NOW that we’ve developed the basic formalism of statistical mechanics (,!Chapter 4), we can proceed to applications. Measurable quantities can be calculated once the partition function is known, which requires a specification of 1) the Hamiltonian1 and 2) the defining macrovariables associated with the type of ensemble, TVN, etc. In this chapter, we consider ideal systems composed of noninteracting constituents. Statistical mechanics naturally incorporates interactions through the potential energy part of the Hamiltonian, although as we’ll see in Chapter 6, evaluating the partition function for systems of interacting particles is a challenging problem.
The Hamiltonian of a gas of N noninteracting particles is ∑i=1Npi2/(2m). The partition function for this system (volume V, temperature T) is found from Eqs. (4.47) and (4.53),
Zcan(N,V,T)=1N!VλT3N≡1N!Z(N,V,T),
(5.1)
where λT is the thermal wavelength, Eq. (1.65), which results from integrating over the momentum variables. With Zcan one can calculate the equation of state and the entropy using Eq. (4.58) (Exercise 5.1). The phase-space probability density is, from Eq. (4.54),
where ρi is a one-particle distribution function. Because the Hamiltonian is separable, the N-particle distribution occurs as the product of N, single-particle distributions, i.e., the particles are independently distributed.2 Note that ρi is normalized on a one-particle phase space:
Another way to calculate the entropy is through the distribution function, Eq. (4.60). One can show that Eq. (4.60) yields the Sackur-Tetrode formula when combined with Eq. (5.2) (see Exercise 5.3).
We can express ρ1 (the index denotes a single-particle distribution) as a probability density of the speeds of the particle. Start from the normalization integral, Eq. (5.3):
is the Maxwell speed distribution3,4,5 for v≥0, the probability f(v)dv of finding a particle with speed between v and v+dv. We cited the Maxwell distribution as an example of a probability density function in Eq. (3.27). Equation (5.5) shows there is a distribution in molecular speeds in a gas in equilibrium at temperature T—a big conceptual discovery by Maxwell.6 A gas in thermal equilibrium—seemingly a quiescent system—actually consists of a collection of molecules having a range of speeds: a few slow ones, a few fast ones, with most having speeds near the mean.7 The shape of the distribution is shown in Fig. 5.1. The speed distribution confirms our physical expectation that as the temperature is lowered, progressively more of the molecules have slower speeds. We’ll see that such an expectation can fail when quantum mechanics is brought into account.
Figure 5.1Maxwell speed distribution f(v) for T1<T2<T3. The area under the curve is the same for each temperature.
We can calculate the mean speed using the rules of probability,
v¯=∫0∞vf(v)dv=8kTmπ.
(5.6)
There are other ways, however, to characterize the speed of atoms in a gas. What’s the “rms” (root-mean-square) speed? By definition,
vrms=v2¯=∫0∞v2f(v)dv1/2=3kTm.
(5.7)
Note that we don’t actually need to do the integral in Eq. (5.7); it follows from the equipartition theorem (Section 4.1.2.11), 〈mv2/2〉=(3/2)kT —why the factor of 3? We can also ask for the most probable speed vmp at which the distribution has a maximum. This is readily found to have the value
Figure 5.2Characteristic molecular speeds of a gas in thermal equilibrium: vmp, v¯, vrms.
Example. Nitrogen is the largest component of air (approximately 78%, with oxygen comprising 21%).8 What is the mean speed v¯ of a nitrogen molecule at room temperature, T=293 K, given that nitrogen occurs as a diatomic molecule N2 at this temperature? To apply Eq. (5.6), we need the molecular mass. The mass number of a nitrogen atom is approximately 14 grams/mole (consult a periodic table of the elements, 14.007 when isotopic variances are taken into account). The mass of the molecule is therefore 28 grams/mole; Avogadro’s number of N2 molecules has a mass of 28 grams. The mass of one molecule is therefore m=28g/(6.02×1023)=4.65×10−23g=4.65×10−26 kg. Using Eq. (5.6), we find v¯=471 m/s. That’s fast! The speed of a bullet fired from a gun is ≈500 m/s. Keep in mind that Eq. (5.5) is a distribution of speeds, not velocities. In equilibrium, the molecules of a gas have their velocities directed at random, implying the net velocity is zero. Table 5.1 lists the mean speed v¯ for various gases at room temperature.
Table 5.1 Mean speed v¯ of selected gases at T=293 K
Some of the most successful applications of statistical mechanics involve the magnetic properties of materials. Under the general banner of magnetism there are different types of magnetic phenomena: ferromagnetism, antiferromagnetism, paramagnetism, diamagnetism, and others. In the limited space of this book we can only offer a cursory treatment of the subject. Ferro- and antiferromagnetism are cooperative effects produced by interactions among the magnetic dipoles of the atoms in a solid. Paramagnetism is the “ideal gas” of magnetism, in which magnetic moments interact only with an applied magnetic field and not with each other.
For a collection of magnetic moments {μi} that interact only with the external field, we need treat only the statistical mechanics of a single magnetic moment. The partition function for N identical, noninteracting particles ZN=(Z1)N, where Z1 is the single-particle partition function. The energy of interaction between a magnetic dipole moment μ and a magnetic field9B is E=−μ·B.
