Chapter 14. Ideal Bose gas Equation of state

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1 Chapter 14 Ideal Bose gas In this chapter, we shall study the thermodynamic properties of a gas of non-interacting bosons. We will show that the symmetrization of the wavefunction due to the indistinguishability of particles has important consequences on the behavior of the system. The most important consequence of the quantum mechanical symmetrization is the Bose- Einstein condensation, which is in this sense a special phase transition as it occurs in a system of non-interacting particles. We shall consider, as an example, a gas of photons and a gas of phonons Equation of state We consider a gas of non-interacting bosons in a volume at temperature T and chemical potential µ. The system is allowed to interchange particles and energy with the surroundings. The appropriate ensemble to treat this many-body system is the grand canonical ensemble. Our bosons are non-relativistic particles with spin s, whose one-particle energies ε( k) include only the kinetic energy term and are given by ε( k) = ε(k) = h2 k 2 2m. Since the lowest one-particle energy ε 0 is zero, ε 0 = ε(0) = 0, the chemical potential must fulfill the condition: as we discussed in chapter 12. < µ < 0, We consider a situation when the gas is in a box with volume = L x L y L z and subject to periodic boundary conditions. In the thermodynamical limit (N,, with 177

2 178 CHAPTER 14. IDEAL BOSE GAS n = N = const), the sums over the wavevector k can be replaced by integrals as in the case of the Fermi gas. However, here we have to be careful when µ happens to approach the value 0. In order to see what kind of trouble we then might get into, let us calculate the ground state occupation. The expectation value of the occupation number is given by < ˆn r >= f + (ε r ) = 1 e β(εr µ) 1. For βµ << 1, the occupation of the lowest energy state ε(0) = 0 can be approximated as 1 < ˆn 0ms >= e βµ 1 = 1 1 βµ βµ, which means that < ˆn 0ms > can assume any macroscopic value. On the other hand, the density of states Ω(E), which has the same expression as for fermionic systems (eq. 13.8), Ω(E) E 0 as E 0. This is where we are going to encounter a problem: if we replace r by de Ω(E), we will get that the ground state has zero weight even though, as we have just shown, it can be macroscopically occupied. In the case of a fermionic system, we would not have this problem due to the Pauli principle: < ˆn r >: 0 < ˆn r > 1. In order to avoid the problem in the bosonic system, we treat the ground state separately. The grand canonical potential Ω(T,,z) (do not confuse it with the DOS Ω(E)) can be written then as βω(t,,z) = (2s+1) (2π) 34π + (2s+1)ln(1 z) }{{} term corresponding to the ground state ε(0) = 0 With the substitution β x = hk 2m and the definition of the thermal de Broglie wavelength 2πβ h 2 λ = m, expression (14.1) becomes βω(t,,z) = 2s+1 4 λ 3 π We use the Taylor series expansion of the ln, 0 ln(1 y) = 0 dk k 2 ln ( 1 ze βε(k)). (14.1) ( dxx 2 ln 1 ze x2) +(2s+1)ln(1 z). (14.2) n=1 y n n ( y < 1),

3 14.1. EQUATION OF STATE 179 for the integral: We define the functions and 0 ( dxx 2 ln 1 ze x2) π = 4 g 5/2 (z) = 4 π g 3/2 (z) = z d dz g 5/2(z) = 0 n=1 z n n 5/2. ( dxx 2 ln 1 ze x2) = n=1 z n n 3/2. n=1 z n n 5/2 With these definitions, the grand canonical potential (eq. (14.2) takes the form βω(t,,z) = 2s+1 λ 3 g 5/2 (z)+(2s+1)ln(1 z). (14.3) Note that except for the extra term on the right-hand side, it has the same form as the expression for the Fermi gas. From Ω = P, we get βp = 2s+1 g λ 3 5/2 (z) 2s+1 ln(1 z). (14.4) We express the fugacity z in terms of the density of particles n: n = < ˆN > ( ) = z z βp T, = 2s+1 g λ 3 3/2 (z)+ 2s+1 z 1 z (14.5) and then, by combining eq. (14.4) and (14.5), we obtain the thermal equation of state. The term n 0 = 2s+1 z 1 z in eq. (14.5) describes the contribution of the ground state to the particle density n as can be seen from with 1 < ˆn 0m s >= 1 1 z 1 1 = 1 m s = s, s+1,...,+s. z 1 z n 0 2s+1, n 0 can be macroscopically large, which corresponds to the Bose-Einstein condensation.

