N. W. Ashcroft and N. D. Mermin, Solid State Physics, chapter 13. P. L. Taylor and O. Heinonen, Condensed Matter Physics, chapter 8.

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1 Chapter Boltzmann Transport. References H. Smith and H. H. Jensen, Transport Phenomena N. W. Ashcroft and N. D. Mermin, Solid State Physics, chapter 3. P. L. Taylor and O. Heinonen, Condensed Matter Physics, chapter 8. J. M. Ziman, Principles of the Theory of Solids, chapter 7..2 Introduction Transport is the phenomenon of currents flowing in response to applied fields. By current we generally mean an electrical current j, or thermal current j q. By applied field we generally mean an electric field E or a temperature gradient T. The currents and fields are linearly related, and it will be our goal to calculate the coefficients (known as transport coefficients) of these linear relations. Implicit in our discussion is the assumption that we are always dealing with systems near equilibrium.

2 2 CHAPTER. BOLTZMANN TRANSPORT.3 Boltzmann Equation in Solids.3. Semiclassical Dynamics and Distribution Functions The semiclassical dynamics of a wavepacket in a solid are described by the equations dr dt dk dt = v n (k) = n (k) k (.) = e E(r, t) e c v n(k) B(r, t). (.2) Here, n is the band index and ε n (k) is the dispersion relation for band n. The wavevector is k ( k is the crystal momentum ), and ε n (k) is periodic under k k + G, where G is any reciprocal lattice vector. These formulae are valid only at sufficiently weak fields. They neglect, for example, Zener tunneling processes in which an electron may change its band index as it traverses the Brillouin zone. We also neglect the spin-orbit interaction in our discussion. We are of course interested in more than just a single electron, hence to that end let us consider the distribution function f n (r, k, t), defined such that f nσ (r, k, t) d3 r d 3 k (2π) 3 # of electrons of spin σ in band n with positions within d 3 r of r and wavevectors within d 3 k of k at time t. (.3) Note that the distribution function is dimensionless. By performing integrals over the distribution function, we can obtain various physical quantities. For example, the current density at r is given by j(r, t) = e d 3 k (2π) 3 f nσ(r, k, t) v n (k). (.4) n,σ The symbol in the above formula is to remind us that the wavevector integral is performed only over the first Brillouin zone. We now ask how the distribution functions f nσ (r, k, t) evolve in time. To simplify matters, we will consider a single band and drop the indices nσ. It is clear that in the absence of collisions, the distribution function must satisfy the continuity equation, f + (uf) = 0. (.5) t This is just the condition of number conservation for electrons. Take care to note that and u are six-dimensional phase space vectors: u = ẋ, ẏ, ż, kx, ky, kz ) (.6) ( = x, y, ) z,,,. (.7) k x k y k z We will assume three space dimensions. The discussion may be generalized to quasi-two dimensional and quasi-one dimensional systems as well.

3 .3. BOLTZMANN EQUATION IN SOLIDS 3 Now note that as a consequence of the dynamics (.,.2) that u = 0, i.e. phase space flow is incompressible, provided that ε(k) is a function of k alone, and not of r. Thus, in the absence of collisions, we have f t + u f = 0. (.8) The differential operator D t t + u is sometimes called the convective derivative. EXERCISE: Show that u = 0. Next we must consider the effect of collisions, which are not accounted for by the semiclassical dynamics. In a collision process, an electron with wavevector k and one with wavevector k can instantaneously convert into a pair with wavevectors k + q and k q (modulo a reciprocal lattice vector G), where q is the wavevector transfer. Note that the total wavevector is preserved (mod G). This means that D t f 0. Rather, we should write f t + ṙ f r + k f k = ( ) f I k {f} (.9) t coll where the right side is known as the collision integral. The collision integral is in general a function of r, k, and t and a functional of the distribution f. As the k-dependence is the most important for our concerns, we will write I k in order to make this dependence explicit. Some examples should help clarify the situation. First, let s consider a very simple model of the collision integral, I k {f} = f(r, k, t) f 0 (r, k) τ(ε(k)). (.0) This model is known as the relaxation time approximation. Here, f 0 (r, k) is a static distribution function which describes a local equilibrium at r. The quantity τ(ε(k)) is the relaxation time, which we allow to be energy-dependent. Note that the collision integral indeed depends on the variables (r, k, t), and has a particularly simple functional dependence on the distribution f. A more sophisticated model might invoke Fermi s golden rule, Consider elastic scattering from a static potential U(r) which induces transitions between different momentum states. We can then write I k {f} = 2π = 2π V k U k 2 (f k f k ) δ(ε k ε k ) (.) k d 3 k (2π) 3 Û(k k ) 2 (f k f k ) δ(ε k ε k ) (.2) where we abbreviate f k f(r, k, t). In deriving the last line we ve used plane wave wave-

4 4 CHAPTER. BOLTZMANN TRANSPORT functions 2 ψ k (r) = exp(ik r)/ V, as well as the result d 3 k A(k) = V A(k) (.3) (2π) 3 k for smooth functions A(k). Note the factor of V in front of the integral in eqn..2. What this tells us is that for a bounded localized potential U(r), the contribution to the collision integral is inversely proportional to the size of the system. This makes sense because the number of electrons scales as V but the potential is only appreciable over a region of volume V 0. Later on, we shall consider a finite density of scatterers, writing U(r) = N imp i= U(r R i ), where the impurity density n imp = N imp /V is finite, scaling as V 0. In this case Û(k k ) apparently scales as V, which would mean I k {f} scales as V, which is unphysical. As we shall see, the random positioning of the impurities means that the O(V 2 ) contribution to Û(k k ) 2 is incoherent and averages out to zero. The coherent piece scales as V, canceling the V in the denominator of eqn..2, resulting in a finite value for the collision integral in the thermodynamic limit (i.e. neither infinite nor infinitesimal). Later on we will discuss electron-phonon scattering, which is inelastic. An electron with wavevector k can scatter into a state with wavevector k = k q mod G by absorption of a phonon of wavevector q or emission of a phonon of wavevector q. Similarly, an electron of wavevector k can scatter into the state k by emission of a phonon of wavevector q or absorption of a phonon of wavevector q. The matrix element for these processes depends on k, k, and the polarization index of the phonon. Overall, energy is conserved. These considerations lead us to the following collision integral: I k {f, n} = 2π V g λ (k, k ) 2{ ( f k ) f k ( + n q,λ ) δ(ε k + ω qλ ε k ) k,λ +( f k ) f k n qλ δ(ε k ω qλ ε k ) f k ( f k ) ( + n qλ ) δ(ε k ω qλ ε k ) } f k ( f k ) n qλ δ(ε k + ω qλ ε k ) δ q,k k mod G, (.4) which is a functional of both the electron distribution f k as well as the phonon distribution n qλ. The four terms inside the curly brackets correspond, respectively, to cases (a) through (d) in fig... While collisions will violate crystal momentum conservation, they do not violate conservation of particle number. Hence we should have 3 d 3 r d 3 k (2π) 3 I k{f} = 0. (.5) 2 Rather than plane waves, we should use Bloch waves ψ nk (r) = exp(ik r) u nk (r), where cell function u nk (r) satisfies u nk (r + R) = u nk (r), where R is any direct lattice vector. Plane waves do not contain the cell functions, although they do exhibit Bloch periodicity ψ nk (r + R) = exp(ik R) ψ nk (r). 3 If collisions are purely local, then R d 3 k I (2π) 3 k {f} = 0 at every point r in space.