Should we adopt a classical or a quantum treatment of this problem? It turns out that a quantum treatment leads to excellent agreement with experimental results. Thus, we consider the energy of interaction between μ and B as the Hamiltonian operator,
H^=−B·μ^=gμBℏB·J^=gBμBℏJ^z,
(5.9)
where we’ve used Eq. (E.4), μ=−gμBJ/ℏ, where μB≡eℏ/(2m) is the Bohr magneton, g is the Land e´ g-factor (see Appendix E), and the operator J^z is the z-component of the total angular momentum (the B-field defines the z-direction). To use Eqs. (4.123) or (4.125) (quantum statistical mechanics in the canonical ensemble), we require the eigenfunctions and eigenvalues of the Hamiltonian operator, which in this case is proportional to J^z (Eq. (5.9)). As is well known, J^2 and J^z have a common set of eigenfunctions |J,m〉 (a complete orthonormal set), such that
J^2|J,m〉=J(J+1)ℏ2|J,m〉J^z|J,m〉=mℏ|J,m〉,
where the quantum number J has the values J=0,1,2,⋯ or J=12,32,52,⋯, and m=−J,−J+1,⋯,J−1,J so that there are (2J+1) values of m. The energy eigenvalues are therefore Em=gμBmB. From Eq. (4.123),10
Z1=∑m=−JJe−βmμBgB=sinhy(J+12)sinhy/2,
(5.10)
where y≡βμBgB. The summation in Eq. (5.10) is simple because it’s a finite geometric series.
To calculate the average value of11μz, we use Eq. (4.125), (where |m〉≡|J,m〉)
By evaluating the derivative indicated in Eq. (5.11), we find, after some algebra,
〈μz〉=μBgJBJ(βμBgBJ),
(5.12)
where BJ(x) is the Brillouin function of order J, defined as
BJ(x)≡2J+12Jcoth2J+12Jx−12Jcoth12Jx.
(5.13)
Graphs of these functions are shown in Fig. 5.3. They demonstrate the characteristic feature of saturation, that limx→∞BJ(x)=1 for all J. At a fixed temperature, for increasing values of the magnetic field, μ becomes (on average) progressively more aligned with the direction of the field. For strong enough fields, the moments are ostensibly all aligned with the field; increasing the field further can only keep the moments at their maximum alignment with the field—saturation.
Figure 5.3Brillouin functions BJ(x) for J=1/2,3/2,5/2, and J=∞.
Figure 5.4 shows measured values of 〈μz〉 as a function of B/T at several values of T for three paramagnetic salts which contain an ion for which g=2, but for which the values of J are different, J=32,52,72. The data are presented as 〈μz〉/μB, which, from Eq. (5.12), because g=2, saturate at the values of 3, 5, 7, precisely what is found experimentally. Furthermore, the data fall almost perfectly on the Brillouin functions, validating the predictions of Eq. (5.12).
Figure 5.4Plot of 〈μz〉/μB versus B/T for three paramagnetic ions. The solid lines are the predictions of Eq. (5.12). Curve I is for potassium chromium alum ( J=32,g=2), curve II is for iron ammonium alum ( J=52,g=2), and curve III is for gadolinium sulfate octahydrate ( J=72,g=2). Reprinted figure with permission from W.E. Henry, Phys. Rev. 88, p.559 (1952).[48] Copyright (2020) American Physical Society.
Another measured quantity is the magnetic susceptibility,
χ≡(∂M∂H)|H=0,
(5.14)
where M is the magnetization, M=N〈μz〉. To calculate the susceptibility, we could differentiate the Brillouin function BJ(x) and let x→0, but that’s an unnecessary step. Using the result of Exercise 5.9, combined with Eq. (5.12), and setting B=μ0H, we have for small H,
M~H→0N3kT(μBg)2J(J+1)μ0H≡CHT,
(5.15)
where
C=N3kμ0μBg2J(J+1).
(5.16)
Equation (5.15), Curie’s law, is the equation of state of a paramagnet—the magnetization is linear with H for small M. The constant C is the Curie constant, the value of which is material specific.12 From Eq. (5.15), as H→0, M→0, the hallmark of paramagnetism—the system acquires a magnetization in an applied magnetic field, which vanishes in zero field.13 For small fields, the larger the susceptibility, the larger is the magnetization obtained for the same field strength.
Paramagnetism can be treated classically if we consider the magnetic moment a vector (not an operator): μ=μe^, where e^ is a unit vector that can point in any direction. The energy E=−μBcosθ, where θ is the angle between B and μ. We use Eq. (4.15) for the partition function, Z=∫e−βEΩ(E)dE. The density of states function Ω(E) is found by differentiating the formula E(θ)=−μBcosθ, implying Ω(θ)=sinθ. (We leave off the factor of μB to keep the density of states dimensionless, the number of states in the range [θ,θ+dθ]). Thus,
Z=∫0πeβμBcosθsinθdθ=2βμBsinh(βμB).
(5.17)
To obtain the average value of μz, we use Eq. (5.11),
〈μz〉=∂∂(βB)lnZ=μL(x),
(5.18)
where x≡βμB and L(x) is the Langevin function,
L(x)≡cothx−1x.
(5.19)
From Exercise 5.9, L(x) is the limiting form of BJ(x) as J→∞.
