, 1/2), hence the smallest integers having the same ratio are 2,3,3, i.e., the Miller in- dices are (233). For a negative intercept the convention is 1.¯ FIG. 5: Pyrite, F eS crystal with cubic symmetry. 2 (Picture from Ashcroft and Mermin)

Illustrations SCATTERING

See ppt notes. See supplement on diffraction.

In order to determine the structure of a crystal it Summary is possible to observe the interference pattern produced by scattering particles with wave-length comparable to Periodic solids can be classified into two main classes: the lattice spacing, i.e., of the order if 1A˚. There are three main classes of particles, which can be used: • Bravais lattices: Every point of the lattice R~ can be reached from a a linear combination of the primitive • Photons, in particular X-rays, whose wave lengths ~ vectors: R = n1aˆ1 + n2aˆ2 + n3aˆ3, where ni are are around 1A˚. The probability to scatter off the integers. crystal is not that large, hence they can penetrate quite deeply into the crystal. The main scattering • Lattices with basis: Here every point in a primitive occurs with the electrons. Hence, what is really cell is described by a basis vector B~ so that any i observed with X-rays is the periodic distribution of point of the lattice can be reached through: R~ = electrons. n1aˆ1 + n2aˆ2 + n3aˆ3+B~i. • Neutrons are also extensively used since they In 3D there are 14 Bravais lattices and 230 symmetry mainly interact with the nuclei and the magnetic groups for lattices with basis. In 2D there are 5 Bravais moments. lattices and in 1D only 1. This is the zoology of crystals and they all have names. • Electrons, have a very scattering probability with It is important to remember that many of the physical anything in the crystal, hence they do not penetrate properties cannot be deduced from the crystal structure very deep, but they are therefore an interesting tool directly. The same crystal structure could be a metal to probe the surface structure. (Cu) or an insulator (Ca). All forms of scattering share very similar basic prin- ciples and can be applied to the scattering’s theory of everything described in the next section. 7

Scattering theory of everything Physics is hidden in n(r), which contains the informa- tion on the position of scatterers and their individual scattering distribution and probability. In the following we discuss the most important implications.

dV ← crystal k r 1D scattering pattern Incident ϕϕϕ k’ Outgoing beam beam O Let’s suppose we have a 1D crystal along directiony ˆ ik~ ·~r ⋅ ik'⋅r with lattice spacing a and that the incoming wave e in eik r e is perpendicular to the crystal alongx ˆ. We want to cal- ~ culate the scattered amplitude along direction kout. By The total scattered wave off V is ~ ~ defining ~q = kout − kin, we can write the scattering am- [ ⋅ ] A(q) ∝ ∫∫∫ dr 3n(r) ⋅e iq r plitude as V Z i~q·~r 3 q = k - k' A(~q) = n(~r)e dr (24) V where n( r) is the distribution of scatterers If we assume that the crystal is composed of point-like scatterers we can write:

C NX−1 FIG. 7: Diffraction set-up (picture from G. Frossati) n(x, y, z) = δ(y − an)δ(x)δ(z), (25) N n=0

This formulation can describe any scattering process where C is simply a constant. Hence, by inserting 25 in terms of the scattering amplitude A(~q). The entire into 24, and defining ~q = (0, q, 0) we have

N−1 ½ C X C 1 − eiaqN C when q = 2πm/a A(q) = eiqan = = = Cδ (26) N N 1 − eiaq N→∞ 0 when q 6= 2πm/a q,G~ n=0

G~ =y ˆ2πm/a is the reciprocal lattice and m an integer. and If we had a screen alongy ˆ we would see a diffraction 2 ½ pattern along I(q) = |A(q)| , since what is measured C NX−1 ~ i~q·R~n C when ~q ∈ G experimentally is the intensity. In this simple case the A(~q) = e =N→∞ = Cδ ~ N 0 when ~q ∈/ G~ ~q,G reciprocal lattice is the same as the real lattice, but with n=0 lattice spacing 2π/a instead. (28) In this case the Bravais lattice (or Real-Space) is

R~n = n1 ~a1 + n2 ~a2 + n3 ~a3 (29)

and the reciprocal space G~m (or k-space) can be deduced Point-like scatterers on a Bravais lattice in 3D ~ ~ from eiRn·Gm = 1, hence

We start again with the general form of A(~q) from 24 ~ ~ ~ G~m = m1b1 + m2b2 + m3b3, (30) and assume that our 3D crystal is formed by point-like scatterers on a Bravais lattice R~. Hence, where

N−1 ~ 2π ~a2 × ~a3 ~ 2π ~a3 × ~a1 ~ 2π ~a1 × ~a2 C X b1 = , b2 = and b3 = n(~r) = δ3(~r − R~ ), (27) | ~a1 · ~a2 × ~a3| | ~a1 · ~a2 × ~a3| | ~a1 · ~a2 × ~a3| N n n=0 (31) 8

As a simplification we suppose that f1 = f2 = f, then

S(~q) = (1 + e(ia/2)~q·(ˆx+ˆy+ˆz))fe(~q), (37)

where fe(~q) is the Fourier transform of f. Hence, ½ e 2δ~q,G~ if q1 + q2 + q3 is even A(~q) = S(~q)·δ~q,G~ = f(~q) 0 if q1 + q2 + q3 is odd General case of a Bravais lattice with basis (38)

The most general form for n(~r), when the atoms sits on the basis { ~uj} along the Bravais lattice R~n is

M,N−1 C X n(~r) = f (~r − R~ − ~u ), (32) N j n j j=1,n=0 where fj is the scattering amplitude for the atoms on site ~uj and is typically proportional to the number of electrons centered on ~uj. Now

M,N−1 Z C X A(~q) = f (~r − R~ − ~u )ei~q·~rd3r N j n j j=1,n=0 M,N−1 Z C X ~ = f (~r)ei~q·~rd3r · ei~q· ~uj · ei~q·Rn N j j=1,n=0

with ~r → ~r + R~n + ~uj

= CS(~q) · δ~q,G~ , (33) where we have defined the structure factor S(~q) as

XM Z i~q·~r 3 i~q· ~uj S(~q) = fj(~r)e d r · e (34) j=1

This is the most general form. It is interesting to re- mark that in most case, experiments measure the inten- sity I(~q) = |A(~q)|2, rather than the amplitude.

Example: the structure factor of a BCC lattice

The BCC crystal can be viewed as a cubic crystal with lattice a and a basis. Therefore, all lattice sites are de- scribed by

R~n = n1axˆ + n2ayˆ + n3azˆ + ~uj, (35) where ~u1 = 0(ˆx +y ˆ +z ˆ) and ~u2 = (a/2)(ˆx +y ˆ +z ˆ). The structure factor then becomes: X Z i~q·~r i~q· ~uj 3 S(~q) = fj(~r)e e d r. (36) j=1,2 9

Bragg’s law nuclear sites. The scattered wave amplitude with wave number ~k is then Equivalence between Bragg’s law for Miller planes and Z ~ ~ ~ ~ 3 i(k−kin)·~r the reciprocal lattice. A(k − kin) ∼ d rn(~r)e (43) V This leads to three typical cases:

• Mono crystal diffraction: Point-like Bragg peaks (1,0,0)

(0,1,0) Miller index in out out q2=k -k2 b k1

in out θ q1=k -k1 a kin FIG. 8: Point-like Bragg peaks from a single crystal. (Ref: lassp.cornell.edu/lifshitz) y z x • Powder diffraction:

From Bragg’s law we know that for the planes perpen- dicular tox ˆ or Miller index (1,0,0) the following diffrac- tion condition applies:

nλ = 2a sin(θ). (39) hence,

2π n = 2 sin(θ) · |k~in| = |~q |, (40) a 1 since k = 2π/λ and we supposed that k~in is perpendicular FIG. 9: Circle-like Bragg peaks from a powder with different grain sizes toz ˆ. This condition is equivalent to ~q1 =x ˆ2π/a ∈ G~ . The same relation applies for the scattering of the plane perpendicular toy ˆ or Miller index (0,1,0) the following diffraction condition applies: • Liquid diffraction: nλ = 2b cos(θ). (41) hence,

2π n = 2 cos(θ) · |k~in| = |~q |, (42) b 2 or ~q2 =y ˆ2π/b ∈ G~ .

FIG. 10: Pattern evolution for a complex molecule evolving Summary of scattering from a crystal-like structure to an isotropic liquid

~ We have an incoming wave eikin·~r diffracting on some sample with volume V and with a scattering probability The liquid diffraction is essentially the limiting case n(~r) inside V . For X-ray n is typically given by the elec- of the powder diffraction when the grain size be- tron distribution whereas for neutrons it is typically the comes comparable to the size of an atom. 10

PROPERTIES OF SOLIDS AND LIQUIDS where the pre-factor 2 comes from the spin degeneracy. Hence, The Theory of everything discussed in the first section k3 4π can serve as a guideline to illustrate which part of the n3D = F in D = 3 since V D = k3 (48) e 3π2 kF 3 F Hamiltonian is important for a given property. For in- 2 stance, when interested in the mechanical properties, the 2D kF D 2 ne = in D = 2 since Vk = πkF (49) terms containing the electrons can be seen as a pertur- 2π F bation. However, when considering thermal conductivity, 1D 2kF D ne = in D = 1 since Vk = 2kF , (50) for example, the kinetic terms of the ions and the elec- π F trons are important. where the maximum energy of the electrons is In the following we will start by considering a few cases 2 2 and we will start with the single electron approximation. ~ kF EF = the Fermi energy (51) 2me This defines the Fermi energy: it is the highest energy single electron approximation when all possible states with energy lower than EF are occupied, which corresponds to the ground state of the The single electron approximation can be used to de- system. This is one of the most important defi- rive the energy and density of the electrons. This sim- nitions in condensed matter physics. The electron plest model will always serve as a reference and in some density can now be rewritten as a function of the Fermi cases the result is very close to the experimental value. energy, through eqs. (51) and (48) in 3D: Alkali metals (Li, Na, K,..) are reasonably well described by this model, when we suppose that the outer shell elec- (2mE )3/2 n(E ) = F . (52) tron/atom (there is 1 for Li, Na, K) is free to move inside F 3π2~3 the metal. This picture leads to the simple example of N electrons in a box. This box can be viewed as a uniform We now want to define the energy density of states and positively charge background due to the atomic ions. D(E) as We suppose that this box with size L×L×L has periodic Z EF ∂ boundary conditions, i.e., n(EF ) ≡ D(E)dE =⇒ D(E) = n(E), (53) 0 ∂E XN ~2∇2 H = − n with Ψ(0) = Ψ(L) (44) which leads to 2me √ n=1 m 2mE D(E) = in 3D (54) ~3π2 For one electron the solutions can be written as m D(E) = in 2D (55) 2 ~ 2 ~2π i~k·~r ~ |k| ~ 2π r Ψ(~r) = e with E = and k = (n1, n2, n3), 2m 2me L D(E) = in 1D (56) (45) ~2π2E where n are the quantum numbers, which are positive i and illustrated below. or negative integers. Since electrons are Fermions we cannot have two electrons in the same state, except for 3D the spin degeneracy. Hence each electron has to have different quantum numbers. This implies that 2π/L is 2D the minimum difference between two electrons in k-space, D(E) which means that 1 electron uses up 1D

