5. Exam Winter 2010 A 5.1 Problems 1. Two identical non-interacting particles each having mass M are confined in a one dimensional parabolic poten tial given by V (x)= 1 2 Mω 2 x 2 , (5.1) where the angular frequency ω is a constant. a) Calculate the canonical partition function of the system Z c,B for the case where the particles are Bosons. b) Calculate the canonical partition func tion of t he system Z c,F for the case where the particles are Fermions. 2. Consider a one dimensional gas con taining N non-interacting electrons mo ving along the x direction. The e lectrons are con fined to a section of length L. A t zero temperature τ =0calculatetheratioU/ε F between the total energy of the system U and the Fermi energy ε F . 3. Consider an ideal classical gas at temperature τ.Thesetofinternal eigenstates of each p article in the gas, when a magnetic field H is applied, contains 2 states having energies ε − = −µ 0 H and ε + = µ 0 H,wherethe magnetic moment µ 0 is a constant. Calculate the magnetization of the system, which is defined b y M = − µ ∂F ∂H ¶ τ , (5.2) where F is the Helmholtz free energy. 4. (Note: replace this with Ex. 3.9 in the lecture notes) Consider an ideal gas made of N electrons in the extreme relativistic limit. The gas is contained in a box having a cube shape with a volume V = L 3 . In the extreme relativistic limit the dispersion relation ε (k)ismodified: the energy ε of a single particle quantum state having a wavefunction ψ given by ψ (x, y, z)= µ 2 L ¶ 3/2 sin (k x x)sin(k y y)sin(k z z) , (5.3) Chapter5. ExamWinter2010A is given by ε (k)=~kc , (5.4) where c is the speed of ligh t and where k = q k 2 x + k 2 y + k 2 z (contrary to the non-r elativistic case where it is given by ε (k)=~ 2 k 2 /2M). The system is in thermal equilibrium at zero temperature τ =0.Calculate the ratio p/U between the pressure p and the total energy of the system U. 5. Consider a mixture of two classical ideal gases, consisting of N A particles of type A and N B particles of type B. T he heat capacities c p,A and c V,A (c p,B and c V,B )atconstantpressureandatconstantvolumerespectively of gas A (B) are assumed to be temperature independent. The volume of the mixture is initially V 1 and the pressure is initially p 1 . The mixture undergoes an adiabatic (slow) and isentropic (at a constant en t ropy) process leading to a final volume V 2 . Calculate the final pressure p 2 . 5.2 Solutions 1. The single pa rticle eigen en ergies are given by  n = ~ω µ n + 1 2 ¶ , (5.5) where n =0, 1, 2, ···. a) For Bosons Z c,B = 1 2 ∞ X n=0 ∞ X m=0 exp [−β ( n +  m )] + 1 2 ∞ X n=0 exp (−2β n ) = 1 2 à ∞ X n=0 exp (−β n ) ! 