Phys463.nb 33
6.1.1. Electrons in one atom One electron in an atom (a hydrogen-like atom): the nucleon has charge +Z e, where Z is the atomic number, and there is one electron moving around this nucleon
Four quantum number: n, l and lz, sz.
Energy levels En with n = 1, 2, 3 ... Z2 Μ e4 1 En = - (6.1) 2 2 2 2 32 Π Ε0 Ñ n where Μ = me mN me + mN » me where me is the mass of an electron and mN is the mass of the nucleon. 13.6 eV Z2 En = - H L (6.2) n2 H L For each n, the angular momentum quantum number [L2 Ψ = l l + 1 Ψ] can take the values of l = 0, 1, 2, 3, 4 … , n - 1. These states are known as the s, p, d, f , g,… states
For each l, the quantum number for Lz can be any integer betweenH -Ll and +l lz = -l, -l + 1, … , l - 1, l .
For fixed n, l and lz, the spin quantum number sz can be +1/2 or -1/2 (up or down). H L At each n, there are 2 n2 quantum states. In a real atom In a hydrogen-like atom, for the same n, all the 2 n2 quantum number has the same energy. In a real atom, the energy level splits according to the total angular momentum quantum number l. Typically, the state with lower l has lower energy (not always true). Due to the rotational symmetry, states with the same l has the same energy (if we ignore the spin-orbital coupling, etc.). … … … … n = 4 l = 1 4 s 2 n = 3 l = 2 3 d 10 n = 3 l = 1 3 p 6 n = 3 l = 0 3 s 2 n = 2 l = 1 2 p 6 n = 2 l = 0 2 s 2 n = 1 l = 0 1 s 2 34 Phys463.nb
Many electrons at T = 0 Electrons are fermions. The Pauli exclusive principle requires that we can have at most 1 electron per quantum state.
At T = 0, to minimize the total energy, the electrons want to state at the lowest Ne quantum state.
Ne Element No. of electrons on each shell 1 H 1 s1 2 He 1 s2 3 Li 1 s2 2 s1 4 Be 1 s2 2 s2 5 B 1 s2 2 s2 2 p1 6 C 1 s2 2 s2 2 p2 … … … 10 Ne 1 s2 2 s2 2 p6 11 Na 1 s2 2 s2 2 p2 3 s1 12 Mg 1 s2 2 s2 2 p2 3 s2 ... … ... Valence electrons Electrons in the low energy states (inner layers) are bonded tightly to the nucleon. (It costs about 1-10 eV to remove one of these electrons from an atom. 1eV is 104 K. It is a very high energy cost in solid state physics). Electrons in high energy states (outer layers) are loosely bonded to the nucleon (easy to remove). These electrons are called the valence electrons and these energy states are called the valence shells. Valence electrons are the electrons in a atom which can participate in the formation of chemical bonds. They are typically electrons in the outermost (or second-outer-most) shell.
6.1.2. Electrons in a metal In a metal, an atom may lose some or all of its valence electrons and thus turns into an ion (nucleon+inner electrons). These ions form a crystal and their motions are phonons. The valence electrons are no longer bonded to nucleons. They can move freely in a crystal. A crystal = A lattice of ions + valence electrons (6.3)
Motions in a crystal = phonons + valence electrons (6.4)
At low temperature, the interactions between phonons are typically very weak. So we can consider treat them as a quantum gas (a Bose gas). In a metal, because valence electrons can move around, we can treat them as a quantum fluid (a fermion fluid). Typically, we call this fluid a Fermi liquid. It is called a Fermi liquid, instead of a Fermi gas, because the interactions between electrons (Coulomb interactions) are typically pretty strong (comparable to the kinetic energy in most metals). A solid can be considered as the mixture of two type of fluids:
A Bose gas of phonons + A Fermi liquid of electrons (6.5)
6.1.3. Free electrons in 1D at zero temperature T = 0 Here, we start from the simplest situation: a free Fermi gas The word “free” here means two things. 1. We ignore interactions between electrons 2. We ignore interactions between electrons and ions (nucleons) Within this approximation, electrons are free particles. So their Hamiltonian is (for a 1D system)
p2 1 â 2 Ñ2 â2 H = = -ä Ñ = - (6.6) 2 m 2 m â x 2 m â x2 One electron in 1D The Schrodinger equation is Phys463.nb 35
Ñ2 â2 - Ψn x = Ε Ψn x (6.7) 2 m â x2 The solutions of thisH LequationH areL plane waves
Ψn x = A exp ä kn x (6.8) The eigen-energy Ε is H L Hn L 2 2 Ñ kn Εn = (6.9) 2 m
For a 1D system with length L and periodic boundary conditions, Ψn x = Ψn x + L , we know that the wavevector (momentum) is quantized
kn L = 2 n Π (6.10) H L H L with n being an integer, n = … , -2, -1, 0, 1, 2, … Or say, Π kn = 2 n (6.11) L The eigen-energy here is also quantized
2 2 2 2 Ñ kn Ñ 2 n Π Εn = = (6.12) 2 m 2 m L For a large system L ® ¥, the energy and momentum become continuous variables. Many electrons in 1D at T = 0 At T = 0, the electrons want to stay in the lowest energy states to minimize the energy. But electrons are fermions, which means that they cannot all just go to the lowest energy state Ε0. Due to the Pauli exclusive principle, one quantum state can host at most one electron. These means that for each n, we can have at most two electrons (one with spin up and one with spin down). If we have N electrons, at T = 0, the electron occupies the lowest N 2 states. The energy of the highest filled state is known as the Fermi energy ΕF . The momentum of this state is known as the Fermi momentum PF . The wavevector of this state is known as the Fermi wavevector kF . Obviously, 2 2 2 PF Ñ kF PF = Ñ kF and ΕF = = (6.13) 2 m 2 m
For a large system (L ® ¥), ΕF = ?
