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Advanced Statistical Mechanics: Examples of quantum statistics
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Advanced Statistical Mechanics: Examples of quantum statistics


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  • 1. 7 Examples of quantum statisticsQuantum statistics involves either Bose–Einstein or Fermi–Dirac counting in the evaluation of phys-ical quantities. Examples have been covered extensively in the statistical physics course of the thirdyear (G. Bauer). Therefore, we restrict ourselves here to a brief review of the major applications. 7.1 Thermodynamics of free quantum gasesAs we have seen in the exercises, we can write Nλ 3 1 1 = 3/2 d3x , V π ex2 −β µ ±1where the + sign corresponds to Fermi, and the − sign to Bose statistics. Furthermore h λ=√ . 2π mkB TIn the classical limit for which µ is strongly negative, this leads to 1 nλ 3 = eβ µ ∓ e2β µ + · · · . 23/2This equation shows that when we keep the density fixed, strongly negative µ corresponds to hightemperatures. For the pressure, we have the expression P 1 1 2 +β µ = ln Z = ± 3 3/2 d 3 x ln 1 ± e−x , kB T V λ π 2 +β µwhere the + sign is for fermions, and the − sign for bosons. In the classical limit, e−x ≪ 1, thisyields, after expanding to second order in exp(β µ ): P 1 1 1 d 3 x e−x eβ µ ∓ e2β µ eβ µ ∓ 5/2 e2β µ . 2 2 = 3 3/2 d 3 x e−2x = kB T λ π λ 3 2If we now substitute eβ µ by the expansion obtained above, we see that P = nkB T 1 ± 2−5/2 nλ 3The first term is the classical results; the second term gives the quantum correction. For Fermions wehave the plus-sign, which indicates that the particles seem to repel each other as a result of the Pauliprinciple. For Bosons (minus-sign) the pressure becomes smaller, indicating an effective attraction. 42
  • 2. 43 Finally we can derive the entropy: ∂ kB T ln Z ∂P S=− =V . ∂T µ ,V ∂T µ ,VNow we copy the lowest order term in the expansion found above for the pressure P: ∂ m 3/2 S =V (kB T )5/2 eβ µ ∂T 2π ℏ2 V 5 µ = 3 eβ µ kB − . λ 2 TCombining this with the classical relation eβ µ = nλ 3 gives 5 S = NkB + ln(λ 3 /n) , 2which, for λ = h2 /(2π mkB T ) and E/N = 3kB T /2 can be written in the form: 3 E V 3 4π m S = NkB ln + ln + ln + 5/2 . 2 N N 2 3h2This is again the Sackur–Tetrode formula for the entropy of an ideal gas. The quantum correctionscan be evaluated analogous to the case of the pressure. The result is 1 S = SClass ± √ λ 3 NkB . 8 2 7.2 Bose-Einstein systems 7.2.1 Planck distributionTake an empty box in contact with a reservoir of temperature T . The reservoir can interact withthe box by emitting electromagnetic (EM) field waves into it. In quantum language we say thatphotons can travel into the box. The photons carry energy ℏω , with ω = c|k|. Which k vectors areaccessible is determined by the shape of the box. If we take a cube of size L × L × L, the k vectors are2π /L(nx , ny , nz ) with Ni integer. From the quantum theory of the elactromagnetic field, it follows thatthe photons are created and annihilated freely, so that there number cannot be controlled by a chemicalpotential. Creating a new photon in particular does not involve any cost except for its energy, so µ = 0.Therefore, the number of modes available at frequency ω is given by ω 2dω N(k)dk = 2 · 4π k2 dkL3 /(2π )3 = V . c3 π 2In this expression, the factor of 2 arises from the fact that there are two transverse modes (only trans-verse modes are allowed by Maxwell’s equations), and we have divided the volume of the sphericalshell with thickness dk in k-space by the volume (2π /L)3 of each k-point, and used ω = ck. Now weuse the fact that the photons are bosons (they are spin-1 particles) and we have for the energy radiatedat frequencies between ω and ω + d ω : ℏ ω 3dω u(ω )d ω = n(ω )ℏω d ω = . π 2 c3 eℏω /kB T − 1where the BE distribution is recognised, multiplied by the energy of the mode and the occupancy.