5 .3. BOLTZMANN EQUATION IN SOLIDS 5 Figure.: Electron-phonon vertices. The total particle number, N = d 3 r d 3 k f(r, k, t) (.6) (2π) 3 is a collisional invariant - a quantity which is preserved in the collision process. Other collisional invariants include energy (when all sources are accounted for), spin (total spin), and crystal momentum (if there is no breaking of lattice translation symmetry) 4. Consider a function F (r, k) of position and wavevector. Its average value is Taking the time derivative, d F dt F (t) = = F t = d 3 r = d 3 r d 3 k (2π) 3 d 3 r d 3 k F (r, k) f(r, k, t). (.7) (2π) 3 d 3 { k F (r, k) (2π) 3 r (ṙf) } k ( kf) + I k {f} {[ F r dr dt + F k dk ] } f + F I k {f} dt. (.8) 4 Note that the relaxation time approximation violates all such conservation laws. Within the relaxation time approximation, there are no collisional invariants.

6 6 CHAPTER. BOLTZMANN TRANSPORT Hence, if F is preserved by the dynamics between collisions, then d F dt = d 3 d 3 k r (2π) 3 F I k{f}, (.9) which says that F (t) changes only as a result of collisions. If F is a collisional invariant, then F = 0. This is the case when F =, in which case F is the total number of particles, or when F = ε(k), in which case F is the total energy..3.2 Local Equilibrium The equilibrium Fermi distribution, f 0 (k) = { exp ( ε(k) µ k B T ) + } (.20) is a space-independent and time-independent solution to the Boltzmann equation. Since collisions act locally in space, they act on short time scales to establish a local equilibrium described by a distribution function f 0 (r, k, t) = { exp ( ε(k) µ(r, t) k B T (r, t) ) + } (.2) This is, however, not a solution to the full Boltzmann equation due to the streaming terms ṙ r + k k in the convective derivative. These, though, act on longer time scales than those responsible for the establishment of local equilibrium. To obtain a solution, we write f(r, k, t) = f 0 (r, k, t) + δf(r, k, t) (.22) and solve for the deviation δf(r, k, t). We will assume µ = µ(r) and T = T (r) are timeindependent. We first compute the differential of f 0, df 0 = k B T f 0 ( ) ε µ d k B T { } = k B T f 0 dµ (ε µ) dt k B T k B T 2 + dε k B T { } = f 0 µ r dr + ε µ T dr T r k dk, (.23) from which we read off f 0 r f 0 k = { µ r + ε µ T T r } ( f 0 ) (.24) = v f 0. (.25)

7 .4. CONDUCTIVITY OF NORMAL METALS 7 We thereby obtain δf t + v δf e [E + c ] v B δf [ k + v e E + ε µ ] ( T T f 0 ) = I k {f 0 + δf} (.26) where E = (φ µ/e) is the gradient of the electrochemical potential ; we ll henceforth refer to E as the electric field. Eqn (.26) is a nonlinear integrodifferential equation in δf, with the nonlinearity coming from the collision integral. (In some cases, such as impurity scattering, the collision integral may be a linear functional.) We will solve a linearized version of this equation, assuming the system is always close to a state of local equilibrium. Note that the inhomogeneous term in (.26) involves the electric field and the temperature gradient T. This means that δf is proportional to these quantities, and if they are small then δf is small. The gradient of δf is then of second order in smallness, since the external fields φ µ/e and T are assumed to be slowly varying in space. To lowest order in smallness, then, we obtain the following linearized Boltzmann equation: δf t e c v B δf [ k + v e E + ε µ ] ( T T f 0 ) = L δf (.27) where L δf is the linearized collision integral; L is a linear operator acting on δf (we suppress denoting the k dependence of L). Note that we have not assumed that B is small. Indeed later on we will derive expressions for high B transport coefficients..4 Conductivity of Normal Metals.4. Relaxation Time Approximation Consider a normal metal in the presence of an electric field E. We ll assume B = 0, T = 0, and also that E is spatially uniform as well. This in turn guarantees that δf itself is spatially uniform. The Boltzmann equation then reduces to δf t f 0 ev E = I k{f 0 + δf}. (.28) We ll solve this by adopting the relaxation time approximation for I k {f}: I k {f} = f f 0 τ = δf τ, (.29) where τ, which may be k-dependent, is the relaxation time. In the absence of any fields or temperature and electrochemical potential gradients, the Boltzmann equation becomes δf = δf/τ, with the solution δf(t) = δf(0) exp( t/τ). The distribution thereby relaxes to the equilibrium one on the scale of τ.

8 8 CHAPTER. BOLTZMANN TRANSPORT Writing E(t) = E e iωt, we solve and obtain δf(k, t) t δf(k, t) = iωt f 0 e v(k) E e e E v(k) τ(ε(k)) iωτ(ε(k)) = δf(k, t) τ(ε(k)) (.30) f 0 e iωt. (.3) The equilibrium distribution f 0 (k) results in zero current, since f 0 ( k) = f 0 (k). Thus, the current density is given by the expression j α d 3 k (r, t) = 2e δf vα (2π) 3 = 2e 2 E β e iωt d 3 k τ(ε(k)) v α (k) v β (k) (2π) 3 iωτ(ε(k)) ( f 0 ). (.32) In the above calculation, the factor of two arises from summing over spin polarizations. The conductivity tensor is defined by the linear relation j α (ω) = σ αβ (ω) E β (ω). We have thus derived an expression for the conductivity tensor, σ αβ (ω) = 2e 2 d 3 k τ(ε(k)) v α (k) v β (k) (2π) 3 iωτ(ε(k)) ( f 0 ) (.33) Note that the conductivity is a property of the Fermi surface. For k B T ε F, we have f 0 / δ(ε F ε(k)) and the above integral is over the Fermi surface alone. Explicitly, we change variables to energy ε and coordinates along a constant energy surface, writing d 3 k = dε ds ε / k = dε ds ε v, (.34) where ds ε is the differential area on the constant energy surface ε(k) = ε, and v(k) = k ε(k) is the velocity. For T T F, then, σ αβ (ω) = e2 4π 3 τ(ε F ) iωτ(ε F ) ds F v α (k) v β (k) v(k). (.35) For free electrons in a parabolic band, we write ε(k) = 2 k 2 /2m, so v α (k) = k α /m. To further simplify matters, let us assume that τ is constant, or at least very slowly varying in the vicinity of the Fermi surface. We find ( f 0 ) σ αβ (ω) = δ αβ 2 3m where g(ε) is the density of states, g(ε) = 2 e 2 τ iωτ dε g(ε) ε, (.36) d 3 k δ (ε ε(k)). (.37) (2π) 3

