Solutions Manual for Introduction to Statistical Physics (draft)

PrefaceThis is page iiiPrinter: Opaque thisWe give some schematic solutions of exercises from chapters 1to 10 of "Introduction to Statistical Physics", by Silvio R. A.Salinas, …rst published by Springer, New York, in 2001. We alsoadd a number of corrections and some new exercises. Additionalcorrections and suggestions are warmly welcomed.Silvio SalinasInstitute of Physics, University of São Paulosrasalinas@gmail.com

ivPreface

1Introduction to Statistical PhysicsThis is page 1Printer: Opaque this1- Obtain the probability of adding up six points if we toss threedistinct dice.*** Let´s consider an easier problem, two dice, for example.In this (simpler) case, there are 6 6 = 36 con…gurations(events), but only 5 of them correspond to 6 points. Since all ofthe con…gurations are equally probable, we have P (6) = 5=36.2- Consider a binomial distribution for a one-dimensional randomwalk, with N = 6, p = 2=3, and q = 1 p = 1=3.(a) Draw a graph of P N (N 1 ) versus N 1 =N.(b) Use the values of hN 1 i and hN 2 1 i to obtain the correspondingGaussian distribution, p G (N 1 ). Draw a graph of p G (N 1 ) versusN 1 =N to compare with the previous result.(c) Repeat items (a) and (b) for N = 12 and N = 36. Are thenew answers too di¤erent?*** The "equivalent Gaussian" distribution has the same …rstand second moments as the binomial distribution,"# "#(N 1 hN 1 i) 2p G (N 1 ) = C exp2 (N 1 hN 1 i) 2 (N 1 pN) 2= C exp;2Npq

2 1. Introduction to Statistical Physicswhere the "normalization factor" C comes from+1 Z "#(N 1 pN) 2C expdN 1 = 1 =) C = (2Npq) 1=2 :2Npq1It is instructive to draw graphs of p G (N 1 ) versus N 1 for somevalues of N.In the …gure, we show some graphs of P N (N 1 ) versus N 1 =N.3- Obtain an expression for the third moment of a binomialdistribution. What is the behavior of this moment for large N?*** Using the tricks introduced in the text, it is easy to seethat (N1 hN 1 i) 3 = Npq (q p) :Note that (N 1 ) 3 = 0 for p = q (same probabilities). Also,note the dependence of (N 1 ) 3 on N, so thatfor large N.(N1 ) 3 1=3(N1 ) 2 1=2 1N 1=6 ;4- Consider an event of probability p. The probability of noccurrences of this event out of N trials is given by the binomialdistribution,W N (n) =N!n!(N n)! pn (1 p) N n :If p is small (p

P(N1)PN(N1)1. Introduction to Statistical Physics 3A)0.160.140.120.10.080.060.040.0200 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1N1/NB)0.060.050.040.030.020.0100 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1N1/NGraphs of P N(N 1) (dots) versus N 1/N, and the corresponding Gaussian distribution (solidline), for N=36 (top figure) and N=216 (at bottom).

4 1. Introduction to Statistical Physics*** It is easy to see that1X1XP (n) = exp ( )n=0hni =n=0n=0 nn! = 1;1X1XnP (n) = exp ( ) n nn! == exp ( )n=0 @ X 1 n@ n! = ;n=0and (n)2 = (n hni) 2 = :5- Consider an experiment with N equally likely outcomes,involving two events A and B. Let N 1 be the number of eventsin which A occurs, but not B; N 2 be the number of events inwhich B occurs, but not A; N 3 be the number of events in whichboth A and B occur; and N 4 be the number of events in whichneither A nor B occur.(i) Check that N 1 + N 2 + N 3 + N 4 = N.(ii) Check thatP (A) = N 1 + N 3Nand P (B) = N 2 + N 3N ;where P (A) and P (B) are the probabilities of occurrence of Aand B, respectively.(iii) Calculate the probability P (A + B) of occurrence of eitherA or B.(iv) Calculate the probability P (AB) of occurrence of bothA and B.(v) Calculate the conditional probability P (A j B) that Aoccurs given that B occurs.(vi) Calculate the conditional probability P (B j A) that Boccurs given that A occurs.(vii) Show thatP (A + B) = P (A) + P (B)P (AB)

and1. Introduction to Statistical Physics 5P (AB) = P (B) P (A j B) = P (A) P (B j A) :(viii) Considering a third event C, show thatP (B j A)P (C j A) = P (B) P (A j B)P (C) P (A j C) ;which is an expression of Bayes’theorem.6- A random variable x is associated with the probabilitydensityp (x) = exp ( x) ;for 0 < x < 1.(a) Find the mean value hxi.(b) Two values x 1 and x 2 are chosen independently. Findhx 1 + x 2 i and hx 1 x 2 i.(c) What is the probability distribution of the random variabley = x 1 + x 2 ?*** Note thathxi = 1; hx 1 + x 2 i = 2; hx 1 x 2 i = 1;and thatZ Zp (y) =dx 1 dx 2 p (x 1 ) p (x 2 ) (y x 1 x 2 ) :Using an integral representations of the delta-function (see theAppendix), it is easy to see thatp (y) = y exp ( y) :7- Consider a random walk in one dimension. After N stepsfrom the origin, the position is given byx =NXs j ;j=1

6 1. Introduction to Statistical Physicswhere fs j g is a set of independent, identical, and identically distributedrandom variables, given by the probability distribution" #w(s) = 2 2 1=2 (s l) 2exp ;2 2where and l are positive constants. After N steps, what isthe average displacement from the origin? What is the standarddeviation of the random variable x? In the large N limit, whatis the form of the Gaussian distribution associated with thisproblem?*** It is easy to see thathxi = N hsi = Nl;and (x hxi)2 = N (s hsi) 2 = N 2 ;from which we write the Gaussian form"#p G (x) = 2 2 1=2 (x Nl) 2exp :2N 2*** Try to solve a similar problem with8< 0; s < 1=2;w(s) = 1; 1=2 < s < +1=2;:0; s > +1=2:Calculate hxi, hx 2 i, and the limiting Gaussian distribution p G (x)(for large N). Note that+1 Zw(s)ds = 1; hxi = N hsi = 0;(x)2 = N 12 :18- Consider again problem 7, with a distribution w (s) of theLorentzian formw(s) = 1 a s 2 + a ; 2

1. Introduction to Statistical Physics 7with a > 0. Obtain an expression for the probability distributionassociated with the random variable x. Is it possible to write aGaussian approximation for large N? Why?*** You should be careful. It is immediate to see that hsi = 0,but hs 2 i is associated with a diverging integral! The Lorentzianform does not obey the conditions for the validity of the centrallimit theorem.Additional exercises9- The Ehrenfest “urn model”provides an excellent illustrationof statistical ‡uctuations, the role of large numbers, andthe direction of the “time arrow”. Take a look at Section 1 ofChapter 15. The “stochastic equation”associated with the simpleurn model is linear (and exactly soluble). There are manyworks on the urn model. See, for example, the relatively recentwork by C. Godrèche and J. M. Luck, J. Phys.: Condens. Matter14, 1601-1615 (2002), which contains a number of historicalreferences.In the simple urn model, we consider two boxes, N numberedballs, and a generator of N random numbers. Initially, there areN 1 balls in urn 1, and N 2 = N N 1 balls in urn 2. Each timeunit, we draw a random number, between 1 and N, and changethe position (urn location) of the corresponding ball.Choose a reasonable generator of random numbers, and performtime simulations for this simple urn model. Draw graphs ofN 1 (number of balls in urn 1) as a function of time t (in uniformdiscrete steps t), from an initial situation in which N 1 = N(all the balls are in urn 1), using two values of the total numberof balls: (a) N = 10, and (b) N = 100. What can you say aboutthe ‡uctuations of the value of N 1 ? What happens at long times,t ! 1?It is reasonable to assume the “stochastic equation”P (N 1 ; t + t) = P (N 1 1; t) W 1 + P (N 1 + 1; t) W 2 ;

N1 (t)1. Introduction to Statistical Physics 91009080706050400 50 100 150 200 250 300 350 400 450 500t (passos)Figura 1: Simulation for the Ehrenfest urn model. Graph of N 1 versus the discrete timet, for N=100 and the initial condition N 10=N. The solid line represents the theoreticalresult for the time evolution of the average value of N 1.

10 1. Introduction to Statistical Physics

2Statistical Description of a PhysicalSystemThis is page 11Printer: Opaque this1. Neglect the complexities of classical phase space, and considera system of N distinguishable and noninteracting particles,which may be found in two states of energy, with = 0and > 0, respectively. Given the total energy U of this system,obtain an expression for the associated number of microscopicstates.*** Suppose that there are N 0 particles in the ground stateand N e particles in the excited state. The number of accessiblemicroscopic states of this system is given by (U; N) = N!N 0 !N e ! ;where N 0 + N e = N and U = N e .2. Calculate the number of accessible microscopic states ofa system of two localized and independent quantum oscillators,with fundamental frequencies ! o and 3! o , respectively, and totalenergy E = 10~! o .*** There are three microscopic states:(n 1 = 2; n 2 = 2), (n 1 = 5; n 2 = 1), and (n 1 = 8; n 2 = 0).

12 2. Statistical Description of a Physical System3. Consider a classical one-dimensional system of two noninteractingparticles of the same mass m: The motion of theparticles is restricted to a region of the x axis between x = 0and x = L > 0. Let x 1 and x 2 be the position coordinates of theparticles, and p 1 and p 2 be the canonically conjugated momenta.The total energy of the system is between E and E + E. Drawthe projection of phase space in a plane de…ned by position coordinates.Indicate the region of this plane that is accessible tothe system. Draw similar graphs in the plane de…ned by themomentum coordinates.4. The position of a one-dimensional harmonic oscillator isgiven byx = A cos (!t + ') ;where A; !; and ' are positive constants. Obtain p (x) dx, thatis, the probability of …nding the oscillator with position betweenx and x+dx. Note that it is enough to calculate dT=T , where Tis a period of oscillation, and dT is an interval of time, within aperiod, in which the amplitude remains between x and x + dx.Draw a graph of p(x) versus x.*** It is easy to show thatp (x) = 1 12A xA 2 1=2:Now consider the classical phase space of an ensemble of identicalone-dimensional oscillators with energy between E andE + E. Given the energy E, we have an ellipse in phase space.So, the accessible region in phase space is a thin elliptical shellbounded by the ellipses associated with energies E and E + E,respectively. Obtain an expression for the small area A of thiselliptical shell between x and x + dx. Show that the probabilityp(x)dx may also be given by A=A, where A is the total area ofthe elliptical shell. This is one of the few examples where we cancheck the validity of the ergodic hypothesis and the postulateof equal a priori probabilities.

