Go to Topics Autumn 010 Electrons in metals revised 0.1.010 PHYS08
Topics 0.1.010 Classical Models The Drude Theory of metals Conductivity - static electric field Thermal conductivity Fourier Law Wiedemann-Franz Law Lorenz number Quantum Models Fermi-Dirac distribution Fermi level quantities Fermi Gas Available quantum states - density of states Estimate of Fermi temperature Heat capacity of electrons in metals is small Sommerfeld method to evaluate Fermi function integrals Heat capacity of the Fermi electron gas Heat conductivity of Fermi Gas Lorenz number revisited An extra note on Fermi gas - not for presentation Two factor 100 errors in classical theory Alternative density derivations Click on the topics above Links to topics slide / 37
Go to Topics Classical part slide 3/ 37 Drude Theory of electrons in metals (1900) The Drude model is an application of classical ideal gas theory to electrons in a solid. Paul Drude (1863-1906) Enrico Fermi (1901-1954) Fermi gas model for electrons by others (late 190s)
Go to Topics Classical part slide 4/ 37 Basic assumptions of Drude s model Electrons move in static lattice of ions Free electron approximation (neglect ion-electron interaction) Independent electron approximation (neglect electron-electron interaction) Some scattering mechanism exists, erasing all information of the electron previous to the scattering event, giving each a random velocity determined by the temperatur at the given position. Probability of scattering during time interval t: t/τ, where τ is known as the relaxation time. Alternatively: the electrons move a distance l before colliding (mean free path).
Go to Topics Classical part slide 5/ 37 Conductivity - static electric field If the electrons in a metal are subjected to a static electric field, they will tend to drift in the opposite direction to the field, with a drift velocity determined by the field. Since an electron has approximately a time τ to be accelerated by the field before a collision occurs, it will gain a velocity v drift. This is added to all electrons own thermal velocity v e v e = 0 v e + v drift = v drift
Go to Topics Classical part slide 6/ 37 The drift velocity v drift given by v drift = τ 0 ee dt = eeτ m m. (1) This net displacement of electrons results in a nonzero current density j, which is charge density velocity = charge velocity () unit volume With N electrons of charge e in a volume V, and an average drift velocity v drift, we arrive at Ohm s law gives us: j = en V eeτ m = Ne τ E. (3) mv j = σe. (4) Where the proportionality constant is called the conductivity σ. Written in terms of electrons per unit volume n = N/V, the conductivity becomes σ = ne τ m (5)
o to Topics Classical part slide 7/ 37 Thermal conductivity Heat conduction caused by temperature gradient will induce a thermal current Fourier s law applies for most cases Generally: j u(x) v j = κ T. (6) u(x) - energy density (per volume), v - velocity of carriers: We derive thermal conductivity κ of metals for electron gas model We set the z-direction along the temperature gradient. Mean free path l - we assume that all really travel l An electron arriving at a point r will represent the temperature at its last collision site r + l, since T = T (z) this is u(t (z l cos θ) θ is the direction of l