9 .4. CONDUCTIVITY OF NORMAL METALS 9 The (three-dimensional) parabolic band density of states is found to be g(ε) = (2m ) 3/2 ε Θ(ε), (.38) 2π 2 3 where Θ(x) is the step function. In fact, integrating (.36) by parts, we only need to know about the ε dependence in g(ε), and not the details of its prefactor: dε ε g(ε) ( f 0 ) = dε f 0 (ε) (ε g(ε)) dε g(ε) f 0 (ε) = 3 2 n, (.39) = 3 2 where n = N/V is the electron number density for the conduction band. The final result for the conductivity tensor is σ αβ (ω) = ne2 τ m δ αβ iωτ (.40) This is called the Drude model of electrical conduction in metals. The dissipative part of the conductivity is Re σ. Writing σ αβ = σδ αβ and separating into real and imaginary parts σ = σ + iσ, we have σ (ω) = ne2 τ m + ω 2 τ 2. (.4) The peak at ω = 0 is known as the Drude peak. Here s an elementary derivation of this result. Let p(t) be the momentum of an electron, and solve the equation of motion to obtain p(t) = eτe iωτ e iωt + dp dt = p τ e E e iωt (.42) [ p(0) + eτe ] e t/τ. (.43) iωτ The second term above is a transient solution to the homogeneous equation ṗ + p/τ = 0. At long times, then, the current j = nep/m is j(t) = ne 2 τ m ( iωτ) E e iωt. (.44) In the Boltzmann equation approach, however, we understand that n is the conduction electron density, which does not include contributions from filled bands. In solids the effective mass m typically varies over a small range: m (0. ) m e. The two factors which principally determine the conductivity are then the carrier density n and the scattering time τ. The mobility µ, defined as the ratio σ(ω = 0)/ne, is thus (roughly) independent of carrier density 5. Since j = nev = σe, where v is an average carrier velocity, we have v = µe, and the mobility µ = eτ/m measures the ratio of the carrier velocity to the applied electric field. 5 Inasmuch as both τ and m can depend on the Fermi energy, µ is not completely independent of carrier density.

10 0 CHAPTER. BOLTZMANN TRANSPORT Figure.2: Frequency-dependent conductivity of liquid sodium by T. Inagaki et al, Phys. Rev. B 3, 560 (976)..4.2 Optical Reflectivity of Metals and Semiconductors What happens when an electromagnetic wave is incident on a metal? Inside the metal we have Maxwell s equations, H = 4π c j + ( D 4πσ = ik B = iω ) E c t c c (.45) E = B c t = ik E = iω c B (.46) E = B = 0 = ik E = ik B = 0, (.47) where we ve assumed µ = ɛ = inside the metal, ignoring polarization due to virtual interband transitions (i.e. from core electrons). Hence, k 2 = ω2 c 2 + 4πiω c 2 σ(ω) (.48) = ω2 c 2 + ω2 p c 2 iωτ iωτ ɛ(ω) ω2 c 2, (.49) where ω p = 4πne 2 /m is the plasma frequency for the conduction band. The dielectric function, ɛ(ω) = + 4πiσ(ω) ω = + ω2 p ω 2 iωτ iωτ (.50)

11 .4. CONDUCTIVITY OF NORMAL METALS determines the complex refractive index, N(ω) = ɛ(ω), leading to the electromagnetic dispersion relation k = N(ω) ω/c. Consider a wave normally incident upon a metallic surface normal to ẑ. In the vacuum (z < 0), we write while in the metal (z > 0), E(r, t) = E ˆx e iωz/c e iωt + E 2 ˆx e iωz/c e iωt (.5) B(r, t) = c iω E = E ŷ e iωz/c e iωt E 2 ŷ e iωz/c e iωt (.52) E(r, t) = E 3 ˆx e inωz/c e iωt (.53) B(r, t) = c iω E = N E 3 ŷ e inωz/c e iωt (.54) Continuity of E ˆn gives E + E 2 = E 3. Continuity of H ˆn gives E E 2 = N E 3. Thus, E 2 = N E + N, E 3 E = 2 + N (.55) and the reflection and transmission coefficients are R(ω) = E 2 2 E = N(ω) 2 + N(ω) (.56) T (ω) = E = + N(ω) 2. (.57) We ve now solved the electromagnetic boundary value problem. E Typical values For a metal with n = 0 22 cm 3 and m = m e, the plasma frequency is ω p = s. The scattering time varies considerably as a function of temperature. In high purity copper at T = 4 K, τ s and ω p τ 0 7. At T = 300 K, τ s and ω p τ 00. In either case, ω p τ. There are then three regimes to consider. ωτ ω p τ: We may approximate iωτ, hence N 2 (ω) = + N(ω) + i 2 ( ω 2 p τ ω i ωpτ 2 ω( iωτ) i ω2 pτ ω ) /2 = R 2 2ωτ ω p τ. (.58) Hence R and the metal reflects. ωτ ω p τ: In this regime, N 2 (ω) ω2 p ω 2 + i ω2 p ω 3 τ (.59)

12 2 CHAPTER. BOLTZMANN TRANSPORT Figure.3: Frequency-dependent absorption of hcp cobalt by J. Weaver et al., Phys. Rev. B 9, 3850 (979). which is almost purely real and negative. Hence N is almost purely imaginary and R. (To lowest nontrivial order, R = 2/ω p τ.) Still high reflectivity. ω p τ ωτ: Here we have N 2 (ω) ω2 p ω 2 = R = ω p 2ω and R the metal is transparent at frequencies large compared to ω p. (.60).4.3 Optical Conductivity of Semiconductors In our analysis of the electrodynamics of metals, we assumed that the dielectric constant due to all the filled bands was simply ɛ =. This is not quite right. We should instead have written k 2 ω 2 = ɛ c 2 + 4πiωσ(ω) c 2 (.6) { } ɛ(ω) = ɛ + ω2 p iωτ ω 2, (.62) iωτ

13 .4. CONDUCTIVITY OF NORMAL METALS 3 Figure.4: Frequency-dependent conductivity of hcp cobalt by J. Weaver et al., Phys. Rev. B 9, 3850 (979). This curve is derived from the data of fig..3 using a Kramers- Krönig transformation. A Drude peak is observed at low frequencies. At higher frequencies, interband effects dominate. where ɛ is the dielectric constant due to virtual transitions to fully occupied (i.e. core) and fully unoccupied bands, at a frequency small compared to the interband frequency. The plasma frequency is now defined as ( ) 4πne 2 /2 ω p = (.63) m ɛ where n is the conduction electron density. Note that ɛ(ω ) = ɛ, although again this is only true for ω smaller than the gap to neighboring bands. It turns out that for insulators one can write ɛ + ω2 pv ω 2 g (.64) where ω pv = 4πn v e 2 /m e, with n v the number density of valence electrons, and ω g is the energy gap between valence and conduction bands. In semiconductors such as Si and Ge, ω g 4 ev, while ω pv 6 ev, hence ɛ 7, which is in rough agreement with the experimental values of 2 for Si and 6 for Ge. In metals, the band gaps generally are considerably larger. There are some important differences to consider in comparing semiconductors and metals:

14 4 CHAPTER. BOLTZMANN TRANSPORT The carrier density n typically is much smaller in semiconductors than in metals, ranging from n 0 6 cm 3 in intrinsic (i.e. undoped, thermally excited at room temperature) materials to n 0 9 cm 3 in doped materials. ɛ 0 20 and m /m e 0.. The product ɛ m thus differs only slightly from its free electron value. Since n semi < 0 4 n metal, one has ω semi p 0 2 ω metal p 0 4 s. (.65) In high purity semiconductors the mobility µ = eτ/m > 0 5 cm 2 /vs the low temperature scattering time is typically τ 0 s. Thus, for ω > s in the optical range, we have ωτ ω p τ, in which case N(ω) ɛ and the reflectivity is R = ɛ + ɛ 2. (.66) Taking ɛ = 0, one obtains R = 0.27, which is high enough so that polished Si wafers appear shiny..4.4 Optical Conductivity and the Fermi Surface At high frequencies, when ωτ, our expression for the conductivity, eqn. (.33), yields σ(ω) = ie2 2π 3 dε ( f 0 ) ds ε v(k), (.67) ω where we have presumed sufficient crystalline symmetry to guarantee that σ αβ = σ δ αβ is diagonal. In the isotropic case, and at temperatures low compared with T F, the integral over the Fermi surface gives 4πk 2 F v F = 2π 3 n/m, whence σ = ine 2 /m ω, which is the large frequency limit of our previous result. For a general Fermi surface, we can define σ(ω τ ) ine2 m opt ω (.68) where the optical mass m opt is given by m opt = 2π 3 n dε ( f 0 ) ds ε v(k). (.69) Note that at high frequencies σ(ω) is purely imaginary. What does this mean? If E(t) = E cos(ωt) = 2 E ( e iωt + e +iωt) (.70)

15 .5. CALCULATION OF THE SCATTERING TIME 5 then j(t) = 2 E { σ(ω) e iωt + σ( ω) e +iωt} = ne2 E sin(ωt), (.7) m opt ω where we have invoked σ( ω) = σ (ω). The current is therefore 90 out of phase with the voltage, and the average over a cycle j(t) E(t) = 0. Recall that we found metals to be transparent for ω ω p τ. At zero temperature, the optical mass is given by = m opt 2π 3 ds n F v(k). (.72) The density of states, g(ε F ), is g(ε F ) = 4π 3 ds F v(k), (.73) from which one can define the thermodynamic effective mass m th, appealing to the low temperature form of the specific heat, where c V = π2 3 k2 B T g(ε F ) m th m e c 0 V, (.74) c 0 V m e k 2 B T 3 2 (3π 2 n) /3 (.75) is the specific heat for a free electron gas of density n. Thus, m th = 4π(3π 2 n) /3 ds F v(k) (.76).5 Calculation of the Scattering Time.5. Potential Scattering and Fermi s Golden Rule Let us go beyond the relaxation time approximation and calculate the scattering time τ from first principles. We will concern ourselves with scattering of electrons from crystalline impurities. We begin with Fermi s Golden Rule 6, I k {f} = 2π k U k 2 (fk f k ) δ(ε(k) ε(k )), (.77) k 6 We ll treat the scattering of each spin species separately. We assume no spin-flip scattering takes place.

16 6 CHAPTER. BOLTZMANN TRANSPORT m opt/m e m th /m e Metal thy expt thy expt Li Na K Rb Cs Cu Ag Au Table.: Optical and thermodynamic effective masses of monovalent metals. (Taken from Smith and Jensen). where U(r) is a sum over individual impurity ion potentials, U(r) = N imp j= k U k 2 = V 2 Û(k k ) 2 U(r R j ) (.78) N imp j= e i(k k ) R 2 j, (.79) where V is the volume of the solid and Û(q) = d 3 r U(r) e iq r (.80) is the Fourier transform of the impurity potential. Note that we are assuming a single species of impurities; the method can be generalized to account for different impurity species. To make progress, we assume the impurity positions are random and uncorrelated, and we average over them. Using N imp j= e iq R j 2 = N imp + N imp (N imp ) δ q,0, (.8) we obtain k U k 2 = N imp V 2 Û(k k ) 2 + N imp(n imp ) V 2 Û(0) 2 δ kk. (.82) EXERCISE: Verify eqn. (.8). We will neglect the second term in eqn..82 arising from the spatial average (q = 0 Fourier component) of the potential. As we will see, in the end it will cancel out. Writing

17 .5. CALCULATION OF THE SCATTERING TIME 7 f = f 0 + δf, we have I k {f} = 2πn imp ( ) d 3 k (2π) 3 Û(k k ) 2 2 k 2 δ 2m 2 k 2 2m (δf k δf k ), (.83) where n imp = N imp /V is the number density of impurities. Note that we are assuming a parabolic band. We next make the Ansatz δf k = τ(ε(k)) e E v(k) f 0 (.84) ε(k) and solve for τ(ε(k)). The (time-independent) Boltzmann equation is e E v(k) f 0 = 2π ( ) n imp e E d 3 k (2π) 3 Û(k k ) 2 2 k 2 δ 2m 2 k 2 2m ( τ(ε(k )) v(k ) f 0 ε(k ) τ(ε(k)) v(k) f 0 ε(k) ). (.85) Due to the isotropy of the problem, we must have τ(ε(k)) is a function only of the magnitude of k. We then obtain 7 whence k m = n imp 4π 2 τ(ε(k)) 0 dk k 2 τ(ε F ) = m k F n imp 4π 2 3 dˆk Û(k k ) 2 δ(k k ) 2 k/m m (k k ), (.86) dˆk U(k Fˆk kfˆk ) 2 ( ˆk ˆk ). (.87) If the impurity potential U(r) itself is isotropic, then its Fourier transform Û(q) is a function of q 2 = 4k 2 F sin 2 2 ϑ where cos ϑ = ˆk ˆk and q = k k is the transfer wavevector. Recalling the Born approximation for differential scattering cross section, we may finally write ( ) m 2 σ(ϑ) = 2π 2 Û(k k ) 2, (.88) τ(ε F ) = 2πn imp v F π 0 dϑ σ F (ϑ) ( cos ϑ) sin ϑ (.89) where v F = k F /m is the Fermi velocity 8. The mean free path is defined by l = v F τ. 7 We assume that the Fermi surface is contained within the first Brillouin zone. 8 The subscript on σ F (ϑ) is to remind us that the cross section depends on k F as well as ϑ.

