Graduate Statistical Mechanics
Leon Hostetler
December 16, 2018 Version 0.5 Contents
Preface v
1 The Statistical Basis of Thermodynamics1 1.1 Counting States...... 1 1.2 Connecting to Thermodynamics...... 4 1.3 Intensive and Extensive Parameters ...... 7 1.4 Averages...... 8 1.5 Summary: The Statistical Basis of Thermodynamics...... 9
2 Ensembles and Classical Phase Space 13 2.1 Classical Phase Space ...... 13 2.2 Liouville’s Theorem ...... 14 2.3 Microcanonical Ensemble ...... 16 2.4 Canonical Ensemble ...... 20 2.5 Grand Canonical Ensemble ...... 29 2.6 Summary: Ensembles and Classical Phase Space ...... 32
3 Thermodynamic Relations 37 3.1 Thermodynamic Potentials ...... 37 3.2 Fluctuations...... 44 3.3 Thermodynamic Response Functions...... 48 3.4 Minimizing and Maximizing Thermodynamic Potentials...... 49 3.5 Summary: Thermodynamic Relations ...... 51
4 Quantum Statistics 55 4.1 The Density Matrix ...... 55 4.2 Indistinguishable Particles...... 57 4.3 Thermodynamic Properties ...... 59 4.4 Chemical Potential...... 63 4.5 Summary: Quantum Statistics ...... 65
5 Boltzmann Gases 67 5.1 Kinetic Theory...... 68 5.2 Equipartition and Virial Theorems...... 70 5.3 Internal Degrees of Freedom...... 71 5.4 Chemical Equilibrium ...... 74 5.5 Summary: Boltzmann Gases ...... 76
6 Ideal Bose Gases 79 6.1 General Bose Gases...... 79 6.2 Bose-Einstein Condensation...... 80 6.3 Photons and Black-body Radiation...... 84 6.4 Phonons...... 85 Contents iii
6.5 Summary: Ideal Bose Gases...... 88
7 Ideal Fermi Gases 91 7.1 General Fermi Gases...... 91 7.2 Pauli Spin Susceptibility...... 93 7.3 Sommerfeld Expansion...... 95 7.4 Summary: Ideal Fermi Gases ...... 98
8 Interacting Gases 101 8.1 Correlation Functions ...... 101 8.2 Virial Expansion ...... 104 8.3 Summary: Interacting Gases ...... 111
9 Phase Transitions 113 9.1 Liquid-Gas Phase Transition ...... 116 9.2 Ising Model...... 119 9.3 Critical Exponents...... 124 9.4 Summary: Phase Transitions ...... 128
10 Landau-Ginzburg Field Theory 133 10.1 Introduction...... 133 10.2 Phase Transitions...... 135 10.3 Fluctuations and Correlations...... 137 10.4 Symmetry Breaking ...... 139 10.5 The Ginzburg Criterion ...... 143 10.6 Scaling Hypothesis...... 145 10.7 Summary: Landau-Ginzburg Field Theory...... 149
11 Non-Equilibrium Statistical Mechanics 153 11.1 BBGKY...... 153 11.2 Hydrodynamics...... 156 11.3 Summary: Non-Equilibrium Statistical Mechanics ...... 159
12 Review 161 12.1 Entropy...... 161 12.2 Ensembles...... 161 12.3 Classical Limit ...... 162 12.4 Thermodynamic Relations...... 164 12.5 Quantum Statistics...... 165 12.6 Interacting Gases...... 167 12.7 Phase Transitions...... 167 12.8 Landau-Ginzburg Field Theory...... 167 Preface
About These Notes
These are my class notes from the graduate statistical mechanics class I took at Michi- gan State University with Professor Luke Roberts. The primary textbook we used was Statistical Mechanics by Pathria and Beale. My class notes can be found at www.leonhostetler.com/classnotes Please bear in mind that these notes will contain errors. Any errors are certainly my own. If you find one, please email me at [email protected] with the name of the class notes, the page on which the error is found, and the nature of the error. If you include your name, I will probably list your name on the thank you page if I decide to compile and sell my notes. This work is currently licensed under a Creative Commons Attribution-NonCommercial- NoDerivatives 4.0 International License. That means you are free to copy and distribute this document in whole for noncommercial use, but you are not allowed to distribute derivatives of this document or to copy and distribute it for commercial reasons.
Updates
Last Updated: December 16, 2018 Version 0.5: (Dec. 16, 2018) First upload. Chapter 1
The Statistical Basis of Thermodynamics
1.1 Counting States Tip In general, we often want to know how to go from the Hamiltonian H of a system to the physical, macroscopic properties of the system. These properties typically include Often we will work in units where kB = 1. • N: The number of particles the system is confined to. This is typically a large number ∼ 1023 • V : The volume of the system • E: The energy of the system
46 The classical Hamiltonian H({qi, pi}) will have a huge number of variables, e.g. a 10 - dimensional space. Generally, we don’t know all the coordinates qi and momenta pi of the particles in our system, but we may know the overall volume V of the system and its energy E. There are typically many microstates of a system each with the same overall energy E. In fact, the quantity Ω(N,V,E), denotes the number of microstates that correspond to the single macrostate with particle number N, volume V , and energy E. The number Ω tells us something about our uncertainty of the microscopic state of the system. For example, if Ω = 1, i.e. there is only a single microstate with N,V,E, then there is no uncertainty because we know exactly which microstate the system must be in. Typically, however, Ω ∼ 1023, which means we are very uncertain about exactly which microstate the system is in. Another useful quantity is the Boltzmann entropy
S = kB ln Ω(N,V,E).
The entropy is also a measure of the uncertainty of the system. Example 1.1.1: Paramagnet
Suppose we have N spin-1/2 particles locked on a lattice so there are no translational degrees of freedom. Assume there are no interactions between the different particles. We will assume a magnetic field of the form
B~ = Bzˆ. 2 The Statistical Basis of Thermodynamics
The energy of a single particle is
εi = −2µBSz,iB,
where µB is the Bohr magneton, and Sz,i = ±1/2 is the spin of the particle in the z-direction. Our notation can be made more compact by defining the constant b = gµbB/2, and then the total energy of the system is
N N X X E = εi = −b σz,i, i=1 i=1
where σz,i = ±1. For our system, there is a spectrum of energies
E = {−Nb, −(N − 2)b, . . . , (N − 2)b, Nb} ,
where, −Nb corresponds to all spins being down and Nb corresponds to all spins being up. There are a total of 2N states. However, there are only N + 1 different energy levels. The question then becomes, how many states are there per energy level? For example, there is only a single state with energy −Nb, and that is the state with all spins being down ↓↓↓ · · · . Likewise, there is a single state with energy Nb, and that is the state with all spins being up ↑↑↑ · · · . Since 2N is generally much larger than N + 1, clearly, the states are not distributed uniformly among the energy levels. For example, the energy level −(N − 2)b is the state with all spins but one being down. There are N ways to arrange N spins if only one spin is up. For a given energy E, how many microstates are there with this energy? We know that E = −b (N↑ − N↓) ,
where, for example, N↑ is the number of spin-up particles. How can we enumerate the number of microstates with this energy? That is, how many states are there with N↑ up spins and N↓ down spins? In general, if you have N = N1 + N2 + ··· total objects, consisting of N1 objects of one kind, N2 objects of another kind, and so on, then the total number of permutations is N! . N1!N2! ··· In our case, N! N! Ω(N↑) = = . N↑!N↓! N↑!(N − N↑)! Then N! Ω(E) = N E N E . 2 + 2b ! 2 − 2b ! Often, we write N N N = + s, N = − s, ↑ 2 ↓ 2 then the total energy is
E = −b (N↑ − N↓) = −b(2s),
where 2s is called the spin excess. 1.1. Counting States 3
The entropy of the paramagnet is S kT S(E) = k ln Ω = k [ln N! − ln N↑! − ln N↓!] .
In the large N limit, we can apply the lowest order Stirling approximation x ln N! = N ln N − N.
If we define Figure 1.1: A plot of the “net N − N N magnetization” x versus inverse x ≡ ↑ ↓ ,N ≡ (1 ± x), N ↑/↓ 2 temperature for a paramagnet. then E = −bNx, Tip and the entropy can be written as When counting mi- 1 + x 1 + x 1 − x 1 − x crostates, make sure S = −kN ln + ln . 2 2 2 2 you are counting the states and not just the energy If we plot S/kN as a function of x, we see that it is positive in [−1, 1], symmetric levels. In general, because about x = 0, and it peaks at x = 0 with a value of ln 2. That is of degeneracy, the number of states is greater than the S(x = 0) number of energy levels. = ln 2. kN This implies that S(x = 0) = k ln 2N , but we know that S = k ln Ω, so Ω(x = 0) = 2N . This is the total number of microstates of all energies. This is due to our approximation using Stirling’s formula, but it illustrates the fact that the probabilities are dominated by a single term.
When counting states, it is often useful to use Stirling’s approximation
√ 1 1 N! ' 2πN N N exp −N + + O . 12N N 2
Often this is more useful in the logarithmic form (and then just taking the exponential in the end)
1 1 1 1 ln N! ' ln (2π) + N + ln N − N + + O . 2 2 12N N 2
The lowest order Stirling approximation can be derived as follows
N! = N(N − 1)(N − 2) ··· 2 · 1 ln N! = ln N + ln(N − 1) + ln(N − 2) + ··· + ln 2 + ln 1 N X = ln n n=1 N ' ln n dn ˆ0 = N ln N − N.
One approach to the continuum limit is to make N! continuous via the gamma func- tion. The gamma function Γ(N), where N is a continuous variable in the real numbers, 4 The Statistical Basis of Thermodynamics
is the function with the properties
N! = Γ(N + 1) Γ(N + 1) = NΓ(N) 0! = Γ(1) = 1
To satisfy these properties, the gamma function is defined as
∞ Γ(z) = dz xz−1e−x. ˆ0
The gamma function is the continuous version of the factorial function. Useful values to know, in addition to all the non-negative integer values, include 1 √ Γ = π 2 3 1√ Γ = π. 2 2 The gamma function will come up repeatedly in statistical mechanics. We can also define the digamma function ∂ Ψ(z) = ln Γ(z). ∂z The digamma function has a nice asymptotic expansion. For large z,
Ψ(z + 1) ' ln z.
Example 1.1.2
Using the Gamma function, we can write the entropy of a paramagnet
S(E) = k [ln N! − ln N↑! − ln N↓!] ,
as S(E) = k [ln Γ(N + 1) − ln Γ(N↑ + 1) − ln Γ(N↓ + 1)] . Then we can write ∂S = −k [Ψ(N + 1) − Ψ(N + 1)] , ∂x ↑ ↓
where x ≡ (N↑ − N↓)/N. Applying the asymptotic expansion of the digamma function gives us ∂S 1 1 + x = − kN ln . ∂x 2 1 − x
1.2 Connecting to Thermodynamics
Previously, we defined the Boltzmann entropy. We want to connect this with Clausius’ thermodynamic entropy. Recall the laws of thermodynamics: Zeroth law: If systems A and B are individually in equilibrium with C, then A and B are in equilibrium with each other 1.2. Connecting to Thermodynamics 5
First law: Energy is conserved Tip dE = δQ − dW. That is, the energy increase in a system is due to the heat added to the system minus Thermal contact between the work done by the system. Keep in mind that the work done by the system could two systems means that en- be a more complicated expression like dW = −P dV + µ dN + B~ · dS~ + ··· . ergy can be exchanged be- tween the two systems. Second law: Entropy never decreases. That is, dS ≥ 0. In general,
δQ C dS ≥ = V dT, T T Tip which implies that A general principle is that Tf CV equilibrium occurs when S − S ≥ dT. f 0 the accessible states Ω is ˆT0 T maximized. Third law: The entropy of a system approaches a constant value as the temperature of the system approaches absolute zero (T = 0).
Suppose we have two systems in equilibrium. The first system has the macroscopic properties N1,V1,E1, and the second has N2,V2,E2. We put the two systems in thermal contact with each other such that energy can be exchanged between the two systems but not particle number N or volume V . The energy of the combined system is
0 E = E1 + E2 = const.
The number of microstates of the system is multiplicative. For each individual system, the number of microstates depends only on the energy and not the particle number or volume, so for the combined system, the number of microstates is
0 Ω = Ω1Ω2, or more explicitly,
0 0 Ω(E ,E1) = Ω1(E1)Ω2(E2) = Ω1(E1)Ω2(E − E1).
Equilibrium in the combined system is achieved when Ω0 is maximized. This makes sense statistically because an equilibrium system is by definition in a highly probable mi- crostate, and the maximum of Ω0 corresponds to the macrostate with the largest number of microstates. An equilibrium system is then most likely in a microstate associated with 0 0 the maximum of Ω . To maximize Ω , we differentiate with respect to E1 and set it equal to zero
0 ∂Ω ∂Ω1 ∂Ω2 ∂Ω1 ∂Ω2 ∂E2 ∂Ω1 ∂Ω2 0 = = Ω2 + Ω1 = Ω2 + Ω1 = Ω2 − Ω1 . ∂E1 ∂E1 ∂E1 ∂E1 ∂E2 ∂E1 ∂E1 ∂E2 We can write this result in the form ∂ ln Ω ∂ ln Ω 1 = 2 ≡ β, ∂E ∂E 1 V,N,E1=E1 2 V,N,E2=E2 where the subscript on the first term means V and N are held constant, and the result is evaluated at E1 equal to the equilibrium energy E1. In general, then
∂ ln Ω ≡ β. (1.1) ∂E V,N,E=E 6 The Statistical Basis of Thermodynamics
Previously, we defined the Boltzmann entropy as
SB = kB ln Ω. This implies that ∂ ln Ω 1 ∂S = B . ∂E kB ∂E Recall the thermodynamic identity
dE = T dST − P dV + µ dN.
The entropy, ST used here is the thermodynamic entropy. This implies that ∂S 1 T = . (1.2) ∂E V,N T Comparing Eq. (1.1) and (1.2), we conclude that dS 1 T = = const. d ln Ω βT
−1 This implies that β ∝ T . In order to have ST = SB, we need
1 β = . kT This essentially defines the Boltzmann constant k. Instead of exchanging E, we can have our two systems in contact exchange N or V to get other relations ∂ ln Ω ∂ ln Ω 1 = 2 ≡ η ∂V ∂V 1 N,E,V1=V 1 2 N,E,V2=V 2 ∂ ln Ω ∂ ln Ω 1 = 2 ≡ ζ. ∂N ∂N 1 V,E,N1=N 1 2 V,E,N2=N 2 These relations define pressure ∂ ln Ω P = , ∂V kT and chemical potential ∂ ln Ω µ = − . ∂N kT
This implies that in equilibrium, T1 = T2, P1 = P2, and µ1 = µ2 if the systems are allowed to exchange energy, volume, and particles. In general, if we have two systems A and B in contact and they are capable of exchanging quantities X,Y,..., then the number of microstates will depend on these quantities Ω(X,Y,...) = ΩA(XA,YA,...)ΩB(XB,YB,...),
where X = XA + XB, Y = YA + YB, and so on. Then,
∂Ω ∂ΩA ∂ΩB ∂ΩA ∂ΩB = ΩB + ΩA = ΩB − ΩA ∂XA ∂XA ∂XA ∂XA ∂XB ∂Ω ∂ΩA ∂ΩB ∂ΩA ∂ΩB = ΩB + ΩA = ΩB − ΩA ∂YA ∂YA ∂YA ∂YA ∂YB . . 1.3. Intensive and Extensive Parameters 7
The equilibrium of the combined system occurs when Ω is maximized with respect to each of these quantities. I.e. equilibrium occurs when the above equations are equal to zero, and then
∂ΩA ∂ΩB ΩB = ΩA ∂XA ∂XB ∂ΩA ∂ΩB ΩB = ΩA ∂YA ∂YB . .
1.3 Intensive and Extensive Parameters
Remember, an extensive parameter depends on the size (or extent) of the system. The way in which it depends on the size of the system is additive. In general, the parameters N,V,E are extensive. Suppose you chop a system with parameters E,N,V into two pieces—one with E1,N1,V1 and the other with E2,N2,V2, then E = E1 +E2, V = V1 +V2, and N = N1 + N2. This is what it means for these quantities to be extensive. In contrast, intensive (or inEXtensive) variables do not depend on the size (i.e. extent) of the system. Examples of intensive parameters include T , P , µ, and ratios of extensive quantities like n = N/V . The intensive quantities in terms of the extensive derivatives of the entropy are
∂S 1 ∂S P ∂S µ = , = , = − . ∂E N,V T ∂V N,E T ∂N V,E T
These can be derived by looking at the derivatives when a pair of systems in thermal contact is allowed to exchange E, V , or N. They can be remembered by comparing the differential ∂S ∂S ∂S dS = dE + dV + dN, ∂E ∂V ∂N with the thermodynamic identity
dE = T dS − P dV + µ, dN, which can be rearranged as 1 P µ dS = dE + dV − dN. T T T The intensive quantities in terms of the derivatives of the energy are
∂E ∂E ∂E = T, = −P, = µ. ∂S N,V ∂V N,S ∂N V,S
For the extensive quantities, we can write down scaling variables λ, as in,
λS(E,N,V ) = S(λE, λN, λV ).
This is essentially a definition of a system whose variables are extensive. For an extensive system, E = TS + µN − PV. Taking the total derivative, gives us the Gibbs-Duhem relation
S dT = V dP − N dµ. 8 The Statistical Basis of Thermodynamics
If we want to find the intensive entropy s = S/V , we let λ = 1/V
x S E N s = = S , , 1 /V = s(, n). V V V
1/T All of this is only true if our system experiences no surface effects. Example 1.3.1: Paramagnet Figure 1.2: A plot of the “net magnetization” x versus inverse We now continue the paramagnet example. temperature for a paramagnet. For a paramagnet, we know that
∂S kN 1 + x = − ln . ∂x 2 1 − x
From E = −bNx, we get
∂S 1 k 1 + x = = − ln . ∂E T 2b 1 − x
We can invert to get b x = tanh . kT Note that x is kind of a net magnetization. See Fig. (1.2). We cannot define a pressure P for the paramagnet because the system has no volume dependence.
1.4 Averages
If the probability to have state i is Pi and the value of a variable A in the state i is Ai, then the expectation or mean of A is X hAi = AiPi. i
In the continuous case, if you have the probability p(x) of a particular value of x, then
∞ hAi = A p(x) dx. ˆ−∞ 1.5. Summary: The Statistical Basis of Thermodynamics 9
1.5 Summary: The Statistical Basis of Thermodynamics
Skills to Master • Be able to count states and calculate the number of microstates Ω accessible to a system • Calculate the average value of quantities • Calculate the Boltzmann entropy • Understand the paramagnet model system • Know the laws of thermodynamics • Derive relations for two systems in equilibrium which are exchanging some quantity • Understand extensive vs. intensive variables
Counting States Equilibrium Exchanges The Boltzmann entropy is Given two systems individually in equilibrium with N1,V1,E1 and N2,V2,E2, we are often interested in S(N,V,E) = kB ln Ω(N,V,E), relations between quantities when the two systems are where Ω(N,V,E) is the number of microstates which placed in some kind of contact (allowing the exchange correspond to the single macrostate with properties of something, e.g. energy) and the combined system N,V,E. reaches equilibrium. The energy of the combined sys- When counting microstates, make sure you are tem is just the sum counting the states and not just the energy levels. In general, because of degeneracy, the number of states is E = E1 + E2 = const. greater than the number of energy levels. The number of microstates of the combined system is When counting states, it is often useful to use Stir- multiplicative. Given that Ω (E ) microstates of sys- ling’s approximation. To lowest order, it is 1 1 tem 1 correspond to E1, and so on, the number of ln N! = N ln N − N. microstates of the total system is Ω(E,E ) = Ω (E )Ω (E ) = Ω (E )Ω (E − E ). Laws of Thermodynamics 1 1 1 2 2 1 1 2 1 Zeroth law: If systems A and B are individually in Equilibrium of the combined system occurs when Ω is equilibrium with C, then A and B are in equilib- maximized. To find this value, we partially differenti- rium with each other. ate Ω with respect to E1 and set it equal to zero. The result can be written in terms of the partial derivatives First law: Energy is conserved of ln Ω1 and ln Ω2 equal to each other and constant. In general, if we have two systems A and B in dE = δQ − dW. contact and they are capable of exchanging extensive I.e., the energy increase in a system is due to the quantities X,Y,..., then the number of microstates will heat added to the system minus the work done depend on those quantities by the system. Ω(X,Y,...) = ΩA(XA,YA,...)ΩB(XB,YB,...), Second law: Entropy never decreases. That is, dS ≥ 0. In general, where X ≡ XA + XB, Y ≡ YA + YB, and so on. Then,
δQ CV ∂Ω ∂ΩA ∂ΩB ∂ΩA ∂ΩB dS ≥ = dT, = ΩB + ΩA = ΩB − ΩA T T ∂XA ∂XA ∂XA ∂XA ∂XB which implies that ∂Ω ∂ΩA ∂ΩB ∂ΩA ∂ΩB = Ω + Ω = Ω − Ω ∂Y B ∂Y A ∂Y B ∂Y A ∂Y Tf A A A A B CV Sf − S0 ≥ dT. . . ˆT0 T . . Third law: The entropy of a system approaches a This is just the partial derivative with the product rule constant value as the temperature of the system followed by simplification. The equilibrium of the com- approaches absolute zero (T = 0). bined system occurs when Ω is maximized with respect 10 The Statistical Basis of Thermodynamics to each of these quantities. I.e. equilibrium occurs These can be derived by looking at the derivatives when when the above equations are equal to zero, and then a pair of systems in thermal contact is allowed to ex- change E, V , or N. They can be remembered by com- ∂ΩA ∂ΩB ΩB = ΩA paring the differential ∂XA ∂XB ∂ΩA ∂ΩB ∂S ∂S ∂S ΩB = ΩA dS = dE + dV + dN, ∂YA ∂YB ∂E ∂V ∂N . . with the fundamental thermodynamic relation . . dE = T dS − P dV + µdN, A general principle is that equilibrium occurs when the accessible states Ω is maximized. For the extensive quantities, we can write down Using this approach in combination with the ther- scaling variables λ, as in, modynamic differential identity gives us the following λS(E,N,V ) = S(λE, λN, λV ). relations This is essentially a definition of a system whose vari- ∂ ln Ω 1 ≡ β = . ables are extensive. For an extensive system, ∂E kT V,N,E=E E = TS + µN − PV. These relations define pressure Taking the total derivative, gives us the Gibbs-Duhem ∂ ln Ω P ≡ , relation ∂V N,E,V =V kT S dT = V dP − N dµ. and chemical potential Model: Paramagnet ∂ ln Ω µ ≡ − . For a paramagnet, we have N non-interacting spin-1/2 ∂N V,E,N=N kT particles on a lattice in an external magnetic field B~ = This implies that in equilibrium, T1 = T2, P1 = P2, Bzˆ. The energy of a single particle is εi = −2µBSz,iB, and µ1 = µ2 if the systems are allowed to exchange where µB is the Bohr magneton, and Sz,i = ±1/2 is energy, volume, and particles. the spin of the particle in the z-direction. The total energy of the system is
Intensive and Extensive Parameters N N X X An extensive parameter depends on the size (or ex- E = εi = −b σz,i = −b (N↑ − N↓) , tent) of the system. The way in which it depends on i=1 i=1 the size of the system is additive. In general, the pa- where σz,i = ±1, and b = gµbB/2. rameters N,V,E are extensive. Suppose you chop a There are a total of 2N states. The energy of the system with parameters E,N,V into two pieces—one lattice depends only on the number of up spins ver- with E1,N1,V1 and the other with E2,N2,V2, then sus the number of down spins. But given a specified E = E1 + E2, V = V1 + V2, and N = N1 + N2. This is number of up and down spins, there are many different what it means for these quantities to be extensive. possible arrangements of those spins. The number of In contrast, intensive (or inEXtensive) variables arrangements of spins which add up to the energy E is do not depend on the size (i.e. extent) of the system. N! N! Examples of intensive parameters include T , P , µ, and Ω(E) = = , N !N ! N N ratios of extensive quantities like n = N/V . ↑ ↓ 2 + s ! 2 − s ! The intensive quantities in terms of the extensive where N↑ = N/2 + s, N↓ = N/2 − s, and the quantity derivatives of the entropy are 2s is called the spin excess. Now the total energy can ∂S 1 be written as = E = −b(2s). ∂E N,V T ∂S P The entropy of the paramagnet is S(E) = k ln Ω. = Applying the lowest order Stirling approximation, we ∂V T N,E get that in the large N limit, ∂S µ = − . 1 + x 1 + x 1 − x 1 − x ∂N T S = −kN ln + ln , V,E 2 2 2 2 1.5. Summary: The Statistical Basis of Thermodynamics 11 where The expectation value of a quantity A is N − N x ≡ ↑ ↓ , N X hAi = AiPi, is kind of a net magnetization. Now, i
E = −bNx. where Pi is the probability of getting Ai. In the con- tinuous case, if your probability distribution is p(x), The temperature of the paramagnet is given by then ∞ 1 ∂S kB 1 + x hAi = Ap(x) dx. = = − ln . ˆ−∞ T ∂E 2b 1 − x The gamma function Inverting gives us ∞ z−1 −x b Γ(z) = dz x e , x = tanh . ˆ0 kT is the continuous version of the factorial, and its prop- Miscellaneous erties include
In general, if you have N = N1 + N2 + ··· total ob- N! = Γ(N + 1) = NΓ(N). jects, consisting of N1 objects of one kind, N2 objects of another kind, and so on, then the total number of Useful values to know include permutations is N! 1 √ . Γ = π N !N ! ··· 2 1 2 √ The higher order Stirling approximation is 3 π Γ = . 2 2 √ 1 1 N! ' 2πN N N exp −N + + O . 12N N 2 The digamma function is
In logarithmic form it is ∂ Ψ(z) = ln Γ(z). ∂z 1 1 1 ln N! ' ln (2π) + N + ln N − N + + ··· 2 2 12N It has a nice asymptotic expansion. For large z,
The general Gaussian integral is Ψ(z + 1) ' ln z. ∞ r 2k −ax2 (2k − 1)!! π A useful approximation if x << 1 is x a dx = k+1 k , ˆ0 2 a a 1 ln(1 + x) ' x − x2. provided that k ∈ Z, and a > 0. 2 Chapter 2
Ensembles and Classical Phase Space
2.1 Classical Phase Space
Consider an N particle system with coordinates and momenta q1, . . . , q3N , p1, . . . , p3N . Using the Hamiltonian H({qi, pi}), we can write down Hamilton’s equations
∂H ∂H q˙i = , p˙i = − . ∂p ∂qi
Our system corresponds to a single point in a 6N-dimensional phase space. As our system evolves, its corresponding point in phase space traces out some trajectory. Given some function of time f({qi(t), pi(t)}), we want to know how this function evolves in time. The total derivative is
df ∂f X ∂f ∂f = + q˙i +p ˙ dt ∂t ∂qi ∂pi ∂f X ∂f ∂H ∂f ∂H = + − ∂t ∂qi ∂pi ∂pi ∂qi ∂f = + {f, H} , ∂t This can be written as df ∂f = + {f, H} , dt ∂t where X ∂f ∂H ∂f ∂H {f, H} = − , ∂qi ∂pi ∂pi ∂qi
df is called a Poisson bracket. The total derivative dt is a convective time derivative since it tells us how the system is evolving in phase space. Now instead of a single system corresponding to a single point in phase space, let’s think about an entire ensemble of systems (with the same H) corresponding to an entire ensemble of points moving through our phase space. An ensemble is simply many different copies of our system. The phase space density is a function of the form
ρ = ρ (qi, . . . , q3N , pi, . . . , p3N ) .