Go to Topics Classical part slide 8/ 37 Thermal conductivity l 1 One direction u( r+ l 1 ) v1 u( r ) v 1 The current density due to volume i is j i = J in J out = ( v i ) z (u r+li u r ) (7) Where in represents the incomming current and out is the outgoing current, j i = ( v i ) z u, (8) 1 All directions 4 v 4 l l v 4 1 1 v v 3 v 3 u is the difference in energy density between two sites separated by l. The z-component of the velocity from volume i is ( ) li ( v i ) z = v = v cos θ i e z, (9) l z v l 3 l v 4 v 1 3 v is the average velocity of the carriers. v i points against l i
Go to Topics Classical part slide 9/ 37 Thermal conductivity u( r+ l 1 ) l 1 1 v1 One direction l 1 u( r ) v 1 All directions v 3 v1 v 4 l 4 v 4 ( j i = v cos θ i u( r + li ) u( r) ) e z (10) ( ) ( ) = v cos θ i u u li = v cos θi z l cos θ i e z = u z lv cos θ i e z To obtain the total current density we sum over i, j = i u z lv cos θ i e z. (11) v 3 v l 3 l v 4 v 1 3 The sum must be converted to integral The relative contribution - it must be averaged
Go to Topics Classical part slide 10/ 37 Instead of summing, we integrate over all solide angles dω = sin θdθdφ, 1 u j = e z dω z lv cos θdω = 1 4π e u z z lv cos θdω, (1) where Thus cos θdω = 4π 3 e z u z = u z u = u T T = c v T ( ) 1 u 1 j = e z dω z lv cos θdω = 3 c v vl T (13) Comparing this with Fouriers law: The thermal conductivity becomes j = κ T. (14) κ = 1 3 lvc v = 1 3 v τc v (15)
Wiedemann-Franz Law The Wiedemann-Franz law states that in a metal the ratio of thermal to electrical conductivity is proportional to the temperature, We have derived κ = 1 3 v τc v and σ = n e τ m κ σ T, (16) Ideal gas theory yields 1 mv = 3 k BT (i.e. v = 3 m k BT ) σ = N V c v = 3 N V k B κ = 1 3 3 m k BT 3 N V k Bτ κ = 3 T N V From these relations, the Drude model gives the Lorenz number as (relaxation time τ cancelled) κ σt = 3 e m τ k B m τ (17) ( ) kb. (18) e Go to Topics Classical part slide 11/ 37
Go to Topics Classical part slide 1/ 37 Wiedemann-Franz law - Lorenz number Calculated Lorenz number: κ σt = 3 ( kb e ) = 1.11 10 8 W Ω K, (19) which is about halv of the experimental values. Experimental Lorenz numbers: Drude published value. 10 8 W Ω K Figure: From Tony Harker, Chiranijib Mitra and Andrew Horsfield, Solid State Physics Notes
Go to Topics Quantum part slide 13/ 37 Fermi-Dirac distribution The distribution of electrons in the quantum gas is given by the Fermi- Dirac distribution, 1 e (ε µ)/k B T + 1 f(ε) = 1 e (ε µ)/k B T + 1 (0) Special case: T = 0. Then f(ε) = 1 for ε ε F, zero elsewhere. Very different from classical Boltzmann distribution e ε/k B T for low temperatures!
Go to Topics Quantum part slide 14/ 37 Fermi level quantities Fermi energy: all states bellow occupied to give place to the N electrons in the gas For free (gas) electrons we can define related quantities Fermi energy (typically 1.5-15 ev i metals) ε F = k F m (1) It can be shown that the Fermi energy is proportional to the density of free electrons in the metal. Fermi wave vector, k F, the wave vector of the electron in the highest energy state (T = 0). Fermi momentum Fermi temperature (of the order 10 5 for many metals) p F = k F () T F = ε F k B (3)
Go to Topics Quantum part slide 15/ 37 The ideal Fermi gas Electrons are Fermions. One electron in each state. Ideal gas: Free electrons; The states to be occupied are plane waves ( ) m x + y + z ψ(r i ) = εψ(r i ) m ψ(r i ) = εψ(r i ) (4) The total wavefunction for all particles is an (antisymmetrized) product of single particle states The solutions are plane waves, ψ k (r) e ik r, (5) To count the states, we proceed by considering boxes of size L in all 3 directions The allowed solutions are required to be periodic with L (faces of the box) ψ(x, y, z) = ψ(x+l, y, z) = ψ(x, y+l, z) = ψ(x, y, z+l) = ψ(x, y+l, z+l) =... Thus only some values of k are allowed.
