Solid State Theory Solutions Sheet 11. Exercise 1. FS 2014 Prof. Manfred Sigrist Relaxation time approximation In this exercise we will show that the so-called single-relaxation-time approximation, Z ∂f (k) f (k) − f0 (k) dd k 0 W (k, k0 )[f (k) − f (k0 )] −→ − =− , d ∂t (2π) τ coll (1) is a true solution to the Boltzmann equation under certain conditions. We consider a spatially homogeneous two-dimensional metal with an isotropic Fermi surface (εk = ~2 k 2 /2m) at zero temperature. The impurity scattering responsible for a finite resistivity is described by a delta potential in real space, Vimp (r) = V0 δ(r). (2) The system is subject to a homogeneous and time-independent electric field along the x-axis. a) Show that the transition rates W (k, k0 ) for the impurity potential (2) are constant and non-zero only for scattering events conserving the energy of the incoming state. b) Write down the static Boltzmann transport equation for this setup in the form “drift-term” = “collision-integral” (3) and take advantage of the zero-temperature limit and the symmetries of the system to eliminate all but angular variables. c) In a case with only angular dependence, it turns out to be useful to expand the drift term ∇k f ·(eE) and δf = f − f0 in Fourier modes X X δf = fl eilϕ , ∇k f · (eE) = dl eilϕ . (4) l l Rewrite the Boltzmann equation as a set of algebraic equations for the coefficients in the expansion (4) X dm = Lm,n fn . (5) n d) What are the eigenvalues of the so-called collision operator Lm,n and what is their meaning? How can one interpret vanishing eigenvalues? e) Find a solution to equation (5) and compare δf to the single-relaxation-time approximation, equation (1). Solution. a) The transition rates are computed by using Fermi’s golden rule, relating the probability per unit of time for an initial state labeled by k to scatter to some final state labeled by k0 due to a collision with an impurity (cf. section 6.3 of the lecture notes). W (k, k0 ) = 2π Nimp 0 |hk |Vimp |ki|2 δ(εk − εk0 ), ~ 1 (S.1) where Nimp denotes the total number of impurities in the system. The matrix elements hk0 |V |ki are computed using the explicit form of the eigenstates for the free electron Hamiltonian, i.e. Z 0 1 0 V (k − k ) = d2 r ei(k−k )·r V0 δ(r) Ω 1 (S.2) = V0 . Ω Where we used ψk (r) = rates are given as √1 eik·r , Ω with Ω being the area of our system. Thus, the transition W (k, k0 ) = 2πnimp 2 V0 δ(εk − εk0 ), | Ω~{z } (S.3) W0 with nimp = Nimp /Ω giving the density of impurities. Due to the isotropy of the energy εk ∼ k 2 they do not depend on k or k0 as long as the energy is conserved in the collision. b) In the static limit, the left hand side of the Boltzmann equation ∂f (r, k, t) dr eE + ∇r f (r, k, t) · + ∇k f (r, k, t) · ∂t dt ~ (S.4) reduces with f (r, k, t) = f (k) to the third term proportional to the driving external field. We restrict our considerations to linear order in E since we are interested in linear response to the driving force. In consequence, we must replace ∇k f with ∇k f0 because δf is already of order O(E). This term represents the so-called drift term. The right hand side of the Boltzmann equation is the collision integral, which for impurity scattering acquires the form (cf. Eq. (6.22) in the lecture notes) Z ∂f (k) Ω =− d2 kW (k, k0 )(f (k) − f (k0 )) 2 ∂t (2π) coll Z Ω 0 0 2 0 ) f0 (k) + δf (k) − f0 (k ) − δf (k ) =− d kW δ(ε − ε 0 k k (2π)2 Z Ω m =− dθ dk kW0 2 δ(k − k 0 ) f0 (k) + δf (k) − f0 (k0 ) − δf (k0 ) 2 (2π) ~ k Z Ω m =− dθW0 [δf (θ) − δf (θ0 )]. (S.5) (2π)2 ~2 Taking (S.4) and (S.5) together we arrive at Z ∼ eE ∇k f0 · = − dθ0 W 0 [δf (θ) − δf (θ0 )], ~ (S.6) ∼ where W 0 = mΩW0 /(2π~)2 . c) We start with the Boltzmann equation, X dl eilθ = − Z ∼ dθ0 W 0 l ! X l 2 fl eilθ − X l ilθ0 fl e , (S.7) which we project onto the m-th Fourier mode by multiplying on both sides with exp(−imθ) and integrating over θ ! Z Z Z X X X ∼ 0 dθe−imθ dl eilθ = − dθe−imθ dθ0 W 0 fl eilθ − fl eilθ . (S.8) l l l This equation can be easily evaluated by using the relation Z dθeinθ = 2πδn,0 , (S.9) and leaves us with ∼ 2πdm = −W 0 XZ dθdθ 0 e −imθ ilθ e −imθ ilθ0 −e e fl l ∼ = −(2π)2 W 0 X (δm,l − δm,0 δl,0 )fl . (S.10) l With the expression (S.10) we have reduced the problem of finding a solution to the Boltzmann equation to solving a system of linear equations. In general, this system of equations may be infinitely large and quite often only manageable numerically. In our case, relating the equation (S.10) to the matrix equation on the exercise sheet, we find that the matrix L has the following diagonal form   .. .     