For equilibrium,
∂tρ = 0. 2.2. Liouville’s Theorem 13
Phase-space averaged quantities are calculated as
d3N q d3N p f ({q , p }) ρ hf(t)i = i i . ´ d3N q d3N p ρ ´ The question then becomes, how does ρ evolve through phase space?
2.2 Liouville’s Theorem
We can define some generalized phase space velocity
~vps = ({q˙i}, {p˙i}) .
To get the total number of systems in a volume V , we integrate ρ over the volume
NV = dω ρ. ˆV The flux of points through a surface is given by
F~ = ρ~vps.
Then ∂ ρ = − ρ~v · d~s. ∂t ˆV ˆ So the rate of change of the number of systems in the volume equals the flux through the surface. This implies a continuity equation in the phase space
∂tρ + ∇ · (ρ ~vps) = 0.
This tells us how the phase space evolves if we know the phase space velocities. Note, the divergence here has many dimensions. We can expand the divergence as
∇ · (ρ ~v) = ρ ∇ · ~v + ~v · ∇ρ.
The first term on the right can be written as
2 2 X ∂q˙i ∂p˙i X ∂ H ∂ H ρ ∇ · ~v = + = − = 0. ∂q ∂p ∂q ∂p ∂p ∂q i i i i i i i i To simplify the above result, we plugged in Hamilton’s equations
∂H ∂H q˙i = , p˙i = − . ∂pi ∂qi 14 Ensembles and Classical Phase Space
So our divergence simplifies to
X ∂ρ ∂ρ ∇ · (ρ ~v) = ~v · ∇ρ = q˙ +p ˙ . i ∂q i ∂p i i i Again, we apply Hamilton’s equations to get
X ∂ρ ∂H ∂ρ ∂H ∇ · (ρ ~v) = − = {ρ, H} . ∂q ∂p ∂p ∂q i i i i i So, we have that
∂tρ + {ρ, H} = 0. This is Liouville’s theorem. Recall the convective derivative df ∂f = {f, H} + . dt ∂t The convective derivative of ρ is dρ = 0. dt This has to be true for any ensemble of systems that is described by Hamilton’s equations. For a stationary ensemble,
∂tρst = 0 =⇒ {ρst, H} = 0. This is satisfied if ρ = ρ[H], i.e. if ρ is a functional of H. One possibility (a microcanonical ensemble) is when
ρst = const. This is consistent with the Boltzmann hypothesis. I.e. the Boltzmann hypothesis is plausible. Another example (a canonical ensemble) is when
ρ ∝ e−βH.
The Master Equation is another approach—a discrete approach to justify all this. We have an ensemble of N systems that can take on some number of states. There are Ni states with energy Ei in the overall ensemble of N systems. The rate of transition from state i to state j, denoted ωij is
dNi X = (ω N − ω N ) . dt ji j ij i j
dNi We want to find an equilibrium state where dt = 0. The most obvious way to find such a state is to say that ωjiNj = ωijNi. Note that this is a particular solution—not a general solution. If we assume time-reversal invariance, then ωij = ωji.
This implies that Ni = Nj, which implies that N i = P = P . N i j That is, we are justified to assume that the probabilities are equal. 2.3. Microcanonical Ensemble 15
2.3 Microcanonical Ensemble
In the microcanonical ensemble, the macrostate is defined by a fixed number of particles 1 1 N, a fixed volume V , and a fixed energy E or a fixed energy range E − 2 ∆,E + 2 ∆ . Then the chief problem is to calculate the number of microstates Ω(N,V,E) that are accessible to the system. Unfortunately, for most physical systems, calculating Ω is very difficult. Furthermore, we typically do not know the energy E for a physical system. Because of such problems, later we will look at other approaches, e.g. the canonical ensemble. What does our ensemble tells us about real systems? For the microcanonical ensemble, the phase density is constant ( const if E − 1 ∆ ≤ H(q, p) ≤ E + 1 ∆ ρ = 2 2 . 0 otherwise.
In other words, the phase volume of the microcanonical ensemble is the volume of the “shell” defined by 1 1 E − ∆ ≤ H(q, p) ≤ E + ∆. (2.1) 2 2 The volume enclosed by this shell is
ω = dω = d3N q d3N p, ˆ ˆ where the integral goes over the region of phase space for which Eq. (2.1) is true. The ensemble average of a function f is defined as dω fρ hfi = . ´ dω ρ ´ The time average is defined as
N 1 X f = f ({q(t ), p(t )}) . N j j j=1 We want to assume that hfi = f. This is essentially the ergodic hypothesis. Note, there are systems which do not satisfy this. A system that satisfies it must trace out all of its available phase space over some time. We define the classical weight function as integrating over the shell in phase space d3q ··· d3q d3p ··· d3p 1 1 Ω(E) = 1 N 1 N θ H − E − ∆E θ H − E + ∆E , ˆ ω0 2 2 (2.2) where ( 1 if x > 0 θ(x) = . 0 if x < 0. is the Heaviside step function. All that the product of Heaviside functions is doing is selecting the shell. It multiplies the integral by 1 if the condition in Eq. (2.1) is true, and otherwise, it multiplies the integral by 0. In the limit ∆E → 0, the Heaviside functions become a delta function dω Ω(E) = ∆E δ(E − H). ˆ ω0 16 Ensembles and Classical Phase Space
We will later derive that 3N ω0 = h , where h is Planck’s constant, and 3N are the degrees of freedom for N particles in 3D space. The quantity ω0, called the fiducial element of phase space, can be thought of as the infinitesimal phase space volume corresponding to a single microstate. For discrete states, as in our paramagnet example, we would have X Ω(E) = δEi,E. i
Example 2.3.1: Classical Ideal Gas
We will now apply the microcanonical machinery to an example—the classical (i.e. non-quantum) non-relativistic ideal gas. For the ideal gas, we have N particles in a volume V = L3 for a cubic box. The total Hamiltonian is N 2 X ~pj H = . 2m j=1 We can split Eq. (2.2) into two pieces. We start by calculating the volume of phase space with energy less than E, denoting it Σ(E) and then we can find the density of states around E as ∂Σ(E) Ω(E) = ∆E, ∂E where ∆E << E. For the number of states with energy less than E, we get
N N N 2 1 Y 1 Y X ~pj Σ(E) = d3q d3p θ (E − H) = d3q d3p θ E − . ω ˆ i i ω ˆ i i 2m 0 i 0 i j=1
This is the volume of phase space below energy E. This calculation assumes that the particles are distinguishable. In reality, our particles are indistinguishable, and so we are overcounting with this formula. To remove the overcounting, we need to divide by N!
N N 2 1 Y X ~pj Σ(E) = d3q d3p θ E − . ω N! ˆ i i 2m 0 i j=1
We can write this as N ∞ N 2 V 1 X ~pj Σ (E) = d3p ··· d3p θ E − , CL N! ω ˆ 1 N 2m 0 −∞ j=1
where the subscript “CL” indicates that this is for the classical gas. Note, the √integral here is the volume of a 3N-dimensional hypersphere with radius R = 2mE. What about for the quantum-mechanical case? For a QM ideal gas, the wave function of the system is
23N/2 Y ψ(~x ,... ~x ) = sin (kxx ) sin (kyy ) sin (kzz ) . 1 N V N/2 i i i i i i i 2.3. Microcanonical Ensemble 17
The energy eigenvalues are N 2 X 2 E = ~ ~k . α 2m j j where π ~k = nx + ny + nz , nl = 1, 2,... j L j j j j l The subscript α simply takes the place of the quantum numbers {nj} that the system depends on. Note that this is simply the wave function and energy for N particles in an infinite square well. Then the number of states with energy less than E is ∞ N 1 X 2 X 2 Σ (E) = θ E − ~ ~k . QM N! 2m j ~n1,...,~nN j Schematically, we expect our energy to have the form
2/3 1/3 Eα = Nε + N c1 + N c2 + ··· ,
where ε is the average energy per particle. The first term is the bulk energy, the second term is a correction due to surface effects, and the third term is a correction due to edge effects. In the large N limit, only the bulk (first) term survives. For large ~k, we can approximate the sum over one of the quantum numbers as a momentum integral
∞ X L ∞ L ∞ L ∞ → dk = dk = dp. π ˆ 2π ˆ 2π ˆ n=1 0 −∞ ~ −∞
We can do this since the functions depend only on k2. In the first equality, we double the integration region and divide by two. In the second, we use the fact that pi = ~ki. Then we end up with N ∞ N 2 1 V X ~pj Σ (E) = d3p ··· d3p θ E − . QM N! (2π )3N ˆ 1 N 2m ~ −∞ j
This is the same as the classical result provided that we define
3N ω0 = h .
Generically, this is the value of the phase space normalization factor ω0 and it can be shown using a variety of approaches and a variety of systems. It is also the smallest resolvable volume that is predicted by the Heisenberg uncertainty principle ∆x∆p ≥ ~. 2 So far we have not made any assumptions about the form of the single particle energies, so we expect the classical ideal gas law
PV = NkT
to hold true. We can write this relation as ∂S P ∂k ln Ω kN = = = , ∂V E,N T ∂V E,N V 18 Ensembles and Classical Phase Space
which implies that ln Ω = N ln V + ln Ωp(E,N). It can be shown that the volume of a 3N-dimensional hypersphere of radius R is 3N/2 π 3N 3N Λ3N (R) = 3N R = C3N R . Γ 2 + 1 √ In our case, R = 2mE, so we have that
3N/2 Λ3N (R) = C3N (2mE) .
Then ∂Σ(E) V N 3N 2mE 3N/2 ∆E Ω(E) = ∆E = C . ∂E N! 3N 2 h2 E Now we just have to take the log and apply Stirling’s approximation to get the entropy using S/k = ln Ω
S 3N 3N 2mE ∆E = N ln V − ln N! + ln C + ln + ln + ln . k 3N 2 2 h2 E
Using Stirling’s approximation,
N ln N! ' N ln N − N = N ln N − N ln e = N ln , e
and it turns out that 3N 2πe ln C ' ln . 3N 2 3N With these approximations, we get " # S V e emE 3/2 3N ∆E ' N ln + N ln + ln + ln . k N 3π~2N 2 E
The last two terms go to zero as N → ∞, so we get
" # S V mE 3/2 5 = N ln + N. k N 3π~2N 2
This is the Sackur-Tetrode equation for the entropy of an ideal gas. Now that we have the entropy, we can extract some thermodynamic quanti- ties. For example, ∂S 1 = , ∂E V,N T gives us 3 E = NkT, 2 and ∂S µ = − , ∂N E,V T 2.4. Canonical Ensemble 19
gives us Tip " 3/2# V mE An “adiabatic process” is µ = T ln 2 . N 3π~ N often taken to mean a pro- cess that occurs at constant For adiabatic (i.e. constant entropy) changes, we have entropy.
VE3/2 = const VT 3/2 = const PV 5/3 = const.
We imposed no symmetry requirements on the wave function of the system. This is okay to do if our gas is a very low density. So our results here are only really valid in the low density N/V limit. The main purpose of this example was to illustrate that calculations done in the microcanonical ensemble are long and tiresome. This motivates our develop- ment of more useful ensembles such as the canonical ensemble.
The Gibb’s Paradox If the particles are distinguishable, then we have to drop the 1/N! factor from Σ(E). Then the entropy gets an extra contribution of N ln(N/e), which would make it non-extensive. Suppose we have two boxes with the same volume, same number of particles, and with entropies S1 and S2. Suppose these two boxes are next to each other and we remove the barrier between them. The total entropy of the combined system is now " # 2V m2E 3/2 2N Stot = k2N ln + 5kN + 2N ln . 2N 3π~2 e
We can write this as
Stot = S1 + S2 + Smix, where S1 and S2 are just the Sackur-Tetrode entropies of the individual boxes, and
2E S = 2N ln , mix e is the additional entropy term that appears when the two boxes are mixed together. Apparently, removing the barrier and allowing distinguishable particles to mix creates entropy. Such a process is irreversible.
2.4 Canonical Ensemble
In the microcanonical ensemble, we had N, V , and E as fixed quantities. Then we count states to calculate the entropy, and from that we calculate thermodynamic quantities. In the canonical ensemble, the macrostate is defined by the parameters N, V , and T . The energy E of the system can vary (in principle from 0 to ∞). In the canonical ensemble, we split a very large system into two pieces which can exchange energy. We assume that one of the systems is very large (i.e. N ∼ 1046), and we call it the reservoir and the other system called the subsystem is just large (i.e. N ∼ 1023). 20 Ensembles and Classical Phase Space
Considering the total system, we can neglect the multiplicity of the subsystem and write the multiplicity of the total system as the multiplicity of the reservoir
Ωt = Ωr.
Then, assuming the total system has energy E0, the probability of finding the subsystem in a state j with energy Ej is
P (Ej) ∝ Ωr(E0 − Ej),
or in terms of entropy Sr(E0 − Ej) P (Ej) ∝ exp . kB Expanding this gives us " # S (E ) ∂S E E r 0 r j j −βEj P (Ej) ≈ exp − ≈ exp − = e . kB ∂E V,N kB kBT
∂Sr since ∂E V,N = 1/T . Here, we define
1 β ≡ . kBT
Thus, the normalized probability is
e−βEj P (Ej) = , ZN
−1 where ZN is the normalization factor called the partition function. The subscript N denotes this as the partition function of the canonical ensemble. It is a sum over all the possible microstates of the system
X −βEj ZN = e . j
Knowing this partition function tells us a lot about the properties of our subsystem. We can also write the canonical partition function as a sum over energies
X −βE X −β(E−TS(E)) −βFH ZN = Ωs(E)e = e = e . E E
In the first sum, Ωs(E) is the possible states in our subsystem which consists of all states with energy E. We can convert to the second sum by using the fact that
S/kB βST S = kB ln Ω =⇒ Ω = e = e . 2.4. Canonical Ensemble 21
The quantity in the exponent, FL = E − TS(E), is called the Landau free energy. It is calculated in the microcanonical ensemble by holding E constant and calculating S(E) in the microcanonical ensemble. In the final equality, we have the Helmholtz free energy
FH = hEi − T S, which is calculated in the canonical ensemble and uses the average energy hEi. The final equality comes from the fact that in the thermodynamic limit, the sum is completely dominated by a single term, and that term is
∗ ∗ e−βFH = e−β(E −TS(E )),
∗ where E is the energy that minimizes FL = E − TS(E). In other words, in the ther- modynamic limit, FL = FH . In general, to find the dominant term, we simply minimize FL = E − TS(E) with respect to E. The thermal average of any quantity O is
1 X hOi = O e−βEj . Z j N j
For example, the average energy is calculated as ∂ 1 X −βEj ln ZN = − Eje = − hEi . ∂β ZN Thus, ∂ ∂ hEi = − ln Z = T 2 ln Z . ∂β N ∂T N The natural variables for the canonical ensemble are T , N, and V . The Helmholtz free energy is
F = E − T S.
In the thermodynamic limit, ∂F ∂F ∂βF ∂ ln Z E = F + TS = F − T = F + = = N , ∂T N,V ∂ ln β N,V ∂β N,V ∂β N,V which implies that
F = −kBT ln ZN . The differential of the Helmholtz free energy is
dF = dE − T dS − S dT.
Plugging in the fundamental thermodynamic identity and simplifying gives us
dF = −S dT − P dV + µ dN.
Then taking partial derivatives gives us the relations
∂F ∂F ∂F = −S, = −P, = µ, ∂T N,V ∂V T,N ∂N T,V which allows us to calculate S, P , and µ. 22 Ensembles and Classical Phase Space
Example 2.4.1: Paramagnet Revisited
For the paramagnet, the energy of a particular configuration is
X 1 E = −b σ , σ = ±1, b = gµ B . j j 2 B ext j
The partition function is then
X βb P σ Z = e i i configurations X X X = eβbσ1 eβbσ2 ··· eβbσN
σ1=±1 σ2=±1 σN =±1 N Y X N = eβbσi = [2 cosh(βb)] .
i σi=±1 We can then calculate
1 X P hσ i = σ eβb σi k Z k configurations 1 = [2 cosh(βb)]N−1 eβb − e−βb Z = tanh(βb).
The average total spin of the system is
N hσki = N↑ − N↓ = Nx,
which is the same result we found using the microcanonical ensemble.
Classical Limit
The phase space density ρ(ξi), where ξi denotes both position xi and momentum pi (i.e. the point in phase space) is just proportional to the probability of finding an ensemble member at the phase space point ξi. So in the classical limit, the canonical ensemble is given by ρ = e−βH.
Since this satisfies {ρ, H} = 0, this ensemble is stationary, and so the continuous version of the canonical partition function is
1 dω −βH ZN = e . N! ˆ ω0
Note, that this is for indistinguishable particles. If the particles are distinguishable, then we don’t need the 1/N!, and then
dω Z = e−βH. ˆ ω0 2.4. Canonical Ensemble 23
Example 2.4.2
For example, if we have a gas of N particles interacting with a two-body potential 1 X U = u(~r − ~r ), 2 i j i,j then 3 3 3 3 2 1 d x1 ··· d xN d p1 ··· d pN X p 1 X Z = exp −β i + u(~r − ~r ) . N N! ˆ h3N 2m 2 i j i i,j
This is separable, so we can write
N 3 3 " 2 # V d p1 ··· d pN X p Z = exp −β i N N! ˆ h3N 2m i 3 3 d x1 ··· d xN β X × exp − u(~r − ~r ) . ˆ V N 2 i j i,j
The first part is easy since it is just a product of Gaussians N 3 3 V d x1 ··· d xN β X Z = `−3N × exp − u(~r − ~r ) , N N! Q ˆ V N 2 i j i,j
where s 2π~2 `Q = , mkBT
is the thermal de Broglie wavelength. The second part is hard to calculate in general.
Gibbs Entropy We want an alternative (as opposed to the Boltzmann formula) method for calculating the entropy. Given that ∂F S = − , ∂T V,N we have that
∂T ln ZN ∂ ln ZN kBT ∂ZN S = kB = kB ln ZN + kBT = kB ln ZN + ∂T ∂T ZN ∂T 1 1 ∂ZN hEi = kB ln ZN − = kB ln ZN + . T ZN ∂β T
But we also know that e−βEi Pi = , ZN which implies
ln Pi = −βEi − ln ZN . 24 Ensembles and Classical Phase Space
Then hln Pii = −β hEi − ln ZN .
The ZN comes outside the average since it is just some number. We can write our result as 1 k ln Z = −k hln P i − β hEi . B N B i T Plugging this into our result for S gives us
S = −kB hln Pii .
Remember, the ensemble average of a quantity O is calculated as 1 X hOi = O e−βEi . Z i N i So we can write our result as
1 X X e−βEi S = −k ln P e−βEi = −k ln P . B Z i B Z i N i i N This simplifies to X S = −kB Pi ln Pi. i This is Gibb’s expression for the entropy. It is often more useful than the Boltzmann entropy S = kB ln Ω since you don’t necessarily have to calculate Ω. We should check that the Gibb’s entropy is consistent with the Boltzmann entropy. Recall the Boltzmann democracy hypothesis, which states that all microstates of a given energy are equally probable 1 P = . i Ω(E) If we assume this is true, then the Gibb’s entropy is
Ω(E) Ω(E) X 1 1 ln Ω(E) X S = −k ln = k 1 = k ln Ω(E) = S . Gibbs B Ω(E) Ω(E) B Ω(E) B Boltz i=1 i=1 We can also derive the Gibb’s entropy by thinking about uncertainty. Suppose we have an ensemble of Ns systems. This ensemble consists of ni systems in state i (but P we don’t know which is which). Of course i ni = Ns. Then the number of possible configurations is N! W = . n1!n2! ··· This is a very large quantity and it quantifies our ignorance since larger values mean there are more configurations and we know less about its state. Since W is very large, one might want to consider its logarithm instead. Then k S = B ln W, Ns is also a quantification of the uncertainty of our ensemble. Applying Stirling’s approxi- mation gives us ! kB X X kB X S' N ln N − n ln n − N + n = N ln N − n ln n , N s s i i s i N s s i i s i s 2.4. Canonical Ensemble 25
P since i ni = Ns. This gives us X ni ni S = −k ln . B N N i s s
Then since Pi = ni/Ns, we get X S = −kB Pi ln Pi = SBoltz. i
Boltzmann’s H-theorem What can we say about the time evolution of the entropy? From the master equation, the rate of change of the probability of finding a state i in an ensemble of systems is X P˙i = (ωjiPj − ωijPi) . j
Now consider the time derivative of the Gibbs entropy X X X X S˙ = −kB P˙i [1 + ln Pi] = −kB P˙i − kB P˙i ln Pi = −kB P˙i ln Pi. i i i i P ˙ Notice that i Pi goes away because we can pull the time derivative outside the sum, d P d then dt Pi = dt 1 = 0. Applying the master equation, X S˙ = −kB (ωjiPj − ωijPi) ln Pi. i,j
Since ωji = ωij for processes that are invariant under time reversal, this simplifies to X S˙ = kB ωij (Pi − Pj) ln Pi. i,j
1 You can convince yourself that ln Pi = 2 (ln Pi − ln Pj). Then 1 X Pi S˙ = k ω (P − P ) ln . 2 B ij i j P i,j j
The last two pieces will always have the same sign, so this quantity is never negative. That is, dS S˙ = ≥ 0. dt This is just the second law of thermodyamics (entropy never decreases) from a statistical mechanics approach.
Optimization using Lagrange Multipliers In statistical mechanics, we often need to maximize a multivariable function given some constraints. Constrained optimization problems like that can be solved using the method of Lagrange multipliers. Given a multivariable function f with no constraints, we would maximize f by taking the gradient and setting it equal to zero
∇f(~x) = 0. ~x=~x0 26 Ensembles and Classical Phase Space
Just like we would set the derivative equal to zero if it was a single-variable function. Now suppose we are on a hypersurface defined by the constraints
g(~x) = 0.
That is, in our n-dimensional space, we are constrained to move only on the (n − 1)- dimensional hypersurface defined by our constraints. Suppose we are at a point ~y on this hypersurface. Now consider a small displacement δ~y. To stay on the hypersurface, the only directions we’re allowed to move in are those which are parallel to the hypersurface. That is, we are only allowed to move in directions that are perpendicular to the gradient of the hypersurface. In other words, we are allowed to move in directions δ~y which satisfy
δ~y · ∇g(~x) = 0. ~x=~y
We know that f is stationary in directions that are orthogonal to the gradient of f. So a necessary condition for ~y to be an extremum of f subject to the constraint g(~x) = 0, is that the gradients of f and g point in the same direction. So
∇f(~x) = λ ∇g(~x) , ~x=~y ~x=~y
for an arbitrary scalar λ. The scalar is there since we don’t care about the magnitudes of the gradients but only their directions. The constraint g = 0 implies that to find the extremum of f subject to g, we need to solve the system of equations
∇ (f − λg) = 0
∂λ (f − λg) = 0.
If we define L(~x, ~λ) = f(~x) − λg(~x), or in the case of multiple constraints by X L(~x, ~λ) = f(~x) − λigi(~x), i Then finding the extremum of f subject to g is equivalent to finding the extremum of L
∇~x,~λL = 0.