2. Statistical Description of a Physical System 135. Consider a classical system of N localized and weakly interactingone-dimensional harmonic oscillators, whose Hamiltonianis written asNX 1H =2m p2 j + 1 2 kx2 j ;j=1where m is the mass and k is an elastic constant. Obtain theaccessible volume of phase space for E H E + E, withE 0 and the spin variable S j may assume the values 1or 0, for all j = 1; 2; 3:::. This spin Hamiltonian describes thee¤ects of the electrostatic environment on spin-1 ions. An ionin states 1 has energy D > 0, and an ion in state 0 has zeroj=1

14 2. Statistical Description of a Physical Systemenergy. Show that the number of accessible microscopic statesof this system with total energy U can be written as (U; N) =N!NU D!XN1UN !N ! ;Dfor N ranging from 0 to N, with N > U=D and N < U=D.Thus, we have (U; N) = N!2 U=D NU D 1 U! !:DUsing Stirling’s asymptotic series, show that1N ln ! u D ln 21u ln 1DuDuD ln u D ;for N; U ! 1, with U=N = u …xed. This last expression is theentropy per particle in units of Boltzmann’s constant, k B .*** Let´s write the number of microscopic con…gurations withN 0 ions with spin S = 0, N + ions with spin +1, and N ionswith spin 1, (N 0 ; N + ; N ) =N!N !N 0 !N + ! :It is easy to see that the number of microscopic con…gurations,with energy U and total number of ions N, is given by the sumwith the restrictions = (U; N) =XN 0 ;N + ;NN!N !N 0 !N + ! ;N 0 + N + + N= NandD (N + + N ) = U:

2. Statistical Description of a Physical System 15We then use these restrictions to eliminate two variables, andwrite (U; N) =U=DXN =0N!N U D!UDN !N ! ==U=DN! XN D U !UD !N =0UD!UN !N ! :DNow it is easy to calculate the (binomial) sum and obtain the(exact) answer.*** We usually look for results in the thermodynamic limitonly. It is then acceptable to replace the sum by its maximumterm. In fact, we can write (U; N) =XN 0 ;N + ;NN!N !N 0 !N + ! N!eN ! N e 0 ! N e + ! ;which is the asymptotic result in the limit N; U ! 1, withU=N = u …xed. In order to …nd the “occupation numbers” e N ,eN 0 , and e N + , we use the technique of Lagrange multipliers. Letus de…ne the functionf (N ; N 0 ; N + ; 1 ; 2 ) = lnN!N !N 0 !N + ! + 1 (N 0 + N + + N N) + 2 (DN + + DN U) :Using Stirling´s expansion, we take derivatives with respect toall of the arguments. It is easy to eliminate the Lagrange multipliers.The maximum is given byeN= e N + = U2D ; e N0 = N U D ;from which we have the same asymptotic expression1N ln u D ln 21u ln 1DuDuD ln u D ;

16 2. Statistical Description of a Physical Systemin agreement with the limiting result from the previously obtainedexact expression for . In slightly more complicated problems,without an exact solution, we will be forced to resort tosimilar maximization techniques.7. In a simpli…ed model of a gas of particles, the system isdivided into V cells of unit volume. Find the number of ways todistribute N distinguishable particles (with 0 N V ) withinV cells, such that each cell may be either empty or …lled upby only one particle. How would your answer be modi…ed forindistinguishable particles?*** If we consider distinguishable particles, we have d =V !(V N)! :This result, however, does not make sense in the thermodynamiclimit (see that ln d =N does not exist in the limit V; N ! 1,with v = V=N …xed).If the particles are indistinguishable, we have i =V !(V N)!N! ;so that1N ln i v ln v (v 1) ln (v 1) ;in the thermodynamic limit (note that v = V=N 1). This isa simpli…ed, and fully respectable, model of a non-interactinglattice gas. The entropy per particle is given bys = s (v) = k B [v ln v (v 1) ln (v 1)] ;from which we obtain the pressure,pT = @s@v = k vB lnv 1 :It is more instructive to write the pressure in terms of the particledensity, = 1=v,pT = k B ln (1 ) = k B + 1 2 2 + ::: :

2. Statistical Description of a Physical System 17This is a virial expansion. The low-density limit gives the wellknownexpression for the ideal gas.8. The atoms of a crystalline solid may occupy either a positionof equilibrium, with zero energy, or a displaced position,with energy > 0. To each equilibrium position, there correspondsa unique displaced position. Given the number N ofatoms, and the total energy U, calculate the number of accessiblemicroscopic states of this system.*** It easy to see that (U; N) =UN!! N :U !Additional exercises9. Obtain an expression for the volume of a hypersphere of radiusR in d dimensions. Use this expression for obtaining t i=1 hevolume (E; V; N; E) in phase space associated with a gas ofN non-interacting classical monatomic particles, inside a box ofvolume V , with energy between E and E +E (with E

18 2. Statistical Description of a Physical Systemand make = 0 at the end of the calculations. Using an integralrepresentation of the delta-function (see the Appendix), we have0 1+1dYZ!NXA = 2R @ dx iA exp 1i=11 1+1 Z "dk exp2= R +1 Z1ikRNXi=1i=1x 2 ix 2 i!#dk exp ikR 2 d=2:d ikChanging variables, kR 2 = z, and resorting to the method ofresidues in the complex plane, we haveA = 1 Z Rd 1 (i) d=2 exp ( iz)dz[z + iR 2 ] : d=2Closing the contour around a pole of order d=2, we obtain the…nal resultA =2dd1 ! Rd 1 :2Now it is easy to write the volume in phase space, andto see that we have to divide this volume by the Boltzmanncounting factor N! in order to obtain an extensive entropy inthe thermodynamic limit.10. Stirling´s asymptotic expansion, given byln N! = N ln N N + O (ln N) ;which works very well for large N (N ! 1), is a most usefultrick in statistical mechanics, in connection with the thermodynamiclimit.(i) Show that Z 1x n e x dx = n!0=

2. Statistical Description of a Physical System 19for n = 0; 1; 2; ::: (and assuming an analytic continuation relatedto the Gamma function).(ii) Using this integral and the Laplace method of asymptoticintegration, derive the …rst two terms of Stirling´s expansion.(ii) Prove that1limn!1n ln Z baexp [nf (x)] dx = f (x 0 ) ;where x 0 is the maximum of a continuous function f (x) in theinterval between a and b > a. This results gives a good degreeof con…dence in the usual replacements of certain sums by theirmaximum term!*** For integer n, we can used the induction method to provethe that Z 1x n e x dx = n!0It is simple to check that it works for n = 1 (and n = 0, with0! = 1). Supposing that if holds for n 1, it is easy to show thatit holds for n as well.*** The use of Laplace´s method to …nd the …rst few terms inthe asymptotic expansion of ln n! is fully described in AppendixA1.*** The mathematical proof that has been asked is based on asequence of very reasonable steps. Since f (x 0 ) is the maximumof f (x) in the interval between a and b, it is immediate to seethatI n =Z baZ baexp fn [f (x)f (x 0 )]g dx exp f[f (x) f (x 0 )]g dx = C;where C is a well-de…ned constant value.Let us …nd an inequality in the reversed direction. It is alwayspossible to writeI n =Z baexp fn [f (x)f (x 0 )]g dx

20 2. Statistical Description of a Physical SystemZ xo+ 1 2 "x o12 " exp fn [f (x) f (x 0 )]g dx:Supposing that f (x) is a continuous function, it is clear that,given > 0, there exists " > 0 such that jf (x) f (x 0 )j < ,for all . Thus,Therefore,I n which leads toZ xo12 "x o12 " exp fn [f (x) f (x 0 )]g dx Z xo12 "" exp [ n] exp [nf (x 0 )] x o12 " exp [ n] dx = " exp [ n] :" exp [ n] I n C;Z baexp [nf (x)] dx C exp [nf (x 0 )] :Taking the logarithm and dividing by n, we have1n ln "+f (x 0) 1 n ln Z baexp [nf (x)] dx 1 n ln C+f (x 0) :In the limit n ! 1, and taking into account that " is …xed, wehaveZ1 b + f (x 0 ) limn!1 n ln exp [nf (x)] dx f (x 0 ) :aSince > 0 is arbitrary, we can take the limit ! 0, which leadsto the expected result. Note that the only requirements are theexistence of the integral and the continuity of the function f (x).

