Statistical Mechanics

Equilibrium statistical mechanics is the basic ingredient to compute thermodynamic properties of macroscopic systems based on the knowledge of their microscopic dynamics.

The equilibrium probability distribution of different microstates in any system at thermal equilibrium is determined uniquely by the energy of these states, in an extremely simple way: at a given temperature T, the probability P(E) of a given state of energy E is proportional to exp[-E/(k_BT)].

Boltzmann, the discoverer (inventor?) of this statistical distribution, leaves his name to the constant k_B=1.38064852×10^-23 J/K, which is basically the conversion factor between temperature and energy. The quantity β = 1/(k_BT) provides the angular coefficient -β of the straight line representing Boltzmann's equilibrium probability distribution in a lin-log plot.

The Boltzmann probability distribution is normalized to unity by dividing each exp(-E_n/k_BT) factor by the sum

This basic idea finds applications in all problems of equilibrium statistical mechanics. For an arbitrary system, one is supposed to carry out the complete calculation of the whole spectrum, and then use it to get the partition function, and eventually all the thermodynamic averages. In practice this project can be carried out analytically on fairly small/simple problems, such as, e.g., two or three coupled spins.

A lot can be done also for the wide category of non-interacting systems. A good example of such kind of system is the electromagnetic fields at equilibrium in a cavity: this is likely the only properly non-interacting system in physics: its exact thermodynamic properties can be computed with the techniques of statistical mechanics. Other examples include "isolated" spin systems, or internal excitations of isolated atoms, or isolated molecules. Here the "isolation" property is an approximation. In highly rarefied systems, this approximation is usually quite good. For all these non-interacting systems, the Hamiltonian is the sum of several (often identical) Hamiltonians, one for each isolated quantum system (IQS). The Hamiltonians really are the same for the rotation of molecules or the spin states of atoms in a uniform magnetic field. Instead, when the normal modes of the electromagnetic field are considered, they are all quantum harmonic oscillators but with generally different frequencies. Each IQS reaches equilibrium separately, independently of all other IQSs: its states are occupied according to the Boltzmann probability distribution P(E) governed exclusively by the "single-IQS" energy E. When the IQSs are all the same, as in the case of a gas of equal atoms or molecules, this implies that the average number of IQSs (also called the "population") in a given quantum state n is the total number N of IQSs multiplied by the probability of that state: N×P(E_n).

Ideal fermion and boson systems

Other popular non-interacting systems are collections of identical fermions or bosons. They are relevant to existing physical systems to the extent that their particle-particle interactions can be neglected.

Boltzmann (grancanonical) statistics applied to the total energy of a set of non-interacting fermions (governed by Pauli's antisymmetrization requirement) gives a single-particle probability distribution, the Fermi-Dirac distribution, which turns out different from the one occurring without the antisymmetrization requirement.

Similarly, the single-particle Bose-Einstein statistics derived from the Boltzmann distribution of total energies of a set of bosons is also quite different from the way it would be without the symmetrization requirement.

Note: the justification of Bose-Einstein statistics given in Sect. 11-2 of Eisberg-Resnick Quantum Physics is incorrect. In particular, the comparison of Eq. (11-1) (wrong, unnormalized) with Eq. (11-2) (OK) is unfair. Consequently, the enhancement factors n! of (11-3) and previous formulas are pure fantasy, and the conclusion if there are already n bosons in a quantum state, the probability of one more joining them is larger by an enhancement factor of (1+n) than it would be if there where no quantum mechanical indistinguishableness requirements is plain wrong. An extra boson joins the quantum states available with a given probability which only depends on the overlap of its initial state with the Hamiltonian eigenstates available (and is normalized so that the sum over all the states available is one). It does not care of the possible presence of other bosons (as long as they are not interacting of course!). If it did, the probability could not be normalized to unity.

Sadly, also "The Feynman" Lectures in Physics Vol. III Chap. 4 falls in this same pitfall.