Generating Ensembles by Maximizing the Entropy We will now consider an algorithmic approach to calculating the entropy in a specified ensemble. We know that for two systems in thermal contact, equilibrium occurs when the Boltzmann entropy is maximized. In equilibrium, the system will most likely be in the most probable state, and this is the state associated with the equilibrium. Since S ∝ ln W , maximizing the entropy is equivalent to maximizing the possible configurations W of the ensemble. For example, for the microcanonical ensemble, the allowed states have fixed energy E, particle number N, and volume V . To find the probabilities of different states in the microcanonical ensemble, we need to maximize the entropy X f = S = −kB Pi ln Pi, i 2.4. Canonical Ensemble 27 with the constraint X g = Pi − 1 = 0 i i.e. that the sum of all probabilities must be 1. Here, i runs over only the allowed states. Applying the method of Lagrange multipliers, X L = λ − (kBPj ln Pj + λPj) . j
Taking the gradients with respect to the Pi and λ gives us the system of equations ∂ X λ − (k P ln P + λP ) = 0 ∂P B j j j i j ∂ X λ − (k P ln P + λP ) = 0. ∂λ B j j j j
This simplifies to
−kB ln Pi − kB − λ = 0 X 1 − Pj = 0. j The second equation doesn’t give us anything that we didn’t already know—that the sum of probabilities is equal to 1. The first equation tells us that
−1−λ/kB const Pi = e = e = const. That is, all the probabilities are equal. This implies the Boltzmann democracy hypothesis 1 P = . i Ω For the canonical ensemble, we have fixed N and V but our microstates can have any energy. However, in the end, we want our system to have some defined average energy X hEi = PiEi. i So our constraint is X g = hEi − EiPi = 0, j so we include an additional Lagrange multiplier β and then X L = λ + β hEi − (kBPj ln Pj + λPj + βEjPj) . j Again the gradient with respect to the Lagrange multipliers is trivial. For the other one we get ∂ L = −kB ln Pj − kB − λ − βEj = 0. ∂Pi This implies that −βEj Pj ∝ e . Normalizing gives us the probability expression for the canonical ensemble
e−βEj Pj = . ZN 28 Ensembles and Classical Phase Space
2.5 Grand Canonical Ensemble
In the grand canonical ensemble, both energy and particle number are variable. This ensemble is appropriate when the system is in contact with a large reservoir with which it can exchange energy and particles. In the thermodynamic limit, we can use any ensemble—they all give the same macro- scopic results. The grand canonical ensemble is often the most convenient one to do calculations in. We now have a third constraint in addition to the two from the microcanonical and canonical ensembles. This constraint is the particle number has some expectation value X X hNi = NiPi =⇒ hNi − NiPi = 0. i i Since we have an additional constraint, we need an additional Lagrange multiplier. We will use α. Now, X L = λ + β hEi + α hNi − [kBPj ln Pj + λPj + βPjEj + αPjNj] . j
Taking the interesting gradient gives us ∂ L = −kB ln Pi − kB − λ − βEi − αNi = 0, ∂Pi which implies that −βEi−αNi Pi ∝ e . This is the probability to find the system in a particular state in the grand canonical ensemble. Our normalization factor is the grand canonical partition function
X X Z = e−βEi−αNi = e−βEi+βµNi , i i
where µ α = − = −µβ. kBT
Note that we will use Z to denote the grand canonical partition function and ZN to denote the canonical partition function. Then the probability to find the system in state i is e−βEi−αNi P = . i Z We can easily calculate the averages of E and N. For example,
1 X hEi = E e−βEi−αNi . Z i i This can be written as ∂ ln Z hEi = − . ∂β α For particle number, we get
1 X hNi = N e−βEi−αNi . Z i i 2.5. Grand Canonical Ensemble 29
This can be written as Tip ∂ ln Z hNi = − . In the thermodynamic ∂α β limit, expectation val- ues are just equal to the To connect back to thermodynamics, we calculate the Gibb’s entropy quantities. For example, X X hEi = E. S = −kB Pi ln Pi = kB Pi (ln Z + βEi + αNi) . i i This simplifies to
S = kB ln Z + kBβ hEi + α hNi . We could also have calculated the entropy by using the reservoir method like we did for the canonical ensemble. This approach would have shown us that 1 β = , kBT and ∂S µ α = = − . ∂N E,V T With these, we can write the entropy as 1 µ S = k ln Z + hEi − hNi , B T T and rearrange to get
Ω ≡ −kBT ln Z = hEi − TS − µ hNi , where Ω is the grand potential. For an extensive system, the equation above is also equal to −PV . That is,
E − TS − µN = −PV, and ln Z = βP V. Note: In the thermodynamic limit, expectation values are just equal to the quantities. For example, hEi = E and hNi = N. Example 2.5.1: Classical Ideal Gas
For the grand canonical ensemble, the partition function is X Z = e−βEi+βµNi . i We can split this into two pieces. Think of it as a sum over microcanonical ensembles ∞ X X Z = eβµNj e−βE(Nj ), j=0 m=1 where the first sum is over the different possibilities of particle number, and the second sum is over the energies. I.e. the second sum is the sum over the energies of all the microstates which have Nj particles. We can think of the second sum as 30 Ensembles and Classical Phase Space
a canonical partition function. That is,
∞ X βµNj Z = e ZNj . j=0
Recall for a classical ideal gas, the canonical partition function is
!N 1 V 1 N ZN = 3 = ζ , N! `Q N!
where r 2π 2 ` = ~ , Q mT is the thermal de Broglie wavelength, and
V ζ = 3 , `Q
is the 1-particle canonical partition function. Then
∞ N X 1 X (λζ) Z = eβµNj ζNj = . Nj! N! j=0 N
where λ = eβµ, is the fugacity. But now the sum is just the series expression for the exponential function Z = eλζ . Then PV = ln Z = λζ = hNi , T which is the ideal gas law since
∂ ln Z hNi = = λζ. ∂α β,V If we grab a system from our ensemble, the probability of getting one with N particles is λN Z λN ζN P (N) = N = e−λζ . Z N! This simplifies to hNiN e−λhNi P (N) = . N! Notice that this is a Poisson distribution with mean hNi and standard deviation phNi. 2.6. Summary: Ensembles and Classical Phase Space 31
2.6 Summary: Ensembles and Classical Phase Space
Skills to Master • Understand the difference between the three named ensembles • Calculate the number of microstates Ω for a microcanonical ensemble • Calculate the partition function in the canonical and grand canonical ensembles • Calculate various thermodynamic quantities from Ω, ZN , and Z
The goal of statistical mechanics is to relate mi- Canonical ensemble: Energy E is variable and a croscopic quantities of a large number of particles to system is characterized by the parameters N, V , the macroscopic (i.e. thermodynamical) quantities of and T . the whole system. In general, we have an N particle system with co- Grand canonical ensemble: Energy E and particle number N are both variable. ordinates and momenta q1, . . . , q3N , p1, . . . , p3N . Our system corresponds to a single point in a 6N- dimensional phase space. As our system evolves, its Microcanonical Ensemble corresponding point in phase space traces out some tra- The microcanonical ensemble is essentially the brute jectory. Now instead of a single system corresponding force approach. to a single point in phase space, let’s think about an en- In the microcanonical ensemble, the macrostate is tire ensemble of systems (with the same H) correspond- defined by a fixed number of particles N, a fixed vol- ing to an entire ensemble of points moving through our ume V , and a fixed energy E or a fixed energy range phase space. An ensemble is simply many different E − 1 ∆,E + 1 ∆. Then the chief problem is to cal- copies of our system. 2 2 culate the number of microstates Ω(N,V,E) that are The phase space density is a function of the form accessible to the system. For the microcanonical ensemble, the phase den- ρ = ρ (q , . . . , q , p , . . . , p ) . i 3N i 3N sity ρ is constant, and the phase volume is the volume of the “shell” defined by For equilibrium, ∂tρ = 0. The volume of a region of phase space is 1 1 E − ∆E ≤ H(q, p) ≤ E + ∆E. 2 2 ω = dω, ˆ The number of microstates with energy E ± ∆E is ∂Σ where Ω(E) = ∆E. 3 3 3 3 dω = d q1 ··· d qN d p1 ··· d pN , ∂E and the integral goes over the region. where the number of microstates with energy less than E is The ensemble average of a function f is defined as dω Σ(E) = θ (E − H) . dω fρ ˆ ω0 hfi = . ´ dω ρ Here, θ is the Heaviside step function, and in general ´ the infinitesimal phase space volume corresponding to 3N Liouville’s theorem tells us that a single microstate is, ω0 = h . For discrete states, e.g. for a paramagnet, ∂ ρ + {ρ, H} = 0. t X Ω(E) = δEi,E. Different ensembles can be generated by maximiz- i ing the Gibbs entropy and using the method of La- The Boltzmann democracy hypothesis, grange multipliers. Think of the named ensembles as a hierarchy: 1 P = , i Ω(E) Microcanonical ensemble: A system is character- ized by the parameters N, V , and E (or an energy is that all microstates of a given energy are equally range). probable. 32 Ensembles and Classical Phase Space
Calculations in the microcanonical ensemble are From this, we can obtain S, P , and µ by taking appro- long and tiresome. Hence, we prefer more convenient priate derivatives ensembles. ∂F Note: If the particles are distinguishable, then we = −S have to drop the 1/N! factor from Σ(E). ∂T N,V 1. Starting from the Hamiltonian of the system, cal- ∂F = −P culate the number of microstates Ω(N,V,E) ac- ∂V T,N cessible to the system ∂F 2. Calculate the entropy S = kB ln Ω = µ. ∂N 3. Calculate µ, P , and T by differentiating S or ln Ω T,V In the classical limit, the canonical ensemble is Canonical Ensemble given by ρ = e−βH. The canonical ensemble takes a different approach from the brute calculation of Ω done in the microcanonical The continuous version of the canonical partition func- ensemble. In the canonical ensemble, our system is tion is 1 dω in thermal contact with a reservoir, with which it ex- Z = e−βH. N N! ˆ ω changes energy until they reach a mutual equilibrium. 0 In the canonical ensemble, the macrostate is defined Note, that this is for indistinguishable particles. If the by the parameters N, V , and T . The energy E of the particles are distinguishable, then we don’t need the system can vary. 1/N!. Reasoning in this manner, we conclude that the The Gibb’s expression for the entropy is probability of finding our system in a state with en- X S = −kB Pi ln Pi. ergy Ej is i e−βEj P (Ej) = , It is often more useful than the Boltzmann entropy ZN S = kB ln Ω since you don’t necessarily have to calcu- where the normalization given by the canonical parti- late Ω. tion function is a sum over all the possible microstates In the canonical ensemble, the Gibb’s entropy sim- of the system X plifies to Z = e−βEj . N ∂ j S = k (T ln Z ) = k ln Z + k β hEi . B ∂T N B N B Knowing this partition function tells us a lot about the properties of our subsystem. The general approach in the canonical ensemble is The thermal average of any quantity O is 1. Calculate the partition function ZN 2. Calculate various thermodynamic quantities 1 X from ZN and ln ZN hOi = O e−βEj . Z j N j Grand Canonical Ensemble For the energy, In the grand canonical ensemble, both energy and par- ∂ ∂ ticle number are variable. This ensemble is appropri- hEi = − ln Z = T 2 ln Z . ∂β N ∂T N ate when the system is in contact with a large reservoir with which it can exchange energy and particles. The Helmholtz free energy is In the thermodynamic limit, we can use any ensemble—they all give the same macroscopic results. F = E − T S. The grand canonical ensemble is often the most conve- nient one to do calculations in. In the thermodynamic limit, The grand canonical partition function is X X F = −kBT ln ZN . Z = e−βEi−αNi = e−βEi+βµNi , i i The differential associated with the Helmholtz free en- where ergy is µ dF = −S dT − P dV + µ dN. α = − = −µβ. kBT 2.6. Summary: Ensembles and Classical Phase Space 33
The probability to find the system in state i is Model: Classical Ideal Gas
e−βEi−αNi The ideal gas law is P = . i Z PV = NkT To calculate averages, we again use X The Sackur-Tetrode equation for the entropy hOi = OiPi, " # S(N,V,E) V mE 3/2 5 but now Pi is of the form given above. The average = N ln 2 + N, energy and particle number simplify to k N 3π~ N 2 ∂ ln Z hEi = − can be derived in the microcanonical ensemble. ∂β α 3 ∂ ln Z E = NkT hNi = − . 2 ∂α β " # V mE 3/2 In the grand canonical ensemble, the Gibb’s en- µ = T ln . N 3π 2N tropy simplifies to ~ For adiabatic (i.e. constant entropy) changes, we S = k ln Z + k β hEi + α hNi . B B have Plugging α and β in and rearranging gives us the grand 3/2 potential VE = const VT 3/2 = const Ω ≡ −kBT ln Z = hEi − TS − µ hNi . PV 5/3 = const. For an extensive system, the grand potential is equal to −PV . That is, The canonical partition function is E − TS − µN = −PV = −k T ln Z. 1 B Z = ζN , N N! The general approach in the grand canonical en- semble is where 3 1. Calculate the partition function Z mT 2 ζ = V , 2. Calculate various thermodynamic quantities 2π~2 from Z and ln Z is the canonical partition function for a single particle. The grand canonical partition function is Model: Paramagnet λζ The energy of a particular configuration is Z = e . X 1 where E = −b σ , σ = ±1, b = gµ B . j j 2 B ext λ = eβµ, j is the fugacity. The partition function is In the grand canonical ensemble, the expected par- P X βb σi N ZN = e i = [2 cosh(βb)] . ticle number is configurations ∂ ln Z hNi = = λζ. We can then calculate ∂α β,V
hσki = tanh(βb). and the probability of our system having N particles simplifies to a Poisson distribution The average total spin of the system is hNiN e−λhNi N hσki = N↑ − N↓ = Nx, P (N) = , N! which is the same result we found using the micro- canonical ensemble. with mean hNi and standard deviation phNi. 34 Ensembles and Classical Phase Space
Miscellaneous Thermal de Broglie wavelength is
The full time derivative of a function f({xi}, t) is s 2π~2 df ∂f X ∂f `Q = . = + x˙ . mkBT dt ∂t ∂x i i i Note: In the thermodynamic limit, expectation The Poisson bracket is values are just equal to the quantities. For example, X ∂f ∂g ∂f ∂g {f, g} = − , hEi = E and hNi = N. ∂qi ∂pi ∂pi ∂qi Remember: where qi are the generalized coordinates and pi are the generalized momenta. adiabatic =⇒ S = const The total time derivative of a function =⇒ N = const f({qi(t), pi(t)}) of the coordinates and momenta can isothermal =⇒ T = const. be written in the form of a Poisson bracket of the Hamiltonian as df ∂f = + {f, H}. dt ∂t Chapter 3
Thermodynamic Relations
3.1 Thermodynamic Potentials
The thermodynamic identity relates small changes in energy of a system in equilibrium to small changes in its entropy and volume as dE = T dS − P dV. This is the fundamental thermodynamic relation, and it can be derived from the first and second laws of thermodynamics. If the particle number can change, then an additional term is added dE = T dS − P dV + µ dN. The internal energy E is an example of what is sometimes called a thermodynamic potential. By applying Legendre transformations, we can transform the fundamental thermodynamic identity into other forms. Our partition functions are related to the energy of the system and its derivatives. So when we go from an extensive independent variable to an intensive one, we are going from an expression in terms of a particular quantity to the derivative of that quantity. Intensive quantities are essentially derivatives of extensive quantities. Consider the Helmholtz free energy F = E − T S, for example. It comes from the canonical ensemble which has natural variables T , N, and V , and so, derivatives of F with respect to these natural variables give us the thermody- namic properties of the system. To make the transformation from one set of variables to derivatives in another set of variables, we do Legendre transforms. Suppose we have a generic function Y (X) with differential dY = x dX, where ∂Y x = . ∂X We want an expression that gives the information in Y (X), but we want the expression to ∂Y be in terms of x instead of X. The naive approach is to calculate x(X) = ∂X as a function of X, and then invert it to get X(x), which we plug into Y to get Y (X) = Y (X(x)) = Y (x). However, this gives a solution that is not unique. We lose a normalization constant when we take this approach. Another approach is to express the information in Y (X) in terms of lines which are tangent to Y (X). We can use this approach provided that Y (X) is a convex function. The equation for the line that is tangent to Y (X) at the point X = X0 is ∂Y y(X,X0) = (X − X0) + Y (X0). ∂X X0 36 Thermodynamic Relations
We can write this as y(X,X0) = x (X − X0) + Y (X0). The vertical intercept of this line is
y(0,X) = Y (X0) − xX0, so we can write the tangent line in terms of its vertical intercept as y(X) = xX + y(0,X). This expression contains all of the same information as Y , so as long as Y is concave or convex, we can find a unique transformation X(x) such that we can write y(x) = Y (X(x)) − xX, without losing any information. The associated differential is dy = −X dx. In summary, suppose we have Y (X,A), and we want y(x, A), where ∂Y x = , ∂X A and A stands for whatever variables are being held constant. Then we take the Legendre transform ∂Y y(x, A) = Y − xX = Y − X. ∂X A
Example 3.1.1
Suppose we want a potential that has the independent variables P , S, and N, where ∂E P = − . ∂V S,N We would start with E(N, S, V ) and transform. We can define a new potential that is the Legendre transform of E as
∂E H(P, S, N) = E − V = E + PV. ∂V S,N Incidentally, this potential is the enthalpy. In this example, we had Y = E, x = P , X = V , and A = S, N.
Useful thermodynamic potentials include the internal energy E, the Helmholtz free energy F , the Gibbs free energy G, the enthalpy H, and the grand potential Ω. These are defined as
E(N,V,S) = E, Energy F (N,V,T ) = E − TS Helmholtz free energy G(N,P,T ) = F + PV = E − TS + PV = µN Gibb’s free energy H(N,P,S) = E + PV Enthalpy Ω(µ, T, V ) = F − µN = E − TS − µN Grand potential
You need to know how to get to these via Legendre transforms, need to know their natural variables, and need to know how to get other thermodynamic quantities from them. 3.1. Thermodynamic Potentials 37
Example 3.1.2
Derive the grand potential
Ω = F − µN,
via a Legendre transform of another thermodynamic potential. With Ω written in this form, it is easy to note the relevant variables. On the right side we have a lone variable F and a pair of variables −µN. So we know that Y = F . I.e. Ω is a Legendre transform of the Helmholtz free energy F . Further, ∂F we know that the pair of variables −µN must be −xX = − ∂X X. But which is which? To figure that out, we need to look at the differential of F
dF = −S dT − P dV + µ dN.
From this, we see that ∂F ∂F ∂F = −S, = −P, = µ. ∂T ∂V ∂N
∂F By comparison, we want the third one. So in our Legendre transform, x = µ = ∂N , and the variables being held constant when taking this derivative are A = V,T .
Thermodynamic Potential Y X x A
Helmholtz Free Energy, F E S T N,V
Gibbs Free Energy, G F V P N,T
Enthalpy, H E V P S, N
Grand Potential, Ω F N µ T,V Table 3.1: Legendre transformations between thermodynamic potentials.
Example 3.1.3
Write down the differential of the enthalpy, which is defined as
H = E + PV.
We write down the total differential of H in terms of its displayed variables E, P , and V as ∂H ∂H ∂H dH = dE + dP + dV = dE + V dP + P dV. ∂E ∂P ∂V From the fundamental relation dE = T dS − P dV + µ dN, we can rewrite this as
dH = T dS + V dP + µ dN.
Compare this with the total differential of H in terms of its natural variables P , S, and V ∂H ∂H ∂H dH = dS + dP + dN, ∂S N,P ∂P N,S ∂N S,P 38 Thermodynamic Relations
and we see that ∂H ∂H ∂H = T, = V, = µ. ∂S N,P ∂P N,S ∂N S,P
The differentials of these quantities are
dE = T dS − P dV + µ dN dF = −S dT − P dV + µ dN dG = −S dT + V dP + µ dN dH = T dS + V dP + µ dN dΩ = −S dT − P dV − N dµ.
Note, for systems with an intensive grand potential, i.e. the grand potential can be written in the form Ω(µ, T, V ) = V ω(µ, T ), then Ω = −PV. The first four of these relations can be rememembered with the help of the ther- modynamic square shown below. Note that U needs to be replaced with E. Then to get the first relation dE, we note that −P and T are on the opposite corners. These are the variables to include. They are multiplied by differentials of the variables on the opposite corners. For example, −P is multiplied by dV . In the end, always add µ dN. To remember the square itself, you can use a mnemonic like “Some Hard Problems Go To Finish Very Easy”.
Next, we can easily derive three partial derivative relations from each of these iden- tities. Consider the first form of the identity dE = T dS − P dV + µ dN, noting that the right side has differentials with respect to S, V , and N. These are the variables we differentiate the whole thing with respect to. For example, if we differentiate with respect to S, we would hold V and N constant (which means dV = dN = 0), and then dS/∂S = 1, and we would get the relation ∂E = T. ∂S N,V We can get two more such relations from the first form of the identity by differentiat- ing with respect to V and N—each time holding the other two variables constant. All together, from the first form of the identity, we get the relations ∂E ∂E ∂E = T, = −P, = µ. ∂S N,V ∂V S,N ∂N N,V 3.1. Thermodynamic Potentials 39
From the second form of the identity, we get
∂F ∂F ∂F = −S, = −P, = µ. ∂T N,V ∂V N,T ∂N V,T
From the third form of the identity, we get
∂G ∂G ∂G = −S, = V, = µ. ∂T P,N ∂P T,N ∂N T,P
Finally, from the fourth form of the identity, we get
∂H ∂H ∂H = T, = V, = µ. ∂S P,N ∂P S,N ∂N S,P
A useful rule of partial derivatives is as follows. If x, y, z are mutually related, then
∂x 1 = . ∂y ∂y z ∂x z
So we can invert all of the above relations to get another twelve partial derivative relations
∂S 1 ∂V 1 ∂N 1 = , = − , = ∂E N,V T ∂E S,N P ∂E N,V µ ∂T 1 ∂V 1 ∂N 1 = − , = − , = ∂F N,V S ∂F N,T P ∂F V,T µ ∂T 1 ∂P 1 ∂N 1 = − , = , = ∂G P,N S ∂G T,N V ∂G T,P µ ∂S 1 ∂P 1 ∂N 1 = , = , = . ∂H P,N T ∂H S,N V ∂H S,P µ
Another useful partial derivative relation is
∂ ∂y ∂ ∂y = . ∂z ∂x z x ∂x ∂z x z
This comes from the fact that the partial derivatives of functions with perfect differentials (functions like the thermodynamic potentials) must commute. This rule allows us to derive the Maxwell relations. Example 3.1.4
Derive the following Maxwell relation
∂T ∂N = − . ∂µ N,P ∂S T,P The first thing we note is that on the left we have N and P being held constant, and on the right, T and P are being held constant. So we look for a form of the fundamental identity that includes differentials dN, dP, dT . We find this in the form dG = −S dT + V dP + µ dN. 40 Thermodynamic Relations
Tip We can immediately see that
The trick in proving a ∂G ∂G ∂G = −S, = V, = µ. Maxwell relation is figur- ∂T P,N ∂P T,N ∂N T,P ing out which thermody- namic potential needs to be Next we use the partial derivative rule which allows us to switch the order used. To do this, we look at of partial derivatives. Taking the partial derivative with respect to N of the first the three different quanti- relation above and applying the rule gives us ties that are being held con- ! ! stant (two on each side) and ∂ ∂G ∂ ∂G = . then look for the potential ∂N ∂T P,N ∂T ∂N T,P whose natural variables are T,P P,N those quantities. But we know what the middle partial derivative is on both sides, so we have that
∂S ∂µ − = . ∂N T,P ∂T P,N Next we use the partial derivative rule which allows us to flip partial deriva- tives. This allows us to say that
1 ∂S ∂µ 1 − = − = = . ∂N ∂T ∂N T,P ∂T P,N ∂S T,P ∂µ P,N Equating the outside pieces and simplifying gives us
∂N ∂T − = . ∂S T,P ∂µ P,N
Suppose we have a function f that can be expressed as f(x, y) or as f(y, z) where z = z(x, y). Then its total differential in terms of x and y is
∂f ∂f df = dx + dy. ∂x y ∂y x
But we can also write the total differential in terms of y and z as
∂f ∂f df = dy + dz, ∂y z ∂z y
where ∂z ∂z dz = dx + dy. ∂x y ∂y x
This implies two more rules
∂ ∂f ∂z f(y, z) = ∂x y ∂z y ∂x y ∂ ∂f ∂f ∂z f(x, z) = + . ∂y x ∂y z ∂z y ∂y x 3.1. Thermodynamic Potentials 41
If f = x, we also have
∂x ∂z −1 = ∂z y ∂x y ∂x ∂x ∂z = − . ∂y z ∂z y ∂y x
∂x For example, the second rule allows us to calculate ∂y when we know z(x, y) but not x(y, z). Combining several of these rules gives us another rule
∂f ∂f ∂f ∂x ∂x−1 = − . ∂y x ∂y z ∂z y ∂y z ∂z y
This rule is particularly useful when we have f = f(y, z) and x = x(y, z). Example 3.1.5
Derive the Maxwell relation ∂T ∂N = − . ∂µ S,V ∂S µ,V We will approach this step-by-step.
Step 1: Look at the quantities that are being held constant—µ, S, V . We need a thermodynamic potential with these natural variables. I.e. in differential form, this thermodynamic differential needs to be in terms of dµ, dS, and dV . Step 2: We don’t have such a thermodynamic potential yet, so we need to con- struct it using a Legendre transform. Let’s call this new potential K, then we are looking for K(µ, S, V ). We can create this via a Legendre transform if it differs from another potential by a single variable and that difference is related as a derivative. In our case, we can make the Legendre transform ∂E E(N, S, V ) → K(µ, S, V ), with µ = . ∂N
Step 3: The recipe for the Legendre transform tells us that
K(µ, S, V ) = E − µN.
Step 4: We write down the differential form of K. Above, we wrote down K in terms of E, µ, and N. Using that, its differential form is ∂K ∂K ∂K dK = dE + dN + dµ = dE − µ dN − N dµ. ∂E ∂N ∂µ To get this in the right variables, we substitute in the differential form dE = T dS − P dV + µ dN to get
dK = T dS − P dV − N dµ.
Notice that we now have the differential form that we were looking for. 42 Thermodynamic Relations
Step 5: From this, we can easily generate the following derivative relations
∂K ∂K ∂K = T, = −P, = −N. ∂S µ,V ∂V µ,S ∂µ S,V
Step 6: Differentiating one of these pairs with respect to one of the variables being held constant, should lead us to the right Maxwell equation. The ∂T Maxwell equation we seek includes ∂µ with S, and V being held constant. The first relation above has T on the right side, so let’s differentiate both sides with respect to µ while holding S and V constant. ! ∂ ∂K ∂T = . ∂µ ∂S ∂µ µ,V S,V S,V Applying the partial derivative rule that lets us switch derivatives gives us ! ∂ ∂K ∂T = . ∂S ∂µ ∂µ S,V µ,V S,V
∂K But we know what the inside derivative is. We found earlier that ∂µ = S,V −N. Plugging this in gives us
∂N ∂T − = , ∂S µ,V ∂µ S,V which is precisely what we were looking for.
3.2 Fluctuations
In general, the fluctuation of a quantity A is
D E h∆A2i = (A − hAi)2 = hA2i − hAi2 .
Suppose we have two systems in equilibrium with each other. Let system 1 be much smaller than system 2. How will the total system fluctuate away from equilibrium if they are allowed to exchange E and V ? We know that Etot = E1 + E2 and Vtot = V1 + V2. The ratio of probabilities is
Pf Ωf Ω1(E1,V1)Ω2(Etot − E1,Vtot − V1) = = ∗ ∗ ∗ ∗ , Peq Ωeq Ω1(E1 ,V1 )Ω2(E2 ,V2 ) where subscript f denotes fluctuation (i.e. away from equilibrium) and eq denotes equi- librium. The asterisk denotes an equilibrium value. The denominator of the right-hand side contains maximum values, so we expect Pf /Peq < 1, which implies ∆Stot < 0. Since S ∆S S −Seq S = ln Ω, we know that e = Ω and e = e f = Ωf /Ωeq, and so
Pf = e∆Stot . Peq Here, ∆Stot = Stot,f − Stot,eq = ∆S1 + ∆S2, 3.2. Fluctuations 43 since S is additive. Then since dE P dS = + dV, T T we can write f dE2 + P2dV2 ∆Stot = ∆S1 + , ˆeq T2 where the second term is ∆S2. By Etot = E1 + E2 and Vtot = V1 + V2, we can write dE2 = −dE1 and dV2 = −dV1, and so
f dE1 + P2dV1 ∆Stot = ∆S1 − . ˆeq T2 If we assume that system 2 is much larger than 1, then we expect fluctuations won’t change system 2 much, and so we can treat P2 and T2 as roughly constant. Then 1 P ∗ ∆S = ∆S − ∆E − ∆V . tot 1 T ∗ 1 T ∗ 1 Now we have the change in total entropy expressed only in terms of the fluctuations of the small system. Can we eliminate ∆E1 from the equation above? We should be able to since we know that E, S, and V are not independent. Expanding ∆E1 in a Taylor series gives us ∂E ∂E ∗ ∗ ∆E1 = ∆S1 + ∆V1 = T ∆S1 − P ∆V1. ∂S V ∂V S,N
If we plug this into the equation for ∆Stot everything cancels. So we need to take our Taylor expansion to second order
∂E ∂E 1 ∂2E ∂2E ∂2E ∆E = ∆S + ∆V + ∆S2 + 2 ∆V ∆S + ∆V 2 . 1 ∂S 1 ∂V 1 2 ∂S2 1 ∂S∂V 1 1 ∂V 2 1 Using known partial derivative relations, we can simplify this as 1 ∂T ∂T ∂P ∆E = T ∗∆S − P ∗∆V + ∆S2 + 2 ∆V ∆S − ∆V 2 . 1 1 1 2 ∂S 1 ∂V 1 1 ∂V 1 We can also expand ∂T ∂T ∆T1 = ∆S1 + ∆V1. ∂S N ∂V S With this, we can write 1 ∂T ∂P ∆E = T ∗∆S − P ∗∆V + ∆T ∆S + ∆V − ∆V 2 . 1 1 1 2 1 1 ∂V 1 ∂V 1
Plugging this into ∆Stot, we find that we can write the entropy fluctuation as 1 ∆S ' − [∆S ∆T − ∆P ∆V ] . 2T 1 1 1 1 We’ve dropped the “*” on the T out front. Just assume that everything with no “∆” in front is an equilibrium value. We now have the entropy in terms of four fluctuations. We want to express two of those fluctuations in terms of the other two. Consider when the independent variables are T and V . We could just as easily have expanded ∆S as ∂S ∂S ∆S = ∆T + ∆V, ∂T ∂V 44 Thermodynamic Relations
and similarly to expand ∆P . Then we find that
∂S ∂S ∂P ∂P −2T ∆S = ∆T 2 + ∆T ∆V − ∆T ∆V − ∆V 2. ∂T N,V ∂V ∂T ∂V
But we also have the Maxwell relation ∂P ∂2F ∂S = − = , ∂T V ∂T ∂V ∂V so ∂S ∂P ∆T ∆V = ∆T ∆V, ∂V ∂T which implies ∂S ∂P −2T ∆S = ∆T 2 − ∆V 2. ∂T N,V ∂V This implies that
" ( )# 1 ∂S 2 ∂P 2 Pf ∝ exp − ∆T − ∆V . (3.1) 2T ∂T N,V ∂V T,N
We see that these are essentially Gaussian distributions. For a Gaussian,
2 2 p(x) ∝ e−x /2hx i.
