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5 Quantum Statistics : Worked Examples

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5 Quantum Statistics : Worked Examples (5.1) For a system of noninteracting S =0 bosons obeying the dispersion ε(k)= ~v k . | | (a) Find the density of states per unit volume g(ε). (b) Determine the critical temperature for Bose-Einstein condensation in three dimensions. (c) Find the condensate fraction n0/n for T <Tc. (d) For this dispersion, is there a finite transition temperature in d = 2 dimensions? If not, explain why. If so, (d=2) compute Tc . Solution : (a) The density of states in d dimensions is ddk Ω εd−1 g(ε)= δ(ε ~vk)= d . (2π)d − (2π)d (~v)d Z (b) The condition for T = Tc is to write n = n(Tc,µ = 0): ∞ ∞ 2 3 g(ε) 1 ε ζ(3) kBTc n = dε = 2 3 dε = 2 . eε/kBTc 1 2π (~v) eε/kBTc 1 π ~v Z0 − Z0 − Thus, π2 1/3 k T = ~vn1/3 . B c ζ(3) (c) For T <Tc, we have ζ(3) k T 3 n T 3 n = n + B 0 =1 . 0 π2 ~v ⇒ n − T (n) c (d) In d =2 we have ∞ 1 ε ζ(2) k T 2 2πn n = dε = B c k T (d=2) = ~v . ~ 2 ε/k T ~ B c 2π( v) e B c 1 2π v ⇒ sζ(2) Z0 − 1 (5.2) Consider a three-dimensional Fermi gas of S = 1 particles obeying the dispersion relation ε(k)= A k 4. 2 | | (a) Compute the density of states g(ε). (b) Compute the molar heat capacity. (c) Compute the lowest order nontrivial temperature dependence for µ(T ) at low temperatures. I.e. compute the (T 2) term in µ(T ). O Solution : (a) The density of states in d =3, with g =2S+1=2, is given by ∞ 1 1 dk ε−1/4 g(ε)= dk k2 δ ε ε(k) = k2(ε) = . π2 π2 dε 2 3/4 − 1 4 4π A Z0 k=(ε/A) / (b) The molar heat capacity is 2 2 π π R kBT cV = Rg(εF) kBT = , 3n 4 · εF ~2 2 2 1/3 where εF = kF/2m can be expressed in terms of the density using kF = (3π n) , which is valid for any isotropic dispersion in d =3. In deriving this formula we had to express the density n, which enters in the denominator in the above expression, in terms of εF. But this is easy: εF 1 ε 3/4 n = dε g(ε)= F . 3π2 A Z0 (c) We have (Lecture Notes, 5.8.6) § 2 ′ 2 2 π 2 g (εF) π (kBT ) δµ = (kBT ) = . − 6 g(εF) 24 · εF Thus, 2 2 π (kBT ) 4 µ(n,T )= εF(n)+ + (T ) , 24 · εF(n) O ~2 2 2/3 where εF(n)= 2m (3π n) . 2 (5.3) A bosonic gas is known to have a power law density of states g(ε)= Aεσ per unit volume, where σ is a real number. (a) Experimentalists measure Tc as a function of the number density n and make a log-log plot of their results. They find a beautiful straight line with slope 3 . That is, T (n) n3/7. Assuming the phase transition they 7 c ∝ observe is an ideal Bose-Einstein condensation, find the value of σ. (b) For T <Tc, find the heat capacity CV . βµ (c) For T >Tc, find an expression for p(T,z), where z = e is the fugacity. Recall the definition of the polylog- arithm (or generalized Riemann zeta function)1, ∞ ∞ 1 tq−1 zn Li (z) dt = , q ≡ Γ(q) z−1et 1 nq n=1 Z0 − X ∞ where Γ(q)= dt tq−1 e−t is the Gamma function. 0 R (d) If these particles were fermions rather than bosons, find (i) the Fermi energy εF(n) and (ii) the pressure p(n) as functions of the density n at T =0. Solution : (a) At T = Tc, we have µ =0 and n0 =0, hence ∞ g(ε) 1+σ n = dε = Γ(1+ σ) ζ(1 + σ) A (kBTc) . eε/kBTc 1 −∞Z − 1 4 Thus, T n 1+σ = n3/7 which means σ = . c ∝ 3 (b) For T <Tc we have µ =0, but the condensate carries no energy. Thus, ∞ εg(ε) 2+σ E = V dε = Γ(2+ σ) ζ(2 + σ) A (kBT ) eε/kBT 1 −∞Z − 10 10 10/3 =Γ 3 ζ 3 A (kBT ) . Thus, 13 10 7/3 CV =Γ 3 ζ 3 A (kBT ) , where we have used z Γ(z)=Γ(z + 1). (c) The pressure is p = Ω/V , which is − ∞ ∞ − − p(T,z)= k T dε g(ε) ln 1 ze ε/kBT = A k T dε εσ ln 1 ze ε/kBT − B − − B − −∞Z Z 0 ∞ 1+σ A ε 2+σ = dε = Γ(1+ σ) A (kBT ) Li2+σ(z) 1+ σ z−1 eε/kBT 1 Z0 − 7 10/3 =Γ 3 A (kBT ) Li10/3(z) . 