List of Tables. List of Figures. 4 Statistical Ensembles The microcanonical distribution function... 2

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1 Contents Contents i List of ables iii List of Figures iv 4 Statistical Ensembles References Microcanonical Ensemble (µce he microcanonical distribution function Density of states Arbitrariness in the definition of S(E Ultra-relativistic ideal gas Discrete systems he Quantum Mechanical race he density matrix Averaging the DOS Coherent states hermal Equilibrium wo systems in thermal contact hermal, mechanical and chemical equilibrium Gibbs-Duhem relation i

2 ii CONENS 4.5 Ordinary Canonical Ensemble (OCE Canonical distribution and partition function he difference between P (E n and P n Additional remarks Averages within the OCE Entropy and free energy Fluctuations in the OCE hermodynamics revisited Generalized susceptibilities Grand Canonical Ensemble (GCE Grand canonical distribution and partition function Entropy and Gibbs-Duhem relation Generalized susceptibilities in the GCE Fluctuations in the GCE Gibbs ensemble Statistical Ensembles from Maximum Entropy µce OCE GCE Ideal Gas Statistical Mechanics Maxwell velocity distribution Equipartition Selected Examples Spins in an external magnetic field Negative temperature (! Adsorption Elasticity of wool Noninteracting spin dimers

3 4.10 Statistical Mechanics of Molecular Gases Separation of translational and internal degrees of freedom Ideal gas law he internal coordinate partition function Rotations Vibrations wo-level systems : Schottky anomaly Electronic and nuclear excitations Appendix I : Additional Examples hree state system Spins and vacancies on a surface Fluctuating interface Dissociation of molecular hydrogen List of ables 4.1 Rotational and vibrational temperatures of common molecules iii

4 iv LIS OF FIGURES 4.2 Nuclear angular momentum states for homonuclear diatomic molecules List of Figures 4.1 Complex integration contours C for inverse Laplace transform L 1[ Z(β ] = D(E. When the product dn is odd, there is a branch cut along the negative Re β axis A system S in contact with a world W. he union of the two, universe U = W S, is said to be the universe Averaging the quantum mechanical discrete density of states yields a continuous curve wo systems in thermal contact Microscopic, statistical interpretation of the First Law of hermodynamics Maxwell distribution of speeds ϕ(v/v 0. he most probable speed is v MAX = 2 v 0. he 8 average speed is v AVG = π v 0. he RMS speed is v RMS = 3 v When entropy decreases with increasing energy, the temperature is negative. ypically, kinetic degrees of freedom prevent this peculiarity from manifesting in physical systems he monomers in wool are modeled as existing in one of two states. he low energy undeformed state is A, and the higher energy deformed state is B. Applying tension induces more monomers to enter the B state Upper panel: length L(τ, for k B / ε = 0.01 (blue, 0.1 (green, 0.5 (dark red, and 1.0 (red. Bottom panel: dimensionless force constant k/n( l 2 versus temperature A model of noninteracting spin dimers on a lattice. Each red dot represents a classical spin for which σ j = ± Heat capacity per molecule as a function of temperature for (a heteronuclear diatomic gases, (b a single vibrational mode, and (c a single two-level system

5 Chapter 4 Statistical Ensembles 4.1 References F. Reif, Fundamentals of Statistical and hermal Physics (McGraw-Hill, 1987 his has been perhaps the most popular undergraduate text since it first appeared in 1967, and with good reason. A. H. Carter, Classical and Statistical hermodynamics (Benjamin Cummings, 2000 A very relaxed treatment appropriate for undergraduate physics majors. D. V. Schroeder, An Introduction to hermal Physics (Addison-Wesley, 2000 his is the best undergraduate thermodynamics book I ve come across, but only 40% of the book treats statistical mechanics. C. Kittel, Elementary Statistical Physics (Dover, 2004 Remarkably crisp, though dated, this text is organized as a series of brief discussions of key concepts and examples. Published by Dover, so you can t beat the price. M. Kardar, Statistical Physics of Particles (Cambridge, 2007 A superb modern text, with many insightful presentations of key concepts. M. Plischke and B. Bergersen, Equilibrium Statistical Physics (3 rd edition, World Scientific, 2006 An excellent graduate level text. Less insightful than Kardar but still a good modern treatment of the subject. Good discussion of mean field theory. E. M. Lifshitz and L. P. Pitaevskii, Statistical Physics (part I, 3 rd edition, Pergamon, 1980 his is volume 5 in the famous Landau and Lifshitz Course of heoretical Physics. hough dated, it still contains a wealth of information and physical insight. 1

6 2 CHAPER 4. SAISICAL ENSEMBLES 4.2 Microcanonical Ensemble (µce he microcanonical distribution function We have seen how in an ergodic dynamical system, time averages can be replaced by phase space averages: ergodicity f(ϕ t = f(ϕ S, (4.1 where and f(ϕ t = lim 1 0 dt f ( ϕ(t. (4.2 f(ϕ S = dµ f(ϕ δ ( E Ĥ(ϕ/ dµ δ ( E Ĥ(ϕ. (4.3 Here Ĥ(ϕ = Ĥ(q, p is the Hamiltonian, and where δ(x is the Dirac δ-function1. hus, averages are taken over a constant energy hypersurface which is a subset of the entire phase space. We ve also seen how any phase space distribution ϱ(λ 1,..., Λ k which is a function of conserved quantitied Λ a (ϕ is automatically a stationary (time-independent solution to Liouville s equation. Note that the microcanonical distribution, ϱ E (ϕ = δ ( E Ĥ(ϕ/ dµ δ ( E Ĥ(ϕ, (4.4 is of this form, since Ĥ(ϕ is conserved by the dynamics. Linear and angular momentum conservation generally are broken by elastic scattering off the walls of the sample. So averages in the microcanonical ensemble are computed by evaluating the ratio A = r A δ(e Ĥ r δ(e Ĥ, (4.5 where r means trace, which entails an integration over all phase space: r A(q, p 1 N d d p i d d q i A(q, p. (4.6 N! (2π d i=1 Here N is the total number of particles and d is the dimension of physical space in which each particle moves. he factor of 1/N!, which cancels in the ratio between numerator and denominator, is present for indistinguishable particles 2. he normalization factor (2π Nd renders the trace dimensionless. Again, this cancels between numerator and denominator. hese factors may then seem arbitrary in the definition of the trace, but we ll see how they in fact are required from quantum mechanical considerations. So we now adopt the following metric for classical phase space integration: dµ = 1 N! N d d p i d d q i (2π d. (4.7 i=1 1 We write the Hamiltonian as Ĥ (classical or quantum in order to distinguish it from magnetic field (H or enthalpy (H. 2 More on this in chapter 5.

7 4.2. MICROCANONICAL ENSEMBLE (µce Density of states he denominator, D(E = r δ(e Ĥ, (4.8 is called the density of states. It has dimensions of inverse energy, such that D(E E = de dµ δ(e Ĥ = dµ (4.9 E+ E E E<Ĥ<E+ E = # of states with energies between E and E + E. Let us now compute D(E for the nonrelativistic ideal gas. he Hamiltonian is Ĥ(q, p = N i=1 p 2 i 2m. (4.10 We assume that the gas is enclosed in a region of volume V, and we ll do a purely classical calculation, neglecting discreteness of its quantum spectrum. We must compute D(E = 1 N! N i=1 d d p i d d q i (2π d ( δ E N i=1 p 2 i. (4.11 2m We shall calculate D(E in two ways. he first method utilizes the Laplace transform, Z(β: Z(β = L [ D(E ] he inverse Laplace transform is then D(E = L 1[ Z(β ] 0 de e βe D(E = r e βĥ. (4.12 c+i c i dβ 2πi eβe Z(β, (4.13 where c is such that the integration contour is to the right of any singularities of Z(β in the complex β-plane. We then have Z(β = 1 N! = V N N! = V N N! N i=1 d d x i d d p i (2π d e βp2 i /2m dp /2m 2π e βp2 ( m 2π 2 Nd/2 β Nd/2. Nd (4.14