Splitting Eq. (3.1) into two Gaussians, we can simply read off the mean square values
∂S −1 h∆T 2i = T ∂T N,V ∂P −1 h∆V 2i = −T . ∂V N,T
Note: The averages, ∆T and ∆V are just zero, since these are fluctuations from equilib- rium. Doing a similar process for S and P , we find that
∂T ∂V −2∆T ∆S = ∆S2 − ∆P 2, ∂S N,P ∂P N,S
we find that
∂T −1 h∆S2i = T ∂S N,P ∂V −1 h∆P 2i = −T . ∂P N,S
Notice that the fluctuations of an intensive quantity squared goes as the derivative of an intensive quantity with respect to an extensive quantity. So we expect 1 1 h∆T 2i ∼ =⇒ ph∆T 2i ∼ √ . N N So for macroscopic (i.e. large N) systems, the fluctuations are small. 3.2. Fluctuations 45
Recall that the probability of finding a system away from its equilibrium when con- nected to an energy and volume reservoir is 1 P ∝ exp − (∆S∆T − ∆P ∆V ) . f 2T What about energy fluctuations? Since the fluctuations are small, we can expand ∂E ∂E ∆E = ∆T + ∆V. ∂T V,N ∂V T,N The ensemble average of ∆E2 is ∂E 2 ∂E 2 h∆E2i = h∆T 2i + h∆V 2i . ∂T V,N ∂V T,N Notice that when we square ∆E and then take the expectation value, the partial deriva- tives don’t get the angle brackets since they are evaluated at equilibrium and are thus constant. Normally there would also be a pair of cross-terms of the form h∆V ∆T i mul- tiplied by coefficients, but these are zero in this case since the fluctations in V and T are uncorrelated. In terms of heat capacity, we can write this as 2 2 2 ∂E 2 h∆E i = T CV + h∆V i . ∂V T,N On the other hand, if the system is connected to just an energy reservoir (and no volume is exchanged), then the probability of finding the system away from equilibrium is 1 P ∝ exp − ∆S∆T . f 2T In this case, we just get 2 2 h∆E i = T CV . ∂E 2 So when volume fluctutations are included, a correction term ∂V T,N h∆V i must be added. In the canonical ensemble, we had found that
∂2 h∆E2i = ln Z . ∂β2 N In the grand canonical ensemble, we get
∂2 h∆E2i = ln Z . ∂β2 α Remember that α = −βµ. The different fluctuations for the different ensembles all have a different character because they are connected to different reservoirs. Similarly, for number fluctuations,
∂2 ∂N h∆N 2i = ln Z = T . 2 ∂α β ∂µ T,V We also know that ∂ ln Z = − hEi = −E. ∂β α So in the grand canonical ensemble, we can write 2 2 ∂E 2 ∂E 2 ∂E 2 h∆E i = − = T = T CV + h∆N i . ∂β α,V ∂T α,V ∂N T,V 46 Thermodynamic Relations
3.3 Thermodynamic Response Functions
The heat capacity tells us how much the heat changes for a given energy input. We know that
dE = δQ + δW = δQ − P dV ∂E ∂E = dT + dV. ∂T V,N ∂V T,N We are leaving out the dN term since we are considering the case in which N is constant. For constant N and V , ∂E δQ = dT, ∂T N,V then the heat capacity at constant volume and particle number is
δQ ∂E ∂S T 2 CV = = = T = 2 . dT ∂T V,N ∂T N,V h∆T i
We can also look at the heat capacity at constant pressure. We start with dE = δQ − P dV . Then, since P is constant, we can write
d(E + PV ) = δQ.
Put E + PV = H, so dH = δQ. Then δQ ∂H ∂S 2 CP = = = T = h∆S i . dT P ∂T P,N ∂T N,P
From the chain rule, we know that ∂H ∂H ∂T = = T. ∂S P,N ∂T P,N ∂S P,N Other response functions include compressibilities, and thermal expansion. The isothermal compressibility is
1 ∂V h∆V 2i κT ≡ − = . V ∂P T,N VT
Adiabatic (i.e. isentropic) compressibility is
1 ∂V T κS ≡ − = 2 . V ∂P S,N V h∆P i
Thermal expansion at constant pressure and entropy is
1 ∂V αP = V ∂T P,N 1 ∂V αS = . V ∂T S,N
The big picture is that the response of a system to some input is intimately related to the fluctuations of the system. 3.4. Minimizing and Maximizing Thermodynamic Potentials 47
3.4 Minimizing and Maximizing Thermodynamic Potentials
In general, we want to find equilibrium states of our system by maximizing S. Sometimes we have a model entropy that depends on some parameter x, so that
S = S(E, V, N, x), where x is some non-fluctuating parameter. The natural thing to do is maximize S with respect to x to get the equilibrium state. Suppose you have a two-phase system—one is high density and the other is low density. Then x might be, for example, the fraction of material in the high density phase. Or maybe the parameter is an energy associated with the surface area between the two phases. Or with the paramagnet, for example, x could be the average magnetization. In that example, we would minimize S with respect to x to get the equilibrium state. What about the general case? Connect your system to a large energy reservoir, then
dStot = dSa + dSr, where a denotes our system, and r denotes the reservoir. Energy conservation implies that dSr = dEr/T , and so dE dS = dS − a . tot a T We can move T inside the differential since it is constant, then 1 1 dS = d (TS − E ) = − dF . tot T a a T a
So if we wanted to maximize Stot, with respect to some parameter, we would want to minimize Fa with respect to x to find the equilibrium value. We would minimize Fa since dSa ∝ −dFa. We can do something similar if we’re exchanging both particles and energy. Then dE µ 1 1 dS = dS − a + dN = d (TS − E + µN ) = − dΩ . tot a T T a T a a a T a
Now we would minimize Ωa with respect to x. Example 3.4.1: Paramagnet
Recall that we wrote the entropy of a paramagnet as 1 + x 1 + x 1 − x 1 − x S = −N ln − ln , 2 2 2 2 where 1 X x = σ = hσi , N i i is the average number of particles with spin in a particular direction. Now we treat x as a variational parameter in S. We can do this since we expect fluctuations in x to be small and not to change the results. We start by writing down the Helmholtz free energy
F (T, N, x) = −Nbx − TS(x).
Then we minimize F with respect to x ∂F NT 1 + x = −Nb + ln = 0, ∂x 2 1 − x 48 Thermodynamic Relations
which gives us 2b 1 + x = ln . T 1 − x Solving for x gives us b x = tanh . T This is the same result we got in the microcanonical ensemble. 3.5. Summary: Thermodynamic Relations 49
3.5 Summary: Thermodynamic Relations
Skills to Master • Know the fundamental thermodynamic relation • Be able to derive various potentials by performing Legendre transforms of other potentials • Given a thermodynamic potential, be able to write down its natural differential form • Given the differential form of a thermodyamic potential, write down the partial derivative relations • Be able to derive the Maxwell relations
Thermodynamic Potentials 1. Start by writing the differential in terms of the variables shown—E, T , and S The fundamental thermodynamic relation ∂F ∂F ∂F dE = T dS − P dV + µ dN, dF = dE + dT + dS. ∂E ∂T ∂S can be derived from the first and second laws of ther- Then plugging in the partial derivatives gives us modynamics. If the particle number is constant, then the last term is zero. By applying Legendre transfor- mations, we can transform the fundamental thermody- dF = dE − S dT − T dS. namic identity into other forms. Suppose we have Y (X,A), and we want y(x, A), 2. Simplify by plugging in the differential of the re- where lated potential. In this case the related potential ∂Y is E, meaning that F was originally obtained as x = , ∂X the Legendre transform of E. More simply, E is A the potential that appears on the right-hand side and A stands for whatever variables are being held con- of F . So plugging in dE = T dS − P dV + µ dN stant. Then we take the Legendre transform gives us
∂Y y(x, A) = Y − xX = Y − X. dF = −P dV − S dT + µ dN. ∂X A
By applying appropriate Legendre transforms of Note the differentials are now of V , T and N. E and F , we can easily derive the following thermody- These are the natural variables of F . namic potentials: Given a thermodynamic potential in differential form, it is very easy to write down several partial E(N,V,S) = E, Energy derivative relations. Look at the differentials on the right-hand side. These are your variables to alter- F (N,V,T ) = E − TS Helmholtz free energy nately divide by or hold constant. For example, with G(N,P,T ) = F + PV Gibb’s free energy dF = −P dV − S dT + µ dN, suppose we divide by H(N,P,S) = E + PV Enthalpy dT and hold V and N constant. Since V and N are Ω(µ, T, V ) = F − µN Grand potential constant, we should really be using partial derivative notation. Since V and N are constant, we know that The differentials of these quantities are dV = dN = 0. So only one term remains. So we have that dE = T dS − P dV + µ dN dF dT ∂F dF = −S dT − P dV + µ dN = 0 − S + 0 =⇒ = −S. dT const N,V dT ∂T dG = −S dT + V dP + µ dN N,V dH = T dS + V dP + µ dN From each differential, you can get many relations. For dΩ = −S dT − P dV − N dµ. example, from dF , you can not only get partial deriva- tives of F , you can also rearrange to get another differ- Writing down the differentials is easy. For exam- ential on the left side and then take partial derivatives ple, to find the differential of F = E − TS: of that. 50 Thermodynamic Relations
If x, y, z are mutually related, then
∂x 1 = ∂y ∂y z ∂x z ∂ ∂y ∂ ∂y = . ∂z ∂x z x ∂x ∂z x z The first one allows us to invert our partial derivative relations. The second allows us to derive the Maxwell relations. To derive a given Maxwell relation, 1. Start by noting the three quantities that are be- ing held constant. Let’s call these variables a, b, c. Find the differential form of the thermodynamic potential that has those variables in its differen- tials 2. Write down the three partial derivative relations that follow from this potential. Let’s call these partial derivatives ∂X , ∂Y , ∂Z . 3. Take the partial derivative with respect to a, b, or c of ∂X , ∂Y , or ∂Z (One of these should give the right result, and you should be able to deduce without having to do all nine possibilities) 4. Use the partial derivative rule that allows you to switch the order of a double partial derivative 5. Simplify to get the Maxwell relation Suppose we have a function f that can be ex- pressed as f(x, y) or as f(y, z) where z = z(x, y). Then we get two more rules:
∂ ∂f ∂z f(y, z) = ∂x y ∂z y ∂x y ∂ ∂f ∂f ∂z f(x, z) = + . ∂y x ∂y z ∂z y ∂y x In particular, we get useful simpler cases when f = x. Another rule is
∂f ∂f ∂f ∂x ∂x−1 = − . ∂y x ∂y z ∂z y ∂y z ∂z y This rule is particularly useful when we have f = f(y, z) and x = x(y, z).
Miscellaneous The heat capacity at constant volume is
∂2F C = −T . V ∂T 2 3.5. Summary: Thermodynamic Relations 51
Fluctuations Remember that α = −βµ. When a problem involves the change δf of a func- In general, the fluctuation of a quantity A is tion like f = f(x, y), think of it in terms of h∆A2i = h(A − hAi)2i = hA2i − hAi2 . ∂f ∂f Various fluctuations can be calculated as follows: δf = δx + δy. ∂x ∂y ∂S −1 h∆T 2i = T ∂T N,V Thermodynamic Response Functions ∂P −1 The heat capacities at constant volume and pressure h∆V 2i = −T are ∂V N,T −1 2 2 ∂T ∂E ∂S T h∆S i = T CV = = T = 2 ∂S N,P ∂T V,N ∂T N,V h∆T i −1 ∂H ∂S 2 ∂V C = = T = h∆S2i . h∆P i = −T P ∂T ∂T ∂P N,S P,N N,P ∂2 ∂N h∆N 2i = ln Z = T The isothermal and adiabatic (isentropic) com- 2 ∂α β ∂µ T,V pressibilities are ∂2 h∆E2i = ln Z 1 ∂V h∆V 2i ∂β2 N κ = − = T V ∂P VT ∂E 2 T,N = T 2C + h∆V 2i . 1 ∂V T V ∂V T,N κS = − = 2 . V ∂P S,N V h∆P i Note, if no volume is exchanged, i.e. the system is connected only to an energy reservoir, then the second Thermal expansion at constant pressure and en- term in the last line above is zero. tropy is We can also calculate the energy fluctuations in 1 ∂V grand canonical ensemble as α = P V ∂T ∂2 P,N h∆E2i = ln Z ∂β2 α 1 ∂V αS = . ∂E V ∂T S,N = − ∂β α,V ∂E = T 2 ∂T α,V 2 2 ∂E 2 = T CV + h∆N i . ∂N T,V Chapter 4
Quantum Statistics
4.1 The Density Matrix
Now we look at a more concrete way of talking about ensembles. The new approach is similar to the classical phase space density. Now we talk about an ensemble in Hilbert or Fock space. The density matrix is defined as
1 X ρˆ = |θi hθ| , N θ where N is the number of systems in our ensemble, and the |θi are normalized states which are members of those ensembles. For a generic operator O, the ensemble average (expectation value) is
1 X hOi = hθ|O|θi . N θ We can insert a complete set of orthogonal basis states to get
1 X X hOi = hθ|ii hi|O|θi N θ i 1 X X = hi|O|θi hθ|ii N θ i X X = hi| Oρˆ|ii . θ i
But this is just the trace of the operator Oˆρˆ, so
h i hOi = Tr Oˆρˆ .
What properties does the density matrix have? Since the |θi are normalized, we know that 1 X h1i = hθ|θi = 1. N θ This implies that Tr [ˆρ] = 1. This is useful for interpreting this stuff in terms of probability. 4.1. The Density Matrix 53
How can we expressρ ˆ in a complete set of states? We can expand X |θi = cθ,i |ii . i Then the elements ofρ ˆ in this basis are
X ∗ X ∗ X ∗ ρij = hi|ρˆ|ji = cθ,kcθ,` hi|`i hk|ji = cθ,kcθ,`δi`δkj = cθ,jcθ,i. θ,k,` θ,k,` θ
This implies that ρ is a Hermitian operator, and
∗ ρij = ρji. Since it is a Hermitian operator,ρ ˆ is an observable and has real eigenvalues. We can write the density matrix as
X ρˆ = wi |ρii hρi| , i where |ρii are the eigenstates ofρ ˆ, and wi are the real eigenvalues. Note that X Tr [ˆρ] = 1 = wi. i
This suggests that we can interpret the wi as probabilities. Let’s double-check that they are all non-negative. Consider
Tr [ˆρ |ρji hρj|] = wj 1 X X = c∗ c hρ |ρ`i hρ |ρ i hρ |ρ i N θ,m θ,` k m j j k θ k,`,m 1 X X = c∗ c δ δ δ N θ,m θ,` k` mj jk θ k,`,m 1 X X = c∗ c N θ,j θ,j θ k,`,m ≥ 0.
Since the wi are real, positive, and sum to 1, we can indeed interpret them as probabilities. The density matrix is for any ensemble—not just for stationary ensembles. For a pure state, we have only one system in the ensemble
ρˆ = |θi hθ| .
This implies that ρˆ2 =ρ, ˆ and Tr ρˆ2 = 1. For a non-pure state, i.e. any density matrix with more than one state in the ensem- ble, 2 X 2 ρˆ = wi |ρii hρi| .
2 Since each wi < 1, we know that wi < wi, and so Tr ρˆ2 < 1. 54 Quantum Statistics
To do equilibrium statistical mechanics, we know we want a stationary ensemble. Recall that for a stationary ensemble, dρˆ ∂ρˆ i h i = + Hˆ, ρˆ = 0. dt ∂t ~ The first term on the right is zero, assuming there is no explicit time dependence. So for stationary ensembles, we need Hˆ andρ ˆ to commute with each other h i Hˆ, ρˆst = 0,
which implies X ρˆst = wi |Eii hEi| . i That is, we can express ρ of stationary ensemble in the energy eigenstate basis. For the microcanonical ensemble,
X δEj ,E ρˆ (E) = |E i hE | . micro Ω(E) j j j In the canonical ensemble, we know that
e−βEn wn = . P −βEn n e So for the canonical ensemble, e−βHˆ ρˆcan = h i. Tr e−βHˆ Then the canonical partition function is
h −βHˆ i ZN = Tr e .
We can perform this trace in whatever basis we like. In the grand canonical ensemble, we have to consider the different particle num- bers. Then e−β(H−ˆ µNˆ) ρˆGC = h i, Tr e−β(H−ˆ µNˆ) and the partition function is h i Z = Tr e−β(H−ˆ µNˆ) . In general, the entropy can be calculated from the density matrix as
S = −kBTr [ˆρ lnρ ˆ] .
4.2 Indistinguishable Particles
Previously, we didn’t account for the symmetry or antisymmetry of the underlying wave functions. In higher density gases, we have to account for this. We will use the single particle basis ψα(x) where α denotes the single particle quan- tum numbers. For a system of N particles, the many-particle wave function can be written very generally as X X Ψ(x1, . . . , xN ) = ··· c(α1, . . . , αN ; t)ψα1 (x1) ··· ψαN (xN ).
α1 αN 4.2. Indistinguishable Particles 55
This wave function imposes no symmetry requirements, so it’s appropriate only for dis- tinguishable particles. For indistinguishable particles, if you exchange particle indices, the wave function shouldn’t change except for a minus sign in the case of fermions
Ψ(. . . , xi, . . . , xj,...) = ±Ψ(. . . , xj, . . . , xi,...).
So we need c(. . . , αi, . . . , αj,...) = ±c(. . . , αj, . . . , αi,...). This limits the number of states in the Hilbert space the particles can be in. For the symmetric case (i.e. with the plus sign), the order of the α’s in the coefficients does not matter. So we can write
c(α1, . . . , αN ) → c(n1, n2,...).
That is, the coefficients can be written in terms of the number of particles in each single particle state. I.e. n1 is the number of particles in state 1. Normalization requires that
X 1 = dτ|Ψ|2 = |c(α , . . . , α )|2, ˆ 1 N α1,...,αN where the integral is over all particle coordinates. We can just as well write this as X 2 X 2 N! |c(α1, . . . , αN )| = |c(n1, n2,...)| Q , n1! α1,...,αN n1,n2,... i P where the sum on the right has the constraint that N = ni. So we can write it in the new basis
X 2 Ψ = a(n1, n2, . . . , t)Φn1,n2,...(x1, x2,...)|a(n1, n2,...)| ,
n1,n2,... P where the sum has the constraint N = ni, and pQ ni! X Φ (x , . . . , x ) = i ψ (x ) ··· ψ (x ), n1,n2,... 1 N N! α1 1 αN N α1,...,αN where this sum is constrained to have n1 particles with α1. The symmetric case corre- sponds to bosons. For the antisymmetric case (i.e. with the minus sign), we get fermionic basis states of the form
F 1 X P Φ (x1, . . . , xN ; t) = (−1) ψα (x1) ··· ψα (xN ), n1,n2,... N! 1 N α1,...,αN where P is the number of particle permutations. The antisymmetric case corresponds to fermions. The above is all general for both interacting and noninteracting gases. Now we focus specifically on noninteracting gases. For the total energy, we have X En1,n2,... = niεi, i where, ni is the number of particles in state i, and εi is the energy of a single particle in state i. 56 Quantum Statistics
4.3 Thermodynamic Properties
The grand canonical partition function becomes h i Z = Tr e−β(H−ˆ µNˆ
X −β(H−ˆ µNˆ = hΦn1,n2,...| e |Φn1,n2,...i
n1,n2,... X −β P n (ε −µ) = e i i i
n1,n2,... Y X = e−niβ(εi−µ).
i ni We can write this as Y Z = zi, i where the product goes over the single-particle states, and
X −niβ(εi−µ) zi = e .
ni For fermions we get −β(εi−µ) zi = 1 + e . For bosons we get ∞ 1 X −niβ(εi−µ) zi = e = . 1 − e−β(εi−µ) ni=0 This is just the sum of a geometric series. So
( 1 (Bosons) 1−e−β(εi−µ) zi = . 1 + e−β(εi−µ) (Fermions)
The thermal average of ni is 1 X hn i = n e−β(Ej −µNj ) i Z i j 1 X −βmj (εi−µ) = mje zi mj
= ∂βµ ln zi 1 = . exp[β(εi − µ)] ∓ 1 This is what we typically call the distribution function
1 f∓(ε) = hnii = . exp[β(εi − µ)] ∓ 1
The minus sign is taken for bosons, and the plus sign is taken for fermions. Next we look at ln Z since that will eventually get us to Ω. X h i ln Z = ∓ ln 1 ∓ e−β(εi−µ) , i 4.3. Thermodynamic Properties 57 where the upper sign is taken for bosons, and the lower sign is taken for fermions. We can just as well convert this to a sum over energies of the system as X h i ln Z = ∓ N(ε) ln 1 ∓ e−β(ε−µ) , ε wehre N(ε) is the number of single-particle states that have energy ε. If Σ(ε) is the number of single-particle states with energy less than ε, then
X ∆ε X ∆ε N(ε) = ε + − ε − , 2 2 and so X Σ+ − Σ− h i ln Z = ∓ ∆ε ln 1 + e−β(ε−µ) . ∆ε ε Converting to an integral gives us
∞ h i ln Z ' ∓ dε g(ε) ln 1 ∓ e−β(ε−µ) , (4.1) ˆε0 where the density of single-particle states at ε is dΣ(ε) g(ε) = . dε Recall that we found previously that
PV ln Z = . T
Since Z is related to the pressure, we can simplify the integral above by parts—integrating g(ε) and differentiating the logarithm to get
∞ ln Z = β dε Σ(ε) f∓(ε). ˆε0 Recall that F = −T ln Z, from which we get that ∂F S = − . ∂T V,N Then from Eq. (4.1), we get that
∞ S = − dε g(ε)[f∓ ln f∓ ∓ (1 ± f∓) ln(1 ± f∓)] . ˆ0
Some other thermodynamic properties include ∂ ∞ N = T ln Z = dε g(ε) f∓(ε) ∂µ ˆε0 ∂ ∂ ln g P = T ln Z = dε Σ(ε)f (ε) ∂V ∂V ˆ ∓ ∂ ln Z ∞ E = = dε g(ε) ε f (ε). ˆ ∓ ∂β βµ=α ε0 58 Quantum Statistics
We can easily calculate Σ(ε) for d-dimensional, homogeneous, isotropic gases X X Σ(ε) = θ [ε − ε(ν1, . . . , νd, α)]
ν1,...,νd α d L X ' ddp θ [ε − ε (~p)] . 2π ˆ α ~ α Here, the ν are the quantum numbers of states in our box. For a 1D gas, we would only sum over ν1. The sum over α is for other states, e.g. spin. Differentiating gives us
d dΣ L X g(ε) = = ddp δ(ε − ε (~p)). dε 2π ˆ α ~ α Here we used the fact that the derivative of the Heaviside step function θ is a delta function. Using standard relations for the delta-function, we can write this as d L X δ(p − p0) g(ε) = ddp . 2π ˆ ~ α ∂εα ∂p p0
Example 4.3.1: 3D Gas Density of State P Let’s evaluate the density of state for a 3D non-relativistic gas. Suppose α is the sum over spin states, then from that, we get a factor of (2s + 1), and
3 ∞ L 2 δ(p − p0) g(ε) = (2s + 1) 4π dp p . 2π~ ˆ0 ∂ε ∂p p0
2 ∂ε If ε = p /2m, then ∂p = p/m. So
3 ∞ L m 2 g(ε) = (2s + 1) 4π dp p δ(p − p0) 2π~ p0 ˆ0 L 3 = (2s + 1) 4πmp0 2π~ (2s + 1)V 3 √ = m 2 2ε. 2π2~3 √ 3 In the end we made the substitutions L = V and p0 = 2mε.
Example 4.3.2: 2D Fermion Gas
In the previous example, we derived g(ε) for a 3D gas. The process is similar for a 2D gas, and we get A g2d(ε) = (2s + 1)m, 2π~2 where A denotes the area. Notice that there’s no energy dependence in the density of states for 2D. Then the number of fermions with energy ε is A ∞ 1 N = (2s + 1)m dε. 2d 2 β(ε−µ) 2π~ ˆ0 e + 1 4.3. Thermodynamic Properties 59
If we make the substitution x = βµ, we can write this as A N2d = (2s + 1)mT h(µ), 2π~2 where ∞ 1 h(µ) = x−βµ dx, ˆ0 e + 1 is a function of µ. The energy is 1 ∞ x E = NT x−βµ dx. h(µ) ˆ0 e + 1 If βµ << 0, we can drop the “1” in the denominator of the integrals, then they become trivial to evaluate, and we get
h(µ) ' eβµ
E2d ' NT
A βµ N2d ' (2s + 1)mT e . 2π~2
For a d-dimensional, noninteracting quantum gas, the density of state simplifies to
d −1 L d−1 ∂ε g(p) = (2s + 1) sd p , 2π~ ∂p where sD = 2, 2π, or 4π is the surface area you’re integrating over after the angles drop out. We can also write this in terms of k = p/~ as
V ∂ε −1 g(k) = (2s + 1) s kd−1 . (2π)d d ∂k
For a non-relativistic gas i.e. ε = p2/2m, g(p) becomes
d L d/2−1 g(ε) = (2s + 1) sd2m(2mε) . 2π~
Following are the densities of state for quantum gases composed of free non-relativistic particles
(2s + 1)L 2 g(ε) = √ m1/2ε−1/2, (1D) (2π~) 2 (2s + 1)A g(ε) = 2πm, (2D) (2π~)2 √ (2s + 1)V 3 1/2 g(ε) = 4 2π m 2 ε , (3D) (2π~)3
For an ultra-relativistic gas (i.e. ε = pc), g(p) becomes
L d s εd−1 g(ε) = (2s + 1) d . 2π~ c c 60 Quantum Statistics
Following are the densities of state for quantum gases composed of free ultra-relativistic particles (2s + 1)L 2 g(ε) = , (1D) (2π~) c (2s + 1)A 2π g(ε) = ε (2D) (2π~)2 c2 (2s + 1)V 4π g(ε) = ε2, (3D) (2π~)3 c3 For a Bose gas, we have 1 hn(ε)i ∼ . ex − 1 This is illustrated by the blue curve in the figure below. For a Fermi gas, we have 1 hn(ε)i ∼ , ex + 1 which is illustrated by the red curve. For large x = β(ε − µ), we can approximate both boson and fermion gases as Boltzmann gases with 1 hn(ε)i ∼ . ex
4.4 Chemical Potential
This section is a sidebar on chemical potential. If two systems capable of exchanging particles are in equilibrium with each other, then the chemical potential is the same for the two systems. The chemical potential is defined in terms of the derivative of various energies with respect to particle number ∂E ∂F ∂G ∂H µ = = = = . ∂N S,V ∂N T,V ∂N T,P ∂N S,P You can think of chemical potential as the change in the energy of the system if you add one particle to it while holding S and V constant. However, since adding a particle to the system significantly increases the entropy of the system, in order to keep the entropy constant while adding the particle, the energy must decrease to compensate. Thus, the chemical potential of most gases is negative because in order to keep S constant as N is increased, the energy E must decrease. In fact, • Classical gas: For a classical aka Boltzmann gas, µ < 0 4.4. Chemical Potential 61
• Bose gas: For a Bose gas, µ ≤ 0, which implies you can add any number of particles to the ground state of the system. For a photon gas, µ = 0. • Fermi gas: For a Fermi gas, µ > 0, which implies that the system resists the addition of more particles.
