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Unformatted text preview: Chapter 6 ManyParticle Systems
c 2010 by Harvey Gould and Jan Tobochnik 2 December 2010 We apply the general formalism of statistical mechanics to systems of many particles and discuss the semiclassical limit of the partition function, the equipartition theorem for classical systems, and the general applicability of the Maxwell velocity distribution. We then consider noninteracting quantum systems and discuss the single particle density of states, the FermiDirac and BoseEinstein distribution functions, the thermodynamics of ideal Fermi and Bose gases, blackbody radiation, and the speciﬁc heat of crystalline solids among other applications. 6.1 The Ideal Gas in the Semiclassical Limit We ﬁrst apply the canonical ensemble to an ideal gas in the semiclassical limit. Because the thermodynamic properties of a system are independent of the choice of ensemble, we will ﬁnd the same thermal and pressure equations of state as we found in Section 4.5. Although we will not obtain any new results, this application will give us more experience in working with the canonical ensemble and again show the subtle nature of the semiclassical limit. In Section 6.6 we will derive the classical equations of state using the grand canonical ensemble without any ad hoc assumptions. In Sections 4.4 and 4.5 we derived the thermodynamic properties of the ideal classical gas1 using the microcanonical ensemble. If the gas is in thermal equilibrium with a heat bath at temperature T , it is more natural and convenient to treat the ideal gas in the canonical ensemble. Because the particles are not localized, they cannot be distinguished from each other as were the harmonic oscillators considered in Example 4.3 and the spins in Chapter 5. Hence, we cannot simply focus our attention on one particular particle. The approach we will take here is to treat the particles as distinguishable, and then correct for the error approximately. As before, we will consider a system of noninteracting particles starting from their fundamental description according to quantum mechanics. If the temperature is suﬃciently high, we expect
1 The theme music for this section can be found at <www.classicalgas.com/> . 292 CHAPTER 6. MANYPARTICLE SYSTEMS 293 that we can treat the particles classically. To do so we cannot simply take the limit →0 wherever it appears because the counting of microstates is diﬀerent in quantum mechanics and classical mechanics. That is, particles of the same type are indistinguishable according to quantum mechanics. So in the following we will consider the semiclassical limit, and the particles will remain indistinguishable even in the limit of high temperatures. To take the semiclassical limit the mean de Broglie wavelength λ of the particles must be smaller than any other length in the system. For an ideal gas the only two lengths are L, the linear dimension of the system, and the mean distance between particles. Because we are interested in the thermodynamic limit for which L ≫ λ, the ﬁrst condition will always be satisﬁed. As shown in Problem 6.1, the mean distance between particles in three dimensions is ρ−1/3 . Hence, the semiclassical limit requires that λ ≪ ρ−1/3 or ρλ ≪ 1 Problem 6.1. Mean distance between particles (a) Consider a system of N particles conﬁned to a line of length L. What is the deﬁnition of the particle density ρ? The mean distance between particles is L/N . How does this distance depend on ρ? (b) Consider a system of N particles conﬁned to a square of linear dimension L. How does the mean distance between particles depend on ρ? (c) Use similar considerations to determine the density dependence of the mean distance between particles in three dimensions. To estimate the magnitude of λ we need to know the typical value of the momentum of a particle. For a nonrelativistic system in the semiclassical limit we know from (4.65) that p2 /2m = 3kT /2. (We will rederive this result more generally in Section 6.2.1.) Hence p2 ∼ mkT and √ λ ∼ h/ p2 ∼ h/ mkT . We will ﬁnd it is convenient to deﬁne the thermal de Broglie wavelength λ as 1/2 h2 2π 2 1/2 λ≡ (thermal de Broglie wavelength). (6.2) = 2πmkT mkT √ This form of λ with the factor of 2π will allow us to express the partition function in a convenient form [see (6.11)]. The calculation of the partition function of an ideal gas in the semiclassical limit proceeds as follows. First, we assume that λ ≪ ρ−1/3 so that we can pick out one particle if we make the additional assumption that the particles are distinguishable. (If λ ∼ ρ−1/3 , the wavefunctions of the particles overlap.) Because identical particles are intrinsically indistinguishable, we will have to correct for the latter assumption later. With these considerations in mind we now calculate Z1 , the partition function for one particle, in the semiclassical limit. As we found in (4.40), the energy eigenvalues of a particle in a cube of side L are given by h2 (nx 2 + ny 2 + nz 2 ), (6.3) ǫn = 8mL2
3 (semiclassical limit). (6.1) CHAPTER 6. MANYPARTICLE SYSTEMS 294 where the subscript n represents the set of quantum numbers nx , ny , and nz , each of which can be any nonzero, positive integer. The corresponding partition function is given by
∞ ∞ ∞ Z1 =
n e−βǫn =
nx =1 ny =1 nz =1 e−βh 2 (nx 2 +ny 2 +nz 2 )/8mL2 . (6.4) Because each sum is independent of the others, we can rewrite (6.4) as
∞ ∞ ∞ Z1 =
nx =1 e −α2 nx 2 ny =1 e −α2 ny 2 nz =1 e −α 2 nz 2 = S 3, (6.5) where
∞ S=
nx =1 e −α 2 nx 2 . (6.6) and α2 = π λ2 βh2 = . 2 8mL 4 L2 (6.7) It remains to evaluate the sum over nx in (6.6). Because the linear dimension L of the container is of macroscopic size, we have λ ≪ L and α in (6.6) is much less than 1. Hence, because the diﬀerence between successive terms in the sum is very small, we can convert the sum in (6.6) to an integral:
∞ ∞ S=
nx =1 e −α2 nx 2 = e −α
nx =0 2 nx 2 ∞ −1 → e −α
0 2 n2 x dnx − 1. (6.8) We have accounted for the fact that the sum over nx in (6.6) is from nx = 1 rather than nx = 0. We next make a change of variables and write u2 = α2 n2 . We have that x S= 1 α
∞ 0 e−u du − 1 = L 2 2πm βh2 1/2 − 1. (6.9) The Gaussian integral in (6.9) gives a factor of π 1/2 /2 (see the Appendix). Because the ﬁrst term in (6.9) is order L/λ ≫ 1, we can ignore the second term, and hence we obtain Z1 = S 3 = V 2πm βh2
3/2 . (6.10) The result (6.10) is the partition function associated with the translational motion of one particle in a box. Note that Z1 can be conveniently expressed as Z1 = V . λ3 (6.11) It is straightforward to ﬁnd the mean pressure and energy for one particle in a box. We take the logarithm of both sides of (6.10) and ﬁnd ln Z1 = ln V − 3 3 2πm ln β + ln 2 . 2 2 h (6.12) CHAPTER 6. MANYPARTICLE SYSTEMS The mean pressure due to one particle is given by p= and the mean energy is e=− ∂ ln Z1 ∂β
V ,N 295 1 ∂ ln Z1 β ∂V T ,N = 1 kT = , βV V 3 3 = kT. 2β 2 (6.13) = (6.14) The mean energy and pressure of an ideal gas of N particles is N times that of the corresponding quantities for one particle. Hence, we obtain for an ideal classical gas the equations of state P= and E= 3 N kT. 2 (6.16) N kT V (6.15) In the following we will usually omit the overbar on mean quantities. The heat capacity at constant volume of an ideal gas of N particles is CV = ∂E ∂T = 3 N k. 2 (6.17) V We have derived the mechanical and thermal equations of state for an ideal classical gas for a second time! The derivation of the equations of state is much easier in the canonical ensemble than in the microcanonical ensemble. The reason is that we were able to consider the partition function of one particle because the only constraint is that the temperature is ﬁxed instead of the total energy. Problem 6.2. Independence of the partition function on the shape of the box The volume dependence of Z1 should be independent of the shape of the box. Show that the same result for Z1 is obtained if the box has linear dimensions Lx , Ly , and Lz with V = Lx Ly Lz . Problem 6.3. Semiclassical limit of the single particle partition function We obtained the semiclassical limit of the partition function Z1 for one particle in a box by writing it as a sum over single particle states and then converting the sum to an integral. Show that the semiclassical partition function Z1 for a particle in a onedimensional box can be expressed as Z1 = dp dx −βp2/2m e . h (6.18) The integral over p in (6.18) extends from −∞ to +∞. The entropy of an ideal classical gas of N particles . Although it is straightforward to calculate the mean energy and pressure of an ideal classical gas by considering the partition function for one particle, the calculation of the entropy is more subtle. To understand the diﬃculty, consider the calculation of the partition function of an ideal gas of two particles. Because there are no CHAPTER 6. MANYPARTICLE SYSTEMS microstate s 1 2 3 4 5 6 7 8 9 red ǫa ǫb ǫc ǫa ǫb ǫa ǫc ǫb ǫc blue ǫa ǫb ǫc ǫb ǫa ǫc ǫa ǫc ǫb Es 2 ǫa 2 ǫb 2 ǫc ǫa + ǫb ǫa + ǫb ǫa + ǫc ǫa + ǫc ǫb + ǫc ǫb + ǫc 296 Table 6.1: The nine microstates of a system of two noninteracting distinguishable particles (red and blue). Each particle can be in one of three microstates with energy ǫa , ǫb , or ǫc . interactions between the particles, we can write the total energy as a sum of the single particle energies ǫ1 + ǫ2 , where ǫi is the energy of the ith particle. The partition function Z2 is Z2 =
all states e − β ( ǫ 1 +ǫ 2 ) . (6.19) The sum over all microstates in (6.19) is over the microstates of the twoparticle system. If the two particles were distinguishable, there would be no restriction on the number of particles that could be in any single particle microstate, and we could sum over the possible microstates of each particle separately. Hence, the partition function for a system of two distinguishable particles has the form 2 Z2, distinguishable = Z1 . (6.20) It is instructive to show the origin of the relation (6.20) for a speciﬁc example. Suppose the two particles are red and blue and are in equilibrium with a heat bath at temperature T . For simplicity, we assume that each particle can be in one of three microstates with energies ǫa , ǫb , and ǫc . The partition function for one particle is given by Z1 = e−βǫa + e−βǫb + e−βǫc . (6.21) In Table 6.1 we list the 32 = 9 possible microstates of this system of two distinguishable particles. The corresponding partition function is given by Z2, distinguishable = e−2βǫa + e−2βǫb + e−2βǫc + 2 e − β ( ǫ a +ǫ b ) + e − β ( ǫ a +ǫ c ) + e − β ( ǫ b +ǫ c ) . (6.22) It is easy to see that Z2 in (6.22) can be factored and expressed as in (6.20). In contrast, if the two particles are indistinguishable, many of the microstates shown in Table 6.1 cannot be counted as separate microstates. In this case we cannot assign the microstates of the particles independently, and the sum over all microstates in (6.19) cannot be factored as in (6.20). For example, the microstate a, b cannot be distinguished from the microstate b, a. As discussed in Section 4.3.6, the semiclassical limit assumes that microstates with multiple occupancy such as a, a and b, b can be ignored because there are many more single particle states CHAPTER 6. MANYPARTICLE SYSTEMS 297 than there are particles (see Problem 4.14, page 190). (In our simple example, each particle can be in one of only three microstates, and the number of microstates is comparable to the number of particles.) If we assume that the particles are indistinguishable and that microstates with multiple occupancy can be ignored, then Z2 is given by Z 2 = e − β ( ǫ a +ǫ b ) + e − β ( ǫ a +ǫ c ) + e − β ( ǫ b +ǫ c ) (indistinguishable, no multiple occupancy). (6.23) We see that if we ignore multiple occupancy there are three microstates for indistinguishable particles and six microstates for distinguishable particles. Hence, in the semiclassical limit we can 2 write Z2 = Z1 /2! where the factor of 2! corrects for overcounting. For three particles (each of which can be in one of three possible microstates) and no multiple occupancy, there would be one microstate of the system for indistinguishable particles and no multiple occupancy, namely, the microstate a, b, c. However, there would be six such microstates for distinguishable particles. Thus if we count microstates assuming that the particles are distinguishable, we would overcount the number of microstates by N !, the number of permutations of N particles. We conclude that if we begin with the fundamental quantum mechanical description of matter, then identical particles are indistinguishable at all temperatures. If we make the assumption that single particle microstates with multiple occupancy can be ignored, we can express the partition function of N noninteracting identical particles as ZN = Z1 N N! (ideal gas, semiclassical limit). (6.24) We substitute for Z1 from (6.10) and obtain the partition function of an ideal gas of N particles in the semiclassical limit: V N 2πmkT 3N/2 . (6.25) ZN = N! h2 If we take the logarithm of both sides of (6.25) and use Stirling’s approximation (3.102), we can write the free energy of an ideal classical gas as F = −kT ln ZN = −kT N ln 2πmkT 3 V + ln N 2 h2 +1 . (6.26) In Section 6.6 we will use the grand canonical ensemble to obtain the entropy of an ideal classical gas without any ad hoc assumptions such as assuming that the particles are distinguishable and then correcting for overcounting by including the factor of N !. That is, in the grand canonical ensemble we will be able to automatically satisfy the condition that the particles are indistinguishable. Problem 6.4. Equations of state of an ideal classical gas Use the result (6.26) to ﬁnd the pressure equation of state and the mean energy of an ideal gas. Do the equations of state depend on whether the particles are indistinguishable or distinguishable? Problem 6.5. Entropy of an ideal classical gas CHAPTER 6. MANYPARTICLE SYSTEMS (a) The entropy can be found from the relations F = E − T S or S = −∂F/∂T . Show that S (T, V, N ) = N k ln 3 2πmkT V + ln N 2 h2 + 5 . 2 298 (6.27) The form of S in (6.27) is known as the SackurTetrode equation (see Problem 4.20, page 197). Is this form of S applicable for low temperatures? (b) Express kT in terms of E and show that S (E, V, N ) can be expressed as S (E, V, N ) = N k ln V 5 3 4πmE +, + ln N 2 3 N h2 2 (6.28) in agreement with the result (4.63) found using the microcanonical ensemble. Problem 6.6. The chemical potential of an ideal classical gas (a) Use the relation µ = ∂F/∂N and the result (6.26) to show that the chemical potential of an ideal classical gas is given by µ = −kT ln V 2πmkT N h2
3/2 . (6.29) (b) We will see in Chapter 7 that if two systems are placed into contact with diﬀerent initial chemical potentials, particles will go from the system with higher chemical potential to the system with lower chemical potential. (This behavior is analogous to energy going from high to low temperatures.) Does “high” chemical potential for an ideal classical gas imply “high” or “low” density? (c) Calculate the entropy and chemical potential of one mole of helium gas at standard temperature and pressure. Take V = 2.24 × 10−2 m3 , N = 6.02 × 1023 , m = 6.65 × 10−27 kg, and T = 273 K. Problem 6.7. Entropy as an extensive quantity (a) Because the entropy is an extensive quantity, we know that if we double the volume and double the number of particles (thus keeping the density constant), the entropy must double. This condition can be written formally as S (T, λV, λN ) = λS (T, V, N ). (6.30) Although this behavior of the entropy is completely general, there is no guarantee that an approximate calculation of S will satisfy this condition. Show that the SackurTetrode form of the entropy of an ideal gas of identical particles, (6.27), satisﬁes this general condition. (b) Show that if the N ! term were absent from (6.25) for ZN , S would be given by S = N k ln V + Is this form of S extensive? 2πmkT 3 ln 2 h2 + 3 . 2 (6.31) CHAPTER 6. MANYPARTICLE SYSTEMS 299 (a) (b) Figure 6.1: (a) A composite system is prepared such that there are N argon atoms in container A and N argon atoms in container B . The two containers are at the same temperature T and have the same volume V . What is the change of the entropy of the composite system if the partition separating the two containers is removed and the two gases are allowed to mix? (b) A composite system is prepared such that there are N argon atoms in container A and N helium atoms in container B . The other conditions are the same as before. The change in the entropy when the partition is removed is equal to 2N k ln 2. (c) The fact that (6.31) yields an entropy that is not extensive does not indicate that identical particles must be indistinguishable. Instead the problem arises from our identiﬁcation of S with ln Z as mentioned in Section 4.6, page 199. Recall that we considered a system with ﬁxed N and made the identiﬁcation that [see (4.106)] dS/k = d(ln Z + βE ). It is straightforward to integrate (6.32) and obtain S = k (ln Z + βE ) + g (N ), (6.33) (6.32) where g (N ) is an arbitrary function only of N . Although we usually set g (N ) = 0, it is important to remember that g (N ) is arbitrary. What must be the form of g (N ) in order that the entropy of an ideal classical gas be extensive? Entropy of mixing . Consider two containers A and B each of volume V with two identical gases of N argon atoms each at the same temperature T . What is the change of the entropy of the combined system if we remove the partition separating the two containers and allow the two gases to mix [see Figure 6.1)(a)]? Because the argon atoms are identical, nothing has really changed and no information has been lost. Hence, ∆S = 0. In contrast, suppose that one container is composed of N argon atoms and the other is composed of N helium atoms [see Figure 6.1)(b)]. What is the change of the entropy of the CHAPTER 6. MANYPARTICLE SYSTEMS 300 combined system if we remove the partition separating them and allow the two gases to mix? Because argon atoms are distinguishable from helium atoms, we lose information about the system, and therefore we know that the entropy must increase. Alternatively, we know that the entropy must increase because removing the partition between the two containers is an irreversible process. (Reinserting the partition would not separate the two gases.) We conclude that the entropy of mixing is nonzero: ∆S > 0 (entropy of mixing). (6.34) In the following, we will derive these results for the special case of an ideal classical gas. Consider two ideal gases at the same temperature T with NA and NB particles in containers of volume VA and VB , respectively. The gases are initially separated by a partition. We use (6.27) for the entropy and ﬁnd VA + f (T, mA ) , NA VB SB = NB k ln + f (T, mB ) , NB SA = NA k ln (6.35a) (6.35b) where the function f (T, m) = 3/2 ln(2πmkT /h2 ) + 5/2, and mA and mB are the particle masses in system A and system B , respectively. We then allow the particles to mix so that they ﬁll the entire volume V = VA + VB . If the particles are identical and have mass m, the total entropy after the removal of the partition is given by S = k (NA + NB ) ln VA + VB + f (T, m) , NA + NB (6.36) and the change in the value of S , the entropy of mixing, is given by ∆S = k (NA + NB ) ln VA VB VA + VB − NA ln − NB ln NA + NB NA NB (identical gases). (6.37) Problem 6.8. Entropy of mixing of identical particles (a) Use (6.37) to show that ∆S = 0 if the two gases have equal densities before separation. Write NA = ρVA and NB = ρVB . (b) Why is the entropy of mixing nonzero if the two gases initially have diﬀerent densities even though the particles are identical? If the two gases are not identical, the total entropy after mixing is S = k NA ln VA + VB VA + VB + NB ln + NA f (T, mA ) + NB f (T, mB ) . NA NB VA + VB VA VB VA + VB . + NB ln − NA ln − NB ln NA NB NA NB ∆S = 2N k ln 2. (6.38) Then the entropy of mixing becomes ∆S = k NA ln (6.39) For the special case of NA = NB = N and VA = VB = V , we ﬁnd (6.40) CHAPTER 6. MANYPARTICLE SYSTEMS Problem 6.9. More on the entropy of mixing (a) Explain the result (6.40) for nonidentical particles in simple terms. 