University:
Massachusetts Institute of Technology
Course:
8.333 | Statistical Mechanics I: Statistical Mechanics of Particles
Academic year:

2020
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68

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49
Author:

Nathan Ferguson

8.333: Statistical Mechanics I Re: 2007 Final Exam Review Problems The enclosed exams (and solutions) from the previous years are intended to help you review the material. ******** Note that the ﬁrst parts of each problem are easier than its last parts. Therefore, make sure to proceed to the next problem when you get stuck. You may ﬁnd the following information helpful: Physical Constants Electron mass Electron Charge Speed of light Boltzmann’s constant me ≈ 9.1 × 10−31 Kg Proton mass e ≈ 1.6 × 10−19 C Planck’s constant/2π c ≈ 3.0 × 108 ms−1 Stefan’s constant −23 −1 kB ≈ 1.4 × 10 JK Avogadro’s number Conversion Factors ˚ ≡ 10−10 m 1A 1atm ≡ 1.0 × 105 N m−2 mp ≈ 1.7 × 10−27 Kg h̄ ≈ 1.1 × 10−34 Js1 σ ≈ 5.7 × 10−8 W m−2 K −4 N0 ≈ 6.0 × 1023 mol−1 1eV ≡ 1.1 × 104 K Thermodynamics dE = T dS+dW For a gas: dW = −P dV For a ﬁlm: dW = σdA Mathematical Formulas limx→∞ coth x = 1 + 2e−2x + O e−4x limx→0 coth x = R∞ 0 dx xn e−αx = αn+1 h dx exp −ikx − −∞ R∞ e−ikx = η fm (z) = P∞ n=1 1 (m−1)! R∞ − limz→∞ fm (z) = + ζ m ≡ fm (1) x2 2σ 2 (−ik)n n! 0 ( 21 )! = n! i = h 2 2i √ 2πσ 2 exp − σ 2k �xn � m−1 (ln z)m m! h 1+ α+1 z α=1 η αm π2 m(m 6 ζ3/2 ≈ 2.612 α P∞ π2 6 1 x 3 + O x2 limN→∞ ln N ! = N ln N − N η dfm dz P∞ n=1 (−ik)n n! �xn �c η = 1z fm−1 i − 1)(ln z)−2 + · · · f2− (1) = ζ2 = + √ π 2 ln e−ikx = dx z−x1 ex −η = 1 x ζ5/2 ≈ 1.341 π2 12 f4− (1) = ζ3 ≈ 1.202 7π 4 720 ζ4 = π4 90 8.333: Statistical Mechanics I Fall 1998 Final Exam 1. Exciton dissociation in a semiconductor: By shining an intense laser beam on a semicon ductor, one can create a metastable collection of electrons (charge −e, and eﬀective mass me ) and holes (charge +e, and eﬀective mass mh ) in the bulk. The oppositely charged particles may pair up (as in a hydrogen atom) to form a gas of excitons, or they may dissociate into a plasma. We shall examine a much simpliﬁed model of this process. (a) Calculate the free energy of a gas composed of Ne electrons and Nh holes, at temper ature T , treating them as classical non-interacting particles of masses me and mh . • The canonical partition function of gas of non-interacting electrons and holes is the product of contributions from the electron gas, and from the hole gas, as Ze−h 1 = Ze Zh = Ne ! V λ3e Ne 1 · Nh ! V λ3h Nh , √ where λα = h/ 2πmα kB T (α =e, h). Evaluating the factorials in Stirling’s approximation, we obtain the free energy Nh 3 Ne 3 Fe−h = −kB T ln Ze−h = Ne kB T ln λ + Nh kB T ln λ . eV e eV h (b) By pairing into an excition, the electron hole pair lowers its energy by ε. [The binding energy of a hydrogen-like exciton is ε ≈ me4 /(2h̄2 ǫ2 ), where ǫ is the dielectric constant, −1 and m−1 = m−1 e + mh .] Calculate the free energy of a gas of Np excitons, treating them as classical non-interacting particles of mass m = me + mh . • Similarly, the partition function of the exciton gas is calculated as 1 Zp = Np ! V λ3p Np e−β(−Np ǫ) , leading to the free energy Fp = Np kB T ln where λp = h/ p Np 3 λ eV p − Np ǫ, 2π (me + mh ) kB T . (c) Calculate the chemical potentials µe , µh , and µp of the electron, hole, and exciton states, respectively. • The chemical potentials are derived from the free energies, through µe = ∂Fe−h ∂Ne T,V = kB T ln ne λ3e , 2 µh = µp = where nα = Nα /V (α =e, h, p). ∂Fe−h ∂Nh ∂Fp ∂Np T,V T,V = kB T ln nh λh3 , = kB T ln np λ3p − ǫ, (d) Express the equilibrium condition between excitons and electron/holes in terms of their chemical potentials. • The equilibrium condition is obtained by equating the chemical potentials of the electron and hole gas with that of the exciton gas, since the exciton results from the pairing of an electron and a hole, electron + hole ⇀ ↽ exciton. Thus, at equilibrium µe (ne , T ) + µh (nh , T ) = µp (np , T ) , which is equivalent, after exponentiation, to ne λ3e · nh λ3h = np λ3p e−βǫ . (e) At a high temperature T , ﬁnd the density np of excitons, as a function of the total density of excitations n ≈ ne + nh . • The equilibrium condition yields λe3 λ3h βǫ np = ne nh 3 e . λp At high temperature, np ≪ ne = nh ≈ n/2, and n 2 λ3 λ3 h3 np = ne nh e 3 h eβǫ = 2 (2πkB T )3/2 λp me + mh me mh 3/2 eβǫ . ******** 2. The Manning Transition: When ionic polymers (polyelectrolytes) such as DNA are immersed in water, the negatively charged counter-ions go into solution, leaving behind a positively charged polymer. Because of the electrostatic repulsion of the charges left behind, the polymer stretches out into a cylinder of radius a, as illustrated in the ﬁgure. While thermal ﬂuctuations tend to make the ions wander about in the solvent, electrostatic attractions favor their return and condensation on the polymer. If the number of counter ions is N , they interact with the N positive charges left behind on the rod through the 3 potential U (r) = −2 (N e/L) ln (r/L), where r is the radial coordinate in a cylindrical geometry. If we ignore the Coulomb repulsion between counter-ions, they can be described by the classical Hamiltonian N 2 r X pi 2 + 2e n ln , H= L 2m i=1 where n = N/L. 2a R + + + − − − − z − r + − L + − − + − + − − + (a) For a cylindrical container of radius R, calculate the canonical partition function Z in terms of temperature T , density n, and radii R and a. • The canonical partition function is ( ) N 2 r 3 3 X p d q d p i i i i Z= exp −β + 2e2 n ln L N !h3N 2m i=1 "Z #N N R 2 2 2πLe = LN·β2e n rdr · r −2e n/kB T N λ3 a " #N N 2(1−e2 n/kB T ) 2(1−e2 n/kB T ) 2 a − R 2πe = L2Ne nβ . nλ3 2 (1 − e2 n/kB T ) Z Q (b) Calculate the probability distribution function p (r) for the radial position of a counter ion, and its ﬁrst moment �r�, the average radial position of a counter-ion. 4 • Integrating out the unspeciﬁed N momenta and N − 1 positions from the canonical distribution, one obtains the distribution function re−(2e p (r) = R R a 2 n/kB T ) ln(r/L) drre−(2e2 n/kB T ) ln(r/L) (Note the normalization condition �r� = Z R rp (r) dr = a RR a 2 r 1−2e n/kB T e2 n . =2 1− kB T R2(1−e2 n/kB T ) − a2(1−e2 n/kB T ) drp(r) = 1.) The average position is then 2kB T − 2e2 n 3kB T − 2e2 n 2 2 R3−2e n/kB T − a3−2e n/kB T R2−2e2 n/kB T − a2−2e2 n/kB T ! . (c) The behavior of the results calculated above in the limit R ≫ a is very diﬀerent at high and low temperatures. Identify the transition temperature, and characterize the nature of the two phases. In particular, how does �r� depend on R and a in each case? • Consider ﬁrst low temperatures, such that e2 n/kB T > 1. In the R ≫ a limit, the distribution function becomes 1−2e2 n/kB T e2 n r , p (r) = 2 1 − kB T a2(1−e2 n/kB T ) and �r� ∝ a. To see this, either examine the above calculated average �r� in the R ≫ a limit, or notice that 2 e2 n x1−2e n/kB T dx, p (r) dr = 2 1 − kB T R∞ 2 where x = r/a, immediately implying �r� ∝ a (as 1 dxx1−2e n/kB T < ∞ if e2 n/kB T > 1). On the other hand, at high temperatures (e2 n/kB T < 1), the distribution function reduces to 1−2e2 n/kB T e2 n r , p (r) = 2 1 − kB T R2(1−e2 n/kB T ) and �r� ∝ R, from similar arguments. Thus, at temperature Tc = e2 n/kB there is a transition from a “condensed” phase, in which the counter-ions are stuck on the polymer, to a “gas” phase, in which the counter-ions ﬂuctuate in water at typical distances from the polymer which are determined by the container size. (d) Calculate the pressure exerted by the counter-ions on the wall of the container, at r = R, in the limit R ≫ a, at all temperatures. • The work done by the counter-ions to expand the container from a radius R to a radius R + dR is dW = dF = (force) dR = −P (2πRL) dR, 5 leading to 1 ∂F kB T ∂lnZ = . 