3Overview of ClassicalThermodynamicsThis is page 21Printer: Opaque this1. The chemical potential of a simple ‡uid of a single componentis given by the expressionp = o (T ) + k B T lnp o (T ) ;where T is the temperature, p is the pressure, k B is the Boltzmannconstant, and the functions o (T ) and p o (T ) are well behaved.Show that this system obeys Boyle’s law, pV = Nk B T .Obtain an expression for the speci…c heat at constant pressure.What are the expressions for the thermal compressibility,the speci…c heat at constant volume, and the thermal expansioncoe¢ cient? Obtain the density of Helmholtz free energy,f = f (T; v).**** Note that (T; p) = g (T; p), where g (T; p) is the Gibbsfree energy per particle. Thus, @v = = k BT@p p ;which is the expression of Boyle’s law, and @s = = d o@T dT + k pB lnp o (T )pTk B T dp op o (T ) dT ;

22 3. Overview of Classical Thermodynamicsfrom which we obtain the speci…c heat at constant pressure. Allother expressions are straightforward. In particular,f = gpv:You should give f in terms of T and v, f = f (T; v).2. Consider a pure ‡uid of one component. Show that @cV @ 2 p= T :@vT@T 2 vUse this result to show that the speci…c heat of an ideal gas doesnot depend on volume. Show that @@T=:@p@TT;N*** From the de…nition of the speci…c heat, we have @sc V = T =)@T @cV@v =) = T @2 sT@T @v = T @2 s @ @s@v@T = T :@T @vT vNote that s = s (T; v) is an equation of state in the Helmholtzvp;Nrepresentation. Then, we write @fdf = sdT pdv =) s = ; p =@Tv @s @p=) = ;@vT@Tv @f@vTwhich leads to the …rst identity.The proof of the second identity requires similar tricks.3. Consider a pure ‡uid characterized by the grand thermodynamicpotential = V f o (T ) exp ;k B T

3. Overview of Classical Thermodynamics 23where f o (T ) is a well-behaved function. Write the equations ofstate in this thermodynamic representation. Obtain an expressionfor the internal energy as a function of T , V , and N. Obtainan expression for the Helmholtz free energy of this system. Calculatethe thermodynamic derivatives T and as a function oftemperature and pressure.*** From Euler’s relation, we haveThus, we can writep = V = f o (T ) exp = k B T lnpf o (T ) ; :k B Twhich is identical to the expression for the chemical potentialin the …rst exercise, if we make o = 0 and p o (T ) = f o (T ).Therefore, we have Boyle’s law and the usual expressions for Tand .4. Obtain an expression for the Helmholtz free energy perparticle, f = f (T; v), of a pure system given by the equationsof stateu = 3 2 pv and p = avT 4 ;where a is a constant.*** These equations of state can be explicitly written in theentropy representation, 1=41 3aT = v 1=2 u 1=4 pand2T = 2 1=4 3av 1=2 u 3=4 ;3 2from which we obtain the fundamental equations = 4 3 3a2 1=4v 1=2 u 3=4 + c;where c is a constant. The Helmholtz free energy per particle isgiven byf (T; v) = u T s =

24 3. Overview of Classical Thermodynamics"= 3a 1=4 3=42 v2 T 4 4 3a 3aTv 1=23 22 v2 T 4 + c#:*** Let us consider a similar problem, with a slight modi…cationin one of the equations of state,u = 3 2 pv and p = avT n :Note that, instead of T 4 , we are writing T n , where n is an arbitraryinteger. Is this a bona …de thermodynamic system? Is itpossible to have n 6= 4?Again, we rewrite the equations of state in the entropy representation, 1=n1 3aT = v 2=n u 1=n2andpT = 2 1=n 3av 1+2=n u 1 1=n :3 2In this representation we have 1 @sT = ; p @s@u T = @vvu @ 1 @=) =@v Tu@ufrom which we obtain 1=n 3a 22 n v 1+2=n u 1=n = 2 1=n 3av 1+2=n 13 2pTv;1u 1=n ;nleading to the only thermodynamic bona …de solution, n = 4.5. Obtain an expression for the Gibbs free energy per particle,g = g (T; p), for a pure system given by the fundamentalequation SN c 4= a V U 2N 3 ;where a and c are constants.*** From the fundamental equations = a 1=4 v 1=4 u 1=2 + c;

3. Overview of Classical Thermodynamics 25we write the equations of state 1 @sT = = 1 @u 2 a1=4 v 1=4 u 1=2vand p @sT = = 1 @vu4 a1=4 v 3=4 u 1=2 :The Gibbs free energy per particle is given by the LegendretransformationgT = s 1T u pT v;where u and v come from the equations of state. Note that ghas to be given in terms of T and p.6. Consider an elastic ribbon of length L under a tension f.In a quasi-static process, we can writedU = T dS + fdL + dN:Suppose that the tension is increased very quickly, from f tof + f, keeping the temperature T …xed. Obtain an expressionfor the change of entropy just after reaching equilibrium. Whatis the change of entropy per mole for an elastic ribbon thatbehaves according to the equation of state L=N = cf=T , wherec is a constant?*** Using the Gibbs representation, we have the Maxwell relation @S @L= :@f @TFrom the equation of state, L=N = cf=T , we haveTSN = cfT 2 f:f7. A magnetic compound behaves according to the Curie law,m = CH=T , where C is a constant, H is the applied magnetic

26 3. Overview of Classical Thermodynamics…eld, m is the magnetization per particle (with corrections dueto presumed surface e¤ects), and T is temperature. In a quasistaticprocess, we havedu = T ds + Hdm;where u = u (s; m) plays the role of an internal energy. For anin…nitesimal adiabatic process, show that we can writeT = CHc H T H;where c H is the speci…c heat at constant magnetic …eld.*** We have to calculate the partial derivative (@T=@H) at…xed entropy. Using Jacobians, it is easy to write @T=@Hs@ (T; s)@ (H; s) @ (T; s) @ (T; H) @s=@ (T; H) @ (H; s) = @HTAll derivatives are written in terms of the independent variablesT and H. We then introduce the Legendre transformationg = u T s Hm =) dg = sdT mdH;from which we have @gs = ; m =@TH @g@HT=) @s@HT=@s@T1H @m:@THInserting the equation of state in this Maxwell relation, it is easyto complete the proof.8. From stability arguments, show that the enthalpy of a pure‡uid is a convex function of entropy and a concave function ofpressure.*** The entalpy per particle is given byh = u + pv;:

3. Overview of Classical Thermodynamics 27from which we havedh = T ds + vdp =) T = @h@spand v = @h:@psIt is easy to show that @ 2 hAlso, we have @ 2 h@s 2 p@p 2 s== @T= T > 0:@spc p @v= v s < 0:@psIt is straightforward to use standard tricks (Jacobians, for example)to write an expression for the adiabatic modulus of compressibility, s =1 @v;v @psin terms of positive quantities.*9. Show that the entropy per mole of a pure ‡uid, s =s (u; v), is a concave function of its variables. Note that we haveto analyze the sign of the quadratic formd 2 s = 1 @ 2 s2 @u 2 (du)2 + @2 s@u@v dudv + 1 @ 2 s2 @v 2 (dv)2 :*** This quadratic form can be written in the matrix notationdu dv @ 2 s@u du2:dvd 2 s = 1 2@ 2 s@u@v@ 2 s@u@v@ 2 s@v 2The eigenvalues of the 22 matrix are the roots of the quadraticequation 2 @ 2 s@u + @2 s + @2 s @ 2 s2 @v 2 @u 2 @v 2 @ 2 2s= 0:@u@v

28 3. Overview of Classical ThermodynamicsFor a concave function, the eigenvalues are negative, that is,@ 2 s @ 2 s@u 2 @v 2 @ 2 2s> 0@u@vand@ 2 s@u + @2 s2 @v > 0: 2Now it is straightforward to relate these derivatives of the entropywith positive physical quantities (as the compressibilitiesand the speci…c heats).

4Microcanonical EnsembleThis is page 29Printer: Opaque this1. Consider a model of N localized magnetic ions, given by thespin HamiltonianNXH = D Sj 2 ;where the spin variable S j may assume the values 1; 0; or +1;for all j (see exercise 6 of Chapter 2). Given the total energyE, use the expression for the number of accessible microstates, (E; N), to obtain the entropy per particle, s = s (u), whereu = E=N. Obtain an expression for the speci…c heat c in termsof the temperature T . Sketch a graph of c versus T . Check theexistence of a broad maximum associated with the Schottkye¤ect. Write an expression for the entropy as a function of temperature.What are the limiting values of the entropy for T ! 0and T ! 1?*** The number of accessible microscopic states, (E; N),has already been calculated in exercise 6 of Chapter 2. Theentropy per magnetic ion is given byj=1s = s (u) = k B lim 1 ln (E; N) ;N

30 4. Microcanonical Ensemblein the thermodynamic limit, E; N ! 1, with u = E=N …xed.We thus have11k Bs = u D ln 2u ln 1DuDfrom which we obtains the equation of state1k B T = 1 D2 (1ln u=D)u=D:uD ln u D ;The inversion of this equation leads to the dependence of theenergy on the temperature,u =2D exp ( D)1 + 2 exp ( D) ;where = 1= (k B T ). The speci…c heat c = c (T ) is given by thederivative of u with respect to T . Check the broad maximum inthe graph of c versus T .We now write the entropy s in terms of the temperature, s =s (T ). Check that c = T (@s=@T ). Draw a graph of s (T ) versusT . Check that s (T ) ! 0 as T ! 0, and that s (T ) ! k B ln 3 asT ! 1 (for D > 0). What happens if D < 0?2. In the solid of Einstein, we may introduce a volume coordinateif we make the phenomenological assumption that thefundamental frequency ! as a function of v = V=N is given by v! = ! (v) = ! o A ln ;v owhere ! o ; A, and v o are positive constants. Obtain expressionsfor the expansion coe¢ cient and the isothermal compressibilityof this model system.*** Taking ! as a function of v, ! = ! (v), the entropy of Einstein’ssolid can be written as a function of energy and volume,s = s (u; v). From the equations of state, it is straightforward toobtain the expansion coe¢ cient and the compressibility T .

4. Microcanonical Ensemble 313. Consider the semiclassical model of N particles with twoenergy levels (0 and > 0). As in the previous exercise, supposethat the volume of the gas may be introduced by the assumptionthat the energy of the excited level depends on v = V=N, = (v) = a v ;where a and are positive constants. Obtain an equation ofstate for the pressure, p = p (T; v), and an expression for theisothermal compressibility (note that the constant plays therole of the Grüneisen parameter of the solid).*** Again, as = (v), we can write s = s (u; v). From theequations of state, it is easy to obtain the isothermal compressibility.4. The total number of the accessible microscopic states ofthe Boltzmann gas, with energy E and number of particles N,may be written as (E; N) =XN 1 ;N 2 ;:::N!N 1 !N 2 ! ;with the restrictionsXN j = NjandX j N j = E:jExcept for an additive constant, show that the entropy per particleis given by! !X ~N j~N js = k B ln ;N Njn owhere ~Nj is the set of occupation numbers at equilibrium.Using the continuum limit of the Boltzmann gas, show that theentropy depends on temperature according to a term of the formk B ln T (note the correction!).