8 4 CHAPER 4. SAISICAL ENSEMBLES Figure 4.1: Complex integration contours C for inverse Laplace transform L 1[ Z(β ] = D(E. When the product dn is odd, there is a branch cut along the negative Re β axis. he inverse Laplace transform is then D(E = V N N! = V N N! ( m Nd/2 2π 2 C dβ 2πi eβe β Nd/2 ( m Nd/2 E 1 2 Nd 1 2π 2 Γ(Nd/2, (4.15 exactly as before. he integration contour for the inverse Laplace transform is extended in an infinite semicircle in the left half β-plane. When Nd is even, the function β Nd/2 has a simple pole of order Nd/2 at the origin. When Nd is odd, there is a branch cut extending along the negative Re β axis, and the integration contour must avoid the cut, as shown in fig One can check that this results in the same expression above, i.e. we may analytically continue from even values of N d to all positive values of Nd. For a general system, the Laplace transform, Z(β = L [ D(E ] also is called the partition function. We shall again meet up with Z(β when we discuss the ordinary canonical ensemble. Our final result, then, is D(E, V, N = V N N! ( m Nd/2 E 1 2 Nd 1 2π 2 Γ(Nd/2. (4.16 Here we have emphasized that the density of states is a function of E, V, and N. Using Stirling s approximation, ln N! = N ln N N ln N ln(2π + O( N 1, (4.17 we may define the statistical entropy, ( E S(E, V, N k B ln D(E, V, N = Nk B φ N, V + O(ln N, (4.18 N

9 4.2. MICROCANONICAL ENSEMBLE (µce 5 where ( E φ N, V = d ( ( E V N 2 ln + ln + d ( m N N 2 ln dπ 2 + ( d. (4.19 Recall k B = erg/k is Boltzmann s constant. Second method he second method invokes a mathematical trick. First, let s rescale p α i 2mE u α i. We then have ( Nd D(E = V N 2mE 1 d M u δ ( u u u 2 M 1. (4.20 N! h E Here we have written u = (u 1, u 2,..., u M with M = Nd as a M-dimensional vector. We ve also used the rule δ(ex = E 1 δ(x for δ-functions. We can now write d M u = u M 1 du dω M, (4.21 where dω M is the M-dimensional differential solid angle. We now have our answer: 3 D(E = V N N! ( 2m h Nd E 1 2 Nd Ω Nd. (4.22 What remains is for us to compute Ω M, the total solid angle in M dimensions. We do this by a nifty mathematical trick. Consider the integral where s = u 2, and where I M = d M u e u2 = Ω M = 1 2 Ω M 0 0 du u M 1 e u2 ds s 1 2 M 1 e s = 1 2 Ω M Γ( 1 2 M, Γ(z = 0 (4.23 dt t z 1 e t (4.24 is the Gamma function, which satisfies z Γ(z = Γ(z On the other hand, we can compute I M in Cartesian coordinates, writing I M = M du 1 e u2 1 = ( π M. ( he factor of 1 preceding Ω 2 M in eqn appears because δ(u 2 1 = 1 δ(u δ(u + 1. Since u = u 0, the 2 2 second term can be dropped. 4 Note that for integer argument, Γ(k = (k 1!

10 6 CHAPER 4. SAISICAL ENSEMBLES herefore Ω M = 2πM/2 Γ(M/2. (4.26 We thereby obtain Ω 2 = 2π, Ω 3 = 4π, Ω 4 = 2π 2, etc., the first two of which are familiar Arbitrariness in the definition of S(E Note that D(E has dimensions of inverse energy, so one might ask how we are to take the logarithm of a dimensionful quantity in eqn We must introduce an energy scale, such as E in eqn. 4.9, and define D(E; E = D(E E and S(E; E k B ln D(E; E. he definition of statistical entropy then involves the arbitrary parameter E, however this only affects S(E in an additive way. hat is, ( E1 S(E, V, N; E 1 = S(E, V, N; E 2 + k B ln. (4.27 E 2 Note that the difference between the two definitions of S depends only on the ratio E 1 / E 2, and is independent of E, V, and N Ultra-relativistic ideal gas Consider an ultrarelativistic ideal gas, with single particle dispersion ε(p = cp. We then have Z(β = V N Ωd N N N! h N dp p d 1 e βcp d = V N N! 0 ( Γ(d N Ωd c d h d β d. (4.28 he statistical entropy is S(E, V, N = k B ln D(E, V, N = Nk B φ ( E N, V N, with ( E φ N, V ( ( ( E V Ωd Γ(d = d ln + ln + ln N N N (dhc d + (d + 1 ( Discrete systems For classical systems where the energy levels are discrete, the states of the system σ are labeled by a set of discrete quantities {σ 1, σ 2,...}, where each variable σ i takes discrete values. he number of ways of configuring the system at fixed energy E is then Ω(E, N = σ δĥ(σ,e, (4.30 where the sum is over all possible configurations. Here N labels the total number of particles. For example, if we have N spin- 1 2 particles on a lattice which are placed in a magnetic field H, so the individual particle energy is ε i = µ 0 Hσ, where σ = ±1, then in a configuration in which N particles have

11 4.3. HE QUANUM MECHANICAL RACE 7 σ i = +1 and N = N N particles have σ i = 1, the energy is E = (N N µ 0 H. he number of configurations at fixed energy E is ( N N! Ω(E, N = = ( N N 2 E ( 2µ 0 H! N 2 + E, (4.31 2µ 0 H! since N / = N 2 E 2µ 0 H. he statistical entropy is S(E, N = k B ln Ω(E, N. 4.3 he Quantum Mechanical race hus far our understanding of ergodicity is rooted in the dynamics of classical mechanics. A Hamiltonian flow which is ergodic is one in which time averages can be replaced by phase space averages using the microcanonical ensemble. What happens, though, if our system is quantum mechanical, as all systems ultimately are? he density matrix First, let us consider that our system S will in general be in contact with a world W. We call the union of S and W the universe, U = W S. Let N denote a quantum mechanical state of W, and let n denote a quantum mechanical state of S. hen the most general wavefunction we can write is of the form Ψ = Ψ N,n N n. (4.32 N,n Now let us compute the expectation value of some operator  which acts as the identity within W, meaning N  N =  δ NN, where  is the reduced operator which acts within S alone. We then have Ψ Â Ψ = Ψ N,n Ψ N,n δ NN n  n N,N n,n (4.33 where = r (ˆϱ Â, ˆϱ = N n,n Ψ N,n Ψ N,n is the density matrix. he time-dependence of ˆϱ is easily found: ˆϱ(t = Ψ N,n Ψ N,n n (t n(t N n,n = e iĥt/ ˆϱ e +iĥt/, where Ĥ is the Hamiltonian for the system S. hus, we find n n (4.34 (4.35 i ˆϱ t = [ Ĥ, ˆϱ ]. (4.36

12 8 CHAPER 4. SAISICAL ENSEMBLES Figure 4.2: A system S in contact with a world W. he union of the two, universe U = W S, is said to be the universe. Note that the density matrix evolves according to a slightly different equation than an operator in the Heisenberg picture, for which Â(t = e +iht/ A e iĥt/ = i  t = [ Â, Ĥ] = [ Ĥ, Â]. (4.37 For Hamiltonian systems, we found that the phase space distribution ϱ(q, p, t evolved according to the Liouville equation, i ϱ t = L ϱ, (4.38 where the Liouvillian L is the differential operator Nd L = i j=1 { Ĥ p j Ĥ q j q j p j }. (4.39 Accordingly, any distribution ϱ(λ 1,..., Λ k which is a function of constants of the motion Λ a (q, p is a stationary solution to the Liouville equation: t ϱ(λ 1,..., Λ k = 0. Similarly, any quantum mechanical density matrix which commutes with the Hamiltonian is a stationary solution to eqn he corresponding microcanonical distribution is ˆϱ E = δ ( E Ĥ. ( Averaging the DOS If our quantum mechanical system is placed in a finite volume, the energy levels will be discrete, rather than continuous, and the density of states (DOS will be of the form D(E = r δ ( E Ĥ = l δ(e E l, (4.41

