The state of a quantum ideal gas is uniquely specified by the occupancy of singleparticle

Transcription

1 Ideal Bose gas The state of a quantum ideal gas is uniquely specified by the occupancy of singleparticle states, the set {n α } where α denotes the quantum numbers of a singleparticles state such as k and m the z-component of the spin and n α the number of particles in the state. The reason this is possible is, of course, that the particles are indistinguishable and the interchange of particles does not yield a new state. The total number of particles and the total energy of a given quantum state are given by = α α and E = α α ɛ α. (.) For bosons α =,, j,. Z = e βµ Z = e βµ α α = e β α ɛα α = {α} e β α (ɛα µ) α. (.2) In the last equality we just have an unrestricted sum over α ; summing with the restriction that the sum is and then allowing to vary over all possible values is the same as performing an unconstrained sum over all α. Please convince yourself of this. ote that you have one sum for each α and therefore we have Z = e β (ɛα µ) α. (.3) α α The sum for each α can be done easily since it is a geometric series. For bosons we have e β (µ ɛα ) α =. (.4) eβ(µ ɛα) Thus we obtain This leads to α= Z B = α pv = k B T log Z = α. (.5) eβ(µ ɛα) log ( e β(µ ɛα)). (.6) We can write log Z as a sum convert it to an integral and analyze the behavior of thermodynamic functions. The mathematics is non-trivial 2 and we have to develop One can define the number operator ˆ α = a α a α where the creation and annihilation operators obey commutation or ant-commutation relations. 2 Two useful references are Kerson Huang, Statistical Mechanics R. K. Pathria (and Paul Beale) Statistical Mechanics.

2 intuition for the bosonic case. The sum on α is the sum on the spin states and over the momenta or wave vectors. We will assume that ɛ does not depend on the spin state and include the dependence in a homework problem. Thus we have g V λ k d 3 k (2π) 3 = g V d 3 p (2π ) 3 (.7) This is the fundamental formula you should start from, mutatis mutandis for other dimensions. Given the dispersion relation and the dimension for isotropic cases we can convert this into a one-dimensional integral over energy: dɛ D(ɛ). (.8) We have for given ɛ(k) (the energy depends only on k d d k D(ɛ) = g V δ(ɛ ɛ(k)) = g V Ω d (2π) d (2π) d α k d dɛ(k) dk. (.9) k = k(ɛ) ow it is a matter of elementary substitution to obtain D(ɛ) for ɛ = 2 k 2 in three 2m dimensions. Often one quotes the density of states per unit volume and multiplies by V explicitly. We will do that. D(ɛ) = g 4π k 2 = g m 2mɛ. (.) 8π 3 2 k 2π 2 2 m k = 2mɛ/ 2 Recall that λ T h 2πmkB T. So we can write in three dimensions V D(ɛ) dɛ = 2gV π (2πm) 3/2 h 3 ɛ /2 dɛ = 2g π V Let x = β ɛ and e βµ z find p = k B T 2g π ɛ k B T dɛ k B T. (.) x /2 log ( z e x ). (.2) 2 3/2 It is convenient to rewrite is using integration by parts: x =. The integrand 3 vanishes at both limits (It is exponentially small at infinity.) So we obtain(setting g = to avoid confusion with the other g ν introduced below) pv k B T = 4 3 V ( ) x 3/2 V π e x z g 5/2 (z). (.3) 2

3 We have defined g ν (z) x ν e x z. (.4) The mean number is obtained by summing over the average number of particles in each state. The average number in each state was obtained last quarter (you must remember the result) α = e β(ɛα µ). (.5) Choosing the ground state energy ɛ = we see that µ <. We have = α e β(ɛα µ) = 2g V π x /2 z e x = V g 3/2 (z). (.6) The energy can be obtained from U = α ɛ α e β(ɛα µ) ± = k BT 2 V π x 3/2 z e x = 3 2 k BT V g 5/2 (z). (.7) This yields pv = 2 U. One need not do the algebra completely to obtain this. 3 The mathematical discussion relies on understanding the behavior of g ν (z) as z. We have g ν (z) x ν e x z = Expanding the denominator in a power series since e x z < x ν e x z e x z. (.8) g ν (z) = z l x ν e lx. (.9) l= We let l x y and obtain g ν (z) = l= z l l ν dy y ν e y (.2) The integral yields and therefore, g ν (z) the function diverges. l= z l. Clearly, for l as z lν 3