2πRL ∂R 2πRL ∂R At low temperatures, T < Tc , the pressure vanishes, since the partition function is inde pendent of R in the limit R ≫ a. At T > Tc , the above expression results in P =− kB T P = 2N 2πRL i.e. e2 n 1− kB T 1 , R e2 n P V = N kB T 1 − . kB T (e) The character of the transition examined in part (d) is modiﬁed if the Coulomb in teractions between counter-ions are taken into account. An approximate approach to the interacting problem is to allow a fraction N1 of counter-ions to condense along the polymer rod, while the remaining N2 = N − N1 ﬂuctuate in the solvent. The free counter-ions are again treated as non-interacting particles, governed by the Hamiltonian N 2 r X pi 2 H= + 2e n2 ln , 2m L i=1 where n2 = N2 /L. Guess the equilibrium number of non-interacting ions, N2∗ , and justify your guess by discussing the response of the system to slight deviations from N2∗ . (This is a qualitative question for which no new calculations are needed.) • Consider a deviation (n2 ) from n∗2 ≡ N2∗ /V ≡ kB T /e2 , occuring at a temperature lower than Tc (i.e. e2 n/kB T > 1). If n2 > n∗2 , the counter-ions have a tendency to condensate (since e2 n/kB T > 1), thus decreasing n2 . On the other hand, if n2 > n∗2 , the counter-ions tend to “evaporate” (since e2 n/kB T < 1). In both cases, the system drives the density n2 to the (equilibrium) value of n∗2 = kB T /e2 . If the temperature is higher than Tc , clearly n∗2 = n and there is no condensation. ******** 3. Bose gas in d dimensions: Consider a gas of non-interacting (spinless) bosons with an energy spectrum ǫ = p2 /2m, contained in a box of “volume” V = Ld in d dimensions. (a) Calculate the grand potential G = −kB T ln Q, and the density n = N/V , at a chemical + potential µ. Express your answers in terms of d and fm (z), where z = eβµ , and + fm (z) 1 = Γ (m) Z ∞ 0 xm−1 dx. z −1 ex − 1 (Hint: Use integration by parts on the expression for ln Q.) 6 • We have Q= = ∞ X P eNβµ N=0 n =N i X i exp −β {ni } YX eβ(µ−ǫi )ni = i {ni } Y i X ni ǫi i ! , 1 1 − eβ(µ−ǫi ) P P i) . Replacing the summation whence ln Q = − i ln 1 − eβ(µ−ǫ i with a d dimensional h i R d d R d−1 d k dk, where Sd = 2π d/2 / (d/2 − 1)!, leads integration V d k/ (2π) = V Sd / (2π) to Z V Sd d−1 −β¯ h2 k2 /2m ln Q = − k dk ln 1 − ze . d (2π) p p The change of variable x = βh̄2 k 2 /2m (⇒ k = 2mx/β/h̄ and dk = dx 2m/βx/2h̄) results in d/2 Z V Sd 1 2m xd/2−1 dx ln 1 − ze−x . ln Q = − 2 d 2 h̄ β (2π) Finally, integration by parts yields V Sd 1 ln Q = d (2π) d 2m h̄2 β d/2 Z x d/2 ze−x Sd dx = V 1 − ze−x d i.e. Sd G = −kB T ln Q = −V d 2m h2 β d/2 kB T Γ 2m h2 β d/2 Z dx d (z) , + 1 f+ d 2 +1 2 which can be simpliﬁed, using the property Γ (x + 1) = xΓ (x), to G=− V (z) . kB T f + d 2 +1 λd The average number of particles is calculated as d/2 Z ∂ Sd 2m ze−x d/2−1 x N= ln Q =V dx 1 − ze−x ∂ (βµ) d h2 β , d/2 d V Sd 2m f+ =V Γ f+ d (z) = d (z) 2 2 λd 2 2 h2 β i.e. n= 1 + f d (z) . λd 2 (b) Calculate the ratio P V /E, and compare it to the classical value. 7 xd/2 , z −1 ex − 1 • We have P V = −G, while E=− ∂ d ln Q d ln Q = + = − G. ∂β 2 β 2 Thus P V /E = 2/d, identical to the classical value. (c) Find the critical temperature, Tc (n), for Bose-Einstein condensation. • The critical temperature Tc (n) is given by n= 1 + 1 f d (1) = d ζ d d λ 2 λ 2 for d > 2, i.e. h2 Tc = 2mkB n ζd 2 !2/d . (d) Calculate the heat capacity C (T ) for T < Tc (n). • At T < Tc , z = 1 and ∂E C (T ) = ∂T z=1 d ∂G =− 2 ∂T z=1 d =− 2 d +1 2 d G = T 2 d +1 2 V kB ζ d +1 . 2 λd (e) Sketch the heat capacity at all temperatures. • . (f) Find the ratio, Cmax /C (T → ∞), of the maximum heat capacity to its classical limit, and evaluate it in d = 3 8 • As the maximum of the heat capacity occurs at the transition, Cmax d = C (Tc ) = 2 d +1 2 V ζ d /n 2 kB f + d 2 +1 Thus Cmax = C (T → ∞) d +1 2 d (1) = N kB 2 ζ d +1 2 ζd d +1 2 ζ d +1 2 ζd . 2 , 2 which evaluates to 1.283 in d = 3. (g) How does the above calculated ratio behave as d → 2? In what dimensions are your results valid? Explain. + • The maximum heat capacity, as it stands above, vanishes as d → 2! Since fm (x → 1) → ∞ if m ≤ 2, the fugacuty z is always smaller than 1. Hence, there is no macroscopic occupation of the ground state, even at the lowest temperatures, i.e. no Bose-Einstein condensation in d ≤ 2. The above results are thus only valid for d ≥ 2. ******** 9 8.333: Statistical Mechanics I Fall 1999 Final Exam 1. Electron Magnetism: The conduction electrons in a metal can be treated as a gas of fermions of spin 1/2 (with up/down degeneracy), and density n = N/V . (a) Ignoring the interactions between electrons, describe (in words) their ground state. Calculate the fermi wave number kF , and the ground-state energy density E0 /V in terms of the density n. • In the ground state, the fermi sea is ﬁlled symmetrically by spin up and spin down particles up to kF , where kF is related to the density through N =V 2 d3 k Z k αc the electron gas can lower its total energy by spontaneously developing a magnetization. (This is known as the Stoner instability.) • The interaction energy density is n n U n2 = αn+ n− = α +δ −δ = α − αδ 2 . V 2 2 4 The total energy density is now given by 2 −1/3 4 E0 + αn2 /4 E 2 2/3 h̄ n + 3π = − α δ2 + O δ4 . V V 3 2m When the second order term in δ is negative, the electron gas has lower energy for ﬁnite δ, i.e. it acquires a spontaneous magnetization. This occurs for 2 −1/3 4 2 2/3 h̄ n α > αc = . 3π 3 2m (e) Explain qualitatively, and sketch the behavior of the spontaneous magnetization as a function of α. • For α > αc , the optimal value of δ is obtained by expanding the energy density to fourth order in δ. The coeﬃcient of the fourth order term is positive, and the minimum energy is obtained for a value of δ 2 ∝ (α − αc ). The magnetization is proportional to δ, and hence √ grows in the vicinity of αc as α − αc , as sketched below 11 ******** 2. Boson magnetism: Consider a gas of non-interacting spin 1 bosons, each subject to a Hamiltonian p~ 2 H1 (p, ~ sz ) = − µ0 sz B , 2m where µ0 = e¯h/mc, and sz takes three possible values of (-1, 0, +1). (The orbital eﬀect, ~ , has been ignored.) p~ → p~ − eA (a) In a grand canonical ensemble ofochemical potential µ, what are the average occupation n ~ h? numbers �n+ (~k)�, �n0 (~k)�, �n−(~k)� , of one-particle states of wavenumber ~k = p/¯ • Average occupation numbers of the one-particle states in the grand canonical ensemble of chemical potential µ, are given by the Bose-Einstein distribution ns (~k) = = 1 eβ [H(s)−µ] − 1 , (for s = −1, 0, 1) 1 h 2 i 2 k exp β h¯2m − µ0 sB − βµ − 1 (b) Calculate the average total numbers {N+ , N0 , N− }, of bosons with the three possible + values of sz in terms of the functions fm (z). • Total numbers of particles with spin s are given by Z X V 1 ~ h 2 i . ns (k), =⇒ Ns = d3 k Ns = 3 h̄ k2 (2π) exp β − µ sB − βµ − 1 0 2m {� k} After a change of variables, k ≡ x1/2 √ Ns = where + (z) fm 1 ≡ Γ(m) Z 0 ∞ 2mkB T /h, we get V + f3/2 zeβµ0 sB , 3 λ dx xm−1 , z −1 ex − 1 λ≡ √ 12 h , 2πmkB T z ≡ eβµ . (c) Write down the expression for the magnetization M (T, µ) = µ0 (N+ − N− ), and by expanding the result for small B ﬁnd the zero ﬁeld susceptibility χ(T, µ) = ∂M/∂B |B=0 . • Magnetization is obtained from M (T, µ) = µ0 (N+ − N− ) i V h + + βµ0 B −βµ0 sB − f3/2 ze . = µ0 3 f3/2 ze λ Expanding the result for small B gives ∂ + + + + f3/2 (z[1 ± βµ0 B]) ≈ f3/2 (z) ± z · βµ0 B f3/2 (z). ze±βµ0 B ≈ f3/2 ∂z + + (z)/dz = fm−1 (z), we obtain Using zdfm M = µ0 V 2µ20 V + (2βµ B) · f (z) = · B · f1+/2 (z), 0 1/2 λ3 kB T λ3 and χ≡ ∂M ∂B = B=0 2µ20 V · f + (z). kB T λ3 1/2 To ﬁnd the behavior of χ(T, n), where n = N/V is the total density, proceed as follows: (d) For B = 0, ﬁnd the high temperature expansion for z(β, n) = eβµ , correct to second order in n. Hence obtain the ﬁrst correction from quantum statistics to χ(T, n) at high temperatures. + • In the high temperature limit, z is small. Use the Taylor expansion for fm (z) to write the total density n(B = 0), as 3 N+ + N0 + N− = 3 f3+/2 (z) V λ B=0 3 2 3 z z ≈ 3 z + 3/2 + 3/2 + · · · . λ 3 2 n(B = 0) = Inverting the above equation gives 3 3 2 nλ 1 nλ z= − 3/2 + ···. 3 3 2 The susceptibility is then calculated as 2µ20 V · f + (z), kB T λ3 1/2 z2 2µ20 1 χ/N = z + 1/2 + · · · kB T nλ3 2 3 2µ20 1 1 nλ 2 = 1 + − 3/2 + 1/2 +O n . 3 3kB T 2 2 χ= 13 (e) Find the temperature Tc (n, B = 0), of Bose-Einstein condensation. What happens to χ(T, n) on approaching Tc (n) from the high temperature side? • Bose-Einstein condensation occurs when z = 1, at a density n= 3 + f (1), λ3 3/2 or a temperature h2 Tc (n) = 2πmkB n 3 ζ 3/2 2/3 , where ζ 3/2 ≡ f3+/2 (1) ≈ 2.61. Since limz→1 f1+/2 (z) = ∞, the susceptibility χ(T, n) diverges on approaching Tc (n) from the high temperature side. (f) What is the chemical potential µ for T < Tc (n), at a small but ﬁnite value of B? Which one-particle state has a macroscopic occupation number? −1 � • Chemical potential for T < Tc : Since ns (~k, B) = z −1 eβ E s (k,B) − 1 is a positive number for all ~k and sz , µ is bounded above by the minimum possible energy, i.e. for T < Tc , and zeβµ0 B = 1, B ﬁnite, =⇒ µ = −µ0 B. Hence the macroscopically occupied one particle state has ~k = 0, and sz = +1. (g) Using the result in (f), ﬁnd the spontaneous magnetization, M (T, n) = lim M (T, n, B). B →0 • Spontaneous magnetization: Contribution of the excited states to the magnetization vanishes as B → 0. Therefore the total magnetization for T < Tc is due to the macroscopic occupation of the (k = 0, sz = +1) state, and M (T, n) = µ0 V n+ (k = 0) = µ0 V n − nexcited 3V = µ0 N − 3 ζ 3/2 . λ ******** 3. The virial theorem is a consequence of the invariance of the phase space for a system of N (classical or quantum) particles under canonical transformations, such as a change of scale. In the following, consider N particles with coordinates {~qi }, and conjugate momenta {p~i } (with i = 1, · · · , N ), and subject to a Hamiltonian H ({p~i } , {~qi }). (a) Classical version: Write down the expression for classical partition function, Z ≡ Z [H]. Show that it is invariant under the rescaling ~q1 → λ~q1 , p~1 → p~1 /λ of a pair of conjugate 14 variables, i.e. Z [Hλ ] is independent of λ, where Hλ is the Hamiltonian obtained after the above rescaling. • The classical partition function is obtained by appropriate integrations over phase space as ! Z Y 1 Z= d3 pi d3 qi e−β H . N !h3N i The rescaled Hamiltonian Hλ = H (p~1 /λ, {~pi=1 } , λ~q1 , {~qi=1 }) leads to a rescaled partition � � function ! Z Y 1 Z [Hλ ] = d3 pi d3 qi e−β Hλ , N !h3N i which reduces to 1 Z [Hλ ] = N !h3N Z λ3 d3 p′1 λ−3 d3 q1′ Y 3 3 d pi d qi i ! e−β H = Z, under the change of variables ~q1 ′ = λ~q1 , p~1 ′ = p~1 /λ. (b) Quantum mechanical version: Write down the expression for quantum partition func tion. Show that it is also invariant under the rescalings ~q1 → λ~q1 , p~1 → p~1 /λ, where p ~i and ~qi are now quantum mechanical operators. (Hint: start with the time-independent Schrödinger equation.) • Using the energy basis X −β H Z = tr e = e−βEn , n where En are the energy eigenstates of the system, obtained from the Schrödinger equation H ({~pi } , {q~i }) |ψn � = En |ψn � , where |ψn � are the eigenstates. After the rescaling transformation, the corresponding equation is E E (λ) (λ) (λ) H (p~1 /λ, {p~i=1 } , λ~ q , {q ~ }) ψ = E . ψ � 1 i=1 � n n n In the coordinate representation, the momentum operator is p~i = −ih∂/∂~ ¯ qi , and therefore (λ) ψλ ({~qi }) = ψ ({λ~qi }) is a solution of the rescaled equation with eigenvalue En = En . Since the eigen-energies are invariant under the transformation, so is the partition function which is simply the sum of corresponding exponentials. (c) Now assume a Hamiltonian of the form H= X p~i 2 i 2m + V ({~ qi }) . 15 Use the result that Z [Hλ ] is independent of λ to prove the virial relation p~1 2 m = ∂V · ~q1 , ∂~q1 where the brackets denote thermal averages. (You may formulate your answer in the classical language, as a possible quantum derivation is similar.) • Diﬀerentiating the free energy with respect to λ at λ = 1, we obtain ∂ ln Zλ 0= ∂λ λ=1 = −β i.e., ∂Hλ ∂λ p~1 2 m = λ=1 ∂V p~1 2 = −β − + · ~q1 , m ∂~q1 ∂V · ~q1 . ∂~q1 (d) The above relation is sometimes used to estimate the mass of distant galaxies. The stars on the outer boundary of the G-8.333 galaxy have been measured to move with velocity v ≈ 200 km/s. Give a numerical estimate of the ratio of the G-8.333’s mass to its size. • The virial relation applied to a gravitational system gives mv 2 = GM m R . Assuming that the kinetic and potential energies of the starts in the galaxy have reached some form of equilibrium gives M v2 ≈ ≈ 6 × 1020 kg/m. R G ******** 16 8.333: Statistical Mechanics I Fall 2000 Final Exam 1. Freezing of He3 : At low temperatures He3 can be converted from liquid to solid by application of pressure. A peculiar feature of its phase boundary is that (dP/dT )melting is negative at temperatures below 0.3 o K [(dP/dT )m ≈ −30atm o K −1 at T ≈ 0.1 o K]. We will use a simple model of liquid and solid phases of He3 to account for this feature. (a) In the solid phase, the He3 atoms form a crystal lattice. Each atom has nuclear spin of 1/2. Ignoring the interaction between spins, what is the entropy per particle ss , due to the spin degrees of freedom? • Entropy of solid He3 comes from the nuclear spin degeneracies, and is given by ss = kB ln(2N ) Ss = = kB ln 2. N N ˚3 per atom. What (b) Liquid He3 is modelled as an ideal Fermi gas, with a volume of 46A is its Fermi temperature TF , in degrees Kelvin? • The Fermi temperature for liquid 3 He may be obtained from its density as h2 εF TF = = 2mkB kB 3N 8πV 2/3 (6.7 × 10−34 )2 ≈ 2 · (6.8 × 10−27 )(1.38 × 10−23 ) 3 8π × 46 × 10−30 2/3 ≈ 9.2 o K. (c) How does the heat capacity of liquid He3 behave at low temperatures? Write down an expression for CV in terms of N, T, kB , TF , up to a numerical constant, that is valid for T ≪ TF . • The heat capacity comes from the excited states at the fermi surface, and is given by CV = kB π2 2 3N π2 T π2 kB T D(εF ) = kB T = N kB . 