1     . .   .     1 ∼   .  0 (S.11) Lm,n = −2π W 0     1     ..   .     1   .. . d) We have reduced the problem to a simple matrix equation dl = Ll,m fm , where summation over repeated indices is implied. In our case, since the scattering rates are constant, and in particular independent of the scattering angle, the matrix turns out to be diagonal. The eigenvectors of L represent the Fourier decomposition of eigenfunctions of the collision operator and the eigenvalues represent inverse relaxation times. A vanishing eigenvalue represents and infinite relaxation time, so that the corresponding eigenfunction defines a conserved quantity. Any quantity that cannot be changed due to collisions will be preserved. In our case, we have only one conserved quantity left, because the energy, that can also not be changed within collisions has explicitly been projected out. This conserved quantity, as can be seen from (S.11), corresponds to the zeroth Fourier component. This constant shift of δf corresponds to a change in the particle number of the system and can of course not be changed during collisions. e) Since we know the exact form of the drift term, we can easily find its decomposition in Fourier modes. The drift-term to linear order in E is defined as ∇k f0 · 3 (eE) ~ (S.12) which at zero temperature reduces to ~ e~kE k · (eE) = −δ(εk − µ) cos θ, (S.13) m m where θ is the angle between the k and the electric field, which points into the x-axis. −δ(εk − µ) In the above analysis, we have integrated out the energy variables, so we can drop the delta function in (S.13) and express the Fourier decomposition as   − e~kE , l = ±1 X e~kE iθ −iθ ilθ 2m . (S.14) (e + e ) ⇒ dl = dl e = −  2m 0 , else l Because the matrix L is diagonal, it is straightforward to find a solution to the Boltzmann equation,  e~kE 1   , l = ±1 ∼ 2m 2π W . (S.15) fl = 0   0 , else Finally, by comparing the coefficients from the expansion of the drift term in (S.14) to those of the relaxation time approximation ∂f δf =− , (S.16) ∂t coll τ with the coeficients for δf from (S.15) we can identify the value of the relaxation time to be 1 τ= (S.17) ∼ . 2π W 0 Exercise 2. Residual Resistivity of Copper Recapitulate Section 6.3.1 and try to find an explanation for the data in Table 1. What is the major reason for the increase of the resisitivity? Hint: In analogy to doped semiconductors, assume that an impurity atom will adjust itself corresponding to its neighbourhood. This means that it rejects all electrons from its outer shells in order to match the occupancy of its surrounding atoms. Hence the impurity is left with an effective nuclear charge Z. The explanation of the above table lies in this fact. Solution. In the lecture notes it is shown that 1 Z2 ∝ N (F )nimp 4 τ kTF such that ρ∝ and ρ ∝ N (F ) nimp 2 Z . 4 n kTF 1 nτ (S.18) (S.19) The first fraction only depends on the Fermi surface which is hardly affected by the impurities and the second fraction does not change much because the doping increases the number of free charge carriers only slightly (very small impurity density). Therefore we can conclude that ρ ∝ Z 2. 4 (S.20) As it was mentioned in the hint we assume that the optimal electronic configuration of the impurity atom valence shell is identical with the valence shell of the surrounding atoms. In this spirit we define the effective charge of the impurity as the number of electrons which have to be removed from the impurity in order to have the same valence shell configuration as the surrounding atoms, because those electrons are fully delocalized and can therefore no longer screen the nucleus of the impurity atom (apart from Thomas-Fermi screening). A look at the periodic table gives us the electronic configuration, e.g. copper has the valence band configuration 4s1 . By calculating the difference of the valence electrons between copper and the impurity we find the following table: Impurity Be Mg B Al In Si Ge Sn As Sb Resistivity ρ/(10−8 Ωm) 0.64 0.6 1.4 1.2 1.2 3.2 3.7 2.8 6.5 5.4 Electronic configuration Eff. charge Z 2s2 3s2 2s2 2p1 3s2 3p1 5s2 5p1 3s2 3p2 4s2 4p2 5s2 5p2 4s2 4p3 5s2 5p3 1 1 2 2 2 3 3 3 4 4 Table 1: Electronic configuration and the effective charge of the impurities. Indeed, by comparing the values of the residual resistivity with the electronic configuration, we find a good correlation in agreement with scaling (S.20). Note that this agreement is only qualitative, not quantitative, as impurities with identical effective charges do not lead to precisely the same residual resistivity. 5