301 (b) Consider the special case NA = NB = N and VA = VB = V and show that if we use the result (6.31) instead of (6.27), the entropy of mixing for identical particles is nonzero. This incorrect result is known as Gibbs paradox. Does it imply that classical physics, which assumes that particles of the same type are distinguishable, is incorrect? 6.2 Classical Statistical Mechanics From our discussions of the ideal gas in the semiclassical limit we found that the approach to the classical limit must be made with care. Planck’s constant appears in the expression for the entropy even for the simple case of an ideal gas, and the indistinguishability of the particles is not a classical concept. If we work entirely within the framework of classical mechanics, we would replace the sum over microstates in the partition function by an integral over phase space, that is, ZN, classical = CN e−βE (r1 ,...,rN ,p1 ,...,pN ) dr1 . . . drN dp1 . . . dpN . (6.41) The constant CN cannot be determined from classical mechanics. From our counting of microstates for a single particle and the harmonic oscillator in Section 4.3 and the arguments for including the factor of 1/N ! on page 295 we see that we can obtain results consistent with starting from quantum mechanics if we choose the constant CN to be CN = 1 . N !h3N (6.42) Thus the partition function of a system of N particles in the semiclassical limit can be written as ZN, classical = 1 N! e−βE (r1 ,...,rN ,p1 ,...,pN ) dr1 . . . drN dp1 . . . dpN . h3N (6.43) We obtained a special case of the form (6.43) in Problem 6.3. In the following three subsections we integrate over phase space as in (6.43) to ﬁnd some general properties of classical systems of many particles. 6.2.1 The equipartition theorem We have used the microcanonical and canonical ensembles to show that the energy of an ideal classical gas in three dimensions is given by E = 3kT /2. Similarly, we have found that the energy of a onedimensional harmonic oscillator is given by E = kT in the high temperature limit. These results are special cases of the equipartition theorem which can be stated as follows: CHAPTER 6. MANYPARTICLE SYSTEMS For a classical system in equilibrium with a heat bath at temperature T , the mean value of each contribution to the total energy that is quadratic in a coordinate equals 1 2 kT . 302 Note that the equipartition theorem holds regardless of the coeﬃcients of the quadratic terms and is valid only for a classical system. If all the contributions to the energy are quadratic, the mean energy is distributed equally to each term (hence the name “equipartition”). To see how to calculate averages according to classical statistical mechanics, we ﬁrst consider a single particle subject to a potential energy U (r) in equilibrium with a heat bath at temperature T . Classically, the probability of ﬁnding the particle in a small volume dr about r with a momentum in a small volume dp about p is proportional to the Boltzmann factor and the volume dr dp in phase space: 2 p(r, p)dr dp = Ae−β (p /2m+U (r)) dr dp. (6.44) To normalize the probability and determine the constant A we have to integrate over all the possible values of r and p. We next consider a classical system of N particles in the canonical ensemble. The probability density of a particular microstate is proportional to the Boltzmann probability e−βE , where E is the energy of the microstate. Because a microstate is deﬁned classically by the positions and momenta of every particle, we can express the average of any physical quantity f (r, p) in a classical system as f= f (r1 , . . . , rN , p1 , . . . , pN ) e−βE (r1 ,...,rN ,p1 ,...,pN ) dr1 . . . drN dp1 . . . dpN . e−βE (r1 ,...,rN ,p1 ,...,pN ) dr1 . . . drN dp1 . . . dpN (6.45) Note that the sum over quantum states has been replaced by an integration over phase space. We could divide the numerator and denominator by h3N so that we would obtain the correct number of microstates in the semiclassical limit, but this factor cancels in calculations of average quantities. We have already seen that the mean energy and mean pressure do not depend on whether the factors of h3N and 1/N ! are included in the partition function. Suppose that the total energy can be written as a sum of quadratic terms. For example, the kinetic energy of one particle in three dimensions in the nonrelativistic limit can be expressed as (p2 + p2 + p2 )/2m. Another example is the onedimensional harmonic oscillator for which the total x y z energy is p2 /2m + kx2 /2. For simplicity let’s consider a onedimensional system of two particles, x and suppose that the energy of the system can be written as ˜ E = ǫ1 (p1 ) + E (x1 , x2 , p2 ), (6.46) CHAPTER 6. MANYPARTICLE SYSTEMS 303 where ǫ1 = ap2 with a equal to a constant. We have separated out the quadratic dependence of 1 the energy of particle one on its momentum. We use (6.45) and express the mean value of ǫ1 as ǫ1 = =
∞ ǫ e−βE (x1 ,x2 ,p1 ,p2 ) dx1 dx2 dp1 dp2 −∞ 1 ∞ e−βE (x1 ,x2 ,p1 ,p2 ) dx1 dx2 dp1 dp2 −∞ ∞ ˜ −β [ǫ1 +E (x1 ,x2 ,p2 )] dx1 dx2 dp1 dp2 −∞ ǫ1 e ∞ ˜ −β [ǫ1 +E (x1 ,x2 ,p2 ,p2 )] dx dx dp dp e 1 2 1 2 −∞ ∞ ˜ −βǫ1 dp1 e−β E dx1 dx2 dp2 −∞ ǫ1 e . ∞ ˜ e−βǫ1 dp1 e−β E dx1 dx2 dp2 −∞ (6.47a) (6.47b) = (6.47c) The integrals over all the coordinates except p1 cancel, and we have ǫ1 =
∞ −βǫ1 dp1 −∞ ǫ1 e . ∞ −βǫ 1 dp 1 −∞ e (6.48) As we have done in other contexts [see (4.84), page 202] we can write ǫ1 as ǫ1 = − ∂ ln ∂β
∞ −∞ e−βǫ1 dp1 . (6.49) If we substitute ǫ1 = ap2 , the integral in (6.49) becomes 1
∞ I (β ) =
−∞ e−βǫ1 dp1 =
∞ −∞
2 ∞ −∞ e−βap1 dp1 2 (6.50a) (6.50b) = (βa)−1/2 e−u du, where we have let u2 = βap2 . Note that the integral in (6.50b) is independent of β , and its numerical value is irrelevant. Hence ǫ1 = − ∂ 1 ln I (β ) = kT. ∂β 2 (6.51) Equation (6.51) is an example of the equipartition theorem of classical statistical mechanics. The equipartition theorem is applicable only when the system can be described classically, and is applicable only to each term in the energy that is proportional to a coordinate squared. This coordinate must take on a continuum of values from −∞ to +∞. Applications of the equipartition theorem . A system of particles in three dimensions has 3N quadratic contributions to the kinetic energy, three for each particle. From the equipartition theorem, we know that the mean kinetic energy is 3N kT /2, independent of the nature of the interactions, if any, between the particles. Hence, the heat capacity at constant volume of an ideal classical monatomic gas is given by CV = 3N k/2 as we have found previously. Another application of the equipartition function is to the onedimensional harmonic oscillator in the classical limit. In this case there are two quadratic contributions to the total energy and CHAPTER 6. MANYPARTICLE SYSTEMS 304 hence the mean energy of a onedimensional classical harmonic oscillator in equilibrium with a heat bath at temperature T is kT . In the harmonic model of a crystal each atom feels a harmonic or springlike force due to its neighboring atoms (see Section 6.9.1). The N atoms independently perform simple harmonic oscillations about their equilibrium positions. Each atom contributes three quadratic terms to the kinetic energy and three quadratic terms to the potential energy. Hence, in the high temperature limit the energy of a crystal of N atoms is E = 6N kT /2, and the heat capacity at constant volume is CV = 3N k (law of Dulong and Petit). (6.52) The result (6.52) is known as the law of Dulong and Petit. This result was ﬁrst discovered empirically and is valid only at suﬃciently high temperatures. At low temperatures a quantum treatment is necessary and the independence of CV on T breaks down. The heat capacity of an insulating solid at low temperatures is discussed in Section 6.9.2. We next consider an ideal gas consisting of diatomic molecules (see Figure 6.5 on page 345). Its pressure equation of state is still given by P V = N kT , because the pressure depends only on the translational motion of the center of mass of each molecule. However, its heat capacity diﬀers from that of a ideal monatomic gas because a diatomic molecule has additional energy associated with its vibrational and rotational motion. Hence, we expect that CV for an ideal diatomic gas to be greater than CV for an ideal monatomic gas. The temperature dependence of the heat capacity of an ideal diatomic gas is explored in Problem 6.47. 6.2.2 The Maxwell velocity distribution So far we have used the tools of statistical mechanics to calculate macroscopic quantities of interest in thermodynamics such as the pressure, the temperature, and the heat capacity. We now apply statistical mechanics arguments to gain more detailed information about classical systems of particles by calculating the velocity distribution of the particles. Consider a classical system of particles in equilibrium with a heat bath at temperature T . We know that the total energy can be written as the sum of two parts: the kinetic energy K (p1 , . . . , pN ) and the potential energy U (r1 , . . . , rN ). The kinetic energy is a quadratic function of the momenta p1 , . . . , pN (or velocities), and the potential energy is a function of the positions r1 , . . . , rN of the particles. The total energy is E = K + U . The probability density of a microstate of N particles deﬁned by r1 , . . . , rN , p1 , . . . , pN is given in the canonical ensemble by p(r1 , . . . , rN ; p1 , . . . , pN ) = A e−[K (p1 ,p2 ,...,pN )+U (r1 ,r2 ,...,rN )]/kT = Ae
−K (p1 ,p2 ,...,pN )/kT −U (r1 ,r2 ,...,rN )/kT (6.53a) , (6.53b) e where A is a normalization constant. The probability density p is a product of two factors, one that depends only on the particle positions and the other that depends only on the particle momenta. This factorization implies that the probabilities of the momenta and positions are independent. The probability of the positions of the particles can be written as f (r1 , . . . , rN ) dr1 . . . drN = B e−U (r1 ,...,rN )/kT dr1 . . . drN , and the probability of the momenta is given by f (p1 , . . . , pN ) dp1 . . . dpN = C e−K (p1 ,...,pN )/kT dp1 . . . dpN . (6.55) (6.54) CHAPTER 6. MANYPARTICLE SYSTEMS 305 For notational simplicity, we have denoted the two probability densities by f , even though their meaning is diﬀerent in (6.54) and (6.55). The constants B and C in (6.54) and (6.55) can be found by requiring that each probability be normalized. We stress that the probability distribution for the momenta does not depend on the nature of the interaction between the particles and is the same for all classical systems at the same temperature. This statement might seem surprising because it might seem that the velocity distribution should depend on the density of the system. An external potential also does not aﬀect the velocity distribution. These statements do not hold for quantum systems, because in this case the position ˆˆ ˆ ˆ and momentum operators do not commute. That is, e−β (K +U ) = e−β K e−β U for quantum systems, where we have used carets to denote operators in quantum mechanics. Because the total kinetic energy is a sum of the kinetic energy of each of the particles, the probability density f (p1 , . . . , pN ) is a product of terms that each depend on the momenta of only one particle. This factorization implies that the momentum probabilities of the various particles are independent. These considerations imply that we can write the probability that a particle has momentum p in the range dp as f (px , py , pz ) dpx dpy dpz = c e−(px +py +pz )/2mkT dpx dpy dpz . The constant c is given by the normalization condition
∞ ∞ ∞
2 2 2 (6.56) c
−∞ −∞ −∞ ∞ e−(px +py +pz )/2mkT dpx dpy dpz = c
2 2 2 2 ∞ −∞ e −p 2 /2mkT dp 3 = 1. (6.57) If we use the fact that −∞ e−αx dx = (π/α)1/2 (see the Appendix), we ﬁnd that c = (2πmkT )−3/2 . Hence the momentum probability distribution can be expressed as f (px , py , pz ) dpx dpy dpz =
2 2 2 1 e−(px +py +pz )/2mkT dpx dpy dpz . 3/2 (2πmkT ) (6.58) The corresponding velocity probability distribution is given by f (vx , vy , vz ) dvx dvy dvz = m 2πkT
3/2 e−m(vx +vy +vz )/2kT dvx dvy dvz . 2 2 2 (6.59) Equation (6.59) is known as the Maxwell velocity distribution. Note that its form is a Gaussian. The probability distribution for the speed is discussed in Section 6.2.3. Because f (vx , vy , vz ) is a product of three independent factors, the probability of the velocity of a particle in a particular direction is independent of the velocity in any other direction. For example, the probability that a particle has a velocity in the xdirection in the range vx to vx + dvx is 2 m 1/2 −mvx /2kT f (vx ) dvx = e dvx . (6.60) 2πkT Many textbooks derive the Maxwell velocity distribution for an ideal classical gas and give the misleading impression that the distribution applies only if the particles are noninteracting. We stress that the Maxwell velocity (and momentum) distribution applies to any classical system regardless of the interactions, if any, between the particles. CHAPTER 6. MANYPARTICLE SYSTEMS Problem 6.10. Is there an upper limit to the velocity? 306 The upper limit to the velocity of a particle is the velocity of light. Yet the Maxwell velocity distribution imposes no upper limit to the velocity. Does this contradiction lead to diﬃculties? Problem 6.11. Simulations of the Maxwell velocity distribution (a) Program LJ2DFluidMD simulates a system of particles interacting via the LennardJones potential (1.1) in two dimensions by solving Newton’s equations of motion numerically. The program computes the distribution of velocities in the xdirection among other quantities. Compare the form of the velocity distribution to the form of the Maxwell velocity distribution in (6.60). How does its width depend on the temperature? (b) Program IdealThermometerIdealGas implements the demon algorithm for an ideal classical gas in one dimension (see Section 4.9). All the particles have the same initial velocity. The program computes the distribution of velocities among other quantities. What is the form of the velocity distribution? Give an argument based on the central limit theorem (see Section 3.7) to explain why the distribution has the observed form. Is this form consistent with (6.60)? 6.2.3 The Maxwell speed distribution We have found that the distribution of velocities in a classical system of particles is a Gaussian and is given by (6.59). To determine the distribution of speeds for a threedimensional system we need to know the number of microstates between v and v + ∆v . This number is proportional to the volume of a spherical shell of width ∆v or 4π (v + ∆v )3 /3 − 4πv 3 /3 → 4πv 2 ∆v in the limit ∆v → 0. Hence, the probability that a particle has a speed between v and v + dv is given by f (v )dv = 4πAv 2 e−mv
2 /2kT dv, (6.61) where A is a normalization constant, which we calculate in Problem 6.12. Problem 6.12. Maxwell speed distribution (a) Compare the form of the Maxwell speed distribution (6.61) with the form of the Maxwell velocity distribution (6.59). (b) Use the normalization condition f (v )dv = 4πv 2 m 2πkT
3/2 ∞ 0 f (v )dv = 1 to calculate A and show that (Maxwell speed distribution). (6.62) e−mv 2 /2kT dv (c) Calculate the mean speed v , the most probable speed v , and the rootmeansquare speed vrms ˜ and discuss their relative magnitudes. (d) Make the change of variables u = v/ (2kT /m) and show that √ 2 f (v )dv = f (u)du = (4/ π )u2 e−u du, (6.63) where we have again used the same notation for two diﬀerent, but physically related probability densities. The (dimensionless) speed probability density f (u) is shown in Figure 6.2. CHAPTER 6. MANYPARTICLE SYSTEMS
1.0 umax 0.8 u urms 307 f(u)
0.6 0.4 0.2 0.0 0.0 0.5 1.0 1.5 2.0 u 2.5 3.0 √ 2 Figure 6.2: The probability density f (u) = 4/ πu2 e−u that a particle has a dimensionless speed u. Note the diﬀerence between the most probable speed u = 1, the mean speed u ≈ 1.13, and the ˜ rootmeansquare speed urms ≈ 1.22. The dimensionless speed u is deﬁned by u ≡ v/(2kT /m)1/2 . Problem 6.13. Maxwell speed distribution in one or two dimensions Find the Maxwell speed distribution for particles restricted to one and two dimensions. 6.3 Occupation Numbers and Bose and Fermi Statistics We now develop the formalism for calculating the thermodynamic properties of ideal gases for which quantum eﬀects are important. We have already noted that the absence of interactions between the particles of an ideal gas enables us to reduce the problem of determining the energy levels of the gas as a whole to determining ǫk , the energy levels of a single particle. Because the particles are indistinguishable, we cannot specify the microstate of each particle. Instead a microstate of an ideal gas is speciﬁed by the occupation number nk , the number of particles in the single particle state k with energy ǫk .2 If we know the value of the occupation number for each single particle microstate, we can write the total energy of the system in microstate s as Es =
k nk ǫ k . (6.64) The set of nk completely speciﬁes a microstate of the system. The partition function for an ideal gas can be expressed in terms of the occupation numbers as P e−β k nk ǫk , (6.65) Z (V, T, N ) =
{nk } relation of k to the quantum numbers labeling the single particle microstates is given in (4.35) and in (6.93). In the following we will use k to label single particle microstates. 2 The CHAPTER 6. MANYPARTICLE SYSTEMS where the occupation numbers nk satisfy the condition N=
k 308 nk . (6.66) The condition (6.66) is diﬃcult to satisfy in practice, and we will later use the grand canonical ensemble for which the condition of a ﬁxed number of particles is relaxed. As discussed in Section 4.3.6, one of the fundamental results of relativistic quantum mechanics is that all particles can be classiﬁed into two groups. Particles with zero or integral spin such as 4 He are bosons and have wavefunctions that are symmetric under the exchange of any pair of particles. Particles with halfintegral spin such as electrons, protons, and neutrons are fermions and have wavefunctions that are antisymmetric under particle exchange. The Bose or Fermi character of composite objects can be found by noting that composite objects that have an even number of fermions are bosons and those containing an odd number of fermions are themselves fermions. For example, an atom of 3 He is composed of an odd number of particles: two electrons, two protons, 1 and one neutron each of spin 2 . Hence, 3 He has halfintegral spin, making it a fermion. An atom 4 of He has one more neutron so there are an even number of fermions and 4 He is a boson. It is remarkable that all particles fall into one of two mutually exclusive classes with diﬀerent spin. It is even more remarkable that there is a connection between the spin of a particle and its statistics. Why are particles with halfintegral spin fermions and particles with integral spin bosons? The answer lies in the requirements imposed by Lorentz invariance on quantum ﬁeld theory. This requirement implies that the form of quantum ﬁeld theory must be the same in all inertial reference frames. Although many physicists believe that the relation between spin and statistics must have a simpler explanation, no such explanation yet exists.3 The diﬀerence between fermions and bosons is speciﬁed by the possible values of nk . For fermions we have nk = 0 or 1 (fermions). (6.67) The restriction (6.67) is a statement of the Pauli exclusion principle for noninteracting particles – two identical fermions cannot be in the same single particle microstate. In contrast, the occupation numbers nk for identical bosons can take any positive integer value: nk = 0 , 1 , 2 , · · · (bosons). (6.68) We will see in the following sections that the nature of the statistics of a many particle system can have a profound eﬀect on its properties. Example 6.1. Calculate the partition function of an ideal gas of N = 3 identical fermions in equilibrium with a heat bath at temperature T . Assume that each particle can be in one of four possible microstates with energies, ǫ1 , ǫ2 , ǫ3 , and ǫ4 . Solution. The possible microstates of the system are summarized in Table 6.2. The spin of the fermions is neglected. Is it possible to reduce this problem to a onebody problem as we did for a noninteracting classical system?