2 + 1 2 ∞ X n=0 exp (−2β n ) = exp (−β~ω) 2(1− exp (−β~ω)) 2 + exp (−β~ω) 2(1− exp (−2β~ω)) . (5.6) Note that the av erage energy U B is given by U B = − ∂ log Z c,B ∂β = ~ω 1+2e −2β~ω + e −β~ω 1 −e −2β~ω . (5.7) b) For Fermions Z c,F = 1 2 ∞ X n=0 ∞ X m=0 exp [−β ( n +  m )] − 1 2 ∞ X n=0 exp (−2β n ) = exp (−β~ω) 2(1− exp (−β~ω)) 2 − exp (−β~ω) 2(1− exp (−2β~ω)) . (5.8) Eyal Buks Thermodynamics and Statistical Physics 148 5.2. Solutions Note that for this case the a verage energy U F is given by U F = − ∂ log Z c,F ∂β = ~ω 2+e −2β~ω + e −β~ω 1 −e −2β~ω . (5.9) 2. Th e orbital eigenen ergies are given by ε n = ~ 2 2m ³ π L ´ 2 n 2 , (5.10) where n =1, 2, 3, ···. The grandcanonical par tition functio n of the gas is given by Z gc = Y n ζ n , (5.11) where ζ n = Y l (1 + λ exp (−βε n )exp(−βE l )) (5.12) is the orbital grandcanonical Fermionic partition function where, λ =exp(βµ)=e −η , (5.13) is the fugacity, β =1/τ and {E l } are the eigenenergies of a particle due to internal degrees of freedom. For electrons, in the absen ce of magnetic field both spin states have the same energy, which is taken to be zero. Thus, log Z gc can be written as log Z gc = ∞ X n=1 log ζ n =2 ∞ X n=1 log (1 + λ exp(−βε n )) ' 2 ∞ Z 0 dn log µ 1+λ exp µ −β ~ 2 2m ³ π L ´ 2 n 2 ¶¶ . (5.14) By employing the variable transformation ε = ~ 2 2m ³ π L ´ 2 n 2 , (5.15) one has log Z gc = 1 2 ∞ Z 0 dεD(ε)log(1+λ exp (−βε)) , (5.16) Eyal Buks Thermodynamics and Statistical Physics 149 Chapter5. ExamWinter2010A where D (ε)= ( 2L π q 2m ~ 2 ε −1/2 ε ≥ 0 0 ε<0 (5.17) is the 1D density of states. Using Eqs. (1.80) and (1.94) for the energy U and the number of particles N ,namelyusing U = − µ ∂ log Z gc ∂β ¶ η , (5.18) N = λ ∂ log Z gc ∂λ , (5.19) one finds that U = ∞ Z −∞ dεD(ε) εf FD (ε) , (5.20) N = ∞ Z −∞ dεD(ε) f FD (ε) , (5.21) where f FD is the Fermi-Dirac distribution function [see Eq. (2.35)] f FD ()= 1 exp [β ( − µ)] + 1 . (5.22) At zero temperature, where µ = ε F one has U = D (ε F ) ε −1/2 F ε F Z 0 dεε 1/2 = 2D (ε F ) 3 ε 2 F , (5.23) N = D (ε F ) ε −1/2 F ε F Z 0 dεε −1/2 =2D (ε F ) ε F , (5.24) thus U ε F = N 3 . (5.25) 3. The Helmholtz free energy is given by F = Nτ µ log n n Q − log Z int − 1 ¶ , (5.26) where Z int =exp(βµ 0 H)+exp(−βµ 0 H)=2cosh(βµ 0 H)(5.27) Eyal Buks Thermodynamics and Statistical Physics 150 5.2. Solutions is the internal partition function. Thus the magnetization is given b y M = − µ ∂F ∂H ¶ τ = Nµ 0 tanh (βµ 0 H) . (5.28) 4. The grandcanonical partition function of the gas is giv en by Z gc = Y n ζ n , (5.29) where ζ n = Y l (1 + λ exp (−βε n )exp(−βE l )) (5.30) is a grandcanonical Fermionic partition function of an orbital having energy ε n given by ε n = π~cn L , (5.31) where n = q n 2 x + n 2 y + n 2 z , n x ,n y ,n z =1, 2, 3, ···, λ =exp(βµ)=e −η (5.32) is the fugacity, β =1/τ and {E l } are the eigenenergies of a particle due to internal degrees of freedom. For electrons, in the absen ce of magnetic field both spin states have the same energy, which is taken to be zero. Thus, log Z gc can be written as log Z gc = X l ∞ X n x =1 ∞ X n y =1 ∞ X n z =1 log (1 + λ exp (−βε n )exp(−βE l )) . (5.33) For a macroscopic system the sum over n can