µ ¶ E 2π D (46) L of volume in k-space, if D is the dimension of the space. FIG. 11: Density of states in 1D, 2D and 3D If we now want to compute the electron density (number D of electrons per unit volume ne = N/L ) in the ground state, which have k < kF (Fermi sphere with radius kF ), we obtain: Properties of the free electron model µ ¶D D L D ne = 2 · V · /L , (47) kF 2π Physical quantities at T=0: 11

hEi 3 • Average energy per electron: n = 5 EF ¶ ∂(hEiV ) 2 1 ∂hEi π2 • Pressure: P = − = nEF , where hEiV is 2 ∂V 5 CV = = k TD(EF ) (64) the total energy. V ∂T µ,V 3

−1 ∂P 2 • Compressibility κ = −V = nEF CV ∂V 3 Hence, T = γ, which is the Sommerfeld parameter. Case 2: T6= 0: In equilibrium Periodic potentials 1 fFD = , (57) e(E−µ)/kT + 1 The periodicity of the underlying lattice has important consequences for many of the properties. We will walk where the chemical potential is the energy to add one through a few of them by starting with the simplest case electron µ = F − F . µ = E at T=0 and F is the N+1 N F in 1D. free energy. The Sommerfeld expansion is valid for kT << µ and

Z ∞ Kronig-Penney model hHi = dEH(E)FFD(E) (58) 0 Z Let us first consider the following simple periodic po- µ π2 ' H(E)dE + (kT )2H0(µ) + ... (59) tential in 1D. 6 Z0 ∞ ~2∇2 X hEi = dEED(E)FFD(E) (60) H = − − V δ(x − na) (65) 0 2m Z n EF π2 ' ED(E)dE + (kT )2D(E ) + ...(61) The solutions for na < x < na+a are simply plane waves 6 F 0 and can be written as hni = n(T = 0) (62) 2 0 ikx −ikx π 2 D (EF ) ψ(x) = Ane + Bne . (66) ⇒ µ(T ) ' EF − (kT ) + ... (63) 6 D(EF ) Now the task is to use the boundary conditions in order Physical quantities at T6=0: Specific heat to determine An and Bn. We have two conditions:

(1) ψ(na − ²) = ψ(na + ²) when ² → 0 and (67) R na+² ~ 0 0 (2) na−² (H − E)ψ(x)dx = 0 =⇒ − 2m (ψ (na + ²) − ψ (na − ²)) = V ψ(na). Inserting 66 into condition (1) yields

½ ikna −ikna ikna −ikna An−1e + Bn−1e = Ane + Bne = ψ(na) = ψn ikna −ika −ikna ika (68) An−1e e + Bn−1e e = ψn−1

½ ¡ ¢ −ika −ikan −ika ika e ψn − ψn−1 = Bn−1e ¡ e − e¢ =⇒ ika ikan ika −ika (69) e ψn − ψn−1 = An−1e e − e

½ ¡ ¢ −ika −ikan −ika −ika ika e ψn+1 − ψn = Bne e¡ e −¢e and ika ikan ika ika −ika (70) e ψn+1 − ψn = Ane e e − e

When taking the derivative of 66 then condition (2) implies     ~   − ik A eikan −ik B e−ikan −ik A eikan +ik B e−ikan  = V ψ , (71) 2m  | n{z } | n {z } | n−{z1 } | n−1{z }  n −ika ika ika −ika ψn+1−e ψn ψn+1−e ψn e ψn−ψn−1 e ψn−ψn−1 2i sin(ka) −2i sin(ka) 2i sin(ka) −2i sin(ka) 12 hence,

  ~2 k   − 2ψ + 2ψ + ψ (−e−ika − eika − eika − e−ika) = V ψ (72) 2m 2 sin(ka) n+1 n−1 n | {z } n −4 cos(ka)

and finally, This equation is often called the tight binding equation µ ¶ as we will see in the next section. Finding a solution for 2m sin(ka) ψ + ψ = − V + 2 cos(ka) ψ , (73) this equation is trivial since n+1 n−1 ~2k n | {z } ipan W ψn = e (75) Which can be rewritten as is a solution. This is easily verified by plugging (75) into ψn+1 + ψn−1 = W ψn (74) (74), which yields

eipaneipa + eipane−ipa = W eipan =⇒ W = eipa + e−ipa = 2 cos(pa). (76)

This equation has a solution only if Tight binging approximation

−2 ≤ W ≤ 2 (77) Let us consider the general potential due to the ar- We now recall that the Eigenstate of the original Hamil- rangement of the atoms on a lattice: 2 2 tonian is given by E(k) = ~ k , where eikx is the plane 2m ~2∇2 X wave between two δ functions, hence the dispersion re- H = − + V0(~r − R~n) (78) 2 2 2m ~ k n lation is equal to E(k) = 2m as long as −2 ≤ W ≤ 2. | {z } This condition will create gaps inside the spectrum as V (~r) illustrated in the graph below: This is a very general form for a periodic potential assum- ing that we only have one type of atoms. The periodicity is given by the lattice index R~n. For a general periodic -2 < ∆E W potential Bloch’s theorem (see A& M for proof) tells us that a solution to this Hamiltonian can be written as E(k) W<2 < -2 ~ ψ (~r) = eik·~ru (~r), (79) -2 k k W< 0 1 2 3 ~ π/a where uk is a periodic function such that uk(~r + Rn) = uk(~r). This implies that

~ ~ ~ ik·Rn FIG. 12: Dispersion curve for the Kronig-Penney model ψk(~r + Rn) = e ψk(~r) (80)

Let us now suppose that the solution of the Hamiltonian ~2∇2 ~ ~ Conclusion: a periodic potential creates gaps, H1atom = − 2m +V0(~r−Rn) with one atom is φl(~r−Rn) which leads to the formation of a band structure. corresponding to the energy level ²l, i.e., H1atomφl = ²lφl. This is a very general statement which is true for almost Now comes the important assumption, which allows us any periodic potential. to simplify the problem: We suppose that 13

  X   hφl(r − Rm)|H|φl(r − Rn)i = −tl δ ~m,~n+~i + δ ~m,~n−~i + ²lδ ~m,~n (81) ~i=ˆx,y,ˆ zˆ

Hence, only the nearest neighbor in every direction is Combining Bloch’s theorem with the tight binding taken to be non-zero. Further, we assume that there is no approximation overlap between levels of the one atom potential, which allows us to look for a general solution of the following The tight binding approximation is very general and form for each energy level ²l. can be applied to almost any system, including non- periodic ones, where the tight binding elements can be X l assembled in an infinite matrix. For the periodic case, ψl(r) = c φl(r − R~n) (82) ~n on the other hand, it is possible to describe the system ~n with a finite matrix in order to obtain the full disper- with eigenvalue E (k). To calculate E (k) we plug-in l l sion relation. This is obtained by combining the Bloch this Ansatz into eq. (82) and obtain an equation for theorem for periodic potentials, where the wave-function the coefficients cl , which leads to the following equation ~n from (79) is again: when using the tight binding approximation given in eq. (81). i~k·~r ψk(~r) = e uk(~r) X cl ² − t cl = E (k)cl (83) Instead of writing (82) we write the Bloch-tight- ~m l l ~m+~i l ~m ~i=±x,ˆ y,ˆ zˆ binding solution as

The solution of this equation are plane waves, which X ~ k ik·R~n l ~ ~ ψ (r) = e c φl(r − R~n), can be verified readily by taking cl = eik·Rm and plug- l ~n ~m ~n ging it into the equation to obtain which now depends explicitly on the wavevector ~k. Using

El(k) = ²l − tl(2 cos(kxax) + 2 cos(kyay) + 2cos(kzaz), Bloch’s theorem this implies that (84) l l where ~a is the lattice constant in all 3 space directions: c~n = c ~m, ~a = R ~m+ˆa − R ~m. The energy diagram is illustrated in fig. . The degeneracy of each original single atomic en- whenever ~n and ~m are related by a linear combination of ergy level ²l is lifted by the coupling to the neighboring Bravais vectors. Moreover, the tight binding equation in ~ l l ik·R~n atoms and leads to a dispersion curve or electronic band (83) is the same but with c~n replaced by c~ne structure.

E a ε 3 B

A ε 2

ε 1

-1.0 -0.5 0.0 0.5 1.0 −π /a k π/a FIG. 14: Diatomic square crystal

FIG. 13: Dispersion curve for the tight binding model We apply this to the simple example of a diatomic square lattice of lattice constant a with alternating atoms A and B shown in figure . We will further assume that This tight binding approximation is very successful we have only one band l. Hence, the Bloch-tight-binding in describing the electrons which are strongly bound to solution is written as the atoms. In the opposite limit where the electrons or X ~ more plane-wave like, the weak potential approximation k ik·R~n ψ (r) = e c~nφ(r − R~n), is more accurate: ~n 14

0 i~k·~r 0 ~2k2 where c~n takes on only to possible values due to Bloch’s ψk(~r) = e , with corresponding energies ²k = 2m and 0 0 0 theorem: cA or cB. This leads to the following simplified Hamiltonian H0, i.e., H0ψk(~r) = ²kψk(~r). Since the ori- tight binding equations (assuming hφl(r − Rn)|H|φl(r − gin of this energy dispersion relation can be chosen from 0 0 Rn)i = ²A or ²B and t = −hφl(r − Rn)|H|φl(r − Rm)i any site of the reciprocal lattice, we have ²k+K = ²k, when n and m are nearest neighbors): hence these energies are degenerate. This implies that P we have to use a degenerate perturbation theory. The ia~k·~i ~ cA²A − t ~ cBe = E(k)cA mathematical procedure is very similar to the tight bind- Pi=±x,ˆ yˆ ia~k·~i ~ ing approximation, but we now expand the solution ψ(~r) cB²B − t ~i=±x,ˆ yˆ cAe = E(k)cB of the full Hamiltonian H = H0 +V (~r) in terms of a sum 0 0 0 It is now quite straightforward to rewrite these equa- of plane waves ψk(~r). Since ²k+K = ²k we will only use 0 0 tions in matrix form: two plane waves in this expansion: ψk(~r) and ψk+K (~r). µ ¶ µ ¶ µ ¶ Hence, ²A −t · g cA ~ cA ∗ = E(k) , −t · g ²B cB cB 0 0 ψ(~r) = αψk(~r) + βψk+K (~r), (85) with g = eia~k·xˆ + e−ia~k·xˆ + eia~k·yˆ + e−ia~k·yˆ. The disper- sion relation or band structure is then simply given by where the coefficients α and β have to be determined in obtaining the eigenvalues of HBTB, where order to solve Schr¨odinger’sequation: µ ¶ ²A −t · g HBTB = ∗ . (H − E)ψ(~r) = 0. (86) −t · g ²B