The number of quantum state between -PF to PF is (the factor 2 comes from the spin degrees of freedom):
PF â P 2 L PF L 2 L 2 = â P = 2 PF = PF 2 Π Ñ (6.14) -PF 2 Π Ñ -PF Π Ñ Π Ñ L Numberà of electrons in theseà states 2 L Ne = PF (6.15) Π Ñ So
Π Ñ Ne Π Ñ Ne PF = = (6.16) 2 L 2 L
Here Ne L is the density of electrons (number of electrons per length).
2 2 2 2 PF Π Ñ Ne ΕF = = (6.17) 2 m 8 m L
The bottom line: PF , ΕF and kF are determined by the density of electrons. 36 Phys463.nb
6.1.4. Free electrons in 1D at finite temperature Distribution function The distribution function f Ε measures the average number of electrons on a quantum state with energy Ε.
At zero temperature, the number of electrons on a quantum state is 1 if Ε < ΕF and 0 if Ε > ΕF . In other words, this function is a step function H L 1 Ε < ΕF f Ε = (6.18) 0 Ε > ΕF
At finiteH L T,: the electron number on a quantum state is NOT fixed. Sometimes this state is occupied (1 electron) and sometimes it is empty (0 electron). So the average electron number is some number between 0 and 1. 0 < f Ε < 1 at finite T (6.19)
Fermi distribution:H L For non-interacting fermions, at finite temperature, the distribution function takes this form 1 f Ε = Ε-Μ (6.20) exp + 1 kB T H L where is knownJ as theN Fermi-Dirac distribution. Let’s compare it with the Planck distribution (for phonons) we learned in the previous chapter. 1 fphonon Ε = (6.21) exp Ε - 1 kB T H L We found two differences:J N (1) -1 turns into +1 and (2) we have an extra parameter Μ for electrons. The ±1 is determined by the nature of the particles. For bosons (photons, phonons, etc.), it is -1. For fermions (electrons, protons), it is always +1. The parameter Μ is called the chemical potential. It is a very important parameter in thermodynamics and statistical physics (it is as important as the concept of temperature). It controls the density of particles in a system. Here, for a Fermi gas, we can consider it as a generalization of ΕF at finite T. In fact, we will show below that when T ® 0, Μ ® ΕF . How to determine Μ? The chemical potential Μ can be determined by counting the total number of electrons
¥ â P L ¥ 1 Ne = 2 f Εk = â P k2 -¥ 2 ΠÑ L Π Ñ -¥ -Μ (6.22) exp 2 m + 1 à H L à kB T Divided by L on both sides:
Ne 1 ¥ 1 = â P k2 L Π Ñ -¥ -Μ (6.23) exp 2 m + 1 à kB T
If we know the density of electrons Ne L, we can solve this equation to find Μ. In general, Μ depends on T and the electron density. As T ® 0, Μ ® ΕF . Some properties of f Ε (a) f Ε turns into a step function at T ® 0. Ε-Μ At low T, becomes veryH large.L Depending on the sign of Ε - Μ H L kB T Ε - Μ -¥ Ε < Μ Lim = T®0 +¥ Ε > Μ (6.24) kB T Therefore, : Phys463.nb 37
1 = 1 = 1 Ε < Μ 1 exp -¥ +1 0+1 LimT®0 f Ε = LimT®0 = (6.25) Ε-Μ 1 = 1 = 1 Ε > Μ exp + 1 expH +¥L+1 0+1 kB T H L : This result recovers the zero temperatureJ N limit weH studiedL above. And it is easy to notice that the chemical potential Μ plays the role of ΕF here. In other words, Μ = ΕF at T = 0. 1 (b) At any T, f Μ = 2 1 1 1 f Μ = H L = = Μ-Μ (6.26) exp + 1 exp 0 + 1 2 kB T H L (c) At T < Μ kBJ , f ΕN is close toH aL step function. Only the part with Ε~Μ shows strong deviation from the step function fT=0 Ε