18 8 CHAPTER. BOLTZMANN TRANSPORT Notice the factor ( cos ϑ) in the integrand of (.89). This tells us that forward scattering (ϑ = 0) doesn t contribute to the scattering rate, which justifies our neglect of the second term in eqn. (.82). Why should τ be utterly insensitive to forward scattering? Because τ(ε F ) is the transport lifetime, and forward scattering does not degrade the current. Therefore, σ(ϑ = 0) does not contribute to the transport scattering rate τ (ε F ). Oftentimes one sees reference in the literature to a single particle lifetime as well, which is given by the same expression but without this factor: { } τ sp τtr = 2πn imp v F π 0 { dϑ σ F (ϑ) ( cos ϑ) } sin ϑ (.90) Note that τ sp = (n imp v F σ F,tot ), where σ F,tot is the total scattering cross section at energy ε F, a formula familiar from elementary kinetic theory. To derive the single particle lifetime, one can examine the linearized time-dependent Boltzmann equation with E = 0, δf k = n t imp v F dˆk σ(ϑ kk ) (δf k δf k ), (.9) where v = k/m is the velocity, and where the kernel is ϑ kk = cos (k k ). We now expand in spherical harmonics, writing σ(ϑ kk ) σ tot ν L Y LM (ˆk) YLM(ˆk ), (.92) where as before Expanding σ tot = 2π L,M the linearized Boltzmann equation simplifies to A LM t whence one obtains a hierarchy of relaxation rates, π 0 dϑ sin ϑ σ(ϑ). (.93) δf k (t) = L,M A LM (t) Y LM (ˆk), (.94) + ( ν L ) n imp v F σ tot A LM = 0, (.95) τ L = ( ν L ) n imp v F σ tot, (.96) which depend only on the total angular momentum quantum number L. These rates describe the relaxation of nonuniform distributions, when δf k (t = 0) is proportional to some spherical harmonic Y LM (k). Note that τ L=0 = 0, which reflects the fact that the total particle number is a collisional invariant. The single particle lifetime is identified as τ sp τ L = n imp v F σ tot, (.97) corresponding to a point distortion of the uniform distribution.

19 .5. CALCULATION OF THE SCATTERING TIME Screening and the Transport Lifetime For a Coulomb impurity, with U(r) = Ze 2 /r we have Û(q) = 4πZe2 /q 2. Consequently, ( ) 2 Ze 2 σ F (ϑ) = 4ε F sin 2 2 ϑ, (.98) and there is a strong divergence as ϑ 0, with σ F (ϑ) ϑ 4. diverges logarithmically! What went wrong? The transport lifetime What went wrong is that we have failed to account for screening. Free charges will rearrange themselves so as to screen an impurity potential. At long range, the effective (screened) potential decays exponentally, rather than as /r. The screened potential is of the Yukawa form, and its increase at low q is cut off on the scale of the inverse screening length λ. There are two types of screening to consider: Thomas-Fermi Screening : This is the typical screening mechanism in metals. A weak local electrostatic potential φ(r) will induce a change in the local electronic density according to δn(r) = eφ(r)g(ε F ), where g(ε F ) is the density of states at the Fermi level. This charge imbalance is again related to φ(r) through the Poisson equation. The result is a self-consistent equation for φ(r), 2 φ = 4πe δn = 4πe 2 g(ε F ) φ λ 2 TF φ. (.99) The Thomas-Fermi screening length is λ TF = ( 4πe 2 g(ε F ) ) /2. Debye-Hückel Screening : This mechanism is typical of ionic solutions, although it may also be of relevance in solids with ultra-low Fermi energies. From classical statistical mechanics, the local variation in electron number density induced by a potential φ(r) is δn(r) = n e eφ(r)/k B T n neφ(r), (.00) k B T where we assume the potential is weak on the scale of k B T/e. Poisson s equation now gives us 2 φ = 4πe δn = 4πne2 k B T A screened test charge Ze at the origin obeys the solution of which is φ λ 2 DH φ. (.0) 2 φ = λ 2 φ 4πZeδ(r), (.02) U(r) = eφ(r) = Ze2 r e r/λ = 4πZe2 Û(q) = q 2. (.03) + λ 2

20 20 CHAPTER. BOLTZMANN TRANSPORT The differential scattering cross section is now σ F (ϑ) = ( Ze 2 4ε F sin 2 2 ϑ + (2k Fλ) 2 ) 2 (.04) and the divergence at small angle is cut off. The transport lifetime for screened Coulomb scattering is therefore given by with τ(ε F ) = 2πn imp v F ( Ze 2 4ε F = 2πn imp v F ( Ze 2 2ε F ) 2 π 0 ( ) 2 dϑ sin ϑ ( cos ϑ) sin 2 2 ϑ + (2k Fλ) 2 πζ }, (.05) + πζ ) 2 { ln( + πζ) ζ = 4 π k2 Fλ 2 = 2 k F m e 2 = k Fa B. (.06) Here a B = ɛ 2 /m e 2 is the effective Bohr radius (restoring the ɛ factor). The resistivity is therefore given by where ρ = m ne 2 τ = Z2 h e 2 a B F (ζ) = { ζ 3 ln( + πζ) πζ } + πζ n imp n F (k F a B), (.07). (.08) With h/e 2 = 25, 83 Ω and a B a B = Å, we have ρ = Ω cm Z 2 n imp n F (k F a B). (.09) Impurity ρ per % Impurity ρ per % Ion (µω-cm) Ion (µω-cm) Be 0.64 Si 3.2 Mg 0.60 Ge 3.7 B.4 Sn 2.8 Al.2 As 6.5 In.2 Sb 5.4 Table.2: Residual resistivity of copper per percent impurity.

21 .6. BOLTZMANN EQUATION FOR HOLES 2 Figure.5: Residual resistivity per percent impurity..6 Boltzmann Equation for Holes.6. Properties of Holes Since filled bands carry no current, we have that the current density from band n is j n (r, t) = 2e d 3 k (2π) 3 f n(r, k, t) v n (k) = +2e d 3 k (2π) 3 f n (r, k, t) v n (k), (.0) where f f. Thus, we can regard the current to be carried by fictitious particles of charge +e with a distribution f(r, k, t). These fictitious particles are called holes.. Under the influence of an applied electromagnetic field, the unoccupied levels of a band evolve as if they were occupied by real electrons of charge e. That is, whether or not a state is occupied is irrelevant to the time evolution of that state, which is described by the semiclassical dynamics of eqs. (.,.2). 2. The current density due to a hole of wavevector k is +e v n (k)/v. 3. The crystal momentum of a hole of wavevector k is P = k. 4. Any band can be described in terms of electrons or in terms of holes, but not both simultaneously. A mixed description is redundant at best, wrong at worst, and confusing always. However, it is often convenient to treat some bands within the electron picture and others within the hole picture.