No physics book can be complete without treating the harmonic oscillator, as it’s among the few exactly solved problems in physics. We’re fortunate this problem can be solved exactly, because it occurs widely in physics. For a potential energy function V(r) that has a minimum at r=r0, its Taylor series about r0 is V(r)=V(r0)+(r−r0)dV/dr|r0+12(r−r0)2d2V/d2r|r0+…. If the system is in equilibrium at r=r0 (no force acting), dV/dr|r0=0, and assuming d2V/dr2|r0>0 (stable equilibrium), then small excursions about r=r0 map onto the harmonic oscillator with mω2≡d2V/dr2|r0. In what follows, assume we have a harmonic oscillator of mass m and angular frequency ω in equilibrium with a heat bath at temperature T. We consider the problem from the quantum and the classical perspectives.
5.3.1 Quantum treatment
Harmonic oscillators have quantized energy levels14En=(n+12)ℏω, n=0,1,2,⋯. The energy associated with n=0, 12ℏω, is the zero-point energy, the lowest possible energy that a quantum system may have (which, we note, is not zero).15 The canonical partition function for a single oscillator is, from Eq. (4.123),16
Z1(β)=∑n=0∞e−β(n+12)ℏω=12sinh(βℏω/2).
(5.20)
The partition function specifies the number of states a system has available to it at temperature T. As β→0 (high temperature), we have from Eq. (5.20),
Z1(β)~β→01βℏω,
(5.21)
that all of the infinite number of energy states of the harmonic oscillator become thermally accessible, that Z diverges as we (formally) allow T→∞. Compare with the β→0 limit of the partition function for a paramagnetic ion, Eq. (5.17), Z(β→0)=2. In that case there are only two states available to the system: aligned or antialigned with the direction of the magnetic field. Consider the other limit of Eq. (5.20),
Z1(β)~β→∞e−βℏω/2.
(5.22)
For temperatures such that kT≲ℏω/2, Z1≪1; the number of states available to the system is exponentially smaller than unity. As T→0 there are no states available to the system: Z→0.
Applying Eq. (5.20) to Eq. (4.40), we have the average energy of the oscillator,
〈E〉=ℏω2coth12βℏω=ℏω1eβℏω−1+12≡ℏω〈n〉+12.
(5.23)
Let’s look at the limiting forms of Eq. (5.23):
〈E〉=ℏω2(T→0)〈E〉=kT.(T→∞)
(5.24)
At low temperature, the system occupies its ground state, with energy E=ℏω/2. At sufficiently high temperatures, the system behaves classically, with energy given by the equipartition theorem. We’ve written Eq. (5.23) in the form 〈E〉=(〈n〉+12)ℏω, where 〈n〉 denotes the occupation number specifying the effective average state that the system is “in” (or occupies) in thermal equilibrium,
〈n〉=1eβℏω−1.
(5.25)
The occupation number 〈n〉 is not an integer. The system (oscillator) is continually exchanging energy with its environment, causing it to momentarily occupy the allowed states of the system labeled by integer n. The occupation number is the average value in thermal equilibrium of the quantum number n. We can either say that the system has energy 〈E〉=(〈n〉+12)ℏω, or, equivalently (because the energy of an ideal system is the sum of the energies of its components), that the system consists of 〈n〉quantara17 each of energy ℏω.
We note that Eq. (5.25) is the Bose-Einstein distribution function for bosons with μ=0 (see Eq. (5.61)), which is also the Planck distribution for the number of thermally excited photons of energy E=ℏω (see Section 5.8.2). Is there a connection between bosons and harmonic oscillators? There is indeed, as we now show.18
For a system of N independent oscillators, the partition function ZN(β)=Z1(β)N. Thus, using Eq. (5.20),
ZN(β)=2sinh(βℏω/2)−N=e−Nβℏω/21−e−βℏωN.
(5.26)
Equation (5.26) is a closed-form expression for the partition function of N independent harmonic oscillators, from which one could calculate the heat capacity—see Eq. (5.41). We can, however, express Eq. (5.26) in another way. Apply the binomial theorem Eq. (3.10) to Eq. (5.26):
ZN(β)=e−Nβℏω/2∑k=0∞(N+k−1k)e−kβℏω.
(5.27)
Writing ZN(β) in the form of Eq. (4.15) (the Laplace transform of the density of states), ZN(β)=∫0∞ΩN(E)e−βEdE, we infer from Eq. (5.27) that the density of states for N independent harmonic oscillators has the form
ΩN(E)=∑k=0∞(k+N−1k)δ(E−(k+N/2)ℏω),
(5.28)
where δ(x) is the Dirac delta function. What combinatorial problem does the binomial coefficient N+k−1k pertain to? Consider, starting from N oscillators each in its ground state (so E=Nℏω/2 is the zero-point energy of the system), how many ways can k quanta, each of energy ℏω, be added to N oscillators so that the energy of the system is E=(k+N/2)ℏω ? Quanta are indistinguishable; we can’t say which quantum of energy is added to an oscillator, all we can say is that k quanta have been added to the system. The combinatorial problem is therefore how many distinct ways can k indistinguishable quanta be added to Ndistinguishable oscillators? Figure 5.5 shows 17 “dots,” where the dots represent energy quanta, distributed among 7 oscillators, where the oscillators have been conceptually arranged in a line, separated by vertical lines. N oscillators are delineated by (N−1) vertical lines. There would be (N+k−1)! permutations of the (N+k−1) symbols in Fig. 5.5, but that would overcount the number of distinct configurations of the system. We should divide (N+k−1)! by k! for permutations of the k dots, and by (N−1)! for permutations of the oscillators. Thus, the number of distinct permutations of k dots among the N−1 lines is k+N−1k.