We can find the solution by first multiplying (86) by Weak potential approximation 0 ∗ (ψk) (~r) and then integrating the equation over the whole space which will lead to one equation, and we obtain a 0 ∗ In this case we consider the effect of the periodic po- second equation by multiplying (86) by (ψk+K ) (~r) and tential V (~r) as a perturbation on the plane wave solution then integrating of the whole space. This leads to

½ R R 3 0 ∗ 0 3 0 ∗ 0 R α d r(ψk) (~r)(H − E)ψk(~r) +Rβ d r(ψk) (~r)(H − E)ψk+K (~r) = 0 3 0 ∗ 0 3 0 ∗ 0 , (87) α d r(ψk+K ) (~r)(H − E)ψk(~r) + β d r(ψk+K ) (~r)(H − E)ψk+K (~r) = 0 where (assuming normalized plane waves in the integrals)  R R 3 0 ∗ 0 3 −ikr ikr 0  R d r(ψk) (~r)(H − E)ψk(~r) = dRre (H0 + V (r) − E)e = ²k + 0 −RE  3 0 ∗ 0 3 −ikr i(k+K)r 3 iKr 0 R d r(ψk) (~r)(H − E)ψk+K (~r) = d re (H0 + V (r) − E)e = d re (²k+K + V (r) − E) = VK 3 0 ∗ 0 0  R d r(ψk+K ) (~r)(H − E)ψk+K (~r) = ²k+K − E  3 0 ∗ 0 d r(ψk+K ) (~r)(H − E)ψk(~r) = V−K , (88) R R R 3 iK~ ·~r 3 iKr 3 and VK~ = d re V (~r) (the Fourier transform), d re = 0 (for K 6= 0), and d rV (~r) = 0. With coefficients (88), equation (87) leads to the following couple of equations: ½ ¯ ¯ 0 0 q ¯ ¯ ²0 +²0 (²0 −²0 )2 α(²k − E) + βVK = 0 ¯ ²k − EVK ¯ k k+K k k+K 0 ⇒ ¯ 0 ¯ = 0 ⇒ E = 2 ± 4 + VK V−K . (89) αV−K + β(²k+K − E) = 0 V−K ²k+K − E

0 0 0 Finally if ²k = ²k+K , we have E = ²k ± |VK |, which the wavevector which corresponds to the two dispersion leads to a splitting 2|VK | of the energy levels at these curves which led to the degenerate energy level. degenerate energies. For the example in figure , this weak potential approximation would give us a splitting of 2|VK=2π/a| at k = −π/a and k = K − π/a = π/a. This Localization implies that the first order calculation of the energy split- ting due to the weak periodic potential V (r) is equal to When, instead of having a purely periodic potential twice the fourier transform of this potential evaluated at disorder is included into the system, we no more have 15

Bloch wave solutions but localization of the wave func- equation Hψ = E(k)ψ it follows that uk is a solution of tions occur. This is particularly important in low dimen- µ ¶ ~2 sional systems and tends to suppress transport. (k − i∇)2 + V − E(k) u = 0. (93) 2m k | {z } Hk−E(k) Electronic properties due to periodic potential In order to calculate hvi we use first a first order pertur- Density of states bation in k + q, where q is very small. Therefore, the eigenvalue corresponding to Hk+q is E(k + q), which is to first order Density of states in 1D:

E(k + q) = E(k) + huk|Hk+q − Hk|uki ∂n δn ~2 D(E) = ' = E(k) + hψeikr| ( q2 +2q(k − i∇))|e−ikrψi ∂E δE 2m |{z} ¯ ¯ ¯ ¯−1 →0 δn ¯∂E(k)¯ 2 = ¯ ¯ × |{z}2 × |{z}2 ikr ~ −ikr δk ∂k P = E(k) + hψe | q(k − i∇))|e ψi spin ±k m ¯ ¯ ~ ¯ ¯−1 = E(k) + hψ|q~ (−i∇)|ψi 1 δN ¯∂E(k)¯ = ¯ ¯ × 4 m L δk ∂k = E(k) + q~hψ|v|ψi ¯ ¯ 2 ¯∂E(k)¯−1 D(E) = ¯ ¯ , (90) = E(k) + q∇kE(k) (Taylor expansion) π ¯ ∂k ¯ 1 ⇒ hvi = ∇ E(k). (94) ~ k where we used that δE = (∂E/∂k)δk and δk = 2π/L for one electron, i.e., δN = 1. Hence, the expectation value of the velocity is determined In three dimensions (D=3) we have: by the slope of the dispersion relation. This also implies that the sign of the average velocity depends on the sign of ∂E/∂k. ∂n δn D(E) = ' ∂E δE X Response to an external field and existence of holes and δn −1 = |∇ E(k)| × 2 electrons δk k |{z} E(k)=E spin Z 2 The idea is to describe the average motion of an elec- = d2k |∇ E(k)|−1 , (91) (2π)3 k tron in the presence of an external field (electric, Eel, or E(k)=E magnetic, B) in a semiclassical way. Hence, we want P 2 where we used δE = (∇kE(k)) · δk, (δk) → R E(k)=E ∗ d 2 3 3 m hvi = F = (qEel or qhvi × B), (95) E(k)=E d k, and δn/(δk) = 1/(2π) . This result shows dt that the density of state in the presence of a periodic ∗ potential, i.e., for a crystal depends only on the slope of where m is an affective mass and q the charge. Using the dispersion relation or the band structure. (95) and (94) we have

∂hvi m∗ k˙ = qE Average velocity ∂k el 1 ∂2E m∗ k˙ = qE The average velocity of an electron in a periodic po- ~ ∂k2 el ˙ tential is given by the expectation value of the velocity, ⇒ ~k = qEel = F, (96) i.e., Where we defined the effective mass m∗ as hvi = hψ|v|ψi, (92) ¯ ¯ ¯∂2E ¯−1 m∗ = ~2 ¯ ¯ . (97) where ψ is the wavefunction from the Hamiltonian with ¯ ∂k2 ¯ ~2∇2 periodic potential V , i.e., H = 2m + V . From Bloch’s 2 ikr If ∂ E is negative we need to change the sign of q in order theorem (79) we can write ψ(r) = e uk(r), where uk(r) ∂k2 ∂2E has the same periodicity as V (r). From the Schr¨odinger to remain consistent. Hence, when ∂k2 > 0, the charge 16

∂2E ˙ of an electron is q = −e but if ∂k2 < 0 then q is positive Using (101) and (94) and since k is perpendicular to hvi ˙ (+e). In this case we describe the particles as holes. and B, (k) ∼ ∂E/∂kk we obtain They represent missing electrons. With these definitions Z ∗ 2 k2 of q and m , which is also called the band mass, the ~ dk⊥ semiclassical equations of motion of single electrons in a δt = qB k1 ∂E/∂kk periodic potential are simply given by eqs. (95) and (96). Z ~2 d k2 In general, the effective mass is given by a tensor de- = k dk . (103) qB dE k ⊥ fined as k1 ¯ ¯ ¯ ∂2E ¯−1 For a complete turn this leads to m∗ = ~2 ¯ ¯ , (98) αβ ¯∂k ∂k ¯ I α β ~2 d T = k dk . (104) where α and β are the spatial directions. An important qB dE k ⊥ consequence of this semiclassical description of the mo- | {z } tion of electrons is the dependence of the effective mass S on the energy and the band structure. In some cases the Here S is the area enclosed by an orbit in k-space. This ∂2E effective mass can even diverge (when ∂k2 = 0). Sim- orbit corresponds to an equipotential line perpendicular ilarly the sign of the carriers also depends on the band to the magnetic field. structure and the energy of the carriers. By definition we Let’s suppose for simplicity that the effective mass ten- call the bottom of an energy band and electronic band sor m∗ is diagonal and given by ∂2E when ∂k2 > 0 and a hole band when at the top of the   ∂2E mx 0 0 energy band 2 < 0. ∗ ∂k m =  0 my 0  (105) 0 0 mz Bloch oscillations and that the energy dispersion is harmonic (which is usu- ally true at a band extremum, i.e., In the presence of an electric field and a periodic po- tential we can use the equation of motion (96), i.e., ~2k2 ~2k2 ~2k2 E(k) = x + y + z . (106) eE 2mx 2my 2mz ~k˙ = −eE ⇒ k = − el t, (99) el ~ If we assume that the magnetic field is along z and that but in a periodic potential and in the tight binding ap- the average effective mass perpendicular to B is given by proximation the energy is given by E(k) = −2t cos(ka), 0 m⊥, we can rewrite (106) as where a is the lattice constant and t0 the nearest neighbor overlap integral. Hence, since v =r ˙ and hvi = ~−1∂E/∂k 2 2 ~2k2 ~ k⊥ k we have E(k) = + , (107) 2m⊥ 2mk 2t0 aeEelt hri = cos( ). (100) 2 2 2 eEel ~ where k⊥ = kx +ky. Using (104) and (107 we then obtain This means that the average position of the electrons S = πk2 = π(2m E/~2) − πk2m /m oscillates in time (Bloch oscillations). In artificial struc- ⊥ ⊥ k ⊥ k dS π2m tures these Bloch oscillations are typically of the order of ⇒ = ⊥ 1THz. dE ~2 2π qB ⇒ ωc = = , (108) T m⊥ Semiclassical motion in a magnetic field which is the cyclotron frequency. Hence, the cyclotron In the presence of a magnetic field (B), we can describe frequency depends on the average effective mass perpen- the semiclassical trajectories in k-space using (96), i.e., dicular to the magnetic field. This allows us to measure the effective mass along different directions, simply by ~k˙ = qhvi × B. (101) changing the direction of the magnetic field and by mea- Hence, only the values of k perpendicular to the magnetic suring the cyclotron frequency. field will change, which we denote by k⊥. The component parallel to the field, kk is not affected by B. During a Quantization of the cyclotron orbit: Landau levels small time difference

Z t2 Z k2=k(t2) ˙ In quantum mechanics the energies of these cyclotron δt = t2 − t1 = dt = dk⊥/|k|. (102) t1 k1=k(t1) orbits become quantized. To see this we can write the 17

500 450 300

Hamiltonian of an electron in a magnetic field in the har- ] 400 Ω 200 monic approximation (107) as 350 [ Rxx 100 0 ] 300