be approximately replaced by an integral ∞ X n x =0 ∞ X n y =0 ∞ X n z =0 → 4π 8 ∞ Z 0 dnn 2 , (5.34) thus, one has log Z gc =2 4π 8 ∞ Z 0 dnn 2 log µ 1+λ exp µ −β π~cn L ¶¶ . (5.35) By employing the variable transformation Eyal Buks Thermodynamics and Statistical Physics 151 Chapter5. ExamWinter2010A ε = π~cn L . (5.36) one has log Z gc = ∞ Z 0 dε Vε 2 π 2 ~ 3 c 3 log (1 + λ exp (−βε)) . (5.37) The energy U and the number of particles N are given b y U = − µ ∂ log Z gc ∂β ¶ η = ∞ Z 0 dε Vε 3 π 2 ~ 3 c 3 f FD () , (5.38) N = λ ∂ log Z gc ∂λ = ∞ Z 0 dε Vε 2 π 2 ~ 3 c 3 f FD () , (5.39) where f FD is the Fermi-Dirac distribution function [see Eq. (2.35)] f FD ()= 1 exp [β ( − µ)] + 1 . (5.40) At zero temperature U = ε F Z 0 dε Vε 3 π 2 ~ 3 c 3 = V π 2 ~ 3 c 3 ε 4 F 4 , (5.41) N = ε F Z 0 dε Vε 2 π 2 ~ 3 c 3 = V π 2 ~ 3 c 3 ε 3 F 3 , (5.42) and therefore U = 3N 4 ε F . (5 .43) The energy U can be expressed as a function of V and N as U = (3N) 4/3 ¡ π 2 ~ 3 c 3 ¢ 1/3 V −1/3 4 . At zero temperature the Helmholtz free energy F equals the energy U, thus the pressure p is given by p = − µ ∂F ∂V ¶ τ,N = − µ ∂U ∂V ¶ τ,N = 1 3 ¡ 3N V ¢ 4/3 ¡ π 2 ~ 3 c 3 ¢ 1/3 4 , (5.44) thus p U = 1 3V . (5.45) Eyal Buks Thermodynamics and Statistical Physics 152 5.2. Solutions 5. First, consider the case of an ideal gas made of a unique type of particles. Recall that the entropy σ, c V and c p are given by [see Eqs. (2.87), (2.88) and (2.89)] σ = N µ 5 2 +log n Q n + ∂ (τ log Z int ) ∂τ ¶ , (5.46) c V = N µ 3 2 + h int ¶ , (5.47) c p = c V + N, (5.48) where n = N/V is the density, n Q = µ Mτ 2π~ 2 ¶ 3/2 (5.49) is the quantum density, M is the mass of a particle in the gas, and h int = τ ∂ 2 (τ log Z int ) ∂τ 2 = c V N − 3 2 . (5.50) The requirement that h int is temperature independent leads to ∂ (τ lo g Z int ) ∂τ = g int + h int log τ τ 0 , (5.51) where both g int and τ 0 are constants. Using this notation, the change in entrop y due to a change in V from V 1 to V 2 and a change in τ from τ 1 to τ 2 is given by ∆σ = σ 2 − σ 1 = N à log V 2 τ 3/2 2 V 1 τ 3/2 1 + µ c V N − 3 2 ¶ log τ 2 τ 1 ! = N log à V 2 V 1 µ τ 2 τ 1 ¶ c V N ! . (5.52) Thus the total change in the entropy of the mixture is given by ∆σ = ∆σ A + ∆σ B = N A log à V 2 V 1 µ τ 2 τ 1 ¶ c V,A N A ! + N B log à V 2 V 1 µ τ 2 τ 1 ¶ c V,B N B ! (5.53) =(N A + N B )log   µ V 2 V 1 ¶µ τ 2 τ 1 ¶ c V,A +c V,B N A +N B   , (5.54) and the requirement ∆σ =0leadsto Eyal Buks Thermodynamics and Statistical Physics 153 Chapter5. ExamWinter2010A µ V 2 V 1 ¶µ τ 2 τ 1 ¶ c V,A +c V,B N A +N B =1. (5.55) Alternatively, by employing the equation of state pV = Nτ , (5.56) this can be rewritten as µ V 2 V 1 ¶ c V,A +c V,B N A +N B +1 µ p 2 p 1 ¶ c V,A +c V,B N A +N B =1, (5.57) or with the help of Eq. (5.48) as p 2 = p 1 µ V 2 V 1 ¶ − c V,A +c V,B N A +N B +1 c V,A +c V,B N A +N B = p 1 µ V 1 V 2 ¶ c p,A +c p,B c V,A +c V,B . (5.58) Eyal Buks Thermodynamics and