H L 1 1 H L Plot . T ® 0.0000001, . T ® .01, Ε- Ε- Exp 1 + 1 Exp 1 + 1 T T B: 1 1 1 A E . T ® .1, . AT ® 1E, . T ® 100 , Ε- Ε- Ε- Exp 1 + 1 Exp 1 + 1 Exp 1 + 1 T T T Ε, 0, 2 , PlotStyle ® Thick, AxesLabel ® "Ε Μ", "f Ε " ,LabelStyle> ® Medium, A E A E A E 1 Μ 1 Μ Μ Μ 8PlotLegends< ® "T=0", "T= ", "T= 8 ","T= "H,L"T<=100 " 100 kB 10 kB kB kB : >F f Ε 1.0 H L T=0 0.8 Μ T= 1 100 kB 0.6 Μ T= 1 10 kB Μ 0.4 T= kB Μ T=100 0.2 kB
Ε Μ 0.5 1.0 1.5 2.0 6.1.5. Free electrons in 3D at zero temperature For free electrons, the Hamiltonian in 3D is
2 2 2 2 2 2 2 2 2 2 2 p px + py + pz 1 ¶ ¶ ¶ Ñ ¶ ¶ ¶ H = = = -ä Ñ + -ä Ñ + -ä Ñ = - + + (6.27) 2 m 2 m 2 m ¶ x ¶ y ¶ z 2 m ¶ x2 ¶ y2 ¶ z2 One electron in 3D B F The Schrodinger equation is Ñ2 ¶2 ¶2 ¶2 - + + Ψ® x, y, z = Ε Ψ® x, y, z (6.28) 2 m ¶ x2 ¶ y2 ¶ z2 k k The solutions for this equation areH 3D planeL wavesH L
Ψ® x, y, z = exp ä kx x + ky y + kz z (6.29) k A
H L A I ME 38 Phys463.nb
Considering periodic boundary conditions
Ψ® x + Lx, y, z = Ψ® x, y, z (6.30) k k
Ψ®Hx + Ly, y, zL = Ψ® Hx, y, zL (6.31) k k
Ψ®Ix + Lz, y, zM = Ψ® Hx, y, zL (6.32) k k
We findH that the momentaL areH quantizedL 2 l Π 2 m Π 2 n Π kx = and ky = and kz = (6.33) Lx Ly Lz
where l, m, and n are integers. The eigen-energy is also quantized:
2 2 2 2 2 2 2 2 k kx + ky + kz Ñ 4 Π 4 Π 4 Π 2 2 2 2 2 Εn = Ñ = Ñ = l + m + n (6.34) 2 2 2 2 m 2 m 2 m Lx Ly Lz
For a very large system L ® ¥ , the discrete energy and momenta turn into continuous variables. Many electrons in 3D at T = 0 H L At T = 0, the electrons occupy the lowest N quantum states to save energy. In other words, the quantum states with Ε £ ΕF are occupied, while states with Ε > ΕF are empty.
2 2 Since Ε = Ñ k 2 m, this means that states with momentum k £ kF are occupied and states with k > kF are empty. Here kF = 2 ΕF m Ñ. In
other words, in the k-space, the occupied states form a sphere with radius kF . This sphere is known as the Fermi sphere (or the Fermi sea). Q: ΕF = ? kF = ? A: It is determined by the density of electrons. H L 4 Π 3 3 The total volume of the Fermi sphere is kF . Each quantum state occupies the volume 2 Π V, which comes from the uncertainty relation. So 3 the total number of quantum states is H L 4 Π 3 kF 3 3 V kF = (6.35) 2 Π 3 V 6 Π2 There are two electrons per state (because we have electrons with spin up and spin down), so the total number of electrons is H L 3 3 V kF V kF N = 2´ = (6.36) 6 Π2 3 Π2 So
N 1 3 2 kF = 3 Π (6.37) V
kF is determined by the electron density N V. The Fermi energy Ñ2 Ñ2 N 2 3 2 2 ΕF = kF = 3 Π (6.38) 2 m 2 m V The Fermi velocity
â Ε Ñ Ñ N 1 3 2 vF = k=k = kF = 3 Π (6.39) Ñ â k F m m V
Here, the formula v = âΕ is nothing but the definition of the group velocity. Notice that Ε Ñ = Ω and v = â Ω â k is precisely the definition of the Ñ âk group velocity. Phys463.nb 39
Another equivalent definition of the Fermi velocity is
pF vF = (6.40) m If the energy Ε is a quadratic function of k Ε = Ñ2 k2 2 m , these two definitions are identical. If Ε is NOT a quadratic function of k (which could âΕ happen as will be discussed in the next chapter), we use the first definition v = = F Ñ âk k kF H L 2 2 The bottom line: the first definition is more generic. The second one pF m only works for Ε = Ñ k 2 m.