Ω 1.1 1.3 1.5 1.8 2.0 2.2 2.4 2.6 2.8 1 2 1 2 250 1/B [T -1 ] H = P + (P⊥ + qA) , (109) k 200 2mk 2m⊥ Rxx[ 150 where B = ∇ × A ⇒ A = −Byxˆ in the Landau Gauge if 100 B is alongz ˆ. In analogy to the harmonic oscillator, the 50 0 eigenvalues of (109) are then given by 0.0 0.5 1.0 B [T] ~2k2 qB E = k + (n + 1/2)~ . (110) n,kk 2m m k |{z}⊥ FIG. 15: Magneto-resistance oscillations in a GaAs/AlGaAs ωc quantum well. These eigenvalues can be found by writing ikxx ik z the wavefunction as ψ = e φn(y − y0)e k ~kx with y0 = − , which leads to Hψ = µ qB ¶ PHONONS: LATTICE VIBRATIONS 2 2 2 ³ ´2 ~ kk Py 1 qB 2 + + m⊥ (y − y0) φn(y − y0). 2mk 2m⊥ 2 m⊥ In the y direction this is simply the harmonic oscillator In general: with eigenvalues (n + 1/2)~ωc and in the direction parallel to the field we have a plane wave so that the X Mu¨l = − φlmum, (112) total energy is given by (110). The quantized levels m (n + 1/2)~ωc due to the magnetic field are called the Landau levels. where ul are the deviations from the original lattice sites and φlm areP the elastic constants who have to obey this Magneto-oscillations sum rule m φlm = 0 (translation invariance). In words, equ. 112 simply means that the force producing the devi- ation on lattice site l only depends on the deviations from The quantization of the energy levels in the presence the other lattice sites. No deviation=no force (equilib- of a magnetic field will lead to oscillations of almost any rium). This equation is very general and does not assume experimental quantity (resistance, thermal conductivity, that we have a periodic lattice, but in order to calculate magnetization, ...) as a function of B. These oscillations things we will use a periodic lattice and start with 1D. can be understood by looking at the density of states In fig. 16 we illustrate two typical displacement waves in due to the quantization (110). Indeed, (110) leads to a 2D. peak in the density of states whenever E = (n+1/2)~ωc. Therefore, whenever the component of the Fermi energy u u − u u + u perpendicular to the magnetic field is equal to one of s−2 s 1 s s 1 s+2 these quantized levels there is an extremum in the quan- tity measured. The distance between two of these ex- trema is given by K

qB1 ⊥ qB2 ~ (n1 + 1/2) = EF = ~ (n2 + 1/2) m⊥ m⊥ 1 1 q ⊥ −1 ⇒ − = ~ ( EF ) (n2 − n1) u u s+4 B1 B2 m⊥ |{z} us−1 us us+1 us+2 s+3 ~2 K 2πm ·S µ ¶ ⊥ 1 2πq −1 ⇒ ∆ = S , (111) s −1 s s + 1 s + 2 s + 3 s + 4 B ~ H F 2 where the maximum S is S = kkdk⊥ = π(kx ) = FIG. 16: Two types of lattice displacement waves in 2D, trans- ~2(kF )2 2 π(kF )2 and E⊥ = x = ~ S . Hence, the Magneto- verse and longitudinal modes y F 2m⊥ 2πm⊥ oscillations are periodic in 1/B and the period depends on S. An example of these oscillations in resistance is shown in figure 15 as a function of B and 1/B. 18

Mono-atomic phonon dispersion in 1D 1. Acoustic mode: s C PIn 1D equ. 112 for nearest neighbors and using ω ' ka (117) 2(M1 + M2) m φlm = 0 simply reduces to

Mu¨l = K(ul+1 − 2ul + ul−1) (113) 2. Optical mode: in the simplest approximation, where only the nearest s 2C(M + M ) neighbors are important and where the elastic constants ω ' 1 2 (118) are the same. Further we assume that the equilibrium M1M2 case has a lattice constant a. In this case the solution can be written as

ikla−iωt ul = e 2 2 =⇒ Mω =q 2K(1 − cos(ka)) = 4K sin (ka/2) (114) K =⇒ ω = 2 M | sin(ka/2)| which is illustrated below In this case there is only a

1.2 ω Mono-atomic dispersion 1.0 2 (K M ) 0.8

0.6

0.4

0.2

0 k π π - /a 0 /a 2π/a

FIG. 17: Phonon dispersion for a mono-atomic lattice single dispersion mode, which is called the acoustic mode or branch.

Optical branch

The situation is quite different when there is an addi- tion symmetry breaking in the problem. In the 1D case FIG. 18: Phonon dispersion for a di-atomic lattice and the more general 3D case this occurs when there is a second species with a different mass, for example, or when there are two different elastic constants. Let us discuss the case where there are two different masses but only one elastic constant C: Experimental determination of the phonon ½ dispersion M u¨ = C(u − 2u + u ) 1 l l+1 l l−1 (115) M2v¨l = C(vl+1 − 2vl + vl−1) In order to determine the dispersion curve of the To solve this set of equation we use a solution of the phonons there is essentially one trick. The idea is to ikla−iωt ikla−iωt form ul = ue and vl = ve , which inserted look for inelastic scattering of other particles, which are into eq. 115leads to typically photons or neutrons. The incoming particles scatter with phonons, thereby transferring some of their s p energy and momentum to phonons. From the energy and 2 2 √ M1 + M2 ± M − 2M1M2 cos(ka) + M momentum of the out-coming particles it is then possible ω = C 1 2 M1M2 to infer the phonon absorbtion. This process can also be (116) inverted in that the energy of the out-coming particle in- The following two limiting cases can easily be obtained creases by absorbtion of a phonon. In general, the energy in the limit of small k. conservation can be expressed as 19

FIG. 19: Inelastic neutron scattering experiment in Chalk River used to determine the phonon dispersion ~ω (~k) − ~ω0 (k~0) = ±~Ω(K~ ) (119) | in {z out } | {z } in/out particle phonon

and the momentum conservation as

~~k − ~k~0 = ±~K~ + ~G,~ (120)

where G~ is a reciprocal lattice vector. A neutron scat- tering experiment is illustrated below.

Origin of the elastic constant

From the TOE we have the potential term due to the interactions between ions.

NXions ~2∇2 NXions q q H = − i + i j (121) ions 2m |r − r | i i i

This leads to a potential of the form

 ! displacment X z}|{ X 2 0 0 ∂φ 1 ∂ φ  φ(r1, . . . , rn) = φ(r1, . . . , rn) + ui + uiuj + ··· (122) | {z } ∂ri 2 ∂ri∂rj i 0 i,j ri r0 Cohesive energy | {z } | {z i} 0 Equilibrium pos. φi,j

The linear term has to be zero for stability reasons (no trated in fig. 18. From 123 we have minimum in energy otherwise). The classical equation X ~ 2 ik·(R~l−R~m) of motion is then simply given by equ. 112, because mω ~² = φl,me ·~² (124) ~ ~ F = −∇φ. To solve this equation in general, one can |m {z } write a solution of the form φˆ(k) , ˆ where φ(k) is a 3 × 3 matrix, since φlm too. This leads 2 ~ to an eigenvalue equation for ω with three eigenvalues: ik·R~l−iωt ~ul = ~² · e , (123) 2 ~ 2 ~ ωL (for k k ~²) and ωT (1,2) (for k ⊥ ~²). Hence, we have on longitudinal mode and two transverse modes and all phonon modes can be described by a superposition of where R~l are the lattice sites. One then has to solve these. for the dispersion relation ω(~k) in all directions, as illus- If we have two different masses, we have two addi- 20 tional equations of the form (124). This leads to the general case, where one has two branches the optical one Z and the acoustic one and each is divided in longitudinal 1 dSω D(ω) = 3 (127) and transverse modes. The acoustic branch has always (2π) ω=const |∇kω| a dispersion going to zero at k = 0, whereas for the op- tical mode at k = 0 the energy is non-zero. The acous- Let us now suppose that we have 3 low energy modes tic phonons have a gapless excitation spectrum and can with linear dispersion, one longitudinal mode ωL(k) = therefore be created at very low energy as opposed to cLk and two transverse modes ωT (Rk) = cT k, hence 2 optical phonons which require a miniumum energy to be ∇ωL,T (k) = cL,T , which implies that dS = 4πk . This excited. leads to a density of state

2 2 Quantum case k ω DL,T (ω) = 2 = 2 3 (128) 2π cL,T 2π cL,T While many of the properties, like the dispersion rela- or tion, can be explained in classical terms, which are simply vibrations of the crystal ions, others, such as statistical µ ¶ properties need a quantum mechanical treatment. ω2 1 2 Phonons can simply be seen as harmonic oscillators D (ω) = + (129) tot 2π2 c3 c3 which carry no spin (or spin 0). They can, therefore, be | L {z T } 1 accurately described by bosons with energy c3

in the isotropic case. En(k) = ~ω(k)(n + 1/2). (125) From statistics we know that the probability to find a state at E = E is given by the Boltzmann distribution We can now calculate the density of states of phonons, n P P ∼ e−En/kB T , with normalization P = 1. Hence, or the number of states between ω and ω + dω, i.e., n n

Z −n~ω/kB T −~ω/kB T 1 ω+dω Pn = e (1 − e ) (130) D(ω)dω = 3 dk (126) (2π) ω simply from the normalization condition. We can now but dω = ∇kωdk, hence calculate the average energy at ω,

X X∞ −~ω/kB T −~ω/kB T n E(ω) = EnPn = (1 − e )~ω (n + 1/2)(e ) n  n=0    1 1  = ~ω  +  (131) 2 e~ω/kB T − 1 | {z } hni

here hni is the expectation value of quantum number n zero energy state is totally degenerate (this does sup- at T , which is nothing else but the Bose Einstein distri- pose that there are no interactions between bosons, with bution. As is well known, Bosons obey the Bose-Einstein interactions the delta function will be a little broader). statistics, where he Bose-Einstein distribution function is The total average energy can then be calculated from the given by density of states as:

1 Z fBE = (132) eE/kB T − 1 hEi = dωD(ω)E(ω) (133)

Important: at T = 0, fBE is simply a delta function δ(E). This is the Bose-Einstein condensation, where the This also allows us to calculate the specific heat CV = 21

d dT hEi in the isotropic case and in the linear dispersion If the electrons flow without collisions, 2 3 approximation D(ω) ∼ ω /c , hence (137)=g(r, k, t) = fFD. The situation changes if collisions (f. ex. between electrons, impurities, or Z phonons) are included. In this case d ∞ C = dωD(ω)E(ω) V dT Z0 d ∞ ~ω g(r, k, t) − g(r − rdt,˙ k − kdt,˙ t − dt) = δg (138) = dωD(ω) coll. ~ω/k T dT 0 e B − 1 Z where g describes the collisions. g = 0 without d 3 1 ∞ ~ω3 coll. coll. = dω collisions. Expanding (137) to first order and using (138) 2 3 ~ω/k T dT 2π |{z}c 0 e B − 1 we obtain µ ¶ | {z } R 3 4 1 1 + 2 ∞ dω x (kB T ) 3 c3 c3 0 ex−1 3 L T ~ ¶ 4 Z ∞ 3 ∂g ∂g ∂g dg d 3(kBT ) x r˙ + k˙ + = (139) = 2 3 dω x ∂r ∂k ∂t dt dT 2π (c~) 0 e − 1 | {z } coll. | {z } dg π4/15 dt µ ¶3 2π2k k T This is Boltzmann’s equation. = B B (134) 5 ~c where we had defined the variable x = ~ω/kBT and Relaxation time approximation where 1/c3 is the average over the 3 acoustic modes. The Debye model assumes a linear dispersion (ω = ck) Solving Boltzmann’s equation is not easy in general 3 2 and uses the analogy to electrons where (n = kF /3π ) to and therefore approximations are used. The simplest one define kD. Since phonons have no spins one obtains: is the relaxation time (τ) approximation. In this approx- imation: k3 n = D and ¶ 6π2 dg 1 = − (g − f ) k Θ = ~ω = ~ck (Θ : Debye temperature) FD B D D D D dt coll. τ 4 µ ¶3 12π kB T dg 1 =⇒ CV = · n (135) =⇒ = − (g − fFD) (140) 5 ΘD dt τ The typical Debye temperature is of the order of 100K. It is quite intuitive to see from where this approxima- Finally, combining this result with the contributions from tion comes from, since a kick at t = 0 would lead to electrons (64), we obtain the expression valid for low tem- solution of the form peratures: e−t/τ g(t) = + fFD → fFD (when t = ∞) (141) C = γT + βT 3 (136) τ V |{z} |{z} electrons phonons This is the basic framework, which allows us to eval- uate the effect of external fields on the system and to estimate the linear response to them. TRANSPORT (BOLTZMANN THEORY)