Statistical Physics 154 6. Exam Winter 2010 B 6.1 Problems 1. Consider two particles, both having the same mass m,movinginaone- dimensional potential with coordinates x 1 and x 2 respectively. The po- tential energy is given by V (x 1 ,x 2 )= mω 2 x 2 1 2 + mω 2 x 2 2 2 + mΩ 2 (x 1 − x 2 ) 2 , (6.1) where the angular frequen cies ω and Ω are real constan ts Assume that the temperature τ of the system is sufficiently high to allow treating it classically. Calculate the following average values a)  x 2 1 ® for the case Ω =0. b)  x 2 1 ® , however without assuming that Ω =0. c) D (x 1 − x 2 ) 2 E , again without assuming that Ω =0. 2. Consider an ideal classica l gas containing N identical particles having each mass M in the extreme relativistic limit.Thegasiscontainedin avesselhavingacubeshapewithavolumeV = L 3 . In the extreme relativistic limit the dispersion relation ε (k)ismodified: the energy ε of a single particle quantum state having a wavefunction ψ given by ψ (x, y, z)= µ 2 L ¶ 3/2 sin (k x x)sin(k y y)sin(k z z) , (6.2) is given by ε (k)=~kc , (6.3) where c is the speed of ligh t and where k = q k 2 x + k 2 y + k 2 z (contrary to the non-r elativistic case where it is given by ε (k)=~ 2 k 2 /2M). The system is in thermal equilibrium at tem perature τ.Calculate: a) the total energy U of the system. b) the pressure p. [...]... n , V (6.8) where p, V and are the pressure, volume and temperature, respectively, and A and n are both constants Calculate the dierence cp cV between the heat capacities at constant pressure and at constant volume 6.2 Solutions 1 It is convenient to employ the coordinate transformation x1 + x2 x+ = , 2 x1 x2 x = 2 Eyal Buks Thermodynamics and Statistical Physics (6.9) (6 .10) 156 6.2 Solutions... Thermodynamics and Statistical Physics 157 Chapter 6 Exam Winter 2 010 B 2 The k vector is restricted due to boundary conditions to the values k= n , L (6.21) where n = (nx , ny , nz ) , (6.22) and nx , ny , nz = 1, 2, 3, ã ã ã The single particle partition function is given by Z1 = à ả (k) exp =1 X X X nx =1 ny =1 nz (6.23) Approximating the discrete sum by a continuous integral according to. .. the grandcanonical partition function Zgc is given by [see Eq (2.44)] log Zgc = Z1 , (6.26) where = exp (à) is the fugacity In terms of the Lagrange multipliers = à/ and = 1/ the last result can be rewritten as log Zgc = e V 2 ~3 c3 3 (6.27) a) The average energy U and average number of particle N are calculated using Eqs (1.80) and (1.81) respectively Eyal Buks Thermodynamics and Statistical Physics. .. exp (2à0 H)] + 2 exp (J) (6.37) Eyal Buks Thermodynamics and Statistical Physics 159 Chapter 6 Exam Winter 2 010 B and the magnetic susceptibility is given by = 4à2 0 1 + e2J (6.38) Note that in the high temperature limit J 1 ' 2à2 0 +J (6.39) 4 The orbital eigenenergies in this case are given by à ả 1 n = ~ n + , 2 (6.40) where n = 0, 1, 2, ã ã ã The grandcanonical