1 Some sources use the notation ζq (z) for the polylogarithm, but for those of you who have yet to master the scribal complexities of the Greek ζ, you can use the notation Liq(z) instead. 3 (d) The Fermi energy is obtained from ε F 1 3/7 Aε1+σ (1 + σ) n 1+σ 7n n = dε g(ε)= F ε (n)= = . 1+ σ ⇒ F A 3A Z0 We obtain the pressure from p = ∂E . The energy is − ∂V N εF 2+σ 1 AεF − E = V dε g(ε) ε = V V 1+σ . · 2+ σ ∝ Z0 1 E Thus, p = 1+σ V , i.e. · 2+σ Aε 3/7 p(n)= F = 3 7 A−3/7n10/7 . (1 + σ)(2 + σ) 10 3 4 (5.4) At low energies, the conduction electron states in graphene can be described as fourfold degenerate fermions with dispersion ε(k)= ~v k . Using the Sommerfeld expension, F| | (a) Find the density of single particle states g(ε). (b) Find the chemical potential µ(T,n) up to terms of order T 4. (c) Find the energy density (T,n)= E/V up to terms of order T 4. E Solution : (a) The DOS per unit volume is d2k 2ε g(ε)=4 δ(ε ~v k)= . (2π)2 − F π(~v )2 Z F (b) The Sommerfeld expansion is ∞ µ π2 7π4 dε f(ε µ) φ(ε)= dε φ(ε)+ (kT )2 φ′(µ)+ (k T )4 φ′′′(µ)+ .... − 6 360 B −∞Z −∞Z For the particle density, set φ(ε)= g(ε), in which case 1 µ 2 π k T 2 n = + B . π ~v 3 ~v F F The expansion terminates after the (T 2) term. Solving for µ, O 1/2 π k T 2 µ(T,n)= ~v (πn)1/2 1 B F − 3n ~v " F # π k T 2 π2 k T 4 = ~v (πn)1/2 1 B B + ... F − 6n ~v − 72n2 ~v ( F F ) (c) For the energy density , we take φ(ε)= εg(ε), whence E 2µ µ 2 πk T 2 (T,n)= + B E 3π ~v ~v " F F # π k T 2 π2 k T 4 = 2 √π ~v n3/2 1+ B B + ... 3 F 2n ~v − 8n2 ~v ( F F ) 5 1 (5.5) Consider a system of N spin- 2 particles occupying a volume V at temperature T . Opposite spin fermions may bind in a singlet state to form a boson: f + f ⇋ b ↑ ↓ with a binding energy ∆ < 0. Assume that all the particles are nonrelativistic; the fermion mass is m and the − boson mass is 2m. Assume further that spin-flip processes exist, so that the and fermion species have identical ↑ ↓ chemical potential µf . (a) What is the equilibrium value of the boson chemical potential, µb? Hint : the answer is µb =2µf . (b) Let the total mass density be ρ. Derive the equation of state ρ = ρ(µf ,T ), assuming the bosons have not condensed. You may wish to abbreviate ∞ zn Li (z) . p ≡ np n=1 X (c) At what value of µf do the bosons condense? (d) Derive an equation for the Bose condensation temperature T . Solve this equation for T in the limits ε ∆ c c 0 ≪ and ε ∆, respectively, where 0 ≫ π~2 ρ/2m 2/3 ε . 0 ≡ m ζ 3 2 (e) What is the equation for the condensate fraction ρ0(T,ρ)/ρ when T <Tc? Solution : (a) The chemical potential is the Gibbs free energy per particle. If the fermion and boson species are to coexist at the same T and p, the reaction f +f ⇋ b must result in ∆G = µ 2µ =0. ↑ ↓ b − f (b) For T >Tc, − − ρ = 2m λ 3 Li eµf /kBT +2√8 m λ 3 Li e(2µf +∆)/kBT , − T 3/2 − T 3/2 ~2 where λT = 2π /mkBT is the thermal wavelength for particles of mass m. This formula accounts for both −3 fermion spin polarizations, each with number density nf↑ = nf↓ = λT Li3/2( zf ) and the bosons with number −p3 β∆ 2 − − 3/2 density √8 λT Li3/2(zb e ), with zb = zf due to chemical equilibrium among the species. The factor of 2 = √8 arises from the fact that the boson mass is 2m, hence the boson thermal wavelength is λT /√2. (c) The bosons condense when µ = ∆, the minimum single particle energy. This means µ = 1 ∆. The b − f − 2 equation of state for T <Tc is then − − ρ = 2m λ 3 Li e−∆/2kBT +4√2 ζ 3 m λ 3 + ρ , − T 3/2 − 2 T 0 where ρ0 is the condensate mass density.
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