  • 3. 44 3 2.5 2 g(β µ ) 1.5 1 0.5 0 -3 -2.5 -2 -1.5 -1 -0.5 0 βµ Figure 7.1: The Bose–Einstein function g. 7.2.2 Bose–Einstein condensationThe BE distribution is well defined only for values of µ below the ground state energy – otherwise,the occupancy becomes negative. Let us, for such an acceptable µ , evaluate the number of particlesas a function of µ : 1 V 1 N =∑ ≈ 4π k2 dk. k eβ (ℏ2 k2 /(2m)−µ ) − 1 (2π )3 eβ (ℏ2 k2 /(2m)−µ ) − 1Reparametrising ℏ2 k2 /(2mkB T ), we obtain for the particle density 4 ∞ x2 nλ 3 = √ dx ≡ g(β µ ). π 0 ex2 −β µ − 1Note that the integral depends on β µ only. In function 7.1, we plot the function g(β µ ) as a functionof β µ . For β µ → 0 this function approaches the value 2.61. This imposes a temperature-dependent maximum on the particle density – beyond this value, theanalysis fails, and the only way out is by questioning the transition from the sum over k to an integral.In fact, this transition is not justified if ε0 − µ really approaches the value 0, where ε0 is the groundstate energy (i.e. the energy corresponding to the longest wavelength). In that case we must split offthe term corresponding to the ground state energy, which now is macroscopically occupied (that is, afinite fraction of the particles is in the ground state). What happens is that the gas splits up into two parts. The normal part fills the energy levelsaccording to the BE distribution in the usual way. This fraction corresponds to nnormal λ 3 = 2.61.If the total particle density is greater than prescribed by this limit, the rest of the particles occupied theground state. We therefore know that nG = n − nnormal particles per unit volume will be in the groundstate. The occupation of the ground state is 1 V nG = . eβ (εG −µ ) − 1
  • 4. 45 1 0.8 0.6 P 0.4 0.2 0 0 0.5 1 1.5 2 2.5 λ 3/n Figure 7.2: Pressure versus λ 3 /n. The critical density corresponds to 1/2.61 on the horizontal axis.We now can deduce that εG − µ ≈ 1/(β V nG )i.e., the larger the system, the closer the chemical potential will be to the ground state. As we have seen in section 7.1, the pressure is given by kB T 4 2 +β µ P=− √ dx x2 ln(1 − e−x ). λ3 πThe x = 0 term no longer contributes, even when µ is close to ε , as the denominator in the formulaefor the density is now replaced by the logarithm. Now we keep the temperature constant, and let the density vary. For densities lower than thecritical density, µ will vary and assume negative values. For densities higher than the critical value,µ = 0 and the pressure remains constant. Figure 7.2 shows the pressure as a function of the inversedensity. 7.2.3 Phonons and the specific heatPhonons are lattice vibrations, and they can be understood by realising that the system of interactingatomic nuclei can be approximated by a harmonic system, i.e. a system of particles connected by har-monic springs. Close to the configuration of minimum potential energy, any system can be describedin terms of harmonic interactions, and the excitations can be described in terms of a collection ofindependent harmonic oscillators (see the classical mechanics course). The energy for such a sys-tem can easily be found: we simply add up the expectation value of the energies of the oscillators atfrequencies ωi and at temperature T : ℏωi ℏωi U (T ) = Φ0 + ∑ + ∑ ℏω /(k T ) . i 2 i e i B −1The first term on the right hand side is the energy-offset, the second is the zero-point energy of theharmonic oscillator, and the rightmost term is the average energy due to the energy quanta nℏωi .