Intuitively, if µ < 0, the system will happily accept more particles, and if µ > 0, the system will resist the addition of new particles. The chemical potential governs the flow of particles between systems just like tem- perature governs the flow of energy between systems. The chemical potential increases as particle concentration increases, and particles tend to flow from regions of high chemical potential to regions of low chemical potential. 62 Quantum Statistics
4.5 Summary: Quantum Statistics
Skills to Master • Given a Hamiltonian, calculate the density matrix in various ensembles • Calculate the density of states for a non-relativistic or ultra-relativistic gas in various dimensions
If x = β(ε − µ) is large (i.e. if βµ << 0), then we where N is the number of systems in our ensemble, can treat a gas as a classical Boltzmann gas. However, |θi are normalized states which are members of those if this is not true, we have to account for quantum ensembles, |ρii are the eigenstates ofρ ˆ, and wi are the effects arising from the indistinguishability of parti- real eigenvalues. cles. In particular, boson and fermion gases will be- Properties of the density matrix include have differently from each other at high density. In • Tr [ˆρ] = 1 other words, we have to treat them as quantum gases. • ρ is Hermitian In the next several sections, we are only consid- • ρ is an observable and has real eigenvalues ering non-interacting quantum gases. For these gases, the Hamiltonian contains no interaction terms. Inter- For a pure state, there’s only one system in the acting gases are more complicated to deal with because ensemble the Hamiltonian contains additional terms. The term ρˆ = |θi hθ| . ideal refers to non-interacting gases. This implies thatρ ˆ2 =ρ ˆ, and Tr ρˆ2 = 1. For an im- • Ideal Boltzmann gases: This is the classical limit pure state, i.e. any density matrix with more than one (i.e. βµ << 0) in which quantum gases (both state in the ensemble, Bose and Fermi) can be treated classically. Key- X ρˆ2 = w2 |ρ i hρ | . words include low density i i i • Ideal Bose gases: Non-interacting gas of bosons Then Tr ρˆ2 < 1. • Ideal Fermi gases: Non-interacting gas of The density matrix is for any ensemble—not just fermions for stationary ensembles. However, to study equilib- We also consider three different regimes with respect to rium statistical mechanics, we look only at stationary special relativity. Each regime has a different disper- ensembles. For those, we can write the density matrix sion relation ε(p) for the single-particle energies. Only in the energy eigenstate basis for the non-relativistic and ultra-relativistic limits can X ρˆ = w |E i hE | . we usually write down analytical results. st i i i i • Non-relativistic: ε = p2/2m For a generic operator O, the ensemble average p • Relativistic: ε = p2c2 + m2c4 (expectation value) is • Ultra-relativistic: ε = pc. If the gas is composed 1 X h i of massless particles, then treat it as an ultra- hOi = hθ|O|θi = Tr Oˆρˆ . N relativistic gas θ Finally, we will often consider the behavior of gases in In the canonical ensemble, different dimensions—1D, 2D, and 3D. −βHˆ e h −βHˆ i Note, if you’re given a dispersion relation ε(p), i.e. ρˆcan = ,ZN = Tr e . h ˆ i single-particle energies, then the total energy is just Tr e−βH X In the grand canonical ensemble, H = εi. i −β(H−ˆ µNˆ) e h −β(H−ˆ µNˆ)i ρˆGC = h i,Z = Tr e . Density Matrix Tr e−β(H−ˆ µNˆ)
The density matrix is defined as In general, the entropy can be calculated from the density matrix as 1 X X ρˆ = |θi hθ| = w |ρ i hρ | , N i i i θ i S = −kBTr [ˆρ lnρ ˆ] . 4.5. Summary: Quantum Statistics 63
Thermodynamic Properties Following are the densities of state for quantum gases composed of free non-relativistic particles (i.e. For non-interacting quantum gases, the grand canoni- ε = p2/2m) cal partition function is
h ˆ ˆ i Y (2s + 1)L 2 Z = Tr e−β(H−µN = z , g(ε) = √ m1/2ε−1/2, (1D) i (2π ) i ~ 2 (2s + 1)A where i goes over all single-particle states, and g(ε) = 2πm, (2D) (2π~)2 ( √ 1 (Bosons) (2s + 1)V 3 1/2 1−e−β(εi−µ) g(ε) = 4 2π m 2 ε , (3D) zi = . (2π )3 1 + e−β(εi−µ) (Fermions) ~ Following are the densities of state for quantum The thermal average of ni, also called the distri- bution function, is gases composed of free ultra-relativistic particles (i.e. ε = pc) 1 hn i = f (ε) = , i ∓ exp[β(ε − µ)] ∓ 1 (2s + 1)L 2 i g(ε) = , (1D) (2π ) c where the minus sign is taken for bosons, and the plus ~ (2s + 1)A 2π sign is taken for fermions. g(ε) = ε (2D) For a quantum gas we can write the grand canon- (2π~)2 c2 ical partition function in the following ways (2s + 1)V 4π 2 g(ε) = 3 3 ε , (3D) X h i (2π~) c ln Z = ∓ N(ε) ln 1 ∓ e−β(ε−µ) ε From the free energy ∞ h i ln Z ' ∓ dε g(ε) ln 1 ∓ e−β(ε−µ) F = −T ln Z, ˆε0 PV ln Z = T we get the following thermodynamic properties for ∞ quantum gases ln Z = β dε Σ(ε) f (ε) ˆ ∓ ε0 ∂F ∂ S = − = T ln Z For a d-dimensional, noninteracting quantum gas, ∂T V,N ∂T V,N the density of state simplifies to ∞ = − dε g(ε)[f∓ ln f∓ ∓ (1 ± f∓) ln(1 ± f∓)] d −1 ˆ X L ∂ε 0 g(ε) = s pd−1 , ∂F ∞ 2π d ∂p α ~ N = − = dε g(ε) f∓(ε) ∂µ ˆε0 where s = 2, 2π, or 4π is the surface area you’re inte- ∂F ∂ ln g d P = − = dε Σ(ε)f (ε) grating over after the angles drop out. Use the disper- ∂V ∂V ˆ ∓ sion relation ε(p) (which is often specified only by key- ∂ ln Z ∞ E = = dε g(ε) ε f (ε). words like “non-relativistic”, “relativistic”, or “ultra- ∂β ˆ ∓ relativistic”) to write all p’s as ε in the end. The sum βµ=α ε0 over α is for other states. Typically, this counts the In general, for a d-dimensional quantum gas (Bose spin states, and then P = 2s + 1. α or Fermi) with a dispersion relation ε = Aps, Note that g(ε) is the same for bosons and fermions. The nature of the quantum particle matters only when s E you calculate ln Z and the thermodynamic properties. P = . d V Chapter 5
Boltzmann Gases
For a Boltzmann or classical gas, µ < 0. In fact, a quantum gas in the high temperature or low density limit where βµ << 0 =⇒ eβµ << 1, should reduce to the classical Boltzmann gas. For a quantum gas, recall that
∞ h i ln Z = ∓ dε g(ε) ln 1 ∓ e−β(ε−µ) . ˆε0
In the limit eβµ << 1, we can use the approximation
ln(1 + x) ≈ x, for x << 1, to write ∞ −β(ε−µ) ln ZB = dε g(ε)e , ˆε0 where we’ve labeled ZB with the subscript B to indicate this as the classical limit. In this limit, the ±1 in the denominator of
1 hnii = f∓(ε) = , exp[β(εi − µ)] ∓ 1 is negigible, so
−β(εi−µ) hnii = f∓(ε) ≈ e .
Keep in mind that in this classical limit, the quantum effects are negligible, so these are true for both Bose and Fermi gases in this limit. We expect all thermodynamic quantities calculated in this limit to agree with the results we derived earlier for classical gases in the canonical ensemble. We find that
∂ Ω N = T ln Z = ln Z = − B ∂µ B B T ∂ ∂ ln g NT P = T ln Z = T ln Z = ∂V B ∂V B V E = dε g(ε)εe−β(ε−µ). ˆ 5.1. Kinetic Theory 65
5.1 Kinetic Theory
How does this all connect to the kinetic theory of gases? Kinetic theory derives thermodynamic quantities such as temperature and pressure by treating a gas as a large number of atoms in constant, random motion. In the kinetic theory, pressure comes from the momenta that the particles exert on the walls of the container. Can we find that relation between pressure and individual particle momenta? In general, for a 3D, non-relativistic (i.e. ε = p2/2m) quantum gas,
V δ(p − p ) V δ(p − p ) V g(ε) = d3p 0 = d3p 0 = 4πmp. h3 ˆ h3 ˆ p /m h3 ∂ε 0 ∂p p0
Then since ln Z = P V/T , we can write
T T h i T h i P = ln Z = ∓ dε g(ε) ln 1 ∓ e−β(ε−µ) = ∓ 4π dp p2 ln 1 ∓ e−β(ε(p)−µ) . V V ˆ h3 ˆ Integrating by parts gives us
4π ∞ p3 dε P = 3 dp hnpi . h ˆ0 3 dp We also have that the number per volume is
∞ N 4π 2 = 3 dp p hnpi . V h ˆ0 This implies that 1 N ∂ε 1 P = p = n hpvi , 3 V ∂p 3 where n is the number density, and pv is like the momentum flux. We used the fact that 2 ∂ε from ε = p /2m, the velocity in the non-relativistic case is v = ∂p . For a relativistic gas, the only difference is that ε = pp2 + m2. This suggests we can write thermal averages as
hA(p)i = d3p f˜(p)A(p), ˆ where hn i f˜(p) = p , d3p h3 hnpi ´ is some normalized distribution function. Then f˜(p) d3p = 1. In the case of Boltzmann particles, i.e. low´ density gases with βµ << 0, then
−β(ε(p)−µ) hnpi ≈ e , and 3 2 ˜ 3 m −βmv2/2 3 fβµ(v) d v = e d v. 2πkBT This is the familiar Maxwell-Boltzmann distribution. It is normalized such that
3 2 ∞ ˜ 3 m 2 −βmv2/2 1 = fβµ(v) d v = 4π v e dv. ˆ 2πkBT ˆ0 66 Boltzmann Gases
Remember, if a variable m can take on any values from −∞ to ∞ and the integrand is spherically symmetric, then spherical coordinates allows us to make the transformation
2π π ∞ ∞ d3m → m2 sin θdm dθ dφ → 4π m2 dm. ˆ ˆ0 ˆ0 ˆ0 ˆ0 With the Maxwell-Boltzmann distribution, we find that
3 2 ∞ m 2 3k T hv2i = 4π v4e−βmv /2dv = B , 2πkBT ˆ0 m which implies that the average kinetic energy of a gas particle is
p2 m 3 = hv2i = k T. 2m 2 2 B
To connect with kinetic theory, consider when a gas particle hits the wall of the container. What’s the force per unit area exerted by the particles on the walls of the container?
Assuming the wall is in the x direction from the particle’s initial position, then at the wall,
0 px = −px
py = py
pz = pz.
Think of it without the wall in place. What is the flux of momentum through the wall?
F = n d3p v p f˜(p)θ(v ). p,x ˆ x x x The Heaviside function θ picks up only the particles moving in the x-direction as opposed to also those moving in the −x direction. Then the kinetic pressure, i.e. the force per unit area, exerted on the wall is
P = 2F = n d3p v p f(p) = n hv p i . k p,x ˆ x x x x Since our system is isotropic, our result generalizes to n n P = h~v · ~pi = hvpi . k 3 3 This is the same as the result we got via the grand canonical partition function. The number flux in the x-direction is
π ∞ 2 3 ˜ 2 ˜ n Fn,x = n d p vxf(p)θ(vx) = n dp p f(p) dθ cos θ sin θ = hvi . ˆ ˆ0 ˆ0 4 5.2. Equipartition and Virial Theorems 67
5.2 Equipartition and Virial Theorems
Recall that the equipartition theorem states that every quadratic degree of freedom in the Hamiltonian contributes kT/2 to the energy. Consider for classical Boltzmann gas
p2 2 x −βpx/2m p2 dpx 2m e 1 x = = k T. ´ −βp2 /2m b 2m dpx e x 2 ´ This is the simple equipartition theorem result. Now look at it more generally
∂H dω ρξi ∂H ∂ξj ξi = ´ . ∂ξj dω ρ ´ −βH Here, the ξi are any phase space coordinate, and ρ = e . Then
∂ρ ∂ξi dω ξi dω ρ ∂H ∂ξj ∂ξj ξi = −T ´ = T ´ + surface terms. ∂ξj dω ρ dω ρ ´ ´ But the surface terms are zero, and
∂ξi = δij, ∂ξj so this simplifies to ∂H ξi = T δij. ∂ξj
This is the generalized equipartition theorem. The ξi and ξj are any pair of phase space coordinates. Example 5.2.1
For example, if ξi = ξj = pi, then
∂H pi = T. ∂pi Then for p2 H = i + other terms, 2m we have that p2 T i = . 2m 2
Note: The equipartition theorem only holds in the classical and low density limit. The kinetic pressure is
∂H Pk = n hvxpxi = n hq˙xpxi = n px = nT. ∂px The virial theorem of Clausius is basically the thermal version of the virial theorem from classical mechanics. * 3N + X V ≡ p˙iqi . i=1 68 Boltzmann Gases
The quantity V is called the virial. Applying the equipartition theorem gives us
V = −3NT.
We can also think about an interacting system where there are internal forces F~ ij and external forces ˙ X ~ ~pi = F ij + Fext,i, j Then D E 3PV ~q · F~ = −2 d3p d~s · ~q (nˆ · ~p )(nˆ · ~v ) f˜(p) = − . i ext,i ˆ ˛ i i i N If we plug these all back into the virial theorem, we find * + 1 X PV = NT + ~q − ~q · F~ . 3 i j ij i 5.3 Internal Degrees of Freedom Often gas particles have internal degrees of freedom. For example, the gas particles might have rotational degrees of freedom which can be excited. These internal degrees of freedom contribute to the partition function of the gas, and change its thermodynamic properties. It is easiest to consider this in the canonical ensemble. Suppose we have a system with single-particle energies εα(p) = ε(p) + εα, 2 where ε(p) = p /2m is the energy of motion of the the particle, and εα is the internal excitation energy. Then the partition function for a single particle is d3p X Z = V e−β(ε(p)+εα), 1 ˆ h3 α where the sum over α is the sum over internal states. This is separable, and we get 3 d p 2 X V X V Z = V e−βp /2m e−βεα = e−βεα = Z (T ), 1 ˆ h3 `3 `3 int α Q α Q where the internal partition function of the particle is defined to be X −βεα Zint ≡ e , α and the thermal de Broglie wavelength is r 2π 2 ` = ~ . Q mT Then the overall partition function for all the particles is 1 V N ZN = 3 Zint. N! `Q So the partition function has contribution from motion and from internal degrees of freedom. 5.3. Internal Degrees of Freedom 69 The logarithm of the grand canonical partition function becomes V X ln Z = e−βεα d3p e−β(ε(p)−µ). B (2π )3 ˆ ~ α P −βεα Here again we have the internal partition piece Zint = α e . Note, this is only for classical gases. We don’t get this nice separation for Bose or Fermi gases. With the presence of internal degrees of freedom, we have to add corrections to all of our thermodynamic quantities. Fint = −NT ln Zint ∂ E = −N ln Z int ∂β int ∂ S = −N (T ln Z ) int ∂T int µint = −T ln Zint. For example, the total energy would now be the kinetic energy plus the internal energy E = Ek + Eint. Example 5.3.1: Diatomic Molecules For a diatomic molecule, we get two new kinds of internal energy states— vibrational and rotational. Suppose the internal states are just harmonic oscillator levels εα = ~ωα, where α = 0,..., ∞. Then X −β ωα 1 Zint = e ~ = . 1 − e−β~ω α Here we used the formula for the sum of a geometric series. Then the internal energy per particle is Eint ∂ ~ω = − ln Zint = . N ∂β eβ~ω − 1 In the low and high temperature limits, we get ( E ωe−β~ω if T << ω int = ~ ~ . N T if T >> ~ω. The high temperature limit gives us exactly what we expect from the equipar- tition theorem. This is a somewhat reasonable model for vibrations in diatomic 3 molecules. For most molecules, ~ω/kB ∼ 10 K, so we can ignore those internal degrees of freedom at room temperature. We can also consider rotational degrees of freedom. For diatomic molecules with two equal principle moments of inertia and the third one zero (i.e. if they can be treated as rotors), then 2 ε = ~ `(` + 1), rot 2I 70 Boltzmann Gases where I is the moment of inertia. This is 2` + 1 degenerate in `, so the rotational part of the internal partition function is 2 X −β ~ `(`+1) Zint,rot = (2` + 1)e 2I . ` For large T , we can write this as an integral ∞ 2 ~ 2IkBT −β 2I `(`+1) Zint,rot ' d` (2` + 1)e = 2 . ˆ0 ~ Then E ∂ rot = − ln Z = T. N ∂β rot This matches what we would get from the equipartition theorem for the two ro- tational degrees of freedom. The temperature scale for exciting rotational modes is ∼ 10 K. Recall that the specific heat at constant volume is 1 ∂E CV = . N ∂T N,V Since at low temperature, a diatomic molecule has only the three translational degrees of freedom, we get CV = 3/2. At higher temperatures ∼ 10 K the two ro- tational degrees of freedom are excited, and now CV = 5/2. At high temperatures, two vibrational degrees of freedom are also excited, and we get CV = 7/2. So the properties of the gas can change dramatically as you go up the temperature scale. If the gas is low enough density to be treated as a Boltzmann gas, then the free energy of a gas containing internal degrees of freedom is ! N`3 F (N,V,T ) = Nε + NT ln Q − NT, VZint where ε is the ground state energy of the molecule, `Q is its thermal de Broglie wavelength, and Zint is the internal partition function of the molecule. Then taking a derivative gives 5.4. Chemical Equilibrium 71 the chemical potential ! ∂F N`3 µ = = ε + T ln Q . ∂N T,V VZint 5.4 Chemical Equilibrium Suppose you have a system containing multiple gases. We can think of equilibrium of the mixture on a time scale much larger than the time scale at which the chemical reactions in the system occur. The total free energy for a system containing different kinds of gases is systems X dFtot = − [Si dTi + Pi dVi − µi dNi] , i where i runs over the different species in the system. In general, we can assume the different species occupy the same volume (then V is the same for all), and the different particles exchange energy (then T is the same for all species), and then systems X dFtot = −Stot dT − Ptot dV + µi dNi. i We assume chemical or nuclear reactions can occur between the different species of particles. For example, if there are two species of reactants—A and B, then we may have a reaction like νAA + νBB ↔ νC C + νDD, where the νi are the stoichiometric coefficients which give the number of each type of particle involved in this reaction. The reactions impose constraints, i.e. by conservation of particle numbers. For example, we know that ∆N ∆N ∆N ∆N A = B = − C = − C . νA νB νC νD Then the change in the free energy is ∆F = (νAµA + νBµB − νC µC − νC µD) ∆NA. After a long time, we expect the system to reach equilibrium. The equilibrium condition can be found by minimizing the Helmholtz or Gibbs free energy (they both give the same result), subject to the constraints imposed by the reactions. The change in free energy noted above implies that F is minimized when νAµA + νBµB − νC µC − νC µD = 0. Or more generally, if we have more than two reactants and two products, then F is minimized when X X νiµi = νjµj, i j where the sum on the left is over the reactants, and the sum on the right is over the products. This is the condition for chemical equilibrium. Now we have an equation involving the chemical potentials of the different species. 72 Boltzmann Gases If we have a system containing multiple species, we can write the chemical potential as 3 ! ni`Q,i µi = mi + T ln , Zint,i where mi is the rest mass of a particle, ni = Ni/V is the number density of the species, Zint,i is the internal partition function of the species, and s 2π~2 `Q,i = , miT is the thermal de Broglie wavelength for the species. Note, particles that have no con- served charge associated with them have zero chemical potential. An example is photons. 5.5. Summary: Boltzmann Gases 73 5.5 Summary: Boltzmann Gases Skills to Master • Know the classical limit of a quantum gas • Be able to calculate all of the thermodynamic quantities of a quantum gas in the classical limit • Show that various quantities for a quantum gas in the classical limit agree with predictions from kinetic theory • Calculate quantities using the generalized equipartition theorem • Calculate the canonical internal partition function for a molecule • Calculate corrections to thermodynamic quantities due to internal degrees of freedom • Calculate the chemical potentials for a gas containing multiple species • For a gas containing multiple species, calculate particle number ratios in chemical equilibrium For a Boltzmann (i.e. classical) gas, remember is some normalized distribution function. Then that µ < 0. f˜(p) d3p = 1. In the classical limit ´ In the classical limit, βµ << 1, βµ βµ << 0 =⇒ e << 1, −β(ε(p)−µ) hnpi ≈ e . quantum gases reduce to the classical Boltzmann gas. In this limit, Then we can write the Maxwell-Boltzmann speed dis- tribution ∞ −β(ε−µ) 3 ln ZB = dε g(ε)e 2 ˆ m 2 ε0 ˜ 3 −βmv /2 3 fβµ(v) d v = e d v. −β(εi−µ) 2πkBT hnii = f∓(ε) ≈ e ∂ Ω N = T ln Z = ln Z = − B With this distribution, we find that ∂µ B B T 3 2 ∞ ∂ ∂ ln g NT 2 m 4 −βmv2/2 3kBT P = T ln ZB = T ln ZB = hv i = 4π v e dv = , ∂V ∂V V 2πkBT ˆ0 m E = dε g(ε)εe−β(ε−µ). ˆ which implies that the average kinetic energy of a gas particle is Low density is a keyword for classical gases. p2 m 3 = hv2i = k T. Kinetic Theory 2m 2 2 B In kinetic theory, pressure is the result of individual particle momenta due to particle impacts on the walls Equipartition Theorem of the container. In the classical limit βµ << 0, we ex- The equipartition theorem for classical gases states pect a quantum gas to give us the behavior predicted that every quadratic degree of freedom in the Hamil- by kinetic theory. tonian contributed kT/2 to the energy of the gas. The For a 3D, non-relativistic gas, we can relate the generalized equipartition theorem states that pressure to the individual particle momenta as 1 N ∂ε 1 ∂H P = p = n hpvi . ξi = T δij, 3 V ∂p 3 ∂ξj We can write thermal averages as momenta integrals where the ξi and ξj are any pair of phase space coor- dinates. hA(p)i = d3p f˜(p)A(p), For example, if H = p2/2m+ other terms, then we ˆ i could have ξi = ξj = pi, then this theorem tells us that where hnpi 2 f˜(p) = , pi T d3p = . h3 hnpi 2m 2 ´ 74 Boltzmann Gases Internal Degrees of Freedom where ε is the ground state energy of the molecule, `Q is its thermal de Broglie wavelength, and Zint is the in- If a gas consists of molecules with internal degrees of ternal partition function of the molecule. Then taking freedom, then those degrees of freedom will contribute a derivative gives the chemical potential to the overall partition function, and the thermody- ! namic properties of the gas will be affected. ∂F N`3 For a classical gas, we can separate the two contri- µ = = ε + T ln Q . ∂N VZ butions. Then the overall canonical partition function T,V int becomes 1 V Chemical Equilibrium Z = ZN , N N! `3 int Q In a gas containing multiple species, we can have chem- where the thermal de Broglie wavelength is ical or nuclear reactions like X X r νiXi ↔ νjXj, 2π~2 ` = , i j Q mT where the left side is reactants and the right side is and products. The ν are the stoichiometric coefficients, and X −βεα Zint ≡ e , the X are the different species of gas particles. Then α chemical equilibrium between the species occurs when where the sum over α is the sum over internal energy X X states. Note, this is only for classical gases. We don’t νiµi = νjµj. get this nice expression for Bose or Fermi gases. i j With the presence of internal degrees of freedom, Remember that for photons, µ = 0. we have to add corrections to all of our thermodynamic For a system containing multiple species, we can quantities. write the chemical potential of a given species as Fint = −NT ln Zint 3 ! ni`Q,i ∂ µi = mi + T ln , E = −N ln Z Zint,i int ∂β int ∂ where m is the rest mass of a particle, n = N /V is Sint = −N (T ln Zint) i i i ∂T the number density of the species, Zint,i is the internal µint = −T ln Zint. partition function of the species, and s For example, the total energy would now be the kinetic 2 2π~ energy plus the internal energy E = E + E . `Q,i = , k int m T The free energy of a Boltzmann gas containing in- i ternal degrees of freedom is is the thermal de Broglie wavelength for the species. ! Note that Zint = 2 for spin-1/2 fermions like pro- N`3 F (N,V,T ) = Nε + NT ln Q − NT, tons, neutrons, and electrons since they have two in- VZint ternal spin states. Chapter 6 Ideal Bose Gases 6.1 General Bose Gases A Bose gas is a special case of a quantum gas in which the wave functions are sym- metric. An ideal Bose gas is a Bose gas with no interparticle interactions, and so the Hamiltonian does not contain terms for particle interactions. Bose gases exhibit a unique phenomenon—Bose-Einstein condensation—at sufficiently low temperature. Previously, we considered low density cases where the wave function symmetry didn’t matter, although indistinguishability did. Now we look at temperature regimes where the underlying wave function symmetry does matter. In 3D, for a non-interacting quantum gas, the density of states is √ 3 2m 2 V √ g(ε) = ε. π2~3 In the limit where we can integrate over momenta instead of summing over discrete states, recall that we can calculate the number of particles as ∂F ∂ N = − = T ln Z ' dε g(ε)f (ε). ∂µ ∂µ ˆ ∓ For bosons in 3D, this becomes √ 3 ∞ 2m 2 V √ 1 N = ε dε. 2 3 β(ε−µ) π ~ ˆ0 e − 1 Substituting x = βµ, leads us, for the 3D case, to V βµ N = g 3 (e ), vQ 2 3 where vQ = `Q with `Q being the thermal de Broglie wavelength, and 1 ∞ xν−1 gν (z) = −1 x dx, Γ(ν) ˆ0 z e − 1 is called the Bose-Einstein function. Note, this is a different g from the density of state. We can express most properties of Bose-Einstein gases in terms of such Bose-Einstein functions with different ν and z. A useful property of the Bose-Einstein functions is that ∂ 1 g (z) = g (z), ∂z ν z ν−1 76 Ideal Bose Gases where z = eβµ. However, as we will soon see, this expression for N is not completely correct—it misses the number of bosons which are in the ground state, and this becomes significant at low temperatures. We can also invert the expression for N to get 3 2 2 2π~ N V (N, T, z) = 3 . m T 2 g 3 (z) 2 We can calculate the isothermal compressibility as 1 ∂V V g1/2(z) κT = − = , V ∂P N,T NT g3/2(z) and the adiabatic compressibility as 1 ∂V 3V g3/2(z) κS = − = . V ∂P N,z 5NT g5/2(z) 6.2 Bose-Einstein Condensation For bosons, which are limited to µ < 0, consider N at µ = 0 ∞ D/2−1 1 N(µ = 0) ∝ dx x x = g D (1). ˆ0 e − 1 2 We can rearrange 1 e−x = , ex − 1 1 − e−x in order to use the geometric series formula. Then ∞ X ∞ N(µ = 0) ∝ dx xD/2−1 e−(n+1)x. ˆ n=0 0 Making the substitution y = nx gives us ∞ X ∞ N(µ = 0) ∝ n−D/2 dy yD/2−1 e−y. ˆ n=1 0 The sum over n gives us a Riemann zeta function since ∞ X 1 ∞ xs−1 ζ(m) = n−m = dx , Γ(s) ˆ ex − 1 n=1 0 which converges for m > 1. The integral over y gives us a gamma function so that D D N(µ = 0) ∝ ζ Γ . 