22 22 CHAPTER. BOLTZMANN TRANSPORT Figure.6: Two states: ΨA = e k h k 0 and ΨB = e more current? What is the crystal momentum of each state? k h k 0. Which state carries It is instructive to consider the exercise of fig..6. The two states to be analyzed are Ψ A = ψ c,k ψ v,k Ψ 0 = e k h k 0 (.) Ψ B = ψ Ψ 0 = e 0, (.2) c,k ψ v, k k h k where e k ψ c,k is the creation operator for electrons in the conduction band, and h k ψ v,k is the creation operator for holes (and hence the destruction operator for electrons) in the valence band. The state Ψ 0 has all states below the top of the valence band filled, and all states above the bottom of the conduction band empty. The state 0 is the same state, but represented now as a vacuum for conduction electrons and valence holes. The current density in each state is given by j = e(v h v e )/V, where V is the volume (i.e. length) of the system. The dispersions resemble ε c,v ± 2 E g ± 2 k 2 /2m, where E g is the energy gap. State Ψ A : The electron velocity is v e = k/m ; the hole velocity is v h = k/m. Hence, the total current density is j 2e k/m V and the total crystal momentum is P = p e + p h = k k = 0. State ΨB : The electron velocity is v e = k/m ; the hole velocity is v h = ( k)/m. The

23 .6. BOLTZMANN EQUATION FOR HOLES 23 total current density is j 0, and the total crystal momentum is P = p e + p h = k ( k) = 2 k. Consider next the dynamics of electrons near the bottom of the conduction band and holes near the top of the valence band. (We ll assume a direct gap, i.e. the conduction band minimum is located directly above the valence band maximum, which we take to be at the Brillouin zone center k = 0, otherwise known as the Γ point.) Expanding the dispersions about their extrema, ε v (k) = ε v m v αβ k α k β (.3) ε c (k) = ε c m c αβ k α k β. (.4) The velocity is v α (k) = = ± m kα αβ kβ, (.5) where the + sign is used in conjunction with m c and the sign with m v. We compute the acceleration a = r via the chain rule, a α = vα k β dkβ dt = e m αβ [E β + c ] (v B)β F α = m αβ a β = e [E β + c ] (v B)β (.6). (.7) Thus, the hole wavepacket accelerates as if it has charge +e but a positive effective mass. Finally, what form does the Boltzmann equation take for holes? Starting with the Boltzmann equation for electrons, f t + ṙ f r + k f k = I k{f}, (.8) we recast this in terms of the hole distribution f = f, and obtain f t + ṙ f r + k f k = I k{ f} (.9) This then is the Boltzmann equation for the hole distribution f. Recall that we can expand the collision integral functional as I k {f 0 + δf} = L δf +... (.20) where L is a linear operator, and the higher order terms are formally of order (δf) 2. Note that the zeroth order term I k {f 0 } vanishes due to the fact that f 0 represents a local equilibrium. Thus, writing f = f 0 + δ f I k { f} = I k { f 0 δ f} = L δ f +... (.2)

24 24 CHAPTER. BOLTZMANN TRANSPORT and the linearized collisionless Boltzmann equation for holes is δ f t e c v B δ f [ k v ee + ε µ ] T T f 0 = L δ f (.22) which is of precisely the same form as the electron case in eqn. (.27). Note that the local equilibrium distribution for holes is given by f 0 (r, k, t) = { exp ( µ(r, t) ε(k) k B T (r, t) ) + } (.23).7 Magnetoresistance and Hall Effect.7. Boltzmann Theory for ρ αβ (ω, B) In the presence of an external magnetic field B, the linearized Boltzmann equation takes the form 9 δf t e v E f 0 e c v B δf = L δf. (.24) k We will obtain an explicit solution within the relaxation time approximation L δf = δf/τ and the effective mass approximation, ε(k) = ± 2 2 m αβ kα k β = v α = ± m αβ kβ, (.25) where the top sign applies for electrons and the bottom sign for holes. With E(t) = E e iωt, we try a solution of the form δf(k, t) = k A(ε) e iωt δf(k) e iωt (.26) where A(ε) is a vector function of ε to be determined. Each component A α is a function of k through its dependence on ε = ε(k). We now have (τ iω) k µ A µ e c ɛ αβγ vα B β k γ (kµ A µ ) = e v E f 0, (.27) where ɛ αβγ is the Levi-Civita tensor. Note that ɛ αβγ v α B β 9 For holes, we replace f 0 f 0 and δf δ f. ) k γ (kµ A µ ) = ɛ αβγ v α B (A β γ + k µ Aµ k ( γ ) = ɛ αβγ v α B β A γ + k µ v γ Aµ = ɛ αβγ v α B β A γ, (.28)

25 .7. MAGNETORESISTANCE AND HALL EFFECT 25 owing to the asymmetry of the Levi-Civita tensor: ɛ αβγ v α v γ = 0. Now invoke k α = ±m αβ v β, and match the coefficients of v α in each term of the Boltzmann equation. This yields, [(τ iω) m αβ ± e c ɛ αβγ Bγ] A β = ± e f 0 Eα. (.29) Defining Γ αβ (τ iω) m αβ ± e c ɛ αβγ Bγ, (.30) we obtain the solution δf = e v α m αβ Γ f 0 βγ Eγ. (.3) From this, we can compute the current density and the conductivity tensor. The electrical current density is j α = 2e = +2e 2 E γ d 3 k (2π) 3 vα δf d 3 ( k (2π) 3 vα v ν m νβ Γ βγ (ε) f 0 ), (.32) where we allow for an energy-dependent relaxation time τ(ε). Note that Γ αβ (ε) is energydependent due to its dependence on τ. The conductivity is then { σ αβ (ω, B) = 2 2 e 2 m d 3 k αµ (2π) 3 kµ k ( ν f 0 ) } Γ νβ (ε) (.33) = ± 2 3 e2 ( dε ε g(ε) Γ αβ (ε) f 0 ), (.34) where the chemical potential is measured with respect to the band edge. Thus, where averages denoted by angular brackets are defined by Γ αβ σ αβ (ω, B) = ne 2 Γ αβ, (.35) ( dε ε g(ε) f 0 ( dε ε g(ε) ) f 0 Γ αβ (ε) ). (.36) The quantity n is the carrier density, n = dε g(ε) { f 0 (ε) (electrons) { f 0 (ε) } (holes) (.37)

26 26 CHAPTER. BOLTZMANN TRANSPORT EXERCISE: Verify eqn. (.34). For the sake of simplicity, let us assume an energy-independent scattering time, or that the temperature is sufficiently low that only τ(ε F ) matters, and we denote this scattering time simply by τ. Putting this all together, then, we obtain σ αβ = ne 2 Γ αβ (.38) ρ αβ = ne 2 Γ αβ = [ ne 2 (τ iω)m αβ ± e c ɛ αβγ Bγ]. (.39) We will assume that B is directed along one of the principal axes of the effective mass tensor m αβ, which we define to be ˆx, ŷ, and ẑ, in which case ρ αβ (ω, B) = (τ iω) m x ±eb/c 0 ne 2 eb/c (τ iω) m y 0 (.40) 0 0 (τ iω) m z where m x,y,z are the eigenvalues of m αβ and B lies along the eigenvector ẑ. Note that ρ xx (ω, B) = m x ne 2 ( iωτ) (.4) τ is independent of B. Hence, the magnetoresistance, ρ xx (B) = ρ xx (B) ρ xx (0) (.42) vanishes: ρ xx (B) = 0. While this is true for a single parabolic band, deviations from parabolicity and contributions from other bands can lead to a nonzero magnetoresistance. The conductivity tensor σ αβ is the matrix inverse of ρ αβ. Using the familiar equality ( ) a b = c d ad bc ( ) d b, (.43) c a we obtain ( iωτ)/m x ( iωτ) 2 +(ω c τ) 2 σ αβ (ω, B) = ne 2 τ ± ω cτ/ m x m y ( iωτ) 2 +(ω c τ) 2 ω cτ/ m x m y ( iωτ) 2 +(ω c τ) 2 0 ( iωτ)/m y ( iωτ) 2 +(ω c τ) ( iωτ)m z (.44) where ω c eb m c, (.45)