  • 5. 46We can evaluate the sums by transforming them into integrals, but Einstein avoided this in 1907 byrequiring that there was approximately only one frequency, ωE , at which the oscillators would vibrate.This leads directly to ∂U x2 ex CV (T ) = = 3NkB x , ∂T (e − 1)2where x = ℏωE /(kB T ) = θE /(kB T ). The parameter θE is called the Einstein temperature. For low temperatures, the Einstein result does not match the experimental results very well. Athigher temperatures, however, it approaches the classical result 3NkB (why?). Peter Debye took the actual distribution of modes which we have already encountered above forthe photons. This distribution is however cut off as there cannot be more modes than particles: ωD VCω 2 d ω = N. 0The value of the proportionality constant depends on the sound speeds for transverse and longitudinalwaves. Using the ω 2 distribution, The specific heat is found as CV (T ) = 3NkB D(x0 ),where x0 = ℏωD /(kB T ) = θD /(kB T ). The parameter θD is called the Debye temperature. The Debyefunction D(x) is defined as 3 x0 x4 ex D(x0 ) = 3 dx. x0 0 (ex − 1)2For low temperatures, the Debye result for the specific heat is CV (T ) = C(T /θD )3 . For high temperatures, we can perform a Taylor expansion of the integrand in the expression forD which yields 3 x0 2 D(x0 ) ≈ 3 x dx = 1, x0 0so that we find again CV → 3NkBas it should be for high temperature. 7.3 Fermions 7.3.1 Degenerate Fermi gasThe name ‘degenerate Fermi gas’ is used for a dense system consisting of noninteracting Fermions.Dense means that nλ 3 = eβ µis larger than 1. In that case, the chemical potential µ is positive. For T = 0, the distribution functionhas a square shape, and for small, but positive T , the square shape gets rounded of, as shown infigure 7.3. In the ground state, which is occupied for T = 0, all the one-particle levels are filled forenergies smaller than µ . The chemical potential at T = 0 is called the Fermi energy, εF . For particles
  • 6. 47 1.4 1.2 T= 0 1 0.8 n 0.6 kT 0.4 0.2 T> 0 0 0 0.2 0.4 0.6 0.8 1 1.2 E Figure 7.3: The Fermi distribution function for T = 0 and small, but positive T .moving in a cubic box with side L, the particles fill up a sphere in k-space, since E = ℏ2 k2 /(2m). Theradius of this sphere is as usual related to the particle number: L3 4π 3 N =2 k . (2π )3 3 FThe factor 2 is for the particular case of electrons, and it takes care of the double spin-degeneracy. Weconclude that ℏ2 kF ℏ2 (3nπ 2 )2/3 2 εF = = . 2m 2m From figure 7.3 it is clear that for positive T , the chemical potential will remain more or lessconstant. What happens is that some electrons with ε < εF are excited to ε > εF . These excitedelectrons come from a band of width ≈ kB T below the Fermi energy, and they occupy states in aband kB T above the Fermi energy. We speak of a degenerate Fermi gas when kB T ≪ εF . DegenerateFermi gases are quite familiar: in a metal, the valence electrons have a fermi energy corresponding toabout 50000 K, much larger than room temperature. In a particular kind of stars, the so-called ‘whitedwarfs’, the electrons have a Fermi temeperature of 107 K. To show that the chemical potential for a degenerate electron gas deviates only to order (kB T /εF )2is not easy. We shall take this for granted here. If we do so, we can easily evaluate the specific heat ofthe degenerate electron gas. It is convenient to count the number of states at a specific energy – this iscalled the density of states. The number of k-points which lie in the range k, k + ∆k is given as L3 D(ε )∆ε = 2 4π k2 ∆k; (2π )3therefore 3/2 √ V 2m D(ε )∆ε = ε ∆ε , 2π 2 ℏ2which can also be written as 3 −3/2 √ D(ε ) = N εF ε. 2 The total energy is now given by E = d ε D(ε )ε f (ε , T ).