Figure 5.5One of the k+N−1k ways of distributing k indistinguishable energy quanta (shown here as 17 identical dots) among N oscillators (shown here as 7 boxes, delineated by 6 vertical lines). In this example, a box in the middle has no quanta.
Example. There are 32=3 ways of adding k=2 quanta to N=2 oscillators; see Fig. 5.6.
Figure 5.6The three distinct configurations of two quanta added to two oscillators.
5.3.2 Classical treatment
The Hamiltonian function for the harmonic oscillator is H=p2/(2m)+12mω2x2, implying
Z1(β)=1h∫−∞∞dx∫−∞∞dpe−β[p2/(2m)+12mω2x2]=1βℏω.
(5.29)
Planck’s constant enters the evaluation of Z in classical statistical mechanics (see Section 2.3). The partition function for the quantum harmonic oscillator in the high-temperature limit is the same as that for the classical oscillator.
In Section 5.1, we treated the ideal gas of structureless molecules, the most salient feature of which is the translational kinetic energy of its point particles. Translational motion is present in any gas. The constituents of real gases have internal motions that we have yet to take into account. We consider the ideal diatomic gas,19 i.e., we ignore inter-particle interactions and we assume the conditions for classical behavior apply, nλT3≪1 (see Section 1.11). It should be clear that internal motions must be treated using quantum mechanics—classical statistical mechanics brings with it the equipartition theorem, which we know is insufficient to explain the heat capacity of real gases.
Energies of translational degrees of freedom and those of internal motions can be written in terms of individual Hamiltonians,
H=Htrans+Hrot+Hvib+Helec+⋯,
(5.30)
where Htrans is the Hamiltonian associated with translational motion, Hrot is that for rotations, Hvib for vibrations, Helec for electronic degrees of freedom, and so on. Writing H in this form assumes the degrees of freedom underlying each Hamiltonian are noninteracting, that the various modes of excitation occur independently—an approximation that’s not always true. When H can be written in separable form, the partition function occurs as the product of the partition functions associated with each part of the Hamiltonian:20
Z(T,V)=1N!Ztrans(T,V)·Zrot(T)·Zvib(T)·….
(5.31)
In that case, the heat capacity occurs as the sum of the heat capacities for each of the modes of excitation (because CV is related to lnZ; see Exercise 4.14):
CV(T)=Ctrans+Crot(T)+Cvib(T)+⋯.
(5.32)
For an ideal gas, Ctrans=32Nk is independent of temperature.21 In this section, we calculate the heat capacities for the rotational and vibrational degrees of freedom of the ideal diatomic gas.
5.4.1 Rotatonal motion
The rigid rotor problem treats the two atoms of a diatomic molecule as having a fixed separation distance r0. The allowed rotational energies depend on the moment of inertia I=μr02, where μ is the reduced mass of the two atomic masses, μ=m1m2/(m1+m2). The rotational state is determined by the angular momentum operator, L^. L^2 and L^z have a common set of eigenfunctions,
L^2|l,m〉=l(l+1)ℏ2|l,m〉L^z|l,m〉=mℏ|l,m〉,
where l=0,1,2,⋯ and m=−l,−l+1,⋯,l−1,l so that there are 2l+1 values of m. The Hamiltonian for rotational motion about the center of mass is H^rot=L2/(2I), and thus the rotational energy eigenvalues are El=ℏ2l(l+1)/(2I). Because El is independent of the quantum number m, each state is (2l+1)-fold degenerate. The partition function is, using Eq. (4.123),22
Z1,rot(T)=∑l=0∞(2l+1)e−βEl.
(5.33)
The sum in Eq. (5.33) cannot be evaluated in closed analytic form, and we must introduce approximations. We examine the high and low-temperature limits.
5.4.1.1High-temperature form
As β→0 there are contributions to Eq. (5.33) from large values of the quantum number l, which suggests we approximate the sum in Eq. (5.33) with an integral, using the form of Z in Eq. (4.15). That route requires the density-of-states function, Ω(E), the derivative with respect to energy of the total number of energy states up to and including E. Energy at a specified value E implies a maximum value of l determined by E=ℏ2lmax(lmax+1)/(2I)≈ℏ2lmax2/(2I) because lmax≫1. How many states are there for 0≤l≤lmax ? It can be shown that
∑l=0lmax(2l+1)=lmax+12≈lmax2≈2Iℏ2E.
(5.34)
The density of states is therefore Ω(E)=2I/ℏ2. Thus, we can approximate Eq. (5.33),
Z1,rot(T)=2Iℏ2∫0∞e−βEdE=2Iβℏ2≡TΘr,(T≫Θr)
(5.35)
where Θr=ℏ2/(2Ik) sets a characteristic temperature for rotational motions.23 Using equations that we’ve now used several times (Eqs. (4.40) and (P4.1)), with Z=(Z1)N,
〈E〉rot=NkTCVrot=Nk,(T→∞)
(5.36)
the same as what we obtain from the equipartition theorem.