4 TF 6 6 2kB TF (d) Using the result in (c), calculate the entropy per particle sℓ , in the liquid at low temperatures. For T ≪ TF , which phase (solid or liquid) has the higher entropy? • The entropy can be obtained from the heat capacity as T dS CV = , dT ⇒ 1 sℓ = N Z T 0 CV dT π2 T = kB . T 4 TF As T → 0, sℓ → 0, while ss remains ﬁnite. This is an unusual situation in which the solid has more entropy than the liquid! (The ﬁnite entropy is due to treating the nuclear spins 17 as independent. There is actually a weak coupling between spins which causes magnetic ordering at a much lower temperature, removing the ﬁnite entropy.) (e) By equating chemical potentials, or by any other technique, prove the Clausius– Clapeyron equation (dP/dT )melting = (sℓ − ss )/(vℓ − vs ), where vℓ and vs are the volumes per particle in the liquid and solid phases respectively. • The Clausius-Clapeyron equation can be obtained by equating the chemical potentials at the phase boundary, µℓ (T, P ) = µs (T, P ), and µℓ (T + ΔT, P + ΔP ) = µs (T + ΔT, P + ΔP ). Expanding the second equation, and using the thermodynamic identities ∂µ ∂T results in = S, and P ∂P ∂T melting ∂µ ∂P = sℓ − ss . vℓ − vs T = −V, ˚3 per atom. Using this information, plus (f) It is found experimentally that vℓ − vs = 3A the results obtained in previous parts, estimate (dP/dT )melting at T ≪ TF . • The negative slope of the phase boundary results from the solid having more entropy than the liquid, and can be calculated from the Clausius-Clapeyron relation ∂P ∂T melting s ℓ − ss = ≈ kB vℓ − vs π2 4 T TF − ln 2 vℓ − vs . ˚3 , we estimate Using the values, T = 0.1 o K, TF = 9.2 J o K, and vℓ − vs = 3 A ∂P ∂T melting ≈ −2.7 × 106 Pa ◦ K−1 , in reasonable agreement with the observations. ******** 2. Non-interacting bosons: Consider a grand canonical ensemble of non-interacting bosons with chemical potential µ. The one–particle states are labelled by a wavevector q~, and have energies E(~q). (a) What is the joint probability P ({n�q }), of ﬁnding a set of occupation numbers {nq�}, of the one–particle states, in terms of the fugacities z�q ≡ exp [β(µ − E(~q))]? 18 • In the grand canonical ensemble with chemical potential µ, the joint probability of ﬁnding a set of occupation numbers {n�q}, for one–particle states of energies E(~q) is given by the normalized bose distribution P ({n�q }) = = Y q � Y q � {1 − exp [β(µ − E(q~))]} exp [β(µ − E(~q))nq�] n (1 − zq� ) z�q ~q , with n�q = 0, 1, 2, · · · , for each ~q. (b) For a particular ~q, calculate the characteristic function �exp [ikn�q ]�. n • Summing the geometric series with terms growing as z�q eik ~q , gives �exp [iknq� ]� = 1 − z�q 1 − exp [β(µ − E(~q))] = . 1 − zq� eik 1 − exp [β(µ − E(~q)) + ik] (c) Using the result of part (b), or otherwise, give expressions for the mean and variance of n�q . occupation number �nq� �. • Cumulnats can be generated by expanding the logarithm of the characteristic function in powers of k. Using the expansion formula for ln(1 + x), we obtain ln �exp [ikn�q ]� = ln (1 − z�q ) − ln 1 − zq� 1 + ik − k 2 /2 + · · · z�q k 2 zq� + +··· = − ln 1 − ik 2 1 − z�q 1 − z�q " 2 # z�q z�q z�q k2 = ik + +··· − 1 − z�q 1 − z�q 2 1 − z�q = ik z�q zq� k2 − + ···. 1 − z�q 2 (1 − z�q )2 From the coeﬃcients in the expansion, we can read oﬀ the mean and variance �n�q � = z�q , 1 − zq� and n�q2 c = zq� 2. (1 − z�q ) (d) Express the variance in part (c) in terms of the mean occupation number �n�q �. • Inverting the relation relating nq� to zq� , we obtain z�q = �n�q � . 1 + �n�q� 19 Substituting this value in the expression for the variance gives z�q nq2� c = � � (1 + �n� q �) . 2 = �nq (1 − zq� ) (e) Express your answer to part (a) in terms of the occupation numbers {�n�q�}. • Using the relation between zq� and nq� , the joint probability can be reexpressed as i Yh n −1−nq~ P ({n�q}) = (�n�q �) ~q (1 + �n�q �) . q � (f) Calculate the entropy of the probability distribution for bosons, in terms of {�n�q �}, and comment on its zero temperature limit. • Quite generally, the entropy of a probability distribution P is given by S = −kB �ln P �. Since the occupation numbers of diﬀerent one-particle states are independent, the corre sponding entropies are additive, and given by X S = −kB [�nq� � ln �nq� � − (1 + �n�q �) ln (1 + �nq��)] . q � In the zero temperature limit all occupation numbers are either 0 (for excited states) or inﬁnity (for the ground states). In either case the contribution to entropy is zero, and the system at T = 0 has zero entropy. ******** 3. Hard rods: A collection of N asymmetric molecules in two dimensions may be modeled as a gas of rods, each of length 2l and lying in a plane. A rod can move by translation of its center of mass and rotation about latter, as long as it does not encounter another rod. Without treating the hard-core interaction exactly, we can incorporate it approximately by assuming that the rotational motion of each rod is restricted (by the other rods) to an angle θ, which in turn introduces an excluded volume Ω (θ) (associated with each rod). The value of θ is then calculated self consistently by maximizing the entropy at a given density n = N/V , where V is the total accessible area. θ excluded volume 2l 20 (a) Write down the entropy of such a collection of rods in terms of N , n, Ω, and A (θ), the entropy associated to the rotational freedom of a single rod. (You may ignore the momentum contributions throughout, and consider the large N limit.) • Including both forms of entropy, translational and rotational, leads to " # N 1 N Ω(θ) Ω(θ) N −1 ≈ N kB ln n − + 1 + ln A(θ) . V − A(θ) S = kB ln 2 N! 2 (b) Extremizing the entropy as a function of θ, relate the density to Ω, A, and their derivatives Ω′ , A′ ; express your result in the form n = f (Ω, A, Ω′ , A′ ). • The extremum condition ∂S/∂θ = 0 is equivalent to Ω′ A′ = , 2n−1 − Ω A where primes indicate derivatives with respect to θ. Solving for the density gives n= 2A′ . ΩA′ + Ω′ A (c) Express the excluded volume Ω in terms of θ and sketch f as a function of θ ∈ [0, π], assuming A ∝ θ. • Elementary geometry yields Ω = l2 (θ + sin θ) , so that the equilibrium condition becomes 2 −1 [θ (2 + cos θ) + sin θ] , l2 n = f (θ) = with the function f (θ) plotted below: f(θ) n nc 0 θc θ 21 π (d) Describe the equilibrium state at high densities. Can you identify a phase transition as the density is decreased? Draw the corresponding critical density nc on your sketch. What is the critical angle θc at the transition? You don’t need to calculate θc explicitly, but give an (implicit) relation deﬁning it. What value does θ adopt at n < nc ? • At high densities, θ ≪ 1 and the equilibrium condition reduces to N≈ V ; 2θl2 the angle θ is as open as allowed by the close packing. The equilibrium value of θ increases as the density is decreased, up to its “optimal” value θc at nc , and θ (n < nc ) = θc . The transition occurs at the minimum of f (θ), whence θc satisﬁes d [θ (2 + cos θ) + sin θ] = 0, dθ i.e. 2 (1 + cos θc ) = θc sin θc . Actually, the above argument tracks the stability of a local maximum in entropy (as density is varied) which becomes unstable at θc . There is another entropy maximum at θ = π, corresponding to freely rotating rods, which becomes more advantageous (i.e. the global equilibrium state) at a density slightly below θc . ******** 22 8.333: Statistical Mechanics I Fall 2003 Final Exam 1. Helium 4: 4 He at low temperatures can be converted from liquid to solid by application of pressure. An interesting feature of the phase boundary is that the melting pressure is reduced slightly from its T = 0K value, by approximately 20Nm−2 at its minimum at T = 0.8K. We will use a simple model of liquid and solid phases of 4 He to account for this feature. (a) By equating chemical potentials, or by any other