3 In spite of its fundamental importance, it is only a slight exaggeration to say that “everyone knows the spinstatistics theorem, but no one understands it.” See Duck and Sudarshan (1998). CHAPTER 6. MANYPARTICLE SYSTEMS n1 0 1 1 1 n2 1 0 1 1 n3 1 1 0 1 n4 1 1 1 0 309 Table 6.2: The possible states of a threeparticle fermion system with four single particle energy microstates (see Example 6.1). The quantity n1 represents the number of particles in the single particle microstate labeled 1, etc. Note that we have not speciﬁed which particle is in a particular microstate. From Table 6.2 we see that the partition function is given by Z 3 = e − β ( ǫ 2 +ǫ 3 +ǫ 4 ) + e − β ( ǫ 1 +ǫ 3 +ǫ 4 ) + e − β ( ǫ 1 +ǫ 2 +ǫ 4 ) + e − β ( ǫ 1 +ǫ 2 +ǫ 3 ) . (6.69) ♦ Problem 6.14. Calculate n1 , the mean number of fermions in the single particle microstate 1 with energy ǫ1 , for the system in Example 6.1. Problem 6.15. Mean energy of a toy model of an ideal Bose gas (a) Calculate the mean energy of an ideal gas of N = 2 identical bosons in equilibrium with a heat bath at temperature T , assuming that each particle can be in one of three microstates with energies 0, ∆, and 2∆. (b) Calculate the mean energy for N = 2 distinguishable particles assuming that each particle can be in one of three possible microstates. ¯ ¯ (c) If E1 is the mean energy for one particle and E2 is the mean energy for the twoparticle system, ¯2 = 2E1 for either bosons or distinguishable particles? ¯ is E 6.4 Distribution Functions of Ideal Bose and Fermi Gases The calculation of the partition function for an ideal gas in the semiclassical limit was done by choosing a single particle as the system. This choice is not possible for an ideal gas at low temperatures where the quantum nature of the particles cannot be ignored. So we need a diﬀerent strategy. The key idea is that it is possible to distinguish the subset of all particles in a given single particle microstate from the particles in all other single particle microstates. For this reason we divide the system of interest into subsystems each of which is the set of all particles that are in a given single particle microstate. Because the number of particles in a given microstate varies, we need to use the grand canonical ensemble and assume that each subsystem is coupled to a heat bath and a particle reservoir independently of the other single particle microstates. Because we have not yet applied the grand canonical ensemble, we review it here. The thermodynamic potential in the grand canonical ensemble is denoted by Ω(T, V, µ) and is equal to −P V CHAPTER 6. MANYPARTICLE SYSTEMS 310 [see (2.168)]. The relation of thermodynamics to statistical mechanics is given by Ω = −kT ln ZG , where the grand partition function ZG is given by ZG =
s e−β (Es −µNs ) . (6.70) Es is the energy of microstate s and Ns is the number of particles in microstate s. The goal is to calculate ZG , then Ω and the pressure equation of state −P V (in terms of T , V , and µ), and then determine S from the relation S=− ∂Ω ∂T
V ,µ , (6.71) and the mean number of particles from the relation ∂Ω N =− ∂µ T ,V . (6.72) The probability of a particular microstate is given by Ps = 1 −β (Es−µNs ) e ZG (Gibbs distribution). (6.73) Because we can treat an ideal gas as a collection of independent subsystems where each subsystem is a single particle microstate, ZG reduces to the product of ZG, k for each subsystem. Thus, the ﬁrst step is to calculate the grand partition function ZG, k for each subsystem. We write the energy of the nk particles in the single particle microstate k as nk ǫk and write ZG, k as ZG, k =
nk e−βnk (ǫk −µ) , (6.74) where the sum is over the possible values of nk . For fermions this sum is straightforward because nk = 0 and 1 [see (6.67)]. Hence ZG, k = 1 + e−β (ǫk −µ) . (6.75) The corresponding thermodynamic or Landau potential Ωk is given by Ωk = −kT ln ZG, k = −kT ln[1 + e−β (ǫk −µ) ]. (6.76) We use the relation nk = −∂ Ωk /∂µ [see (6.72)] to ﬁnd the mean number of particles in microstate k. The result is nk = − or nk = 1 eβ (ǫk −µ) + 1 (FermiDirac distribution). (6.78) ∂ Ωk e−β (ǫk −µ) = , ∂µ 1 + e−β (ǫk −µ) (6.77) The result (6.78) for the mean number of particles in single particle microstate k is known as the FermiDirac distribution. CHAPTER 6. MANYPARTICLE SYSTEMS The integer values of nk are unrestricted for bosons. We write (6.74) as
∞ 311 ZG, k = 1 + e−β (ǫk −µ) + e−2β (ǫk −µ) + · · · = e−β (ǫk −µ)
nk =0 nk . (6.79) The geometric series in (6.79) is convergent for e−β (ǫk −µ) < 1. Because this condition must be satisﬁed for all values of ǫk , we require that eβµ < 1 or µ<0 (bosons). (6.80) In contrast, the chemical potential may be either positive or negative for fermions. The summation of the geometric series in (6.79) gives ZG, k = and hence we obtain 1 1− e−β (ǫk −µ) , (6.81) The mean number of particles in single particle microstate k is given by nk = − or nk = e−β (ǫk −µ) ∂ Ωk = , ∂µ 1 − e−β (ǫk −µ) 1 eβ (ǫk −µ) −1 (BoseEinstein distribution). (6.83) Ωk = kT ln 1 − e−β (ǫk −µ) . (6.82) (6.84) The form (6.84) is known as the BoseEinstein distribution. It is frequently convenient to group the FermiDirac and BoseEinstein distributions together and to write nk = 1 eβ (ǫk −µ) ±1 . + FermiDirac distribution − BoseEinstein distribution . (6.85) The convention is that the upper sign corresponds to Fermi statistics and the lower sign to Bose statistics. Because the (grand) partition function ZG is a product, ZG = k ZG, k , the Landau potential for the ideal gas is given by Ω(T, V, µ) =
k Ωk = ∓kT k ln 1 ± e−β (ǫk −µ) . (6.86) The classical limit . The FermiDirac and BoseEinstein distributions must reduce to the classical limit under the appropriate conditions. In the classical limit nk ≪ 1 for all k; that is, the mean number of particles in any single particle microstate must be small. Hence ǫβ (ǫk −µ) ≫ 1 and in this limit both the FermiDirac and BoseEinstein distributions reduce to nk = e−β (ǫk −µ) (MaxwellBoltzmann distribution). (6.87) This result (6.87) is known as the MaxwellBoltzmann distribution. CHAPTER 6. MANYPARTICLE SYSTEMS 312 6.5 Single Particle Density of States To ﬁnd the various thermodynamic quantities we need to calculate various sums. For example, to obtain the mean number of particles in the system we need to sum (6.85) over all single particle states: 1 . (6.88) N (T, V, µ) = nk = eβ (ǫk −µ) ± 1 k k For a given temperature T and volume V , (6.88) is an implicit equation for the chemical potential µ in terms of the mean number of particles. That is, the chemical potential determines the mean number of particles just as the temperature determines the mean energy. Similarly, we can write the mean energy of the system as E (T, V, µ) =
k nk ǫ k . (6.89) For a macroscopic system the number of particles and the energy are well deﬁned, and we will usually replace N and E by N and E , respectively. Because we have described the microscopic states at the most fundamental level, that is, by using quantum mechanics, the macroscopic averages of interest such as (6.88), (6.89), and (6.86) involve sums over the microscopic states. However, because the systems of interest are macroscopic, the volume of the system is so large that the energies of the discrete microstates are very close together and for practical purposes indistinguishable from a continuum. As usual, it is easier to do integrals than to do sums over a very large number of microstates, and hence we will replace the sums in (6.88)–(6.86) by integrals. For example, we will write for an arbitrary function f (ǫ)
∞ k f (ǫk ) → f (ǫ) g (ǫ)dǫ,
0 (6.90) where g (ǫ) dǫ is the number of single particle microstates between ǫ and ǫ + dǫ. The quantity g (ǫ) is known as the density of states, although a better term would be the density of single particle microstates. Although we have calculated the density of states g (ǫ) for a single particle in a box (see Section 4.3), we review the calculation here to emphasize its generality and the common aspects of the calculation for blackbody radiation, elastic waves in a solid, and electron waves. For convenience, we choose the box to be a cube of linear dimension L and assume that there are standing waves that vanish at the faces of the cube. The condition for a standing wave in one dimension is that the wavelength satisﬁes the condition λ= 2L n (n = 1, 2, . . .), (6.91) where n is a nonzero positive integer. It is useful to deﬁne the wave number k as k= 2π , λ (6.92) and write the standing wave condition as k = nπ/L. Because the waves in the x, y , and z directions satisfy similar conditions, we can treat the wave number as a vector whose components satisfy the CHAPTER 6. MANYPARTICLE SYSTEMS condition 313 π (6.93) k = (nx , ny , nz ) , L where nx , ny , nz are positive nonzero integers. Not all values of k are permissible and each combination of {nx , ny , nz } corresponds to a diﬀerent microstate. In the “number space” deﬁned by the three perpendicular axes labeled by nx , ny , and nz , the possible values of the microstates lie at the centers of cubes of unit edge length. Because the energy of a wave depends only on the magnitude of k, we want to know the number of microstates between k and k + dk . As we did in Section 4.3, it is easier to ﬁrst ﬁnd Γ(k ), the number of microstates with wave number less than or equal to k . We know that the volume in nspace of a single particle microstate is 1, and hence the number of single particle microstates in number space that are contained in the positive octant of a sphere of radius n is given by Γ(n) = 1 (4πn3 /3), 8 where n2 = n2 + n2 + n2 . Because k = π n/L, the number of single particle microstates with wave x y z vector less than or equal to k is 1 4πk 3 /3 . (6.94) Γ(k ) = 8 (π/L)3 If we use the relation g (k ) dk = Γ(k + dk ) − Γ(k ) = we obtain g (k ) dk = V k 2 dk , 2π 2 dΓ(k ) dk, dk (6.95) (6.96) where the volume V = L3 . Equation (6.96) gives the density of states in k space between k and k + dk . Although we obtained the result (6.96) for a cube, the result is independent of the shape of the enclosure and the nature of the boundary conditions (see Problem 6.58). That is, if the box is suﬃciently large, the surface eﬀects introduced by the box do not aﬀect the physical properties of the system. Problem 6.16. Single particle density of states in one and two dimensions Find the form of the density of states in k space for standing waves in a twodimensional and in a onedimensional box. 6.5.1 Photons The result (6.96) for the density of states in k space holds for any wave in a threedimensional enclosure. We next determine the number of states g (ǫ) dǫ as a function of the energy ǫ. For simplicity, we adopt the same symbol to represent the density of states in k space and in ǫspace because the meaning of g will be clear from the context. The nature of the dependence of g (ǫ) on the energy ǫ is determined by the form of the function ǫk . For electromagnetic waves of frequency ν we know that λν = c, ω = 2πν , and k = 2π/λ. Hence, ω = 2πc/λ or ω = ck. (6.97) CHAPTER 6. MANYPARTICLE SYSTEMS The energy ǫ of a photon of frequency ω is ǫ = ω = ck. Because k = ǫ/ c, we ﬁnd from (6.96) that g (ǫ) dǫ = V ǫ2 2 π 2 3 c3 dǫ. 314 (6.98) (6.99) The result (6.99) requires one modiﬁcation. The state of an electromagnetic wave or photon depends not only on its wave vector or momentum, but also on its polarization. There are two mutually perpendicular directions of polarization (right circularly polarized and left circularly polarized) for each electromagnetic wave of wave number k.4 Thus the number of photon microstates in which the photon has an energy in the range ǫ to ǫ + dǫ is given by g (ǫ) dǫ = V ǫ2 dǫ π 2 3 c3 (photons). (6.100) We will use (6.100) frequently in the following. 6.5.2 Nonrelativistic particles
p2 . 2m For a nonrelativistic particle of mass m we know that ǫ= (6.101) From the relations p = h/λ and k = 2π/λ, we ﬁnd that the momentum p of a particle is related to its wave vector k by p = k . Hence, the energy can be expressed as ǫ= and thus k dk. (6.103) m If we use (6.96) and the relations (6.102) and (6.103), we ﬁnd that the number of microstates in the interval ǫ to ǫ + dǫ is given by dǫ = V (2m)3/2 ǫ1/2 dǫ. (6.104) 4π 2 3 We have included a factor of ns , the number of spin states for a given value of k or ǫ. Because electrons have spin 1/2, ns = 2, and we can write (6.104) as g (ǫ) dǫ = ns V (2m)3/2 ǫ1/2 dǫ (electrons). (6.105) 2π 2 3 It is common to choose units such that = 1, and we will express most of our results in the remainder of this chapter in terms of instead of h. g (ǫ) dǫ =
4 In the language of quantum mechanics we say that the photon has spin 1 and two helicity states. The fact that the photon has spin S = 1 and two helicity states rather than (2S + 1) = 3 states is a consequence of special relativity for massless particles. k , 2m
2 22 (6.102) CHAPTER 6. MANYPARTICLE SYSTEMS Problem 6.17. Density of states in one and two dimensions 315 Calculate the density of states g (ǫ) for a nonrelativistic particle of mass m in in one and two dimensions (see Problem 6.16). Sketch g (ǫ) on one graph for d = 1, 2, and 3 and comment on the diﬀerent dependence of g (ǫ) on ǫ for diﬀerent spatial dimensions. Problem 6.18. Relativistic particles Calculate the density of states g (ǫ) in three dimensions for a relativistic particle of rest mass m for which ǫ2 = p2 c2 + m2 c4 . Don’t try to simplify your result. Problem 6.19. Relation between the energy and pressure equations of state for a nonrelativistic ideal gas The mean energy E is given by
∞ E=
0 ǫn(ǫ) g (ǫ) dǫ V 4π 2 (2m)3/2 3
∞ 0 (6.106a) ǫ3/2 dǫ . eβ (ǫ−µ) ± 1 (6.106b) = ns Use (6.86) for the Landau potential and (6.104) for the density of states of nonrelativistic particles in three dimensions to show that Ω can be expressed as
∞ Ω = ∓kT = ∓kT 0 g (ǫ) ln[1 ± e−β (ǫ−µ) ] dǫ
∞ 0 (6.107) (6.108) ns V (2m)3/2 4π 2 3 ǫ1/2 ln[1 ± e−β (ǫ−µ) ] dǫ. Integrate (6.108) by parts with u = ln[1 ± e−β (ǫ−µ) ] and dv = ǫ1/2 dǫ and show that V 2 Ω = − ns 2 3 4π (2m)3/2 3
