PHY 14A: Solid State Physics Solution to Homework #7 Xun Jia 1 December 5, 26 1 Email: jiaxun@physics.ucla.edu
Fall 26 Physics 14A c Xun Jia (December 5, 26) Problem #1 Static magnetoconductivity tensor. For the drift velocity theory of (51), show that the static current density can be written in matrix form as: j x 1 ω j y σ c τ E x = ω 1 + (ω j c τ) 2 c τ 1 E y (1) z 1 + (ω c τ) 2 E z In the high magnetic field limit of ω c τ 1, show that σ yx = nec/b = σ xy (2) In this limit σ xx =, to order 1/ω c τ. The quantity σ yx is called the Hall conductivity. From Eqn. (51) in Kittel, in the static case, the velocity is not varying with time, so dv i /dt = for i = x, y, z, therefore, in the matrix form, we have: m/τ eb/c v x E x eb/c m/τ v y = e E y () m/τ v z E z solve this equation by inverting the coefficient matrix, and let ω c = eb/mc, we get: v x 1 ω v y eτ c τ E x = ω m[1 + (ω v c τ) 2 c τ 1 E y (4) ] z 1 + (ω c τ) 2 E z since j = nev and σ = ne 2 τ/m, it follows that: j x v x j y j z = ne v y v z = σ 1 + (ω c τ) 2 1 ω c τ ω c τ 1 1 + (ω c τ) 2 E x E y E z (5) and: In the high magnetic field limit, ω c τ 1, then: σ yx = σ xy = σ xx = thus σ xx =, to order 1/ω c τ. σ ω c τ 1 + (ω c τ) σ ω c τ 2 (ω c τ) = σ 2 ω c τ = nec B σ 1 + (ω c τ) σ ( ) 1 2 (ω c τ) = + σ (7) 2 ω c τ (ω c τ) 2 1 (6)
Fall 26 Physics 14A c Xun Jia (December 5, 26) Problem #2 Kinetic energy of electron gas. Show that the kinetic energy of a three dimensional gas of N free electrons at K is U = 5 Nɛ F (8) For the Fermi-Dirac distribution at K, we have: { 1 f(ɛ) = 1 : ɛ < ɛf e (ɛ ɛ F )/k BT + 1 = (9) T = : ɛ > ɛ F and the density of states of electrons in -D is : D(ɛ) = V dk 4πk2 (2π) dɛ = V 4π 2 ( ) /2 2m h 2 ɛ 1/2 (1) since for free electrons, the kinetic energy equals to the total energy ɛ, the average kinetic energy per electron is simply: ɛ = dɛd(ɛ)f(ɛ)ɛ = dɛd(ɛ)f(ɛ) ɛ F dɛd(ɛ)ɛ = 5 ɛ F (11) dɛd(ɛ) ɛ F therefore the kinetic energy of total N electrons is: U = N ɛ = 5 Nɛ F (12) Problem # Chemical potential in two dimensions. Show that the chemical potential of a Fermi gas in two dimensions is given by: µ(t ) = k B T ln[exp(πn h 2 /mk B T ) 1] (1) for n electrons per unit area. Note: the density of orbitals of a free electron gas in two dimensions is independent of energy: D(ɛ) = m/, per unit area of specimen. At finite temperature T, the chemical potential is determined by the condition: n = dɛd(ɛ)f(ɛ) = dɛ m 1 e (ɛ µ)/k BT + 1 (14) 2
Fall 26 Physics 14A c Xun Jia (December 5, 26) the right hand side can be computed explicitly as: dɛ m 1 e (ɛ µ)/k BT + 1 = m = mk BT = mk BT = mk BT e (ɛ µ)/k BT dɛ 1 + e (ɛ µ)/k BT so from above two equations, we can solve for µ as: [ ( πn h 2 ) µ(t ) = k B T ln exp mk B T d(e (ɛ µ)/kbt ) 1 + e (ɛ µ)/k BT [ ( ln 1 + exp ɛ µ )] k B T [ ( )] µ ln 1 + exp k B T ] 1 ɛ= (15) (16) Problem #4 Fermi gases in astrophysics. (a). Given M = 2 1 g for the mass of the Sun, estimate the number of electrons in the Sun. In a white dwarf star this number of electrons may be ionized and contained in a sphere of radius 2 1 9 cm; find the Fermi energy of the electrons in electron volts. (b). The energy of an electron in the relativistic limit ɛ mc 2 is related to the wavevector as ɛ pc = hkc. Show that the Fermi energy in this limit is ɛ F hc(n/v ) 1/, roughly. (c). If the above number of electrons were contained within a pulsar of radius 1km, show that the Fermi energy would be 1 8 ev. This explains why pulsars are believed to be composed largely of neutrons rather than of protons and electrons, for the energy release in the reaction n p + e is only.8 1 6 ev, which is not large enough to enable many electrons to form a Fermi sea. The neutron decay proceeds only until the electron concentration builds up enough to create a Fermi level of.8 1 6 ev, at which point the neutron, proton, and electron concentrations are in equilibrium. (a). The Sun is mainly composed by hydrogen atoms, whose mass is approximately equal to the mass of proton m p. Since each hydrogen atom contains one electron, the number of electrons in the Sun is: N = M m p = 1.2 1 57 (17)
Fall 26 Physics 14A c Xun Jia (December 5, 26) if this number of electrons is contained in a white dwarf star with a radius R wd = 2 1 9 cm, the Fermi energy is: 2/ ( ɛ F = h2 π 2 N ) 2/ = h2 N 2m V 2m π2 4πRwd (18) = 6. 1 15 J = 4 1 4 ev (b). In the relativistic case with a dispersion relation of ɛ pc = hkc, the density of states of electrons is: D(ɛ) = 2 V dk 4πk2 (2π) dɛ = V ɛ 2 π 2 h (19) c then the Fermi energy is determined by the condition: which gives: ɛ F = ɛ F D(ɛ)dɛ = V ɛ F π 2 h c (2) ( ɛ F = hc π 2 N ) 1/ ( ) 1/ N hc (21) V V (c). If above number of electrons is contained in a pulsar of radius R p = 1km, the Fermi would be: 1/ ( ) 1/ N ɛ F = hc = hc N V 4πRp = 2.1 1 11 J = 1. 1 8 ev (22) Problem #5 Liquid He. The atom He has spin 1 and is a fermion. The density of liquid 2 He is.81g/cm near absolute zero. Calculate the Fermi energy ɛ F and the Fermi temperature T F. The mass of a single He is about: m He = a.m.u. = 1.67 1 27 kg = 5.1 1 27 kg (2) therefore the number density of He is n = ρ m He = 1.62 1 28 m (24) 4
Fall 26 Physics 14A c Xun Jia (December 5, 26) so the Fermi energy is: ɛ F = h2 2m He ( π 2 N ) 2/ = ɛ F = h2 (π 2 n) 2/ V 2m He = 6.74 1 2 J = 4.2 1 4 ev (25) and the Fermi temperature is: T F = ɛ F k B = 4.88K (26) 5