technique, prove the Clausius– Clapeyron equation (dP/dT )melting = (sℓ − ss )/(vℓ − vs ), where (vℓ , sℓ ) and (vs , ss ) are the volumes and entropies per atom in the liquid and solid phases respectively. • Clausius-Clapeyron equation can be obtained by equating the chemical potentials at the phase boundary, µℓ (T, P ) = µs (T, P ), and µℓ (T + ΔT, P + ΔP ) = µs (T + ΔT, P + ΔP ). Expanding the second equation, and using the thermodynamic identities ∂µ ∂µ = S, and = −V, ∂T P ∂P T results in ∂P ∂T = melting s ℓ − ss . vℓ − vs (b) The important excitations in liquid 4 He at T < 1◦ K are phonons of velocity c. Cal culate the contribution of these modes to the heat capacity per particle CVℓ /N , of the liquid. • The important excitations in liquid 4 He at T < 1◦ K are phonons of velocity c. The corresponding dispersion D E relation is ε(k) = h̄ck. From the average number of phonons in −1 ~ ~ mode k, given by n(k) = [exp(β¯ hck) − 1] , we obtain the net excitation energy as ¯hck exp(β¯hck) − 1 � k Z 4πk 2 dk h̄ck ¯ ) =V × (change variables to x = βhck 3 (2π) exp(βh̄ck) − 1 4 Z ∞ 4 V kB T 6 x3 π2 kB T = ¯hc dx x = V hc ¯ , h̄c 3! 0 e −1 30 h̄c 2π 2 Ephonons = X where we have used 1 ζ4 ≡ 3! Z 0 ∞ dx x3 π4 = . 90 ex − 1 23 The corresponding heat capacity is now obtained as dE 2π 2 CV = = V kB dT 15 kB T ¯ hc 3 , resulting in a heat capacity per particle for the liquid of CVℓ 2π 2 = kB vℓ N 15 kB T ¯ hc 3 . (c) Calculate the low temperature heat capacity per particle CVs /N , of solid 4 He in terms of longitudinal and transverse sound velocities cL , and cT . • The elementary excitations of the solid are also phonons, but there are now two trans verse sound modes of velocity cT , and one longitudinal sound mode of velocity cL . The contributions of these modes are additive, each similar inform to the liquid result calculated above, resulting in the ﬁnal expression for solid heat capacity of CVs 2π 2 = kB vs N 15 kB T h̄ 3 × 2 1 + 3 3 cL cT . (d) Using the above results calculate the entropy diﬀerence (sℓ − ss ), assuming a single sound velocity c ≈ cL ≈ cT , and approximately equal volumes per particle vℓ ≈ vs ≈ v. Which phase (solid or liquid) has the higher entropy? • The entropies can be calculated from the heat capacities as sℓ (T ) = ss (T ) = Z Z T 0 T 0 CVℓ (T ′ )dT ′ 2π 2 = kB vℓ T′ 45 2π 2 CVs (T ′ )dT ′ = kB vs 45 T′ kB T ¯ hc kB T h̄ 3 3 , × 2 1 3 + 3 cT cL . Assuming approximately equal sound speeds c ≈ cL ≈ cT ≈ 300ms−1 , and speciﬁc volumes ˚3 , we obtain the entropy diﬀerence vℓ ≈ vs ≈ v = 46A 4π 2 sℓ − ss ≈ − kB v 45 kB T ¯ hc 3 . The solid phase has more entropy than the liquid because it has two more phonon excitation bands. (e) Assuming a small (temperature independent) volume diﬀerence δv = vℓ − vs , calculate the form of the melting curve. To explain the anomaly described at the beginning, which phase (solid or liquid) must have the higher density? 24 • Using the Clausius-Clapeyron equation, and the above calculation of the entropy diﬀer ence, we get 3 4π 2 ∂P s ℓ − ss v kB T =− kB = . ∂T melting vℓ − vs 45 δv h̄c Integrating the above equation gives the melting curve v π2 Pmelt (T ) = P (0) − kB 45 δv kB T ¯ hc 3 T. To explain the reduction in pressure, we need δv = vℓ − vs > 0, i.e. the solid phase has the higher density, which is expected. ******** 2. Surfactant Condensation: N surfactant molecules are added to the surface of water over an area A. They are subject to a Hamiltonian H= N X p~i 2 i=1 2m + 1X V(~ri − ~rj ), 2 i,j where ~ri and p~i are two dimensional vectors indicating the position and momentum of particle i. (a) Write down the expression for the partition function Z(N, T, A) in terms of integrals over ~ri and p~i , and perform the integrals over the momenta. • The partition function is obtained by integrating the Boltzmann weight over phase space, as   Z QN 2 2 N 2 X X ~i d ~qi pi i=1 d p exp −β −β V(q~i − ~qj ) , Z(N, T, A) = 2N 2m N !h i=1 i a such that a 2πrdrV(r) = −u0 . (b) Estimate the total non–excluded area available in the positional phase space of the system of N particles. • To estimate the joint phase space of particles with excluded areas, add them to the system one by one. The ﬁrst one can occupy the whole area A, while the second can 25 explore only A − 2Ω, where Ω = πa2 . Neglecting three body eﬀects (i.e. in the dilute limit), the area available to the third particle is (A − 2Ω), and similarly (A − nΩ) for the n-th particle. Hence the joint excluded volume in this dilute limit is A(A − Ω)(A − 2Ω) · · · (A − (N − 1)Ω) ≈ (A − N Ω/2)N , where the last approximation is obtained by pairing terms m and (N − m), and ignoring order of Ω2 contributions to their product. (c) Estimate the total potential energy of the system, assuming a constant density n = N/A. Assuming this potential energy for all conﬁgurations allowed in the previous part, write down an approximation for Z. ¯ , is • Assuming a uniform density n = N/A, an average attractive potential energy, U estimated as Z 1X 1 Ū = V attr. (~qi − q~j ) = d2~r1 d2~r2 n(~r1 )n(~r2 )V attr. (~r1 − ~r2 ) 2 i,j 2 Z n2 N2 u0 . ≈ A d2~r V attr. (~r ) ≡ − 2A 2 Combining the previous results gives βu0 N 2 1 1 N . Z(N, T, A) ≈ (A − N Ω/2) exp 2A N ! λ2N (d) The surface tension of water without surfactants is σ0 , approximately independent of temperature. Calculate the surface tension σ(n, T ) in the presence of surfactants. • Since the work done is changing the surface area is dW = σdA, we have dF = −T dS + σdA + µdN , where F = −kB T ln Z is the free energy. Hence, the contribution of the surfactants to the surface tension of the ﬁlm is σs = − ∂ ln Z ∂A T,N =− N kB T u0 N 2 + , A − N Ω/2 2A2 which is a two-dimensional variant of the familiar van der Waals equation. Adding the (constant) contribution in the absence of surfactants gives σ(n, T ) = σ0 − ∂ ln Z ∂A T,N =− N kB T u0 N 2 . + A − N Ω/2 2A2 (e) Show that below a certain temperature, Tc , the expression for σ is manifestly incorrect. What do you think happens at low temperatures? 26 • Thermodynamic stability requires δσδA ≥ 0, i.e. σ must be a monotonically increasing function of A at any temperature. This is the case at high temperatures where the ﬁrst term in the equation for σs dominates, but breaks down at low temperatures when the term from the attractive interactions becomes signiﬁcant. The critical temperature is obtained by the usual conditions of ∂σs /∂A = ∂ 2 σs /∂A2 = 0, i.e. from  u0 N 2 N kB T  ∂σs   = =0 −  ∂A (A − N Ω/2)2 A3 T , 2 2  ∂ T 3u N σ 2N k B 0 s   =− + =0  A4 ∂A2 T (A − N Ω/2)3 The two equations are simultaneously satisﬁed for Ac = 3N Ω/2, at a temperature Tc = 8u0 . 