Go to Topics Quantum part slide 16/ 37 The ideal Fermi gas e ik r = e ikx x the periodicity must be true for each of x, y, z For all it must be valid: This is true for k x ( and k y, k x ) if e iky y e ikz z e ikx x = e ikx (x+l) e ikx L = 1 k x (nx ) L = πn x k x (nx ) = π L n x kx nx = k n x where n x is an integer (positive or negative, or zero) The allowed discrete values of k are thus obtained by combinations of components k x (nx ) = k n x k y (ny ) = k n y k z (nz ) = k n z k = π L where the numbers n x,n y,n z are positive and negative integers. Thus the allowed values are on a discrete mesh of distance between points k In a little cube of K-space volume k 3 there is one allowed state. (picture follows) (6)
L L π L L L L π L L For larger L we get finer division of the K-space, k = π/l Go to Topics Quantum part slide 17/ 37
Go to Topics Quantum part slide 18/ 37 Density of states In a little cube of K-space volume k 3 there is one allowed state. The density of allowed states in the K-space is - doubled by spin ρ(k) = k 3 = L3 (π) 3 = V (π) 3 N electrons will occupy lowest energy states up to Fermi energy ɛ F. This means the states k, such that k k F, k F = mɛ F /. i.e. a sphere of radius k F in the K-space. Volume of this sphere is V K = 4π 3 k3 F This gives density ρ(k) = N = 3 N V K 4π kf 3 The two densities 3 N 4π kf 3 = V (π) 3 kf 3 = 3 (π) 3 N 4π V k F = ( 3π N ) 1 3 ( = 3π n ) 1 3 n = N V V Thus Fermi energy is given by (a power of) the spatial density of electrons n = N V
Go to Topics Quantum part slide 19/ 37 Density of states We re-use the two equations, but indicate that our N is related to our ɛ F ( mɛf k F = ) 1 N(ɛ F ) = V 3π k3 F (8) Rearranging, and replacing ɛ F by variable ɛ N(ɛ) = V 3π ( mɛ ) 3 (9) This means that up to general energy ɛ we have N(ɛ) states Thus the density of states g(ɛ) is obtained as g(ɛ) = dn(ɛ) dɛ g(ɛ) = dn(ɛ) dɛ = V π (30) ( ) 3 m 1 ɛ (31)
Go to Topics Quantum part slide 0/ 37 The density of states for electrons g(ɛ) = V π is a consequence of constant density of states in the K-space and the parabolic relation of energy and momentum ( ) 3 m 1 ɛ (3) This relation is important for applications of band theory of solids, semiconductor physics etc. For the phonons of vibrations we have obtained entirely different density of states g(ω) ω or g(ɛ) ɛ
Estimate of Fermi Temperature - from electron density k F = ( ) 1 3π N 3 ( ) = 3π 1 3 n n = N V V Thus Fermi energy is given by (a power of) the spatial density of electrons n = N V As the density increases, i.e. the distances between electrons decrease, the Fermi momentum and Fermi energy increase. The density of electrons in metallic solid estimate: assume that the linear distance between atoms is roughly 3 a 0, with a 0 the Bohr radius n 1 (3a 0) 3 k F ( 3π (3a 0) 3 ) 1 3 1 a 0 ɛ F ma 0 ( = 13.6 ev ) The Fermi Temperature We know: for room temperature k B T room = E 300K = 1/40 ev, Define T F by ɛ F = k B T F, we see how huge the Fermi temperature is: (k B T F )/(k B T room ) (13.6)/(1/40), i.e. T F (300 40 13.6) K or T F 160 000 K Go to Topics Quantum part slide 1/ 37
Go to Topics Quantum part slide / 37 Why is the heat capacity of electrons in metals small? Unless temperature is very high, only relatively few electrons near the Fermi surface can change their state. The remaining (majority) does not contribute to the heat capacity c electr v. In the classical gas, all electrons contribute, yielding a higher heat capacity. The energy of the excited electrons is about ε F.
Go to Topics Quantum part slide 3/ 37 Sommerfeld integrals For integrals involving the step-like Fermi function f(ε) 0 γ(ε)f(ε)dε (33) we use the Sommerfeld expansion. Based on: f(ε) is piecewise constant except around ε = µ. we introduce: γ(ε) = dγ(ε) dε. (34) Integrating by parts, we find Link to application here Γ(ε)f(ε) df(ε) 0 Γ(ε)dε (35) 0 dε The first part is zero since f (ε ) = 0, and since our case Γ(ε 0) = 0. Expanding Γ(ε) around ε = µ, and retaining only one term results in 0 dγ(ε) dε f(ε)dε = Γ(µ) + π (k B T ) 3 d Γ(ε) dε (36) ε=µ
Go to Topics Quantum part slide 4/ 37 Heat capacity of the Fermi electron gas Both the phonons and the free electrons will contribute to the heat capacity of metals. Cv total = Cv electr + Cv phonons (37) We evalulate the specific heat capacity of an electron gas ( ) U C v =. (38) T from the total energy of the gas U = 0 V dε g(ε) f(ε) ε. (39) here g(ε) is the density of quantum states and f(ε) Fermi distribution The approximation method for the integral is called Sommerfeld s expansion.