13 4.3. HE QUANUM MECHANICAL RACE 9 Figure 4.3: Averaging the quantum mechanical discrete density of states yields a continuous curve. where {E l } are the eigenvalues of the Hamiltonian Ĥ. In the thermodynamic limit, V, and the discrete spectrum of kinetic energies remains discrete for all finite V but must approach the continuum result. o recover the continuum result, we average the DOS over a window of width E: D(E = 1 E+ E de D(E. (4.42 E If we take the limit E 0 but with E δe, where δe is the spacing between successive quantized levels, we recover a smooth function, as shown in fig We will in general drop the bar and refer to this function as D(E. Note that δe 1/D(E = e Nφ(ε,v is (typically exponentially small in the size of the system, hence if we took E V 1 which vanishes in the thermodynamic limit, there are still exponentially many energy levels within an interval of width E. E Coherent states he quantum-classical correspondence is elucidated with the use of coherent states. Recall that the onedimensional harmonic oscillator Hamiltonian may be written Ĥ 0 = p2 2m m ω2 0 q 2 ( = ω 0 a a + 1 (4.43 2, where a and a are ladder operators satisfying [ a, a ] = 1, which can be taken to be a = l q + q 2l, a = l q + q 2l, (4.44 with l = /2mω 0. Note that q = l ( a + a, p = ( a a. (4.45 2il

14 10 CHAPER 4. SAISICAL ENSEMBLES he ground state satisfies a ψ 0 (q = 0, which yields he normalized coherent state z is defined as he overlap of coherent states is given by ψ 0 (q = (2πl 2 1/4 e q2 /4l 2. (4.46 z = e 1 2 z 2 e za 0 = e 1 2 z 2 n=0 z n n! n. (4.47 z 1 z 2 = e 1 2 z 1 2 e 1 2 z 2 2 e z 1 z 2, (4.48 hence different coherent states are not orthogonal. Despite this nonorthogonality, the coherent states allow a simple resolution of the identity, 1 = d 2 z 2πi z z ; d 2 z d Rez d Imz 2πi π (4.49 which is straightforward to establish. o gain some physical intuition about the coherent states, define z Q 2l + ilp (4.50 and write z Q, P. One finds (exercise! ψ Q,P (q = q z = (2πl 2 1/4 e ip Q/2 e ip q/ e (q Q2 /4l 2, (4.51 hence the coherent state ψ Q,P (q is a wavepacket Gaussianly localized about q = Q, but oscillating with average momentum P. For example, we can compute Q, P q Q, P = z l (a + a z = 2l Re z = Q (4.52 as well as Q, P p Q, P = z 2il (a a z = Im z = P (4.53 l Q, P q 2 Q, P = z l 2 (a + a 2 z = Q 2 + l 2 (4.54 Q, P p 2 Q, P = z 2 4l 2 (a a 2 z = P l 2. (4.55 hus, the root mean square fluctuations in the coherent state Q, P are q = l =, p = 2mω 0 2l = m ω0, (4.56 2

15 4.4. HERMAL EQUILIBRIUM 11 and q p = 1 2. hus we learn that the coherent state ψ Q,P (q is localized in phase space, i.e. in both position and momentum. If we have a general operator Â(q, p, we can then write Q, P Â(q, p Q, P = A(Q, P + O(, (4.57 where A(Q, P is formed from Â(q, p by replacing q Q and p P. Since d 2 z d Rez d Imz = 2πi π we can write the trace using coherent states as r  = 1 2π dq dq dp 2π, (4.58 dp Q, P  Q, P. (4.59 We now can understand the origin of the factor 2π in the denominator of each (q i, p i integral over classical phase space in eqn Note that ω 0 is arbitrary in our discussion. By increasing ω 0, the states become more localized in q and more plane wave like in p. However, so long as ω 0 is finite, the width of the coherent state in each direction is proportional to 1/2, and thus vanishes in the classical limit. 4.4 hermal Equilibrium wo systems in thermal contact Consider two systems in thermal contact, as depicted in fig he two subsystems #1 and #2 are free to exchange energy, but their respective volumes and particle numbers remain fixed. We assume the contact is made over a surface, and that the energy associated with that surface is negligible when compared with the bulk energies E 1 and E 2. Let the total energy be E = E 1 + E 2. hen the density of states D(E for the combined system is D(E = de 1 D 1 (E 1 D 2 (E E 1. (4.60 he probability density for system #1 to have energy E 1 is then P 1 (E 1 = D 1 (E 1 D 2 (E E 1 D(E. (4.61 Note that P 1 (E 1 is normalized: de 1 P 1 (E 1 = 1. We now ask: what is the most probable value of E 1? We find out by differentiating P 1 (E 1 with respect to E 1 and setting the result to zero. his requires 1 dp 0 = 1 (E 1 = ln P P 1 (E 1 de 1 E 1 (E 1 1 = ln D E 1 (E 1 + ln D 1 E 2 (E E 1. 1 (4.62

16 12 CHAPER 4. SAISICAL ENSEMBLES Figure 4.4: wo systems in thermal contact. We conclude that the maximally likely partition of energy between systems #1 and #2 is realized when S 1 E 1 = S 2 E 2. (4.63 his guarantees that S(E, E 1 = S 1 (E 1 + S 2 (E E 1 (4.64 is a maximum with respect to the energy E 1, at fixed total energy E. he temperature is defined as ( 1 S = E V,N, (4.65 a result familiar from thermodynamics. he difference is now we have a more rigorous definition of the entropy. When the total entropy S is maximized, we have that 1 = 2. Once again, two systems in thermal contact and can exchange energy will in equilibrium have equal temperatures. According to eqns and 4.29, the entropies of nonrelativistic and ultrarelativistic ideal gases in d space dimensions are given by Invoking eqn. 4.65, we then have ( E S NR = 1 2 Nd k B ln N ( E S UR = Nd k B ln N ( V + Nk B ln N ( V + Nk B ln N + const. ( const.. (4.67 E NR = 1 2 Nd k B, E UR = Nd k B. (4.68 We saw that the probability distribution P 1 (E 1 is maximized when 1 = 2, but how sharp is the peak in the distribution? Let us write E 1 = E1 + E 1, where E 1 is the solution to eqn We then have ln P 1 (E1 + E 1 = ln P 1 (E S 1 2k ( E S 2 B E 2k ( E , (4.69 B 1 E 2 E 2 1 E 2 2

17 4.4. HERMAL EQUILIBRIUM 13 where E2 = E E 1. We must now evaluate 2 S E 2 = E ( 1 = 1 ( 2 = 1 E V,N 2, (4.70 C V where C V = ( E/ V,N is the heat capacity. hus, P 1 = P 1 e ( E 1 2 /2k B 2 C V, (4.71 where C V = C V,1 C V,2 C V,1 + C V,2. (4.72 he distribution is therefore a Gaussian, and the fluctuations in E 1 can now be computed: ( E1 2 = k B 2 CV = ( E 1 RMS = k B CV /k B. (4.73 he individual heat capacities C V,1 and C V,2 scale with the volumes V 1 and V 2, respectively. If V 2 V 1, then C V,2 C V,1, in which case C V C V,1. herefore the RMS fluctuations in E 1 are proportional to the square root of the system size, whereas E 1 itself is extensive. hus, the ratio ( E 1 RMS /E 1 V 1/2 scales as the inverse square root of the volume. he distribution P 1 (E 1 is thus extremely sharp hermal, mechanical and chemical equilibrium We have ds V,N = 1 de, but in general S = S(E, V, N. Equivalently, we may write E = E(S, V, N. he full differential of E(S, V, N is then de = ds p dv +µ dn, with = ( E S V,N and p = ( E V S,N and µ = ( E N. As we shall discuss in more detail, p is the pressure and µ is the chemical potential. We S,V may thus write the total differential ds as ds = 1 de + p dv µ dn. (4.74 Employing the same reasoning as in the previous section, we conclude that entropy maximization for two systems in contact requires the following: If two systems can exchange energy, then 1 = 2. his is thermal equilibrium. If two systems can exchange volume, then p 1 / 1 = p 2 / 2. his is mechanical equilibrium. If two systems can exchange particle number, then µ 1 / 1 = µ 2 / 2. his is chemical equilibrium.