32 4. Microcanonical Ensemble*** In the thermodynamic limit, we replace the sum by itslargest term,N! (E; N) eN 1 ! N e 2 ! ;so the entropy per particle is written as1s = k B ln (E; N) N kBXwhich resembles the H-function of Boltzmann (see chapter 15).Also, note the similarity with Shannon´s entropy of informationtheory (see chaptern 5). oIn order to …nd eNj , we resort to the method of Lagrangemultipliers. Let´s introduce the multipliers 1 and 2 , and minimizethe functionj~N jN!N!f (fN j g ; 1 ; 2 ) = ln YN j ! ++ 1 N X jN j!+ 2 E X jjln j N j!~N jN:!;It is straightforward to see thatso thateN j = exp ( 1 2 j ) ;eN jN = exp ( 2 j )Xexp ( 2 j ) :The second Lagrange multiplier comes form the energy,Xjj je NjN = E N = u;

4. Microcanonical Ensemble 33from which we haveu =@@ 2Z 1 ;Z 1 = X jexp ( 2 j ) :In the continuum limit, we haveZ 1 = X Zexp ( 2 j ) d 3! p expj p 23=2 2m 2 =;2m 2which leads to the energy,u = 3 1= 3 2 2 2 k BT;and to the identi…cation of the Lagrange multiplier 2 withthe inverse of the temperature.Now it is easy to write the continuum form of the entropy perparticle,s k BZd 3! pexpZ 1 p22mlnexp p22m=Z 13= k B ln T + constant;2which should be compared with the entropy per particle for theideal gas (pv = k B T , u = 3k B T=2),s = k B32 ln T + k B ln v + constant.Note that s ! 1 for T ! 0, which is a well-known di¢ cultyof classical statistical mechanics.5. Consider a lattice gas of N particles distributed among Vcells (with N V ). Suppose that each cell may be either emptyor occupied by a single particle. The number of microscopicstates of this system will be given by (V; N) =V !N! (V N)! :

34 4. Microcanonical EnsembleObtain an expression for the entropy per particle, s = s (v),where v = V=N. From this fundamental equation, obtain anexpression for the equation of state p=T . Write an expansion ofp=T in terms of the density = 1=v. Show that the …rst termof this expansion gives the Boyle law of the ideal gases. Sketcha graph of =T , where is the chemical potential, in terms ofthe density . What is the behavior of the chemical potential inthe limits ! 0 and ! 1?*** Look at the solution of exercise 7 of Chapter 2. The entropyparticle of this lattice gas model is given bys = k B [v ln v (v 1) ln (v 1)] ;from which we have the equation of state pk B T = ln 11 = 1 v v + 12v + 12 3v + ::: 3Note that Bolyle’s law is already given by the …rst term is thisexpansion.In order to …nd =T , we write the thermodynamic entropyS = Ns = S (V; N), and take the partial derivative with respectto N. Note that we can …nd ratios, as p=T and =T , butwe cannot …nd an independent expression for the temperature(since there is no mention to the energy in the de…nition of thisvery simple and schematic model).It is instructive to draw a graph of =T versus the particledensity = 1=v (remember that 0 < < 1). Note that is anincreasing function of . Also, note that =T ! 1 for ! 0; 1(in the vacuum).*** This lattice gas model can be alternatively de…ned interms of a set variables ft i g, i = 1; :::; V , such that t i = 1 if celli is occupied, and t i = 0 if cell i is empty. The total number ofaccessible states (V; N) is given by a (restricted) sum over thecon…gurations of these occupation variables,X (V; N) =1;ft i g;restriction

with the restriction4. Microcanonical Ensemble 35VXt i = N:i=1Of course, we obtain the same result, (V; N) = V !=N! (V N)!This problem becomes more di¢ cult (and much more realistic)if we include interactions between particles. For example,we can mimic the short-range attraction between particles byassigning an energy , with > 0, to each nearest-neighborpair of occupied lattice sites. We then write the HamiltonianH = X (i;j)t i t j ;where the sum is over nearest-neighbor pairs of lattice sites.Given the total energy U, the calculation of the number of accessiblemicrostates, (U; V; N), becomes a formidable mathematicalproblem (related to the well-known Ising model of magnetism,as we shall see in the forthcoming chapters).

36 4. Microcanonical Ensemble

5Canonical EnsembleThis is page 37Printer: Opaque this“We consider especially ensembles of systems in which the index(or logarithm) of probability of phase is a linear function of energy.The distribution, on account of its unique importance in the theoryof statistical equilibrium, I have ventured to call canonical, andthe divisor of the energy, the modulus of the distribution. The moduliof ensembles have properties analogous to temperature, in thatequality of the moduli is a condition of equilibrium with respect toexchange of energy, when such exchange is made possible.” - J. W.Gibbs, “Elementary Principles In Statistical Mechanics: DevelopedWith Especial Reference To The Rational Foundation Of Thermodynamics”,Scribner´s, New York, 1902.1. The energy of a system of N localized magnetic ions, attemperature T , in the presence of a …eld H, may be written asH = DNXi=1S 2 i o HNXS i ;i=1where the parameters D, o , and H are positive, and S i = +1; 0;or 1, for all sites i. Obtain expressions for the internal energy,the entropy, and the magnetization per site. In zero …eld (H =

38 5. Canonical Ensemble0), sketch graphs of the internal energy, the entropy, and thespeci…c heat versus temperature. Indicate the behavior of thesequantities in the limits T ! 0 and T ! 1. Calculate theexpected value of the “quadrupole moment,”Q = 1 N* NXi=1S 2 ias a function of …eld and temperature.*** The canonical partition function is given byZ = X fS i gexp= X"S 1 ;S 2 ;:::;S Nexp"DD+NXSi 2 + o Hi=1NXSi 2 + o Hi=1;#NXS i =i=1#NXS i =i=1( XS=exp DS 2 + o HS ) N== f2 exp ( D) cosh ( o H) + 1g N :In this simple case, the thermodynamic limit is trivial. The internalenergy per ion comes fromu =1 N@ln Z:@The magnetization per spin is given bym = 1 @ln Z;N @Hand the “quadrupolar moment”byQ =1 @ln Z:D @D

5. Canonical Ensemble 39In zero …eld, we have u (T = 0) = 0, u (T ! 1) = 2D=3,s (T = 0) = 0, and s (T ! 1) = k B ln 3. What happens forD < 0?2. Consider a one-dimensional magnetic system of N localizedspins, at temperature T , associated with the energyH =XJ i i+1i=1;3;5;:::;N 1 o HNX i ;i=1where the parameters J, o , and H are positive, and i = 1for all sites i. Assume that N is an even number, and note thatthe …rst sum is over odd integers.(a) Obtain an expression for the canonical partition functionand calculate the internal energy per spin, u = u (T; H). Sketcha graph of u (T; H = 0) versus temperature T . Obtain an expressionfor the entropy per spin, s = s (T; H). Sketch a graphof s (T; H = 0) versus T .(b) Obtain expressions for the magnetization per particle,*m = m (T; H) = 1 N+XN o i ;i=1and for the magnetic susceptibility, @m = (T; H) =@HSketch a graph of (T; H = 0) versus temperature.*** The canonical partition function is given byZ = X "exp J X #NX i i+1 + o H i =f i g i oddT:i=1=( X 1 ; 2exp [J 1 2 + o H ( 1 + 2 )]) N=2=

40 5. Canonical Ensemble= [2 exp (J) cosh (2 o H) + 2 exp ( J)] N=2 :Note the factorization of this sum.The magnetization per particle is given by!NX im = limN!11N1 XZf i gi=1exp ( H) = 1N= osinh (2 o H)cosh (2 o H) + exp ( 2J) ;@@H ln Z =whose derivative with respect to H yields the susceptibility.From the free energy,we calculate the entropy,f =s = s (T; H) =1ln Z;N @f:@THCheck that, in zero …eld, H = 0, the entropy is given bys (T; H = 0) = 1 2 k B ln (4 cosh J)with the limiting valuesfor T ! 0 (J ! 1), ands (T; H = 0) ! 1 2 k B ln 2;s (T; H = 0) ! k B ln 2;12 k BJ tanh J;for T ! 1 (J ! 0). How do you explain the residual entropyat T = 0?3. Consider a system of N classical and noninteracting particlesin contact with a thermal reservoir at temperature T . Eachparticle may have energies 0, > 0, or 3. Obtain an expression

5. Canonical Ensemble 41for the canonical partition function, and calculate the internalenergy per particle, u = u (T ). Sketch a graph of u versus T(indicate the values of u in the limits T ! 0 and T ! 1). Calculatethe entropy per particle, s = s (T ), and sketch a graph ofs versus T . Sketch a graph of the speci…c heat versus temperature.*** The canonical partition function is given byZ = [1 + exp ( ) + exp ( 3)] N :For T ! 1, we have u = 4=3 and s = k B ln 3. Indicate theslopes of the graph of u (T ) versus T , for T ! 0 and T ! 1.Calculate the asymptotic form of the speci…c heat c (T ) for T !0 and T ! 1. Note that the graph of c (T ) versus T displays abroad maximum at the typical value k B T .4. A system of N localized and independent quantum oscillatorsis in contact with a thermal reservoir at temperature T .The energy levels of each oscillator are given by n = ~! o n + 1 ; with n = 1; 3; 5; 7; ::::2Note that n is an odd integer.(a) Obtain an expression for the internal energy u per oscillatoras a function of temperature T . What is the form of u inthe classical limit (~! o

42 5. Canonical EnsembleThe internal energy is given byu =@@ ln Z 1 = 3 2 ~! 2~! oo +exp (2~! o ) 1 :Note that u ! k B T in the classical limit (the classical speci…c,c = k B , behaves according to the law of Dulong and Petit).5. Consider a system of N noninteracting classical particles.The single-particle states have energies n = n, and are n timesdegenerate ( > 0; n = 1; 2; 3; :::). Calculate the canonical partitionfunction and the entropy of this system. Obtain expressionsfor the internal energy and the entropy as a function of temperature.What are the expressions for the entropy and the speci…cheat in the limit of high temperatures?*** The thermodynamic functions come from canonical partitionfunction, given by Z = Z N 1 , whereZ 1 =Xn=1;2;3;:::n exp ( n) =exp ()[exp () 1] 2 :6. A set of N classical oscillators in one dimension is givenby the HamiltonianNX 1H =2m p2 i + 1 2 m!2 qi2 :i=1Using the formalism of the canonical ensemble in classical phasespace, obtain expressions for the partition function, the energyper oscillator, the entropy per oscillator, and the speci…c heat.Compare with the results from the classical limit of the quantumoscillator. Calculate an expression for the quadratic deviation ofthe energy as a function of temperature.*** The thermodynamic functions come from canonical partitionfunction,Z =+1 Zdq 1 :::+1 Z+1 Zdq Ndp 1 :::+1 Zdp N exp [ H] =1111