∞ 0 ǫ3/2 dǫ . eβ (ǫ−µ) ± 1 (6.109) 2 The form (6.106b) for E is the same as the general result (6.109) for Ω except for the factor of − 3 . Use the relation Ω = −P V [see (2.168)] to show that PV = 2 E. 3 (6.110) The relation (6.110) is exact and holds for an ideal gas with any statistics at any temperature T , and depends only on the nonrelativistic relation ǫ = p2 /2m. Problem 6.20. Relation between the energy and pressure equations of state for photons Use similar considerations as in Problem 6.19 to show that for photons: PV = 1 E. 3 (6.111) Equation (6.111) holds at any temperature and is consistent with Maxwell’s equations. Thus, the pressure due to electromagnetic radiation is related to the energy density by P = u(T )/3. CHAPTER 6. MANYPARTICLE SYSTEMS 316 6.6 The Equation of State of an Ideal Classical Gas: Application of the Grand Canonical Ensemble We have already seen how to obtain the equations of state and other thermodynamic quantities for the ideal classical gas in the microcanonical ensemble (ﬁxed E , V , and N ) and in the canonical ensemble (ﬁxed T , V , and N ). We now discuss how to use the grand canonical ensemble (ﬁxed T , V , and µ) to ﬁnd the analogous quantities under conditions for which the MaxwellBoltzmann distribution is applicable. The calculation in the grand canonical ensemble will automatically satisfy the condition that the particles are indistinguishable. As an example, we ﬁrst calculate the chemical potential given that the mean number of particles is N . We use the MaxwellBoltzmann distribution (6.87) and the density of states (6.104) for particles of mass m and set ns = 1 for simplicity. The result is
∞ N=
k nk → n(ǫ) g (ǫ) dǫ
0 3/2 0 ∞ (6.112a) (6.112b) = V 2m 2 4π 2 e−β (ǫ−µ) ǫ1/2 dǫ. We make the change of variables u = βǫ and write (6.112b) as N= 2m V 4π 2 2 β
3/2 eβµ
0 ∞ e−u u1/2 du. (6.113) The integral in (6.113) can be done analytically (make the change of variables u = y 2 ) and has the value π 1/2 /2 (see the Appendix). Hence, the mean number of particles is given by N (T, V, µ) = V m 2π 2 β
3/2 eβµ . (6.114) Because we cannot easily measure µ, it is of more interest to ﬁnd the value of µ that yields the desired value of N . The solution of (6.114) for the chemical potential is µ = kT ln N 2π 2 β V m
3/2 . (6.115) What is the diﬀerence, if any, between (6.114) and the result (6.29) for µ found in the canonical ensemble? Problem 6.21. The chemical potential (a) Estimate the chemical potential of one mole of an ideal monatomic classical gas at standard temperature and pressure and show that µ ≪ 0. (b) Show that N can be expressed as [see (6.114)] N= V βµ e, λ3 (6.116) CHAPTER 6. MANYPARTICLE SYSTEMS and hence µ(T, V ) = −kT ln where ρ = N /V . 317 1 , ρλ3 (6.117) (c) In Section 6.1 we argued that the semiclassical limit λ ≪ ρ−1/3 [see (6.1)] implies that nk ≪ 1; that is, the mean number of particles in any single particle energy state is very small. Use the expression (6.117) for µ and (6.87) for nk to show that the condition nk ≪ 1 implies that λ ≪ ρ−1/3 . As we saw in Section 2.21, the chemical potential is the change in any of the thermodynamic potentials when a particle is added. It might be expected that µ > 0, because it should cost energy to add a particle. But because the particles do not interact, perhaps µ = 0? So why is µ ≪ 0 for an ideal classical gas? The reason is that we have to include the contribution of the entropy. In the canonical ensemble the change in the free energy due to the addition of a particle at constant temperature is ∆F = ∆E − T ∆S ≈ kT − T ∆S . The number of places where the additional particle can be located is approximately V /λ3 , and hence ∆S ∼ k ln V /λ3 . Because V /λ3 ≫ 1, ∆S ≫ ∆E , and thus ∆F ≪ 0, which implies that µ = ∆F/∆N ≪ 0. The example calculation of N (T, V, µ) leading to (6.114) was not necessary because we can calculate all thermodynamic quantities directly from the Landau potential Ω. We calculate Ω from (6.86) by noting that eβµ ≪ 1 and approximating the logarithm as ln (1 ± x) ≈ ±x. We ﬁnd that Ω = ∓kT → −kT
k ln 1 ± e−β (ǫk −µ) e−β (ǫk −µ) (semiclassical limit). (6.118a) (6.118b) k As expected, the form of Ω in (6.118b) is independent of whether we started with Bose or Fermi statistics. As usual, we replace the sum over the single particle states by an integral over the density of states and ﬁnd Ω = −kT eβµ = −kT =− If we substitute λ = (2πβ
2 ∞ 0 3 g (ǫ) e−βǫ dǫ 2m β
3/2 3/2 ∞ (6.119a) u1/2 e−u du
0 V 4π 2 eβµ (6.119b) (6.119c) m V β 5/2 2π 2 eβµ . /m)1/2 , we ﬁnd Ω = −kT V βµ e. λ3 (6.120) From the relation Ω = −P V [see (2.168)], we obtain P= kT βµ e. λ3 (6.121) CHAPTER 6. MANYPARTICLE SYSTEMS If we use the thermodynamic relation (6.72), we obtain N =− ∂Ω ∂µ
V ,T 318 = V βµ e. λ3 (6.122) The classical equation of state, P V = N kT , is obtained by using (6.122) to eliminate µ. The simplest way of ﬁnding the energy is to use the relation (6.110). We can ﬁnd the entropy S (T, V, µ) using (6.120) and (6.71): S (T, V, µ) = − ∂Ω ∂T
V ,µ = kβ 2 = V kβ 2 5 2β 7/2 ∂Ω ∂β µ − 5/2 β (6.123a) m 2π 2
3/2 eβµ . (6.123b) We eliminate µ from (6.123b) using (6.115) and obtain the SackurTetrode expression for the entropy of an ideal gas: S (T, V, N ) = N k N 2π 2 5 − ln − ln 2 V mkT
3/2 . (6.124) We have written N rather than N in (6.124). Note that we did not have to introduce any extra factors of N ! as we did in Section 6.1, because we already correctly counted the number of microstates. Problem 6.22. Ideal gas equations of state Show that E = (3/2)N kT and P V = N kT from the results of this section. 6.7 Blackbody Radiation We can regard electromagnetic radiation as equivalent to a system of noninteracting bosons (photons), each of which has an energy hν , where ν is the frequency of the radiation. If the radiation is in an enclosure, equilibrium will be established and maintained by the interactions of the photons with the atoms of the wall in the enclosure. Because the atoms emit and absorb photons, the total number of photons is not conserved. If a body in thermal equilibrium emits electromagnetic radiation, this radiation is described as blackbody radiation and the object is said to be a blackbody. This statement does not mean that the body is actually black. The word “black” indicates that the radiation is perfectly absorbed and reradiated by the object. The frequency spectrum of light radiated by such an idealized body is described by a universal spectrum called the Planck spectrum, which we will derive in the following [see (6.133)]. The nature of the spectrum depends only on the temperature T of the radiation. We can derive the Planck radiation law using either the canonical or grand canonical ensemble because the photons are continuously absorbed and emitted by the walls of the container and hence their number is not conserved. This lack of a conservation law for the number of particles implies that the chemical potential vanishes. Hence the BoseEinstein distribution in (6.85) reduces to nk = 1 eβǫk − 1 (Planck distribution) (6.125) CHAPTER 6. MANYPARTICLE SYSTEMS for blackbody radiation. 319 The result (6.125) can be understood by simple considerations. As we have mentioned, equilibrium is established and maintained by the interactions between the photons and the atoms of the wall in the enclosure. The number N of photons in the cavity cannot be imposed externally on the system and is ﬁxed by the temperature T of the walls and the volume V enclosed. Hence, the free energy F for photons cannot depend on N because the latter is not a thermodynamic variable, and we have µ = ∂F/∂N = 0. If we substitute µ = 0 into the general result (6.84) for the BoseEinstein distribution, we ﬁnd that the mean number of photons in single particle state k is given by 1 nk = βǫ , (6.126) e k −1 in agreement with (6.125). To see how (6.126) follows from the canonical ensemble, consider a system in equilibrium with a heat bath at temperature T . Because there is no constraint on the total number of photons, the number of photons in each single particle microstate is independent of the number of photons in all the other single particle microstates. Thus, the partition function is the product of the single particle state partition functions Zk (T, V ) for each state in the same way as the partition function for a collection of noninteracting spins is the product of the partition functions for each spin. We have
∞ Zk (T, V ) =
nk =0 e−βnk ǫk . (6.127) Because the sum in (6.127) is a geometric series, we obtain Zk (T, V ) = 1 . 1 − e−βǫk (6.128) In the canonical ensemble the mean number of photons in the single particle microstate k is given by nk = = We have from (6.128) and (6.129b) nk = ∂ − ln (1 − e−βǫk ) ∂ (−βǫk ) e−βǫk 1 = = βǫ . 1 − e−βǫk e k −1 (6.130a) (6.130b)
∞ −βnk ǫk nk =0 nk e ∞ −βnk ǫk nk =0 e (6.129a) (6.129b) ∂ ln Zk . ∂ (−βǫk ) Planck’s theory of blackbody radiation follows from the form of the density of states for photons found in (6.100). The number of photons with energy in the range ǫ to ǫ + dǫ is given by N (ǫ) dǫ = n(ǫ)g (ǫ) dǫ = V π2 ǫ2 dǫ . 3 c3 eβǫ − 1 (6.131) CHAPTER 6. MANYPARTICLE SYSTEMS 320 If we substitute ǫ = hν on the righthand side of (6.131), we ﬁnd that the number of photons in the frequency range ν to ν + dν is given by N (ν ) dν = 8πV ν 2 dν . c3 eβhν − 1 dν 8πhV ν 3 . 3 βhν − 1 c e (6.132) The distribution of radiated energy is obtained by multiplying (6.132) by hν : E (ν )dν = hνN (ν ) dν = (6.133) Equation (6.133) gives the energy radiated by a blackbody of volume V in the frequency range between ν and ν + dν . The energy per unit volume u(ν ) is given by u(ν ) = 1 8πhν 3 3 βhν − 1 c e (Planck’s radiation law). (6.134) We can change variables to ǫ = hν and write the energy density as u(ǫ) = 8π ǫ3 . (hc)3 eǫ/kT − 1 (6.135) The physical system that most closely gives the spectrum of a black body is the spectrum of the cosmic microwave background, which ﬁts the theoretical spectrum of a blackbody better than the best blackbody spectrum that can be made in a laboratory. In contrast, a piece of hot, glowing ﬁrewood is not really in thermal equilibrium, and the spectrum of glowing embers is only a crude approximation to the blackbody spectrum. The existence of the cosmic microwave background spectrum and its ﬁt to the blackbody spectrum is compelling evidence that the universe experienced a Big Bang.5 Problem 6.23. Wien’s displacement law The maximum of u(ν ) shifts to higher frequencies with increasing temperature. Show that the maximum of u can be found by solving the equation (3 − x)ex = 3, where x = βhνmax . Solve (6.136) numerically for x and show that hνmax = 2.822 kT
5 The (6.136) (Wien’s displacement law). (6.137) universe is ﬁlled with electromagnetic radiation with a distribution of frequencies given by (6.133) with T ≈ 2.725 K. This background radiation is a remnant from a time when the universe was composed primarily of electrons and protons at a temperature of about 3000 K. This plasma of electrons and protons interacted strongly with the electromagnetic radiation over a wide range of frequencies, so that the matter and radiation reached thermal equilibrium. As the universe expanded, the plasma cooled until it became energetically favorable for electrons and protons to combine to form hydrogen atoms. Atomic hydrogen interacts with radiation only at the frequencies of the hydrogen spectral lines. As a result most of the radiation energy was eﬀectively decoupled from matter so that its temperature is independent of the temperature of the hydrogen atoms. The background radiation is now at about 2.725 K because of the expansion of the universe. This expansion causes the radiation to be redshifted. The temperature of the cosmic radiation background will continue to decrease as the universe expands. CHAPTER 6. MANYPARTICLE SYSTEMS Problem 6.24. Derivation of the RayleighJeans and Wien’s laws 321 (a) Make a change of variables in (6.134) to ﬁnd the energy emitted by a blackbody at a wavelength between λ and λ + dλ. (b) Determine the limiting behavior of your result in part (a) for long wavelengths. This limit is called the RayleighJeans law and is given by u(λ)dλ = 8πkT dλ. λ4 (6.138) Does this form involve Planck’s constant? The result in (6.138) was originally derived from purely classical considerations. (c) Classical theory predicts what is known as the ultraviolet catastrophe, namely, that an inﬁnite amount of energy is radiated at high frequencies or short wavelengths. Explain how (6.138) would give an inﬁnite result for the total radiated energy, and thus the classical result cannot be correct for all wavelengths. (d) Determine the limiting behavior of u(λ) for short wavelengths. This behavior is known as Wien’s law, after Wilhelm Wien who found it by ﬁnding a functional form to ﬁt the experimental data. Problem 6.25. Thermodynamics of blackbody radiation Use the various thermodynamic relations to show that
∞ E=V
0 u(ν ) dν = 4σ V T 4, 3c 4σ V T 4, c (6.139a) (6.139b) (6.139c) (6.139d) (6.139e) Ω=F =− S= 16σ V T 3, 3c 4σ 4 1 E P= T= , 3c 3V G = F + P V = 0. The free energy F in (6.139b) can be calculated from Z starting from (6.128) and using (6.100). The StefanBoltzmann constant σ is given by σ= The integral
0 ∞ 2π 5 k 4 . 15h3 c2 (6.140) is evaluated in the Appendix. π4 x3 dx = . x−1 e 15 (6.141) The relation (6.139a) between the total energy and the temperature is known as the StefanBoltzmann law. It was derived based on thermodynamic considerations in Section 2.21. CHAPTER 6. MANYPARTICLE SYSTEMS Problem 6.26. Mean number of photons Show that the total mean number of photons in an enclosure of volume V is given by N= V π 2 c3
∞ 0 322 ω 2 dω V (kT )3 = 233 πc e −1
ω /kT ∞ 0 x2 dx . ex − 1 (6.142) The integral in (6.142) can be expressed in terms of known functions (see the Appendix). The result is ∞2 x dx = 2 × 1.202. (6.143) x 0 e −1 Hence N depends on T as N = 0.244V kT c
3 . (6.144) 6.8 The Ideal Fermi Gas The low temperature properties of metals are dominated by the behavior of the conduction electrons. Given that there are Coulomb interactions between the electrons as well as interactions between the electrons and the positive ions of the lattice, it is remarkable that the free electron model in which the electrons are treated as an ideal gas of fermions near zero temperature is an excellent model of the conduction electrons in a metal under most circumstances.6 In the following, we investigate the properties of an ideal Fermi gas and brieﬂy discuss its applicability as a model of electrons in metals. As we will see in Problem 6.27, the thermal de Broglie wavelength of the electrons in a typical metal is much larger than the mean interparticle spacing, and hence we must treat the electrons using Fermi statistics. When a system is dominated by quantum mechanical eﬀects, it is said to be degenerate. 6.8.1 Ground state properties We ﬁrst discuss the noninteracting Fermi gas at T = 0. From (6.78) we see that the zero temperature limit (β → ∞) of the FermiDirac distribution is n(ǫ) = 1 0 for ǫ < µ, for ǫ > µ. (6.145) That is, all states whose energies are below the chemical potential are occupied, and all states whose energies are above the chemical potential are unoccupied. The Fermi distribution at T = 0 is shown in Figure 6.3.