18 14 CHAPER 4. SAISICAL ENSEMBLES Gibbs-Duhem relation he energy E(S, V, N is an extensive function of extensive variables, i.e. it is homogeneous of degree one in its arguments. herefore E(λS, λv, λn = λe, and taking the derivative with respect to λ yields ( ( ( E E E E = S + V + N S V N (4.75 V,N = S pv + µn. aking the differential of each side, using the Leibniz rule on the RHS, and plugging in de = ds p dv + µ dn, we arrive at the Gibbs-Duhem relation 5, S,N S d V dp + N dµ = 0. (4.76 his, in turn, says that any one of the intensive quantities (, p, µ can be written as a function of the other two, in the case of a single component system. S,V 4.5 Ordinary Canonical Ensemble (OCE Canonical distribution and partition function Consider a system S in contact with a world W, and let their union U = W S be called the universe. he situation is depicted in fig he volume V S and particle number N S of the system are held fixed, but the energy is allowed to fluctuate by exchange with the world W. We are interested in the limit N S, N W, with N S N W, with similar relations holding for the respective volumes and energies. We now ask what is the probability that S is in a state n with energy E n. his is given by the ratio P n = lim E 0 D W (E U E n E D U (E U E = # of states accessible to W given that E S = E n total # of states in U. (4.77 hen ln P n = ln D W (E U E n ln D U (E U = ln D W (E U ln D U (E U E n ln D W (E E α βe n E=EU (4.78 he constant β is given by β = ln D W(E E = 1 E=EU k B. ( See

19 4.5. ORDINARY CANONICAL ENSEMBLE (OCE 15 hus, we find P n = e α e βe n. he constant α is fixed by the requirement that n P n = 1: P n = 1 Z e βe n, Z(, V, N = n e βe n = r e βĥ. (4.80 We ve already met Z(β in eqn it is the Laplace transform of the density of states. It is also called the partition function of the system S. Quantum mechanically, we can write the ordinary canonical density matrix as ˆϱ = e βĥ, (4.81 r e βĥ which is known as the Gibbs distribution. Note that [ˆϱ, Ĥ] = 0, hence the ordinary canonical distribution is a stationary solution to the evolution equation for the density matrix. Note that the OCE is specified by three parameters:, V, and N he difference between P (E n and P n Let the total energy of the Universe be fixed at E U. he joint probability density P (E S, E W for the system to have energy E S and the world to have energy E W is P (E S, E W = D S (E S D W (E W δ(e U E S E W / D U (E U, (4.82 where D U (E U = de S D S (E S D W (E U E S, (4.83 which ensures that de S dew P (E S, E W = 1. he probability density P (E S is defined such that P (E S de S is the (differential probability for the system to have an energy in the range [E S, E S + de S ]. he units of P (E S are E 1. o obtain P (E S, we simply integrate the joint probability density P (E S, E W over all possible values of E W, obtaining as we have in eqn P (E S = D S(E S D W (E U E S D U (E U, (4.84 Now suppose we wish to know the probability P n that the system is in a particular state n with energy E n. Clearly P n = lim E 0 probability that E S [E n, E n + E] # of S states with E S [E n, E n + E] = P (E n E D S (E n E = D W(E U E n. (4.85 D U (E U Additional remarks he formula of eqn is quite general and holds in the case where N S /N W = O(1, so long as we are in the thermodynamic limit, where the energy associated with the interface between S and W may be

20 16 CHAPER 4. SAISICAL ENSEMBLES neglected. In this case, however, one is not licensed to perform the subsequent aylor expansion, and the distribution P n is no longer of the Gibbs form. It is also valid for quantum systems 6, in which case we interpret P n = n ϱ S n as a diagonal element of the density matrix ϱ S. he density of states functions may then be replaced by E U E n + E D W (E U E n E e S W (E U E n, E ra de δ(e ĤW W E U E n E U + E D U (E U E e S U (E U, E ra de δ(e ĤU. U E U (4.86 he off-diagonal matrix elements of ϱ S are negligible in the thermodynamic limit Averages within the OCE o compute averages within the OCE,  = r (ˆϱ  = n n  n e βe n n e βe n, (4.87 where we have conveniently taken the trace in a basis of energy eigenstates. In the classical limit, we have ϱ(ϕ = 1 Z e βĥ(ϕ, Z = r e βĥ = dµ e βĥ(ϕ, (4.88 with dµ = 1 N N! j=1 (dd q j d d p j /h d for identical particles ( Maxwell-Boltzmann statistics. hus, A = r (ϱa = dµ A(ϕ e βĥ(ϕ dµ e βĥ(ϕ. ( Entropy and free energy he Boltzmann entropy is defined by S = k B r (ˆϱ ln ˆϱ = k B P n ln P n. (4.90 he Boltzmann entropy and the statistical entropy S = k B ln D(E are identical in the thermodynamic limit. We define the Helmholtz free energy F (, V, N as 6 See.-C. Lu and. Grover, arxiv n F (, V, N = k B ln Z(, V, N, (4.91

21 4.5. ORDINARY CANONICAL ENSEMBLE (OCE 17 hence herefore the entropy is which is to say F = E S, where P n = e βf e βe n, ln P n = βf βe n. (4.92 S = k B n P n ( βf βen = F + Ĥ, (4.93 E = n P n E n = r Ĥ e βĥ r e βĥ (4.94 is the average energy. We also see that Z = r e βĥ = n e βe n = E = n E n e βe n n e βe n = β ln Z = ( βf. (4.95 β hus, F (, V, N is a Legendre transform of E(S, V, N, with df = S d p dv + µ dn, (4.96 which means ( F S = V,N ( F, p = V,N ( F, µ = + N,V. ( Fluctuations in the OCE In the OCE, the energy is not fixed. It therefore fluctuates about its average value E = Ĥ. Note that E β = k B 2 E = 2 ln Z β 2 ( = r Ĥ2 e βĥ r e r Ĥ e βĥ βĥ r e βĥ = Ĥ 2 Ĥ 2. 2 (4.98 hus, the heat capacity is related to the fluctuations in the energy, just as we saw at the end of 4.4: ( E C V = = 1 ( Ĥ2 k B 2 Ĥ 2 (4.99 V,N For the nonrelativistic ideal gas, we found C V energy to the energy itself is ( Ĥ 2 Ĥ = = d 2 Nk B, hence the ratio of RMS fluctuations in the kb 2 C V d 2 Nk B = 2 Nd, (4.100

22 18 CHAPER 4. SAISICAL ENSEMBLES and the ratio of the RMS fluctuations to the mean value vanishes in the thermodynamic limit. he full distribution function for the energy is hus, P (E = δ(e Ĥ r δ(e Ĥ e βĥ = = 1 r e βĥ Z D(E e βe. (4.101 P (E = e β[e S(E] de e β[e S(E ], (4.102 where S(E = k B ln D(E is the statistical entropy. Let s write E = E + δe, where E extremizes the combination E S(E, i.e. the solution to S (E = 1, where the energy derivative of S is performed at fixed volume V and particle number N. We now expand S(E + δe to second order in δe, obtaining Recall that S (E = ( 1 E = 1. hus, 2 C V S(E + δe = S(E + δe ( δe C V +... (4.103 E S(E = E S(E + (δe2 2 C V + O ( (δe 3. (4.104 Applying this to both numerator and denominator of eqn , we obtain 7 [ ] P (E = N exp (δe2 2k B 2, (4.105 C V where N = (2πk B 2 C V 1/2 is a normalization constant which guarantees de P (E = 1. Once again, we see that the distribution is a Gaussian centered at E = E, and of width ( E RMS = k B 2 C V. his is a consequence of the Central Limit heorem hermodynamics revisited he average energy within the OCE is E = n E n P n, (4.106 and therefore de = E n dp n + n n = dq dw, P n de n ( In applying eqn to the denominator of eqn , we shift E by E and integrate over the difference δe E E, retaining terms up to quadratic order in δe in the argument of the exponent.