8< Z=:+11+1 Z1dpdq exp5. Canonical Ensemble 43 9 N m! 2 =2m p2 q 22 ;= N 2:!* 7. Consider again the preceding problem. The canonical partitionfunction can be written as an integral form,1ZZ () = (E) exp ( E) dE;0where (E) is the number of accessible microscopic states of thesystem with energy E. Note that, in the expressions for Z ()and (E), we are omitting the dependence on the number N ofoscillators. Using the expression for Z () obtained in the lastexercise, perform a reverse Laplace transformation to obtainan asymptotic form (in the thermodynamic limit) for (E).Compare with the expression calculated in the framework ofthe microcanonical ensemble.*** First, we use an integral representation of the -function(see Appendix) to write=1Z0+i1 Z1Z () exp (E 0 ) d =2i1dE (E) 1+i1 Zexp [ (E 0 E)] d = (E 0 ) :2i1Inserting the result of the previous exercise, we have (E) = 12+i1 Zi1 N 2exp (E) d;!which can be written in the form of a saddle-point integration(see Appendix), (E) = 1+i1 Z 2exp N ln2!i1ln + u d;

44 5. Canonical Ensemblewhere u = E=N. Using the asymptotic integration techniques ofthe Appendix, we locate the saddle point at = 1=u and writethe asymptotic form (for N ! 1), (E) 2u 2 1=2 2 expN ln + ln u + 1 :!Therefore, we have the entropy per oscillator, 1 2s = ln + ln u + 1;k B !which should be compared with the well-known result for theclassical one-dimensional harmonic oscillator in the microcanonicalensemble.8. A system of N one-dimensional localized oscillators, at agiven temperature T , is associated with the HamiltonianwhereH =8 0;12 m!2 q 2 + ; for q < 0;with > 0.(a) Obtain the canonical partition function of this classicalsystem. Calculate the internal energy per oscillator, u = u (T ).What is the form of u (T ) in the limits ! 0 and ! 1?(b) Consider now the quantum analog of this model in thelimit ! 1. Obtain an expression for the canonical partitionfunction. What is the internal energy per oscillator of this quantumanalog?*** The classical partition function is given by Z = Z N 1 , whereZ 1 = [exp ( ) + 1] :!

5. Canonical Ensemble 45Note that u ! k B T at both limits, ! 0 and ! 1.The quantum results are simple in the limit of an in…nitepotential barrier ( ! 1). in this limit, we have to discard allof the harmonic oscillator states with even values of n (whichare associated with even wave functions; wave functions that donot vanish at the origin, q = 0).

46 5. Canonical Ensemble

6The Classical Gas in the CanonicalFormalismThis is page 47Printer: Opaque this1. A system of N classical ultrarelativistic particles, in a containerof volume V , at temperature T , is given by the HamiltonianNXH = c j~p i j ;i=1where c is a positive constant. Obtain an expression for thecanonical partition function. Calculate the entropy per particleas a function of temperature and speci…c volume. What is theexpression of the speci…c heat at constant volume?*** The thermodynamic quantities are easily obtained fromthe classical canonical partition function, given byZ = 1 Zh 3N N! V N d 3 p exp ( c j ! Np j) = 1 N 8Vh 3N N! (c) 3 :2. Consider a set of N one-dimensional harmonic oscillators,described by the HamiltonianNX 1H =2m ~p i 2 + 1 2 m!2 x n i ;i=1

48 6. The Classical Gas in the Canonical Formalismwhere n is a positive and even integer. Use the canonical formalismto obtain an expression for the classical speci…c heat ofthis system.*** Note that the canonical partition function may be writtenas Z = Z N 1 , where 1=2 n=2 Z 2m 2+1Z 1 =dy exp ( y n ) : m! 21Thus,u =@@ ln Z 1 = n + 22n k BT:3. Consider a classical system of N very weakly interactingdiatomic molecules, in a container of volume V , at a given temperatureT . The Hamiltonian of a single molecule is given byH m = 12m ~p 1 2 + ~p 22 1 +2 j~r 1 ~r 2 j 2 ;where > 0 is an elastic constant. Obtain an expression for theHelmholtz free energy of this system. Calculate the speci…c heatat constant volume. Calculate the mean molecular diameter,D = j~r 1 ~r 2 j 2 1=2:Now consider another Hamiltonian, of the form H m = 12m ~p 1 2 + ~p 22 + jr12 r o j ;where and r o are positive constants, and r 12 = j~r 1are the changes in your previous answers?~r 2 j. What*** Taking into account the thermodynamic limit (V ! 1),the …rst Hamiltonian is associated with the partition function 3 Z 2mZ 1 =d 3 r 1Zd 3 r 2 exp2 j! r 1! r2 j 2

6. The Classical Gas in the Canonical Formalism 49 3 2m V 3=2 2;from which we obtain the Helmholtz free energy (and the valueof D). It interesting to check the constant value of the molecularspeci…c heat, c = (9=2) k B .The second Hamiltonian represents a more complicated modelof a diatomic molecule. The partition function is given by 3 Z Z2mZ 1 =d 3 r 1 d 3 r 2 exp [ j(j ! r!1 r2 j r o )j] whereZI = 4r o0 3 2m V I;1Zr 2 exp [ (r o r)] dr+4 r 2 exp [ (r r o )] dr =r o= 2r2 o + 4() 3 2() 4 exp ( r o) :Now it is interesting to obtain the speci…c heat as a function oftemperature.4. Neglecting the vibrational motion, a diatomic moleculemay be treated as a three-dimensional rigid rotator. The HamiltonianH m of the molecule is written as a sum of a translational,H tr , plus a rotational, H rot , term (that is, H m = H tr + H rot ).Consider a system of N very weakly interacting molecules ofthis kind, in a container of volume V , at a given temperatureT .(a) Obtain an expression for H rot in spherical coordinates.Show that there is a factorization of the canonical partitionfunction of this system. Obtain an expression for the speci…cheat at constant volume.(b) Now suppose that each molecule has a permanent electricdipole moment ~ and that the system is in the presence of an

50 6. The Classical Gas in the Canonical Formalismexternal electric …eld ~ E (with the dipole ~ along the axis ofthe rotor). What is the form of the new rotational part of theHamiltonian? Obtain an expression for the polarization of themolecule as a function of …eld and temperature. Calculate theelectric susceptibility of this system.*** The Lagrangian of a free rotator (two atoms of mass mand a …xed interatomic distance a) is given byL rot = ma24" d 2+dtfrom which we have the Hamiltonian 2 d'sin #2 ;dtTherefore,H rot =p2 ma + p 2 '2 ma 2 sin 2 :u rot = hH rot i = 1 2 k BT + 1 2 k BT = k B T:In the presence on an electric …eld (taken along the z direction),and with the magnetic moment along the axis of the rotator, wehavep2 H =ma + 2 ma 2 sin 2 E cos :The associated partition function is given byp 2 'Z 1 =Z 0Zd20d'+1 Z1+1 Zdp 1dp ' exp [p 2 ma 2p 2 'ma 2 sin 2 +Thus, we have+E cos ] = 42 ma 2h cos i = coth (E)2 sinh (E):E1= L (E) ;E

6. The Classical Gas in the Canonical Formalism 51where L (x) is known as the Langevin function.5. Consider a classical gas of N weakly interacting molecules,at temperature T , in an applied electric …eld ~ E. Since there isno permanent electric dipole moment, the polarization of thissystem comes from the induction by the …eld. We then supposethat the Hamiltonian of each molecule will be given by the sumof a standard translational term plus an “internal term.” Thisinternal term involves an isotropic elastic energy, which tendsto preserve the shape of the molecule, and a term of interactionwith the electric …eld. The con…gurational part of the internalHamiltonian is be given byH = 1 2 m!2 or 2q ~ E ~r:Obtain the polarization per molecule as a function of …eld andtemperature. Obtain the electric susceptibility. Compare withthe results of the last problem. Make some comments aboutthe main di¤erences between these results. Do you know anyphysical examples corresponding to these models?*** First, we calculate the con…gurational partition function1Z Z Z2m!Z 1 = r 2 2dr sin d d' expor 2 + qEr cos =200 3=2 2q 2 E 2=exp :m! 2 o 2m! 2 oThe polarization is given by0hqr cos i = 1 @@E ln Z 1 = q2 E;m! 2 oso the dielectric susceptibility is just a constant (in sharp contrastto the previous result for permanent electric dipoles!)6. The equation of state of gaseous nitrogen at low densitiesmay be written aspvRT = 1 + B (T ) ;v

52 6. The Classical Gas in the Canonical Formalismwhere v is a molar volume, R is the universal gas constant, andB (T ) is a function of temperature only. In the following tablewe give some experimental data for the second virial coe¢ cient,B (T ), as a function of temperature.T (K) B (cm 3 =mol)100 160:0273 10:5373 6:2600 21:7Suppose that the intermolecular potential of gaseous nitrogen isgiven by8< 1; 0 < r < a;V (r) = V 0 ; a < r < b;:0; r > b:Use the experimental data of this table to determine the bestvalues of the parameters a, b, and V 0 .*** According to Section 6.4 (although Nitrogen is a gas ofdiatomic molecules), the virial coe¢ cient is given by1ZB =2r 2 fexp [ V (r)] 1g =0Z a= 2r 2 drZ b2r 2 [exp (V o )1] dr:0Now it is straightforward to …t the parameters a, b and V o .a

7The Grand Canonical and PressureEnsemblesThis is page 53Printer: Opaque this1. Show that the entropy in the grand canonical ensemble canbe written asXS = k B P j ln P j ;with probability P j given by,P j = 1 exp ( E j + N j ) :Show that this same form of entropy still holds in the pressureensemble (with a suitable probability distribution).2. Consider a classical ultrarelativistic gas of particles, givenby the HamiltonianNXH = c j~p i j ;ji=1where c is a positive constant, inside a container of volume V ,in contact with a reservoir of heat and particles (at temperatureT and chemical potential ). Obtain the grand partition functionand the grand thermodynamic potential. From a Legendretransformation of the grand potential, write an expression for