6 The idea that a system of interacting electrons at low temperatures can be understood as a noninteracting gas of quasiparticles is due to Lev Landau (1908–1968), the same Landau for whom the thermodynamic potential in the grand canonical ensemble is named. Landau worked in many ﬁelds including low temperature physics, atomic and nuclear physics, condensed matter physics, and plasma physics. He was awarded the 1962 Nobel Prize for Physics for his work on superﬂuidity. He was also the coauthor of ten widely used graduatelevel textbooks on various areas of theoretical physics. CHAPTER 6. MANYPARTICLE SYSTEMS 323 n(ε)
1 T=0 T>0 εF ε Figure 6.3: The FermiDirac distribution for T = 0 and T ≪ TF . The form of n(ǫ) for T > 0 is based on the assumption that µ is unchanged for T ≪ TF . Note that the area under the curve n(ǫ) at T = 0 is approximately equal to the area under the curve n(ǫ) for T ≪ TF . The consequences of (6.145) are easy to understand. At T = 0, the system is in its ground state, and the particles are distributed among the single particle states so that the total energy of the gas is a minimum. Because we may place no more than one particle in each state, we need to construct the ground state of the system by adding a particle into the lowest available energy state until we have placed all the particles. To ﬁnd the value of µ(T = 0), we write
∞ µ(T =0) µ(T =0) N=
0 n(ǫ)g (ǫ) dǫ −→ T →0 g (ǫ) dǫ = V
0 0 (2m)3/2 1/2 ǫ dǫ. 2π 2 3 (6.146) We have substituted the electron density of states (6.105) in (6.146). The chemical potential at T = 0 is determined by requiring the integral to give the desired number of particles N . Because the value of the chemical potential at T = 0 will have special importance, it is common to denote it by ǫF : ǫF ≡ µ(T = 0), (6.147) where ǫF , the energy of the highest occupied state, is called the Fermi energy. The integral on the righthand side of (6.146) gives N= From (6.148) we have that
2 V 2mǫF 2 3π 2 3/2 . (6.148) ǫF = 2m (3π 2 ρ)2/3 (Fermi energy), (6.149) where the density ρ = N/V . It is convenient to write ǫF = p2 /2m where pF is known as the Fermi F momentum. It follows that the Fermi momentum pF is given by pF = (3π 2 ρ)1/3 (Fermi momentum). (6.150) CHAPTER 6. MANYPARTICLE SYSTEMS 324 The Fermi momentum can be estimated by using the de Broglie relation p = h/λ and taking λ ∼ ρ−1/3 , the mean distance between particles. That is, the particles are “localized” within a distance of order ρ−1/3 . At T = 0 all the states with momentum smaller than pF are occupied and all the states above this momentum are unoccupied. The boundary in momentum space between occupied and unoccupied states at T = 0 is called the Fermi surface. For an ideal Fermi gas in three dimensions the Fermi surface is the surface of a sphere with radius pF . We can understand why the chemical potential at T = 0 is positive by reasoning similar to that given on page 317 for an ideal classical gas. At T = 0 the contribution of T ∆S to the free energy vanishes, and no particle can be added with energy less than µ(T = 0). Thus, µ(T = 0) > 0. In contrast, we argued that µ(T > 0) is much less than zero for an ideal classical gas due to the large change in the entropy when adding (or removing) a particle. We will ﬁnd it convenient in the following to introduce a characteristic temperature, the Fermi temperature TF , by TF = ǫF /k. (6.151) The values of TF for typical metals are given in Table 6.3. A direct consequence of the fact that the density of states in three dimensions is proportional to ǫ1/2 is that the mean energy per particle at T = 0 is 3ǫF /5: E = N = The total energy is given by E=
2 3 3 N ǫF = N (3π 2 )2/3 ρ2/3 . 5 5 2m ǫF ǫ g (ǫ) dǫ 0 = ǫF g (ǫ) dǫ 0 2 5/2 3 5 ǫF = ǫF . 2 3/2 5 3 ǫF ǫF 0 ǫF 0 ǫ3/2 dǫ ǫ1/2 dǫ (6.152a) (6.152b) (6.153) The pressure can be immediately found from the general relation P V = 2E/3 [see (6.110)] for a nonrelativistic ideal gas at any temperature. Alternatively, the pressure can be found from the relation 2E ∂F = , (6.154) P =− ∂V 3V because the free energy is equal to the total energy at T = 0. The result is that the pressure at T = 0 is given by 2 P = ρǫF . (6.155) 5 The fact that the pressure is nonzero even at zero temperature is a consequence of the Pauli exclusion principle, which allows only one particle to have zero momentum (two electrons if the spin is considered). All other particles have ﬁnite momentum and hence give rise to a nonzero pressure at T = 0. Another way to understand the relation (6.155) is to recall the classical pressure equation of state, P = ρkT , which would predict that the pressure is zero at zero temperature. However, if CHAPTER 6. MANYPARTICLE SYSTEMS element Li Na Al Cu Ag ǫF (eV) 4.7 3.2 11.7 7 5.5 TF (104 K) 5.5 3.8 13.6 8.2 6.4 325 Table 6.3: Values of the Fermi energy and Fermi temperature for several metals at room temperature and atmospheric pressure. we replace T by the Fermi temperature TF , then P ∝ ρkTF = ρǫF , which is the same as (6.155) except for a numerical factor. Problem 6.27. Order of magnitude estimates (a) Verify that the values of ǫF given in electon volts (eV) lead to the values of TF in Table 6.3. (b) Compare the values of TF in Table 6.3 to room temperature. What is the value of kT in eV at room temperature? (c) Given the data in Table 6.3 verify that the electron densities for Li and Cu are ρ = 4.7 × 1028 m−3 and ρ = 8.5 × 1028 m−3 , respectively. (d) What is the mean distance between the electrons for Li and Cu? (e) Use the fact that the mass of an electron is 9.1 × 10−31 kg to estimate the de Broglie wavelength corresponding to an electron with energy comparable to the Fermi energy. (f) Compare your result for the de Broglie wavelength that you found in part (e) to the mean interparticle spacing that you found in part (d). Problem 6.28. Landau potential at zero temperature From (6.107) the Landau potential for an ideal Fermi gas at arbitrary T can be expressed as
∞ Ω = −kT g (ǫ) ln[1 + e−β (ǫ−µ) ] dǫ. (6.156) 0 To obtain the T = 0 limit of Ω, we have that ǫ < µ in (6.156), β → ∞, and hence ln[1+ e−β (ǫ−µ) ] → ln e−β (ǫ−µ) = −β (ǫ − µ). Hence, show that Ω= (2m)3/2 V 2π 2 2
ǫF 0 dǫ ǫ1/2 ǫ − ǫF . (6.157) Calculate Ω and determine the pressure at T = 0. Problem 6.29. Show that the limit (6.145) for n(ǫ) at T = 0 follows only if µ > 0. CHAPTER 6. MANYPARTICLE SYSTEMS 326 6.8.2 Low temperature properties One of the greatest successes of the free electron model and FermiDirac statistics is the explanation of the temperature dependence of the heat capacity of a metal. If the electrons behaved as a classical noninteracting gas, we would expect a contribution to the heat capacity equal to 3N k/2 as T → 0. Instead, we typically ﬁnd a very small contribution to the heat capacity which is linear in the temperature, a result that cannot be explained by classical statistical mechanics. Before we derive this result, we ﬁrst give a qualitative argument for the low temperature dependence of the heat capacity of an ideal Fermi gas. As we saw in Table 6.3, room temperature is much less than the Fermi temperature of the conduction electrons in a metal, that is, T ≪ TF . Hence we should be able to understand the behavior of an ideal Fermi gas at room temperature in terms of its behavior at zero temperature. Because there is only one characteristic energy in the system (the Fermi energy), the criterion for low temperature is that T ≪ TF . Hence the conduction electrons in a metal may be treated as if they are eﬀectively near zero temperature even though the metal is at room temperature. For 0 < T ≪ TF , the electrons that are within order kT below the Fermi surface have enough energy to occupy the microstates with energies that are order kT above the Fermi energy. In contrast, the electrons that are deep within the Fermi surface do not have enough energy to be excited to microstates above the Fermi energy. Hence, only a small fraction of order T /TF of the N electrons have a reasonable probability of being excited, and the remainder of the electrons remain unaﬀected as the temperature is increased from T = 0. This reasoning leads us to write the heat capacity of the electrons as CV ∼ Neﬀ k , where Neﬀ is the number of electrons that can be excited by exchanging energy with a heat bath. For a classical system, Neﬀ = N , but for a Fermi system at T ≪ TF , we have that Neﬀ ∼ N (T /TF ). Hence, we expect the temperature dependence of the heat capacity to be given by CV ∼ N k T TF (T ≪ TF ). (6.158) From (6.158) we see that the contribution to the heat capacity from the electrons is much smaller than the prediction of the equipartition theorem and is linear in T , as is found empirically. As an example, the measured speciﬁc heat of copper for T < 1 K is dominated by the contribution of the electrons and is given by CV /kN = 0.8 × 10−4 T . Our qualitative argument for the low temperature behavior of CV implicitly assumes that µ(T ) is unchanged for T ≪ TF . We can understand why µ(T ) remains unchanged as T is increased slightly from T = 0 by the following reasoning. The probability that a single particle state of energy ǫ = µ − ∆ is empty is 1 − n(ǫ = µ − ∆) = 1 − 1 1 = β∆ = n(ǫ = µ + ∆). e −β ∆ + 1 e +1 (6.159) We see from (6.159) that, for a given distance ∆ from µ, the probability that a particle is lost from a previously occupied single particle state below µ equals the probability that a previously empty single particle state is occupied. This property implies that the area under the step function at T = 0 is nearly the same as the area under n(ǫ) for T ≪ TF (see Figure 6.3). That is, n(ǫ) is symmetrical about ǫ = µ. If we make the additional assumption that the density of states changes very little in the region where n departs from a step function, we see that the mean number of CHAPTER 6. MANYPARTICLE SYSTEMS 327 particles lost from the previously occupied states just balances the mean number gained by the previously empty states. Hence, we conclude that for T ≪ TF , we still have the correct number of particles without any need to change the value of µ. Similar reasoning implies that µ(T ) must decrease slightly as T is increased from zero. Suppose that µ were to remain constant as T is increased. Because the density of states is an increasing function of ǫ, the number of electrons with energy ǫ > µ would be greater than the number lost with ǫ < µ. As a result, we would increase the number of electrons by increasing T . To prevent such a nonsensical increase, µ has to decrease slightly. In addition, we know that, because µ ≪ 0 for high temperatures where the system behaves like an ideal classical gas, µ(T ) must pass through zero. At what temperature would you estimate that µ(T ) ≈ 0? In Problem 6.30 we will determine µ(T ) by evaluating the integral in (6.160) numerically. Then we will evaluate the integral analytically for T ≪ TF and show that µ(T ) − µ(T = 0) ∼ (T /TF )2 . Hence, to ﬁrst order in T /TF , µ is unchanged. Problem 6.30. Numerical evaluation of the chemical potential for an ideal Fermi gas To ﬁnd the chemical potential for T > 0, we need to ﬁnd the value of µ that yields the desired mean number of particles. We have
∞ N=
0 n(ǫ)g (ǫ)dǫ = V (2m)3/2 2π 2 3 ∞ 0 ǫ1/2 dǫ , eβ (ǫ−µ) + 1 (6.160) where we have used (6.105) for g (ǫ). It is convenient to let ǫ = xǫF , µ = µ∗ ǫF , and T ∗ = kT /ǫF , and rewrite (6.160) as (2m)3/2 3/2 ∞ x1/2 dx N ǫF = , (6.161) ρ= (x−µ∗ )/T ∗ + 1 V 2π 2 3 0e or 3∞ x1/2 dx 1= , (6.162) 2 0 e(x−µ∗ )/T ∗ + 1 where we have substituted (6.149) for ǫF . To ﬁnd the dependence of µ∗ on T ∗ and hence µ on T use Program IdealFermiGasIntegral to evaluate the integral on the righthand side of (6.162) numerically. (a) Start with T ∗ = 0.2 and ﬁnd µ∗ such that (6.162) is satisﬁed. (Recall that µ∗ = 1 at T ∗ = 0.) Does µ∗ initially increase or decrease as T is increased from zero? What is the sign of µ∗ for T ∗ ≫ 1? (b) At what value of T ∗ is µ∗ ≈ 0? (c) Given the value of µ∗ (T ∗ ), the program computes the numerical value of E (T ). Describe its qualitative T dependence and the T dependence of CV . We now derive a quantitative expression for CV that is applicable for temperatures T ≪ TF .7
7 The following derivation is adapted from Kittel (1996). CHAPTER 6. MANYPARTICLE SYSTEMS The increase ∆E = E (T ) − E (T = 0) in the total energy is given by
∞ ǫF 328 ∆E =
0 ǫF ǫ n(ǫ)g (ǫ) dǫ − ǫ g (ǫ) dǫ,
0 ∞ (6.163a) (6.163b) =
0 ǫ[n(ǫ) − 1]g (ǫ) dǫ +
∞ ǫ n(ǫ)g (ǫ) dǫ.
ǫF ǫF We multiply the identity N=
0 n(ǫ)g (ǫ)dǫ =
0 g (ǫ) dǫ (6.164) by ǫF and write the integral on the lefthand side as a sum of two contributions to obtain
ǫF ∞ ǫF ǫF n(e)g (ǫ) dǫ +
0 ǫF ǫF n(e)g (ǫ) dǫ =
0 ∞ ǫF g (ǫ) dǫ, (6.165a) or
0 ǫF ǫF [n(ǫ) − 1]g (ǫ) dǫ + ǫF n(ǫ)g (ǫ) dǫ = 0.
ǫF (6.165b) We can use (6.165b) to rewrite (6.163b) as
ǫF ∆E =
ǫF (ǫ − ǫF )n(ǫ)g (ǫ)dǫ + 0 (ǫF − ǫ)[1 − n(ǫ)]g (ǫ)dǫ. (6.166) The heat capacity is found by diﬀerentiating ∆E with respect to T . The only temperature dependent quantity in (6.166) is n(ǫ). Hence, we can write CV as
∞ CV =
0 (ǫ − ǫF ) dn(ǫ) g (ǫ)dǫ. dT (6.167) For T ≪ TF , the derivative dn/dT is large only for ǫ near ǫF . Hence it is a good approximation to evaluate the density of states g (ǫ) at ǫ = ǫF and take it outside the integral:
∞ CV = g (ǫF )
0 (ǫ − ǫF ) dn dǫ. dT (6.168) We can also ignore the temperature dependence of µ in n(ǫ) and replace µ by ǫF . With this approximation we have dn dn dβ 1 (ǫ − ǫF )eβ (ǫ−ǫF ) . (6.169) = = dT dβ dT kT 2 [eβ (ǫ−ǫF ) + 1]2 We next let x = (ǫ − ǫF )/kT and use (6.168) and (6.169) to write CV as CV = k 2 T g (ǫF )
∞ −βǫF x2 ex dx. (ex + 1)2 (6.170) We can replace the lower limit by −∞ because the factor ex in the integrand is negligible at x = −βǫF for low temperatures. If we use the integral
∞ −∞ (ex π2 x2 ex dx = , 2 + 1) 3 (6.171) CHAPTER 6. MANYPARTICLE SYSTEMS we can write the heat capacity of an ideal Fermi gas as CV = It is straightforward to show that g (ǫF ) = and we arrive at the desired result CV = T π2 Nk 2 TF (T ≪ TF ). 3N 3N = , 2 ǫF 2kTF 12 π g (ǫF )k 2 T. 3 329 (6.172) (6.173) (6.174) A more detailed discussion of the low temperature properties of an ideal Fermi gas is given in Section 6.11.2. For convenience, we summarize the main results here: 2 21/2 V m3/2 2 5/2 π 2 µ+ (kT )2 µ1/2 , 3 π2 3 5 4 V (2m)3/2 3/2 π 2 ∂Ω µ+ = (kT )2 µ−1/2 . N =− ∂µ 3π 2 3 8 Ω=− (6.175) (6.176) The results (6.175) and (6.176) are in the grand canonical ensemble in which the chemical potential is ﬁxed. However, most experiments are done on a sample with a ﬁxed number of electrons, and hence µ must change with T to keep N ﬁxed. To ﬁnd this dependence we rewrite (6.176) as π2 3π 2 3 ρ = µ3/2 1 + (kT )2 µ−2 , 3/2 8 (2m) (6.177) where ρ = N /V . If we raise both sides of (6.177) to the 2/3 power and use (6.149), we have µ= 32/3 π 4/3 2 ρ2/3 π2 1+ (kT )2 µ−2 2m 8 −2/3 π2 . (kT )2 µ−2 = ǫF 1 + 8
−2/3 (6.178a) (6.178b) In the limit of T → 0, µ = ǫF as expected. From (6.178b) we see that the ﬁrst correction for low temperatures is given by µ(T ) = ǫF 1 − 2 π 2 (kT )2 π2 T = ǫF 1 − 2 38 µ 12 TF
2 , (6.179) where we have made the expansion (1 + x)n ≈ 1 + nx and replaced µ on the righthand side by ǫF = kTF . From (6.179) we see that the chemical potential decreases with temperature to keep N ﬁxed, but the decrease is second order in T /TF (rather than ﬁrst order), consistent with our earlier qualitative considerations. The explanation for the decrease in µ(T ) is that more particles move CHAPTER 6. MANYPARTICLE SYSTEMS 330 from energy states below the Fermi energy to energy states above the Fermi energy as the temperature increases. Because the density of states increases with energy, it is necessary to decrease the chemical potential to keep the number of particles constant. As the temperature becomes larger than the Fermi temperature, the chemical potential changes sign and becomes negative (see Problem 6.30). Problem 6.31. Low temperature behavior (a) Fill in the missing steps in (6.163)–(6.174). (b) Use (6.175) and (6.179) to show that the mean pressure for T ≪ TF is given by P= 5π 2 T 2 ρǫF 1 + 5 12 TF
2 + ... . (6.180) (c) Use the general relation between E and P V to show that E= 3 5π 2 T N ǫF 1 + 5 12 TF
2 + ... . (6.181) (d) For completeness, show that the low temperature behavior of the entropy is given by S= T π2 . Nk 2 TF (6.182) We see from (6.174) that the conduction electrons of a metal contribute a linear term to the heat capacity. In Section 6.9 we shall see that the contribution from lattice vibrations contributes a term proportional to T 3 to CV at low T . Thus, for suﬃciently low temperature, the linear term due to the conduction electrons dominates. Problem 6.32. Eﬀective electron mass From Table 6.3 we see that TF = 8.2 × 104 K for copper. Use (6.174) to ﬁnd the predicted value of C/N kT for copper. How does this value compare with the experimental value C/N kT = 8 × 10−5 ? It is remarkable that the theoretical prediction agrees so well with the experimental result based on the free electron model. Show that the small discrepancy can be removed by deﬁning an eﬀective mass m∗ of the conduction electrons equal to ≈ 1.3 me , where me is the mass of the electron. What factors might account for the eﬀective mass being greater than me ? Problem 6.33. Temperature dependence of the chemical potential in two dimensions Consider a system of electrons restricted to a surface of area A. Show that the mean number of electrons can be written as mA ∞ dǫ N= . (6.183) 2 β (ǫ−µ) + 1 π 0e The integral in (6.183) can be evaluated in closed form using 1 ebx dx = ln + constant. bx 1 + ae b 1 + aebx (6.184) CHAPTER 6. MANYPARTICLE SYSTEMS (a) Show that µ(T ) = kT ln eρπ where ρ = N/A.