Exercises on spin systems

1. Compute the average energy U(T) and magnetization M(T), the zero-field susceptibility per unit volume χ(T) and molar χ_mol(T), and the molar specific heat at fixed field for a gas of non-interacting spin-½ particles of magnetic moment gµ_B and (low) density N/V.
   RESULT: χ(T) = (N/V) (gµ_B)² β/4, χ_mol(T) = N_A (gµ_B)² β/4, where N_A is the number of particles in a mole N_A=6.022×10²³ mol^-1
2. Same as previous, for a gas of spin-1 particles.
   RESULT: Same χ's as previous, replacing 1/4 with 2/3
3. Same as previous, for a gas of particles where there is no quantization of the angular momentum ("infinite spin", the classical limit), and the magnetic moment is µ.
   RESULT: χ(T)=µ²(N/V)β/3
   Note: This same calculation can be carried out for a gas of dipolar molecules of moment p₀, in a uniform electric field E.
4. The Heisenberg model, appropriate for interacting localized spins, can be written
   H^Heis = - J ∑_<i,j> S_i·S_j - b ∑_i S^z_i
   Here: S_i is the spin operator divided by ℏ for site i; <i,j> indicates nearest-neighbor sites; J is the nearest-neighbor exchange energy (assume J>0, ferromagnetic exchange); and b is the energy of coupling to an external magnetic field B, assumed uniform and directed along the positive z direction: b=gµ_BB^z.
   1. List all the quantum states where this model is defined on the basis where the S^z_i operators are diagonal, for a lattice of N=2 sites.
   2. For the case S=½, diagonalize H^Heis on the basis of previous point, i.e. list all eigenstates and eigenenergies. Use symmetry considerations. Consider first the case b=0.
      RESULT: E₁=-J/4+b, E₂=-J/4, E₃=-J/4-b, and E₄=3J/4.
   3. For S=½ write down the exact partition function for N=2 and the average spin z-component at site 1 [S^z₁] as a function of temperature and b. Show that [S^z₁] equals [S^z₂] (of course!).
      RESULT: [S^z₁] = sinh(β b) / [1 + exp(-β J) + 2 cosh(β b)]
   4. Try to repeat the above points for N=3 sites where each site is nearest neighbor of the two others (3 spins on an equilateral triangle).
   5. Describe the ground state for a large number N of sites, b=0, J>0, no disconnected block of spins. Show in particular that its degeneracy is 2 N S + 1, where S is the spin at each site (e.g. S=½, but not necessarily). Discuss this degeneracy in terms of the symmetry of the original model.
   6. For the case of large N (hard to treat exactly: explain why!), build a mean-field approximate theory. Replace the S_j operator with its average value (and assume it is oriented along z, like the external field). Assume further that all sites are equivalent and call m=[S^z_j] (obviously -½<m<½, but at this stage m is still an unknown quantity). Rewrite the approximate Hamiltonian as:
      H^MF = - (m J n_nn/2 + b) ∑_i S^z_i ,
      where n_nn is the number of nearest neighbor sites to any given site i, and the factor 1/2 avoids double counting the J interactions. This mean-field Hamiltonian describes a set of uncorrelated spins in the effective field created by the actual external field plus the average effect of the neighboring atoms.

      Depending on the spin of the original model, this mean field approximation allows us to describe the Heisenberg model as a set of IQSs, whose thermodynamics has been studied in previous problems 1, 2 or 3. Consider e.g. the case S=½: apply the results for the magnetization of exercise 1 and write down the self-consistent MF equation for m=[S^z_i].
      RESULT: m = ½ tanh[½β(½ n_nn J m + b)]