54 7. The Grand Canonical and Pressure Ensemblesthe Helmholtz free energy of this system. To check your result,use the integral in equation (??) for obtaining an asymptoticform for the canonical partition function.*** As we have already calculated in a previous exercise, thecanonical partition function is given byZ N = 1 N 8VN! (hc) 3 :Using this expression, we obtain the grand canonical partitionfunction,1X 8V = exp (N) Z N = exp(hc) 3 exp () ;N=0from which we have the grand potential, =1 8V3exp () : (hc)If we write exp () = z, the canonical partition function isgiven byZ = 1 I (; z; V )dz 1 Iexp [Nf (z)] dz;2i z N+1 2iwithf (z) =8v3z ln z;(hc)where v = V=N. From a saddle-point integration (see Appendix),we have1N ln Z f (z o) = 1ln (hc)38v :It is interesting to note that the same expression comes fromthe asymptotic form(1N ln Z N = 1 ) NN ln 1 8VN! (hc) 3 :

7. The Grand Canonical and Pressure Ensembles 553. Obtain the grand partition function of a classical systemof particles, inside a container of volume V , given by the HamiltonianNX 1H =2m ~p i 2 + u (~r i ) :i=1Write the equations of state in the representation of the grandpotential. For all reasonable forms of the single-particle potentialu (~r), show that the energy and the pressure obey typicalequations of an ideal gas.4. Show that the average quadratic deviation of the numberof particles in the grand canonical ensemble may be written as(N)2 = Nj 2 hN j i 2 = z @ z @ @z @z ln (; z) :Obtain an expression for the relative deviationq (N)2 = hN j iof an ideal gas of classical monatomic particles.and*** For a classical ideal gas of monatomic particles, we havehNi = z @ @z(N)2 = z @ @z 3=2zV 2mln (; z) =h 3 z @ @z ln (; z) The relative deviation is thus of order 1= p N.= zV 3=2 2m= hNi :h 3 5. Show that the average quadratic deviation of energy in thegrand canonical ensemble may be written as(E)2 = @UEj 2 hE j i 2 =;@z;V

56 7. The Grand Canonical and Pressure Ensembleswhere U = hE j i is the thermodynamic internal energy in termsof , z, and V . Hence, show that we may also write(E)2 " @U = k B T 2 + #@U@T;VT @T;V(you may use the Jacobian transformations of Appendix A.5).From this last expression, show that(E)2 = " (E) 2 @U+ k can BT 2 @N+ #@U @N;T @NT;V@T;VT;V @N@TV;where(E)2 can = Nk BT 2 c Vis the average quadratic deviation of energy in the canonicalensemble. Finally, show that" @U @N+ #@U @N@NT;V@TV;T @NT;V@T;Vsince= 1 T " # 2@N @U> 0;@T;V@NT;V(N)2 @N= k B T@T;V> 0:6. At a given temperature T , a surface with N o adsorptioncenters has N N o adsorbed molecules. Suppose that thereare no interactions between molecules. Show that the chemicalpotential of the adsorbed gas is given by = k B T lnN(N o N) a (T ) :

7. The Grand Canonical and Pressure Ensembles 57What is the meaning of the function a (T )?*** Suppose an adsorbed particle has energy . We thenwrite the canonical partition functionZ =N o !exp (N) :N! (N o N)!Now we can use the grand-canonical formalism to write = X NN o !N! (N o N)! exp (N) zN = [z exp () + 1] No ;from which we haveN =N o zz + exp ( ) ;and the expression for the chemical potential with a = exp ().7. The grand partition function for a simpli…ed statisticalmodel is given by the expression (z; V ) = (1 + z) V 1 + z V ;where is a positive constant. Write parametric forms of theequation of state. Sketch an isotherm of pressure versus speci…cvolume (draw a graph to eliminate the variable z). Show thatthis system displays a (…rst-order) phase transition. Obtain thespeci…c volumes of the coexisting phases at this transition. Findthe zeros of the polynomial (z; V ) in the complex z plane, andshow that there is a zero at z = 1 in the limit V ! 1.*** Note the structures of zeroes atz = 1 and z = exp [i (2n + 1) = (V )] ;for n = 0; 1; :::.Also, note the thermodynamic limit,1Vln ! ln (1 + z) + ln z; for z > 1

58 7. The Grand Canonical and Pressure Ensemblesand1ln ! ln (1 + z) ; for z < 1:VFor z > 1, we have the equations of statepk B T = ln (1 + z) + ln z; and 1 v = z2 + (1 + z):z (1 + z)For z < 1, we havepk B T = ln (1 + z) ; and 1 v = z1 + z :There is a …rst-order transition [v = 4= (1 + 2)] at z = 1.

8The Ideal Quantum GasThis is page 59Printer: Opaque this1. Obtain the explicit forms of the ground state and of the …rstexcited state for a system of two free bosons, with zero spin,con…ned to a one-dimensional region of length L. Repeat thisproblem for two fermions of spin 1=2.*** The single-particle states are given by ' (x) = A sin (kx),where k = p 2mE=~ and kL = n, with n = 1; 2; 3:::. Note that' (x) vanishes at x = 0 and x = L.The ground state is given by C sin (x 1 =L) sin (x 2 =L), whereC is a normalization constant.The …rst excited state is given byD [sin (x 1 =L) sin (2x 2 =L) + sin (2x 1 =L) sin (x 2 =L)] :2. Show that the entropy of an ideal quantum gas may bewritten asS =Xk B ff j ln f j (1 f j ) ln (1 f j )g ;j

60 8. The Ideal Quantum Gaswhere the upper (lower) sign refers to fermions (bosons), andf j = hn j i =1exp [ ( j )] 1is the Fermi–Dirac (Bose–Einstein) distribution. Show that wecan still use these equations to obtain the usual results for theclassical ideal gas.3. Show that the equation of statepV = 2 3 Uholds for both free bosons and fermions (and also in the classicalcase). Show that an ideal ultrarelativistic gas, given by theenergy spectrum = c~k, still obeys the same equation of state.*** For fermions, we have = pV = 1 Xln [1 + z exp ( j )] :jIn the thermodynamic limit, for free fermions, we can writeXln [1 + z exp ( j )] V 1Z(2) 3= V (2) 3 f 4 3 k3 ln+ 2 1Z ~4k 2 2 k 2dk32m00 2 3j ~4k 2 2 k 2dk ln 1 + z exp=2mXj1 + z exp z 1 exp ~ 2 k 2 1+2m ~ 2 k 21+ 1g 2m jz 1 exp ( j ) + 1 = 2 3 U;0

8. The Ideal Quantum Gas 61from which we prove that pV = 2U=3. The same manipulationscan be carried out for bosons, but we should pay special attentionto the k = 0 state. For the classical gas, this result is trivial.4. Consider a quantum ideal gas inside a cubic vessel of side L,and suppose that the orbitals of the particles are associated withwave functions that vanish at the surfaces of the cube. Find thedensity of states in ~ k space. In the thermodynamic limit, showthat we have the same expressions as calculated with periodicboundary conditions.5. An ideal gas of N atoms of mass m is con…ned to a vesselof volume V , at a given temperature T . Calculate the classicallimit of the chemical potential of this gas.Now consider a “two-dimensional” gas of N A free particlesadsorbed on a surface of area A. The energy of an adsorbedparticle is given by A = 12m ~p2 o ;where ~p is the (two-dimensional) momentum, and o > 0 is thebinding energy that keeps the particle stuck to the surface. Inthe classical limit, calculate the chemical potential A of theadsorbed gas.The condition of equilibrium between the adsorbed particlesand the particles of the three-dimensional gas can be expressedin terms of the respective chemical potentials. Use this conditionto …nd the surface density of adsorbed particles as a function oftemperature and pressure p of the surrounding gas.*** In the thermodynamic limit, the chemical potential of thethree-dimensional gas is given by = k B T ln"8 3 N V #~2 3=2:2m

62 8. The Ideal Quantum GasFor the adsorbed gas, we have the classical limitln = X jz exp ( j ) = z A(2) 2 1Z0 ~ 2 k 22k exp2m+ o ;which yields the chemical potential 42N A ~2 A = k B T ln A 2m o :From the physical requirement of equilibrium, = A , weobtain N A =A in terms of temperature and pressure.6. Obtain an expression for the entropy per particle, in termsof temperature and density, for a classical ideal monatomic gasof N particles of spin S adsorbed on a surface of area A. Obtainthe expected values of H, H 2 , H 3 , where H is the Hamiltonianof the system. What are the expressions of the second and thirdmoments of the Hamiltonian with respect to its average value?7. Consider a homogeneous mixture of two ideal monatomicgases, at temperature T , inside a container of volume V . Supposethat there are N A particles of gas A and N B particles ofgas B. Write an expression for the grand partition function associatedwith this system (it should depend on T , V , and thechemical potentials A and B ). In the classical limit, obtain expressionsfor the canonical partition function, the Helmholtz freeenergy F , and the pressure p of the gas. Show that p = p A + p B(Dalton’s law), where p A and p B are the partial pressures of Aand B, respectively.8. Under some conditions, the amplitudes of vibration of adiatomic molecule may be very large, with a certain degree ofanharmonicity. In this case, the vibrational energy levels aregiven by the approximate expression n =n + 1 ~! x n + 1 2~!;22

8. The Ideal Quantum Gas 63where x is the parameter of anharmonicity. To …rst order inx, obtain an expression for the vibrational speci…c heat of thissystem.*** Note thatwhereln Z v = ln S () + xS () d S () + O x2 ;2S () =1Xn=0expwith = ~!. We then haveu =d 2 n + 1 h= 2 sinh i 1;22~! @@ ln Z v and c = k B ~!k B T 2@ 2@ 2 ln Z v;from which we calculate the …rst-order correction to the speci…cheat, 2 ~!c = k B x d2 d 2k B T d 2 S () d S () :29. The potential energy between atoms of a hydrogen moleculemay be described by the Morse potential, 2 (r ro )r roV (r) = V oexp2 exp;aawhere V o = 7 10 19 J, r o = 8 10 11 m, and a = 5 10 11 m.Sketch a graph of V (r) versus r. Calculate the characteristictemperatures of vibration and rotation to compare with experimentaldata (see table on Section 8.4).and*** Note thatdVdr = 0 for r = r o d 2 V= 2V odr 2 r oa > 0; 2

64 8. The Ideal Quantum Gaswhich shows the existence of a minimum at r = r o .The characteristic temperature of rotation is given by r = ~22k B I = ~ 2;2k B Mro2where I is the moment of inertia and M is the reduced mass ofthe hydrogen molecule.From the frequency of vibrations,! = 1=2 2Vo;a 2 Mwe write the characteristic temperature of vibrations, v = ~ 1=2 2Vo:k B a 2 MNow it is easy to check the numerical predictions.