23 4.5. ORDINARY CANONICAL ENSEMBLE (OCE 19 where dw = n P n de n (4.108 dq = n E n dp n. (4.109 Finally, from P n = Z 1 e E n /k B, we can write with which we obtain E n = k B ln Z k B ln P n, (4.110 dq = n E n dp n = k B ln Z dp n k B ln P n dp n n n = d ( k B P n ln P n = ds. n (4.111 Note also that dw = n P n de n = n P n ( i E n X i dx i (4.112 = n,i P n n Ĥ X i n dx i i F i dx i, so the generalized force F i conjugate to the generalized displacement dx i is F i = n E P n Ĥ n =. (4.113 X i X i his is the force acting on the system 8. In the chapter on thermodynamics, we defined the generalized force conjugate to X i as y i F i. hus we see from eqn that there are two ways that the average energy can change; these are depicted in the sketch of fig Starting from a set of energy levels {E n } and probabilities {P n }, we can shift the energies to {E n}. he resulting change in energy ( E I = W is identified with the work done on the system. We could also modify the probabilities to {P n} without changing the energies. he energy change in this case is the heat absorbed by the system: ( E II = Q. his provides us with a statistical and microscopic interpretation of the First Law of hermodynamics. 8 In deriving eqn , we have used the so-called Feynman-Hellman theorem of quantum mechanics: d n Ĥ n = n dĥ n, if n is an energy eigenstate.

24 20 CHAPER 4. SAISICAL ENSEMBLES Figure 4.5: Microscopic, statistical interpretation of the First Law of hermodynamics Generalized susceptibilities Suppose our Hamiltonian is of the form where λ is an intensive parameter, such as magnetic field. hen Ĥ = Ĥ(λ = Ĥ0 λ ˆQ, (4.114 β(ĥ0 λ ˆQ Z(λ = r e (4.115 and But then from Z = e βf we have 1 Z Z λ = β 1 ( Z r ˆQ e βĥ(λ = β ˆQ. (4.116 ( Q(λ, = ˆQ F = λ. (4.117 ypically we will take Q to be an extensive quantity. We can now define the susceptibility χ as χ = 1 V Q λ = 1 V he volume factor in the denominator ensures that χ is intensive. 2 F λ 2. (4.118

25 4.6. GRAND CANONICAL ENSEMBLE (GCE 21 It is important to realize that we have assumed here that [ Ĥ 0, ˆQ] = 0, i.e. the bare Hamiltonian Ĥ 0 and the operator ˆQ commute. If they do not commute, then the response functions must be computed within a proper quantum mechanical formalism, which we shall not discuss here. Note also that we can imagine an entire family of observables { ˆQk } satisfying [ ˆQk, ˆQk ] = 0 and [Ĥ0, ˆQk ] = 0, for all k and k. hen for the Hamiltonian Ĥ ( λ = Ĥ0 k λ k ˆQk, (4.119 we have that ( Q k ( λ, = ˆQ F k = (4.120 λ k, N a, λ k k and we may define an entire matrix of susceptibilities, χ kl = 1 V Q k λ l = 1 V 2 F λ k λ l. ( Grand Canonical Ensemble (GCE Grand canonical distribution and partition function Consider once again the situation depicted in fig. 4.2, where a system S is in contact with a world W, their union U = W S being called the universe. We assume that the system s volume V S is fixed, but otherwise it is allowed to exchange energy and particle number with W. Hence, the system s energy E S and particle number N S will fluctuate. We ask what is the probability that S is in a state n with energy E n and particle number N n. his is given by the ratio P n = lim E 0 D W (E U E n, N U N n E D U (E U, N U E = # of states accessible to W given that E S = E n and N S = N n total # of states in U. (4.122 hen ln P n = ln D W (E U E n, N U N n ln D U (E U, N U = ln D W (E U, N U ln D U (E U, N U E n ln D W (E, N E E=EU N=N U N n ln D W (E, N N +... E=EU N=N U (4.123 α βe n + βµn n.

26 22 CHAPER 4. SAISICAL ENSEMBLES he constants β and µ are given by β = ln D W(E, N E = E=EU N=N U µ = k B ln D W(E, N N 1 k B ( (4.125 E=EU N=N U he quantity µ has dimensions of energy and is called the chemical potential. Nota bene: Some texts define the grand canonical Hamiltonian ˆK as ˆK Ĥ µ ˆN. (4.126 hus, P n = e α e β(e n µn n. Once again, the constant α is fixed by the requirement that n P n = 1: P n = 1 Ξ e β(e n µn n, Ξ(β, V, µ = n e β(e n µn n = r e β(ĥ µ ˆN = r e β ˆK. (4.127 hus, the quantum mechanical grand canonical density matrix is given by ˆϱ = e β ˆK. (4.128 r e β ˆK Note that [ˆϱ, ˆK ] = 0. he quantity Ξ(, V, µ is called the grand partition function. It stands in relation to a corresponding free energy in the usual way: Ξ(, V, µ e βω(,v,µ Ω = k B ln Ξ, (4.129 where Ω(, V, µ is the grand potential, also known as the Landau free energy. he dimensionless quantity z e βµ is called the fugacity. If [ Ĥ, ˆN ] = 0, the grand potential may be expressed as a sum over contributions from each N sector, viz. Ξ(, V, µ = N e βµn Z(, V, N. (4.130 When there is more than one species, we have several chemical potentials {µ a }, and accordingly we define ˆK = Ĥ a µ a ˆNa, (4.131 with Ξ = r e β ˆK as before.

27 4.6. GRAND CANONICAL ENSEMBLE (GCE Entropy and Gibbs-Duhem relation In the GCE, the Boltzmann entropy is S = k B P n ln P n n = k B n = Ω + Ĥ P n ( βω βe n + βµn n µ ˆN, (4.132 which says Ω = E S µn, (4.133 where E = n N = n E n P n = r (ˆϱ Ĥ (4.134 N n P n = r (ˆϱ ˆN. (4.135 herefore, Ω(, V, µ is a double Legendre transform of E(S, V, N, with dω = S d p dv N dµ, (4.136 which entails S = ( Ω V,µ, p = ( Ω V,µ, N = ( Ω µ,v. (4.137 Since Ω(, V, µ is an extensive quantity, we must be able to write Ω = V ω(, µ. We identify the function ω(, µ as the negative of the pressure: ( Ω V = k B Ξ = 1 Ξ V,µ Ξ n ( E = = p(, µ. V,µ E n V e β(e n µn n (4.138 herefore, Ω = pv, p = p(, µ (equation of state. (4.139 his is consistent with the result from thermodynamics that G = E S + pv = µn. aking the differential, we recover the Gibbs-Duhem relation, dω = S d p dv N dµ = p dv V dp S d V dp + N dµ = 0. (4.140

28 24 CHAPER 4. SAISICAL ENSEMBLES Generalized susceptibilities in the GCE We can appropriate the results from and apply them, mutatis mutandis, to the GCE. Suppose we have a family of observables { ˆQk } satisfying [ ˆQk, ˆQk ] = 0 and [Ĥ0, ˆQk ] = 0 and [ ˆNa, ˆQk ] = 0 for all k, k, and a. hen for the grand canonical Hamiltonian ˆK( λ = Ĥ0 a µ a ˆNa k λ k ˆQk, (4.141 we have that and we may define the matrix of generalized susceptibilities, ( Q k ( λ, = ˆQ Ω k = (4.142 λ k,µ a, λ k k χ kl = 1 V Q k λ l = 1 V 2 Ω λ k λ l. ( Fluctuations in the GCE Both energy and particle number fluctuate in the GCE. Let us compute the fluctuations in particle number. We have N = ˆN = r ˆN e β(ĥ µ ˆN = 1 ln Ξ. (4.144 r e β(ĥ µ ˆN β µ herefore, 1 β N µ = r ˆN 2 e β(ĥ µ ˆN r e β(ĥ µ ˆN ( r ˆN e 2 β(ĥ µ ˆN r e β(ĥ µ ˆN = ˆN 2 ˆN 2. (4.145 Note now that ˆN 2 ˆN 2 ˆN 2 = k B N 2 ( N µ,v where κ is the isothermal compressibility. Note: ( N (N,, V, V = = (N, µ (µ,, V (V,, µ,v (N,, V = (N,, p = N 2 V 2 ( V p (N,, p (V,, p,n = N 2 V κ. = k B V κ, ( {}}{ (V,, p, µ (N, (N,, µ (V,, µ (4.147