27 .7. MAGNETORESISTANCE AND HALL EFFECT 27 with m m x m y, is the cyclotron frequency. Thus, σ xx (ω, B) = ne2 τ m x σ zz (ω, B) = ne2 τ m z iωτ + (ω 2 c ω 2 )τ 2 2iωτ (.46) iωτ. (.47) Note that σ xx,yy are field-dependent, unlike the corresponding components of the resistivity tensor..7.2 Cyclotron Resonance in Semiconductors A typical value for the effective mass in semiconductors is m 0. m e. From e m e c = Hz/G, (.48) we find that eb/m c =.75 0 Hz in a field of B = kg. In metals, the disorder is such that even at low temperatures ω c τ typically is small. In semiconductors, however, the smallness of m and the relatively high purity (sometimes spectacularly so) mean that ω c τ can get as large as 0 3 at modest fields. This allows for a measurement of the effective mass tensor using the technique of cyclotron resonance. The absorption of electromagnetic radiation is proportional to the dissipative (i.e. real) part of the diagonal elements of σ αβ (ω), which is given by σ xx(ω, B) = ne2 τ m x + (λ 2 + )s 2 + 2(λ 2 + )s 2 + (λ 2 ) 2 s 4, (.49) where λ = B/B ω, with B ω = m c ω/e, and s = ωτ. For fixed ω, the conductivity σ xx(b) is then peaked at B = B. When ωτ and ω c τ, B approaches B ω, where σ xx(ω, B ω ) = ne 2 τ/2m x. By measuring B ω one can extract the quantity m = eb ω/ωc. Varying the direction of the magnetic field, the entire effective mass tensor may be determined. For finite ωτ, we can differentiate the above expression to obtain the location of the cyclotron resonance peak. One finds B = ( + α) /2 B ω, with α = (2s2 + ) + (2s 2 + ) 2 s 2 (.50) = 4s 4 + 8s 6 + O(s 8 ). As depicted in fig..7, the resonance peak shifts to the left of B ω for finite values of ωτ. The peak collapses to B = 0 when ωτ / 3 =

28 28 CHAPTER. BOLTZMANN TRANSPORT Figure.7: Theoretical cyclotron resonance peaks as a function of B/B ω for different values of ωτ..7.3 Magnetoresistance: Two-Band Model For a semiconductor with both electrons and holes present a situation not uncommon to metals either (e.g. Aluminum) each band contributes to the conductivity. The individual band conductivities are additive because the electron and hole conduction processes occur in parallel, exactly as we would deduce from eqn. (.4). Thus, σ αβ (ω) = n σ (n) αβ (ω), (.5) where σ (n) αβ is the conductivity tensor for band n, which may be computed in either the electron or hole picture (whichever is more convenient). We assume here that the two distributions δf c and δ f v evolve according to independent linearized Boltzmann equations, i.e. there is no interband scattering to account for. The resistivity tensor of each band, ρ (n) αβ exhibits no magnetoresistance, as we have found. However, if two bands are present, the total resistivity tensor ρ is obtained from ρ = ρ c + ρ v, and ρ = ( ρ c will in general exhibit the phenomenon of magnetoresistance. + ρ ) v (.52) Explicitly, then, let us consider a model with isotropic and nondegenerate conduction band

29 .7. MAGNETORESISTANCE AND HALL EFFECT 29 minimum and valence band maximum. Taking B = Bẑ, we have ρ c = ( iωτ c)m c I + B 0 0 α c β c = β c α c 0 (.53) n c e 2 τ c n c ec α c where ρ v = ( iωτ v)m v I B = n v e 2 τ v n v ec α v β v 0 β v α v α v α c = ( iωτ c)m c n c e 2 τ c β c = B n c ec, (.54) (.55) α v = ( iωτ v)m v n v e 2 τ v β v = B n v ec, (.56) we obtain for the upper left 2 2 block of ρ: [ ( αv ρ = αv 2 + βv 2 + α ) 2 ( c βv αc 2 + βc 2 + αv 2 + βv 2 + β ) ] 2 c αc 2 + βc 2 α v + αc α 2 v +β2 v α 2 c +β2 c βv α 2 v+βv 2 βc α 2 c+β 2 c from which we compute the magnetoresistance β v + βc α 2 v +β2 v α 2 c +β2 c α v + αc α 2 v+βv 2 α 2 c+βc 2, (.57) where ρ xx (B) ρ xx (0) ρ xx (0) = ( ) 2 σ c σ σc v σ v B 2 n c ec n v ec (σ c + σ v ) 2 + (σ c σ v ) 2 ( n c ec + n v ec σ c = α c = n ce 2 τ c m c ) 2 B 2 (.58) (.59) iωτ c σ v = α v = n ve 2 τ v m v. (.60) iωτ v Note that the magnetoresistance is positive within the two band model, and that it saturates in the high field limit: ( ) 2 ρ xx (B ) ρ xx (0) σ c σ σc v σ v n = c ec n v ec ( ) 2. (.6) ρ xx (0) (σ c σ v ) 2 + n c ec n v ec

30 30 CHAPTER. BOLTZMANN TRANSPORT The longitudinal resistivity is found to be and is independent of B. ρ zz = (σ c + σ v ) (.62) In an intrinsic semiconductor, n c = n v exp( E g /2k B T ), and ρ xx (B)/ρ xx (0) is finite even as T 0. In the extrinsic (i.e. doped) case, one of the densities (say, n c in a p-type material) vanishes much more rapidly than the other, and the magnetoresistance vanishes with the ratio n c /n v..7.4 Hall Effect in High Fields In the high field limit, one may neglect the collision integral entirely, and write (at ω = 0) e v E f 0 e c v B δf dk = 0. (.63) We ll consider the case of electrons, and take E = Eŷ and B = Bẑ, in which case the solution is δf = ce B k f 0 x. (.64) Note that k x is not a smooth single-valued function over the Brillouin-zone due to Bloch periodicity. This treatment, then, will make sense only if the derivative f 0 / confines k to a closed orbit within the first Brillouin zone. In this case, we have j x = 2ec E d 3 k B (2π) 3 k f 0 x (.65) k x = 2ec E B d 3 k (2π) 3 k f 0 x. (.66) k x Now we may integrate by parts, if we assume that f 0 vanishes on the boundary of the Brillouin zone. We obtain j x = 2ecE d 3 k B (2π) 3 f 0 = nec B E. (.67) We conclude that σ xy = σ yx = nec B, (.68) independent of the details of the band structure. Open orbits trajectories along Fermi surfaces which cross Brillouin zone boundaries and return in another zone post a subtler problem, and generally lead to a finite, non-saturating magnetoresistance. For holes, we have f 0 = f 0 and j x = 2ecE B d 3 k (2π) 3 k x f 0 k x = + nec B E (.69)