A more accurate high-temperature form can be obtained using the result of Exercise P5.2:
Z1,rot(T)=TΘr+13+115ΘrT+4315ΘrT2+⋯.(T≫Θr)
(5.37)
From Eq. (5.37) we obtain an expression for the heat capacity more general than Eq. (5.36) (see Exercise 5.12),
CV(T)rot=Nk1+145ΘrT2+16945ΘrT3+⋯.(T≫Θr)
(5.38)
We see that (CV(T))rot exceeds the classical value Nk, a value that it tends to as T→∞.
5.4.1.2Low-temperature form
In the low-temperature regime, T≪Θr, we have, from Eq. (5.33),
Z(T)1,rot=1+3e−2Θr/T+5e−6Θr/T+⋯.
(5.39)
In this case, the variable e−Θr/T is exponentially small as T→0. From Eq. (5.39), we find to lowest order
CV(T)rot≈12NkΘrT2e−2Θr/T.(T≪Θr)
(5.40)
As T→0, CV(T)rot drops to zero exponentially fast; rotational degrees of freedom can’t be excited at sufficiently low temperature—they become “frozen out.”
The two equations, (5.38) and (5.40), are limiting forms of CV(T)rot in the high- and low-temperature regimes. They each show that the heat capacity is temperature dependent. To obtain the complete temperature dependence of CV(T)rot requires the use of a computer to evaluate the sum in Eq. (5.33) at each temperature. A detailed analysis shows there is a maximum value of CV(T)rot≈1.1Nk at T≈0.81Θr. Given that Θr≈10 K, measurements of CV on diatomic gases at room temperature are consistent with the prediction of the equipartition theorem.
5.4.2 Vibrational motion
The partition function for the quantum harmonic oscillator is given in Eq. (5.20), from which may be derived an expression for the heat capacity,
CV(T)vib=NkΘvT2eΘv/TeΘv/T−12,
(5.41)
where Θv≡ℏω/k is a characteristic temperature associated with vibrational energies. For HCl, Θv≈4300 K, and for H2, Θv≈6300 K. Thus, only for temperatures on the order of 104 K would we expect vibrational modes to be sufficiently excited that they contribute to CV at the level required by the equipartition theorem. From Eq. (5.41) we have the high and low-temperature limits:
CVvib=Nk(T≫Θv)CVvib~NkΘvT2e−Θv/T.(T≪Θv)
(5.42)
Given that Θv~103 K, vibrational modes are frozen out at room temperature and don’t appreciably contribute to CV.
We now take into account the indistinguishability of identical particles required by quantum mechanics. We start with the Hamiltonian for a system of N noninteracting identical particles, H^=∑n=1Nh^n, where h^n is a function of the coordinates and momenta associated with an isolated atom or molecule, which can include the translational motion of the center of mass or the degrees of freedom associated with rotation, vibration, intra-atomic electronic structure, and the intrinsic spin, S. The energy of an ideal system is the sum of the energies of its constituents, E=∑n=1NE(n), where E(n) is the energy of the nth particle, which is any one of the eigenvalues belonging to h^. (We don’t have to label h^ with an index—it’s the same function for all particles.) The quantity E(n) is a function of the quantum numbers associated with the aforementioned degrees of freedom, such as the wavevector k introduced in Section 2.1.5, rotational or vibrational quantum numbers, or the z-component of the spin. We denote the collection of relevant quantum numbers associated with the nth particle as mn. Thus, E(n)=E(n)(mn).
5.5.1 Partition function
We might assume that the canonical partition function can be written in the form
ZN=∑m1⋯∑mNexp−β∑n=1NE(n).(wrong!)
(5.43)
Equation (5.43) is incorrect because it overcounts the allowed states of identical particles. The occupation numbernk is the number of particles in the system having the eigenstate associated with eigenvalue Ek (see Appendix D). The energy of the system can therefore be written not as a sum over particles, as in E=∑n=1NE(n), but as a sum over energy levels,
E=∑mnmEm.
(5.44)
Equation (5.44) is an important step in setting up the statistical mechanics of identical particles. The occupation numbers must satisfy the constraint
∑mnm=N.
(5.45)
Equations (5.44) and (5.45) are unrestricted sums over all possible energy states.24
How many ways can the energy E be partitioned over the particles of the system? For a given set of occupation numbers {nk} satisfying Eq. (5.45), there are, using Eq. (3.12), N!/∏k(nk!) ways of permuting the particles, which, by the indistinguishability of identical particles, are equivalent and have to be treated as a single state. To correct for overcounting, Eq. (5.43) should be written
ZN=1N!∑m1⋯∑mN∏k(nk!)exp−β∑n=1NE(n).