6.1.6. Conductivity H L ® Under a static electric field (E), electrons feel a constant force. ® ® F = -e E (6.41) The second Law of Newton tell us that
® ® â P F = (6.42) â t ® If we have a constant force, P increases linearly as a function of t ® ® ® P t = P t = 0 + F t (6.43)
® ® In quantumH L mechanics,H L P = Ñ k, so we have ® ® ® ® F ® E k t = k t = 0 + t = k t = 0 - e t (6.44) Ñ Ñ Here,H ifL weH turn Lon the E fieldH atL t = 0, at t £ 0, the Fermi sea is a sphere centered at k = 0. At t > 0, the Fermi sphere is centered at ® ® ∆k t = -e E t Ñ, because the wavevector of every electron is shifted by this amount at time t. For free electrons in a perfect crystal, ∆k t ® ¥ as t ® ¥. However, in a reals solid, ∆k t will NOT diverge, because there are collisions between electrons,H L between electrons and phonons, and between electrons and impurities. These scatterings will reduce the momentum towards ∆k = 0. They will introduce viscosity to the FermiH L liquid, which reduces the velocity and momentumH L of the liquid. Eventually, the scatterings and the E- field will balance each other and ∆k will turn to a fixed value The total force can be written as
® ® ® m ® ® p F = -e E - v = -e E - (6.45) Τ Τ ® Here the first term is the electric force and the second term is a viscosity term, which is proportional to -v, reducing the speed. The coefficients m and Τ are the electron mass and the collision time (average time between two collisions). Using the second law of Newton:
® ® ® â p ® p - p0 = -e E - (6.46) â t Τ ® ® ® where p0 is the momentum before we turn on the electric field and we assume the initial condition p t = 0 = p0. It is easy to find that the solution to this equation is H L ® ® ® -t Τ p = p0 - e E Τ 1 - ã (6.47)
This means that the change of momentum I M ® ® ® ® -t Τ ∆p = p - p0 = -e E Τ 1 - ã (6.48)
In the static limit t ® ¥ , I M
H L 40 Phys463.nb
® ® ∆p = -e E Τ (6.49) The change in wavevector
® ® e E ∆ k = - Τ (6.50) Ñ This approximation is known as the Drude approximation (the Drude model). The classical theory (ignore the fact that electrons are fermions) The total electric current is
® 2 ® Ñ ∆ k e n Τ ® j = -e n v = -e n = E (6.51) m m This is the Ohm’s Law
® ® j = Σ E (6.52) where the conductivity
e2 n Τ Σ = (6.53) m The quantum theory (taken into account the Pauli exclusive principle) Assuming that the E field is along the z direction. The total velocity along z is
e e 2 2 Π Π 2 j = - e n v = - N v = - â Φ â Θ kF ∆ k cos Θ ´vF cos Θ = V V 2 Π 3 V 0 0 (6.54) 2 m vF X \ 2 [ _ Π 2 2 2 k 2 3 kF vF ∆ k àkF vF ∆àk kF vF e E kF vF F Ñ e kF -e â Θ cosH2 Θ L= -e = -e - Τ = e 2 Τ E = e 2 Τ E = Τ E 2 Π2 0 4 Π2 4 Π2 Ñ 4 Π2 Ñ 4 Π2 m m 4 Π2
Because à N 1 3 2 kF = 3 Π (6.55) V
N 2 3 2 3 Π2 2 e kF e V 3 e n Τ j = Τ E = Τ E = E (6.56) m 4 Π2 m 4 Π2 4 m So
3 e 2 n Τ Σ = (6.57) 4 m Typically, we absorb the extra 3/4 factor into the definition of Τ, so the conductivity turns back to its classical form
e 2 n Τ Σ = (6.58) m which is identical to the classical formula shown above.