o to Topics Quantum part slide 5/ 37 Heat capacity of Fermi electron gas We use the Sommerfeld expansion (link) γ(ε) = dγ(ε) = g(ε) = C ε dε µ(t ) = ε F 1 π ( ) T 1 T F γ(ε) = dγ(ε) = εg(ε) = Cε ε dε εf N = g(ε)dε = ε F g(ε F ) 3 0 U = dγ(ε) dε µ 0 f(ε)dε = Γ(µ) + π (k B T ) 3 εg(ε) + π 3 d Γ(ε) dε ε=µ(40) (k B T ) [µg (µ) + g(µ)] (41) εg (ε) + g(ε) = 3 g(ε) U = U 0 + π (k B T ) g(ε F ), 3 3 ε 1 F g(ε F ) = N, T F and thus for C = U T with the above relations, we get for c v = C V = k B ε F (4) C electr = π k BN T cv electr = π k B N T (43) T F V T F
Go to Topics Quantum part slide 6/ 37 Notice linear temperature dependence. Differs from classical result (ideal gas), c classic which is 0.01 at room temperature. c electr v v = 3 N V k B, by a factor π 3 T T F, has significant contribution at low temperature, when c phonons v T 3.
Go to Topics Quantum part slide 7/ 37 Heat conductivity - Fermi Gas vs Classical Ideal Gas κ = 1 3 v τc v κ = 1 3 vlcv In classical ideal electron gas theory: 1 mv = 3 k BT (i.e. v = 3 m k BT ) v = 3 N V k B κ = 1 3 c class 3 m k BT 3 N V k Bτ κ class = 3 T N V k B m τ In the Fermi gas theory we must estimate the velocities in κ = 1 3 vlτc v from ɛ F = 1 mv F, thus v F = m k BT F c FG v = π k B N V T T F κ FG = 1 3 m k π k B N BT F V T τ κ FG = π T F 3 T N kb V m τ Note: cv FG is about 100 times smaller than cv class transport velocity factor vf is roughly 50 times larger than v
o to Topics Quantum part slide 8/ 37 Lorenz number revisited Lorenz number obtained from Fermi Gas κ FG : κ σt = π 3 ( ) kb =.45 10 8 W Ω e K. about the same as the experimental values. Experimental Lorenz numbers: Lorenz number of Classical Drude model κ σt = 3 ( ) kb. e is about factor smaller. Table from Tony Harker, Chiranijib Mitra and Andrew Horsfield, Solid State Physics Notes
Go to Topics Note for modifying Quantum part slide 9/ 37 Density of states - Fermi Gas Model PHYS08 autumn 010 - work with metal slides 19.11.010
Go to Topics Note for modifying Quantum part slide 30/ 37 Density of States We consider electron states of freely moving electrons enclosed in a box with volume V. We shall impose so called periodic conditions on the plane waves for the electrons. The volume V is realized as a box of length L, i.e. V = L 3. For a finite volume there will be a discrete number of states satisfying the imposed boundary conditions. Therefore the sum over final states is an ordinary sum. As this volume approaches infinite size, the summation over k, will be approaching an integral. The allowed discrete values of k are obtained by combinations of components k (nx ) x = π L n x k (ny ) y = π L n y k (nz ) z = π L n z (44) where the numbers n x,n y,n z are positive and negative integers. The density of states factor is found from performing such a limiting process, using following relations.