(5.46)
Equation (5.46) presents a challenging combinatorial problem because it connects the energy levels of individual particles, E(n), to the occupation numbers nk, which apply to the entire system. To apply Eq. (5.46), we must know the occupation numbers associated with a system in thermal equilibrium, which is what we’re trying to solve for! At high temperature (β→0) the number of energy levels that can make a significant contribution to the sum in Eq. (5.46) becomes quite large. One would expect, for fixed values of N and E, that in this limit the occupation numbers will be predominately 0 or 1 and thus (nk)!=1 for most configurations. If we set (nk)!=1 in Eq. (5.46), we have an approximate expression for the partition function (which factorizes)
where in the last step we’ve used that all particles are identical, where Z1 is the partition function for a single particle. Equation (5.47) is the high-temperature limit of the partition function for bosons or fermions.
It was noted in Section 1.11 that for combinations of temperature and density such that nλT3≳1, the boson or fermion character of the particles of the system must be taken into account. We’ll see in Section 5.5.3 how this criterion emerges from a calculation of the equation of state.25 We consider an ideal gas of point particles in the absence of an external magnetic field, so that the relevant quantum numbers26 are the wavevector k associated with the particle’s kinetic energy (see Eq. (2.14)) and its spin quantum number σ. The energy eigenvalues in this case27 are
Ek,σ=Ek=ℏ22mk2,
(5.48)
where the eigenvalues are g-fold degenerate, with g=2S+1. It’s shown in Appendix D how, in creating wavefunctions displaying the proper symmetries under permutations of particles, that information about particle identity is lost. We can’t say which particle has a given energy, because of the indistinguishability of identical particles. The most we can say is how many particles have a given energy (the occupation number), and not which particles have that energy. We can therefore write the N-particle partition function, not as in Eq. (5.46) (a sum over particles), but in the form
ZN=∑{nk,σ}∑k,σnk,σ=Ne−β∑k,σnk,σEk,σ,
(5.49)
a sum over energy levels, where ∑{nk,σ} denotes a summation over all quantum numbers (k,σ), ∑{nk,σ}≡∏k,σ∑k,σ. Equation (5.49) indicates, for each allowed (k,σ), to sum over the occupation numbers nk,σ, subject to the constraint ∑k,σnk,σ=N.
Equation (5.49) is in general impossible to evaluate, because of the constraint of a fixed number of particles, Eq. (5.45). One has a formidable combinatorial problem of summing over all sets of occupation numbers consistent with the constraint on the number of particles. Without the constraint, Eq. (5.49) would simply factorize. Surprisingly, this problem simplifies in the grand canonical ensemble where, by allowing N to vary, the constraint is eliminated. Referring to Eq. (4.127),
where now the sums over occupation numbers are unrestricted. The transition from Eq. (5.50) to Eq. (5.51) can be shown by the method of staring at it long enough. Consider a system that has only two energy levels, E1 and E2. In that case, we would have from Eq. (5.50)
ZG=∑N=0∞∑n1(n1+n2=N)∑n2eβ(μ−E1)n1+β(μ−E2)n2
(5.52)
Let a≡eβ(μ−E1) and b≡eβ(μ−E2), so that Eq. (5.52) can be expressed
ZG=∑N=0∞∑n1=0N∑n2=0N−n1an1bn2.
Write out the first few terms in the series: For N=0, a0b0=1; for N=1, a0b1+a1b0=a+b; for N=2, a2+ab+b2; and for N=3, a3+a2b+ab2+b3. Add these terms (up to and including N=3):
ZG=(1+b+b2+b3)+a(1+b+b2)+a2(1+b)+a3(1).
We see the pattern: As N→∞, we have the unrestricted sums,
ZG=∑n1=0∞∑n2=0∞an1bn2=∑n1=0∞an1∑n2=0∞bn2.
Generalize to a set of k energy levels E1,E2,⋯,Ek, with ai≡eβ(μ−Ei), i=1,⋯,k. Then, from Eq. (5.50),
the same as Eq. (5.51) when we let k→∞. With Eq. (5.51) established, we can quickly complete the calculation.
The occupation numbers of fermions are restricted to the values n=0,1 for any state (the Pauli principle). In that case, we have from Eq. (5.51)
ZG(β,μ)=∏k,σ1+eβ(μ−Ek,σ).(F)
(5.53)
For bosons there are no restrictions on the occupation numbers. The sum in Eq. (5.51) can be evaluated explicitly as a geometric series, with the result
ZG(β,μ)=∏k,σ11−eβ(μ−Ek,σ).(B)
(5.54)
For the infinite series to converge requires that μ≤0 for bosons. There is no restriction on μ for fermions. These formulas can be combined into a common expression. Let θ=+1 for bosons and θ=−1 for fermions (the same factor of θ introduced in Eq. (4.110)). Equations (5.53) and (5.54) can then be written as a common expression
ZG(β,μ)=∏k,σ1−θeβ(μ−Ek,σ)−θ.(θ=±1)
(5.55)
The partition function for the ideal quantum gas thus factorizes, but the factors don’t pertain to individual particles—they refer to individual energy levels.
5.5.2 The Fermi-Dirac and Bose-Einstein distributions
The average number of particles a system has (associated with given values of μ,T,V —we’re using the grand canonical ensemble) can be found from the derivative (see Eq. (4.78))
〈N〉=kT∂lnZG∂μT,V.