Sometimes, we use the mean free path l = vF Τ in the formula, instead of Τ. The physical meaning of the mean free path is the average distance that an electron travels between two collisions. Using l, we have e 2 n l Σ = (6.59) m vF The resistivity Ρ is Phys463.nb 41
1 m 1 Ρ = = (6.60) Σ e 2 n Τ Collision time Τ The collision time comes from
collisions between electrons Τee
collisions between electrons and phonons: Τep
collisions between electrons and impurities: Τi If we assume that different scatterings are independent, the total Τ satisfies 1 1 1 1 = + + (6.61) Τ Τee Τep Τi
Notice that m 1 Ρ = (6.62) e 2 n Τ So m 1 1 1 m 1 m 1 m 1 Ρ = + + = + + (6.63) 2 2 2 2 e n Τee Τep Τi e n Τee e n Τep e n Τi
If we define m 1 m 1 m 1 Ρee = and Ρep = and Ρi = (6.64) 2 2 2 e n Τee e n Τep e n Τi so we found
Ρ = Ρee + Ρep + Ρi (6.65)
Resistivity in a real metal at finite temperature In a real metal at finite temperature, the approximation used above are still valid, but the collision time Τ will show temperature-dependence. As a result, Ρ is a function of T. Typically, Ρ decreases as T goes down. At T = 0, Ρ goes to a finite value Ρ T = 0 , which is known as the residue resistivity. 2 5 At low T, Ρee and Ρep decreases to 0 as T is reduced down to 0, while Ρi is T independent. Very typically, ΡHee µ TL , Ρep µ T and Ρi µ constant 2 5 Ρ = Ρi + Aee T + Aep T (6.66)
At low temperature, Ρi >> Ρee >> Ρep, so impurity scattering is the dominate contribution for Ρ. At high temperature, Ρ is typically a linear function of T Ρ = A T (6.67)
This is because the number of phonons is proportional to T at high temperature.
1 1 kB T n = » = (6.68) exp Ε - 1 1 + Ε + O T-2 - 1 Ε kB T kB T
J N J H LN 6.1.7. the Hall effect (part I) Consider a thin sample (electrons moving in 2D). We apply a magnetic field perpendicular to the plane (in the z direction)
® ® ® ® F = -e E + v ´ B (6.69)
The secondK term is theO Lorentz force, which is perpendicular to the velocity. If we pass a current along the x direction. The electrons will feel a Lorentz force along the y direction and thus positive (negative) charge will accumulate along the top and bottom edge of the sample. This charge accumulation will induce a electric field along y, perpendicular to the current. If we have a static current, this electric field will balance the :Lorentz force. 42 Phys463.nb The second term is the Lorentz force, which is perpendicular to the velocity. If we pass a current along the x direction. The electrons will feel a Lorentz force along the y direction and thus positive (negative) charge will accumulate along the top and bottom edge of the sample. This charge accumulation will induce a electric field along y, perpendicular to the current. If we have a static current, this electric field will balance the :Lorentz force. ` ® ® ` -e Ey ey = e v ´ B = -e vx B ey (6.70)
So
Ey = vx B (6.71)
The current along x is
jx = -e n vx (6.72)
So the ratio between Ey and jX , which is called the Hall resistance, is
Ey vx B 1 Rxy = = = - B (6.73) jx -e n vx e n
The ratio between Rxy and B is known as the Hall coefficient
Rxy 1 RH = = - (6.74) B e n
RH is proportional to 1 n (inverse of the electron density), this measurement is the standard technique to determine the density of electrons in a material. 1 n = - (6.75) e RH Notice that the density n here is 2D electron density (number of electron per area)
Ne n = n2 D = (6.76) A For a 3D material, the 3D electron density is
Ne Ne n = n3 D = = = n2 D d (6.77) V A d where d is the thickness of the sample.