Transport allows us to calculate transport coefficients Case 1: F~ = −eE~ such as resistances and thermal conductivities. While several transport theories exist, Boltzmann’s approach In this case we look for a response in current density is the most powerful and applies to most situations in and define the conductivity tensor in linear response as condensed matter. The main idea in Boltzmann theory is to describe the electrons by a distribution function g. In equilibrium g is simply the Fermi-Dirac distribution ~j = σE~ Z function f . In general, g(r, k, t) and at t − dt it can be d3k FD ⇒ j = σ E = −2e v g(r, k, t). (142) written as: α αβ β (2π)3 α

Since we are interested in the case of a homogenous g(r − rdt,˙ k − kdt,˙ t − dt) (137) uniform electric field, ∂g/∂t = 0 and ∂g/∂r = 0 and 22

~k˙ = −eE, Boltzmann’s equation (139) in the relaxation Diffusion model of transport (Drude) time approximation becomes In the case where the scattering of electrons is domi- ∂g 1 nated by inelastic diffusion, we can write a very simple k˙ = − (g − f ) FD form for the conductivity or resistivity tensor. Indeed, in |{z∂k} τ ~ a diffusive regime the average velocity is directly propor- −eE ∂g · ∂² ~ ∂² ∂~k tional to the external force and to the inverse effective eEτ~ ∂g ∂² mass. The proportionality coefficient is then simply the ⇒ g = · +f (143) ~ ∂² ~ FD scattering probability 1/τ. Hence, |{z} |{z}∂k '− ∂fFD ~~v ∂µ m∗ ~v = F~ = −eE~ − e~v × B~ (148) Here we used the first order approximation ∂g/∂² ' τ −∂fFD/∂µ, which is justified if the correction g − fFD is smoother than ∂fFD/∂µ, which is close to a delta func- If we suppose that the magnetic field is small and in ~ tion when kT << EF . Hence, using (143) to evaluate thez ˆ direction and E alongx ˆ, then eq. (148) becomes the current (142), we obtain m∗ Z 3 Z eE~ = − ~v −e~v × B~ d k ∂fFD |{z} τ | {z } jα = 2e vαeτE~ · ~v + ∼ ( vαfFD), (144) | {z } 3 eEx 'jy B/n (2π) ∂µ m∗jx | {z } ' τen =0 m∗ B E = j + j (149) which leads to x ne2τ x ne y Z 3 ∂jα 2 d k ∂fFD σαβ = = 2e vαvβ τ(²) (145) Since in general E~ = ρ~j and because the resistivity ∂E (2π)3 ∂µ β along B~ is not affected by the magnetic field we can write This is one of the most important expressions for the for B~ in thez ˆ direction, conductivity. Since ∂fFD/∂µ is almost a delta function for kT << EF this expression shows that only electrons  m∗ B  close to the chemical potential µ will significantly con- ne2τ ne 0  B m∗  −1 tribute to transport. It is possible to evaluate (145) in ρ = − ne ne2τ 0 = σ (150) m∗ simple cases. 0 0 ne2τ Let’s assume that τ does not depend on the energy, then we can rewrite (145) as This is the famous Drude formula in a magnetic field. This formula is in fact equivalent to the relaxation time

Z 3 µ ¶ approximation in the Boltzmann theory (147). 2 d k ∂² ∂fFD σαβ = 2e τ 3 vα − (2π) ~∂kβ ∂² Z µ ¶ 3 Case 2: Thermal inequilibrium 2 d k ∂fFD = e τ 3 vα − (2π) ~∂kβ Z 3 Z We now consider the case, where we also have a spacial 2 d k ∂vα = e τ 3 fFD/~ − fFDvα gradient. Hence (139) and (140) become (2π) ∂kβ |{z} | k⊥β{z } ∗ ∂g ∂g 1 ~δαβ /m =0 ˙ Z r˙ + k = − (g − fFD), (151) e2τ d3k |{z∂r} |{z∂k} τ = 2 fFD δαβ (146) m∗ (2π)3 '~v· ∂f '−eE~ ·~v ∂f | {z } ∂~r ∂² n here we used again that g − f is smooth so that ∂(g − • If τ(²) = τ we therefore obtain the most important f)/∂² << ∂f/∂². Moreover, since formula of conductivity (the Drude formula) e2τn 1 σαβ = δαβ (147) f = m∗ FD e(²−µ(r))/kT (r) + 1 µ ¶ ∂f ∂f ∇ T Using the same approach but in the presence of a mag- ⇒ = −∇ µ − (² − µ) r ~ ~ ~ ∂r ∂² r T netic field, F = −eE−e~v×B, we can derive an equivalent µ ¶ expression for σ , but in this case the off-diagonal com- ∂f ∇ T αβ ⇒ g = ~v · τ eG~ + (² − µ) r + f ,(152) ponent is not zero anymore. (See assignment). ∂² T FD 23

~ ~ ~ where we defined the generalized field eG = eE+∇µ(r) Is obtained from eqs. (153,154,157), by setting³ j´e = and the external forces are now −eE~ , ∇~ µ(r) and ∇~ T (r), 11 ~ 12 ∇~ T 0, hence at low temperatures, L G+L − T = corresponding to an external electrical field, a gradient 0 and in the chemical potential (f.ex. a density gradient), and à ! a temperature gradient. The electrical current density is ∇~ T j = (L12(L11)−1L12 −L22) − the same as before but is now expressed (in the linear Q | {z } T response) in terms of the additional external fields: ∼O(T 4)'0 L22 à ! = − ∇T Z 3 ~ T ~ d k 11 ~ 12 −∇T je = −2e ~vg ' L G + L . (153) π2 k2T (2π)3 T ⇒ κ = σ (158) αβ 3 e2 αβ An expression for the thermal current can be de- This is the well known Wiedemann-Franz law duced from the following thermodynamical relation dQ =

T dS = dU − µdN, hence • Thermopower (Q): E~ = Q∇T with je = 0. In this case Z Ã ! d3k −∇~ T L12 π2 k2T σ0 j~ = 2e (² − µ)~vg ' L21G~ + L22 . Q = = (159) Q (2π)3 T L11T 3 e σ (154) Here Lαβ are the linear transport coefficients and σ = • Peltier effect (∇T = 0): jQ = Πje ⇒ Π = 21 11 L11. We now want to evaluate these expressions in the L /L = TQ low temperature limit, where we can use ∂fFD/∂µ ' Historical note: δ(² − ²F ), hence the expression for σ in (145) can be written as The Seebeck effect: The discovery of thermoelectricity dates back to Seebeck [1] (1770-1831). Thomas Johann See- beck was born in Revel (now Tallinn), the capital of Estonia Z ∂j d3k which at that time was part of East Prussia. Seebeck was a α 2 member of a prominent merchant family with ancestral roots σαβ(x) = = 2e 3 vαvβτ(x)δ(² − x), (155) ∂Eβ (2π) in Sweden. He studied medicine in Germany and qualified as a doctor in 1802. Seebeck spent most of his life involved where σαβ = σαβ(²F ). We will use this new function σ(x) in scientific research. In 1821 he discovered that a compass to express the other transport coefficients. For instance needle deflected when placed in the vicinity of a closed loop in linear response, formed from two dissimilar metal conductors if the junctions were maintained at different temperatures. He also observed that the magnitude of the deflection was proportional to the 2 22 ∂jQ temperature difference and depended on the type of conduct- e Lαβ = ³ ´ −∇~ T ing material, and does not depend on the temperature distri- ∂ T β bution along the conductors. Seebeck tested a wide range of Z d3k ∂f materials, including the naturally found semiconductors ZnSb = 2e2 v v (² − µ)2τ(²) and PbS. It is interesting to note that if these materials had (2π)3 α β ∂µ Z been used at that time to construct a thermoelectric genera- ∂f tor, it could have had an efficiency of around 3% - similar to = dx (² − µ)2σ (²). (156) ∂µ αβ that of contemporary steam engines. The Seebeck coefficient is defined as the open circuit volt- π2 2 0 age produced between two points on a conductor, where a Using that (²F − µ) ' (kT ) D (²F )/D(²F ), we ob- 6 uniform temperature difference of 1K exists between those tain (derivation in assignment). points. The Peltier effect: It was later in 1834 that Peltier[2] de- π2 scribed thermal effects at the junctions of dissimilar conduc- 22 2 tors when an electrical current flows between the materials. Lαβ = 2 (kT ) σαβ(²F ) and similarly 3e Peltier failed however to understand the full implications of π2 his findings and it wasn’t until four years later that Lenz[3] L12 = L21 = − (kT )2σ0 (² ) αβ αβ 3e αβ F concluded that there is heat adsorption or generation at the 11 junctions depending on the direction of current flow. L = σαβ(²F ) (157) αβ The Thomson effect: In 1851, Thomson[4] (later Lord Kelvin) predicted and subsequently observed experimentally the cooling or heating of a homogeneous conductor resulting Physical quantities from the flow of an electrical current in the presence of a temperature gradient. This is know as the Thomson effect • Thermal conductivity (κ): jQ = κ(−∇T ) and is defined as the rate of heat generated or absorbed in a 24

metal semiconductor insulator

FIG. 21: There is no band gap at the Fermi energy in a metal, while there is a band gap in an insulator. Semiconductors on the other hand have a band gap, but it is much smaller than those found in insulators.

