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Lecture 24. Degenerate Fermi Gas (Ch Lecture 24. Degenerate Fermi Gas (Ch. 7) We will consider the gas of fermions in the degenerate regime, where the density n exceeds by far the quantum density nQ, or, in terms of energies, where the Fermi energy exceeds by far the temperature. We have seen that for such a gas μ is positive, and we’ll confine our attention to the limit in which μ is close to its T=0 value, the Fermi energy EF. ~ kBT μ/EF 1 1 kBT/EF occupancy T=0 (with respect to E ) F The most important degenerate Fermi gas is 1 the electron gas in metals and in white dwarf nε()(),, T= f ε T = stars. Another case is the neutron star, whose ε⎛ − μ⎞ exp⎜ ⎟ +1 density is so high that the neutron gas is ⎝kB T⎠ degenerate. Degenerate Fermi Gas in Metals empty states ε We consider the mobile electrons in the conduction EF conduction band which can participate in the charge transport. The band energy is measured from the bottom of the conduction 0 band. When the metal atoms are brought together, valence their outer electrons break away and can move freely band through the solid. In good metals with the concentration ~ 1 electron/ion, the density of electrons in the electron states electron states conduction band n ~ 1 electron per (0.2 nm)3 ~ 1029 in an isolated in metal electrons/m3 . atom The electrons are prevented from escaping from the metal by the net Coulomb attraction to the positive ions; the energy required for an electron to escape (the work function) is typically a few eV. The model assumes that the electrons for an ideal Fermi gas confined within impenetrable walls. Why can we treat this dense gas as ideal? Indeed, the Coulomb interactions between electrons at this density must be extremely strong, and in a a solid, the electrons move in the strong electric fields of the positive ions. The first objection is addressed by the Landau’s Fermi liquid theory. The answer to the second objection is that while the field of ions alters the density of states and the effective mass of the electrons, it does not otherwise affect the validity of the ideal gas approximation. Thus, in the case of “simple” metals, it is safe to consider the mobile charge carriers as electrons with the mass slightly renormalized by interactions. There are, however, examples that the interactions lead to the mass enhancement by a factor of 100-1000 (“heavy fermions”). The density of states per unit 2/3 The Fermi 1 ⎛ 2m ⎞ volume for a 3D free electron gas 3D 2/1 g ()ε = 2⎜ 2 ⎟ ε Energy (m is the electron mass): 2π ⎝ h ⎠ N ∞ The number of electron per unit volume: n≡ =g∫ ε()() f ε d ε V 0 ⎧ ,1 ε ≤ EF At T = 0, all the states up to ε = EF are filled, at ε > EF –empty: f ()ε = ⎨ ⎩ ,0 ε > EF ∞ EF 2/3 EF 2/3 1 ⎛ 2m ⎞ 1 ⎛ 2m ⎞ 2/3 n= gε f ε d ε= g ε= εd εd ε= E ∫()() ∫ () 2⎜ 2 ⎟ ∫ 2⎜ 2 ⎟ ()F 0 0 2π ⎝ h ⎠ 0 3π ⎝ h ⎠ 2 2 3/2 ε T>0 h 2 3/2 h ⎛ 3 ⎞ the Fermi energy (μ of an EF = ()3π n = ⎜ n⎟ 2m 8m⎝π ⎠ ideal Fermi gas at T=0) ε = EF 2 2/3 −34 2 h ⎛ 3 ⎞ 6.6⋅10 2/3 E = ⎜ n⎟ ≈ ()1⋅1029 J ≈1⋅10−18 J ≈ 6 eV F ⎜ ⎟ −31 () 8m ⎝ π ⎠ 8⋅9⋅10 - at room temperature, this 4 TF ≡ EF / kB ≈ few eV ≈ few 10 K Fermi gas is strongly degenerate (EF >> kBT). The total energy of all ε F 3 electrons in the conduction U = ε × g()ε dε = NE 1/ 2 0 ∫ 5 F g(ε )∝ ε band (per unit volume): 0 - a very appreciable zero-point energy! The Fermi Gas of Nucleons in a Nucleus Let’s apply these results to the system of nucleons in a large nucleus (both protons and neutrons are fermions). ons in the In heavy elements, the number of nucle nucleus is large and statistical treatment is a reasonable approximation. We need to estimate the density of protons/neutrons in the nucleus. The radius of the nucleus that contains A nucleons: 1=R . 3( × 10−15 ) A1 ×m / 3 A Thus, the density of nucleons is: n = 1≈ ⋅ 10 44 m- 3 4 3 1π .() 3×− 1015 m ×A 3 For simplicity, we assume that the # of protons = the # of neutrons, hence their density is 44 -3 np=0 n n .≈ 5 ⋅ 10m 2 2 / 3 6 . 6⋅ − 1034 ⎛ 3 ⎞ The Fermi eneryg E0= .( 5 10) ⎜ ⋅J44 ⎟ 4.3= ⋅ 10−12 =J 27 MeV F −27 ⎜ ⎟ 8 1× . 