29 4.6. GRAND CANONICAL ENSEMBLE (GCE 25 hus, which again scales as V 1/2. ( N RMS N = kb κ V, ( Gibbs ensemble Let the system s particle number N be fixed, but let it exchange energy and volume with the world W. Mutatis mutandis, we have hen P n = lim E 0 lim V 0 D W (E U E n, V U V n E V D U (E U, V U E V ln P n = ln D W (E U E n, V U V n ln D U (E U, V U = ln D W (E U, V U ln D U (E U, V U. (4.149 E n ln D W (E, V E E=EU V =V U V n ln D W (E, V V E=EU V =V U +... (4.150 α βe n βp V n. he constants β and p are given by he corresponding partition function is β = ln D W(E, V E = E=EU V =V U p = k B ln D W(E, V V Y (, p, N = r e β(ĥ+pv = 1 V k B ( (4.152 E=EU V =V U dv e βpv Z(, V, N e βg(,p,n, (4.153 where V 0 is a constant which has dimensions of volume. he factor V0 1 in front of the integral renders Y dimensionless. Note that G(V 0 = G(V 0 + k B ln(v 0 /V 0, so the difference is not extensive and can be neglected in the thermodynamic limit. In other words, it doesn t matter what constant we choose for V 0 since it contributes subextensively to G. Moreover, in computing averages, the constant V 0 divides out in the ratio of numerator and denominator. Like the Helmholtz free energy, the Gibbs free energy G(, p, N is also a double Legendre transform of the energy E(S, V, N, viz. G = E S + pv dg = S d + V dp + µ dn, (4.154

30 26 CHAPER 4. SAISICAL ENSEMBLES which entails S = ( G p,n, V = + ( G p,n, µ = + ( G N,p. ( Statistical Ensembles from Maximum Entropy he basic principle: maximize the entropy, S = k B P n ln P n. (4.156 n µce We maximize S subject to the single constraint C = n P n 1 = 0. (4.157 We implement the constraint C = 0 with a Lagrange multiplier, λ k B λ, writing S = S k B λ C, (4.158 and freely extremizing over the distribution {P n } and the Lagrange multiplier λ. hus, We conclude that C = 0 and that δs = δs k B λ δc k B C δλ [ ] = k B ln P n λ δp n k B C δλ 0. (4.159 and we fix λ by the normalization condition n P n = 1. his gives n ln P n = ( 1 + λ, (4.160 P n = 1 Ω, Ω = n Θ(E + E E n Θ(E n E. (4.161 Note that Ω is the number of states with energies between E and E + E OCE We maximize S subject to the two constraints C 1 = n P n 1 = 0, C 2 = n E n P n E = 0. (4.162

31 4.7. SAISICAL ENSEMBLES FROM MAXIMUM ENROPY 27 We now have two Lagrange multipliers. We write 2 S = S k B λ j C j, (4.163 and we freely extremize over {P n } and {C j }. We therefore have j=1 δs ( 2 = δs k B λ1 + λ 2 E n δpn k B C j δλ j = k B n n j=1 [ ln P n λ 1 + λ 2 E n ] δp n k B 2 C j δλ j 0. j=1 (4.164 hus, C 1 = C 2 = 0 and We define λ 2 β and we fix λ 1 by normalization. his yields ln P n = ( 1 + λ 1 + λ 2 E n. (4.165 P n = 1 Z e βe n, Z = n e βe n. ( GCE We maximize S subject to the three constraints C 1 = n P n 1 = 0, C 2 = n E n P n E = 0, C 3 = n N n P n N = 0. (4.167 We now have three Lagrange multipliers. We write 3 S = S k B λ j C j, (4.168 j=1 and hence δs ( 3 = δs k B λ1 + λ 2 E n + λ 3 N n δpn k B C j δλ j = k B n n j=1 [ ln P n λ 1 + λ 2 E n + λ 3 N n ] δp n k B 3 C j δλ j 0. j=1 (4.169 hus, C 1 = C 2 = C 3 = 0 and ln P n = ( 1 + λ 1 + λ 2 E n + λ 3 N n. (4.170 We define λ 2 β and λ 3 βµ, and we fix λ 1 by normalization. his yields P n = 1 Ξ e β(e n µn n, Ξ = n e β(e n µn n. (4.171

32 28 CHAPER 4. SAISICAL ENSEMBLES 4.8 Ideal Gas Statistical Mechanics he ordinary canonical partition function for the ideal gas was computed in eqn We found Z(, V, N = 1 N! = V N N! N i=1 ( V d d x i d d p i (2π d e βp2 i /2m dp /2m 2π e βp2 N, Nd (4.172 = 1 N! λ d where λ is the thermal wavelength: λ = 2π 2 /mk B. (4.173 he physical interpretation of λ is that it is the de Broglie wavelength for a particle of mass m which has a kinetic energy of k B. In the GCE, we have Ξ(, V, µ = = e βµn Z(, V, N N=0 N=1 ( N ( 1 V e µ/k B V e µ/k B N! λ d = exp λ d. (4.174 From Ξ = e Ω/k B, we have the grand potential is Since Ω = pv (see 4.6.2, we have he number density can also be calculated: Ω(, V, µ = V k B e µ/k B / λ d. (4.175 p(, µ = k B λ d eµ/k B. (4.176 n = N V = 1 V ( Ω µ,v Combined, the last two equations recapitulate the ideal gas law, pv = Nk B. = λ d eµ/k B. ( Maxwell velocity distribution he distribution function for momenta is given by 1 g(p = N N i=1 δ(p i p. (4.178

33 4.8. IDEAL GAS SAISICAL MECHANICS 29 Figure 4.6: Maxwell distribution of speeds ϕ(v/v 0. he most probable speed is v MAX = 2 v 0. he 8 average speed is v AVG = π v 0. he RMS speed is v RMS = 3 v 0. Note that g(p = δ(p i p is the same for every particle, independent of its label i. We compute the average A = r ( Ae βĥ / r e βĥ. Setting i = 1, all the integrals other than that over p 1 divide out between numerator and denominator. We then have d 3 p g(p = 1 δ(p 1 p e βp2 1 /2m d 3 p 1 e βp2 1 /2m (4.179 = (2πmk B 3/2 e βp2 /2m. extbooks commonly refer to the velocity distribution f(v, which is related to g(p by Hence, f(v = f(v d 3 v = g(p d 3 p. (4.180 ( m 3/2 e mv2 /2k B. ( πk B his is known as the Maxwell velocity distribution. Note that the distributions are normalized, viz. d 3 p g(p = d 3 v f(v = 1. (4.182 If we are only interested in averaging functions of v = v which are isotropic, then we can define the Maxwell speed distribution, f(v, as ( m 3/2 f(v = 4π v 2 f(v = 4π v 2 e mv2 /2k B. ( πk B