  • 7. 48We have used f (ε , T ) for the Fermi-Dirac distribution function and we will stick to this conventionfrom now on. Taking the temperature derivative of this expectation value yields the specific heat. Using the fact that N= d ε D(ε ) f (ε , T ),we can write E = N εF + d ε D(ε )(ε − εF ) f (ε , T ).If we take the derivative with respect to T and assume that µ is approximately temperature-independent(this is not quite correct, see below), we obtain: ∂ E 1 ∞ ε − εF CV = = d ε D(ε ) ∂T kB T 2 0 e(ε −εF )/(2kB T ) + e−(ε −εF)/(2kB T ) 2Note that the integrand is small everywhere, except in a band kB T around εF . This allows us to takeD(εF ) out of the integral. Changing to the integration variable x = (ε − εF )/(kB T ), we have ∞ x2 CV = kB T D(εF ) 2 2 dx. −εF /(kB T ) ex/2 + e−x/2The lower boundary εF /(kB T ) of the integral is large but negative – we replace it by −∞. Using ∞ x2 dx π2 = , ∞ (ex/2 + e−x/2 )2 3we have π2 2 k T D(εF ). CV = 3 B Substituting the value D(εF ) = 3/2N/(kB T ) we obtain: π2 T CV = NkB 2 TFwhere the Fermi temperature TF is defined by εF = kB TF . We see that the specific heats grows linearlywith T . This growth stops only at the Fermi temperature, which, as we have seen, lies fairly high. Forvery high temperatures (higher than TF ), the specific heat saturates at 3NkB /2. If we compare this withphonons, we see that for low temperature, where the specific heat due to the phonons, grows as T 3 ,the electronic contribution dominates, whereas for temperatures well above the Debye, the phononcontribution saturates at 3NkB , well above the maximum contribution of the electrons. Although this calculation yields the correct result, a few things have been wiped under the carpet,in particular the fact that µ was replaced by εF . We now present a correct calculation, which starts byexpressing N as an integral over the function ∆(ε ) which is an integral of the density of states over theenergy: ∞ ∆(ε ) = D(ε ′ )d ε ′ . 0Then we can write, using partial integration: ∞ ∞ f (ε , T ) N= D(ε ) f (ε , T ) d ε = − ∆(ε ) dε . 0 0 dε
  • 8. 49The partial integration has the advantage that the energy derivative of f is nonzero only in a smallinterval around µ so that we can expand D around µ : (ε − µ )2 ′ 1 1 N= ∆(µ ) + (ε − µ )D(µ ) + D (µ ) 2 dε . 2 kB T e(ε −µ )/(2kB T ) + e−(ε −µ )/(2kB T )We have used the prime ′ to indicate a derivative with respect to T . Again selecting only the evenintegrands, we obtain two terms: π2 N = ∆(µ ) + D′ (µ ) (kB T )2 , 6where in the first term we could integrate directly as the integrand is proportional to the energy-derivative of f , and in the second integral we have used the same result as was used in the simplifiedderivation. The main observation now is to realize that, if we want to calculate the specific heat at constantdensity, the number of particles should be fixed. This implies that its derivative with respect to thetemperature should vanish: dN π2 2 π2 = D(µ )µ ′ + D′ (µ ) kB T + D′′ (µ )µ ′ (kB T )2 . dT 3 6For low temperatures, the rightmost term is much smaller than the first two, so that we have π2 2 D(µ )µ ′ + D′ (µ )k T = 0. 3 B We now perform a similar analysis for the specific heat, along the lines of our simple derivationabove: ∂f ∂f ∂f cV = ε d ε = µ D(ε ) d ε + D(ε )(ε − µ ) dε . ∂T ∂T ∂TWe have ∂f 1 ε − µ + T µ′ = . ∂ε kB T 2 e(ε −µ )/(2kB T ) + e−(ε −µ )/(2kBT ) 2Substituting this into the two integrals appearing in the expression for cV , we obtain: 1 ε − µ + T µ′ cV = µ D(µ ) + (ε − µ )D′ (µ ) + . . . dε + kB T 2 e(ε −µ )/(2kB T ) + e−(ε −µ )/(2kB T ) 2 1 ε − µ + T µ′ D(µ ) + (ε − µ )D′ (µ ) + . . . (ε − µ ) 2 2 dε kB T e(ε −µ )/(2kB T ) + e−(ε −µ )/(2kB T )Carefully analysing these integrals gives three dominant terms, where two arise from the first integral,and the third one from the second integral: π2 2 2 π 2 cV = µ µ ′ D(µ ) + µ D′ (µ ) kB T + D(µ )kB T . 3 3 Using the relation obtained above from the vanishing temperature-derivative of the particle num-ber N then yields: cV = D(µ )kB T π 2 /3. 2Substituting the explicit expression for the density of states yields the result obtained above. Note that in this derivation, no reference to the explicit form of D(ε ) has been made.