4 180 CHAPTER 14. IDEAL BOSE GAS The caloric equation of state can be obtained by calculating the internal energy U: ( ) ( ) U = β lnz(t,z,) = β βω(t,,z) By considering eq. (14.3), one obtains z, U = 3 2 k BT 2s+1 λ 3 g 5/2 (z). (14.6) This expression is the same as the one for the Fermi gas when f 5/2 (z) is substituted by g 5/2 (z). Combining eq. (14.6) and eq. (14.5), one can derive the caloric equation of state. Now, we combine (14.4) and eq. (14.6) to get z, U = 3 2 P k BT(2s+1)ln(1 z). Note that the expression for the Bose gas has an extra term 3 2 k BT(2s+1)ln(1 z), which was absent in the case of the Fermi gas as well as in the case of the ideal gas of particles Classical limit We first investigate the classical limit (non-degenerate Bose gas) corresponding to low particle densities and high temperatures. In this limit, z = e βµ << 1 so that 1 < ˆn r >= z 1 e βεr 1 << 1 ze βεr and the Bose-Einstein distribution function reduces to the Maxwell-Boltzmann one. All energy levels are only weakly occupied. The probability of double occupancy is practically zero. It is therefore to be expected that in this limit the differences between Bose-, Fermiand Boltzmann statistics are negligible. As z << 1, it is sufficient to retain only the first two terms of the series for g 5/2 (z) and g 3/2 (z): g 5/2 (z) z + z2 2 5/2, g 3/2 (z) z + z2 2 3/2.

5 14.2. CLASSICAL LIMIT 181 With that, the particle density takes the following form: n 2s+1 ( z 1+ z ) + 2s+1 λ 3 2 3/2 2s+1 λ 3 z ( 1+ z 2 3/2 ) In the zeroth approximation, we thus have that and therefore z 1 z + 2s+1 z(1 z). ( ) nλ 3 (2s+1)z (0) 1+ λ3 z (0) nλ 3 (2s+1) ( 1+ λ3 As in the case of the Fermi gas, here the classical limit corresponds to nλ 3 << 1, i.e., small particle densities and small thermal de Broglie wavelengths. The condition λ 3 << 1 allows us to neglect the correcting term coming from the explicit consideration of the ground state contribution so that we can write z (0) nλ3 2s+1, which is the same result as we had for fermions. The equation of state reduces then to the equation of state for classical particles: ). βp = 2s+1 nλ 3 λ 3 2s+1, P = < ˆN > k B T. In order to obtain the next order corrections, we make use of the equation n 2s+1 ( z 1+ z ) λ 3 2 3/2 where we neglect the term corresponding to the ground state contribution, which can be rewritten in terms of z (0) as ) nλ 3 2s+1 z(0) z (1+ (1) z(1) 2 3/2 and then as z (1) z (0) ( 1+ z(1) 2 3/2 ). (14.7)