6.1.8. the Hall effect (part II with scatterings) In the calculations above, we didn’t consider the electron collisions. If we treat the collisions using the Drude approximation (assuming that the collisions are described by a single parameter Τ, the collision time), the force on an electron is
® ® ® ® m ® F = -e E + v ´ B - v (6.78) Τ Using the secondK law ofO Newton, we find that
® â v ® ® ® m ® m = -e E + v ´ B - v (6.79) â t Τ So we have two Kequations ofO motion
â vx vx m + + e vy B = -e Ex (6.80) â t Τ
â vy vy m + - e vx B = -e Ey (6.81) â t Τ
Using the initial condition vx = vy = 0 at t = 0, the solution of these differential equations is
B e t B e t e Τ Ex m - B e Ey Τ t e Τ -Ex m + B e Ey Τ cos + Ey m + B e Ex Τ sin - m m vx = - - ã Τ (6.82) m2 + B2 e2 Τ2 m2 + B2 e2 Τ2 I M AI M I M E Phys463.nb 43
B e t B e t e Τ Ey m + B e Ex Τ t e Τ Ey m + B e Ex Τ cos + Ex m - B e Ey Τ sin - m m vy = - + ã Τ (6.83) m2 + B2 e2 Τ2 m2 + B2 e2 Τ2 I M AI M I M E We can define the cyclotron frequency Ωc = e B m, so that
e Τ Ex - Ωc Τ Ey t e Τ -Ex + Ωc Τ Ey cos Ωc t + Ey + Ωc Τ Ex sin Ωc t - vx = - - ã Τ (6.84) 2 2 2 2 m 1 + Ωc Τ m 1 + Ωc Τ I M H L I M H L
e Τ Ey + Ωc Τ Ex t e Τ Ey + Ωc Τ Ex cos Ωc t + Ex - Ωc Τ Ey sin Ωc t - vy = - + ã Τ (6.85) 2 2 2 2 m 1 + Ωc Τ m 1 + Ωc Τ I M H L I M H L The static solution at t ® ¥ is
e Τ Ex - Ωc Τ Ey vx = - (6.86) 2 2 m 1 + Ωc Τ
e Τ Ey + Ωc Τ Ex vy = - (6.87) 2 2 m 1 + Ωc Τ
If the current is along x vy = 0 , we find that
Ey = -Ωc Τ Ex (6.88) I M So
2 2 e Τ Ex - Ωc Τ Ey e Τ Ex + Ωc Τ Ex e Τ vx = - = - = - Ex (6.89) 2 2 2 2 m 1 + Ωc Τ m 1 + Ωc Τ m Therefore,
Ey Ey -Ωc Τ Ex 1 m Ωc 1 m e B 1 R = = = = = - = - = - B xy e Τ (6.90) jx -n e vx e n E e n e e n e m e n m x And thus
Rxy 1 RH = = - (6.91) B e n
6.1.9. Free electrons in 3D at finite temperature Number of electrons on a quantum state with energy Ε (the Fermi-Dirac distribution function) 1 f Ε = Ε-Μ (6.92) exp + 1 kB T H L Notice that 0 £Jf Ε N£ 1, which agrees with the Pauli exclusive principle. Density of states (DOS) H L How many states do we have in the window of Ε ® Ε + â Ε?
D Ε â Ε = ? (6.93)
To answer this question, let’s consider the total number of electrons H L ΕF N = D Ε â Ε (6.94) 0
Total numberà ofH electronsL can also be written as 44 Phys463.nb
® 2 â k kF 4 Π k â k V kF V ΕF â k 2 2 N = 2´ ® = 2 V = k â k = k â Ε (6.95) 3 3 2 2 k £kF 2 Π V 0 8 Π Π 0 Π 0 â Ε
We know thatá á à à à à H L â Ε â Ñ2 k2 Ñ2 = = k (6.96) â k â k 2 m m So
2 V ΕF â k V ΕF k â Ε m V ΕF N = k2 â Ε = = k â Ε 2 2 2 3 2 (6.97) Π 0 â Ε Π 0 Ñ k Π Ñ 0 m à à à Because Ε = Ñ2 k2 2 m, k = 2 m Ε Ñ
3 2 3 2 m V ΕF m V ΕF ΕF 2 m V ΕF V 2 m ΕF N = k â Ε = 2 m Ε Ñ â Ε = Ε â Ε = Ε â Ε = D Ε â Ε (6.98) Π2 Ñ2 0 Π2 Ñ2 0 0 2 Π2 Ñ3 0 2 Π2 Ñ2 0 H L So à à à à à H L V 2 m 3 2 D Ε = Ε (6.99) 2 Π2 Ñ2 AnotherH L way to get D Ε is â N D Ε = H L (6.100) â Ε BecauHseL V k3 V 2 m Ε 3 2 N = = (6.101) 3 Π2 3 Π2 Ñ3 H L 3 3 2 â N 3 V 2 m 2 V 2 m D Ε = = Ε = Ε (6.102) â Ε 2 3 Π2 Ñ3 2 Π2 Ñ2 H L Here, HweL can also prove that D Ε = 3 N 2 Ε,
â N 3 V 2 m 3 2 3 V 2 m Ε 3 2 3 N D Ε = = H L Ε = = (6.103) â Ε 2 3 Π2 Ñ3 2 Ε 3 Π2 Ñ3 2 Ε H L H L TotalH energyL ¥ U T = Ε f Ε D Ε â Ε (6.104) 0
HeatH capacityL à H L H L ¶U ¥ â f Ε CV = V = Ε D Ε â Ε (6.105) ¶T 0 â T H L ¥ H Lâ f Ε First we prove that àCV = Ε - Μ D Ε â Ε 0 âT H L It is easy to prove that Ù H L H L ¥ â f Ε â ¥ â N D Ε â Ε = f Ε D Ε â Ε = = 0 (6.106) 0 â T â T 0 â T H L (becauseà the numberH L of electronsà are Hfixed,L H soL â N = 0) âT Therefore, Phys463.nb 45
¶U ¥ â f Ε CV = v = Ε D Ε â Ε = ¶T 0 â T H L (6.107) ¥ â f Ε ¥ â f Ε ¥ â f Ε ¥ â f Ε Ε à D Ε â Ε - ΜH ´L 0 = Ε D Ε â Ε - Μ´ D Ε â Ε = Ε - Μ D Ε â Ε 0 â T 0 â T 0 â T 0 â T H L H L H L H L We know thatà H L H L à H L à H L à H L H L