27kB Ω As in the van der Waals gas, at temperatures below Tc , the surfactants separate into a high density (liquid) and a low density (gas) phase. (f) Compute the heat capacities, CA and write down an expression for Cσ without explicit evaluation, due to thesurfactants. • The contribution of the surfactants to the energy of the ﬁlm is given by Es = − ∂ ln Z kB T u0 N 2 = 2N × − . ∂β 2 2A The ﬁrst term is due to the kinetic energy of the surfactants, while the second arises from their (mean-ﬁeld) attraction. The heat capacities are then calculated as CA = and Cσ = dQ dT dQ dT = A = σ ∂E ∂T ∂E ∂T σ = N kB , A −σ ∂A ∂T . σ ******** 3. Dirac Fermions are non-interacting particles of spin 1/2. The one-particle states come in pairs of positive and negative energies, p ~ E ± (k) = ± m2 c4 + h̄2 k 2 c2 , independent of spin. 27 (a) For any fermionic system of chemical potential µ, show that the probability of ﬁnding an occupied state of energy µ + δ is the same as that of ﬁnding an unoccupied state of energy µ − δ. (δ is any constant energy.) • According to Fermi statistics, the probability of occupation of a state of of energy E is p [n(E)] = eβ(µ−E )n , 1 + eβ(µ−E ) for n = 0, 1. For a state of energy µ + δ, p [n(µ + δ)] = eβδn , 1 + eβδ p [n(µ + δ) = 1] = =⇒ eβδ 1 = . βδ 1+e 1 + e−βδ Similarly, for a state of energy µ − δ, p [n(µ − δ)] = e−βδn , 1 + e−βδ =⇒ p [n(µ − δ) = 0] = 1 = p [n(µ + δ) = 1] , 1 + e−βδ i.e. the probability of ﬁnding an occupied state of energy µ + δ is the same as that of ﬁnding an unoccupied state of energy µ − δ. (b) At zero temperature all negative energy Dirac states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. Using the result in (a) ﬁnd the chemical potential at ﬁnite temperatures T . • The above result implies that for µ = 0, �n(E)� + �n − E)� is unchanged for an tem perature; any particle leaving an occupied negative energy state goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at zero. Thus, the particle–hole symmetry enfrces µ(T ) = 0. (c) Show that the mean excitation energy of this system at ﬁnite temperature satisﬁes E(T ) − E(0) = 4V Z d3~k E + (~k) (2π )3 exp β E (~k ) + 1 . + • Using the label +(-) for the positive (energy) states, the excitation energy is calculated as X E(T ) − E(0) = [�n+ (k)� E + (k) + (1 − �n− (k)�) E − (k)] k,sz =2 X k 2 �n+ (k)� E + (k) = 4V 28 Z d3~k E + (~k) . (2π)3 exp βE (~k) + 1 + (d) Evaluate the integral in part (c) for massless Dirac particles (i.e. for m = 0). • For m = 0, E + (k) = hc ¯ |k|, and ∞ 4πk 2 dk hck ¯ = (set β¯hck = x) 8π 3 eβ¯hck + 1 0 3 Z ∞ 2V kB T x3 = 2 kB T dx x ¯hc e +1 π 0 3 7π 2 kB T . = V kB T 60 h̄c E(T ) − E(0) = 4V Z For the ﬁnal expression, we have noted that the needed integral is 3!f4− (1), and used the given value of f4− (1) = 7π 4 /720. (e) Calculate the heat capacity, CV , of such massless Dirac particles. • The heat capacity can now be evaluated as ∂E CV = ∂T V 7π 2 = V kB 15 kB T ¯ hc 3 . (f) Describe the qualitative dependence of the heat capacity at low temperature if the particles are massive. • When m �= 0, there is an energy gap between occupied and empty states, and we thus expect an exponentially activated energy, and hence heat capacity. For the low energy excitations, ¯h2 k 2 2 E + (k) ≈ mc + + ···, 2m and thus √ Z ∞ 2V 2 −βmc2 4π π E(T ) − E(0) ≈ 2 mc e dxx2 e−x 3 π λ 0 2 48 V = √ 3 mc2 e−βmc . πλ The corresponding heat capacity, to leading order thus behaves as C(T ) ∝ kB 2 2 V βmc2 e−βmc . 3 λ ******** 29 8.333: Statistical Mechanics I Fall 2004 Final Exam 1. Neutron star core: Professor Rajagopal’s group has proposed that a new phase of QCD matter may exist in the core of neutron stars. This phase can be viewed as a condensate of quarks in which the low energy excitations are approximately E(~k)± =± h̄2 2 ~ | k | − kF 2M . The excitations are fermionic, with a degeneracy of g = 2 from spin. (a) At zero temperature all negative energy states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. By relating occupation numbers of states of energies µ + δ and µ − δ, or otherwise, ﬁnd the chemical potential at ﬁnite temperatures T . • According to Fermi statistics, the probability of occupation of a state of of energy E is p [n(E)] = eβ(µ−E )n , 1 + eβ(µ−E ) for n = 0, 1. For a state of energy µ + δ, eβδn p [n(µ + δ)] = , 1 + eβδ =⇒ eβδ 1 = . p [n(µ + δ) = 1] = βδ 1 + e−βδ 1+e Similarly, for a state of energy µ − δ, p [n(µ − δ)] = e−βδn , 1 + e−βδ =⇒ p [n(µ − δ) = 0] = 1 = p [n(µ + δ) = 1] , 1 + e−βδ i.e. the probability of ﬁnding an occupied state of energy µ + δ is the same as that of ﬁnding an unoccupied state of energy µ − δ. This implies that for µ = 0, �n(E)� + �n(−E)� is unchanged for an temperature; for every particle leaving an occupied negative energy state a particle goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at zero. Thus, the particle–hole symmetry enforces µ(T ) = 0. (b) Assuming a constant density of states near k = kF , i.e. setting d3 k ≈ 4πkF2 dq with q = |~k | − kF , show that the mean excitation energy of this system at ﬁnite temperature is k2 E(T ) − E(0) ≈ 2gV F2 π Z ∞ 0 30 dq E + (q ) exp (β E + (q )) + 1 . • Using the label +(-) for the positive (energy) states, the excitation energy is calculated as X E(T ) − E(0) = [�n+ (k)� E + (k) + (1 − �n− (k)�) E − (k)] k,s =g X k 2 �n+ (k)� E + (k) = 2gV Z d3~k E + (~k) . (2π)3 exp βE (~k) + 1 + The largest contribution to the integral comes for |~k | ≈ kF . and setting q = (|~k | − kF ) and using d3 k ≈ 4πkF2 dq, we obtain Z ∞ Z ∞ E + (q) kF2 E + (q) 4πkF2 = 2gV 2 . 2 dq dq E(T ) − E(0) ≈ 2gV 3 π 0 exp (βE + (q)) + 1 exp (βE + (q)) + 1 8π 0 (c) Give a closed form answer for the excitation energy by evaluating the above integral. • For E + (q) = h̄2 q 2 /(2M ), we have Z ∞ kF2 h̄2 q 2 /2M = (set βh̄2 q 2 /2M = x) E(T ) − E(0) = 2gV 2 dq βh̄2 q 2 /2M π 0 e +1 Z ∞ 1/2 x1/2 gV kF2 2M kB T = k T dx B ex + 1 π2 h̄2 0 1/2 √ ζ3/2 V kF2 gV kF2 2M kB T π 1 1 √ √ 1 − ζ = 1 − kB T. = 2 kB T 3/2 2 π λ π h̄2 2 2 − (1), and introduced the thermal For the ﬁnal expression, we have used the value of fm √ wavelength λ = h/ 2πM kB T . (d) Calculate the heat capacity, CV , of this system, and comment on its behavior at low temperature. • Since E ∝ T 3/2 , √ 3ζ3/2 1 V kF2 ∂E 3E CV = 1− √ kB ∝ T . = = 2T 2π λ ∂T V 2 This is similar to the behavior of a one dimensional system of bosons (since the density of states is constant in q as in d = 1). Of course, for any fermionic system the density of states close to the Fermi surface has this character. The diﬀerence with the usual Fermi systems is the quadratic nature of the excitations above the Fermi surface. ******** 2. Critical point behavior: The pressure P of a gas is related to its density n = N/V , and temperature T by the truncated expansion b c P = kB T n − n 2 + n 3 2 6 31 , where b and c are assumed to be positive temperature independent constants. (a) Locate the critical temperature Tc below which this equation must be invalid, and the corresponding density nc and pressure Pc of the critical point. Hence ﬁnd the ratio kB Tc nc /Pc . • Mechanical stability of the gas requires that any spontaneous change in volume should be opposed by a compensating change in pressure. This corresponds to δP δV < 0, and since δn = −(N/V 2 )δV , any equation of state must have a pressure that is an increasing func tion of density. The transition point between pressure isotherms that are monotonically increasing functions of n, and those that are not (hence manifestly incorrect) is obtained by the usual conditions of dP/dn = 0 and d2 P/dn2 = 0. Starting from the cubic equation of state, we thus obtain dP c =kB Tc − bnc + n2c = 0 dn 2 . d2 P = − b + cnc = 0 dn2 From the second equation we obtain nc = b/c, which substituted in the ﬁrst equation gives kB Tc = b2 /(2c). From the equation of state we then ﬁnd Pc = b3 /(6c2 ), and the dimensionless ratio of kB Tc nc = 3. Pc (b) Calculate the isothermal compressibility κT = − V1 function of T for n = nc . • Using V = N/n, we get κT (n) = − 1 ∂V V ∂P = T 1 ∂P n ∂n −1 T ∂V ∂P T , and sketch its behavior as a −1 = n kB T − bn + cn2 /2 . For n = nc , κT (nc ) ∝ (T − Tc )−1 , and diverges at Tc . (c) On the critical isotherm give an expression for (P − Pc ) as a function of (n − nc ). • Using the coordinates of the critical point computed above, we ﬁnd b2 b c b3 + n − n2 + n3 2 6c 2c 2 6 2 b3 c b 2 b 3 = n − 3 n + 3 2n − 3 c c c 6 c 3 = (n − nc ) . 