2 2 What matters is that D N(µ = 0) ∝ ζ , 2 6.2. Bose-Einstein Condensation 77 and this will blow up when D ≤ 2. I.e. N → ∞ as µ = 0 for D ≤ 2. Tip For D = 3 and higher, this suggests some maximal value for N. In 3D, it is Bosonic gases have negative 3/2 − mT V chemical potential. N(µ = 0 ) = V ζ(3/2) 2 = ζ(3/2) . (6.1) 2π~ vQ However, we know that physically there is no such limit to the number of bosons in our system. So what is the problem? The problem comes from our approximation of the sum as an integral. We are not including the ground state energy in the density of state for D ≥ 3. That is, our expression above is strictly for ε > 0. To fix this, we need to explicitly include the ground state contribution in the partition function. We now write ∞ h i ln Z ' − ln 1 − eβµ − dε g(ε) ln 1 − e−β(ε−µ) , ˆ0 where the first term is the contribution from the ground state. Now we get eβµ ∞ 1 N ' ∂ ln Z = + dε g(ε) , βµ βµ β(ε−µ) 1 − e ˆ0 e − 1 where the first term is the contribution from the ground state and the integral is the N for ε > 0. We can write this as βµ e V βµ 3 N = βµ + g (e ), 1 − e vQ 2 where the first term gives the number of bosons in the ground state, and the second term gives the number of bosons above the ground state. We can interpret the result Eq. (6.1) as the maximal number of particles which are not in the ground state. Inverting that result gives us the critical temperature 2 2π 2 N 3 T = ~ . c m V ζ(3/2) Below this critical temperature, a macroscopic number of particles will be found in the ground state of the system. Above this critical temperature, a macroscopic number of particles will be outside of the ground state of the system. So this temperature gives the temperature at which a system undergoes Bose-Einstein condensation. We assume that below Tc, N = N0 + N(µ = 0). This implies that the fraction of particles in the ground state is 3 N T 2 0 = 1 − . N Tc When the system is condensed, we see that T T N ≈ − =⇒ µ ≈ − ∼ O(T/N). µ N0 This is consistent with our previous assumption that µ = 0. So our approximation is reasonable. 78 Ideal Bose Gases What does this mean for the other thermodynamic properties of the Bose-Einstein condensate? The equation of state of this gas is given by PV V m3/2 ∞ √ = ln Z = − ln 1 − eβµ − √ T 3/2 dx x ln 1 − eβµ−x . 2 3 T 2π ~ ˆ0 Integrating by parts gives us V 4 ∞ 1 βµ √ 3/2 ln Z = − ln 1 − e − dx x x−βµ vQ 3 π ˆ0 e − 1 βµ V 4 = − ln 1 − e − √ Γ(5/2)g5/2(λ), vQ 3 π βµ βµ where λ = e is the fugacity. Using ln[1 − e ] ≈ −1/(2N0) − ln N0, we get V βµ 1 ln Z = g5/2(e ) + + ln N0. vQ 2N0 The last two terms are negligible for large N, so PV V βµ = ln Z = g5/2(e ). T vQ This implies that 3 m 2 5 P (T, z) = T 2 g5/2(z), 2π~2 where z = eβµ. Below Tc we found that µ 1 ' − =⇒ λ = eβµ ≈ 1. T N0 Then if we expand the logarithm ln Z, we get VP −1 V = N0 + ln N0 + g5/2(1). T vQ We can neglect the terms containing N0 and since g5/2(1) = ζ(5/2), we have that 3 m 2 5 P (T, z) = T 2 ζ(5/2). 2π~2 So below Tc, we have no contribution to the pressure from the condensate, P is indepen- dent of V , and so the system must be infinitely compressible. 6.2. Bose-Einstein Condensation 79 Next, we want CV for the system. We start by finding the internal energy PV 2 E = −∂β ln Z = T ∂T βµ T βµ 3 V βµ 3 = T g5/2(e ) = PV. 2 vQ 2 This is consistent with the result we found that for a D dimensional gas with a dispersion relation of ε ∝ ps, then s E P = . (6.2) D V Then the heat capacity is 2 2 ∂E CV = . 3 3 ∂T N,V Using a derivative relation, we can write this as 2 ∂P V ∂P V ∂N ∂N −1 CV = − . 3 ∂T V,λ ∂λ T,V ∂T λ,V ∂λ T,V In the condensed phase, i.e. below Tc, the pressure is independent of λ, so ∂P V = 0, ∂λ T,V which implies that 15 V 3/2 CV (T < Tc) = ζ(5/2) ∝ T . 4 vQ At Tc, we find that CV (Tc) = 1.925N. For temperatures above Tc, " 2 # 15 V 3 g3/2(λ) CV (T > Tc) = g5/2(λ) 1 − . 4 vQ 5 g1/2(λ)g5/2(λ) In the large T limit, this quantity goes to 3/2. By the equipartition theorem, we know that this must be so in the thermodynamic limit. Since g1/2(1) → ∞, CV is actually continuous across Tc. However, the derivative of CV with respect to T is discontinous at Tc Plotting CV (T ) gives us the picture shown below. The critical temperature is the boundary on which for T > Tc, you have a microscopic number of particles in the ground state. On the other side, for T < Tc, you have a macroscopic number of particles in the ground state. 80 Ideal Bose Gases 6.3 Photons and Black-body Radiation There are two ways to think of photons: 1. As a collection of harmonic oscillators with ε = ~ω 2. As an ideal gas of indistinguishable quantum particles We will take the second approach. For photons in a box, 2π ~k = (`, m, n) L ω = c|~k| ε = ~ω. There is no conserved charge associated with photons, and so they can be created or destroyed at will. Since particle number is not conserved for photons, we must use the grand canonical ensemble. However, since we cannot specify the average number of photons, we set the Lagrange multiplier α to zero, and therefore, µ = 0, for photons. Then X d3k h i V ∞ ln Z = −V ln 1 − e−β~ω(k) = − dε ε2 ln 1 − e−βε . ˆ (2π)3 π2 3c3 ˆ λ=± ~ 0 The sum over λ = ± is the sum over the two polarizations. Then the pressure is ∞ T 2 −βε P = − 2 3 3 dε ε ln 1 − e . π ~ c ˆ0 Integrating by parts gives us Γ(4)ζ(4) T 4 π2T 4 P = = V . π2~3c3 3 45~3c3 For the energy, we get V ∞ ε3 VT 4Γ(4)ζ(4) π2T 4 E = 2 3 3 dε βε = 2 3 3 = V 3 3 = 3P. π ~ c ˆ0 e − 1 π ~ c 15~ c The energy density (i.e. differential energy per volume contained in the interval from ω to ω + dω, is ω3 dω 1 u(ω) dω = ~ . π2c3 eβω − 1 At small energies, this gives us the Rayleigh-Jeans law u(ω) ∝ ω2, and for large energies, this gives us Wien’s formula u(ω) ∝ ω3e−β~ω. This is what we expect if we treat the photons as a Boltzmann gas. Now let’s consider photon emission from a black-body. The effusion rate, or number of particles escaping a cavity is dN dN esc = ~v · nˆ Θ(~v · nˆ), dA d3p dt p d3p d3r p 6.4. Phonons 81 where, for photons, dN 2 1 = . d3p d3r (2π~)3 eβε − 1 Integrating out the angular part of the momentum gives us the number of escaping pho- tons per time per momentum dN 4πp2 2 1 esc = c . dA dp dt 4 (2π~)3 eβε − 1 We just replaced d3p → 4πp2 dp and replaced v → c. The rate of total energy emission is 4 ∞ 3 dNesc T x c E 4 Γ = dp ε = 2 3 2 dx x = = σT , ˆ dA dp dt 4π ~ c ˆ0 e − 1 4 V where π2k4 σ = B ≈ 5.67 × 10−8 W/m2K4, 60~3c2 is the Stefan-Boltzmann constant. For example, for stars, the luminosity (total power emitted) is 3 4 L = 4πR∗σT∗ , where R∗ is the radius of the star and T∗ is its temperature. 6.4 Phonons Phonons are sound-like excitations in solids. What kind of excitations are allowed in lattice solids and what are their thermody- namic properties? Think of a solid composed of N atoms in a lattice. For static atoms, i.e. for the atoms of a solid in equilibrium, the potential energy is minimized. Then the Hamiltonian for small perturbations from these equilibrium posi- tions is 3N X 1 X H = Φ + mξ˙2 + α ξ ξ , 0 2 i ij i j i i,j where 1 ∂2Φ αij = , 2 ∂xi∂xj is evaluated at the equilibrium position. This looks like a matrix equation. Think of it as an expansion of a matrix with elements αij in terms of the eigenvalues and eigenvectors (i.e. the normal modes) of the matrix. So we can write 3N X 1 H = Φ + mq˙2 + ω2q2 , 0 2 i i i i where the ωi are the eigenvalues of the matrix. This is now a system of 3N oscillators. There’s no constraint on the number of phonons allowed in the system, so our analysis will look similar to our analysis of photons. The total energy is X E = ~ωini, i 82 Ideal Bose Gases where ωi is the oscillation frequency, and ni can be thought of as the number of phonons at that frequency. The phonons have µ = 0 just like the photons, and the total energy at finite temperature is 3N X ωi E = ~ . eβ~ωi − 1 i This is as far as we can go without knowing the spectrum of the frequencies ωi. We can look at some simple models. The simplest model is to assume that ωi = ωE = const. This is the Einstein model. Then we get 3N ω E = ~ E . eβ~ωE − 1 For T >> 0, this becomes E ' 3NT, as it must to match the equipartition theorem for a system with 6N degrees of freedom. This model does not match experimental results at low temperature, so it is not a very good model. A more realistic model is the Debye model. This model asserts that we can assume there’s some maximum frequency ωD—the Debye frequency—in our system. For example, wavelengths must be larger than the lattice spacing. This gives us the constraint on g(ω) that ωD g(ω) dω = 3N, ˆ0 where ωD is the cutoff frequency. For phonons, ε = ~ω = csp = cs~k, where cs is the speed of sound in the solid. In terms of the wave number, V 12πk2 dk g(k) dk = 3 4πk2 dk = V , (2π)3 (2π)3 where we’ve added the factor of 3 to account for the three possible polarizations—two transverse and one longitudinal—of the waves in our system. We can treat these as sound waves, which have the dispersion relation ω = csk. Then 12π 2 g(ω) dω = V 3 ω dω, (2πcs) and so our constraint becomes ωD 12π 2 V 3 ω dω = 3N. (2πcs) ˆ0 This implies 1 6π2N 3 ω = c . D s V We can also define the Debye wavelength 1 3 2πcs V λD = = 2 ∝ a, ωD 6π N 6.4. Phonons 83 where a is the lattice spacing i.e. the interatomic distance. Following standard procedures, we look at the thermodynamic properties of the Debye solid. The grand canonical partition function for a bosonic gas is given by Eq.(4.1) with the minus signs. With µ = 0 and ε = ~ω, this gives us ωD 12π ωD −β~ω 2 −β~ω ln Z = dω g(ω) ln 1 − e = V 3 dω ω ln 1 − e . ˆ0 (2πcs) ˆ0 Keep in mind that this is specifically for phonons with the dispersion relation ω = csk. We can write this as 9N ωD 2 −β~ω ln Z = 3 dω ω ln 1 − e , ωD ˆ0 the total energy is ∂ ln Z 9N ωD ω3e−β~ω 9N ωD ω3 E = = ~ dω = ~ dω . 3 −β~ω 3 β~ω ∂β ωD ˆ0 1 − e ωD ˆ0 e − 1 Taking a derivative gives us the heat capacity ωD β~ω ∂E 9N~ 3 e ~ω TD CV = = dω ω = 3ND , 3 β ω 2 2 ∂T ωD ˆ0 (e ~ − 1) T T where 3 x y4ey D(x) = 3 dy y 2 , x ˆ0 (e − 1) is the Debye function, and ~ωD TD = , kB is the Debye temperature. For T >> TD, we get x2 D(x) ' 1 − , 20 which recovers the equipartition result. For T << TD, we get 12π4 T 3 CV = N . 5 TD Overall, the Debye model gives a pretty good approximation for the heat capacity of solids. 84 Ideal Bose Gases 6.5 Summary: Ideal Bose Gases Skills to Master • Compare and contrast the qualitative differences between Bose gases and Fermi and Boltzmann gases • Be comfortable working with Bose-Einstein functions • Identify whether Bose condensation can occur in a given system • Identify the critical temperature for a given system • Calculate various thermodynamic quantities for a Bose gas • Calculate the equilibrium temperature of a blackbody • Calculate thermodynamic quantities in the Debye model of phonons with some given dispersion relation Remember, for bosons, µ ≤ 0. A key feature of The critical temperature Tc is obtained by replacing Bose gases is the phenomenon of Bose-Einstein con- Nc with N in this expression and inverting to solve for densation, which occurs at very low temperatures. T . Above this temperature, a microscopic number of Most thermodynamic properties of Bose gases can bosons are in their ground state, and below this tem- be expressed in terms of the Bose-Einstein functions perature a macroscopic number are in the ground state. Below T , the fraction of bosons in the ground state is 1 ∞ xν−1 c gν (z) = dx. 3 Γ(ν) ˆ z−1ex − 1 N T 2 0 0 = 1 − . In the limit z → 1, we have N Tc Note, below T , z = eβµ ≈ 1, then g (z) ≈ g (1) = 1 ∞ xν−1 c ν ν g (1) = dx = ζ(ν), ζ(ν), where ζ(ν) is the Riemann zeta function. ν ˆ x Γ(ν) 0 e − 1 The equation of state can be calculated as which only converges for ν > 1. For small z (i.e. large ∞ PV −βε T ), we can use the expansion = ln Z = − ln(1 − z) − dε g(ε) ln 1 − ze . T ˆ0 z2 z3 Then the energy is gν (z) = z + + + ··· 2ν 3ν 2 ∂ A useful property of the Bose-Einstein functions E = T ln Z . ∂T z is that ∂ 1 The heat capacity at constant volume is gν (z) = gν−1(z). ∂z z ∂E In general, the number of particles is CV = . ∂T V,N ∞ z 1 Remember to use the product and chain rules as nec- N = + dε g(ε) −1 βε , 1 − z ˆ0 z e − 1 essary when differentiating. Number fluctuations can be calculated as where z = eβµ, and the first term gives the number of ∂N ∂N particles N0 in the ground state. At high temperatures h∆N 2i = T = z , (T > Tc), the first term can be neglected. This can also ∂µ T,V ∂z T,V βµ be inverted to get V (N, T, z), where z = e . βµ To see if Bose condensation will occur, we look at where z = e . the limit µ → 0, which is the same as z = eβµ → 1. The isothermal and adiabatic (i.e. isentropic) Clearly, the first term (which is the number of parti- compressibilities are cles in the ground state) blows up in this limit, which 1 ∂V 1 ∂n is what we expect for Bose condensation, but we also κT = − = V ∂P N,T n ∂P N,T have to show that the integral converges in this limit. 1 ∂V 1 ∂n The critical number is the second term when z → 1 κ = − = , S V ∂P n ∂P ∞ 1 N,S N,S N = dε g(ε) . c βε where n = N/V . ˆ0 e − 1 6.5. Summary: Ideal Bose Gases 85 3D Non-Relativistic Bose Gas The rate of energy emission per surface area by a blackbody is For a 3D, non-relativistic, Bose gas we get c E Γ = = σT 4, βµ 4 V e V βµ N = + g 3 (e ) where βµ 2 1 − e vQ 2 4 2 π kB −8 2 4 2 3 σ = ≈ 5.67 × 10 W/m K , 2π~ N 60 3c2 Tc = ~ m V ζ(3/2) is the Stefan-Boltzmann constant. The intensity of PV V βµ photons received at a distance R from a blackbody of = ln Z = g5/2(e ) T vQ radius r emitting at a rate Γ, is 3 m 2 5 r 2 P = T 2 g5/2(z) I = Γ . 2π~2 R ∂ 3 E = T 2 ln Z = PV The radiation pressure exerted by these photons is ∂T 2 βµ I 15 V P = . ζ(5/2) if T < Tc c 4 vQ C = 2 V 15 V 3 g3/2(λ) For a blackbody, equilibrium occurs when the in- g5/2(λ) 1 − if T > Tc. 4 vQ 5 g1/2(λ)g5/2(λ) coming energy equals the outgoing energy. The luminosity (total power emitted) of a star is V g1/2(z) κT = 3 4 NT g3/2(z) L = 4πR∗σT∗ , 3V g3/2(z) where R∗ is the radius of the star and T∗ is its temper- κS = . 5NT g5/2(z) ature. Note that v = `3 with ` being the thermal de Q Q Q Phonons Broglie wavelength. ∂E Phonons are sound-like excitations in solids. The heat capacity, calculated as CV = ∂T N,V is a continuous but not smooth function. For T < Tc At finite temperature, the total energy of the 3/2 CV ∝ T . At the critical temperature, CV peaks phonons in a solid is with a value of CV = 1.925N. In the T → ∞ limit, 3N X ~ωi CV → 3/2. E = . eβ~ωi − 1 i Photons The Einstein model assumes that ωi = ωE = const. This is a good model at high temperatures but For photons, µ = 0. not at low temperatures. The dispersion relation for photons is ε = pc, so A better model is the Debye model, which assumes we get that there exists some maximum ωD, which gives us the ∞ constraint V ωD ln Z = − dε ε2 ln 1 − e−βε π2 3c3 ˆ g(ω) dω = 3N. ~ 0 ˆ0 ∞ 2 4 T 2 −βε π T If given some dispersion relation ω(k), use the fol- P = − 2 3 3 dε ε ln 1 − e = V 3 3 π ~ c ˆ0 45~ c lowing process to analyze phonons: E = 3P. 1. Write down the density of states g(k) for the phonons, keeping in mind that they’re bosons The energy density (i.e. differential energy per and to include a factor of 3 for the three polar- volume contained in the interval from ω to ω + dω, is izations. In 3D, 3V ω3 dω 1 g(k) dk = k2 dk. u(ω) dω = ~ . 2π2 π2c3 eβω − 1 2. Write it in terms of ω using the given dispersion At small energies, this gives us the Rayleigh-Jeans law relation ω(k) 2 u(ω) ∝ ω , and for large energies, this gives us Wien’s dk 3 −β~ω g(ω) dω = g(k) dω. formula u(ω) ∝ ω e . dω 86 Ideal Bose Gases 3. Calculate the Debye frequency ωD by applying and the constraint implies the cutoff frequency the constraint ωD 1 6π2N 3 g(ω) dω = 3N. ω = c . ˆ0 D s V 4. Calculate various thermodynamic quantities by integrating over ω. Keep in mind that the upper Then we can calculate various thermodynamic quanti- limit is now ωD ties ωD −β~ω ln Z = dω g(ω) ln 1 − e ωD ˆ0 ln Z = dω g(ω) ln 1 − e−β~ω ∂ ln Z ˆ0 E = ∂β ∂ ln Z 9N ωD ω3 E = = ~ dω 3 β~ω ∂E ∂β ωD ˆ0 e − 1 CV = . ∂T ωD β~ω ∂E 9N~ 3 e ~ω CV = = dω ω . If the dispersion relation is 3 β ω 2 2 ∂T ωD ˆ0 (e ~ − 1) T ε = ~ω = csp = cs~k, The Debye model gives a pretty good approximation then 12π 2 for the heat capacity, although we can only approxi- g(ω) = V 3 ω , (2πcs) mate the integral. Chapter 7 Ideal Fermi Gases 7.1 General Fermi Gases A Fermi gas is a special case of a quantum gas in which the wave functions are antisym- metric. An ideal Fermi gas is a Fermi gas with no interparticle interactions, and so the Hamiltonian does not contain terms for particle interactions. The interesting thing about fermionic gases is that you can have a finite (degeneracy) pressure as you go to T → 0. This is different from bosonic gases. For fermions, we can have −∞ < µ < ∞. Let’s consider the thermodynamic properties of non-relativistic and ultra-relativistic gases. Non-Relativistic Fermions For a non-relativistic gas (i.e. ε = p2/2m), the density of states is 3 2 V 2m 1 g(ε) = (2S + 1) ε 2 . 4π2 ~2 Then 2 Σ(ε) = ε g(ε), 3 and N ∞ n = = dε g(ε) nF [β(ε − µ)] , V ˆ0 where 1 n (x) ≡ . F ex + 1 We define the Fermi-Dirac function ∞ ν−1 1 x dx −z fν (z) = −1 x = −Liν e , Γ(ν) ˆ0 z e + 1 where Liν is the polylogarithm function. A useful derivative relation is ∂ f (z) = z f (z). ν ∂z ν+1 Another useful relation is z −z f1 (e ) − f1 e = z. 88 Ideal Fermi Gases Then the derivative relations give us N (2S + 1) βµ n = = 3 f3/2 e V `Q P (2S + 1) βµ = 3 f5/2 e T `Q E 3 (2S + 1) βµ = T 3 f5/2 e . V 2 `Q Ultra-Relativistic Fermions For ultra-relativistic fermions (i.e. ε = pc), we get V 1 g(ε) = (2S + 1) ε2, 2π2 (~c)3 and 3 N (2S + 1)T βµ n = = f3 e V π2(~c)3 4 E 3(2S + 1)T βµ = f4 e V π2(~c)3 4 E (2S + 1)T βµ PV = = f4 e . 3 π2(~c)3 For an arbitrary relativistic gas (i.e. ε ∝ pp2 + m2), we would find p g(ε) ∝ ε2 − m2. I.e. we can’t write it in terms of fν (λ). We can only write thermodynamic properties in terms of fν (λ) if the dispersion relation ε(p) is a power law. Fermi Gases at T = 0 It is helpful to note that 1 1 1 x n (x) = = − tanh . F ex + 1 2 2 2 Plotting this function for some T gives us a picture like the one below. As β becomes larger (i.e. T becomes smaller), the width of the transition region shrinks. At β = ∞ (i.e. T = 0), nF becomes a step function nF [β(ε − µ)] = θ(µ − ε), for β → ∞. So, at T = 0, all energy levels up to µ are filled and no energy levels ε > µ will be filled. 7.2. Pauli Spin Susceptibility 89 For T = 0 Fermi gases, ∞ µ N = dε g(ε) θ(µ − ε) = dε g(ε). ˆ0 ˆ0 Then for a non-relativistic gas, 3 V (2S + 1) 2mε 2 N = F , 6π2 ~2 where εF is the Fermi energy which is that final filled energy level at T = 0. The Fermi energy (at any temperature) is 2 p2 2 6π2 N 3 ε = F = ~ . F 2m 2m (2S + 1) V For a T = 0 Fermi gas, the entropy is zero, which implies F = E − TS = E. So εF E = F = dε ε g(ε). ˆ0 If the gas is non-relativistic, this becomes 3 E = F = Nε . 5 F Then the pressure is ∂E 2 E 2 P = − = = nεF . ∂V N 3 V 5 7.2 Pauli Spin Susceptibility What is the spin susceptibility, i.e., the propensity for spin polarization in a Fermi system? We will ignore Landau quantization, i.e., the effect of the magnetic field on the motion of the particles. We only consider the effect of the magnetic fields on the particles via the spin interaction. For electrons in a magnetic field, the single particle energy now depends on the spin εσ(p) = ε(p) − σµBB, where σ = ±1. The number of particles with spin σ is X 1 Nσ(T, µ, B) = . eβ(ε(p)−σµB B−µ) + 1 ~p 90 Ideal Fermi Gases If we define X 1 N0(T, µ) ≡ , eβ(εp−µ) + 1 ~p then we can write Nσ(T, µ, B) = N0 (T, µ + σµBB) . The magnetization of the system is given by M = µB (N↑ − N↓) , where N = N↑ + N↓. For small B, we can do a Taylor expansion β(εp−µ) 1 e 2 ' nF [β(εp − µ)] + βσµBB + O(B ). eβ(εp−σµB B−µ) + 1 (eβ(εp−µ) + 1)2 Then ∂N0 2 Nσ(T, µ, B) = N0(T, µ) + σµBB + O(B ). ∂µ T,B=0 This implies that 2 N = N↑ + N↓ = 2N0(T, µ) + O(B ). So we have the same result (to first order) as when there’s no magnetization. The mag- netization is 2 ∂N0 2 M = 2µBB + O(B ). ∂µ T,B=0 Then the magnetic susceptibility is 1 ∂M µ2 ∂N χ = = B 0 . V ∂B T,µ V ∂µ T At high temperatures, the electrons behave like Boltzmann particles, and we get the Curie susceptibility µ2 N χ = B . V T At low temperatures (T → 0), we can use ∂N = g(ε ), ∂µ F then we get the Pauli susceptibility µ2 χ = B g(ε ). V F For a non-relativistic gas, this becomes 3 N µ2 χ = B . 2 V εF 7.3. Sommerfeld Expansion 91 7.3 Sommerfeld Expansion We will now look at the properties of Fermi gases when 0 < T << µ. How does this affect the equation of state? The thermodynamic properties all have the form of the integral ∞ I = dε φ(ε) nF [β(ε − µ)] . ˆ0 Our goal is to approximate this integral. Recall that 1 1 1 x n (x) = = − tanh . F ex + 1 2 2 2 Another useful fact is that nF (x) = 1 − nF (−x). We will start by splitting the integral into two pieces—one above µ and one below it so that I = I1 + I2 + I3, where µ I1 = dε φ(ε) ˆ0 µ I2 = − dε φ(ε) [1 − nF (β(ε − µ))] ˆ0 ∞ I3 = dε φ(ε)nF (β(ε − µ)) . ˆµ After defining τ ≡ µ − ε and dτ = −dε (since µ is fixed here), we can use our useful fact to write 0 µ I2 = dτ φ(µ − τ) [1 − nF (−βτ)] = − dτ φ(µ − τ)nF (βτ). ˆµ ˆ0 We assume µ >> T here, so βτ is large and nF for large value of its argument goes to zero. So we can let the integral limit go to infinity since the contribution to the integral in the region [µ, ∞) is negligible ∞ −µ/T I2 = − dτ φ(µ − τ)nF (βτ) + O(e ). ˆ0 Similarly, we can define τ 0 ≡ ε − µ, then ∞ 0 0 0 I3 = dτ φ(µ + τ ) nF (βτ ). ˆ0 Combining the results and renaming the integration variables gives us µ ∞ I ' dε φ(ε) + [φ(µ + τ) − φ(µ − τ)] nF (βτ), ˆ0 ˆ0 where the first integral is the T = 0 contribution, and the second integral is the finite temperature correction. Since T << µ, we can do the Taylor expansion 1 φ(µ + τ) − φ(µ − τ) ' 2φ0(µ) τ + φ000(µ) τ 3. 3 92 Ideal Fermi Gases Notice that we only get the odd terms since φ(µ + τ) − φ(µ − τ) is an odd function. Putting it all together, we have that µ 0 2 1 000 4 I ' dε φ(ε) + 2φ (µ) T h1 + φ (µ) T h3, ˆ0 3 where ∞ n hn = dx x nF (x) = some number. ˆ0 2 For example, h1 = π /12. Keep in mind that Sommerfeld expansion is valid for 0 < T << µ, i.e. z = eβµ is large. Typically, we only keep O(T 2), and in that case, Sommerfeld expansion gives us the approximation ∞ µ 2 2 0 π T dε φ(ε) nF [β(ε − µ)] ' dε φ(ε) + φ (µ) . ˆ0 ˆ0 6 We will use this to get some corrections to the low temperature properties of some Fermi systems. The number of particles in the system is ∞ µ 2 2 0 π T N = dε g(ε) nF [β(ε − µ)] ' dε g(ε) + g (µ) . ˆ0 ˆ0 6 But we also have the definition εF N = dε g(ε). ˆ0 This implies that µ π2T 2 − dε g(ε) ' g0(µ) . ˆεF 6 We can approximate these by evaluating them at εF : µ 2 2 0 π T − dε g(εF ) ' g (εF ) . ˆεF 6 This gives us the correction for the chemical potential d (ln g(ε)) π2T 2 µ ' εF − . dε ε=εF 6 For the energy we have µ π2T 2 d (εg(ε)) E ' dε ε g(ε) + , ˆ0 6 dε ε=εF which implies εF µ 2 2 π T d ([ε − εF ]g) E − NεF = dε (ε − εF ) g(ε) + dε (ε − εF ) g(ε) + . ε=ε ˆ0 ˆεF 6 dε F We can drop the second integral since it’s a higher T -order contribution, and there’s some cancellation. We find that εF π2T 2 E = NεF + dε (ε − εF ) g(ε) + g(εF ). ˆ0 6 7.3. Sommerfeld Expansion 93 For example, for an ideal, non-relativistic, 3D gas, we have 3Nε1/2 g(ε) = . 