2 331 /mkT −1 , (6.185) (b) What is the value of the Fermi energy ǫF = µ(T = 0)? What is the value of µ for T ≫ TF ? (c) Plot µ versus T and discuss its qualitative dependence on T . 6.9 The Heat Capacity of a Crystalline Solid The free electron model of a metal successfully explains the temperature dependence of the contribution to the heat capacity from the electrons. What about the contribution from the ions? In a crystal each ion is localized about its lattice site and oscillates due to springlike forces between nearestneighbor atoms. Classically, we can regard each atom of the solid as having six quadratic 1 contributions to the energy, three of which contribute 2 kT to the mean kinetic energy and three 1 contribute 2 kT to the mean potential energy. Hence, the heat capacity at constant volume of a homogeneous isotropic solid is given by CV = 3N k , independent of the nature of the solid. This behavior of CV agrees with experiment remarkably well at high temperatures. (The meaning of high temperature will be deﬁned later in terms of the parameters of the solid.) At low temperatures the classical behavior is an overestimate of the experimentally measured heat capacity for crystalline solids, which is found to be proportional to T 3 . To understand this low temperature behavior, we ﬁrst consider the Einstein model and then the more sophisticated Debye model of a solid. 6.9.1 The Einstein model The reason why the heat capacity decreases at low temperature is that the oscillations of the crystal must be treated quantum mechanically rather than classically. The simplest model of a solid, proposed by Einstein in 1906, is that each atom behaves like three independent harmonic oscillators each of frequency ω . Because the 3N identical oscillators are independent and are associated with distinguishable sites, we need only to ﬁnd the thermodynamic functions of one of them. The partition function for one oscillator in one dimension is [see (4.129)] Z1 = e−β ω/2 . 1 − e −β ω (6.186) Other thermodynamic properties of one oscillator are given by f = −kT ln Z1 = s=− ω + kT ln[1 − e−β 2
ω
, (6.187) (6.188) (6.189) 1 ∂f = −k ln[1 − e−β ω ] + kβ ω β ω , ∂T e −1 e = f + T s = (n + 1/2) ω , where CHAPTER 6. MANYPARTICLE SYSTEMS n= 1 eβ
ω 332 (6.190) Note the form of n, which is identical to the BoseEinstein distribution with µ = 0. We can think of n as the mean number of quanta (phonons). Because the number of phonons is not conserved, µ = 0 in the BoseEinstein distribution. To obtain extensive quantities such as F , S , and E , we multiply the single particle values by 3N . For example, the heat capacity of an Einstein solid is given by eβ ω ∂e ∂E = 3N = 3N k (β ω )2 β ω . (6.191) CV = ∂T V ∂T V [e − 1]2 It is convenient to introduce the Einstein temperature kTE = ω , and express CV as eTE /T . [eTE /T − 1]2 from (6.191) or (6.193) is CV = 3N k TE T
2 −1 . (6.192) (6.193) The limiting behavior of CV CV → 3N k and CV → 3N k (T ≫ TE ), (6.194a) ω 2 − ω/kT e (T ≪ TE ). (6.194b) kT The calculated heat capacity as T → 0 is consistent with the third law of thermodynamics and is not very diﬀerent from the heat capacity actually observed for insulating solids. However, it decreases too quickly at low temperatures and does not agree with the observed low temperature behavior CV ∝ T 3 satisﬁed by all insulating solids. Problem 6.34. Limiting behavior of the heat capacity in the Einstein model Derive the limiting behavior of CV given in (6.194). 6.9.2 Debye theory The Einstein model is based on the idea that each atom behaves like a harmonic oscillator whose motion is independent of the other atoms. A better approximation was made by Debye (1912), who observed that solids can carry sound waves. Because waves are inherently a collective phenomenon and are not associated with the oscillations of a single atom, it is better to think of a crystalline solid in terms of the collective motion rather than the independent motions of the atoms. The collective or cooperative motion corresponds to the 3N normal modes of the system, each with its own frequency. For each value of the wavevector k there are three sound waves in a solid – one longitudinal with velocity cℓ and two transverse with velocity ct . (Note that ct and cℓ are speeds of sound, not light.) The density of states of each mode is determined by the same analysis as for photons. From (6.99) we see that the density of states of the system is given by g (ω )dω = 1 V ω 2 dω 2 +3. 2π 2 c3 cℓ t (6.195) CHAPTER 6. MANYPARTICLE SYSTEMS It is convenient to deﬁne a mean speed of sound c by the relation 1 3 2 = 3 + 3, ct cℓ c3 so that the density of states can be written as g (ω ) dω = The total energy is given by E= ω n(ω )g (ω ) dω, = 3V 2 π 2 c3 ω 3 dω . eβ ω − 1 3V ω 2 dω . 2 π 2 c3 333 (6.196) (6.197) (6.198) Equation (6.198) does not take into account the higher frequency modes that do not satisfy the linear relation ω = kc. For reasons that we will discuss shortly, we will use a high frequency cutoﬀ at ω = ωD such that for the frequencies included ω ≈ kc. Because the low temperature heat capacity depends only on the low frequency modes, which we have treated correctly using (6.197), it follows that we can obtain a good approximation to the heat capacity by extending the integral in (6.197) to a maximum frequency ωD which is determined by requiring that the total number of modes be 3N . That is, we assume that g (ω ) ∝ ω 2 for ω < ωD such that
ωD 3N =
0 g (ω ) dω. (6.199) If we substitute (6.197) into (6.199), we ﬁnd that ωD = 2 π c 3ρ 4π
1/3 . (6.200) It is convenient to relate the maximum frequency ωD to a characteristic temperature, the Debye temperature TD , by the relation ωD = kTD . (6.201) The thermal energy can now be expressed as E= 3V 2 π 2 c3
kTD / 0 eβ ω
3 0 = 9N kT T TD TD /T ω 3 dω −1 (6.202a) (6.202b) x3 dx . ex − 1 In the high temperature limit, TD /T → 0, and the important contribution to the integral in (6.202) comes from small x. Because the integrand is proportional to x2 for small x, the integral is proportional to (T /TD )−3 , and hence the energy is proportional to T . Thus in the high temperature limit, the heat capacity is independent of the temperature, consistent with the law of Dulong and Petit. In the low temperature limit, TD /T → ∞, and the integral in (6.202) is independent of temperature. Hence in the limit T → 0, the energy is proportional to T 4 and the heat capacity is proportional to T 3 , consistent with experimental results at low temperatures. CHAPTER 6. MANYPARTICLE SYSTEMS Problem 6.35. More on the Einstein and Debye theories 334 (a) Determine the wavelength λD corresponding to ωD and show that this wavelength is approximately equal to a lattice spacing. This equality provides another justiﬁcation for a high frequency cutoﬀ because the atoms in a crystal cannot oscillate with a wavelength smaller than a lattice spacing. (b) Show explicitly that the energy in (6.202) is proportional to T for high temperatures and proportional to T 4 for low temperatures. (c) Plot the temperature dependence of the mean energy as given by the Einstein and Debye theories on the same graph and compare their predictions. (d) Derive an expression for the mean energy analogous to (6.202) for one and twodimensional crystals. Then ﬁnd explicit expressions for the high and low temperature dependence of the speciﬁc heat on the temperature. 6.10 The Ideal Bose Gas and Bose Condensation The historical motivation for discussing the ideal Bose gas is that this idealized system exhibits BoseEinstein condensation. The original prediction of BoseEinstein condensation by Satyendra Nath Bose and Albert Einstein in 1924 was considered by some to be a mathematical artifact or even a mistake. In the 1930s Fritz London realized that superﬂuid liquid helium could be understood in terms of BoseEinstein condensation. However, the analysis of superﬂuid liquid helium is complicated by the fact that the helium atoms in a liquid strongly interact with one another. For many years scientists tried to create a Bose condensate in a less complicated system. In 1995 several groups used laser and magnetic traps to create a BoseEinstein condensate of alkali atoms at approximately 10−6 K. In these systems the interaction between the atoms is very weak so that the ideal Bose gas is a good approximation and is no longer only a textbook example.8 Although the forms of the Landau potential for the ideal Bose gas and for the ideal Fermi gas diﬀer only superﬁcially [see (6.86)], the two systems behave very diﬀerently at low temperatures. The main reason is the diﬀerence in the ground states; that is, for a Bose system there is no limit to the number of particles in a single particle state. The ground state of an ideal Bose gas is easy to construct. We can minimize the total energy by putting all the particles into the single particle state of lowest energy: ǫ0 = π2 2 2 3π 2 2 (1 + 12 + 12 ) = . 2 2mL 2mL2 (6.203) The energy of the ground state is given by N ǫ0 . For convenience, we choose the energy scale such that the ground state energy is zero. The behavior of the system cannot depend on the choice of the zero of energy.
8 The 2001 Nobel Prize for Physics was awarded to Eric Cornell, Wolfgang Ketterle, and Carl Wieman for achieving BoseEinstein condensation in dilute gases of alkali atoms and for early fundamental studies of the properties of the condensate. CHAPTER 6. MANYPARTICLE SYSTEMS 335 The behavior of an ideal Bose gas can be understood by considering the temperature dependence of N (T, V, µ): N=
k 1 eβ (ǫk −µ) − 1 (2m)3/2 3
0 ∞ →
∞ n(ǫ)g (ǫ)dǫ
0 (6.204) (6.205) = V 4π 2 ǫ1/2 dǫ . eβ (ǫ−µ) − 1 For simplicity, we assume that the gas of bosons has zero spin, the same value of the spin as the helium isotope 4 He. To understand the nature of an ideal Bose gas at low temperatures, we assume that the mean density of the system is ﬁxed and consider the eﬀect of lowering the temperature. The correct choice of µ gives the desired value of ρ when substituted into (6.206). ρ= (2m)3/2 N = V 4π 2 3
∞ 0 ǫ1/2 dǫ . eβ (ǫ−µ) − 1 (6.206) We know that the chemical potential µ of an ideal Bose gas must be negative at all temperatures [see (6.80)]. We also know that for high temperatures, µ reduces to the semiclassical limit given by (6.29), which is large in magnitude and negative. To see how µ must change in (6.206) to keep the density ρ constant as we decrease the temperature, we make the change of variables βǫ = x and let µ → −µ: (2mkT )3/2 ∞ x1/2 dx . (6.207) ρ= (x+β µ) − 1 4π 2 3 0e As we decrease the temperature, the factor in front of the integral in (6.207) decreases and hence the integral must increase to compensate so that the density remains ﬁxed. Hence β µ must become smaller, which implies that µ must become smaller. Because µ is negative for BoseEinstein statistics, µ becomes less negative. The integral is ﬁnite for all values of β µ and has its maximum value when µ = 0. Thus, there is a minimum value of T such that the righthand side of (6.207) equals the given value of ρ. We denote this temperature by Tc and determine its value by solving (6.207) with µ = 0: (2mkTc )3/2 ∞ x1/2 dx . (6.208) ρ= 4π 2 3 ex − 1 0 Hence, we obtain The deﬁnite integral in (6.208) can be written in terms of known functions (see the Appendix) and has the value ∞ 1/2 π 1/2 x dx = 2.612 . (6.209) ex − 1 2 0 kTc = 2π 2 2/3 1 ρ. 2/3 m 2.612 (6.210) Problem 6.36. Relation of Tc to the zeropoint energy Express (6.210) in terms of the zeropoint energy associated with localizing a particle of mass m in a volume a3 , where a = ρ−1/3 is the mean interparticle spacing. CHAPTER 6. MANYPARTICLE SYSTEMS Problem 6.37. Numerical evaluation of µ for an ideal Bose gas 336 In this problem we study the behavior of µ as a function of the temperature. Program IdealBoseGasIntegral numerically evaluates the integral on the righthand side of (6.206) for particular values of β and µ. The goal is to ﬁnd the value of µ for a given value of β that yields the desired value of ρ. To put (6.206) in a convenient form we introduce dimensionless variables and let ǫ = kTc y , T = Tc T ∗ , and µ = kTc µ∗ and rewrite (6.206) as 1= or
∞ 2 √ 2.612 π ∞ 0 y 1/2 dy e(y−µ∗ )/T ∗ − 1 y 1/2 dy , , (6.211a) 1 = 0.432
0 e(y−µ∗ )/T ∗ − 1 (6.211b) where we have used (6.210). (a) Fill in the missing steps and derive (6.211). (b) The program evaluates the lefthand side of (6.211b). The idea is to ﬁnd µ∗ for a given value of T ∗ such the lefthand side of (6.211b) equals 1. Begin with T ∗ = 10. First choose µ∗ = −10 and ﬁnd the value of the integral. Do you have to increase or decrease the value of µ∗ to make the numerical value of the lefthand side of (6.211b) closer to 1? Change µ∗ by trial and error until you ﬁnd the desired result. You should ﬁnd that µ∗ ≈ −25.2. (c) Next choose T ∗ = 5 and ﬁnd the value of µ∗ so that the lefthand side of (6.211b) equals 1. Does µ∗ increase or decrease in magnitude? You can generate a plot of µ∗ versus T ∗ by clicking on the Plot button each time you ﬁnd an approximately correct value of µ. (d) Discuss the qualitative behavior of µ as a function of T for ﬁxed density. Problem 6.38. Show that the density ρc corresponding to µ = 0 for a given temperature is given by 2.612 ρc = , (6.212) λ3 where λ is given by (6.2). Is it possible for the density to exceed ρc for a given temperature? Problem 6.39. Show that the thermal de Broglie wavelength is comparable to the interparticle spacing at T = Tc . What is the implication of this result? There is no physical reason why we cannot continue lowering the temperature at ﬁxed density (or increasing the density at ﬁxed temperature). Before discussing how we can resolve this diﬃculty, consider a familiar situation in which an analogous phenomenon occurs. Suppose that we put argon atoms into a container of ﬁxed volume at a given temperature. If the temperature is high enough and the density is low enough, the argon atoms will form a gas and obey the ideal gas equation of state, which we write as P = N kT /V . If we now add atoms, we expect that the pressure will increase. However at some density, this dependence will abruptly break down, and P will stop CHAPTER 6. MANYPARTICLE SYSTEMS 337 P ρ
Figure 6.4: Sketch of the dependence of the pressure P on the density N/V at constant temperature for a typical gas and liquid. changing as indicated in Figure 6.4. We will study this behavior of P in Chapter 7, but you will probably recognize this behavior as a signature of the condensation of the vapor and the existence of a phase transition from gas to liquid. That is, at a certain density for a ﬁxed temperature, droplets of liquid argon will begin to form in the container. As the density is increased further, the liquid droplets will grow, but the pressure will remain constant because most of the extra atoms will go into the denser liquid state. We can describe the ideal Bose gas in the same terms, that is, in terms of a phase transition. At a special value of T , the chemical potential stops decreasing in magnitude and reaches its limit of µ = 0. Beyond this point, the relation (6.205) is no longer able to keep track of all the particles. Because the particles cannot appear or disappear when we change the temperature, (6.206) cannot be correct for temperatures T < Tc . The origin of the problem lies with the behavior of the threedimensional density of states g (ǫ), which is proportional to ǫ1/2 [see (6.104)]. Because of this dependence on ǫ, g (ǫ = 0) = 0, and hence our calculation of N has ignored all the particles in the ground state. For the classical and Fermi noninteracting gas, this neglect is of no consequence. In the classical case the mean number of particles in any microstate is much less than one, and in the degenerate Fermi case there are only two electrons in the ground state. However, for the noninteracting Bose gas, the mean number of particles in the ground state is given by N0 = 1 . e−βµ − 1 (6.213) (Remember that we have set the ground state energy ǫ0 = 0.) When T is suﬃciently small, N 0 will be very large. Hence, the denominator of (6.213) must be very small, which implies that e−βµ ≈ 1 and −βµ must be very small. Therefore, we can approximate e−βµ as 1 − βµ, and N 0 becomes N0 = − kT kT = ≫ 1. µ µ (6.214) The chemical potential must be such that the number of particles in the ground state approaches its maximum value, which is of order N . Hence, if we were to use the integral (6.205) to calculate N for T < Tc , we would have ignored the particles in the ground state. We have resolved the CHAPTER 6. MANYPARTICLE SYSTEMS 338 problem – the missing particles are in the ground state! The phenomenon we have described, macroscopic occupation of the ground state, is called BoseEinstein condensation. Macroscopic occupation means that, for T < Tc , the ratio N 0 /N is nonzero in the limit N → ∞. Now that we know where the missing particles are, we can calculate the thermodynamics of the ideal Bose gas. For T < Tc the chemical potential is zero in the thermodynamic limit, and the mean number of particles not in the ground state is given by (6.205): Nǫ = V 4π 2