6 P − Pc = − (d) The instability in the isotherms for T < Tc is avoided by phase separation into a liquid of density n+ and gas of density n− . For temperatures close to Tc , these densities behave 32 as n± ≈ nc (1 ± δ). Using a Maxwell construction, or otherwise, ﬁnd an implicit equation for δ(T ), and indicate its behavior for (Tc − T ) → 0. (Hint: Along an isotherm, variations of chemical potential obey dµ = dP/n.) • According to the Gibbs–Duhem relation, the variations of the intensive variables are related by SdT − V dP + N dµ = 0, and thus along an isotherm (dT = 0) dµ = dP/n = ∂P/∂n|T dn/n. Since the liquid and gas states are in coexistence they should have the same chemical potential. Integrating the above expression for dµ from n− to n+ leads to the so-called Maxwell construction, which reads Z nc (1+δ) Z n+ dP kB T − bn + cn2 /2 0 = µ(n+ ) − µ(n− ) = = dn . n n nc (1−δ) n− Performing the integrals gives the equation c 2 1+δ 1+δ 2 2 0 = kB T ln − 2kB Tc δ, − bnc (2δ) + nc (1 + δ) − (1 − δ) = kB T ln 1−δ 4 1−δ where for the ﬁnd expression, we have used nc = b/c and kB Tc = b2 /(2c). The implicit equation for δ is thus T 1+δ T δ= ln ≈ δ − δ3 + · · · . 2Tc 1−δ Tc The leading behavior as (Tc − T ) → 0 is obtained by keeping up to the cubic term, and given by r Tc δ ≈ 1− . T ******** 3. Relativistic Bose gas in d dimensions: Consider a gas of non-interacting (spinless) bosons with energy ǫ = c |p~ |, contained in a box of “volume” V = Ld in d dimensions. (a) Calculate the grand potential G = −kB T ln Q, and the density n = N/V , at a chemical + potential µ. Express your answers in terms of d and fm (z), where z = eβµ , and Z ∞ 1 xm−1 + fm (z) = dx. (m − 1)! 0 z −1 ex − 1 (Hint: Use integration by parts on the expression for ln Q.) • We have P ! ni =N ∞ iX X X Q= eNβµ exp −β ni ǫi N=0 = YX i {ni } eβ(µ−ǫi )ni = i {ni } Y i 33 1 1 − eβ(µ−ǫi ) , P P i) . Replacing the summation whence ln Q = − i ln 1 − eβ(µ−ǫ i with a d dimensional h iR R∞ ∞ d d integration 0 V dd k/ (2π) = V Sd / (2π) k d−1 dk, where Sd = 2π d/2 / (d/2 − 1)!, 0 leads to Z ∞ V Sd d−1 −β¯ hck ln Q = − k dk ln 1 − ze . d (2π) 0 The change of variable x = βh̄ck results in ln Q = − V Sd d (2π) kB T hc ¯ d Z ∞ 0 xd−1 dx ln 1 − ze−x . Finally, integration by parts yields V Sd 1 ln Q = d (2π) d kB T h̄c d Z ∞ ze−x Sd x dx =V −x 1 − ze d d 0 kB T hc d Z ∞ dx 0 xd , z −1 ex − 1 leading to Sd G = −kB T ln Q = −V d kB T hc d kB T d!fd++1 (z) , which can be somewhat simpliﬁed to V π d/2 d! + f (z) , G = −kB T d λc (d/2)! d+1 where λc ≡ hc/(kB T ). The average number of particles is calculated as N =− ∂G ∂G V π d/2 d! + f (z) , = −βz = d ∂µ ∂z λc (d/2)! d where we have used z∂fd+1 (z)/∂z = fd (z). Dividing by volume, the density is obtained as n= 1 π d/2 d! + f (z) . λdc (d/2)! d (b) Calculate the gas pressure P , its energy E, and compare the ratio E/(P V ) to the classical value. • We have P V = −G, while E=− ∂ ln Q ∂β = +d z ln Q = −dG. β Thus E/(P V ) = d, identical to the classical value for a relativistic gas. 34 (c) Find the critical temperature, Tc (n), for Bose-Einstein condensation, indicating the dimensions where there is a transition. • The critical temperature Tc (n) is given by n= 1 π d/2 d! + 1 π d/2 d! (z = 1) = ζd . f λdc (d/2)! λdc (d/2)! d This leads to hc Tc = kB n(d/2)! π d/2 d!ζd 1/d . However, ζd is ﬁnite only for d > 1, and thus a transition exists for all d > 1. (d) What is the temperature dependence of the heat capacity C (T ) for T < Tc (n)? • At T < Tc , z = 1 and E = −dG ∝ T d+1 , resulting in C (T ) = ∂E ∂T = (d + 1) z=1 E V π d/2 d! G = −d(d + 1) = d(d + 1) d kB ζd+1 ∝ T d . T (d/2)! λc T (e) Evaluate the dimensionless heat capacity C(T )/(N kB ) at the critical temperature T = Tc , and compare its value to the classical (high temperature) limit. • We can divide the above formula of C(T ≤ T c), and the one obtained earlier for N (T ≥ T c), and evaluate the result at T = Tc (z = 1) to obtain d(d + 1)ζd+1 C(Tc ) = . ζd N kB In the absence of quantum eﬀects, the heat capacity of a relativistic gas is C/(N kB ) = d; this is the limiting value for the quantum gas at inﬁnite temperature. ******** 35 8.333: Statistical Mechanics I Fall 2005 Final Exam 1. Graphene is a single sheet of carbon atoms bonded into a two dimensional hexagonal lattice. It can be obtained by exfoliation (repeated peeling) of graphite. The band struc ture of graphene is such that the single particles excitations behave as relativistic Dirac fermions, with a spectrum that at low energies can be approximated by E ± (~k) = ±h̄v ~k . There is spin degeneracy of g = 2, and v ≈ 106 ms−1 . Recent experiments on unusual transport properties of graphene were reported in Nature 438, 197 (2005). In this problem, you shall calculate the heat capacity of this material. (a) If at zero temperature all negative energy states are occupied and all positive energy ones are empty, ﬁnd the chemical potential µ(T ). • According to Fermi statistics, the probability of occupation of a state of of energy E is p [n(E)] = eβ(µ−E )n , 1 + eβ(µ−E ) for n = 0, 1. For a state of energy µ + δ, p [n(µ + δ)] = eβδn , 1 + eβδ =⇒ p [n(µ + δ) = 1] = eβδ 1 = . βδ 1+e 1 + e−βδ Similarly, for a state of energy µ − δ, p [n(µ − δ)] = e−βδn , 1 + e−βδ =⇒ p [n(µ − δ) = 0] = 1 = p [n(µ + δ) = 1] , 1 + e−βδ i.e. the probability of ﬁnding an occupied state of energy µ + δ is the same as that of ﬁnding an unoccupied state of energy µ − δ. At zero temperature all negative energy Dirac states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. The above result implies that for µ = 0, �n(E)� + �n − E)� is unchanged for all temperatures; any particle leaving an occupied negative energy state goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at zero. Thus, the particle–hole symmetry enforces µ(T ) = 0. (b) Show that the mean excitation energy of this system at ﬁnite temperature satisﬁes E(T ) − E(0) = 4A Z E + (~k) d2~k (2π)2 exp β E (~k) + 1 + 36 . • Using the label +(-) for the positive (energy) states, the excitation energy is calculated as X E(T ) − E(0) = [�n+ (k)� E + (k) + (1 − �n− (k)�) E − (k)] k,sz =2 X k 2 �n+ (k)� E + (k) = 4A Z d2~k E + (~k) . (2π)2 exp β E (~k) + 1 + (c) Give a closed form answer for the excitation energy by evaluating the above integral. • For E + (k) = h̄v|k|, and ∞ 2πkdk h̄vk = (set β¯hck = x) 4π 2 eβ¯hvk + 1 0 2 Z ∞ 2A kB T x2 = kB T dx x hv e +1 π ¯ 0 2 3ζ3 kB T . = AkB T π h̄v E(T ) − E(0) = 4A Z For the ﬁnal expression, we have noted that the needed integral is 2!f3− (1), and used f3− (1) = 3ζ3 /4. Z E + (~k ) d2~k . E(T ) − E(0) = A (2π )2 exp β E (~k ) − 1 + (d) Calculate the heat capacity, CV , of such massless Dirac particles. • The heat capacity can now be evaluated as ∂E CV = ∂T V 9ζ3 = AkB π kB T ¯ hv 2 . (e) Explain qualitatively the contribution of phonons (lattice vibrations) to the heat capac ity of graphene. The typical sound velocity in graphite is of the order of 2×104 ms−1 . Is the low temperature heat capacity of graphene controlled by phonon or electron contributions? • The single particle excitations for phonons also have a linear spectrum, with E p = h̄vp |k| and correspond to µ = 0. Thus qualitatively they give the same type of contribution to energy and