4 How does one understand Bose condensation? Start from the expression for the number density n. Continue to let g = to avoid confusion with g ν ; you can restore it when necessary. V = (mk BT ) 3/2 2 π2 3 x /2 z e x = g 3/2 (z). (.2) For fixed T as n increases (or at fixed n as T decreases) the denominator in the integrand has to decrease or z = e βµ has to increase. However, z = e βµ cannot exceed unity since the chemical potential cannot be positive. More mathematically, for fixed T the right-hand side has a maximum value, given by, ζ(3/2). λ 3 t Consider a container with volume at fixed temperature T to which we add particles thus increasing /V. The integral must increase and this occurs as µ increases. At fixed temperature there appears to be a maximum density theoretically. Physically, there is no reason why we should not be able to add particles. To understand this consider an an analogous phenomenon of a classical gas (say 2 ) at somewhat low temperatures, more precisely, below its critical point. For simplicity let us assume that the gas is ideal (not correct but that is not the point) and plot the pressure as a function of the number density both of which can be measured. We get a straight line according to the classical ideal gas law, p = n k B T, (there may be small deviations as the gas is not strictly ideal) but then we will find that the pressure stops changing. What happens when we add more molecules? They condense into droplets of liquid nitrogen. As more molecules are added the number and/or size of the droplets grow but the pressure remains the same. There is coexistence of the gas and liquid phase. We can interpret this as either saying that there is a critical density n c = for a fixed temperature or conversely if we imagine reducing the temperature at a given density there is a critical temperature k B T c = α 2 2m n2/3 where α = 4π given density. Whence the difficulty? Consider α = chosen to be ɛ = we have As µ we find z e βɛα ζ(3/2) 2/3 for a. For the ground state = z z > z. (.22) = eβµ e βµ k BT µ. (.23) 4

5 The number of particles in the lowest energy state becomes macroscopic when µ k BT. This phenomenon, of a macroscopic occupation of a single (one-particle) quantum state, is a consequence of quantum statistics, and known as Bose-Einstein conden- sation. The density of states is proportional to ɛ in three dimensions and hence our integral approximation yields no states at ɛ =. It fails to account for the macroscopic occupation of or condensation into the ground state. The integral only yields the density of particles in the excited states and the ground state occupancy is obtained by subtracting it from the total density. We have n V = n + g 3/2 (z) n (T ) + n ex (T ) (.24) where we have denoted by n V. For a given density the second term, the density of particles in the excited state n ex yields the full contribution down to T c because z varies accordingly and n is negligible in the thermodynamic limit. Below T c we see that the excited state density decreases as T 3/2 and equals n at T = T c. We use the fact that for T T c, z =. Denoting the thermal de Broglie wavelength at T = T c by λ Tc we know that n = g 3/2 ()/ c. Thus below the condensation temperature n ex = g 3/2() = λ3 T c g 3/2 () = T 3 n. (.25) c Tc 3 The last equality uses the fact that n = n (T ) + T 3/2 n T 3/2 c /T 3/2. Thus n (T ) = n ( T ) 3/2. (.26) T 3/2 c It is worth observing that for all T < T c the condensed (in the ground state)phase and the excited phase coexist just as the liquid and gas do in the nitrogen example. ote that at zero temperature in an ideal Bose gas all the particles are in the ground state: n = n = /V or =. The energy of the gas at T T c is entirely determined by the particles in the excited state as is the pressure. Since z = below the condensation temperature we can obtain these quantities more simply than above the transition. We find that as z we have from Equation(.6) V = ζ(3/2) = g. (.27) 5