3/2 2εF Plugging this in gives us " # 3 5π2 T 2 E = NεF 1 + . 5 12 εF We know the pressure should be 2 E P = 3 V for this gas, so we can use above result to get P . We can also find the entropy correction by using F = E − TS = Nµ − PV. The second equality comes because we showed that E = TS − Nµ − PV . Recall that ∂F π2 T − = S = N . ∂T N 2 εF For an ultra-relativistic gas (i.e. ε = pc), we would find " # 3 2 πT 2 E = NεF 1 + . 4 3 εF For a 3D non-relativistic gas, the heat capacity at constant volume is N 2 T CV = π . 2 εF 94 Ideal Fermi Gases 7.4 Summary: Ideal Fermi Gases Skills to Master • Compare and contrast a Fermi gas with a Bose gas • Be comfortable working with Fermi-Dirac functions • Calculate the Fermi energy for a given system • Calculate various thermodynamic quantities for Fermi gases in the non-relativistic, ultra- relativistic, and T = 0 limits • Calculate corrections using Sommerfeld expansion For fermions, we can have −∞ < µ < ∞. 3D Fermi Gases Many properties of Fermi gases can be expressed For a non-relativistic Fermi gas (i.e. ε = p2/2m), in terms of Fermi-Dirac functions 3 ∞ ν−1 2 1 x dx V 2m 1 2 fν (z) = −1 x . g(ε) = (2S + 1) 2 2 ε Γ(ν) ˆ0 z e + 1 4π ~ N (2S + 1) βµ A useful derivative relation is n = = 3 f3/2 e V `Q ∂ P (2S + 1) fν (z) = z fν+1(z). βµ ∂z = 3 f5/2 e T `Q Another useful relation is E 3 (2S + 1) = T f eβµ . V 2 `3 5/2 z −z Q f1 (e ) − f1 e = z. For ultra-relativistic fermions (i.e. ε = pc), For a Fermi gas, V 1 g(ε) = (2S + 1) ε2 f∓ = hni = nF (β(ε − µ)), 2π2 (~c)3 3 N (2S + 1)T βµ where n = = f3 e V π2(~c)3 1 1 1 x E 3(2S + 1)T 4 nF (x) ≡ = − tanh . = f eβµ ex + 1 2 2 2 2 3 4 V π (~c) 4 The Fermi energy is obtained by calculating E (2S + 1)T βµ PV = = f4 e . 3 π2(~c)3 εF N = dε g(ε). ˆ0 3D Fermi Gas at T = 0 and then inverting to get εF . The Fermi energy, at any temperature, is For a general Fermi gas, 2 2 2 2 3 ∞ pF ~ 6π N PV h i εF = = . = ln Z = dε g(ε) ln 1 + e−β(ε−µ) . 2m 2m (2S + 1) V T ˆε0 The Fermi energy is the final filled energy level when Then T = 0. The Fermi energy can be inverted to get N(V ) ∂F ∞ at T = 0. N = − = dε g(ε) nF (β(ε − µ)) At T = 0, the number function nF (β(ε − µ)) is a ∂µ ˆε0 step function. All energy levels up to ε = εF = µ are ∂F ∂ ln g P = − = dε Σ(ε)nF (β(ε − µ)) filled and no energy levels higher than this are filled. ∂V ∂V ˆ As T > 0, the step function turns into a smooth func- ∂ ln Z ∞ tion with a transition region about ε = µ where some E = = dε g(ε) ε n (β(ε − µ)). ˆ F energy levels are filled above and below this point. At ∂β βµ=α ε0 7.4. Summary: Ideal Fermi Gases 95 T = 0, µ 2 2 µ 0 π T N = dε g(ε) N ' dε g(ε) + g (µ) ˆ0 6 ˆ0 d (ln g(ε)) π2T 2 S = 0 µ ' εF − εF dε ε=εF 6 E = F = dε ε g(ε). εF π2T 2 ˆ0 E ' NεF + dε (ε − εF ) g(ε) + g(εF ) ˆ0 6 For a non-relativistic Fermi gas at T = 0, 3 2 For a 3D, non-relativistic Fermi gas, V (2S + 1) 2mεF N = 2 2 6π ~ 3Nε1/2 3 g(ε) = E = F = Nε 2ε3/2 5 F F " 2# ∂E 2 E 2 3 5π2 T P = − = = nε . E = NεF 1 + F 5 12 ε ∂V N 3 V 5 F 2 E P = Sommerfeld Expansion 3 V π2 T S = N Sommerfeld expansion is valid for 0 < T << µ, i.e. 2 εF z = eβµ is large N T C = π2 . V 2 ε ∞ µ 2 2 F 0 π T dε φ(ε) nF [β(ε − µ)] ' dε φ(ε) + φ (µ) . For a 3D, ultra-relativistic Fermi gas, ˆ0 ˆ0 6 " 2# This can be used to get corrections to the low temper- 3 2 πT E = NεF 1 + . ature properties of some Fermi systems. 4 3 εF Chapter 8 Interacting Gases 8.1 Correlation Functions For interacting gases, we are interested in correlation functions—in particular, the density- density correlation function. We will limit our discussion to classical, 3D non-relativistic gases with Hamiltonians of the form N X p2 H = i + U, 2m i=1 where the potential energy has the form 1 X 1 X U = u(|~r − ~r |) = u(r ). 2 i j 2 ij i6=j i6=j The potential specifies the interactions between the particles. We will limit our discussion to two-body interactions that depend only on the distance between the two particles. The canonical partition function is 1 1 3N 3N −βH ZN = d p d r e . N! (2π~)3N ˆ We can do the momentum integrals in the kinetic part of the Hamiltonian, then we get 1 1 ZN = 3N QN , N! `Q where the configuration integral is 3N −βU 3N − β P u(r ) Q = d r e = d r e 2 i6=j ij . N ˆ ˆ 3N If there’s no interaction, then U = 0, and we get QN = d r = V . Interactions will lead to spatial correlations between´ the particles. We would naturally define the single-particle number density (i.e. the number of particles in some small infinitesimal space) as * N + X (3) n1(~r) = δ (~r − ~ri) . i=1 This way when integrated over space V , we get the number of particles in the system 3 d r n1(~r) = N. ˆV 8.1. Correlation Functions 97 By extension, we define a two-particle number density as * + 0 X 0 n2(~r, ~r ) = δ (~r − ~ri) δ (~r − ~rj) . i6=j This tells us the probability of finding a second particle in a small neighborhood of ~r 0 if there is a particle in a small neighborhood of ~r. Note that the sum is a double sum over j and i 6= j. Normalization gives us 3 3 0 0 2 d r d r n2(~r, ~r ) = N − N . ˆV ˆV The first integral gives us “1” N − 1 times and the second integral does the same. If our system is translationally invariant, we can write down the pair correlation function, which tells us the probability of finding another particle a distance R away, as V 3 g(R~ ) = d r n2 (~r, ~r − R) hNi2 ˆ * + V X = d3r δ (~r − ~r ) δ ~r − R~ − ~r 2 ˆ i j hNi i6=j * + V X ~ = 2 δ R + ~ri − ~rj . hNi i6=j It is normalized such that N 2 − N d3R g(R~ ) = V . ˆ hNi2 Remember, the brackets mean the ensemble/thermal average. In the canonical ensemble, where the particle number is fixed, we can write down the pair correlation function as ~ V 1 3N X ~ −βU gN (R) = 2 d r δ R + ~ri − ~rj e N QN ˆ i6=j V (N − 1) 3N −βU = d r δ R~ + ~r1 − ~r2 e NQN ˆ N − 1 V 2 3 3 3 −βU(R~,~r3,~r4,...~rN ) = d r3d r4 ··· d rN e . N QN ˆ The average potential just depends on the pairwise correlations, so we can write 1 3N 1 X −βU hUi = d r u(rij)e QN ˆ 2 i6=j 1 3N 3 1 X −βU = d rd R u(R)δ R~ + ~ri − ~rj e QN ˆ 2 i6=j N 2 = d3R g (R~ ) u(R). 2V ˆ N We can write the pressure in terms of the canonical pair correlation function as N N ∂ V ∂(QN /V ) P = T ln ZN = nT + T . ∂V QN ∂V T,N 98 Interacting Gases 0 −1/3 We can rewrite QN in terms of ~ri = ~r V as 1 0 N 3N 0 β X rij QN = V d r exp − u . ˆ 2 V 1/3 0 i6=j Then ∂(Q /V N ) N 2 Q du N = − N d3R g (R~ ). ∂V 6TV 2 V N ˆ d ln R N Then the virial equation of state becomes n du P = nT 1 − d3R g (R~ ) . 6T ˆ d ln R N Remember, n = N/V . The second term in the virial equation of state gives the correction based on the two-particle correlation function. We can also write du/d ln r = rdu/dr, and then for spherically symmetric g(R), we can write ∞ P 4πn 3 du = 1 − dR R gN (R). nT 6T ˆ0 dR This derivation was specifically for 3D. More generally, in d-dimensions, we have ∞ P n d du = 1 − d RR gN (R). nT 2dT ˆ0 dR In 3D, the relation to fluctuations is given by 2 hNi 3 h i h∆N i T ∂N 1 + d R g(R~ ) − 1 = = = nT κT . V ˆ hNi N ∂µ T,V Example 8.1.1 Calculate the equation of state for a weakly interacting gas which has a cor- relation function g(R) = e−βu(R). The formula for the equation of state is P n du n du = 1 − d3R g (R~ ) = 1 − d3RR g (R~ ). nT 6T ˆ d ln R N 6T ˆ dR N In our case, P n du n ∞ du = 1 − d3RR e−βu(R) = 1 − 4π dR R3 e−βu(R). nT 6T ˆ dR 6T ˆ0 dR du −βu(R) 3 We can integrate by parts (integrating dR e and differentiating R ), then the boundary term vanishes, and we get P n 3 ∞ = 1 − 4π − dR R2 1 − e−βu(R) nT 6T β ˆ0 n = 1 + dR3 1 − e−βu(R) . 2 ˆ 8.2. Virial Expansion 99 Doing a Taylor expansion of the exponential and keeping only the first two terms gives us P n ≈ 1 + dR3 u(R). nT 2T ˆ 8.2 Virial Expansion Classical Gases We want to find an expansion of the equation of state in the form ∞ m−1 vP X vQ = a (T ) , T m v m=1 where V v = , N is the specific volume, and 2π 2 3/2 v = `3 = ~ , Q Q mT is the thermal de Broglie wavelength cubed. For low density or high temperature gases, we expect to only need the first few terms of this expansion. To get the ideal gas law, we use only the first term. Our problem is, how do we find the coefficients am(T ) for the higher-order corrections? Recall that our Hamiltonian has the form N X p2 H = i + U, 2m i=1 where the potential energy has the form 1 X X U = u(|~r − ~r |) = u(r ). 2 i j ij i6=j i We are only considering classical gases here. Recall that the canonical partition function for a classical gas is 1 ZN = 3N QN , N!`Q where the configuration integral, which contains the information about the interaction, is 3 3 −β P u(r ) 3 3 Y −βu(r ) Q = d r ··· d r e i We can also write N QN = N! vQ ZN . We can write the grand canonical partition function as ∞ X QN Z = λN , N!vN N=0 Q 100 Interacting Gases where λ = eβµ. This will help us get to the expression that we want. Since P V/T = ln Z, we can also write Z = eP V/T . We want to expand ln Z in powers of λ ∞ V X ln Z = b (T,V )λm. v m Q m=1 The first few coefficients of this expansion are Q b (T,V ) = 1 = 1 1 V 1 2 b2(T,V ) = Q2 − Q1 2vQV 1 3 3 b3(T,V ) = 2 Q − 3Q1Q2 + 2Q1 . 6vQV We can also expand the particle number as ∞ ∂ ln Z ∂ ln Z V X m N = T = λ = mbmλ . ∂µ T ∂λ v T Q m=1 We can express the am coefficients, which are what we want, in terms of the bm coefficients, which are what we can calculate using the two expansions above. Combining them, we can write ∞ vQ X = mb (T )λm v m m=1 ∞ P v v X = b (T )λm. T v m Q m=1 Then the relation between the am and bm coefficients is m−1 ∞ ∞ ! P∞ m X X bm(T )λ a nb (T )λn = m=1 . m n P∞ mb (T )λm m=1 n=1 m=1 m We can then move everything to the right-hand side and collect like powers of λ, so that we have something like 2 3 0 = c1λ + c2λ + c3λ + ··· Then we set each of the coefficients ci equal to zero to find the coefficients of the virial expansion. The first few are a1 = b1 = 1 a2 = −b2 a3 = 4b2 − 2b3. If the temperature of the gas is high, we expect it to behave almost like an ideal gas. So we expect e−βuij ≈ 1. Thus, we want to expand in terms of −βu(rij ) fij = e − 1. 8.2. Virial Expansion 101 Then Y Q = d3r ··· d3r (1 + f ) N ˆ 1 N ij i This expansion is called Meyer cluster expansion. We will only use it to calculate Q2, ∞ 3 3 −βu(r12) 2 −βu(R) Q2 = d r1 d r2 e = 4πV dR R e . ˆ ˆ0 Then we have ∞ 2π 2 −βu(R) b2 = −a2 = dR R e − 1 . vQ ˆ0 Thus we get our first virial correction to the equation of state P v 2π ∞ = 1 − dR R2 e−βu(R) − 1 . T V ˆ0 Integrating by parts gives us P v 2π ∞ du(R) = 1 − dR R2 e−βu(R). T 3V ˆ0 d ln R This implies that in a weakly interacting gas, the correlation function is g(R) ≈ e−βu(R). Example 8.2.1: Van der Waals Equation To derive the Van der Waals equation of state, we use the potential ( ∞ if r < r0 u(rij) = 6 . r0 −u0 r if r ≥ r0. This potential is attractive at long-range and repulsive at short-range. We can use this to calculate what our coefficient a2 should be. We find that r0 ∞ 2π 6 6 2 2 βu0r /R a2 = dR R + dR R 1 − e 0 . vQ ˆ0 ˆr0 If T >> u0, then 3 2πr0 u0 a2 ≈ 1 − . 3vQ T Then the equation of state is P v 2πr3 u = 1 − 0 1 − 0 . T 3v T If we define 2πr3 a ≡ u b, b ≡ 0 , 0 3 then T bT a P = + − . v v2 v2 102 Interacting Gases This is approximately equal to the van der Waals equation of state: T a P = − . v − b v2 Quantum Gases The virial expansion can be used for quantum or classical gases with or without interac- tions. For Bose or Fermi gases, the virial expansion process is essentially the same. We look for an expansion of the equation of state in the form ∞ m−1 vP X vQ = a (T ) . T m v m=1 To do that, we expand ln Z in powers of λ = eβµ and then compare it with ∞ V X ln Z = b (T,V )λm, v m Q m=1 to get the coefficients bm. Use the form of ln Z that is relevant. For example, if you’re calculating the coefficients for a non-interacting (i.e. ideal) Bose gas, then you would use ∞ h i ∞ ln Z = − dε g(ε) ln 1 − e−β(ε−µ) = − dε g(ε) ln 1 − λe−βε . ˆ0 ˆ0 Expand the logarithm as a Taylor series in λ to as many degrees as you need coefficients of bm. Then you would do the integration and equate coefficients with the series above to get the bm. For a quantum gas with interactions, ln Z will be much more complicated. After obtaining the bm, one relates them to the am in the same manner as for classical gases. So once again, the first few relations are a1 = b1 = 1 a2 = −b2 a3 = 4b2 − 2b3. For a general, interacting quantum system with a central, two-body potential, 1 2 −a2 = b2 = Q2 − Q1 . 2vQV For quantum gases, we don’t get the nice separation between the QN part of the integral and the momentum part of it. In general, Q1 = V , and 2 Q2 = 2!vQZ2, where Z2 is the canonical two-particle partition function. Comparing the interacting case (0) with the non-interacting case, if we let b2 be the b2 coefficient in the non-interacting case, and b2 in the interacting case, then (0) 1 (0) b2 − b2 = Q2 − Q2 , 2vQV (0) since Q1 = Q1 . This becomes (0) vQ (0) vQ h −βHˆ −βHˆ (0) i b − b = Z − Z = Tr e 2 − e 2 . 2 2 V 2 2 V 8.2. Virial Expansion 103 So the problem has been reduced to that of calculating the energy eigenvalues of Hˆ2 ˆ (0) and H2 . These are the Hamiltonians of the two-body system in the interacting and non-interacting cases. The two-body Schrodinger equation is 2 − ~ ∇2 + ∇2 + u(|~r − ~r |) Ψ (~r , ~r ) = E Ψ (~r , ~r ). 2m 1 2 1 2 α 1 2 α α 1 2 We transform to the center-of-mass frame using 1 R~ = (~r + ~r ), ~r = ~r − ~r , 2 1 2 1 2 then Ψα(~r1, ~r2) = ψj(R~ )ψn(~r). Then the energy spectrum is P 2 E = j + ε . j,n 4m n where Pj is the center-of-mass momentum, the total mass is 2m, and εn is the energy coming from the relative coordinates part of the problem: 2 k˜2 ~ ∇2 + u(r) ψ (~r) = ε ψ (~r) = ψ (~r), 2µ n n n 2µ n where µ is the reduced mass of the system. Then P 2 vQ h −βHˆ i vQ X −β j X −βε b = Tr e 2 = e 4m e α . 2 V V j α The sum over j is a sum over single-particle states and we can turn it into an integral. We find that 3/2 X −βεα b2 = 2 e . α So ! (0) (0) 3/2 X −βεα X −βεα b2 − b2 = 2 e − e . α α Now we consider the relative wave function. We have a spherically symmetric po- tential, so we expect rotationally invariant solutions of the form χ (r) ψ = ψ (~r) = kl Y m(θ, φ). n klm r l Assume that the potential is localized, i.e. it goes to zero fast enough as r → ∞, then the asymptotic form of the wave function becomes X m ψ → [Aljl(kr) + Blnl(kr)] Yl (θ, φ), l,m where sin(kr − lπ/2) j (kr) ∼ l kr cos(kr − lπ/2) n (kr) ∼ . l kr We can write Al = Cl cos (ηl(k)) ,Bl = Cl sin (ηl(k)) , 104 Interacting Gases then the radial part of the solution goes as lπ χ (r) ∝ sin kr − + η (k) , kl 2 l where ηl(k) acts like a scattering phase shift. Recall that in quantum scattering, the cross-section is related to the scattering phase shift as ∞ 4π X σ = (2l + 1) sin2(η ). k2 l l=0 Assume that we are confined to a spherical volume with a large radius R. The wave function must be zero at the boundary by the assumption that we are confined therein. This implies a quantization condition lπ kR − + η (k) = nπ, n = 0, 1, 2,... 0 2 l This implies that the number of states below k is kR l η (k) Σ (k) = n(k) = 0 − + l . lk π 2 π The density of states is dΣ R 1 ∂η (k) g (k) = = 0 + l . lm dk π π ∂k So we find that ∞ (0) ! (0) X 2 2 ∂ηl ∂η b − b = 23/2 (2l + 1) dk e−β~ k /m − l . 2 2 ˆ ∂k ∂k l 0 (0) ∂ηl Note that ∂k = 0. For a free gas, 1 b(0) = ± , 2 25/2 where the plus sign is taken for bosons, and the minus sign is taken for fermions. Then bound ∞ 1 X X 2 2 ∂ηl b = ± + e−βεα + 23/2 (2l + 1) dk e−β~ k /m . 2 25/2 ˆ ∂k α l 0 Note, when performing the sum over l, you have to take wave function symmetry into account. For bosons, you only sum over even l, and for fermions, you only sum over odd l. Flipping partical identity results in flipping the sign of the r coordinate. This changes the angular part of the wave function since ~r → −~r, and m l m Yl (−rˆ) = (−1) Yl (ˆr). For bosons, we symmetric wave functions, and for fermions, we need antisymmetric wave functions. So ignoring the spin degrees of freedom, this implies that bosons have only odd l and fermions have only even l. However, it’s more complicated than this since free particles often have spin. I.e. our wave function can pick up another minus sign from the spin wave function. We just have to determine, based on particle type, what happens to the wave function when the 8.2. Virial Expansion 105 particle is interchanged. Recall that the spin wave function for the singlet state with s = 0 is 1 √ (|↑↓i − |↓↑i) , 2 and for the s = 1 triplet states, 1 |↑↑i , √ (|↑↓i + |↓↑i) , |↓↓i . 2 Example 8.2.2: Low Energy Scattering For low energy scattering, η0(k) = −ak. Plugging this in gives us (for fermions) 1 b = − − 4πa`−1. 2 25/2 Q Then vP v 4πa`2 = 1 + Q + Q . T 25/2v v 106 Interacting Gases 8.3 Summary: Interacting Gases Skills to Master • Calculate the virial equation of state given the correlation function of an interacting gas • Calculate number fluctuation given the correlation function of an interacting gas • Calculate virial corrections to the equation of state for various types of gases in various limits Correlation Functions For a spherically symmetric g(R), in d-dimensions, the virial equation of state is Here we only consider classical, 3D, non-relativistic, ∞ gases with Hamiltonians of the form P n d du = 1 − d RR gN (R). N nT 2dT ˆ0 dR X p2 H = i + U, 2m In 3D, the relation to number fluctuations is given i=1 by where the potentials are limited to two-body interac- hNi h i h∆N 2i 1 + d3R g(R~ ) − 1 = . tions that depend only on the distance between the V ˆ hNi particles 1 X U = u(r ). Virial Expansion 2 ij i6=j The method of virial expansion can be used with clas- We can do the momentum integrals in the kinetic part sical or quantum gases, with or without interactions. of the Hamiltonian, then we get the canonical partition For the virial expansion, we write the virial equation function of state in the form 1 1 ZN = QN , 3N ∞ m−1 N! ` vP X vQ Q = a (T ) , T m v where the configuration integral is m=1 Q = d3N r e−βU where N ˆ 2 3/2 β V 2π 3N − P u(r ) 3 ~ = d r e 2 i6=j ij v = , vQ = ` = . ˆ N Q mT 3 3 Y −βu(r ) = d r ··· d r e ij . The goal is to find the coeffcients am. If we keep only ˆ 1 N i For a non-interacting quantum gas, this process In the high temperature limit, is easy because we know what ln Z are for those gases. ∞ All we have to do is expand ln Z and get the relevant b 2π 2 −βu(R) m b2 = −a2 = dR R e − 1 . coefficients. It is helpful to know the Taylor expansion vQ ˆ0 1 1 ln(1 + x) ≈ x − x2 + x3. Then, the virial equation of state with first correction 2 3 is vP 2π ∞ For a classical gas with an interaction, it’s more = 1 − dR R2 e−βu(R) − 1 . complicated because we have to calculate configuration T V ˆ0 integrals QN . The first few bm coefficients are Q b (T,V ) = 1 = 1 1 V 1 2 b2(T,V ) = Q2 − Q1 2vQV 1 3 3 b3(T,V ) = 2 Q − 3Q1Q2 + 2Q1 . 6vQV Chapter 9 Phase Transitions Below is a graphic of the water PT (pressure vs. temperature) phase diagram. Qualita- tively, the PT-diagram for other molecules is similar. The PT phase space can be mapped experimentally. We can also think of phase transitions in terms of Gibbs free energy, which is a function of N, V , and T . We only want a function of P and T , so we write it as G(N,P,T ) = Ng(P,T ), where g(P,T ) is the Gibbs free energy per particle. This function should be smooth and continuous in the vapor phase. Derivatives should all exist. But we expect non-analytic behavior as we cross a phase transition line. I.e., g is non-analytic across phase transitions. This is because on the phase transition line, we have two possible equilibrium states—e.g. vapor and liquid. Note, this is only true in the thermodynamic limit. The Gibbs free energy per particle is the same as chemical potential. We could also plot, for example, the magnetic field versus T . Think of the P in a PT-diagram as an external field that we’re imposing on the system. In general, there are three phases that we will deal with: • The gas phase is characterized by weak interactions and small or no correlations. In the ideal case, PV = NT • The liquid phase is characterized by stronger interactions and short-range spatial correlations • The solid phase is characterized by strong interactions, long range correlations (e.g. it may form a lattice), and broken symmetries (e.g. a lattice breaks rotational symmetry) There may be many different solid phases, e.g. with different crystal symmetries. Phase Transitions 109 For water, we see a critical point (Tc,Pc) around which there is a continuous transition from liquid to gas. Since there exists a smooth path from gas → liquid, this implies there is no symmetry breaking in the gas → liquid transition unlike the liquid → solid transition. Another useful diagram is that of the pressure versus the specific volume v = V/N shown below. In the forbidden region, you have phase coexistence. I.e. if you try to access a point in the forbidden region, the system will separate into multiple phases—e.g. the water might separate into gas and liquid phases as it does when water is boiling. Notice that the triple point is a line in this diagram—where all three phases coexist at a constant pressure. We can also look at ferromagnetic transitions as shown in the diagram below. When it comes to classifying phase transitions, there are two approaches. The older, Ehrenfest classification scheme classifies phase transitions by discontinuity of derivatives. ∂g • 1st order phase transition: Discontinuous ∂x ∂2g • 2nd order phase transition: Discontinuous ∂x2 . I.e. if the system is discontin- uous in the response functions, there is a second order phase transition . • . The more modern scheme is to classify in terms of an order parameter. Examples of possible order parameters include the specific volume V/N of the system, the average magnetization of a ferromagnetic system, or the vacuum expectation of the Higgs field. We could even have complex order parameters. Whenever the order parameter goes from zero to some nonzero value, there is a phase transition. • 1st order phase transition: When the order parameter is discontinuous, as in the example shown here 110 Phase Transitions Tip Whenever the order param- eter goes from zero to some nonzero value, there is a phase transition. • 2nd order phase transition: When the order parameter is continuous, as in the example shown here of the magnetization in the Ising model In the modern classification scheme, there are only two levels defined. What are some things we can measure across a phase transition? One is the latent heat. This is the heat released/required to go across a phase transition. L = T (Sgas − Sliquid) = NT (sgas − sliquid), where Sgas − Sliquid is the entropy difference across the phase transition. Note that temperature T is constant across a phase transition. For first order transitions, generally, L 6= 0. For a second order transition, entropy has to be continuous, then L = 0. The Clausius-Clapeyron relation tells us something about the derivative of pres- sure with respect to temperature in the phase diagram. I.e. it tells us something about the slope of the line in the PT-diagram. Gibbs free energy should be continuous across a phase transition. In the large N limit, G = N g(P, t), then ∂G = g = µ. ∂N T,P At a point on a phase transition line, you have one phase below you and a different phase above you. Then the following relations between the two different phases hold true at this point on the transition line: g1 = g2 =⇒ µ1 = µ2 P1 = P2 T1 = T2. 