∞ 3 (2m)3/2
0 ǫ1/2 dǫ T =N eβǫ − 1 Tc 3/2 (T < Tc ), (6.215) where Tc is deﬁned by (6.210). The remaining particles, which we denote as N 0 , are in the ground state. Another way of understanding (6.215) is that, for T < Tc , µ must be zero because the number of particles not in the ground state is determined by the temperature. We write N = N 0 + N ǫ and N0 = N − Nǫ = N 1 − T Tc
3/2 (T < Tc ). (6.216) Because the energy of the gas is determined by the particles with ǫ > 0, we have for T < Tc
∞ E=
0 V (mkT )3/2 kT ǫ g (ǫ) dǫ = βǫ − 1 e 21/2 π 2 3
∞ 0 ∞ 0 x3/2 dx . ex − 1 (6.217) The deﬁnite integral in (6.217) is given in the Appendix: 3π 1/2 x3/2 dx = 1.341 . ex − 1 4 (6.218) If we substitute (6.218) into (6.217), we can write the mean energy as E=3 1.341 V (mkT )3/2 kT m3/2 (kT )5/2 = 0.1277 V . 3 3 25/2 π 3/2 (6.219) Note that E ∝ T 5/2 for T < Tc . The heat capacity at constant volume is CV = or CV = 1.9N ǫ k, (6.220b) (mkT )3/2 k ∂E = 3.2V , 3 ∂T (6.220a) where we have used (6.215) for N ǫ . Note that the heat capacity has a form similar to an ideal classical gas for which CV ∝ N k . The pressure of the Bose gas for T < Tc can be obtained from the general relation P V = 2E/3 for a nonrelativistic ideal gas. From (6.219) we obtain P= 1.341 m3/2 (kT )5/2 23/2 π 3/2
3 = 0.085 m3/2(kT )5/2
3 . (6.221) Note that the pressure is proportional to T 5/2 and is independent of the density. This independence is a consequence of the fact that the particles in the ground state do not contribute to the pressure. CHAPTER 6. MANYPARTICLE SYSTEMS 339 If additional particles are added to the system at T < Tc , the number of particles in the single particle state ǫ = 0 increases, but the pressure does not. What is remarkable about the phase transition in an ideal Bose gas is that it occurs at all. That is, unlike all other known phase transitions, its occurrence has nothing to do with the interactions between the particles and has everything to do with the nature of the statistics. Depending on which variables are being held constant, the transition in an ideal Bose gas is either ﬁrstorder or continuous. We postpone a discussion of the nature of ﬁrstorder and continuous phase transitions until Chapter 9 where we will discuss phase transitions in more detail. It is suﬃcient to mention here that the order parameter in the ideal Bose gas can be taken to be the fraction of particles in the ground state, and this fraction goes continuously to zero as T → Tc from below at ﬁxed density. Another interesting feature of the Bose condensate is that for T < Tc , a ﬁnite fraction of the atoms are described by the same quantum wavefunction, which gives the condensate many unusual properties. In particular, Bose condensates have been used to produce atomic lasers – laserlike beams in which photons are replaced by atoms – and to study fundamental processes such as superﬂuidity. Problem 6.40. Temperature dependence of the pressure (a) Start from the classical pressure equation of state, P V = N kT , replace N by Neﬀ for an ideal Bose gas, and give a qualitative argument why P ∝ T 5/2 at low temperatures. (b) Show that the ground state contribution to the pressure is given by P0 = kT ln(N 0 + 1). V (6.222) Explain why P0 can be regarded as zero and why the pressure of an Bose gas for T < Tc is independent of the volume. Problem 6.41. Estimate of the Bose condensation temperature (a) What is the approximate value of Tc for an ideal Bose gas at a density of ρ ≈ 125 kg/m3 , the density of liquid 4 He? Take m = 6.65 × 10−27 kg. (b) The value of Tc for a collection of 87 Rb (rubidium) atoms is about 280 nK (2.8 × 10−7 K). What is the mean separation between the atoms? 6.11
6.11.1 Supplementary Notes
Fluctuations in the number of particles It is convenient to express ZG in terms of the canonical partition function ZN for N particles. The sum over all microstates in (6.70) can ﬁrst be done over all possible microstates s for a ﬁxed CHAPTER 6. MANYPARTICLE SYSTEMS number of particles and then over all values of N :
∞ 340 ZG =
N =1 eβµN
s e−βEs , (6.223) where Es is the energy of microstate s with N particles. The latter sum in (6.223) is the canonical partition function for N particles, and we have
∞ ZG =
N =1 eβµN ZN . (6.224) In the following we will derive a relation between the compressibility and the ﬂuctuations of the number of particles. The number of particles ﬂuctuates about the mean number N , which is given by 1 ∂ ln ZG N eβµN ZN . (6.225) = N = kT ∂µ ZG
N Because N ﬂuctuates, we need to reinterpret (6.72) as N = −∂ Ω/∂µ. Recall from (4.88) that the ﬂuctuations in the energy are related to the heat capacity. In the following we show that the ﬂuctuations in the number of particles are related to the isothermal compressibility κ, which is deﬁned as [see (2.172)] κ=− 1 ∂V V ∂P . (6.226) T ,N The ﬁrst step in the derivation is given in Problem 6.42. Problem 6.42. Number ﬂuctuations Use the Gibbs distribution Ps in (6.73) to show that N can be written as N= Then use (6.227) to show that ∂N ∂µ or [N 2 − N ] = kT where N2 =
s 2 s Ns e−β (Es −µNs ) e−β (Es −µNs )
s . (6.227) T ,V = 1 2 [N 2 − N ], kT ∂N ∂µ , (6.228) T ,V (6.229) Ns 2 e−β (Es −µNs ) e−β (Es −µNs )
s . (6.230) CHAPTER 6. MANYPARTICLE SYSTEMS In Problem 6.43 we relate the partial derivatives (∂µ/∂N )T,V to (∂V /∂P )T,N . Problem 6.43. Another Maxwell relation Because the Helmholtz free energy F (T, V, N ) is extensive, it may be expressed in the form F (T, V, N ) = N f (T, ρ), 341 (6.231) where f is the free energy per particle and is a function of the intensive variables T and ρ. (a) Show that µ=f +ρ ∂µ ∂ρ and P = ρ2 ∂P ∂ρ ∂f ∂ρ ∂f = 2ρ ∂ρ
T ∂f ∂ρ
T T , ∂2f ∂ρ2 , (6.232) (6.233) T =2 ∂f ∂ρ +ρ T , + ρ2 ∂2f ∂ρ2 =ρ ∂µ ∂ρ . (6.234) (6.235) T T T T Note that (6.235) is an example of a Maxwell relation (see Section 2.22). (b) Show that ∂P ∂ρ ∂µ ∂ρ (c) Use (6.235) and (6.236) to show that N Hence it follows from (6.228) that κ= 1 (N 2 − N ) . ρkT N
2 T =− T V 2 ∂P , N ∂V T ,N ∂µ =V . ∂N T ,V (6.236a) (6.236b) ∂µ ∂N T ,V = 1 . ρκ (6.237) (6.238) Equation (6.238) is another example of the relation of a response function, the compressibility κ, to the meansquare ﬂuctuations of a thermodynamic variable. From (6.229) we have that N2 − N N
2 = kT ∂ N N ∂µ T ,V . (6.239) CHAPTER 6. MANYPARTICLE SYSTEMS 342 Because µ is an intensive quantity, the righthand side of (6.239) is intensive, that is, independent of N . Hence the lefthand side of (6.239) must also be independent of N . This independence implies that the standard deviation is given by ∆N = N 2 − N relative ﬂuctuations in the number of particles is ∆N −1/2 ∝N → 0 as N → ∞. N
2 1/2 ∝N 1/2 , and therefore the (6.240) That is, in the thermodynamic limit, N → ∞, V → ∞ with ρ = N /V a constant, we can identify the thermodynamic variable N with N . As in our discussion of the canonical ensemble in Section 4.6, we see that the thermodynamic properties calculated in diﬀerent ensembles become identical in the thermodynamic limit.
∗ Problem 6.44. Number ﬂuctuations in a noninteracting classical gas (a) Show that the grand partition function of a noninteracting classical gas can be expressed as
∞ ZG =
N =0 (zZ1 )N = ezZ1 , N! (6.241) where the activity z = eβµ . (b) Show that the mean value of N is given by N = zZ1 , (6.242) and the probability that there are N particles in the system is given by a Poisson distribution: PN z N ZN (zZ1 )N N −N = = = e. ZG N !ZG N!
N (6.243) (c) What is the N dependence of the variance, (N − N )2 ? 6.11.2 Low temperature expansion of an ideal Fermi gas We derive the low temperature expansion of the thermodynamic properties of an ideal Fermi gas. For convenience, we ﬁrst give the formal expressions for the thermodynamic properties of a ideal Fermi gas at temperature T . The mean number of particles is given by N= 21/2 V m3/2 π2 3
∞ 0 eβ (ǫ−µ) ǫ1/2 dǫ , +1 (6.244) and the Landau potential Ω is given by [see (6.109)] Ω=− 2 21/2 V m3/2 3 π2 3
∞ 0 eβ (ǫ−µ) ǫ3/2 dǫ . +1 (6.245) CHAPTER 6. MANYPARTICLE SYSTEMS 343 The integrals in (6.244) and (6.245) cannot be expressed in terms of familiar functions for all T . However, in the limit T ≪ TF (as is the case for almost all metals), it is suﬃcient to approximate the integrals. To understand the approximations, we express the integrals (6.244) and (6.245) in the form ∞ f (ǫ) dǫ , (6.246) I= eβ (ǫ−µ) + 1 0 where f (ǫ) = ǫ1/2 and e3/2 , respectively. The expansion procedure is based on the fact that the FermiDirac distribution function n(ǫ) diﬀers from its T = 0 form only in a small range of width kT about µ. We let ǫ − µ = kT x and write I as I = kT = kT f (µ + kT x) dx ex + 1 −βµ f (µ + kT x) dx + kT ex + 1 −βµ
βµ 0 ∞ 0 ∞ (6.247a) f (µ + kT x) dx. ex + 1 f (µ + kT x) dx. ex + 1 (6.247b) In the ﬁrst integrand in (6.247b) we let x → −x so that I = kT
0 f (µ − kT x) dx + kT e −x + 1 ∞ 0 (6.247c) We next write 1/(e−x + 1) = 1 − 1/(ex + 1) in the ﬁrst integrand in (6.247c) and obtain
βµ βµ 0 I = kT
0 f (µ − kT x) dx − kT f (µ − kT x) dx + kT ex + 1 ∞ 0 f (µ + kT x) dx. ex + 1 (6.248) Equation (6.248) is still exact. Because we are interested in the limit T ≪ TF or βµ ≫ 1, we can replace the upper limit in the second integral by inﬁnity. Then after making the change of variables, w = µ − kT x, in the ﬁrst integrand, we ﬁnd
µ ∞ 0 I=
0 f (w) dw + kT f (µ + kT x) − f (µ − kT x) dx. ex + 1 (6.249) The values of x that contribute to the integrand in the second term in (6.249) are order one, and hence it is reasonable to expand f (µ ± kT x) in a power series in kT x and integrate term by term. The result is
µ I=
0 f (ǫ) dǫ + 2(kT )2 f ′ (µ)
0 ∞ ex 1 xdx dx + (kT )4 f ′′′ (µ) +1 3 ∞ 0 x3 dx dx + · · · . ex + 1 (6.250) The deﬁnite integrals in (6.250) can be evaluated using analytical methods (see the Appendix). The results are x dx π2 = , x+1 12 0e ∞3 7π 4 x dx = . x+1 120 0e
∞ (6.251) (6.252) CHAPTER 6. MANYPARTICLE SYSTEMS If we substitute (6.251) and (6.252) into (6.250), we obtain the desired result
µ 344 I=
0 f (ǫ) dǫ + 7π 4 π2 (kT )2 f ′ (µ) + (kT )4 f ′′′ + · · · . 6 360 (6.253) Note that although we expanded f (µ − kT x) in a power series in kT x, the expansion of I in (6.253) is not a power series expansion in (kT )2 . Instead (6.253) represents an asymptotic series that is a good approximation to I if only the ﬁrst several terms are retained. To ﬁnd Ω in the limit of low temperatures, we let f (ǫ) = ǫ3/2 in (6.253). From (6.245) and (6.253) we ﬁnd that in the limit of low temperatures 2 21/2 V m3/2 2 5/2 π 2 µ+ (kT )2 µ1/2 , 3 π2 3 5 4 ∂Ω V (2m)3/2 3/2 π 2 N =− µ+ = (kT )2 µ−1/2 . ∂µ 3π 2 3 8 Ω=− (6.254) (6.255) A more careful derivation of the low temperature behavior of an ideal Fermi gas has been given by Weinstock (1969). Vocabulary
thermal de Broglie wavelength, λ semiclassical limit equipartition theorem Maxwell velocity and speed distributions occupation numbers, spin and statistics, bosons and fermions BoseEinstein distribution, FermiDirac distribution, MaxwellBoltzmann distribution single particle density of states, g (ǫ) Fermi energy ǫF , Fermi temperature TF , and Fermi momentum pF macroscopic occupation, BoseEinstein condensation law of Dulong and Petit, Einstein and Debye theories of a crystalline solid Additional Problems
Problem 6.45. Explain in simple terms why the mean kinetic energy of a classical particle in 1 equilibrium with a heat bath at temperature T is 2 kT per quadratic contribution to the kinetic energy, independent of the mass of the particle. Problem 6.46. Heat capacity of a linear rigid rotator So far we have considered the thermal properties of an ideal monatomic gas consisting of spherically symmetrical, rigid molecules undergoing translational motion, that is, their internal motion was CHAPTER 6. MANYPARTICLE SYSTEMS 345 x z y
Figure 6.5: A schematic representation of a diatomic molecule. ignored. Real molecules are neither spherical nor rigid, and rotate about two or three axes and vibrate with many diﬀerent frequencies. For simplicity, consider a linear rigid rotator consisting of two point masses m1 and m2 located a ﬁxed distance r from each other. We ﬁrst assume that r is ﬁxed and ignore vibrational motion, which is discussed in Problem 6.47. The rotational energy levels are given by
2 , (6.256) 2I where I is the moment of inertia and j = 0, 1, 2, . . . is the angular momentum quantum number. The degeneracy of each rotational energy level is (2j + 1). (a) Express the partition function Zrot for one molecule as a sum over energy levels. (b) The sum that you found in part (a) cannot be evaluated exactly in terms of wellknown functions. However, for T ≫ Trot = 2 /(2kI ), the energy spectrum of the rotational states may be approximated by a continuum and the sum over j can be replaced by an integral. Show that the rotational heat capacity (at constant volume) of an ideal gas of linear rigid rotators is given by Crot = N k in the high temperature limit T ≫ Trot . Compare this limiting behavior with the prediction of the equipartition theorem. In this case we say that the linear rigid rotator has two quadratic contributions to the energy. Explain. (c) The magnitude of Trot for a typical diatomic molecule such as HCl is Trot ≈ 15 K. Sketch the temperature dependence of Crot , including its behavior for T ≪ Trot and T ≫ Trot . Problem 6.47. Heat capacity of an ideal diatomic gas In addition to translational and rotational motion, a diatomic molecule can exhibit vibrational motion (see Figure 6.5). It is a good approximation to take the rotational and vibrational motion to be independent and to express the total energy of an ideal diatomic gas as a sum of the translational, ǫ(j ) = j (j + 1) CHAPTER 6. MANYPARTICLE SYSTEMS 346 rotational, and vibrational contributions. Hence the total heat capacity (at constant volume) of the gas can be written as C = Ctran + Crot + Cvib . (6.257) The last two terms in (6.257) arise from the internal motion of the molecule. The rotational contribution Crot was discussed in Problem 6.46. (a) The vibrational motion of a diatomic molecule can be modeled by harmonic oscillations about the minimum of the potential energy of interaction between the two molecules. What is the high temperature limit of Cvib ? (b) Let us deﬁne a temperature Tvib = ω /k . The magnitude of Tvib for HCl is Tvib ≈ 4227 K, where ω is the vibrational frequency and ω is the energy diﬀerence between neighboring vibrational energy levels. What do you expect the value of Cvib to be at room temperature? (c) Use the value of Trot given in Problem 6.46 and the value of Tvib given in part (b) for HCl to sketch the T dependence of the total heat capacity C in the range 10 K ≤ T ≤ 10, 000 K. Problem 6.48. Law of atmospheres Consider an ideal classical gas in equilibrium at temperature T in the presence of an uniform gravitational ﬁeld. Find the probability P (z )dz that an atom is at a height between z and z + dz above the Earth’s surface. How do the density and the pressure depend on z ? Problem 6.49. Alternative derivation of the Maxwell velocity distribution The Maxwell velocity distribution can also be derived by making some plausible assumptions. We ﬁrst assume that the probability density f (v) for one particle is a function only of its speed v or equivalently v 2 . We also assume that the velocity distributions of the components vx , vy , vz are independent of each other. (a) Given these assumptions, explain why we can write
2 2 2 2 2 2 f (vx + vy + vz ) = Cf (vx )f (vy )f (vz ), (6.258) where C is a constant independent of vx , vy , and vz . (b) Show that a function that satisﬁes the condition (6.258) is the exponential function f (v 2 ) = c e−αv , where c and α are independent of v. (c) Determine c in terms of α using the normalization condition 1 = nent. Why must α be positive?
∞ −∞
2 (6.259) f (u)du for each compo 1 1 2 (d) Use the fact that 2 kT = 2 mvx to ﬁnd the Maxwell velocity distribution in (6.59). (e) *Show that f (v 2 ) in part (b) is the only function that satisﬁes the condition (6.258). Problem 6.50. Consequences of the Maxwell velocity distribution CHAPTER 6. MANYPARTICLE SYSTEMS 347 (a) What is the probability that the kinetic energy of a classical nonrelativistic particle is in the ˜ range ǫ to ǫ + dǫ? What is the most probable kinetic energy? Is it equal to 1 mv 2 , where v is 2˜ the most probable speed?