heat capacity. The diﬀerence is only in numerical pre-factors. The precise 37 contribution from a single phonon branch is given by ∞ 2πkdk h̄vp k = (set β¯hck = x) 4π 2 eβ¯hvp k − 1 0 2 Z ∞ x2 A kB T = kB T dx x e −1 2π ¯hvp 0 2 2 ζ3 kB T 3ζ3 kB T = AkB T , CV,p = AkB . hv ¯hvp π ¯ p π Ep (T ) − Ep (0) = A Z We see that the ratio of electron to phonon heat capacities is proportional to (vp /v)2 . Since the speed of Dirac fermions is considerably larger than that of phonons, their contribution to heat capacity of graphene is negligible. ******** 2. Quantum Coulomb gas: Consider a quantum system of N positive, and N negative charged relativistic particles in box of volume V = L3 . The Hamiltonian is H= 2N X i=1 c|p~i | + 2N X i 0, i.e. the solid phase has the higher density, which is normal. ******** 2. Squeezed chain: A rubber band is modeled as a single chain of N massless links of ﬁxed length a. The chain is placed inside a narrow tube that restricts each link to point parallel or anti-parallel to the tube. (a) Ignoring any interactions amongst the links, give the number of conﬁgurations of the chain, Ω(L, N ), where L is the end–to–end length of the chain. • A typical conﬁguration will have N+ steps in one direction, and N− steps in the opposite direction, such that L = (N+ − N− )a, and N = N+ + N− , or equivalently N ± L/a . 2 The number of conﬁgurations is given by the binomial formula N± = Ω(L, N ) = N! N! = . [(N + L/a)/2]![(N − L/a)/2]! N+ !N− ! (b) Use Stirling’s approximation to express the result for entropy as a function of N and x = L/(N a). • Using Stirling’s approximation S = ln Ω ≈ N ln N − N+ ln N+ − N− ln N− kB N = N ln 2 − [(1 + x) ln(1 + x) + (1 − x) ln(1 − x)] . 2 45 The tube is now uniformly squeezed such that stretched conﬁgurations are energet ically favored. For this problem, assume a simpliﬁed form of this energy that depends inversely on the local link density, such that the energy of a uniformly stretched state is σ L σa2 E(L, N ) = − L × =− N x2 . 2 N 2 (Use this formula for energy for all states of given L and N in the remainder of this problem.) (c) Calculate the free energy F (T, L, N ), and the force J(T, L, N ) acting on the end points of the chain. (The work done on expanding the chain is dW ¯ = JdL.) • The free energy is σ L 2 kB T + [(N + L/a) ln(N + L/a) + (N − L/a) ln(N − L/a) − N ln 2 − 2N ln N ] . 2 N 2 The force conjugate to L is obtained as F = E−T S = − J= ∂F L kB T Na + L kB T 1+x = −σ + ln = −σax + ln . N 2a Na − L 2a 1−x ∂L (d) Sketch the isotherms J(x, T ) at high and low temperatures, and identify the tempera ture Tc when the behavior changes. • The curves J(x) start linearly at x = 0 with a slope (kB T /a − σa), and go to (plus) inﬁnity as x → 1. The slope changes sign at Tc = (σa2 )/kB , with the curves for T < Tc being non-monotonic in x. (e) What is the condition for stability of the chain? What portion of the above isotherms are inherently unstable. • The condition for stability δJδL ≥ 0, implies that the slope of the isotherm J(x) must be positive. The portion of the isotherms below the minimum for T < Tc violate this condition, and are inherently unstable. (f) Using the form for J obtained in part (c), ﬁnd an expression for the unforced (J = 0) chain length that is valid as T approaches Tc from below. • As T → Tc− , x ∝ L goes to zero. Expanding the logarithms in the expression for J(x) = 0, we ﬁnd 2kB T 3 2kB T x + O(x5 ). x+ 0 = −σa + a 3a The solution to this equation is x2 Tc =1− , 3 T s Tc L ≈ Na 3 1 − . T 46 ******** 3. Graphene bilayer: The layers of graphite can be peeled apart through diﬀerent exfo liation processes. Many such processes generate single sheets of carbon atoms, as well as bilayers in which the two sheets are weakly coupled. The hexagonal lattice of the single layer graphene, leads to a band structure that at low energies can be approximated by E 1± layer (~k) = ±t� (ak), as in relativistic Dirac fermions. (Here k = ~k , a is a lattice spac- ing, and t� is a typical in-plane hopping energy.) A weak hopping energy t⊥ between the two sheets of the bilayer modiﬁes the low energy excitations drastically, to E bilayer (~k) ± =± t2� 2t⊥ (ka)2 , i.e. resembling massive Dirac fermions. In addition to the spin degeneracy, there are two branches of such excitations per unit cell, for an overall degeneracy of g = 4. (a) For the undoped material with one electron per site, at zero temperature all negative energy states are occupied and all positive energy ones are empty. Find the chemical potential µ(T ). • According to Fermi statistics, the probability of occupation of a state of of energy E is eβ(µ−E )n p [n(E)] = , 1 + eβ(µ−E ) for n = 0, 1. For a state of energy µ + δ, p [n(µ + δ)] = eβδn , 1 + eβδ =⇒ p [n(µ + δ) = 1] = eβδ 1 = . βδ 1+e 1 + e−βδ Similarly, for a state of energy µ − δ, p [n(µ − δ)] = e−βδn , 1 + e−βδ =⇒ p [n(µ − δ) = 0] = 1 = p [n(µ + δ) = 1] , 1 + e−βδ i.e. the probability of ﬁnding an occupied state of energy µ + δ is the same as that of ﬁnding an unoccupied state of energy µ − δ. At zero temperature all negative energy Dirac states are occupied and all positive energy ones are empty, i.e. µ(T = 0) = 0. The above result implies that for µ = 0, �n(E)� + �n − E)� is unchanged for all temperatures; any particle leaving an occupied negative energy state goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at zero. Thus, the particle–hole symmetry enforces µ(T ) = 0. 47 (b) Show that the mean excitation energy of this system at ﬁnite temperature satisﬁes E(T ) − E(0) = 2gA Z E + (~k ) d2~k (2π )2 exp β E (~k) + 1 . + • Using the label +(-) for the positive (energy) states, the excitation energy is calculated as X E(T ) − E(0) = [�n+ (k)� E + (k) + (1 − �n− (k)�) E − (k)] k,sz ,α =g X k 2 �n+ (k)� E + (k) = 2gA Z d2~k E + (~k) . (2π)2 exp βE (~k) + 1 + (c) Give a closed form answer for the excitation energy of the bilayer by evaluating the above integral. • Let E + (k) = αk 2 , with α = (t� a)2 /(2t⊥ ), to get ∞ 2πkdk αk 2 = (set βαk2 = x) 2 eβαk2 + 1 4π 0 Z ∞ gA kB T x dx x = kB T e +1 2π α 0 2 gπ kB T π A kB T = t⊥ . = AkB T t� 24 α 3 a2 E(T ) − E(0) = 2gA Z For the ﬁnal expression, we have noted that the needed integral is f2− (1), and used f2− (1) = ζ2 /2 = π 2 /12. (d) Calculate the heat capacity, CA , of such massive Dirac particles. • The heat capacity can now be evaluated as CA = ∂E ∂T = A 2π A kB 3 a2 kB T t ⊥ t2� ! . (e) A sample contains an equal proportion of single and bilayers. Estimate (in terms of the hopping energies) the temperature below which the electronic heat capacity is dominated by the bilayers. • As stated earlier, the monolayer excitations for phonons have a linear spectrum, with E 1 layer = ±t� (ka). Their contribution to energy and heat capacity can be calculated as 48 before. Including the various prefactors (which are not required for the solution), we have E 1 layer ∞ 2πkdk t� ak = (set βt� ak = x) 4π 2 eβtk ak + 1 0 2 Z ∞ x2 gA kB T dx x = kB T e +1 π t� a 0 2 2 kB T kB T A A 1 layer , CA . ∝ 2 kB T ∝ 2 kB t� t� a a (T ) = 2gA Z The bilayer heat capacity, which is proportional to T is more important at lower temper atures. By comparing the two expressions, it is apparent that the electronic heat capacity per particle is larger in the bilayer for temperatures smaller than T ∗ ≈ t⊥ /kB . ******** 49

Statistical Mechanics I: Final Exam Review Solution

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