  • 9. 50 7.3.2 Pauli paramagnetismIn chapter 3 we have already considered paramagnetism (section 3.9 and 10). Here we shall considerthe full quantum description for spin-1/2 fermions – you may think of electrons in a solid. Suppose wehave no magnetic field. Then, all the properties of the electrons are determined by the density of statesD(ε ) (see the previous section). Once we know this function, all relevant physical quantities can bedetermined. The important issue now is that this density of states does not depend on the spin degreesof freedom (this is a direct consequence of the fact that B = 0). If the magnetic field is switched on,the only thing which changes is that the energies are shifted over ±µ ∗ B, where the sign depends onthe spin. Note that µ ∗ is the magnetic moment – it should not be confused with chemical potentialwhich is µ without the asterisk. First we analyse how the chemical potential changes with the magnetic field. We do this bycalculating the total number of particles and then require that this is constant: N = N+ + N− = ∑ [ f+ (i) − f− (i)] iwhere the sum is over the orbital states i; f± are the Dirac functions for the appropriate spin state.A sum over the orbital states can however be replaced by an integral over the energy if we insert thedensity of states: 1 N= d ε [D(ε ) f (ε − µ ∗ B) + D(ε ) f (ε + µ ∗ B)] . 2The integral is over the orbital energies, and the magnetic field only enters in the Fermi distributionfunctions. The factor 1/2 in front of the integral takes into account that D(ε ) includes up- and downspins. Now we assume that the field B is very small (i.e. µ ∗ B smaller than kB T ). Then we can expandthe distribution functions about B = 0: (µ ∗ B)2 ′′ f (ε ± µ ∗ B) ≈ f (ε ) ± µ ∗ B f ′ (ε ) + f (ε ). 2Substituting this back into the integral expression for N, we have N= d ε D(ε ) f (ε ) + O(B2 ) + . . . .We see that to first order in B the density does not change if we keep the chemical potential constant;hence we conclude that µ varies with B only to second order. We are interested in how the system reacts to an applied field, that is, we want to calculate themagnetisation as a function of the field strength B. The magnetisation is given as the differencebetween the number of spin-up and -down electrons: M = µ ∗ (N+ − N− ) = µ ∗ ∑ [ f+ (i) − f− (i)] iwhere the sum is over the orbital states i – f± are the Dirac functions for the appropriate spin state.The sum over the orbital states can again be replaced by an integral over the energy: µ∗ M= d ε [D(ε ) f (εi − µ ∗ B) − D(ε ) f (εi + µ ∗ B)] . 2Substituting the same Taylor expansion for the distribution functions as above, we obtain: ∞ M = µ ∗2B d ε D(ε ) f ′ (ε ). 0
  • 10. 51 For small T , the Fermi function decays rapidly from 1 to 0 near the Fermi energy, hence f ′ (ε ) ≈−δ (ε − εF ). Using this, we obtain M ≈ µ ∗ 2 BD(εF ).The magnetic susceptibility χ tells us how the magnetisation varies with the field: dM χ= = µ ∗ 2 D(εF ). dBWe see that measuring the susceptibility at low temperatures tells us what the density of states nearthe Fermi level is. For the free electron gas, the density of states was found in the previous section – we find for thesusceptibility in this case 3 µ ∗2 χ= . 2 εFFor higher temperature, the susceptibility can be expanded in powers of kB T ; the result is 3 µ ∗2 π 2 kB T χ≈ 1− + ··· . 2 εF 12 εF 7.3.3 Landau diamagnetismElectrons moving in a solid have a magnetic moment not only as a result of their spin, but also as aresult of their orbit. This is called the orbital magnetic moment. If we apply a magnetic field in the z-direction, the particles will have quantized field levels associated with the x- and y degrees of freedom.In addition, they have an energy associated with their motion in the z-direction. The spectrum is givenby eℏB p2 ε ( j, pz ) = ( j + 1/2) + z . m 2mThis problem has been treated in the exercise class of your quantum course (believe it or not). For evaluating particle numbers and magnetisations, we need to know the density of states, in otherwords the multiplicity of these levels. This holds in particular for the x and y degrees of freedom, aswe can simply perform an integral over pz when summing over all states. It can be argued that themultiplicity for the energy levels associated with the orbital motion in the xy plane is given by eB D( j) = Lx Ly . hWe can then evaluate the number of particles and the magnetic moment. This is most easily done byfirst evaluating the grand partition function: Z = ∏ 1 + eβ (µ −εi ) . iWe then have, with z = exp(β µ ): ln Z = ∑ ln 1 + ze−β εi . iThe index i denotes the states, which for our particular problem are defined by pz and j: ∞ Lz eB eℏB p2 ln Z = d pz ∑ Lx Lz ln 1 + z exp −β ( j + 1/2) + z . 2π ℏ j=0 h m 2m
  • 11. 52This partition function can be evaluated in the classical limit, where z ≪ 1. Expanding the argumentof the logarithm, we get ∞ zVeB zVeB 1 e−β pz /(2m) d pz ∑ e−β eℏB( j+1/2)/m = 2 ln Z = (2π mkB T )1/2 . h2 j=0 h2 2 sinh [eℏB/(2mkB T )] Now the desired quantities can easily be evaluated. With λ = h/(2π mkB T ), x = β Beh/(4π m) andµeff = eh/(4π m), we have ∂Z zV x N=z = 3 ∂z λ sinh xand 1 ∂ zV 1 x cosh x M= ln Z = 3 µeff − . β ∂B λ sinh x sinh2 x We can write M = −N µeff L(x),where L is the Langevin function L(x) = coth x − 1/x.The result we have obtained is similar to that of 3.9, except for a minus sign. This means that themagnetisation is now opposite to the field – this effect is called diamagnetism.