2 22 2 (b) Find the values of v x , vx , vx vy , and vx vy for a classical system of particles at temperature T . No calculations are necessary. Problem 6.51. Mean energy of a nonlinear oscillator Consider a classical onedimensional nonlinear oscillator whose energy is given by ǫ= p2 + ax4 , 2m (6.260) where x, p, and m have their usual meanings; the parameter a is a constant. (a) If the oscillator is in equilibrium with a heat bath at temperature T , calculate its mean kinetic energy, mean potential energy, and mean total energy. (It is not necessary to evaluate any integrals explicitly.) (b) Consider a classical onedimensional oscillator whose energy is given by ǫ= 1 p2 + kx2 + ax4 . 2m 2 (6.261) In this case the anharmonic contribution ax4 is very small. What is the leading contribution of this term to the mean potential energy? (Recall that for small u, eu ≈ 1 + u.) Problem 6.52. Granular systems A system of glass beads or steel balls is an example of a granular system. In such a system the beads are macroscopic objects and the collisions between the beads are inelastic. Because the collisions in such a system are inelastic, a gaslike steady state is achieved only by inputting energy, usually by shaking or vibrating the walls of the container. Suppose that the velocities of the particles are measured in a direction perpendicular to the direction of shaking. Do the assumptions we used to derive the MaxwellBoltzmann velocity distribution apply here?9 Problem 6.53. A toy system of two particles Consider a system consisting of two noninteracting particles in equilibrium with a heat bath at temperature T . Each particle can be in one of three states with energies 0, ǫ1 , and ǫ2 . Find the partition function for the cases described in parts (a)–(c) and then answer parts (d)–(f): (a) The particles obey MaxwellBoltzmann statistics and can be considered distinguishable. (b) The particles obey FermiDirac statistics.
9 See for example, the experiments by Daniel L. Blair and Arshad Kudrolli, “Velocity correlations in dense granular gases,” Phys. Rev. E 64, 050301(R) (2001), and the theoretical arguments by J. S. van Zon and F. C. MacKintosh, “Velocity distributions in dissipative granular gases,” Phys. Rev. Lett. 93, 038001 (2004). CHAPTER 6. MANYPARTICLE SYSTEMS (c) The particles obey BoseEinstein statistics. (d) Find the probability in each case that the ground state is occupied by one particle. (e) What is the probability that the ground state is occupied by two particles? (f) Estimate the probabilities in (d) and (e) for kT = ǫ2 = 2ǫ1 . 348 Problem 6.54. Consider a single particle of mass m in a onedimensional harmonic oscillator potential given by V (u) = 1 ku2 . As we found in Example 4.3, the partition function is given by 2 Z1 = e−x/2/(1 − e−x ), where x = β ω . (a) What is the partition function for two noninteracting distinguishable particles in the same potential? (b) What is the partition function for two noninteracting fermions in the same potential assuming the fermions have no spin? (c) What is the partition function for two noninteracting bosons in the same potential? Assume the bosons have spin zero. (d) Calculate the mean energy and entropy in the three cases considered in parts (a)–(c). Plot E and S as a function of T and compare the behavior of E and S in the limiting cases of T → 0 and T → ∞. Problem 6.55. Neutron stars A neutron star can be considered to be a collection of noninteracting neutrons, which are spin1/2 fermions. A typical neutron star has a mass M close to one solar mass, M⊙ ≈ 2 × 1030 kg. The mass of a neutron is about m = 1.67 × 10−27 kg. In the following we will estimate the radius R of the neutron star. (a) Find the energy of a neutron star at T = 0 as a function of R, M , and m assuming that the star can be treated as an ideal nonrelativistic Fermi gas. (b) Assume that the density of the star is uniform and show that its gravitational energy is given by EG = −3GM 2 /5R, where the gravitational constant G = 6.67 × 10−11 N m2 /kg2 . (Hint: from classical mechanics ﬁnd the gravitational potential energy between an existing sphere of radius r and a shell of volume 4πr2 dr coming from inﬁnity to radius r. Then integrate from r = 0 to R.) (c) Assume that gravitational equilibrium occurs when the total energy is minimized and ﬁnd an expression for the radius R. (d) Estimate the actual value of R in kilometers. Estimate the mass density and compare it with the density of material on the surface of the Earth such as water. ∗ CHAPTER 6. MANYPARTICLE SYSTEMS 349 (e) Determine the Fermi energy and Fermi temperature. A typical internal temperature for a neutron star is T = 108 K. Compare this value with the Fermi temperature and determine if the zero temperature approximation that we have used is applicable. (f) Compare the rest energy mc2 of a neutron with the Fermi energy of a neutron star. Is the nonrelativistic approximation valid? Problem 6.56. White dwarfs A white dwarf is a very dense star and can be considered to be a degenerate gas of electrons and an equal number of protons to make it charge neutral. We will also assume that there is one neutron per electron and that the mass of the star is about the same as our Sun. Many of the results of Problem 6.55 can be used here with slight modiﬁcations. (a) Find the mean energy at T = 0 as a function of R, M , and me , where M is the total mass of the star and me is the electron mass. (b) Assume that the star has a uniform density and show that the gravitational energy of the star is given by EG = −3GM 2 /5R. (c) Assume that gravitational equilibrium occurs when the total energy is minimized and ﬁnd an expression for the radius R. (d) Estimate the actual value of R in kilometers. Estimate the mass density and compare it with the density of water on the surface of Earth. (e) For extremely relativistic electrons the relation between the energy ǫ and momentum p of an electron is given by ǫ = cp, where c is the speed of light. Find the Fermi energy and temperature. (f) Find the mean energy at T = 0 for relativistic electrons in a white dwarf. Add this energy to the gravitational energy. Is there a minimum at a ﬁnite value of R? If not, what does this result mean about stability? (g) Compare the rest energy me c2 with the nonrelativistic Fermi energy. Is the nonrelativistic approximation valid? When the rest energy equals the nonrelativistic Fermi energy, we know that the nonrelativistic approximation is not valid. At what value of the total mass does this equality occur? Chandrasekhar obtained a limiting mass of 1.4M⊙ by taking into account the more accurate relation ǫ = (m2 c4 + c2 p2 )1/2 and the variation of density within the star. How e does your crude estimate compare?
∗ Problem 6.57. Toy systems of fermions (a) Consider a system of noninteracting (spinless) fermions such that each particle can be a single particle state with energy 0, ∆, and 2∆. Find an expression for ZG using (6.224). Determine how the mean number of particles depends on µ for T = 0, kT = ∆/2, and kT = ∆. CHAPTER 6. MANYPARTICLE SYSTEMS 350 (b) A system contains N identical noninteracting fermions with 2N distinct single particle states. Suppose that 2N/3 of these states have energy zero, 2N/3 have energy ∆, and 2N/3 have energy 2∆. Show that µ is independent of T . Calculate and sketch the T dependence of the energy and heat capacity.
∗ Problem 6.58. Periodic boundary conditions Assume periodic boundary conditions so that the wavefunction ψ satisﬁes the condition (in one dimension) ψ (x) = ψ (x + L). (6.262) The form of the oneparticle eigenfunction consistent with (6.262) is given by ψ (x) ∝ eikx x . (6.263) What are the allowed values of kx ? How do they compare with the allowed values of kx for a particle in a onedimensional box? Generalize the form (6.263) to a cube and determine the allowed values of k. Find the form of the density of states and show that the same result (6.96) is obtained. Problem 6.59. Chemical potential of a onedimensional ideal Fermi gas Calculate the chemical potential µ(T ) of a onedimensional ideal Fermi gas at low temperatures T ≪ TF . Use the result for µ(T ) found for the twodimensional case in Problem 6.33 and compare the qualitative behavior of µ(T ) in one, two, and three dimensions. Problem 6.60. High temperature limit of the ideal Fermi gas If T ≫ TF at ﬁxed density, quantum eﬀects can be neglected and the thermal properties of an ideal Fermi gas reduce to the ideal classical gas. In the following we will ﬁnd the ﬁrst correction to the classical pressure equation of state. (a) Does the pressure increase or decrease when the temperature is lowered (at constant density)? That is, what is the sign of the ﬁrst quantum correction to the classical pressure equation of state? The pressure is given by [see (6.109)] P= (2m)3/2 3π 2 3
∞ 0 ǫ3/2 dǫ . eβ (x−µ) + 1 (6.264) In the high temperature limit, eβµ ≪ 1, we can make the expansion 1 1 = eβ (µ−ǫ) eβ (ǫ−µ) + 1 1 + e−β (ǫ−µ) ≈ eβ (µ−ǫ) [1 − e−β (ǫ−µ) ]. (6.265a) (6.265b) If we use (6.265b), we obtain eβµ
0 ∞ x3/2 e−x (1 − eβµ e−x ) dx = 1 3 1/2 βµ π e 1 − 5/2 eβµ . 4 2 (6.266) Use (6.266) to show that P is given by P= m3/2 (kT )5/2 βµ 1 e 1 − 5/2 eβµ . 21/2 π 3/2 3 2 (6.267) CHAPTER 6. MANYPARTICLE SYSTEMS 351 (b) Derive an expression for N similar to (6.267). Eliminate µ and show that the leading order correction to the equation of state is given by P V = N kT 1 + ρ3 π 3/2 4 (mkT )3/2 1 = N kT 1 + 7/2 ρλ3 . 2 (6.268a) (6.268b) (c) What is the condition for the correction term in (6.268b) to be small? Note that as the temperature is lowered at constant density, the pressure increases. What do you think would be the eﬀect of Bose statistics in this context (see Problem 6.61)? Mullin and Blaylock (2003) have emphasized that it is misleading to interpret the sign of the correction term in (6.268b) in terms of an eﬀective repulsive exchange “force,” and stress that the positive sign is a consequence of the symmetrization requirement for same spin fermions. Problem 6.61. High temperature limit of ideal Bose gas If T ≫ Tc at ﬁxed density, quantum eﬀects can be neglected and the thermal properties of an ideal Bose gas reduce to those of the ideal classical gas. Does the pressure increase or decrease when the temperature is lowered (at constant density)? That is, what is the ﬁrst quantum correction to the classical equation of state? The pressure is given by [see (6.109)] P= 21/2 m3/2 (kT )5/2 3π 2 3
∞ 0 x3/2 dx . ex−βµ − 1 (6.269) Follow the same procedure as in Problem 6.60 and show that P V = N kT 1 − ρ3 π 3/2 . 2 (mkT )3/2 (6.270) We see that as the temperature is lowered at constant density, the pressure becomes less than its classical value. Problem 6.62. Bose condensation in one and two dimensions? Does Bose condensation occur for a one and twodimensional ideal Bose gas? If so, ﬁnd the transition temperature. If not, explain.
∗ Problem 6.63. Graphene Graphene is a twodimensional sheet of carbon, which was ﬁrst made in the laboratory in 2004 with the help of clear adhesive (Scotch) tape. Graphite, the material used in pencils, consists of many layers of graphene. The gentle stickiness of the tape was used to break apart the many layers of graphite. Among graphene’s many interesting properties is that its low temperature behavior can be understood by treating it as a collection of noninteracting excitations which behave as relativistic Dirac fermions and obey the dispersion relation ǫ± (k ) = ±hvk. The spin degeneracy is g = 2 and v ≈ 106 m/s. (6.271) CHAPTER 6. MANYPARTICLE SYSTEMS 352 (a) What is the chemical potential at T = 0, assuming that all negativeenergy states are occupied and all positiveenergy states are empty? (b) Show that for a system of fermions the probability of ﬁnding an occupied state of energy µ + δ equals the probability of ﬁnding an unoccupied state of energy µ − δ . (c) Use the results of parts (a) and (b) to argue that µ = 0 for all T . (d) Show that the mean energy is given by E (T ) − E (0) = 4A where A is the area of the system. (e) Calculate the temperature dependence of the heat capacity at low temperatures due to these massless Dirac particles. This problem is adapted from Kardar (2007). Problem 6.64. Discuss why Bose condensation does not occur in a system of photons corresponding to blackbody radiation. Problem 6.65. More on the Debye model (a) Show that if the volume of the crystal is N a3 , where a is the equilibrium distance between atoms, then the Debye wave number, kD = ωD /c, is about π/a. (b) Evaluate the integral in (6.202) numerically and plot the heat capacity versus T /TD over the entire temperature range.
∗ ǫ+ (k ) βǫ+ (k) + e 1 d2 k, (6.272) Problem 6.66. BoseEinstein condensation in lowdimensional traps As you found in Problem 6.62, BoseEinstein condensation does not occur in ideal one and twodimensional systems. However, this result holds only if the system is conﬁned by rigid walls. In the following, we will show that BoseEinstein condensation can occur if a system is conﬁned by a spatially varying potential. For simplicity, we will treat the system semiclassically. Let us assume that the conﬁning potential has the form V (r) ∼ rn . (6.273) In this case the region accessible to a particle with energy ǫ has a radius ℓ ∼ ǫ1/n . Show that the corresponding density of states behaves as g (ǫ) ∼ ℓd ǫ(1/2)d−1 ∼ ǫd/n ǫ(1/2)d−1 ∼ ǫα , where α= (6.274) d d + − 1. (6.275) n2 What is the range of values of n for which Tc > 0 for d = 1 and 2? More information about experiments on BoseEinstein condensation can be found in the references. CHAPTER 6. MANYPARTICLE SYSTEMS
∗ 353 Problem 6.67. Number ﬂuctuations in a degenerate ideal Fermi gas (N − N )2 = kT kT V (2m)3/2 2 2π 2 3 3NT 2TF (6.276) Use the relation (6.228) ∂N ∂µ to ﬁnd the number ﬂuctuations in the ideal Fermi gas for ﬁxed T, V , and µ. Show that
∞ 0 (N − N )2 = → eβ (ǫ−µ) ǫ−1/2 dǫ +1 (6.277a) (6.277b) (T ≪ TF ). Explain why the ﬂuctuations in a degenerate Fermi system are much smaller than in the corresponding classical system. Suggestions for further reading
More information about BoseEinstein condensation can be found at <jilawww.colorado.edu/bec/>, <bec.nist.gov/>, and <www.rle.mit.edu/cua_pub/ketterle_group/>. Vanderlei Bagnato and Daniel Kleppner, “BoseEinstein condensation in lowdimensional traps,” Phys. Rev. A 44, 7439–7441 (1991). Ralph Baierlein, “The elusive chemical potential,” Am. J. Phys. 69, 428–434 (2001). Roger Balian and JeanPaul Blaizot, “Stars and statistical physics: A teaching experience,” Am. J. Phys. 67, 1189–1206 (1999). The article, which is part of a theme issue on thermal and statistical mechanics, contains an excellent introduction to the use of statistical physics to understand stars. John J. Brehm and William J. Mullin, Introduction to the Structure of Matter, John Wiley & Sons (1989) or Robert Eisberg and Robert Resnick, Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles, second edition, John Wiley & Sons (1985). See these books, for example, for more complete discussions of blackbody radiation. Ian Duck and E. C. G. Sudarshan, Pauli and the SpinStatistics Theorem, World Scientiﬁc (1998). This graduate level text simpliﬁes and clariﬁes the formal statements of the spinstatistics theorem, and corrects the ﬂawed intuitive explanations that are frequently given. David L. Goodstein, States of Matter, Prentice Hall (1975). An excellent text whose emphasis is on the applications of statistical mechanics to gases, liquids and solids. Chapter 3 on solids is particularly relevant to this chapter. J. D. Gunton and M. J. Buckingham, “Condensation of the ideal Bose gas as a cooperative transition,” Phys. Rev. 166, 152–158 (1968). F. Herrmann and P. W¨ rfel, “Light with nonzero chemical potential,” Am. J. Phys. 73, 717–721 u (2005). The authors discuss thermodynamic states and processes involving light in which the chemical potential of light is nonzero. CHAPTER 6. MANYPARTICLE SYSTEMS 354 Mehran Kardar, Statistical Physics of Particles, Cambridge University Press (2007). An excellent graduate level text. Charles Kittel, Introduction to Solid State Physics, seventh edition, John Wiley & Sons (1996). See Chapters 5 and 6 for a discussion of the Debye model and the free electron gas. W. J. Mullin and G. Blaylock, “Quantum statistics: Is there an eﬀective fermion repulsion or boson attraction?,” Am. J. Phys. 71, 1223–1231 (2003). A. Pais, “Einstein and the quantum theory,” Rev. Mod. Phys. 51, 863–914 (1979). This review article has a fascinating account of how Planck arrived at his famous blackbody radiation formula. Donald Rogers, Einstein’s Other Theory: The PlanckBoseEinstein Theory of Heat Capacity, Princeton University Press (2005). Robert H. Swendsen, “Statistical mechanics of colloids and Boltzmann’s deﬁnition of the entropy,” Am. J. Phys. 74, 187–190 (2006). The author argues that the usual deﬁnition of the entropy for a classical system of particles in terms of the logarithm of a volume in phase space is not fundamental and not the deﬁnition orginally given by Boltzmann. Swendsen’s treatment resolves the Gibbs paradox in a novel way. Robert Weinstock, “Heat capacity of an ideal free electron gas: A rigorous derivation,” Am. J. Phys. 37, 1273–1279 (1969). ...
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This note was uploaded on 01/23/2011 for the course PHYS 123 taught by Professor Smith during the Spring '07 term at UC Davis.
 Spring '07
 SMITH
 mechanics, Statistical Mechanics

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