   7. Study graphically the case b=0, and discover a critical temperature
      T_c=n_nn J/(8 k_B) .
      Show that the MF theory of the previous point describes a nonmagnetic state (m=0) for T > T_c, and a spontaneously-magnetized state (m≠0) for T < T_c. Verify that the nonmagnetic state becomes an unstable solution of the self-consistent equation for T < T_c. Verify that for T→0, m→±½. By expanding the tanh to third order in the self-consistent MF equation, show that when T→T_c from the magnetized state, m→0, with a temperature dependence of the type (T_c-T)^x. Determine the exponent x. Discuss the transition at T_c in terms of properties of magnetic materials.
      RESULT: x = ½
   8. Study the case b>0: show that there exists a single stable solution m(b)>0 for all temperatures. Verify by taking a small b that m(b)≥m(0) for all temperatures.
   9. By expanding the self-consistent MF equation, compute the magnetic susceptibility ∂M/∂B(B=0) of the T>T_c state, thus demonstrating the expected paramagnetic high-temperature behavior.
      RESULT: χ(T)=g²µ_B²(N/V) / (4k_BT - n_nn J/2)
   10. Explain why the calculation of the susceptibility in the low-temperature phase does not make sense.
   11. Redo the same treatment as above for the S=1 case.
5. The Heisenberg model with J<0 describes some properties of a large class of antiferromagnetic compounds, such as NiO, for example. Assume that the lattice is bipartite (it can be divided into two sublattices, such that all the nearest neighbors of any site belong to the other sublattice).
   1. For J<0 revise the results of the N=2-sites case (points a-c of the previous exercise).
   2. Assume that the magnetization is uniform in each of the two sublattices, but not necessarily equal. Assume further that both magnetizations [S_{s.l. 1 or 2}] are directed along z (this assumption is much less reasonable than in the ferromagnetic case, and indeed, if one allows for canted states, (s)he will find that they are lower in energy than the ones considered here), and call them m₁=[S^z₁];, m₂=[S^z₂] (sites 1 and 2 are assumed to belong so sublattices 1 and 2 respectively). Make the same mean-field approximation of the ferro case, separate symmetrically the interactions for the spins at sublattices 1 and 2, and show that the mean-field Hamiltonian can be rewritten as
      H^MF = (m₂ |J| n_nn/2 - b) ∑_{i in s.l. 1} S^z_i + (m₁ |J| n_nn/2 - b) ∑_{i in s.l. 2} S^z_i
      where again the factor 1/2 avoids double counting the |J|=-J interactions.
   3. In analogy to the previous exercise, write down the self-consistent MF equations for m₁ and m₂, for the case S=½.
      RESULT:
      m₁ = ½ tanh[½ β (b -½ n_nn J m₂)]
      m₂ = ½ tanh[½ β (b -½ n_nn J m₁)]
   4. Study the b=0 case: combine the MF equations into a single equation for, say m₁, and study it graphically. Find a transition between a high-temperature nonmagnetic state and a low-temperature antiferromagnetically-ordered state with m₂=-m₁, at the temperature T_N=n_nn |J|/(8 k_B). Verify that for T→0, m₁→±½.
   5. Study the b>0 case: show that now m₂≠-m₁. Conclude that the total magnetization (proportional to (m₁+m₂)/2) is positive all nonzero temperatures.
   6. By expanding the self-consistent MF equations, compute the magnetic susceptibility ∂M/∂B(B=0) of the T>T_c state, thus demonstrating the expected paramagnetic high-temperature behavior.
   7. Explain the technical difficulty of doing the calculation of the susceptibility in the low-temperature phase (which is paramagnetic anyway).
   8. Redo the same treatment as above for the S=1 case.

More exercises of application of the Boltzmann equilibrium distribution to IQSs are collected in the page about molecules.

Exercises on quantum systems in equilibrium with electromagnetic fields

In an excitation transition (E₂>E₁), clearly R_1→2 is proportional to the density of electromagnetic energy per unit volume and per unit energy spectral interval ρ(ε), at the transition energy ε=(E₂-E₁)/h in the region where these systems are placed. Therefore R_1→2 = B₁₂ ρ(ε). In a decay transition the spontaneous and stimulated emission rates add up as follows: R_2→1 = A₂₁ + B₂₁ρ(ε). Both these relations do not require thermodynamic equilibrium to hold. At thermodynamic equilibrium, however, one can take the appropriate equilibrium distribution u(ε,T) = π^-2 c^-3 ℏ^-3 ε³/[exp(βε) - 1] for the radiant energy density ρ(ε).

1. Consider atomic H gas in in thermodynamic equilibrium with radiation. Determine:
   1. The ratio between the probabilities of spontaneous and stimulated emission at T=300K for the radiation produced by the decay of the 2p levels.
   2. The temperature at which these probabilities become equal.
   
   RESULT: A₂₁ / [B₂₁ u(ε,T)]= exp[ε / (k_BT)] -1 = 2.6×10¹⁷¹, T_equal = ε / (k_B ln 2) = 170837 K.
2. Assume that approximately 10% of the mass of the Sun (2×10³⁰ kg) consists of gaseous H in its 1s or 2p states. Imagine that this gas is held all at the temperature T=6000 K. How many atoms are spontaneously decaying and how many are decaying via stimulated emission in a second?
   RESULT: Determine N_1s and N_2p by Boltzmann statistics. Use the rate of spontaneous-emission computed in the exercise on the atomic-H bulb A_2p-1s=6.2649×10⁸ s^-1 to evaluate the spontaneous emission total rate N_2p A_2p-1s = 6.08×10⁵⁶ s^-1 and the result of previous exercise to get N_2p B_2p-1s ρ(ε)=1.632×10⁴⁸ s^-1.
3. An oven (T=1200°C) contains Na vapor in equilibrium with the thermal radiation. Compute the ratio between the stimulated and spontaneous emission rates for the 3p decay. The associated wavelengths are 589,0 and 589,6 nm. Compute also the temperatures at which these probabilities become equal.

A paradox

How do you solve this paradox?