Band Structure FIG. 20: Thermoelectric cooling (left): If an electric current is applied to the thermocouple as shown, heat is pumped from the cold junction to the hot junction. The cold junction will Clearly the band structure of the semiconductors is rapidly drop below ambient temperature provided heat is re- crucial for the understanding of their properties and their moved from the hot side. The temperature gradient will vary device applications. Semiconductors fall into several cat- according to the magnitude of current applied. Thermoelec- egories, depending upon their composition, the simplest, tric generation (right): The simplest thermoelectric generator type IV include silicon and germanium. The type refers consists of a thermocouple, comprising a p-type and n-type thermoelement connected electrically in series and thermally to their valence. in parallel. Heat is pumped into one side of the couple and The band structure is quite rich and shown in figure rejected from the opposite side. An electrical current is pro- (22) for silicon. Germanium is very similar to Si and does duced, proportional to the temperature gradient between the hot and cold junctions. single current carrying conductor subjected to a temperature gradient. [1] Seebeck, T.J., 1822, Magnetische Polarisation der Met- alle und Erzedurch Temperatur-Differenz. Abhand Deut. Akad. Wiss. Berlin, 265-373. [2] Peltier, J.C., 1834, Nouvelles experiences sur la calo- riecete des courans electriques. Ann. Chem., LVI, 371-387. [3] See Ioffe, A.F., 1957, Semiconductor Thermoelements and Thermoelectric Cooling, Infosearch, London. [4] Thomson, W., 1851, On a mechanical theory of thermo- electric currents, Proc.Roy.Soc.Edinburgh, 91-98.

FIG. 22: Band structure of Si. (Figures from ioffe.rssi.ru/SVA/NSM/Semicond) SEMICONDUCTORS not have a direct gap either. In general the dispersion Semiconductors are distinguished from metals in that relation can be approximated with the use of the effective they have a gap at the Fermi surface, and are distin- masses, noting that guished from insulators in that the gap is smaller, which is ambiguous. But generally, if the gap is close to 1eV 1 1 ∂2E(~k) = , (160) it’s a standard semiconductor. If the gap is close to 3eV m∗ ~2 ∂k ∂k it’s a wide band gap semiconductor and above 6eV it’s ij i j usually called an insulator. Typically, the distinction is hence simply made from the temperature dependence of the conductivity. If σ → 0 for T → 0 it’s a semiconductor or à ! k2 + k2 2 an insulator but the material is a metal if σ → σ0 > 0. ~ 2 x y kx E(k) = Ec + ~ ∗ + ∗ (161) However, even at room temperatures the conductivities 2mt 2ml of pure materials differ by many orders of magnitude (σmetal >> σsemi >> σinsu). for the conduction band and 25

FIG. 23: Band structure of Ge.

FIG. 24: Band structure of GaAs. Note the direct, Γ → Γ, minimum gap energy. The nature of the gap can be tuned 2~ 2 ~2~k2 2~ 2 with Al doping. ~ ~ kh lp ~ k E(k) = Ev − ∗ − ∗ ' Ev − ∗ (162) 2mh 2mlp 2mp conduction ω ∼ h Eg phonon ≅ ω for the valence band, where there are typically two bands, vs k photon ck ≅ ω ≅ E / h the heavy holes and the light holes. These dispersion re- valence g lations are generally good approximations to the real sys- tem. In addition, the valence band has another band due FIG. 25: A particle-hole excitation across the gap can readily to spin-orbit interaction. However, at zero wavenumber ~ recombine, emit a photon (which has essentially no momen- (k = 0) this band is not degenerate with the other two tum) and conserve momentum in a direct gap semiconductor light and heavy hole bands. (left) such as GaAs. Whereas, in an indirect gap semiconduc- tor (right), this recombination requires the additional creation ∗ ∗ ∗ ∗ ∗ ∗ mn mp mt ml mh mlp gap [eV] or absorption of a phonon or some other lattice excitation to Si 0.36 0.81 0.19 0.98 0.49 0.16 1.12 conserve momentum. Ge 0.22 0.34 0.0815 1.59 0.33 0.043 0.661 GaAs 0.063 0.53 0.063 0.063 0.51 0.082 1.424 no electrons in the conduction band (n = 0). At non- zero temperatures, the situation is very different and the TABLE I: The effective masses in units of the free electron carrier concentrations are highly T -dependent since all of mass for the conduction band, the valence band, the trans- the carriers in an intrinsic (undoped) semiconductor are verse and longitudinal part in the conduction band and the heavy and light hole mass in the valence band. thermally induced. In this case, the Fermi-Dirac distribution defines the temperature dependent density The situation in III-V semiconductors such as GaAs Z Z is similar but the gap is direct. For this reason GaAs Etop ∞ makes more efficient optical devices than does either Si n = DC (E)fFD(E)dE ' DC (E)fFD(E)dE Ec Ec or Ge. A particle-hole excitation across the gap can read- (163) ily recombine, emit a photon (which has essentially no Z momentum) and conserve momentum in GaAs; whereas, Ev in an indirect gap semiconductor, this recombination re- p = DV (E)(1 − fFD(E))dE (164) −∞ quires the addition creation or absorption of a phonon or some other lattice excitation to conserve momentum. To proceed further we need forms for DC and DV . Recall ~2~k2 For the same reason, excitons live much longer in Si and that in the parabolic approximation Ek ' 2m∗ we found especially Ge than they do in GaAs.

material τexciton Electron and hole densities in intrinsic (undoped) GaAs 1ns(10−9s) semiconductors Si 19µs(10−5s) Ge 1ms(10−3s) At zero temperature, the Fermi energy lies in the gap, hence there are no holes (p = 0) in the valence band and TABLE II: 26

E w = v 2 g k k T = 0 clearly kinetic Thus, assuming that E − µ & 2 À kBT energy increases 1 1 1 ' = e−(E−µ)/kB T (168) e(E−µ)/kB T + 1 e(E−µ)/kB T ie., Boltzmann statistics. A similar relationship holds for 2 2 E h k g 0 ξ = -E + holes where −(E − µ) & À kBT k F 2m 2

1 1 (E−µ)/k T FIG. 26: Partially filled conduction band and hole band at 1 − = ' e B e(E−µ)/kB T + 1 e−(E−µ)/kB T + 1 non-zero temperature (169) since e(E−µ)/kB T is small. Thus, the concentration of

∗ 3 √ electrons n that D(E) = (2m ) 2 E. Thus, 2π2~3 3 Z (2m∗ ) 2 ∞ p n µ/kB T −E/kB T 3 n ' e E − EC e dE (2m∗ ) 2 p 2π2~3 D (E) = n E − E (165) EC C 2 3 C 3 Z 2π ~ ∗ 2 ∞ (2m ) 3 1 n 2 −(EC −µ)/kB T 2 −x = 2 3 (kBT ) e x e dx 2π ~ 0 ¡ ¢ 3 | {z } ∗ 2 √ 2mp p π/2 DV (E) = EV − E (166) µ ¶ 3 2 3 ∗ 2 2π ~ 2πm kBT = 2 n e−(EC −µ)/kB T h2 for E > EC and E < EV respectively, and zero otherwise C −(EC −µ)/kB T EV < E < EC . = Neff e (170) In an intrinsic (undoped) semiconductor n = p, and so EF must lie in the band gap. Physically, this also means Similarly that we have two types of carriers at non-zero temper- µ ¶ 3 ∗ 2 atures. Both contribute actively to physical properties 2πmpkBT p = 2 e(EV −µ)/kB T = N V e(EV −µ)/kB T such as transport. h2 eff (171) C V D where Neff and Neff are the partition functions for a C classical gas in 3-d and can be regarded as ”effective densities of states” which are temperature-dependent. f(E)D (E) E C C E F Within this interpretation, we can regard the holes and E V (1 - f(E))D (E) V electrons statistics as classical. This holds so long as n f(E) and p are small, so that the Pauli principle may be ig- D V nored - the so called nondegenerate limit. In general, in the nondegenerate limit, FIG. 27: The density of states of the electron and hole bands µ ¶3 k T ¡ ¢ 3 B ∗ ∗ 2 −Eg /kB T np = 4 mnmp e (172) ∗ ∗ 2π~2 If mn 6= mp (ie. DC 6= DV ), then the chemical potential, EF , must be adjusted up or down from the this, the law of mass action, holds for both doped and in- center of the gap so that n = p. trinsic semiconductor so long as we remain in the nonde- Furthermore, the carriers which are induced across the generate limit. However, for an intrinsic semiconductor, gap are relatively (to kBT ) high in energy since typically where n = p, it gives us further information. Eg = EC − EV À kBT . µ ¶ 3 k T 2 ¡ ¢ 3 B ∗ ∗ 4 −Eg /2kB T −3 ◦ ni = pi = 2 m m e (173) Eg(eV ) ni(cm )(300 K) 2π~2 n p Ge 0.67 2.4 × 1013 Si 1.1 1.5 × 1010 (See table (II)). However, we already have relationships GaAs 1.43 5 × 107 for n and p involving EC and EV

C −(EC −µ)/kB T V (EV −µ)/kB T TABLE III: Intrinsic carrier densities at room temperature n = p = Neff e = Neff e (174)

V Neff 1eV ◦ ◦ e2µ/k T = e(EV +EC )/kB T (175) ' 10000 K À 300 K (167) B N C kB eff 27 or Carrier Densities in Doped semiconductor à ! N V 1 1 eff The law of mass action is valid so long as the use µ = (EV + EC ) + kBT ln C (176) 2 2 Neff of Boltzmann statistics is valid i.e., if the degeneracy is small. Thus, even for doped semiconductor

µ ∗ ¶ 1 3 mp µ = (EV + EC ) + kBT ln (177) # ionized ∗ E 2 4 mn C + 0 N = N + N ED D D D ∗ ∗ E Thus if mp 6= mn, the chemical potential µ in a semicon- F # un-ionized + ductor is temperature dependent. N = N + N0 EA A A A EV

Doped Semiconductors FIG. 30: Ionization of the dopants Since, σ ∼ nτ, so the conductivity depends linearly upon the doping (it may also effect µ in some materials, C V −βEg 2 2 np = Neff Neff e = ni = pi , (180) leading to a non-linear doping dependence). A typical metal has where β = 1/kBT . Using (177) we can define µ ∗ ¶ 1 3 mp 23 3 µ = (E + E ) + k T ln , (181) nmetal ' 10 /(cm) (178) i V C B ∗ 2 4 mn whereas we have seen that a typical semiconductor has hence, n = n e(−µi−µ)/kB T and p = n e(µi−µ)/kB T (182) 1010 i i n ' at T ' 300◦K (179) i cm3 To a good approximation we can assume that all donors and all acceptors are ionized. Therefore, Thus the conductivity of an intrinsic semiconductor is 2 quite small! n − p = ND − NA and np = ni 18 2 To increase n (or p) to ∼ 10 or more, dopants are ni used. For example, in Si the elements used as dopants ⇒ n = ND − NA + np are typically in the third or fifth column. Thus P or B N − N (N − N )2 + 4n2 = D A + D A i 2 2 2 2 Si 3s 3p Si Si Si Si Si Si Si Si 2 e+ 2 3 P 3s 3p ⇒ n ' ND and p ' ni /ND for ND À NA Si Si Si Si Si B Si 2 1 - Si B 3s 3p 2 e p ' NA and n ' ni /NA for NA À ND (183) Si P Si Si Si Si Si Si 2ε r r = h big! By convention, if n > p we have an n-type semicon- Si * 2 Si Si Si Si Si Si Si m e ductor, n+-Si, or n-doped Si. The same is true for p. Quite generally, at large doping the semiconductor will FIG. 28: The dopant, P, (left) donates an electron and the behave like a metal, since the dopant will spill over into acceptor, B, donates a hole (or equivalently absorbs an elec- tron). the conduction band (or valence band for holes). In this case there is no gap anymore for the carriers and the con- ductivity remains constant to the lowest possible temper- will either donate or absorb an additional electron (with atures. The main difference to a metal is only that the the latter called the creation of a hole). density of carriers is still very much lower (several orders In terms of energy levels of magnitude) than in a metal. At low doping the num-

n - SeC p - SeC ber of carriers in the conduction band (or valence band) vanishes at zero temperature and the semiconductor be- E EC F EC haves like an insulator. ED unoccupied at T= 0 (occupied by holes at T = 0) An effective way to describe a semiconductor at high occupied at T= 0 E eff A temperature or high doping is to consider that n = ND EV EV EF eff and p = NA , where n and p are mobile negative and eff eff positive carriers, whereas ND and NA are effective FIG. 29: Left, the density of states of a n-doped semiconduc- positive and negative fixed charges, respectively. In this tor with the Fermi level close to the conduction band and, approximation, charge neutrality is automatically veri- right, the equivalent for a p-doped semiconductor. fied and we can use it to discuss heterogenous systems, eff where ND depends on position (but is fixed). 28