6 182 CHAPTER 14. IDEAL BOSE GAS In the right-hand side of eq. (14.7), we approximate z (1) by z (0) : ) z (1) z(0) z (1 (0) z(0). 1+ z(0) 2 3/2 2 3/2 We substitute this expression in the pressure equation for the ideal Bose gas, βp 2s+1 λ 3 g 5/2 (z) 2s+1 λ 3 z (1), where the ground state contribution has been neglected. This gives us P =< ˆN > k B T [ 1 ] nλ (2s+1) The last term in this expression are the quantum corrections. Note that if we consider now, for comparison, the equation of state for a real gas (the van der Waals equation, for instance), we would observe that similar additive corrections appear with respect to the ideal case due to the interaction between particles. Here, however, the additive terms originate from the indistinguishability principle and not from the interaction among particles. The correcting term for the ideal Fermi gas was positive and therefore contributed as a repulsion among particles. For the Bose gas, the additive term is negative and therefore contributes as an attraction among particles. Quantitatively, the quantum corrections are much smaller than terms coming from the interaction among particles Bose-Einstein condensation We consider now the limit of high particle densities and low temperatures(quantum limit), where one finds important qualitative differences between bosons, fermions and classical particles. For bosons, we have the restriction < µ 0, which determines the range of value that z assumes: 0 < z < 1. The functions g α (z) = n=1 z n n α

7 14.3. BOSE-EINSTEIN CONDENSATION 183 are related to one another through the recursive formula: g α 1 (z) = z d dz g α(z). In the interval 0 z 1, they are positive and monotonically growing functions of z (see Fig. 14.1). Figure 14.1: g α family. For z = 1, g α coincide with the Riemann function ζ, defined as 1 ζ(n) = p = 1 dy yn 1 n (1 2 1 n )Γ(n) e y +1. Thus, p=1 ( ) 5 g 5/2 (1) = ζ = 1.342, 2 ( ) 3 g 3/2 (1) = ζ = and g 1/2 (z) diverges at z = 1. From the expression for the particle density for bosons n, n = 2s+1 g λ 3 3/2 (z)+ 2s+1 z z 1, we have that the particle density of the (2s+1)-degenerate ground state ε(k = 0) = 0, defined as n 0 = 2s+1 z z 1, can be written in the following form: n 0 = n 2s+1 λ 3 g 3/2 (z). Since, as can be observed in Fig. 14.1, g 3/2 (z) for 0 < z 1 varies in the interval [0, 2.612], there should be temperatures T and densities n where n > 2s+1 λ 3 g 3/2 (1). 0

8 184 CHAPTER 14. IDEAL BOSE GAS In this case, n 0 > 0, that is, a finite fraction of bosons occupies the ground state. When βµ 0, n 0 becomes macroscopically large. This phenomenon is called Bose-Einstein condensation. This macroscopic state would not be that spectacular if it occured only for temperatures k B T < ε 1 ε(k = 0) = ε 1, (14.8) where ε 1 is the first excited state. For macroscopic systems, this condition is fulfilled for temperatures T < K so that at T = 0 all particles are in the ground state. What is so spectacular about the Bose-Einstein condensation is that it sets in at much higher temperatures than those imposed by eq. (14.8). In fact, the phase transition to the condensed state occurs when nλ 3 = (2s+1)g 3/2 (1), whichdefinesthecritical valueforthedebrogliewavelengthλ c foragivenparticledensity n: λ 3 c = 2s+1 ( 2π h 2 ) 3/2 n g 3/2(1) = mk B T c and hence the critical temperature T c at which the phase transition occurs: k B T c = 2π h2 m ( ) 2/3 n. (2s+1)g 3/2 (1) For a given temperature T, we can find the critical density n c : n c (T) = 2s+1 ( ) 3/2 mkb T g λ 3 3/2 (1) = (2s+1) 2π h 2 g 3/2(1). Since the particle density is proportional to the inverse the average interparticle distance r to the third power, n 1 r 3, we can argue that the Bose-Einstein condensation occurs when the thermal de Broglie wavelength λ approaches r : λ r. Comparison of the critical temperature T c of the ideal Bose gas with the Fermi temperature T F of the ideal Fermi gas Consider s = 1/2 fermions with mass m f and s = 0 bosons with mass m b, both system with equal densities n f = n b = n. The T F /T c is then given as T F T c = 1 4π (3π2 g 3/2 (1)) 2/3m b m f 1.45 m b m f.