31 .7. MAGNETORESISTANCE AND HALL EFFECT 3 Figure.8: Nobel Prize winning magnetotransport data in a clean two-dimensional electron gas at a GaAs-AlGaAs inversion layer, from D. C. Tsui, H. L. Störmer, and A. C. Gossard, Phys. Rev. Lett. 48, 559 (982). ρ xy and ρ xx are shown versus magnetic field for a set of four temperatures. The Landau level filling factor is ν = nhc/eb. At T = 4.2 K, the Hall resistivity obeys ρ xy = B/nec (n =.3 0 cm 2 ). At lower temperatures, quantized plateaus appear in ρ xy (B) in units of h/e 2. and σ xy = +nec/b, where n is the hole density. We define the Hall coefficient R H = ρ xy /B and the Hall number z H, (.70) n ion ecr H where n ion is the ion density. For high fields, the off-diagonal elements of both ρ αβ and σ αβ are negligible, and ρ xy = σ xy. Hence R H = /nec, and z H = ±n/n ion. The high field Hall coefficient is used to determine both the carrier density as well as the sign of the charge carriers; z H is a measure of valency. In Al, the high field Hall coefficient saturates at z H =. Why is z H negative? As it turns out, aluminum has both electron and hole bands. Its valence is 3; two electrons go into a filled band, leaving one valence electron to split between the electron and hole bands. Thus

32 32 CHAPTER. BOLTZMANN TRANSPORT Figure.9: Energy bands in aluminum. n = 3n ion The Hall conductivity is σ xy = (n h n e ) ec/b. (.7) The difference n h n e is determined by the following argument. The electron density in the hole band is n e = 2n ion n h, i.e. the total density of levels in the band (two states per unit cell) minus the number of empty levels in which there are holes. Thus, n h n e = 2n ion (n e + n e) = n ion, (.72) where we ve invoked n e + n e = n ion, since precisely one electron from each ion is shared between the two partially filled bands. Thus, σ xy = n ion ec/b = nec/3b and z H =. At lower fields, z H = +3 is observed, which is what one would expect from the free electron model. Interband scattering, which is suppressed at high fields, leads to this result..8 Thermal Transport.8. Boltzmann Theory Consider a small region of solid with a fixed volume V. The first law of thermodynamics applied to this region gives T S = E µ N. Dividing by V gives dq T ds = dε µ dn, (.73) where s is the entropy density, ε is energy density, and n the number density. This can be directly recast as the following relation among current densities: j q = T j s = j ε µ j n, (.74)

33 .8. THERMAL TRANSPORT 33 Figure.0: Fermi surfaces for electron (pink) and hole (gold) bands in Aluminum. where j n = j/( e) is the number current density, j ε is the energy current density, j ε = 2 d 3 k ε v δf, (.75) (2π) 3 and j s is the entropy current density. Accordingly, the thermal (heat) current density j q is defined as j q T j s = j ε + µ e j (.76) d 3 k = 2 (ε µ) v δf. (.77) (2π) 3 In the presence of a time-independent temperature gradient and electric field, linearized Boltzmann equation in the relaxation time approximation has the solution ( δf = τ(ε) v ee + ε µ ) T ( f 0 ). (.78) T We now consider both the electrical current j as well as the thermal current density j q. One readily obtains j = j q = 2 2 e d 3 k (2π) 3 v δf L E L 2 T (.79) d 3 k (2π) 3 (ε µ) v δf L 2 E L 22 T (.80)

34 34 CHAPTER. BOLTZMANN TRANSPORT where the transport coefficients L etc. are matrices: L αβ = e2 4π 3 dε τ(ε) ( f 0 ) v α v β ds ε v L αβ 2 = T Lαβ 2 = e 4π 3 dε τ(ε) (ε µ) ( f 0 ) ( L αβ 22 = 4π 3 dε τ(ε) (ε µ) 2 f 0 ) T If we define the hierarchy of integral expressions v α v β ds ε v ds ε v α v β v Jn αβ ( 4π 3 dε τ(ε) (ε µ) n f 0 ) v α v β ds ε v (.8) (.82). (.83) (.84) then we may write L αβ = e2 J αβ 0 L αβ 2 = T Lαβ αβ 2 = e J L αβ 22 = T J αβ 2. (.85) The linear relations in eqn. (.80) may be recast in the following form: where the matrices ρ, Q,, and κ are given by or, in terms of the J n, E = ρ j + Q T (.86) j q = j κ T, (.87) ρ = L Q = L L 2 (.88) = L 2 L κ = L 22 L 2 L L 2, (.89) ρ = e 2 J 0 Q = e T J = e J J 0 κ = T 0 J (.90) ( J 2 J J 0 J ), (.9) The names and physical interpretation of these four transport coefficients is as follows: ρ is the resistivity: E = ρj under the condition of zero thermal gradient (i.e. T = 0). Q is the thermopower: E = Q T under the condition of zero electrical current (i.e. j = 0). Q is also called the Seebeck coefficient. is the Peltier coefficient: j q = j when T = 0. κ is the thermal conductivity: j q = κ T when j = 0.

35 .8. THERMAL TRANSPORT 35 Figure.: A thermocouple is a junction formed of two dissimilar metals. With no electrical current passing, an electric field is generated in the presence of a temperature gradient, resulting in a voltage V = V A V B. One practical way to measure the thermopower is to form a junction between two dissimilar metals, A and B. The junction is held at temperature T and the other ends of the metals are held at temperature T 0. One then measures a voltage difference between the free ends of the metals this is known as the Seebeck effect. Integrating the electric field from the free end of A to the free end of B gives V A V B = B A E dl = (Q B Q A )(T T 0 ). (.92) What one measures here is really the difference in thermopowers of the two metals. For an absolute measurement of Q A, replace B by a superconductor (Q = 0 for a superconductor). A device which converts a temperature gradient into an emf is known as a thermocouple. The Peltier effect has practical applications in refrigeration technology. Suppose an electrical current I is passed through a junction between two dissimilar metals, A and B. Due to the difference in Peltier coefficients, there will be a net heat current into the junction of W = ( A B ) I. Note that this is proportional to I, rather than the familiar I 2 result from Joule heating. The sign of W depends on the direction of the current. If a second junction is added, to make an ABA configuration, then heat absorbed at the first junction will be liberated at the second. 0 0 To create a refrigerator, stick the cold junction inside a thermally insulated box and the hot junction outside the box.

36 36 CHAPTER. BOLTZMANN TRANSPORT Figure.2: A sketch of a Peltier effect refrigerator. An electrical current I is passed through a junction between two dissimilar metals. If the dotted line represents the boundary of a thermally well-insulated body, then the body cools when B > A, in order to maintain a heat current balance at the junction..8.2 The Heat Equation We begin with the continuity equations for charge density ρ and energy density ε: ρ t + j = 0 (.93) t + j ε = j E, (.94) where E is the electric field. Now we invoke local thermodynamic equilibrium and write t = n n t + T = µ e ρ t + c V where n is the electron number density (n = ρ/e) and c V now write c V T t = t + µ e ρ t T t = j E j ε µ e j T t, (.95) is the specific heat. We may = j E j q. (.96) Note that it is E j and not E j which is the source term in the energy continuity equation.