The energy of this state comes from the dispersion relation:

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Clarence Jennings
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1 Homework 6 Solutions Problem 1: Kittel 7-2 a The method is the same as for the nonrelativistic gas. A particle confined in a box of volume L 3 is described by a the set of wavefunctions ψ n sinn x πx/l sinn y πy/l sinn z πz/l, which corresponds to a state with wavenumber: k n = π L n2 x + n 2 y + n 2 z 1/2 = π L n The energy of this state comes from the dispersion relation: ɛ n = p n c = hk n c = π hc L n Please note that p = hk is a definition which has nothing to do with relativistic vs nonrelativistic. The number of orbitals with energy less than some energy ɛ n is given by: N< ɛ n = 2 1 4π 8 3 n3 = π L 3 3 π 3 h 3 c 3 ɛ3 n The factor 2 is due to the spin degeneracy. To find the Fermi energy, we set N< ɛ n equal to the total number of fermions N: L 3 N = π 3 π 3 h 3 c 3 ɛ3 F 3Nπ3 h 3 c 3 1/3 3N ɛ F = = π hc L 3 π πl 3 1/3 b To obtain the density of states, we differentiate N< ɛ: nɛ = dn< ɛ L3 = π π 3 h 3 c 3 ɛ2 The ground state of the system is the zero temperature state. In this limit, the Fermi-Dirac distribution becomes: fɛ = 1 if ɛ < ɛ F = if ɛ > ɛ F Therefore, the ground state energy is given by: U = ɛf nɛɛ = π L3 π 3 h 3 c 3 = 3N ɛ 3 F ɛ 4 F 4 = 3 4 Nɛ F ɛf ɛ 3 1

2 c In the neutron star calculation, discussed in pages 17-4 to 17-6 of the lecture notes, it was found that the Fermi energy for nucleons and electrons was approximately 86 MeV. The relativistic dispersion relation for a particle of rest mass m is given by: ɛ = p 2 c 2 + m 2 c 4 mc 2 This expression gives the energy of the particle beyond the energy required to create the particle. The energy required to create the particle, mc 2, is interpreted as a chemical potential, as in the lecture notes. The mass of a proton is about 938 MeV, much greater than the Fermi energy of protons, so the nonrelativistic approximation ɛ p 2 /2m is fine for the nucleons. However, the electron mass of.511 MeV is much less than the electron Fermi energy, suggesting the ultrarelativistic limit ɛ pc for the electrons. As in the notes, we consider a neutron star as a Fermi gas of mainly neutrons, with a small concentration of electrons and protons. The concentrations are denoted by n e,n p,and n n. The argument is exactly the same as in the notes refer to pages 17-5 to 17-6 for notation except now the electron Fermi energy is given by: ɛ F,e = π hc 3n e π 1/3 = π hc 3n π 1/3 x 1/3 The energy balance equation becomes: or ɛ F n n 2/3 1 x 2/3 = ɛ F n n 2/3 m n m p x 2/3 + π hc 3n π 1/3 x 1/3 E 1 x 2/3 + E ɛ F n n 2/3 = m n m p x 2/3 + π hc ɛ F n n 2/3 3n π 1/3 x 1/3 The assumption that n e, n p << n n implies that n n and that x << 1. This approximation is consistent with setting n = n in the above expression; ignoring x relative to 1 on the left hand side; and ignoring the x 2/3 term relative to the x 1/3 term on the right hand side. Making the approximation: 1 + E ɛ F π hc ɛ F 3n π 1/3 x 1/3 x 1/3 = x = = 1 + E ɛ F π hc ɛ F 3n π 1/3 1 + E ɛ F π hc ɛ F 3n π 1/ π π 1/

3 So our small x approximation is self-consistent, though x is larger than in the non-relativistic case. Problem 2: Kittel 7-5 a We are given that liquid 3 He has density ρ =.81 g/cm 3 and the atoms obey Fermi statistics. We compute the concentration: the Fermi energy: n = cm 3 = cm 3, 3 ɛ F = h2 2m 3π2 n 2/3 = π /3 J = J = ev, the Fermi velocity: v F = 2ɛ F m 1/2 = and the Fermi temperature: b The heat capacity is given by: /2m/s = 164m/s, T F = ɛ F = K = 4.9K k B C V = π2 2T F Nk B T 1.6Nk B T where T is the temperature in Kelvin. This value is roughly a factor of 3 lower than the experimental value, which indicates something else is going on. Problem 3: Kittel 7-6 a The gravitational self-energy of the star should depend on G, M and R. Dimensional analysis suggests the following estimate: U GM 2 R The negative sign is due to the attractive nature of the force. b The electrons in the star may be treated as a degenerate Fermi gas. The kinetic energy of the electrons may be estimated: K Nɛ F N h2 2m 3π2 n 2/3 N h2 m N R 3 2/3 h2 mr 2 N 5/3 h2 M 5/3 mm 5/3 H R2 Here N M/M H where M is the mass of the star and M H is the mass of a proton; m is the mass of an electron. We have assumed the star is neutral so 3