9 14.3. BOSE-EINSTEIN CONDENSATION 185 For m b m f 1, both temperatures are of the same order of magnitude. Let us consider now a gas of conduction electrons and a gas of 4 He atoms as representative fermionic and bosonic systems. In this case, m b m f , T f 10 4 K and therefore T c 3.13 K. This value of T c is much bigger that what one would expect for a normal occupation of the ground state, T K. In this example, we demonstrate that the Bose-Einstein condensation is a phase transition. Fugacity In order to understand the phenomenon, let us represent n, n = 2s+1 g λ 3 3/2 (z)+ 2s+1 z 1 z, graphically. For that, we plot the function y 1 (z) and the constant y 1 (z) = g 3/2 (z)+ λ3 z (1 z) y 2 = nλ3 2s+1 on the same graph. The intersection between these two curves provides z as a function of T and n. With this graphical representation, we can obtain z(t,n) at a given.

10 186 CHAPTER 14. IDEAL BOSE GAS The solution z b corresponds to the condensate region since there ( ) nλ 3 > g 3/2 (1). 2s+1 In the thermodynamic limit (N,, n = const) the term coming from the ground state contribution λ 3 z (1 z) has to remain finite. This implies that (1 z) has to remain finite, i.e., 1 z has to behave as 1. Therefore, for, we can approximate the fugacity as nλ 3 solution of z = 2s+1 = g nλ 3 3/2(z) for 2s+1 < g 3/2(1), nλ 3 1 for 2s+1 g 3/2(1). Then, since n 0 n 0 = n 2s+1 λ 3 g 3/2 (z), n 0 if nλ 3 2s+1 < g 3/2(1), ( ) 3/2 n 0 2s+1 1 n nλ g 3/2(1) = 1 λ3 c T 3 λ = 1 nλ 3 if 3 2s+1 g 3/2(1). For 0 < T < T c, we have a mixture of two phases (see Fig. 14.2): the condensate phase formed by the macroscopic fraction of N 0 bosons out of N occupying the lowest energy level: [ ( ) ] 3/2 T N 0 = N 1 and the gas phase formed by the rest of the particles N 1 which are in excited ( k 0) states: ( ) 3/2 T N 1 = N N 0 = N. T c T c T c

11 14.4. THERMODYNAMICS FOR THE IDEAL BOSE GAS 187 Figure 14.2: Occupation density of the ground state of the ideal Bose gas as a function of temperature. At T = 0, N 0 = N, N 1 = 0. The Bose-Einstein condensation takes place at T = T c Thermodynamics for the ideal Bose gas With the Bose-Einstein condensation and the concept of a mixed state of gas and condensate, we have a strong analogy with the gas-liquid phase transition that we discussed when studying properties of a real gas. As in that case, the abrupt change of n 0 at T c leads to discontinuities in the thermodynamic potentials of the ideal Bose gas, which have different analytical expressions above and below the critical point (T c, n c ). Thermal equation of state We consider the thermodymanic limit (N,, n = const). In the equation for pressure for the Bose gas given as the second term approaches zero: βp = 2s+1 g λ 3 5/2 (z) 2s+1 ln(1 z), 2s+1 ln(1 z) 0 for z < 1, i.e., n < n c. For the condensation region (n n c ), where z 1, the above limit is not trivial. Here we make use of the fact that as discussed previously, and also get 2s+1 (1 z) 1, ln(1 z) 0.

12 188 CHAPTER 14. IDEAL BOSE GAS Summarizing, we get the following equation for pressure: 2s+1 g λ 3 5/2 (z) for n < n c, βp = 2s+1 g λ 3 5/2 (1) for n > n c. In the condensation region, the pressure is independent of the volume and particle number; it only depends on temperature. This was also the case in the gas-liquid phase transition. Considering this analogy, the Bose-Einstein condensation can be described as a first-order phase transition. The Bose-Einstein condensation was predicted by Satyendra Bose and Albert Einstein in It took almost 70 years to have an experimental corroboration of this phenomenon with the ultracold gas systems. Previous experiments had been done with 4 He as well as with hydrogen.

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