â f Ε â 1 1 2 Ε - Μ Ε - Μ = = exp Ε-Μ Ε-Μ 2 (6.108) â T â T exp + 1 exp + 1 kB T kB T H L kB T kB T B F â f Ε â J 1 N J 1N 2 Ε - Μ 1 = = - exp Ε-Μ Ε-Μ (6.109) â Ε â Ε exp + 1 exp + 1 kB T kB T H L kB T kB T B F So, J N J N â f Ε â f Ε Ε - Μ = - (6.110) â T â Ε T H L H L As a result,
¥ â f Ε Ε - Μ ¥ Ε - Μ 2 â f Ε CV = - Ε - Μ D Ε â Ε = - D Ε â Ε (6.111) 0 â Ε T 0 T â Ε H L H L H L â f Ε At T << Ε àk ,H Lhas a very high peakH L at Ε »àΕ . So we onlyH Lneed to consider the integral near Ε = Ε . Therefore, we can set D Ε » D Ε . F B âΕ F F F H L This approximation is very good for most metals, where ΕF kB » 100 T or larger. So we have 2 2 H L H L ¥ Ε - Μ â f Ε 0 Ε - Μ D ΕF 1 2 CV = -D ΕF â Ε = -D ΕF â f = Ε - Μ â f (6.112) 0 T â Ε 1 T T 0 H L H L H L H L H L à H L à à H L 46 Phys463.nb
- +Ε 2 - +Ε 2 Sech 1 Sech 1 2 t 2 t Plot . t ® .005, . t ® .01, 4B t F 4B t F - +Ε 2 - +Ε 2 SechB: 1 Sech 1 2 t 2 t . t ® .1, . t ® 1 , Ε, 0, 2 , PlotRange ® 0, 60 , 4B t F 4B t F â> f 8Ε < 8 < PlotStyle ® Thick, AxesLabel ® "Ε Μ", "-Μ " , LabelStyle ® Medium, âΕ 1 Μ 1 Μ Μ H L PlotLegends ® "T= ", "T= : ", "T= " , PlotPoints> ® 1000 100 kB 10 kB kB : > F â f Ε -Μ âΕ 60 H L
50
Μ T= 1 40 100 kB Μ T= 1 30 10 kB Μ T= 20 kB
10
Ε Μ 0.0 0.5 1.0 1.5 2.0 Notice that
-1 kB T ln f - 1 = Ε - Μ (6.113)
D ΕF 1 I M 2 CV = Ε - Μ â f = T 0 H L 2 2 (6.114) D ΕF 1 1 Π Π à H L -1 2 2 -1 2 2 2 kB T ln f - 1 â f = D ΕF kB T ln f - 1 â f = D ΕF kB T = D ΕF kB T T 0 0 3 3 H L So CV is a linearà functionA ofI T at T < We know that D ΕF = 3 N 2 ΕF , so 2 2 2 Π Π 3 N Π kB T 2 2 CV = D ΕHF LkB T = kB T = N kB (6.115) 3 3 2 ΕF 2 ΕF Define Fermi temperatureH L TF = ΕF kB 2 2 Π kB T Π T CV = N kB = N kB (6.116) 2 ΕF 2 TF Heat capacity in a real metal In a real metal, the heat capacity takes this form at low T 3 CV = Γ T + A T (6.117) Here, the first term comes form electrons and the second one comes from acoustic phonons. At very low temperature, ΓT >> A T3, so the electron contribution dominates the heat capacity. Phys463.nb 47 2 In experiments, one can plot the curve CV T vs T . This curve is a linear function y = Γ + A x (6.118) We can determine the slope A and the intercept Γ by fitting the experimental data. Thermal effective mass According to the theory of free electron gases, the coefficient Γ is 2 2 Π N kB Γ = (6.119) 2 ΕF Notice that Ñ2 Ñ2 N 2 3 2 2 ΕF = kF = 3 Π (6.120) 2 m 2 m V So 2 2 Π N kB 1 Γ = 2 3 (6.121) 3 Π2 N Ñ2 m V If we measure Γ in an experiment, we can use it to determine the mass of an electron I M Γ N 2 3 m-1 = 3 Π2 Ñ2 (6.122) 2 2 Π N kB V In real metals, this mass is typically different from the real electron mass (by a factor of 0.3-2). To distinguish the mass determined using this formula from the true electron mass, we call this mass the thermal effective mass mth . The reason that this mass is different from the real electron mass is because the electrons in a real metal are not free electrons. Instead, we have H L A lattice background Interactions between electrons Interactions between electrons and phonons Although we ignored all these effects, the free electron theory discussed above turns out to be valid in many solids (most metals and semi- conductors we have in our daily life). It turns out that indeed we can treat the electrons in a metal as a bunch of free fermions. These free fermionic particles are NOT electrons. They are called “quasi-particles”. Landau Fermi liquid theory The electrons in a metal can be considered as a gas formed by non-interacting quasi-particles. The total number of these quasi-particles is identical to the total number of electrons. (Known as the Luttinger theorem). These quasi-particles may