Metal-Insulator transition where the constants (ρ0, A and B) depend on the mate- rial. In most metals and heavily doped semiconductors

In n-type semiconductors, when EF < EC , carriers ex- the temperature dependence of the resistivity is domi- perience a gap ∆ = EC −EF . Hence at low temperatures nated by these three mechanisms, which means that the the system is insulating. Indeed, importance of impurities, electron-phonon and phonon- phonon interactions can be extracted from the tempera- µ−EC n = N e kB T (184) ture dependence of the resistivity. |{z}C A special case is the magnetic impurity case (Kondo), ∼T 3/2 which gives rise to an additional term in 1/τK ∼ 2 Since from Drude −(T/TK ) . ne2τ σ = → 0 when T → 0. (185) m∗ In practice When EF > EC , the semiconductor behaves like a metal. In this case we obtain again from Drude that 2 To determine the density (n-p) the Hall resistance can σ = ne τ > 0 even for T → 0, since the density does m∗ be used. The ration τ/m∗ can then be obtained from no vanish. In general, τ also depends on temperature. the Drude conductivity and m∗ can be obtained from Indeed, in the metallic phase the most important tem- magneto-oscillations due to the Landau levels. This can perature dependence comes from τ. To evaluate the con- in principle be done for all temperatures, hence it is possi- tributions from different scattering mechanism we can ble to extract m∗, n(T ), and τ(T ) simply by using trans- consider 1/τ as the scattering probability. This follows port and to deduce the dominant scattering mechanisms directly from Boltzmann’s equation, where ¶ in the systems under study. dg X = (Γ(k0k) − Γ(kk0)), (186) dt coll k0 p-n junction where (See also pn junction supplement) Γ(kk0) = W (kk0) · g(k) · (1 − g(k0)) | {z } |{z} | {z } prob. k→k0 # of states in k # of empty states in k0 (187) is the transition rate from state k to k0. Local equilibrium implies Γkk0 = Γk0k, hence g = fFD. The most important scattering cases are the following:

• Impurity scattering (for a density of impurities nI ): 1 ∼ nI (188) τe−imp

• Electron-electron scattering:

1 2 ∼ (T/TF ) (189) τe−e

• Electron-phonon scattering: FIG. 31: Formation of a pn junction, with the transfer of 1 5 charges from the n region to the p region and the alignment ∼ (T/TD) (190) of the chemical potential so that ∇~ µ(~r) = 0. The depletion τe−ph and accumulation regions are delimited by xn and −xp, re- In general, the total scattering probability is the sum spectively. of all possible scattering probabilities, hence When two differently doped semiconductors are brought together they form a pn junction. In general, 1 1 1 1 = + + electrons form the more n-type doped region will trans- τtot τe−imp τe−e τe−ph µ ¶ µ ¶ fer to the less doped or p-type region. This leaves a m∗ T 2 T 5 positively charged region on the n side and and accumu- ⇒ ρ = 2 ' ρ0 + A + B (191) e nτ TF TD lation of negative charges on the p side. The potential 29 distribution can then be obtained by solving Poisson’s equation v z}|{ Z kR F dk ~k 2 ∂ V (x) e I = −env = −e 2 · ∗ = − ρ(x) (192) L 2π m 2 kF ∂x ²0 L 2 R 2 ~(kF ) − (kF ) = −e ∗ where ρ(x) = −n(x)+p(x)+ND(x)−NA(x) is the charge 2πm distribution of the mobile carriers (electrons and holes) 2e = − ( µR − µL ) plus the fixed charges (donors and acceptors) and ²0 is h |{z} |{z} the dielectric constant. −eVR −eVL 2e2 = (V − V ) (193) h R L

- + - + - + p type - + n type - + - +

no holes no electrons charge density in depletion region

electric field l a ti n e t Vo=(n-p)/e po FIG. 33: Schematic cross-sectional view of a quantum point contact, defined in a high-mobility 2D electron gas at the in- terface of a GaAs-AlGaAs heterojunction. The point contact FIG. 32: Sketch of the charge density, the electric field and is formed when a negative voltage is applied to the gate elec- the internal potential distribution due the transfer of charges trodes on top of the AlGaAs layer. Transport measurements in a pn junction are made by employing contacts to the 2D electron gas at ei- ther side of the constriction. (Beenakker, PHYSICS TODAY, July 1996). The most important consequence of a pn-junction is the diode behavior. Indeed, when applying a negative Hence I = (2e2/h)∆V ⇒ R = h/2e2 and G = 2e2/h, bias on the n region, the conduction bands tend to align where R and G are the resistances and conductances, more and current can flow (forward bias). If the bias is respectively. This result can seem surprising at first since positive the internal potential is increased and almost no it implies that the resistance does not depend on the current can flow (reverse bias). This leads to the well length of the system. A very short quantum wire has the known asymmetry in the current-voltage characteristics same resistance as an infinitely long wire. This result of a diode. This is one of the most important elements of however, only applies for a perfect conductor. As soon electronics. These ideas can be extended to three termi- as impurities lie in the wire this result has to be modified nal devices such as a pnp junction of npn junction (which to take into account scattering by the impurities and then is equivalent to two pn junction put together). In this R would typically depend on the length of the system. case we have a transistor, i.e., by varying the potential on the center element we can control the current through the device. MORE THAN ONE CHANNEL, THE QUANTUM POINT CONTACT

The previous result is specific to the purely one- ONE DIMENSIONAL CONDUCTANCE dimensional case. In general the system can be extended to a system of finite width, W . In this case the electron Suppose that we have a perfect one-dimensional con- energy is given by ductor (quantum wire) connected by two large electron ~2k2 ~2(πn/W )2 reservoirs. In real life they would be electrical contacts. ² = x + (194) 2m∗ 2m∗ The left reservoir is fixed at chemical potential µL and the right one at µR. The current is then given as usual if the boundary of our narrow wire is assumed to be by sharp, since in this case the wave function has to vanish 30

FIG. 34: Conductance quantization of a quantum point con- FIG. 35: Imaging of the channels using an AFM (1 to 3 chan- tact in units of 2e2/h. As the gate voltage defining the con- nels from left to right). The images were obtained by applying striction is made less negative, the width of the point contact a small negative potential on the AFM tip and then measur- increases continuously, but the number of propagating modes ing the conductance as a function of the tip scan and then at the Fermi level increases stepwise. The resulting conduc- reconstruct the 2D image from the observed change in con- tance steps are smeared out when the thermal energy becomes ductance. (From R.M. Westervelt). comparable to the energy separation of the modes.

-M (magnetization) at the boundary (like for an electron in a box of width W ). In general, if

2 2 2 2 Type H Type HH ~ (π/W ) ~ (2π/W ) H B>O (Vortex phase) < EF < (195) H H 2m∗ 2m∗ C1 C2 B (internal field) then we recover the ideal case of a one-dimensional quan- B=O (Meissner) HH tum wire, or single channel. If, however, H H 2 2 2 2 HC1 HC2 ~ (Nπ/W ) ~ ((N + 1)π/W ) T ∗ < EF < ∗ (196) (Abrikosov 2m 2m TC vortex crystal) we can have n channels, where each channel contributes H HH H H H equally to the total conductance. Hence in this case G = C1 C2 N · 2e2/h, where N is the number of channels. This implies that a system, where we reduce the width of the FIG. 36: Magnetic field dependence in a superconductor. conductor will exhibit jumps in the conductance of step (Vortex picture from AT&T ’95) 2e2/n. Indeed, this is what is seen experimentally.

• Superconductivity can be destroyed by an exter- QUANTUM HALL EFFECT nal magnetic field Hc which is also called the criti- cal field (Kammerlingh-Onnes, 1914). Empirically, 2 The quantum hall effect is a beautiful example, where Hc(T ) = Hc(0)(1 − (T/Tc) ) the concept of a quantized conductance can be applied to (see additional notes on the quantum Hall effect). • The Meissner-Ochsenfeld effect (1933). The mag- netic field does not penetrate the sample, the mag- netic induction is zero, B = 0. This effect distin- SUPERCONDUCTIVITY guishes two types of superconductors, type I and type II. In Type I, no field penetrates the sample, The main aspects of superconductivity are whereas in type II the field penetrates in the form of vortices. • Zero resistance (Kammerlingh-Onnes, 1911) at T < Tc: The temperature Tc is called the critical tem- • Superconductors have a gap in the excitation spec- perature. trum. 31

Due to retardation, the electron-electron Coulomb repul- sion may be neglected! The net effect of the phonons is then to create an attractive interaction which tends to pair time-reversed quasiparticle states. They form an antisymmetric spin

k↑ e

Vanadium ξ ∼ 1000Α°

e

- k↓

FIG. 39: To take full advantage of the attractive potential illustrated in Fig. 38, the spatial part of the electronic pair wave function is symmetric and hence nodeless. To obey the Pauli principle, the spin part must then be antisymmetric or FIG. 37: The critical magnetic field, resistance and specific a singlet. heat as a function of temperature. (Ref: superconductors.org) singlet so that the spatial part of the wave function can The main mechanism behind superconductivity is the be symmetric and nodeless and so take advantage of the existence of an effective attractive force between elec- attractive interaction. Furthermore they tend to pair in trons, which favors the pairing of two electrons of oppo- a zero center of mass (cm) state so that the two electrons site momentum and spin. In conventional superconduc- can chase each other around the lattice. tors this effective attractive force is due to the interaction Using perturbation theory it is in fact possible to show with phonons. This pair of electrons has now effectively that to second order (electron-phonon-electron) the effect zero total momentum and zero spin. In this sense this of the phonons effectively leads to a potential of the form pair behaves like a boson and will Bose-Einsteein conden- sate in a coherent quantum state with the lowest possible (~ω )2 V ∼ q (197) energy. This ground sate is separated by a superconduct- e−ph (²(k) − ²(k − q))2 − (~ω(q))2 ing gap. Electrons have to jump over this gap in order to be excited. Hence when the thermal energy exceeds the This term can be negative, hence effectively produce and gap energy the superconductor becomes normal. attraction between two electrons exceeding the Coulomb repulsion. This effect is the strongest for k = k and q = The origin of the effective attraction between electrons F 2k since ²(k ) = ²(−k ) and ω(2k ) ' ω (the Debye can be understood in the following way: F F F F D frequency). Hence electrons will want to form opposite momentum pairs (kF , −kF ). This will be our starting + e- point for the microscopic theory of superconductivity `a ions e- + + + + + + la BCS (Bardeen, Shockley and Schrieffer). 8 region of + + + + v ∼ 10 cm/s + + F positive charge + attracts a second electron BCS theory