partition function of the gas... à)] + 1 Thermodynamics and Statistical Physics (6.46) 160 6.2 Solutions a) At zero temperature the chemical potential à is the Fermi energy F , and the Fermi-Dirac distribution function becomes a step function, thus with the help of Eq (6.45) one nds that N= 2F , ~ (6.47) thus à = F = N ~ 2 (6.48) b) Using the approximation fFD () ' exp [ ( à)] , (6.49) for the the limit of high temperatures and approximating... where n = Y (1 + exp (n ) exp (El )) (6.42) l is the orbital grandcanonical Fermionic partition function where, = exp (à) = e (6.43) is the fugacity, = 1/ and {El } are the eigenenergies of a particle due to internal degrees of freedom For electrons, in the absence of magnetic eld both spin states have the same energy, which is taken to be zero Thus, log Zgc can be written as log Zgc = X log n... N ~ ~ à = log + 2 2 à ả N ~ ' log 2 5 Using Eq (2.239), which is given by à ả à ả p V , cp cV = V,N p,N (6.51) (6.52) one nds that cp cV = Eyal Buks A2 2 2(n1) n = n2 A n1 pV Thermodynamics and Statistical Physics (6.53) 161 ... x2 1 2 (6.14) and Thus, the kinetic energy T of the system is given by Ă Â Â Ă m x2 + x2 + m x2 + x2 1 2 = , T = 2 2 (6.15) and the potential energy V is given by m 2 x2 m 2 x2 1 2 + + m 2 (x1 x2 )2 2 2 Ă Â m 2 + 4 2 x2 m 2 x2 + + = 2 2 V (x1 , x2 ) = The equipartition theorem yields Ă ư đ Âư đ m 2 + 4 2 x2 m 2 x2 + = = , 2 2 2 thus D E 2 (x1 + x2 )2 = , m 2 (6.16) (6.17) (6.18) and D E (x1 ... lim H0 M , H (6.5) where M = à F H ả (6.6) is the magnetization of the system, and where F is the Helmholtz free energy 4 Consider a one dimensional gas containing N non-interacting electrons moving along the x direction The electrons are conned by a potential given by 1 V (x) = m 2 x2 , 2 (6.7) where m is the electron mass and where is the angular frequency of oscillations Calculate the chemical potential... Statistical Physics 158 6.2 Solutions U = N = thus à à log Zgc log Zgc ả = ả 3 log Zgc , = log Zgc , U = 3N , (6.28) (6.29) (6.30) and = log à V 3 2 N~3 c3 ả b) The entropy is evaluate using Eq (1.86) = log Zgc + U + N = N (1 + 3 + ) ã à ảá V 3 = N 4 + log , 2 N ~3 c3 and the Helmholtz free energy by the denition (1.116) ảá ã à V 3 F = U = N 1 + log 2 N ~3 c3 Thus the pressure p is given by . − 1 2 4Ω 2 ω 2 1+ 4Ω 2 ω 2 ! . (6.20) Eyal Buks Thermodynamics and Statistical Physics 157 Chapter6. ExamWinter2010B 2. The k vector is restricted due to boundary conditions to the values k = πn L , (6.21) where n. N B )log   µ V 2 V 1 ¶µ τ 2 τ 1 ¶ c V,A +c V,B N A +N B   , (5.54) and the requirement ∆σ =0leadsto Eyal Buks Thermodynamics and Statistical Physics 153 Chapter5. ExamWinter2010A µ V 2 V 1 ¶µ τ 2 τ 1 ¶ c V,A +c V,B N A +N B =1 (5.16) Eyal Buks Thermodynamics and Statistical Physics 149 Chapter5. ExamWinter2010A where D (ε)= ( 2L π q 2m ~ 2 ε −1/2 ε ≥ 0 0 ε<0 (5.17) is the 1D density of states. Using Eqs. (1.80) and (1.94)