34 30 CHAPER 4. SAISICAL ENSEMBLES Note that f(v is normalized according to It is convenient to represent v in units of v 0 = k B /m, in which case f(v = 1 v 0 ϕ(v/v 0, ϕ(s = he distribution ϕ(s is shown in fig Computing averages, we have hus, C 0 = 1, C 1 = C k s k = 0 0 dv f(v = 1. (4.184 ds s k ϕ(s = 2 k/2 8 π, C 2 = 3, etc. he speed averages are v k = C k ( kb m 2 π s2 e s2 /2. ( π Γ ( k 2. (4.186 k/2. (4.187 Note that the average velocity is v = 0, but the average speed is v = 8k B /πm. he speed distribution is plotted in fig Equipartition he Hamiltonian for ballistic (i.e. massive nonrelativistic particles is quadratic in the individual components of each momentum p i. here are other cases in which a classical degree of freedom appears quadratically in Ĥ as well. For example, an individual normal mode ξ of a system of coupled oscillators has the Lagrangian L = 1 ξ ω2 0 ξ 2, (4.188 where the dimensions of ξ are [ξ] = M 1/2 L by convention. he Hamiltonian for this normal mode is then Ĥ = p ω2 0 ξ 2, (4.189 from which we see that both the kinetic as well as potential energy terms enter quadratically into the Hamiltonian. he classical rotational kinetic energy is also quadratic in the angular momentum components. Let us compute the contribution of a single quadratic degree of freedom in Ĥ to the partition function. We ll call this degree of freedom ζ it may be a position or momentum or angular momentum and we ll write its contribution to Ĥ as Ĥ ζ = 1 2 Kζ2, (4.190

35 4.9. SELECED EXAMPLES 31 where K is some constant. Integrating over ζ yields the following factor in the partition function: dζ e βkζ2 /2 = ( 2π 1/2. (4.191 Kβ he contribution to the Helmholtz free energy is then ( K F ζ = 1 2 k B ln, ( πk B and therefore the contribution to the internal energy E is E ζ = ( 1 β Fζ = β 2β = 1 2 k B. (4.193 We have thus derived what is commonly called the equipartition theorem of classical statistical mechanics: o each degree of freedom which enters the Hamiltonian quadratically is associated a contribution 1 2 k B to the internal energy of the system. his results in a concomitant contribution of 1 2 k B to the heat capacity. We now see why the internal energy of a classical ideal gas with f degrees of freedom per molecule is E = 1 2 fnk B, and C V = 1 2 Nk B. his result also has applications in the theory of solids. he atoms in a solid possess kinetic energy due to their motion, and potential energy due to the spring-like interatomic potentials which tend to keep the atoms in their preferred crystalline positions. hus, for a three-dimensional crystal, there are six quadratic degrees of freedom (three positions and three momenta per atom, and the classical energy should be E = 3Nk B, and the heat capacity C V = 3Nk B. As we shall see, quantum mechanics modifies this result considerably at temperatures below the highest normal mode (i.e. phonon frequency, but the high temperature limit is given by the classical value C V = 3νR (where ν = N/N A is the number of moles derived here, known as the Dulong-Petit limit. 4.9 Selected Examples Spins in an external magnetic field Consider a system of N S spins, each of which can be either up (σ = +1 or down (σ = 1. he Hamiltonian for this system is N S Ĥ = µ 0 H σ j, (4.194 where now we write Ĥ for the Hamiltonian, to distinguish it from the external magnetic field H, and µ 0 is the magnetic moment per particle. We treat this system within the ordinary canonical ensemble. he partition function is Z = e βĥ = ζ N S, (4.195 σ 1 σ NS j=1

36 32 CHAPER 4. SAISICAL ENSEMBLES where ζ is the single particle partition function: ζ = σ=±1 ( e µ 0 Hσ/k B µ0 H = 2 cosh. (4.196 k B he Helmholtz free energy is then [ ( ] µ0 H F (, H, N S = k B ln Z = N S k B ln 2 cosh. (4.197 k B he magnetization is he energy is ( ( F µ0 H M = = N H S µ 0 tanh, N k B S E = β. (4.198 ( µ0 H βf = NS µ 0 H tanh(. (4.199 k B Hence, E = HM, which we already knew, from the form of Ĥ itself. Each spin here is independent. he probability that a given spin has polarization σ is P σ = e βµ 0 Hσ e βµ 0 H + e βµ 0 H. (4.200 he total probability is unity, and the average polarization is a weighted average of σ = +1 and σ = 1 contributions: ( µ0 H P + P = 1, σ = P P = tanh. (4.201 k B At low temperatures µ 0 H/k B, we have P 1 e 2µ 0 H/k B. At high temperatures > µ 0 H/k B, ( the two polarizations are equally likely, and P σ σµ 0 H k B. he isothermal magnetic susceptibility is defined as χ = 1 ( M N S H ( = µ2 0 µ0 H k B sech2 k B. (4.202 (ypically this is computed per unit volume rather than per particle. At H = 0, we have χ = µ 2 0 /k B, which is known as the Curie law. Aside he energy E = HM here is not the same quantity we discussed in our study of thermodynamics. In fact, the thermodynamic energy for this problem vanishes! Here is why. o avoid confusion, we ll need to invoke a new symbol for the thermodynamic energy, E. Recall that the thermodynamic energy

37 4.9. SELECED EXAMPLES 33 Figure 4.7: When entropy decreases with increasing energy, the temperature is negative. ypically, kinetic degrees of freedom prevent this peculiarity from manifesting in physical systems. E is a function of extensive quantities, meaning E = E(S, M, N S. It is obtained from the free energy F (, H, N S by a double Legendre transform: E(S, M, N S = F (, H, N S + S + HM. (4.203 Now from eqn we derive the entropy [ S = F = N S k B ln 2 cosh ( µ0 H k B ] N S µ 0 H ( µ0 H tanh. (4.204 k B hus, using eqns and 4.198, we obtain E(S, M, N S = 0. he potential confusion here arises from our use of the expression F (, H, N S. In thermodynamics, it is the Gibbs free energy G(, p, N which is a double Legendre transform of the energy: G = E S + pv. By analogy, with magnetic systems we should perhaps write G = E S HM, but in keeping with many textbooks we shall use the symbol F and refer to it as the Helmholtz free energy. he quantity we ve called E in eqn is in fact E = E HM, which means E = 0. he energy E(S, M, N S vanishes here because the spins are noninteracting Negative temperature (! Consider again a system of N S spins, each of which can be either up (+ or down (. Let N σ be the number of sites with spin σ, where σ = ±1. Clearly N + + N = N S. We now treat this system within the microcanonical ensemble.

38 34 CHAPER 4. SAISICAL ENSEMBLES he energy of the system is E = HM, (4.205 where H is an external magnetic field, and M = (N + N µ 0 is the total magnetization. We now compute S(E using the ordinary canonical ensemble. he number of ways of arranging the system with N + up spins is ( NS Ω =, (4.206 N + hence the entropy is } S = k B ln Ω = N S k B {x ln x + (1 x ln(1 x (4.207 in the thermodynamic limit: N S, N +, x = N + /N S constant. Now the magnetization is M = (N + N µ 0 = (2N + N S µ 0, hence if we define the maximum energy E 0 N S µ 0 H, then We therefore have We now have E = M = 1 2x = x = E 0 E. (4.208 E 0 N S µ 0 2E 0 S(E, N S = N S k B [ (E0 E 2E 0 ln ( 1 S = = S E N x S ( E0 E 2E 0 + x E = N S k B 2E 0 ( ( ] E0 + E E0 + E ln. ( E 0 2E 0 ( E0 E ln. (4.210 E 0 + E We see that the temperature is positive for E 0 E < 0 and is negative for 0 < E E 0. What has gone wrong? he answer is that nothing has gone wrong all our calculations are perfectly correct. his system does exhibit the possibility of negative temperature. It is, however, unphysical in that we have neglected kinetic degrees of freedom, which result in an entropy function S(E, N S which is an increasing function of energy. In this system, S(E, N S achieves a maximum of S max = N S k B ln 2 at E = 0 (i.e. x = 1 2, and then turns over and starts decreasing. In fact, our results are completely consistent with eqn : the energy E is an odd function of temperature. Positive energy requires negative temperature! Another example of this peculiarity is provided in the appendix in Adsorption PROBLEM: A surface containing N S adsorption sites is in equilibrium with a monatomic ideal gas. Atoms adsorbed on the surface have an energy and no kinetic energy. Each adsorption site can accommodate at most one atom. Calculate the fraction f of occupied adsorption sites as a function of the gas density n, the temperature, the binding energy, and physical constants. he grand partition function for the surface is N S Ξ surf = e Ω surf /k B = j=0 ( NS j e j(µ+ /k B = ( 1 + e µ/k B e /k B N S. (4.211