Using Eq. (5.55) for ZG,
∂lnZG∂μ|T,V=gβ∑k1eβ(Ek−μ)−θ,
(5.56)
where g=2S+1 is the spin degeneracy (see Eq. (5.48)). Thus,
〈N〉=g∑k1eβ(Ek−μ)−θ.
(5.57)
The sum in Eq. (5.57) can be converted to an integral over k-space, ∑k⟶(V/(8π3))∫d3k (see Section 2.1.5),
where, because for free particles Ek is isotropic in k-space (see Eq. (5.48)), we work in spherical coordinates. Change variables28 in Eq. (5.58); let E=ℏ2k2/(2m). Then,
where we’ve used the free-particle density of states, g(E), Eq. (2.18), and where we’ve introduced the average occupation numbers,
〈n〉θ=1eβ(E−μ)−θ.(θ=±1)
(5.60)
The functions implied by Eq. (5.60) for θ=±1 are called the Bose-Einstein distribution function ( θ=1),
〈n〉BE=1eβ(E−μ)−1,
(5.61)
and the Fermi-Dirac distribution function ( θ=−1),
〈n〉FD=1eβ(E−μ)+1.
(5.62)
For fermions, 0<〈n〉FD≤1, which (as we’ll discuss) reflects the requirements of the Pauli principle. There is no restriction on 〈n〉BE (other than 〈n〉BE>0) because μ≤0 for bosons. These functions (Eqs. (5.61) and (5.62)) are fundamental to any discussion of identical bosons or fermions. We’ve derived them from the partition function in the grand canonical ensemble, the appropriate ensemble if the chemical potential is involved. There’s another method, however, for deriving these functions that’s often presented in textbooks, the method of the most probable distribution, shown in Appendix F.
The structure of Eq. (5.59) is a generic result in statistical mechanics:
〈N〉=∫0∞g(E)n(E)dE.
(5.63)
The average number of particles is obtained from a sum over energy levels, with the density of states g(E) telling us the number of allowed states per energy range, multiplied by n(E), the average number of particles occupying those states in thermal equilibrium. One is from quantum mechanics—the allowed states of the system, the other is from statistical physics, the number of particles actually occupying those states in equilibrium at temperature T and chemical potential μ.
5.5.3 Equation of state, fugacity expansions
From Eq. (4.76), Φ=−kTlnZG. For a system with V as the only relevant external parameter, Φ=−PV, Eq. (4.69). For such a system, PV=kTlnZG. Using Eq. (5.55) for ZG, we have the equation of state:
PV=−θgkT∑kln1−θeβ(μ−Ek).
(5.64)
The sum in Eq. (5.64) can be converted to an integral, ∑k⟶(V/(8π3))∫d3k (see Section 2.1.5). Thus,
PV=−θgkTV8π3∫d3kln1−θeβ(μ−Ek).
(5.65)
Because Ek is isotropic in k-space, work with spherical coordinates:
PV=−gθkTV8π3∫0∞4πk2dkln1−θeβ[μ−ℏ2k2/(2m)].
Change variables: Let x≡βℏ2k2/(2m), a dimensionless variable. Then,
P=−gθkTλT32π∫0∞dxxln1−θeβμ−x.
Integrate by parts:
P(T,μ)=gkTλT31Γ(5/2)∫0∞x3/2ex−βμ−θdx.
(5.66)
Note that Eq. (5.66) provides an expression for P that’s intensive in character; P=P(T,μ) is independent of V. Pressure P, which has the dimension of energy density, is equal to kT, an energy, divided by λT3, a volume, multiplied by a dimensionless function of eβμ (the integral in Eq. (5.66)).
The two equations implied by Eq. (5.66) (one for each of θ=±1) involve a class of integrals known as Bose-Einstein integrals for θ=1, Gn(z), defined in Eq. (B.18), and Fermi-Dirac integrals for θ=−1, Fn(z), defined in Eq. (B.29), where z=eβμ. Written in terms of these functions, we have from Eq. (5.66),
P(T,μ)=gkTλT3{G5/2(z)θ=+1F5/2(z)θ=−1.
(5.67)
Because for bosons 0<z≤1, an expansion in powers of z can be developed for Gn(z), a fugacity expansion (see Eq. (B.21)). For fermions there is no restriction on z (only that z>0). A fugacity expansion can be developed for Fn(z) for 0<z≤1 (see Eq. (B.32)); for z≫1, one has to rely on asymptotic expansions (see Eq. (B.40)), or numerical integration.
Using Eqs. (B.21) and (B.32) for the small-z forms of Gn(z),Fn(z), we have from Eq. (5.67),
P=gkTλT3z1+θz25/2+z235/2+θz345/2+⋯.(θ=±1)
(5.68)
For the classical ideal gas, the chemical potential is such that z=eβμ=nλT3 (see Eq. (P1.3)). In the classical regime, nλT3≪1, implying classical behavior is associated with z≪1. Ignoring the terms in square brackets in Eq. (5.68), we recover the equation of state for the classical ideal gas, P=nkT (the degeneracy factor g “goes away,” for reasons we explain shortly). Fugacity is a proxy for pressure (see page 100), and we see from Eq. (5.68) that z∝P in the classical limit.