6 Use the above expression and Equation(.7) with z = to obtain U = 3 2 k BT The coefficient is V λ 3/2 T ζ(5/2) = 3 ζ(5/2) 2 ζ(3/2) k BT ( ) 3/2 T. (.28) T c p Below T c we have from Equation(.3) k B T = this using the definition of and ζ(5/2) =.3449 as p =.3449 ζ(5/2), and we can rewrite ( m 2π 2 ) 3/2 (kb T ) 5/2. (.29) The pressure increases as T 5/2 below the condensation temperature and is independent of the density n or equivalently the specific volume v = /n. For n > n c the pressure is given above by p (T ) = ζ(5/2) k BT. The critical density as a function of temperature is given by n c = ζ(3/2)/. The boundary of the phase coexistence region (where the excited gas and the condenses phase coexist) is defined p = h2 2πm ζ(5/3) ζ(3/2) 2 n5/3 c. (.3) Equivalently, the critical line is defined by p v 5/3 = constant. The pressure is independent of volume for T < T c. The isotherms in the (p, v) plane are then flat for v < v c. This resembles the coexistence region familiar from our study of the thermodynamics of the liquid-gas transition. We wish to find the behavior above T c and how various quantities behave near T c. We determine the equation of state above above T c from the expressions for p and n: p k B T = n = l= l= z l. (.3) l5/2 z l. (.32) l3/2 The two together define an implicit equation of state. Eliminate z. From the second equation we have [ z n ] 2 3/2 nλ3 T +. (.33) 6

7 Rewriting the pressure in terms of the density we obtain [ p k B T = n ] 2 5/2 nλ3 T +. (.34) The pressure is lower than that of the ideal gas. For fermions we showed that the sign of the second term is positive. The negative sign is loosely said to mimic an effective attractive interaction, an indication of the condensation at low temperatures. Make sure you are clear about the signs. In order to find the non-analyticities in measurable quantities we need some mathematical results quoted below. Analysis of g ν (z) as z One has to understand the mathematical behavior of g ν (z) as z to understand the subtle behavior of Bose-Einstein condensation. It is worth noting that z d dz g ν(z) = g ν (z). We need the behavior of g ν (z) as z. This is approaching T c from above. If the leading term is all we want we can use elementary methods and derive them. It is a good exercise in a mathematical methods course. A useful general result was obtained by J. E. Robinson, Phys. Rev. 83, (95) by computing its Mellin transform and finding an expansion for the inverse. The result for non-integral ν is given by g ν (z) = Γ( ν) α ν + l= ( ) l ζ(n l) α l (.35) l! where α = log z = βµ. The most useful results are given below with z = ɛ. π g /2 (z) ɛ + O(ɛ ). (.36) g 3/2 (z) = l= z l l 3/2 g /2 (z) = ζ ( ) 3 2 l /2 z l l= 2 π ɛ /2 + O(ɛ) (.37) π 2. (.38) ɛ3/2 7

8 Examples: For T > T c we write the behavior in terms of t T Tc T c. n = g 3/2 (z) Γ( /2) ( βµ) /2 + ζ(3/2) + (.39) This implies that for temperatures above the condensation temperature We have for the pressure µ(t ) k BT 4π n2 ( c ) λ 3 2 T t 2. (.4) p = k BT g 5/2 (z) k BT ( ζ(5/2) + ζ(3/2)β µ + ζ( 3/2) ( βµ) 3/2 ). (.4) You showed in the first homework problem set that κ T = n 2 ( 2 p/ µ 2 ) T. Therefore, the leading term is given by κ T ( µ) /2 t. The compressibility diverges. Many other quantities can be worked out with patience and mathematical skill. The specific heat at constant volume per particle can be obtained and is continuous at the transition but its slope is discontinuous. The discontinuity can be calculated given the mathematical formulae. Bose-Einstein condensation combines features of discontinuous (first-order), and continuous (second-order) transitions; there is a finite latent heat while the compressibility diverges. We will discuss some of these issues when we study phase transitions. 8