L L π L L L L π L L For larger L we get finer division of the K-space, k = π/l Go to Topics Note for modifying Quantum part slide 31/ 37
The summation will be replaced by an integral g( k) d k (45) k g( k) is the density of states in the K-space. The above display of allowed states shows that each of the allowed vectors k occupies a small volume of the K-space ( ) 3 π k x k y k z = = (π)3 L V (46) The density of states in the K-space is thus a constant (i.e. one per the above small k-space volume), and the above relation can be written as V (π) 3 d k (47) k Since the derivation of the golden rule assumes integration over frequencies, or energies, we shall transform this integral over momenta (i.e. wave numbers k) to integral over energy, V (π) 3 d k g(e)de [ ] Go to Topics Note for modifying Quantum part slide 3/ 37
Go to Topics Note for modifying Quantum part slide 33/ 37 so that the g(e) can be identified as V dk k (π) 3 de Ω k dω k (49) If there is any dependence on the direction of the wave vector, we must keep the angular information inside of the density of states, but in the present case we can perform the angular integration which gives the surface of sphere with radius one, i.e. 4π.
Go to Topics Note for modifying Quantum part slide 34/ 37 We must now evaluate the above density of states in terms of energy only, using i.e. k = me = E = k m m dk m E de = 1 1 E Setting these relations of k and E, we obtain the expression for g(e) g(e) = V dk k dω (π) 3 k = de Ω k g(e) = V π with the spin caused factor multiplication V (π) 3 (4π)m E dk de = V π g(e) = V π ( ) 3 m 1 ( ) 3 m 1 E (50) E (51) ( ) 3 m E (5)
Go to Topics Note for modifying Quantum part slide 35/ 37 The density of electronic states is proportional to V, for larger volume tighter placed states (smaller energy step) g(e) = V π ( ) 3 m E Now we consider the gas consisting of N fermions in the volume V, i.e. the electron density n is n = N V In accordance with Pauli principle, if we fill all states up to energy ɛ we can accomodate N (ɛ) electrons, i.e. N (ɛ) = ɛ 0 g(ɛ )dɛ N (ɛ) = ɛ 0 V π ( ) 3 m ɛ dɛ since g(ɛ) = C ɛ, we can use g(ɛ)dɛ = Cɛ 1 dɛ = 3 Cɛ3 = 3 ɛg(ɛ)
Go to Topics Note for modifying Quantum part slide 36/ 37 We now define the energy needed to accomodate all Nparticles as Fermi energy ɛ F, so that N N (ɛ F ) = ɛf 0 g(ɛ )dɛ N = ɛf 0 V π Here we can see how the unphysical volume V is absorbed Since clearly N ɛf V = 0 1 π ( ) 3 m ɛ dɛ n = 1 3π k = k F = and here are the final derived relations. ( ) 1 m 1 ɛ ( ) 1 m (ɛf ) 1 ( ) 3 m ɛ dɛ ( ) 3 m (ɛf ) 3 n = 1 3π (k F ) 3 k F = 3 3π n ɛ F = ( 3π n ) 3 = k F m m
Go to Topics Note for modifying Quantum part slide 37/ 37 The derivations thus lead to k F = 3 3π n ɛ F = k F i.e. the Fermi Energy quantities (ɛ F and k F ) are given by the density n of electrons in the ordinary physical space n = N/V As the density increases, i.e. the distances between electrons decrease, the Fermi momentum and Fermi energy increase. The density of electrons in metallic solid estimate: we estimate the density by assuming that the linear distance between atoms is roughly 3 a 0, with a 0 the Bohr radius (in fact the densities are a bit smaller, the distances a bit larger than three Bohr units - but this gives an easy answer) n 1 ( ) 1 3π 3 (3a 0 ) 3 k F (3a 0 ) 3 1 ɛ F a 0 ma 0 m ( = 13.6 ev ) Knowing that the roughly 300 K temperature is k B T room = E 300K = 1/40 ev, and defining T F as ɛ F = k B T F, we see how huge the Fermi temperature is: (k B T F )/(k B T room ) (13.6)/(1/40), i.e. T F (300 40 13.6) K or T F 160 000 K