From Eqs. (5.66) or (5.68) we have the pressure in terms of μ and T. The chemical potential, however, is not easily measured, and P is more conveniently expressed in terms of the density n≡N/V. It can be shown using Eqs. (5.59), (B.18), and (B.29), that
N(T,V,μ)=gVλT3{G3/2(z)θ=+1F3/2(z)θ=−1.
(5.69)
From Eq. (5.69) and Eqs. (B.21) and (B.32), we find a fugacity expansion for the density:
λT3gn=z1+θz23/2+z233/2+θz343/2+⋯.
(5.70)
We can invert the series in Eq. (5.70) to obtain z in terms of a density expansion.29 Working consistently to terms of third order, we find
z=λT3ng1−θ23/2λT3ng+14−133/2λT3ng2+⋯.
(5.71)
When nλT3≪1, the terms in square brackets in Eq. (5.71) can be ignored, leaving us (in this limit) with z=nλT3/g, a generalization of what we found from thermodynamics, Eq. (P1.3), to include spin degeneracy. Thus, from Eq. (5.68), P=nkT in the classical limit.
Substitute Eq. (5.71) into the fugacity expansion for the pressure, Eq. (5.68). We find, working to third order,
P=nkT1+a1θλT3ng+a2λT3ng2+⋯
(5.72)
where
a1=−123/2=−0.1768a2=−235/2−18=−0.0033.
A density expansion of P (as in Eq. (5.72)) is called a virial expansion. For classical gases, the terms in a virial expansion result from interactions between particles (see Section 6.2). The terms in square brackets in Eq. (5.72) are a result of the quantum nature of identical particles and not from “real” interactions. That is, nominally noninteracting quantum particles act in such a way as to be effectively interacting. The requirements of permutation symmetry imply a type of interaction between particles—because of Eq. (4.110), correlations exist among nominally noninteracting particles, a state of matter not encountered in classical physics. We see from Eq. (5.72) that the pressure of a dilute gas of fermions (bosons) is greater (lesser) than the pressure of the classical ideal gas. The Pauli principle effectively introduces a repulsive force between fermions: Configurations of particles in identical states occupying the same spatial position do not occur in the theory, as if there was a repulsive force between particles. For bosons, it’s as if there’s an attractive force between particles.
5.5.4 Thermodynamics
We can now consider the other thermodynamic properties of ideal quantum gases. From Eq. (4.82),
where we’ve used Eq. (5.55). Using the same steps as in previous sections, we find
U(T,V,μ)=32gVλT3kT{G5/2(z)θ=+1F5/2(z)θ=−1.
(5.74)
Equation (5.74) is quite similar to Eq. (5.67), which we may use to conclude that
U=32PV⇒P=23UV.
(5.75)
Equation (5.75) holds for all ideal gases, which is noteworthy because the equations of states of the classical and two quantum ideal gases are different.30
To calculate the heat capacity, it’s useful31 to divide Eq. (5.74) by Eq. (5.69):
where primes indicate a derivative with respect to z, and where (shown in Exercise 5.14)
∂z∂T|n=−32zTA3/2(z)A1/2(z).
(5.78)
Combining Eqs. (5.78) and (5.77), and making use of Eqs. (B.20) and (B.31), we find
CVNk=154A5/2(z)A3/2(z)−94A3/2(z)A1/2(z).
(5.79)
Equation (5.79) applies for fermions and bosons. It’s easy to show, using either Eq. (B.21) or (B.32) that as z→0, CV→Nk(154−94)=32Nk, the classical value. We’ll examine the low-temperature properties in upcoming sections.
An efficient way to calculate the entropy is to use the Euler relation, Eq. (1.53),
S=1TU+PV−Nμ=1T53U−Nμ=Nk52A5/2(z)A3/2(z)−lnz,
(5.80)
where we’ve used Eqs. (5.75) and (5.76) and the definition of z=eβμ. (See Exercise 5.15.) The form of Eq. (5.80) for small z is, using Eqs. (B.21) and (B.32) and Eq. (5.71) at lowest order,
S=Nk52−lnλT3ng,(z≪1)
a generalization of the Sackur-Tetrode formula to include spin degeneracy (see Eq. (1.64)).
Equation (5.75) allows an easy way to calculate the heat capacity in the regime nλT3≪1, without the heavy machinery of Eq. (5.79). From
CV=∂U∂TV=32∂(PV)∂TV=32Nk1−12a1θnλT3g−2a2nλT3g2
(5.81)
where we’ve used Eq. (5.72). Putting in numerical values,
CV=32Nk1+0.0884θnλT3g+0.0066nλT3g2⋯.
(5.82)
As T→∞, CV approaches its classical value, 32Nk. For bosons ( θ=+1) the value of CV for large but finite temperatures, is larger than the classical value, implying that the slope of CV versus T is negative, which a calculation of ∂CV/∂T from Eq. (5.82) shows. We know that heat capacities must vanish as T→0, a piece of physics not contained in Eq. (5.82), a high-temperature result.
As Eqs. (5.72) and (5.82) show, there are minor differences between the two types of quantum gases in the regime nλT3≪1, with the upshot that the two cases can be treated in tandem (as we’ve done in Sections 5.5.3 and 5.5.4). We now consider the opposite regime (nλT3)/g≳1 in which quantum effects are more clearly exhibited, in which Fermi and Bose gases behave differently, and which must be treated separately.