9The Ideal Fermi GasThis is page 65Printer: Opaque this1. What is the compressibility of a gas of free fermions at zerotemperature? Obtain the numerical value for electrons with thedensity of conduction electrons in metallic sodium. Compareyour results with experimental data for sodium at room temperature.*** It is straightforward to show that pressure of a completelydegenerate system of free fermions is given byp = 2 N5 V F = ~2 62 2=3 5=3 N:5m VThe compressibility is given by =1 @V= 3m 2=3 5=3 V:V @p ~ 2 6 2 NUsing the density of sodium (at room temperature), it is easy toshow that 10 5 atm, which is of the order of the experimentalcompressibility of sodium at room temperature. Also, it isinteresting to see that fermions are associated with large valuesof pressure, of the order of 10 4 10 5 atm, at zero temperature(due to the restrictions of the Pauli principle, of course).

66 9. The Ideal Fermi Gas2. An ideal gas of fermions, with mass m and Fermi energy F ,is at rest at zero temperature. Find expressions for the expectedvalues hv x i and hv 2 xi, where ~v is the velocity of a fermion.*** At zero temperature, the expected value of the (singleparticle)energy is calculated in terms of the density of statesD (). With D () = C 1=2 , we haveR FD () d0hi = R FD () d = 3 5 F :0Note that 1hi =2 m! v 2! ! v2 = 6 5 Fm ! ! v2 2 Fx =5 m ;in contrast to the predictions of the classical equipartition theorem.3. Consider a gas of free electrons, in a d-dimensional space,within a hypercubic container of side L. Sketch graphs of thedensity of states D () versus energy for dimensions d = 1 andd = 2. What is the expression of the Fermi energy in terms ofthe particle density for d = 1 and d = 2?*** Given the spectrum of energy j = ! k ;= ~2 k 22m ;we write the thermodynamic number of fermions in d dimensions,ZN = Ldd d 1k(2) d exp ~2 k 2 + 1 :2mDue to the spherical symmetry of the integrand, we haveZZ 1d d k (:::) =) C d k d 1 dk (:::) ;where it can be shown (see Appendix 4) that0C d = 2d=2:d2

9. The Ideal Fermi Gas 67Check that C 2 = 2, C 3 = 4, and so on. Introducing the changeof variables = ~2 k 22m ;and the de…nition of the “Fermi-Dirac distribution”,we havef () =Z 1N = Ld(2) d 0with the “density of states”D () =1exp ( ) + 1 ; d=21 2mC d d=2 1 f () d =2 ~ 2Z 1= L d D () f () d;0 d=21 2m2 (2) C d d d~ 2 2 1 :In two dimensions, note that D () is just a constant.4. Show that the chemical potential of an ideal classical gasof N monatomic particles, in a container of volume V , at temperatureT , may be written as 3 = k B T ln ;vwhere v = V=N is the volume per particle, and = h= p 2mk B Tis the thermal wavelength. Sketch a graph of =k B T versus T .Obtain the …rst quantum correction to this result. That is, showthat the chemical potential of the ideal quantum gas may bewritten as the expansionk B T 3 3 3 2ln = A ln + B ln + : : : ;vvv

68 9. The Ideal Fermi Gasand obtain explicit expressions for the prefactor A for fermionsand bosons. Sketch a graph of =k B T versus 2 (that is, versusthe temperature in convenient units) for fermions, bosons, andclassical particles.*** The grand partition function of the ideal quantum gas isgiven byln = X ln [1 z exp ( j )] :jWe now write an expansion in terms of powers of z,ln = X 1z exp ( j )2 z2 exp ( 2 j ) j== z X jexp ( j ) 1 2 z2 X jexp ( 2 j ) + :The …rst term corresponds to the classical limit (there is nodi¤erence between fermions and bosons). The second term givesthe …rst quantum correction. In the thermodynamic limit, wehaveXexp ( n j ) = V Z 1(2) 3j= V04k 2 exp mkB T2n~ 2 3=2:The classical limit is given by 3=2 mkB Tln cl = zV;2~ 2from which we haveThus,N = z @ @z ln cl = zV mkB T2~ 2 3=2: 2~2 3=2;n ~2 k 2dk =2mz = exp () = N Vmk B T

which leads to the well-known result 3 = k B T ln ;vwithv = V N and = 2~2mk B T9. The Ideal Fermi Gas 69 1=2=hp 2mkB T :Check that the same result can be obtained in the classical formalismof the grand ensemble, with the introduction of the “correctBoltzmann counting”, N!, and of the “quantum dimensionalcorrection" of the classical phase space, h 3N .The …rst quantum correction of this result is given byln = zV mkB T2~ 2 3=2 1 2 z2 V mkB T4~ 2 3=2+ O z 3 :Thus,N = zV mkB T2~ 2 3=2 z 2 V mkB T4~ 2 3=2+ O z 3 ;which can be written asN 1V 3 = z 12 3=2 z2 + O z 3 :Taking into account that z

70 9. The Ideal Fermi GasTherefore,ln z = ln" 3 3v + A vwhich leads to the expression 3lnk B T v 2 3+ #= lnv= 12 3=2 3v + : 12 3=2 3v + ;Try to use the same scheme to show that B = 1=16 (checkthat B assumes the same value for either fermions or bosons).5. Obtain an asymptotic form, in the limit T

Also, we have1Z09. The Ideal Fermi Gas 71f () d ! 1 Z 1exp (x)2 2 f1 [exp (x) + 1] 2 ++ 1 Z 1x 2 exp (x) dx() 2 1 [exp (x) + 1] 2 g = 1 2 2 1 + 1 2() 2 :3From these results, we write a low-temperature expression forthe internal energy (except for vanishingly exponential corrections),"U = 1 # 22 N F 1 + 2 T;3 T Fwhere F = 2~2m NA;which leads to a linear dependence of the speci…c heat withtemperature as in the three-dimensional case.6. Consider a gas of N free electrons, in a region of volumeV , in the ultrarelativistic regime. The energy spectrum is givenby = p 2 c 2 + m 2 c 4 1=2 pc;where ~p is the linear momentum.(a) Calculate the Fermi energy of this system.(b) What is the total energy in the ground state?(c) Obtain an asymptotic form for the speci…c heat at constantvolume in the limit T

72 9. The Ideal Fermi Gaswhere f () is the Fermi-Dirac distribution.At T = 0, we have 62 1=3 N F = ~c V 1=3;andU = 3 4 N F :The asymptotic form of the speci…c heat at low temperaturescomes from a straightforward application of Sommerfeld´smethod. We should note thatandZ 10Z 10 2 f () d = 1 3 3 Z 1 3 f () d = 1 Z 14 4 ! 1 4 4 1 + 6 231 + x 3! 1 3 3 1 + 2 1() 2 ;exp (x) dx[exp (x) + 1] 2 dx !1 + x 4exp (x) dx [exp (x) + 1] 2 dx !1() 2 + 7415where Z 1exp (x) dx 742dx =1 [exp (x) + 1] 15 :1() 4 ;as can be checked in a good integral table (in Gradshteyn andRyzhik, for example).We then write 3 F = 3 1 + 2 1() 2 ;andU = 3 4 N 1 4 1 + 6 2 3 F31() 2 + 74151() 4 :

9. The Ideal Fermi Gas 73From the …rst equation, we write an expansion for the chemicalpotential potential in terms of 1= [ F ] = (T=T F ), = F"12 2 T+ :::#:3 T FInserting this expansion of into the expression for the internalenergy, we …nally have"U = 3 24 N F 1 + 22 T+ :::#:3 T FNote that there is no linear power of (T=T F ).7. At low temperatures, the internal energy of a system offree electrons may be written as an expansion,(U = 3 2 45 N F 1 + 52 T TA + : : :):12 T F T FObtain the value of the constant A, and indicate the order ofmagnitude of the terms that have been discarded.*** This is again a straightforward (although somewhat laborious)application of the method of Sommerfeld to obtain asymptoticlow-temperature results.In the grand-canonical ensemble, the number of particles andthe internal energy of the gas of free fermions are give byandwhereN = V CU = V Cf () =Z 10Z 10 1=2 f () d; 3=2 f () d;1exp [ ( )] + 1 :At low temperatures, f () is almost a step function, so thederivative df=d has a pronounced peak at = , which provides

74 9. The Ideal Fermi Gasthe basis for the development of a low-temperature expansion.Let us write= 2 3Z 10Z 10= 2 3 3=2 Z 1 1=2 1exp [ ( )] + 1 d = 3=2 exp [ ( )]fexp [ ( )] + 1g 2 d =1 + x 3=2exp (x) [exp (x) + 1] 2 dx:At low temperatures, F ! 1, so that we may discardexponential corrections, of order exp ( F ), and writeZ 10 1=2 f () d ! 2 3 3=2 Z +111 + x 3=2exp (x) [exp (x) + 1] 2 dx;which is the main idea of Sommerfeld´s scheme. Except for theseexponential corrections, we have the series expansionwhereZ 10 1=2 f () d = 2 3 3=2 fI 0 + 3 8 I 12() 2 ++ 3 128 I 1 14() 4 + O () 6 g;I 0 =I 2 =I 4 =Z +11Z +11exp (x)2dx = 1;[exp (x) + 1]x 2 exp (x) 22dx =[exp (x) + 1] 3 ;Z x4 exp (x) +1[exp (x) + 1] 2 d1x = 7215 ;and so on (as you can check in the integral table of Gradshteynand Ryzhik, for example). We then have the expansion3N2V C = 3=2 F= 3=2 1 + 32 124 () 2 + 72 1640 () 4 + O( 1() 6 ) ;