4 there are equal numbers of electrons as positive ions which would have a mass on the order of a proton mass and account for the mass of the star. c Kittel argues that the kinetic energy of the electrons primarily due to electrons and the gravitational energy of the ions should be of similar orders of magnitude. This implies: GM 2 R M 1/3 R h 2 M 5/3 mr 2 M 5/3 H h 2 mm 5/3 H G = / g1/3 cm = g 1/3 cm d If the mass of the white dwarf is that of the sun, then part c implies: R /3 cm 8 18 cm ρ M R = 4 16 g/cm 3 e If the Fermi gas is composed of neutrons instead of electrons, then the results are changed by the mass ratio: M 1/3 R n m e m n M 1/3 R e g 1/3 cm 1 18 g 1/3 cm If the neutron star had the mass of the sun, then: R km 8km 1/3 12 Problem 4: Kittel 7-8 In the regime τ < τ E, the chemical potential is very small, µ τ/n. The approximation µ becomes exact in the limit of large N. In this limit: C V = U = U T V Dɛfɛ, τɛ = V 4π 2 2M h 2 3/2 ɛ 3/2 e ɛ/τ 1 = V 4π 2M h 3/2 2 2 dx x3/2 e x 1 = V 4π 2M h 3/2 2 2 τ 5/2 dx x3/2 5 e x 1 2 τ 3/2 4

5 To get the entropy, we note that for constant volume, du = τdσ dσ = du τ. Therefore: σ = 1 U = V T V τ T V 4π 2M h 3/2 2 2 dx x3/2 5 e x 1 2 τ 1/2 We integrate both sides of this equation with respect to temperature keeping the volume constant to get the entropy. τ σ στ σ = T = V V 4π 2M h 3/2 2 2 dx x3/2 5 τ e x dττ 1 1/2 2 = V 4π 2M h 3/2 2 2 dx x3/2 e x τ 3/2 In the case where the temperature τ > τ E, we can no longer take the chemical potential to be small. In this case, the energy is formally given by the expression: U = Dɛfɛ, τ, µɛ = V 4π 2 2M h 2 3/2 ɛ 3/2 e ɛ µ/τ 1 In order to obtain µ, we have to consider the particle number constraint: N = Dɛfɛ, τ, µ = V 4π 2 2M h 2 3/2 ɛ 1/2 e ɛ µ/τ 1 This gives an equation which, at least formally, may be inverted to get µ. Then, we may numerically integrate the previous expression to obtain the energy. Given the energy, we obtain the heat capacity and entropy as in the τ < τ E case. Problem 5: Kittel 7-12 We showed in Homework 4, Problem 6 Kittel 5-1, that if a system has average particle number < N >, the mean squared fluctuation in the particle number is given by < N 2 >= τ <N> µ. In this problem, we are considering a single orbital so the average particle number is given by the Bose-Einstein distribution function: The rest is straightforward: < N >= < N 2 > = τ < N > µ = 1 e ɛ µ/τ 1 = τ e ɛ µ/τ = e ɛ µ/τ 1 2 e ɛ µ/τ e ɛ µ/τ τ 1 e ɛ µ/τ e ɛ µ/τ 1 2 = < N > + < N > 2 =< N > 1+ < N > 5

6 Problem 6 We are consider a gas of 2 atoms of 87 Rb. The observed Bose-Einstein transition occurs at the temperature T E = K. In order to get the concentration, we use the formula for T E given on page 18-4 of the lecture notes: T E = 2π h2 N 2/3 Mk B 2.612V N M 3/2 = V 2π h 2 k BT E /2 = m 3 2π = m 3 = cm 3 To obtain the density, we multiply this by the mass per atom: ρ = NM V = g = 4.14 g 1 9 cm3 cm 3 Assuming a spherical sample, then number of particles is given by N = 4 3 πr3 N/V so that: 3N 1/ /3cm R = = = π N V 4π cm 6

PHY 140A: Solid State Physics. Solution to Homework #7

PHY 140A: Solid State Physics. Solution to Homework #7 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.