39 4.9. SELECED EXAMPLES 35 Figure 4.8: he monomers in wool are modeled as existing in one of two states. he low energy undeformed state is A, and the higher energy deformed state is B. Applying tension induces more monomers to enter the B state. he fraction of occupied sites is f = ˆN surf N S = 1 N S Ω surf µ = e µ/kb e µ/k B + e /k B. (4.212 Since the surface is in equilibrium with the gas, its fugacity z = exp(µ/k B and temperature are the same as in the gas. SOLUION: For a monatomic ideal gas, the single particle partition function is ζ = V λ 3, where λ = 2π 2 /mk B is the thermal wavelength. hus, the grand partition function, for indistinguishable particles, is ( Ξ gas = exp V λ 3 eµ/k B. (4.213 he gas density is n = ˆN gas V = 1 V Ω gas µ = λ 3 eµ/k B. (4.214 We can now solve for the fugacity: z = e µ/k B = nλ 3. hus, the fraction of occupied adsorption sites is f = Interestingly, the solution for f involves the constant. nλ 3 nλ 3 + e /k B. (4.215 It is always advisable to check that the solution makes sense in various limits. First of all, if the gas density tends to zero at fixed and, we have f 0. On the other hand, if n we have f 1, which also makes sense. At fixed n and, if the adsorption energy is (, then once again f = 1 since every adsorption site wants to be occupied. Conversely, taking ( + results in n 0, since the energetic cost of adsorption is infinitely high Elasticity of wool Wool consists of interlocking protein molecules which can stretch into an elongated configuration, but reversibly so. his feature gives wool its very useful elasticity. Let us model a chain of these proteins by assuming they can exist in one of two states, which we will call A and B, with energies ε A and ε B and lengths l A and l B. he situation is depicted in fig We model these conformational degrees of

40 36 CHAPER 4. SAISICAL ENSEMBLES freedom by a spin variable σ = ±1 for each molecule, where σ = +1 in the A state and σ = 1 in the B state. Suppose a chain consisting of N monomers is placed under a tension τ. We then have Ĥ = N j=1 [ ] ε A δ σj,+1 + ε B δ σj, 1. (4.216 Similarly, the length is ˆL = N j=1 [ ] l A δ σj,+1 + l B δ σj, 1. (4.217 he Gibbs partition function is Y = r e ˆK/k B, with ˆK = Ĥ τ ˆL : ˆK = N j=1 [ ] ε A δ σj,+1 + ε B δ σj, 1, (4.218 where ε A ε A τl A and ε B ε B τl B. At τ = 0 the A state is preferred for each monomer, but when τ exceeds τ, defined by the relation ε A = ε B, the B state is preferred. One finds τ = ε B ε A l B l A. (4.219 Once again, we have a set of N noninteracting spins. he partition function is Y = ζ N, where ζ is the single monomer partition function, ζ = r e βĥ, where is the single spin Hamiltonian. hus, It is convenient to define the differences in which case the partition function Y is ĥ = ε A δ σj,1 + ε B δ σj, 1 (4.220 ζ = r e βĥ = e β ε A + e β ε B, (4.221 ε = ε B ε A, l = l B l A, ε = ε B ε A (4.222 Y (, τ, N = e Nβ ε A [1 + e β ε] N (4.223 [ ] G(, τ, N = N ε A Nk B ln 1 + e ε/k B (4.224 he average length is ( G L = ˆL = τ,n N l = Nl A + e ( ε τ l/k B + 1. (4.225

41 4.9. SELECED EXAMPLES 37 Figure 4.9: Upper panel: length L(τ, for k B / ε = 0.01 (blue, 0.1 (green, 0.5 (dark red, and 1.0 (red. Bottom panel: dimensionless force constant k/n( l 2 versus temperature. he polymer behaves as a spring, and for small τ the spring constant is k = τ L = τ=0 4k ( B ε N( l 2 cosh2. ( k B he results are shown in fig Note that length increases with temperature for τ < τ and decreases with temperature for τ > τ. Note also that k diverges at both low and high temperatures. At low, the energy gap ε dominates and L = Nl A, while at high temperatures k B dominates and L = 1 2 N(l A+l B Noninteracting spin dimers Consider a system of noninteracting spin dimers as depicted in fig Each dimer contains two spins, and is described by the Hamiltonian Ĥ dimer = J σ 1 σ 2 µ 0 H (σ 1 + σ 2. (4.227 Here, J is an interaction energy between the spins which comprise the dimer. If J > 0 the interaction is ferromagnetic, which prefers that the spins are aligned. hat is, the lowest energy states are and

42 38 CHAPER 4. SAISICAL ENSEMBLES. If J < 0 the interaction is antiferromagnetic, which prefers that spins be anti-aligned: and. 9 Suppose there are N d dimers. hen the OCE partition function is Z = ζ N d, where ζ(, H is the single dimer partition function. o obtain ζ(, H, we sum over the four possible states of the two spins, obtaining hus, the free energy is ζ = r e Ĥdimer /k B ( = 2 e J/k B + 2 e J/k B 2µ0 H cosh. k B [ ( ] F (, H, N d = N d k B ln 2 N d k B ln e J/k B + e J/k B 2µ0 H cosh. (4.228 k B he magnetization is ( F M = = 2N H d µ 0,N d e J/k B sinh e J/k B + e J/k B cosh ( 2µ0 H k B ( 2µ0 H k B (4.229 It is instructive to consider the zero field isothermal susceptibility per spin, χ = 1 M 2N d H = µ2 0 H=0 k B 2 e J/k B e J/k B + e J/k B. (4.230 he quantity µ 2 0 /k B is simply the Curie susceptibility for noninteracting classical spins. Note that we correctly recover the Curie result when J = 0, since then the individual spins comprising each dimer are in fact noninteracting. For the ferromagnetic case, if J k B, then we obtain χ (J k B 2µ2 0 k B. (4.231 his has the following simple interpretation. When J k B, the spins of each dimer are effectively locked in parallel. hus, each dimer has an effective magnetic moment µ eff = 2µ 0. On the other hand, there are only half as many dimers as there are spins, so the resulting Curie susceptibility per spin is 1 2 (2µ 0 2 /k B. When J k B, the spins of each dimer are effectively locked in one of the two antiparallel configurations. We then have χ ( J k B 2µ2 0 k B e 2 J /k B. (4.232 In this case, the individual dimers have essentially zero magnetic moment. 9 Nota bene we are concerned with classical spin configurations only there is no superposition of states allowed in this model!

43 4.10. SAISICAL MECHANICS OF MOLECULAR GASES 39 Figure 4.10: A model of noninteracting spin dimers on a lattice. Each red dot represents a classical spin for which σ j = ± Statistical Mechanics of Molecular Gases Separation of translational and internal degrees of freedom he states of a noninteracting atom or molecule are labeled by its total momentum p and its internal quantum numbers, which we will simply write with a collective index α, specifying rotational, vibrational, and electronic degrees of freedom. he single particle Hamiltonian is then with he partition function is ĥ = p2 2m + ĥint, (4.233 ĥ k, α ( 2 k 2 = 2m + ε α k, α. (4.234 ζ = r e βĥ = p e βp2 /2m j g j e βε j. (4.235 Here we have replaced the internal label α with a label j of energy eigenvalues, with g j being the degeneracy of the internal state with energy ε j. o do the p sum, we quantize in a box of dimensions L 1 L 2 L d, using periodic boundary conditions. hen ( 2π n1 p =, 2π n 2,..., 2π n d, (4.236 L 1 L 2 L d where each n i is an integer. Since the differences between neighboring quantized p vectors are very tiny, we can replace the sum over p by an integral: p d d p p 1 p d (4.237