9. The Ideal Fermi Gas 75from which we write 2 " 4 #) 6 T T T = F(1 + M 2 + M 4 + O:T F T F T FWe left for the reader the task of obtaining the numerical coef-…cients (M 2 and M 4 ). It is then easy to write an expansion ofthe internal energy in terms of (even) powers of (T=T F ).8. Consider a system of free fermions in d dimensions, withthe energy spectrum ~k; = c ~ k a ;where c > 0 and a > 1.(a) Calculate the prefactor A of the relation pV = AU.(b) Calculate the Fermi energy as a function of volume V andnumber of particles N.(c) Calculate an asymptotic expression, in the limit T

76 9. The Ideal Fermi Gasand note thatZN = Ld F(2) C d d01ac d=a d a1 d:9. Consider again the gas of N ultrarelativistic free electrons,within a container of volume V , at temperature T , in the presenceof a magnetic …eld ~ H. If we neglect the e¤ects of orbitalmagnetism, the energy spectrum is given by ~p; = cp B H;where B is the Bohr magneton and = 1.(a) Show that the Fermi energy of this system may be writtenas F = A + BH 2 + O H 4 :Obtain expressions for the prefactors A and B.(b) Show that the magnetization in the ground state can bewritten in the formM = CH + O H 3 :Obtain an expression for the constant C.(c) Calculate the susceptibility of the ground state in zero…eld.*** The grand partition functions may be written asln = ln + + ln ;withln = X ! kln f1 + z exp [ (c~k B H)]g :We then writeN = z @ @z ln = N + + N ;

9. The Ideal Fermi Gas 77withwhereN = V 31Z 12 2 ~c0f () = 2 f () d;1exp [ ( B H ) + 1] :As the Fermi energy is the chemical potential at zero temperature,we haveN = V 3 Z 1F2 2 ~cfrom which we obtain0 B H 2 d +Z F + B H 3 F + 3 F ( B H) 2 = 3 2 (~c) 3 NV :For B H

78 9. The Ideal Fermi Gas10. In the classical paramagnetic theory of Langevin, proposedbefore the advent of quantum statistics, we assume aclassical Hamiltonian, given byH =NX~ i H ~ =i=1NXH cos i ;i=1where ~ i is the magnetic moment of a localized ion.(a) Show that the canonical partition function of this systemis given byZZ = Z1 N =d exp (H cos ) N;where d is the elementary solid angle of integration.(b) Show that the magnetization (along the direction of the…eld) is given byM = N h cos i = NL (H) ;where1L (x) = coth (x)xis the Langevin function.(c) Show that the susceptibility in zero …eld is given by theCurie law, o = N23kT :*** The magnetization is given by* N+XM = i cos i = 1 i=1@@H ln Z = N @@H ln Z 1;whereZZ 1 =Z d exp (H cos ) = 2 sin exp (H cos ) d =0

Thus,M = N= 4 sinh (H) :H9. The Ideal Fermi Gas 791H + coth (H) = NL (H) ;where L is the Langevin function.11. Obtain an expression for the magnetic susceptibility associatedwith the orbital motion of free electrons in the presenceof a uniform magnetic …eld H, under conditions of strong degeneracy,T

80 9. The Ideal Fermi Gasso that it becomes easy to integrate over k z . In the second integral,we just call = ~ 2 k 2 z=2m. It is now straightforward toobtain an expression for the zero-…eld susceptibility [check theexcellent article by R. B. Dingle, "Some magnetic properties ofmetals. I. General introduction and properties of large systemsof electrons", Proc. Roy. Soc. A211, 500 (1952)].

10Free Bosons: Bose-EinsteinCondensation; Photon GasThis is page 81Printer: Opaque this1. Consider a system of ideal bosons of zero spin ( = 1), withina container of volume V .(a) Show that the entropy above the condensation temperatureT o is given bywhereandS = k BV 3 52 g 5=2 (z) =g (z) =hp 2mkB T1Xn=1k B T g 3=2 (z) ;z nn :(b) Given N and V , what is the expression of the entropybelow T o ? What is the entropy associated with the particles ofthe condensate?(c) From the expression for the entropy, show that the speci…cheat at constant volume above T o is given byc V = 3 4 k B5 g 5=2 (z)g 3=2 (z)3 g 3=2 (z)g 1=2 (z):

82 10. Free Bosons: Bose-Einstein Condensation; Photon Gas(d) Below T o , show that the speci…c heat at constant volumeis given byc V = 154 k vB 3 g 5=2 (1) :(e) Given the speci…c volume v, sketch a graph of c V =k B versus 2 (that is, in terms of the temperature in convenient units).Obtain the asymptotic expressions of the speci…c heat for T ! 0and T ! 1. Obtain the value of the speci…c heat at T = T o .*** Note the asymptotic limits of the speci…c heat,c V = 154 k vB 3 g 5=2 (1) T 3=2 ;for T ! 0, in agreement with Nernst´s third law of thermodynamics,andc V ! 3 2 k B;for T ! 1 (z ! 0), according to the classical theorem ofequipartition of energy. Also, note the value of the speci…c heatat the temperature of the Bose-Einstein condensation,c V (T = T 0 ) = 154 k g 5=2 (1)Bg 3=2 (1) > 3 2 k B:Sketch a graph of c V versus T .2. Consider again the same problem for an ideal gas of twodimensionalbosons con…ned to a surface of area A. What arethe changes in the expressions of item (a)? Show that there isno Bose–Einstein condensation in two dimensions (that is, showthat in this case the Bose–Einstein temperature vanishes).*** Consider a system of N free bosons in d dimensions (ina hypercubic box of volume L d ). The temperature of the Bose-Einstein condensation comes from the expressionN = Ld(2) d 1Z0expC d k d0 ~ 2 k 22m1 dk;1

10. Free Bosons: Bose-Einstein Condensation; Photon Gas 83where C d = 2 d=2 =we havewhere(d=2). With a suitable change of variables,k B T 0 = (2)1=2 ~ 22mI =1Z0 2=d N 2 1;L d C d Ix d=2 dxexp (x) 1 :It is easy to see that the integral I diverges (I ! 1) for d 2,which means that there is no Bose-Einstein condensation belowthree dimensions.3. Consider an ideal gas of bosons with internal degrees offreedom. Suppose that, besides the ground state with zero energy( o = 0), we have to take into account the …rst excitedstate, with internal energy 1 > 0. In other words, assume thatthe energy spectrum is given by j = ~k; = ~2 k 22m + 1;where = 0; 1. Obtain an expression for the Bose–Einsteincondensation temperature as a function of the parameter 1 .*** In the grand-canonical formalism, above the temperatureof condensation, the number of bosons is given byN = X j1exp ( j ) 1 = X 1~~ k;exp2 k 2+ 2m 1 In the thermodynamic limit, we writeNV = 1 3=2 Z 2m1 1=2 df8 2 ~ 2 0 exp ( ) 1 ++Z 10 1=2 dexp ( + 1 ) 1 g == (2mk BT ) 3=2h 3 g3=2 (z) + g 3=2 ze 1 ::1

84 10. Free Bosons: Bose-Einstein Condensation; Photon GasAt the temperature of the Bose-Einstein condensation (z ! 1),we haveg 3=2 (1) + g 3=2expNV = (2mk BT 0 ) 3=2h 3 1;k B T 0from which we may (numerically) obtain T 0 (which is smallerthan the corresponding value with 1 = 0).4. Consider a gas of non-interacting bosons associated withthe energy spectrum = ~c ~ k ;where ~ and c are constants, and ~ k is a wave vector. Calculatethe pressure of this gas at zero chemical potential. What is thepressure of radiation of a gas of photons?*** In the thermodynamic limit we havepV = ln = Z 12 V k 2 dk ln [1 z exp ( c~k)] :20In zero chemical potential, it is easy to show thatp = (k B T ) 4 Z 1x 3 dx6 2 (~c) 3 exp (x) 1 :Note that the pressure of radiation is given by the equation ofstatepV = 1 3 U:Taking into account the law of Stefan-Boltzmann, U = V T 4 ,we obtain an expression for the constant .*5. The Hamiltonian of a gas of photons within an emptycavity of volume V is given by the expressionH = X ~! ~k;j a y ~ k;ja ~k;j + 1 ;2~ k;j0

10. Free Bosons: Bose-Einstein Condensation; Photon Gas 85where j = 1; 2 indicates the polarization, ~ k is the wave vector,! ~k;j = ck;c is the velocity of light, and k = ~ k.(a) Use the formalism of the occupation numbers (secondquantization) to obtain the canonical partition function associatedwith this system.(b) Show that the internal energy is given byU = V T n :Obtain the value of the constants n and .(c) Consider the Sun as a black body at temperature T 5800 K. The solar diameter and the distance between the Sunand the Earth are of the order of 10 9 m and 10 11 m, respectively.Obtain the intensity of the total radiation that reachesthe surface of Earth. What is the value of the pressure of thisradiation?*** The canonical partition function is given byZ = Tr exp ( H) = X= Xnn! k ;joexp2nn! k ;joo oEDnn! k ;jexp ( H)nn! k ;j=4 X ~ k;j~! ~k;ja y ~ k;ja ~k;j + 1 2 3 5 == Y ! k ;j( 1Xn=0exp~! ~k;j n + 1 ) = Y 1exp ~ck 22!1 exp ( ~ck) :k ;jIn order to calculate the internal energy we writeln Z = 1 2X! k ;j~ckXln [1 exp ( ~ck)] ;! k ;j

86 10. Free Bosons: Bose-Einstein Condensation; Photon Gaswhere the …rst term in the right-hand-side leads to well-known(divergent) energy of the quantum vacuum. If we measure theinternal energy with respect to the vacuum, we haveu =1 N@@ ln Z = 1 X ~! ~k;jN~ k;jexp~! ~k;j:1In d dimensions, it is easy to check that this internal energywith respect to the quantum vacuum is given byU = V T n ;where n = d + 1 (n = 4 in three dimension, according to theStefan-Boltzmann law) and =4k d+1B(d + 1) (d + 1) :(4~ 2 c 2 ) d=2 d2