FIG. 38: Origin of the retarded attractive potential. Elec- To describe our pair of electrons (the Cooper pair) we trons at the Fermi surface travel with a high velocity vF . As they pass through the lattice (left), the positive ions respond will use the formalism of second quantization, which is slowly. By the time they have reached their maximum excur- a convenient way to describe a system of more than one sion, the first electron is far away, leaving behind a region of particle. positive charge which attracts a second electron. p2 H = ⇒ H ψ(x) = Eψ(x) (198) When an electron flies through the lattice, the lat- 1particle 2m 1p tice deforms slowly with respect to the time scale of the electron. It reaches its maximum deformation at Let’s define a time τ ' 2π ' 10−13s after the electron has passed. + ∗ ωD c1 (x) |0i = ψ(x) and h0|c1(x) = ψ (x) (199) In this time the first electron has travelled ' vF τ ' |{z} 8 cm −13 ˚ vacuum 10 s · 10 s ' 1000A. The positive charge of the lat- tice deformation can then attract another electron with- With these definitions, |0i is the vacuum (or ground + out feeling the Coulomb repulsion of the first electron. state), i.e., state without electrons. c1 (x)|0i corresponds 32 to one electron in state ψ(x) which we call 1. c+ is also ( called the creation operator, since it creates one elec- + tron from vacuum. c is then the anhilation operator, i.e., c1 = A1 cos(θ) + A2 sin(θ) + + + c c |0i = |0i, which corresponds to creating one electron c2 = −A1 sin(θ) + A2 cos(θ) 1 1 ( from vacuum then anhilating it again. Other properties + A1 = c1 cos(θ) − c2 sin(θ) include ⇒ + + (204) A2 = c1 sin(θ) + c2 cos(θ) + + c1|0i = 0 and c1 c1 |0i = 0 (200) + + It is quite straightforward to see that Ai Aj +AjAi = δij, + + + + The first relation means that we cannot anhilate an AiAj + AjAi = 0, and Ai Aj + Aj Ai = 0 using the electron from vacuum and the second relation is a con- properties of ci. We can now rewrite HMF in terms of sequence of the Pauli principle. We cannot have two our new operators Ai: electrons in the same state 1. + + + + 2 HMF = t(c1 c1 + c2 c2) − ga(c1c2 − c1 c2 ) + ga Hence, + + = t(A1 cos(θ) + A2 sin(θ))(cos(θ)A1 + sin(θ)A2 ) + ··· + + + + + + + c1 c1 = 0 ⇒ (c1 c1 ) = 0 ⇒ c1c1 = 0 = (A1 A1 + A2 A2)(t cos(2θ) − ga sin(2θ)) + + + 2 ⇒ (c1 c1)c1 |0i = c1 |0i (201) + t(1 − cos(2θ)) + ga sin(2θ) + ga + + + + (A1 A2 − A1A2)(t sin(2θ) + ga cos(2θ)) (205) This shows that c1 c1 acts like a number operator. It | {z } counts the number of electrons in state 1. (Either 1 or =0 to diagonalize HMF 0). Hence the diagonalization condition for H fixes the We can now extend this algebra for two electrons in MF + angle θ of our Boguliubov transformation: different states c1 |0i corresponds to particle 1 in state 1 + ga −ga and c2 |0i to particle 2 in state 2. The rule here is that tan(2θ) = − ⇒ sin(2θ) = p (206) + + 2 2 ci cj + cjci = δi,j. t t + (ga) Finally we can write down the two particle hamiltonian Hence H now becomes as MF p p + + 2 2 2 2 2 + + + + HMF = (A1 A1+A2 A2) t + (ga) +(t− t + (ga) )+ga H2p = t1c1 c1 + t2c2 c2 − gc1 c1c2 c2 (202) (207) It is now immediate to obtain the solutions of the where t is the kinetic energy of particle i and g is i Hamiltonian, since we have the ground state |ai and we the attraction between particle 1 and 2. We will also have a diagonal Hamiltonian in terms of A hence the suppose that t = t for the Cooper pair. The job now i 1 2 Ground state energy E is simply given by H |ai = is to find the ground state of this Hamiltonian. Without 0 MF E |ai and A |ai = 0. The first degenerate excited states interactions (g = 0) we would simply have E = t + 0 i 1 are A+|ai with energy E , where H A+|ai = E A+|ai t . The interaction term is what complicates the system i 1 MF i 1 i 2 and the next energy level and state is A+A+|ai, with since it leads to a quadratic term in the Hamiltonian. 1 2 energy E given by H A+A+|ai = E A+A+|ai, hence The idea is to simplify it by getting rid of the quadratic 2 MF 1 2 2 1 2 p term. This is done in the following way. From eq. (202) 2 2 2 E2 = t + t + (ga) + ga we have E = t + ga2 1 p 2 2 2 E0 = t − t + (ga) + ga (208) H = t(c+c + c+c ) − gc+c c+c 1 1 2 2 1 1 2 2 If we take t → 0 and define ∆ = ga we have + + + + = t(c1 c1 + c2 c2) + gc1 c2 c1c2  2 + + + + 2  E2 = ∆ + ga = t(c1 c1 + c2 c2) − ga(c1c2 − c1 c2 ) + ga 2 + + E1 = ga (209) + g(c c + a)(c1c2 − a) (203)  1 2  2 | {z } E0 = −∆ + ga '0 (Mean field approx.) + + + + 2 We can now turn to what a is since, ⇒ HMF = t(c1 c1 + c2 c2) − ga(c1c2 − c1 c2 ) + ga

We used the mean field approximation, which re- a = ha|c1c2|ai places c1c2 by its expectation value ha|c1c2|ai = a ⇒ = − cos(θ) sin(θ)ha|A A+|ai ha|c+c+|ai = −a, where |ai is the ground state of the 1 1 1 2 = − cos(θ) sin(θ) ha|ai Hamiltonian. The idea now is to diagonalize HMF , i.e. | {z } P + =1 a Hamiltonian in the form H = i ci ci. The trick here is to use the Boguliubov transformation: ⇒ a = − sin(2θ)/2 (210) 33

Combining (206) and (210) we obtain another one with momentum −k − q, hence momentum is conserved in this scattering process and a phonon with ga momentum q is exchanged. 2a = p (211) t2 + (ga)2 How can we relate BCS theory to the observed Meiss- ner effect? By including the vector potential A~ into the which is the famous BCS gap (∆) equation. Indeed, it BCS Hamiltonian it is possible to show that the current has two solutions, density is then given by

a = 0 ⇒ ha|c c |ai = 0 ⇒ Normal ne2 1 2 ~j = A~ (216) a 6= 0 ⇒ t2 + (ga)2 = g2/4 ⇒ Superconductor(212) mc | {z } √ ∆=ga= g2/4−t2 The magnetic induction is B~ = ∇~ × A~ as usual. This is in fact London’s equation for superconductivity. We can '=ga now take the rotational on both sides of (216), hence

Normal ne2 ∇~ × ~j = ∇~ × A~ Superconductor |{z} mc ~ ~ 4π~ ∇×B= c j t 2 g/2 4πne ⇒ ∇~ × ∇~ × B~ = − B~ FIG. 40: The gap of a BCS superconductor as function of the mc2 kinetic energy. −x/λL ⇒ Bx ∼ e (217) This gives us the condition for superconductivity g ≥ q mc2 2t. Hence the attraction between our two electrons has with λL = 4πne2 which is the London penetration to be strong enough in order to form the superconducting length. gap ∆. Typically, t is directly related to the temperature, hence there is a superconducting transition as a function of temperature. We now want to find the expression for our supercon- B ducting wavefunction |ai. The most general possible form is -x/ l e L Vacuum + + + + Inside the |ai = α|0i + β1c1 |0i + β2c2 |0i + γc1 c2 |0i (213) superconductor

In addition the condition Ai|ai = 0 has to be verified, x which leads after some algebra to FIG. 41: The decay of the magnetic field inside the supercon- ductor. The decay is characterized by the London penetration + + |ai = α(1 + tan(θ)c1 c2 )|ai (214) length λL. This state clearly describes an electron pair, the Cooper pair and represents the superconducting ground If we had a perfect conductor, this would imply that the current would simply keep on increasing with an ex- state of the Hamiltonian. In our derivation we only con- ~ sidered two electrons, but this framework can be gener- ternal field E, hence alized to N electrons, where the generalized BCS Hamil- ∂~j ne2 tonian can be written as = E~ ∂t mc 2 X X ∂ ~ ne ~ ~ + + + ⇒ ∇ ×~j = ∇ × E HBCS = tkc ck,σ − Vqc c ck,↑c−k,↓ ∂t mc k,|{z}σ k+q,↑ −k+q,↓ k,σ q ∂ ∂ 4πne2 spin ⇒ ∇~ × ∇~ × B~ = B~ (218) (215) ∂t ∂t mc2 Here Vq is positive and represents the effective phonon induced attraction between electrons at the Fermi level. This equation is automatically verified from (217). It’s maximum for q = 0. The second term describes the Hence, (217) implies both the Meissner effect and zero process of one electron with momentum k and another resistance, which are the main ingredients of supercon- ~ ~ 4π~ electron with momentum −k which are anhilated in or- ductivity. (Reminder: Maxwell gives ∇ × B = c j and ~ ~ 1 ∂B~ der to create one electron with momentum k + q and ∇ × E = − c ∂t ). 34

A remarkable aspect of superconductivity is that one where U(1) ⇔ c → ceiα (Gauge invariance), U(2) ⇔ c → of the most fundamental symmetries is broken. Indeed, Ac (A = eiφ/2a) and a+a = 1, SU(2) ⇔ φ = 0. Gauge invariance is broken because ~j ∼ A~. What hap- pens is that below Tc we have a symmetry breaking, which leads to new particles, the Cooper pairs. Math- ematically, we have

U(2) = SU(2) ⊗ U(1) , (219) | {z } | {z } | {z } Normal state SC state Gauge invariance