carry electric charge ±e and they have spin-1/2 (they are charged fermions). The mass of these quasi-particles (the effective mass) differs from the electron mass. The value of the effective mass depends on materials, electron density, temperature, etc. (Not going to be discussed in this course) The quasi-particles have a Fermi residue which is smaller than 1 but larger than 0. Fermi residue is defined as Z = Lim∆®0 f Μ - ∆ - f Μ + ∆ (6.123) where f Ε is the distribution function of the particles. For free fermions, Z = 1, because f Μ - ∆ = 1 and f Μ + ∆ = 0. But for interacting @ H L H LD fermions, this value will be reduced by the interaction effect. However, for a Landau Fermi liquid, the value of Z shall always be positive 0 < Z Non-Fermi liquids It is also worthwhile to notice that there are some other materials, in which these Landau Fermi liquid theory totally fails. This liquids are known as the non-Fermi liquids. Landau Fermi liquid is very powerful, but it also enforce certain limitations to our materials. For example, as will be discussed in latter parts, some materials can turn into a superconductor (a conductor with zero resistivity) at low temperature T < TC. For Landau’s Fermi liquids, the value of Tc cannot exceeds ~10 - 20 K about - 260 degree , which implies that we cannot use these superconductors on an industrial scale, because we cannot run our machines at such a low temperature. In the 80s, people realized that for non-Fermi liquids, such a constrain will no longer exist, so we can have superconductorsH at a much higher temperature.L These materials are known as high temperature superconductors (high Tc). Currently, the highest Tc is about 200 K (about -70 degree). Increase the transition temperature from a few K to 200 K is very important. The reason is because to cool down a system down to a few K, one need to use very expensive methods (e.g. liquid Helium). But to reach the temperature of 100-200K is extremely easy. One just need to dip the sample into liquid Nitrogen. Nitrogen is very easy to obtain (78% of our air is made by Nitrogen). The price for liquid Nitrogen is comparable with beer. The boiling point of Nitrogen is about 77K. Liquid Nitrogen can help us reach 77K without any difficulty at low cost. So raising TC above 77K is a very important revolution, which is achieved in the late 80s, using non-Fermi-liquid materials. 6.1.10. Thermal conductivity In the last chapter, we learned that the thermal conductivity for a gas is: 1 C K = v l (6.124) 3 V We used this formula to compute the thermal conductivity of phonons. Here, we use the same formula to compute the thermal conductivity of electrons. The heat conductivity of electrons is 2 Π kB T Cel = N kB (6.125) 2 ΕF and we also known that the mean free path l = vF Τ. So 1 2 1 C 1 Π2 k T Π2 N k 2 T m vF Π2 n k 2 T B B 2 B (6.126) Kel = vF l = N kB vF vF Τ = Τ = Τ 3 V 3 V 2 ΕF 3 V m ΕF 3 m In a clean metal Kelectron >> Kphonon . In dirty Hmetals,L Kelectron is small because the collision time Τ is small. In dirty metals, Kelectron » Kphonon . In a metal, Kmetal = Kelectron + Kphonon >> Kphonon . In an insulator, Kelectron = 0, so Kinsulator = Kphonon . So we have Kmetal >> Kinsulator . This is why metal is cold to touch in the winter. The Wiedemann–Franz law. The ratio between the heat conductivity and the electric conductivity is 2 2 Π n kB T Τ 2 2 K 3 m Π kB T = = (6.127) 2 2 Σ e n Τ 3 e m We can define the Lorenz number: 2 2 Π kB L = = 2.44´10-8 W W K2 (6.128) 3 e 2 and the ratio between K and Σ is just K = L T (6.129) Σ This relation is known as the Wiedemann–Franz law. In real materials (metals), the value of L is typically very close to 2.44´10-8 W W K2.