6 ⋅ 10⎝ π ⎠ EF >>> kBT – ns are very “cold”enerate. The nucleohe system is strongly degt – they are all in their ground state! The average kinetic energy in a degenerate Feri gas =m 0.6 of the Feri energym 1E 6= MeV- the nucleons are non-rlativistice The Fermi Momentum 3/1 p h ⎛ 3 ⎞ The electrons at the Fermi level are moving at F ⎜ ⎟ 6 vF= = ⎜ n⎟ 1≈ ⋅ 10 m/s a very high velocity, the Fermi velocity: m 2m⎝π ⎠ This velocity is of the same order of magnitude as the orbital velocity of the outer electrons in an atom, and ~10 times the mean thermal velocity that a non-degenerate electron gas would have at room temperature. Still, since vF<<c, we can treat the mobile electrons as non-relativistic particles. 1/3 h ⎛ 3 ⎞ ⎜ ⎟ the Fermi momentum The corresponding momentum: pF = 2mEF = ⎜ n⎟ 2 ⎝ π ⎠ Naturally, the entropy of Fermi gas is zero at T=0 – we are dealing with a single state that can undergo no further ordering process. ε What happens as we raise T, but keep kBT<<EF so that μ≈EF? ε = EF ~ k T B Empty states are available only above (or within ~ kBT ) of the Fermi energy, thus a very small fraction of electrons ncy can be accelerated by an electric field and participate in the current flow. occupa T=0 The electrons with energies ε < E - F g(ε )∝ ε 1/ 2 (few)kBT cannot interact with anything unless this excitation is capable of (with respect to EF) raising them all the way to the Fermi energy. Problem (Final 2005) copper atoms form a crystal lattWhen the ice with the density of atoms of 8.5·1028 m-3, each atom donates 1 electron in the conduction band. (a) Assuming that the effective mass of the conduction electrons is the same as the free electron mass, calculate the Fermi energy. Express your answer in eV. 2 / 3 2 2 / 3 3h2 ⎛ N ⎞ 6 . 6⋅ −34 10 ⎛ 3 ⎞ E = ⎜ ⎟ 8= ( . 5 10) ⎜ ⋅ 28 1⎟ .= 1 ⋅ 10−18 =J 6.7 eV F ⎜ ⎟ −31 ⎜ ⎟ 8m ⎝ π V ⎠8 9× . 1 ⋅ 10⎝ π ⎠ (b) The electronscur parrent ticipate in the flow if their energies correspond to the cy panoccu n(ε) 1 (that is not too close to ble for the lano empty states avai accelerated electrons) and not too small electrons to accelerate). At (no T=300K, calculate the energy interval that is occupied by the electrons that participate in the current flow, assuming that for these electrons the occupancy varies between 0.1 and 0.9. 1 1 ⎛ ε − E ⎞ n()ε = 0 .= 9 exp⎜ 1 F=⎟ 9 ε E = k + ln T 9 ⎛ ⎞ ⎜ ⎟ 1 FB ε − EF ⎛ ε1 − EF ⎞ ⎝ kB T ⎠ exp⎜ ⎟ +1 exp⎜ ⎟ +1 ⎝ kB T ⎠ ⎝ kB T ⎠ 1 ⎛ ε − E ⎞ 1 0 .= 1 exp⎜ 2 F=⎟ ε E = k − ln T 9 2Δε ln =k 9 T 0= . 11 eV ⎜ ⎟ 2 FB B ⎛ ε 2 − EF ⎞ ⎝ kB T ⎠ 9 exp⎜ ⎟ +1 ⎝ kB T ⎠ Problem (Final 2005, cont.) (c)(b) Using the assumptions of (b), calculate the ratio N1/N0 where N1 is the number of “current-carrying” electrons, N0 is the total number of electrons in the conduction band. Assume that within the range where the occupancy varies between 0.1 and 0.9, the occupancy varies linearly with energy (see the Figure), and the density of states is almost energy-independent. The density of states for the three- dimensional Fermi gas: 3N g()ε = 2/3 ε 2EF occupancy ε - EF EF+ Δε 2/ 3N 3 Δε N= nε()() g ε= ε d ×E ×5.0 Δε N = =0N . × 012 1 ∫ 2E 2/3 F 4 E EF− Δε 2/ F F Thus, at T=300K, the ratio of the “current-carrying” electrons to all electrons in the conduction is 0.012 or 1.2 %. The Heat Capacity of a Cold Fermi Gas One of the greatest successes of the free electron model and FD statistics is the explanation of the T dependence of the heat capacity of a metal. So far, we have been dealing with the Fermi gas at T = 0. To calculate the heat capacity, we need to know how the internal energy of the Fermi gas, U(T), depends on the temperature. Firstly, let’s predict the result qualitatively. By heating a Fermi gas, we populate some states above the Fermi energy EF and deplete some states below EF. This modification is significant within a narrow energy range ~ kBT around EF (we assume that kBT the system is cold - strong degeneracy). The fraction of electrons that we “transfer” to higher energies ~ kBT/EF, the energy increase for these electrons ~ kBT. Thus, the increase of the internal energy with 2 temperature is proportional to N×(kBT/EF) ×(kBT) ~ N (kBT) /EF.
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