9.1. Liquid-Gas Phase Transition 111 I.e. our system admits two different equilibrium configurations. These conditions must be satisfied in order for the two phases to coexist. For example, for liquid and gas phase coexistence, the following conditions must be satisfied along the coexistence curve. Tgas = Tliquid Pgas = Pliquid µgas = µliquid. We can use the equation of state (e.g. the van der Waals equation) to set Pgas = Pliquid and obtain the coexistence condition for temperature and pressure in terms of the specific volumes of vgas and vliquid of the different phases. For the chemical potential, we can use the relation TS = E + PV − µN to write E PV ST µ = + − . N N N Then the coexistence condition for the chemical potential is Nµg − Nµl = [Eg + PgVg − SgTg] − [El + PlVl − SlTl] = 0. Consider the variation of G along a phase transition dG1 = −S1 dT + V1 dP = dG2 = −S2 dT + V2 dP. For example, dGliquid = −Sliquid dT + Vliquid dP = dGgas = −Sgas dT + Vgas dP. This implies that along a phase transition, dP S − S S − S L = 1 2 = 1 2 = , dT phase line V1 − V2 v1 − v2 NT (v1 − v2) where, e.g., v1 = V1/N1. This is the Clausius-Clapeyron relation dP L = . dT phase line NT (vgas − vliquid) The slope dP/dT tells us about the relation between latent heat and specific volumes. For example, for water, the phase transition line between solid and liquid phases has a negative slope, but since going from the liquid to the solid phase releases energy, this implies that the specific volume of ice is greater than the specific volume of water. 9.1 Liquid-Gas Phase Transition Suppose we have the exact canonical partition function ZN (V,T ) of an interacting system. In the thermodynamic limit (N,V → ∞), we can write the Helmholtz free energy per particle as T f(v, T ) = − ln Z , N N where v = V/N. What properties will f have? • The pressure should be non-negative. I.e. ∂f ∂ ln Z P (V,T,N) = − = T N > 0. ∂v N,T ∂V N,T Otherwise, if the system had negative pressure, it would be unstable. 112 Phase Transitions • The compressibility should be positive, i.e. 1 ∂v βT = − > 0. v ∂P T Since f(v, T ) is convex everywhere, the derivative ∂P < 0, ∂v T is always negative. ∂P These two conditions need to be met for stability of the system. If ∂v > 0, we have spinodal instability. Example 9.1.1: van der Waals Phase Transition A van der Waals gas, which has the equation of state (EOS) RT a P = − , v − b v2 is an example of an approximate system that exhibits phase transitions. Recall that b is related to the excluded volume resulting from the infinite (hard sphere) potentials. We have to limit ourselves to v > b. 2 ∂P If the attractive term a/v is dominant, we can have ∂v > 0, resulting in spinodal instability and a phase transition. The region where this occurs is shown below where we plot a number of isotherms. This occurs below some critical ∂P temperature Tc, and results in a region where ∂v > 0. Note, this region doesn’t exactly correspond to the phase transition region. This unstable behavior is nonphysical for equilibrium systems, so we need to fix it. To do that, we use a Maxwell construction. We assume the system splits into a high density vH and low density vL phase, which are both stable and in thermodynamic equilibrium with each other. Then TH = TL PH = PL µH = µL. This is for temperatures below Tc. In the phase equilibrium region, v = uvL + (1 − u)vH , 9.1. Liquid-Gas Phase Transition 113 where u is the fraction of volume in the low density phase, and 1 − u is the fraction of volume in the high density phase. Both phases have pressure PH , which implies that the density is independent of the pressure, and the system is infinitely compressible. This implies that the derivative of the pressure with respect to v will be discontinuous when the phase transition begins. This discontinuity is a signal that we can look for when searching for the phase transition. A more useful approach is to consider the grand potential in the thermody- namic limit. Recall that for an extensive system, −PV = E − TS − µN, which implies −V dP = S dT − N dµ. We’re looking at isotherms, i.e. where dT = 0, so we get vH µH − µL = v dP = vH PH − vLPL − P dv = 0, ˆ ˆvL since we know that µH − µL = 0. Also, since PH = PL, we can write v H P (v, T ) − PL/H dv = 0. ˆvL This is our Maxwell construction. This implies that the areas of the P curve above and below the coexistence line of constant pressure PL/H ≡ PL = PH are equal. The critical temperature is the highest temperature for which a phase transition occurs. In the region between vL and vH , we expect to find two places where ∂P = 0. ∂v T One should be a local maximum and another a local minimum. Between these two points, P must change from convex to concave, which implies that there exists a point between the above two points where ∂2P 2 = 0. ∂v T As the temperature is increased, vL and vH approach each other, and all three of the points identified above converge to one another. Thus, at the critical temperature Tc, we 114 Phase Transitions need ∂P ∂2P = = 0. 2 ∂v T Tc,vc ∂v T Tc,vc This gives us two equations and two unknowns, so we can solve for the critical values Tc and vc. Example 9.1.2: van der Waals Critical Temperature For the van der Waals gas, we get the pair of equations Tc 2a 2 = 3 (vc − b) vc 2Tc 6a 3 = 4 . (vc − b) vc Solving for the critical values, we get 8 a v = 3b, T = . c c 27 b Plugging these back into the van der Waals EOS, implies that the pressure at the critical temperature is a P = . c 27b2 If we define P v T Pr ≡ , vr ≡ ,Tr ≡ , Pc vc Tc we can write the van der Waals EOS in terms of the critical numbers as 8Tr 3 Pr = − 2 . 3vr − 1 vr Note that at the phase transition, these parameters are all equal to 1. 9.2 Ising Model The Ising model is a model of interacting spins on a lattice. Below is a graphic illustrating the 2D example, but we can also have 1D and 3D Ising models. This model can be applied to different things like lattice gases and binary alloys. We will only consider the spin-1/2 case, so each lattice site can be in one of two states. The Hamiltonian is N X X H ({σi}) = −µBB σi − J σiσj, i=1 nearest neighbors 9.2. Ising Model 115 where σi = ±1 is the spin at the ith lattice site, J parametrizes the interaction strength between nearest neighbors, µB is the magnetic moment of the spins, and B is the external magnetic field. The second sum is the interaction potential, and it goes over all nearest neighbors. We will only consider nearest neighbor (i.e. local) interactions. I.e. a particle at a given lattice site interacts only with the particles at the adjacent lattice sites. The canonical partition function is X X −βH({σ}) ZN (B,T ) = ··· e . σ1=±1 σN =±1 In general, it’s hard to find an exact solution. However, it is easy to do in 1D, and it’s possible in 2D. The net magnetization is * + X 1 ∂ ln ZN M(B,T ) = µ σ = µ N hσi = . B i B β ∂B i This can be derived by differentiating ln ZN with respect to the magnetic field. In general, we expect the external field V to cause some net magnetization. How- ever, even if B = 0, we want our model to display magnetization because that is what ferromagnetic materials do. Mean Field Approximation In the mean field approximation, we assume that each spin is interacting with some overall mean spin of the lattice rather than the locally fluctuating spins. So we approxi- mate the spin at lattice site i by σi = hσi + ξi, where hσi is the mean spin of the lattice and ξi is some fluctuation. Now the Hamiltonian becomes N X X X q X H ({σ }) = −µ B σ − J σ hσi = −µ B σ − J hσi σ , MF i B i i B i 2 i i=1 nearest i i neighbors where q is approximately the number of nearest neighbors that each lattice site has. We can write this as X HMF ({σi}) = −(B + Bint)µB σi, i where qJ hσi Bint = , µB looks like an internal magnetic field. The mean-field partition function is N X X Y βµB (B+Bint)σi ZN (B,T, hσi) = ··· e σ1=±1 σN =±1 i=1 N Y X = eβµB (B+Bint)σi i=1 σi=±1 N N = 2 cosh (βµB(B + Bint)) . We can calculate 1 ∂ ln ZN = µBN hσi , β ∂B T,hσi 116 Phase Transitions to get the implicit relation hσi = tanh [β (µBB + qJ hσi)] . This is the requirement that needs to be satisfied for self-consistency. Consider the B = 0 case. Then we get hσi = tanh [βqJ hσi] . Here, we see that hσi = 0 is a solution. Are there any other solutions? We can plot both sides of the equation as functions of hσi as shown below. In this plot we see that at sufficiently low temperature T , there are two more solutions hσi= 6 0. So even when B = 0, there is a net magnetization at sufficiently low temperature. In the plot above, we see that the condition for there to be three solutions rather than only the trivial solution hσi = 0, is that the slope of tanh [βqJ hσi] must be greater than 1 at the origin. That is, the point where we go from one to three solutions is 2 1 = βqJ sech [βqJ hσi] = βcqJ. 0 This gives us the critical temperature Tc = qJ. Consider the behavior of hσi versus temperature shown below. For B = 0, the net magnetization drops sharply to zero at Tc. In the B = 0 case and near the critical temperature Tc, hσi is small and we can expand T 1 T 3 hσi ≈ hσi − hσi3 . Tc 3 Tc To leading order, this gives us √ hσi ≈ −3t ∝ t1/2, 9.2. Ising Model 117 where T − T t = c . Tc Note, this is for T < Tc (i.e. for t < 0) since hσi = 0 above Tc. Notice that near the critical point, hσi ∝ t1/2. Recall that the temperature depen- dence of the specific volume v of a van der Waals gas near the critical point has the same exponent. This suggests some universal power law behavior of order parameters like v and hσi near the critical point. The entropy of the system is 1 + hσi 1 + hσi 1 − hσi 1 − hσi S = −N ln + ln . 2 2 2 2 The entropy is continuous, so there is apparently not a first order phase transition. At B = 0, the energy is qJ E(T ) = − hσi2 N. 2 Above Tc, we know hσi = 0, so for T > Tc, the specific heat is ∂E C = = 0. ∂T Note that C = CV = CP . For T < Tc, we can take some implicit derivatives to find 2 2 ∂E Tc hσi C(B = 0, T < Tc) = = N −1 . ∂T T 2 1 − hσi − Tc/T Plotting the specific heat C versus temperature below, we see that C drops sharply to 0 at Tc indicating that this is a second order phase transition. Near the critical point, 3t C(t) ≈ −N(1 + t)−2 , (1 + 3t)−1 − (1 + t)−1 which has the limit 3 C(t → 0) = N, 2 as T approaches Tc from below. The magnetic susceptibility of the mean-field Ising model is 2 2 ∂M NµB 1 − hσi χ = = 2 . ∂B T T 1 − (1 − hσi )Tc/T 118 Phase Transitions Just above the critical temperature, 2 NµB χ0 ≈ , T − Tc since hσi << 1. This is the Curie-Weiss law. Just below the critical temperature, 2 NµB χ0 ≈ . 2(Tc − T ) Thus, the susceptibility is divergent at the critical temperature. 1D Cyclic Spin Chain For a 1D cyclic (meaning periodic boundary conditions), we can actually write down the exact solution—no need to make a mean field approximation. The partition function, which depends on on nearest neighbor interactions is " N !# X X X 1 Z = ··· exp β Jσ σ + µ B [σ + σ ] . N i i+1 2 B i i+1 σ1=±1 σN =±1 i=1 We can also write this in the form X X ZN = ··· hσ1|P|σ2i hσ2|P|σ3i · · · hσN |P|σ1i , σ1=±1 σN =±1 where " # eβ(J+µB B) e−βJ P = , e−βJ eβ(J−µB B) is called the transfer matrix. Then X h N i N N ZN = hσ1|P|σ1i = Tr P = λ+ + λ− , σ1=±1 where βJ −2βJ 2βJ 2 λ± = e cosh (βµBB) ± e + e sinh (βµBB) , are the two eigenvalues of P. N N Since λ+ is always larger than λ−, then for large N, we know that λ+ >> λ− . So we can approximate ln ZN = N ln λ+. Then the magnetization is ∂ ln Z M = N = µ B hσi , ∂B B and the net spin is sinh (βµBB) hσi = q . 2 −4βJ sinh (βµBB) + e This is the exact solution—not a mean field approximation, and it can help us evaluate the validity of mean field approximation. Notice that when B = 0, hσi = 0 for all T . This implies that the exact solution does not have a phase transition at finite temperature, unlike in the mean field case. 9.3. Critical Exponents 119 9.3 Critical Exponents For the mean field Ising model, we found that √ T − T m = hσi ' −3t ∼ t1/2, t ≡ c . Tc The number m is an example of an order parameter. This result is only valid for T < Tc since for T > Tc, we have m = 0. Suppose h is our general external field, then for the free field (i.e. h = B = 0) mean field Ising model, β mh=0 ∼ (T − Tc) , where β = 1/2 is an example of a critical exponent. Don’t confuse this β with the inverse temperature. For the susceptibility of the B = 0 mean field Ising model, we found that ( −γ ∂m (T − Tc) if T > Tc χ0 ∝ ∝ −γ0 , ∂h T,h→0 (T − Tc) if T < Tc where γ = γ0 = 1 is another pair of critical exponents. Note that whereas M is the magnetization, m can be thought of as the local magne- tization density. For the B = 0 mean field Ising model, we also found that ( −α (T − Tc) if T > Tc CV ∼ −α0 , (T − Tc) if T < Tc where α = α0 = 0 is another pair of critical exponents. Finally, for the B = 0 mean field Ising model, it can be shown that m ∼ h1/δ, where h = B, and δ = 3. In general, to characterize the behavior of systems near phase transitions, we look at the critical exponents β, γ, α, and δ, where β m0 ∝ (T − Tc) ( −γ ∂m (T − Tc) if T > Tc χ0 ∝ ∝ −γ0 ∂h T,h→0 (T − Tc) if T < Tc ( −α (T − Tc) if T > Tc CV ∝ −α0 (T − Tc) if T < Tc m ∝ h1/δ, where β = 1/2, γ = γ0 = 1, α = α0 = 0, and δ = 3 for the mean field Ising model. Suppose we allow spatial variation in m so that m = m(~x). Then we can define a connected correlation function D E D E2 Γ(~r) = m(~r)m(~0) − m(~0) , where ~0 denotes the origin. The second term subtracts the expected result when there’s zero correlation. 120 Phase Transitions For the mean field Ising model, 1 ∂ ln ZN 1 hMi = = Tr [Mexp (−βH0 + βBM)] . β ∂B ZN Then ∂ hMi β 2 −βH β −βH 2 χ = = Tr M e − 2 Tr Me , ∂B ZN ZN where H = H0 − BM. This can be written as h i χ = β hM 2i − hMi2 . Treating m as a continuous variable, we can write M = dDx m(~r), ˆ where D is the number of dimensions. Then V χ = β dDx dDx (hm(~x )m(~x )i − hm(~x i hm(~x i) = dDx Γ(~x). ˆ 1 2 1 2 1 2 T ˆ Later, we will show that near the critical point for the mean field Ising model, e−r/ξ Γ(r) ' , r(D−1)/2 for large r (i.e. for r >> ξ) and Γ(r) ' r2−D, for small r. Note that ξ is the correlation length of the system, and it depends on t near the critical point. So we have two new critical exponents ν and η, where ( r2−D−η if r < ξ Γ(r) ∼ r−r/ξ if r > ξ ( −ν |t| if T > Tc ξ ∼ −ν0 . |t| if T < Tc There are known relationships between the critical exponents 2 − α = νD 1 β = − ν(2 − D − η) 2 γ = ν(2 − η) 1 βδ = ν(2 + d − η). 2 Since we have six critical exponents and four relationships between them, there are really only two independent critical exponents. Systems with the same critical exponents are said to be in the same universality class. 9.3. Critical Exponents 121 Example 9.3.1: van der Waals Critical Exponents Recall the van der Waals equation of state T a P = − . v − b v2 If we define the new variables P − P P π = c = − 1 Pc Pc v − v v ψ = c = − 1 vc vc T − T T t = c = − 1, Tc Tc then, recalling that the critical values for the van der Waals gas are 8 a a v = 3b, T = ,P = , c c 27 b c 27b2 we can write the equation of state as 8(t + 1) 3 π + 1 = − . 3ψ + 2 (ψ + 1)2 Expanding this in ψ gives us 3 π ≈ 4t − 6tψ + 9tψ2 − ψ3. 2 For the van der Waals gas, the ordering field is the pressure and the order parameter is the difference in densities of the liquid and gas phases. In the new variables, h = π is the field, and m = ψ is the order parameter. We want to find the critical exponents β, γ, γ0, and δ. The critical exponent δ gives the power relation between the order parameter and the field m ∝ h1/δ, near the critical point. In our case, the critical point T = Tc corresponds to t = 0, which gives us 3 2 1/3 π ≈ − ψ3 =⇒ ψ ≈ − π , 2 3 which tells us that δ = 3. The critical exponent β gives the power relation between the order parameter and the temperature when we approach the critical point from below β mh=0 ∝ (Tc − T ) . To do that, we do a Maxwell construction Pg = P` =⇒ πg = π` Tg = T` =⇒ tg = t`. 122 Phase Transitions With this, we find that πg = π` 3 3 4t − 6tψ + 9tψ2 − ψ3 = 4t − 6tψ + 9tψ2 − ψ3 g g 2 g ` ` 2 ` 3 6t − 9t(ψ + ψ ) + (ψ2 + ψ ψ + ψ2) = 0 g ` 2 g g ` ` We can approximate ψg ≈ −ψ`, p p which gives us ψg ≈ 2 |t| and ψ` ≈ −2 |t|. In terms of the order parameter, 1/2 ψg − ψ` ≈ 4|t| , and so 1 β = . 2 The critical exponents γ and γ0 are obtained by looking at the relationship between susceptibility of the gas and the temperature above and below the critical point. ∂m χ = . ∂h T,h→0 In our case, we look at the isothermal compressibility ∂ψ ∂π −1 1 − = − = . 9 2 ∂π t ∂ψ t 6t − 18tψ + 2 ψ On the critical isodensity contour above the critical temperature, v = vc, so ψ = 0, and we get ∂ψ 1 − ≈ . ∂π t 6t On the critical isodensity contour below the critical temperature, ψ = ±2|t|1/2, so ∂ψ 1 − ≈ . ∂π t 12|t| Thus, γ = γ0 = 1. 9.4. Summary: Phase Transitions 123 9.4 Summary: Phase Transitions Skills to Master • For a given system, determine the critical values at which the phase transition occurs • Classify a phase transition as first or second order • Calculate the critical exponents for a phase transition in a given system • Understand and be able to analyze the liquid-gas phase transition for a van der Waals gas • Understand and be able to analyze the mean-field Ising model A phase transition occurs in a system when some η, and ν, where quantity changes abruptly. More formally, we can de- β scribe a system in terms of a generic ordering field h mh=0 ∝ (T − Tc) that generates order in the system, and a related or- ( −γ ∂m (T − Tc) if T > Tc der parameter m that quantifies the order present in χ0 ∝ ∝ −γ0 ∂h T,h→0 (T − Tc) if T < Tc the system. Below some critical temperature Tc, there ( −α is order present in the system, and so m has some fi- ∂E (T − Tc) if T > Tc nite, nonzero value. The phase transition occurs at the CV = ∝ −α0 ∂T (T − T ) if T < T critical temperature above which the order is no longer c c present and m = 0. m ∝ h1/δ We will examine two model systems—the van der ( r2−D−η if r < ξ Waals gas and the Ising model of ferromagnetism. In Γ(r) ∼ −r/ξ the van der Waals gas, the ordering field is the pres- r if r > ξ sure (i.e. h = P ), and the order parameter is the ( −ν |t| if T > Tc specific volume difference between the two phases (i.e. ξ ∼ −ν0 , |t| if T < Tc m = v` − vg). In the Ising model, the ordering field is the applied magnetic field (i.e. h = B), and the order where t = (T − T )/T , ξ is the correlation length, and parameter is the expected value of the overall spin (i.e. c c m = hσi). D E D E2 ~ ~ To locate the phase transition, we find where the Γ(~r) = m(~r)m(0) − m(0) , order parameter m (as a function of temperature) goes to zero. In the modern classification scheme, there are is the connected correlation function. Note that ~0 de- two kinds of phase transitions: notes the origin. The second term subtracts the ex- pected result when there’s zero correlation. There are known relationships between the critical exponents 2 − α = νD 1 β = − ν(2 − D − η) • First order: When the order parameter jumps 2 to zero in a discontinuous manner. γ = ν(2 − η) • Second order: When the order parameter goes 1 βδ = ν(2 + d − η). to zero in a continuous manner. In this case, the 2 entropy difference between the two phases is zero. Since we have six critical exponents and four relation- ships between them, there are really only two indepen- dent critical exponents. Systems with the same critical exponents are said to be in the same universality class. Liquid-Gas Phase Transition of VDW Gas To characterize the behavior of systems near phase One quantity we can measure across a phase transition transitions, we look at the critical exponents β, γ, α, δ, is the latent heat. This is the heat released/required 124 Phase Transitions to go across a phase transition by using a Maxwell construction to force the areas of the P curve above and below the coexistence line of L = T (Sg − S`) = NT (sg − s`), constant pressure P`/g ≡ P` = Pg to be equal. The condition is where S −S is the entropy difference across the phase g ` vg transition. Note that temperature T is constant across P (v, T ) − P dv = 0. ˆ `/g a phase transition. For first order transitions, gener- v` ally, L 6= 0. For a second order transition, entropy has The Clausius-Clapeyron relation to be continuous, then L = 0. dP L Along the phase transition line, by definition, two = , different phases of matter are in coexistence. For dT phase line NT (vg − v`) the liquid-gas phase transition, along this coexistence gives us the slope of the phase transition line in a PT- curve, we know that diagram. At the critical temperature T , we know that Tg = T` c P = P ∂P ∂2P g ` = = 0. 2 µg = µ`. ∂v T Tc,vc ∂v T Tc,vc We can use the equation of state (e.g. the van der So taking first and second derivatives of the EOS with respect to v and setting them equal to zero gives us Waals equation) to set Pg = P` (with Tg = T` = Tc at the phase transition) and obtain the coexistence con- two equations and two unknowns, so we can solve for dition for temperature and pressure in terms of the the critical values Tc and vc. Plugging these back into the EOS gives us the critical pressure Pc. specific volumes of vg and v` of the different phases. To obtain the phase coexistence condition, we require For the van der Waals gas, the critical values are µ − µ = 0. We can calculate the energy difference 8 a a ` g v = 3b, T = ,P = . between the two phases by integrating over a Maxwell c c 27 b c 27b2 relation We can define new variables v` ∂e P e(T, v ) − e(T, v ) = dv π = − 1 ` g ˆ vg ∂v N,T Pc " # v v` ∂P ψ = − 1 = dv T − P . vc ˆ ∂T vg N,v T t = − 1. T We can calculate the entropy difference similarly c Now the critical point occurs when π = ψ = t = 1. In v` ∂s s(T, v ) − s(T, v ) = dv the new variables, h = π is the field, and m = ψ`−ψg ∝ ` g ˆ vg ∂v N,T ψ is the order parameter. Writing the EOS in terms of v` ∂P the new variables and then expanding in ψ, we get the = dv . new EOS ˆv ∂T g N,v 3 π ≈ 4t − 6tψ + 9tψ2 − ψ3. Then we can use the relation TS = E + PV − µN to 2 write Plugging in t = 0 gives us the critical exponent δ = 3. E PV ST To find the critical exponent β, we set πg = π` µ = + − = e + P v − sT, N N N and tg = t`, and approximate ψg = −ψ` to get 1/2 where e, v, and s are the specific energy, volume, and ψg − ψ` ≈ 4|t| , entropy. which implies β = 1/2. For the van der Waals gas with the equation of Above T , ψ = 0 and below T , ψ = ±2|t|1/2, so state c c T a ( 1 P = − 2 , ∂ψ 6t if T > Tc v − b v − ≈ 1 ∂π if T < Tc, ∂P t 12|t| there are regions of instability where ∂v > 0. This is non-physical, and must be removed from the theory which implies the critical exponent γ = γ0 = 1. 9.4. Summary: Phase Transitions 125 Ising Model and the net spin is The Ising model is a model of interacting spins on a lat- sinh (βµBB) hσi = . tice. We will only consider the spin-1/2 case, so each q sinh2 (βµ B) + e−4βJ lattice site can be in one of two states. The Hamilto- B nian is This is the exact solution. Notice that when B = 0, N hσi = 0 for all T . This implies that the exact solution X X does not have a phase transition at finite temperature. H ({σi}) = −µBB σi − J σiσj, i=1 nearest neighbors Mean-field Approximation where σi = ±1 is the spin at the ith lattice site, J In the mean-field approximation, we assume that each parametrizes the interaction strength between nearest spin is interacting only with some overall mean spin neighbors, µB is the magnetic moment of the spins, of the lattice rather than with the locally fluctuating and B is the external magnetic field is spins of its neighbors. So we approximate the spin at The canonical partition function is lattice site i by σi = hσi + ξi, X X −βH({σ}) ZN (B,T ) = ··· e . where hσi is the mean spin of the lattice, and ξi is some σ =±1 σ =±1 1 N fluctuation. Then The free energy can be calculated as X HMF ({σi}) = −(B + Bint)µB σi, i F = −T ln Z. where qJ hσi The net magnetization, which can be derived by B = , int µ differentiating ln ZN with respect to the magnetic field, B is looks like an internal magnetic field, and q is approxi- mately the number of nearest neighbors that each lat- * + X 1 ∂ ln ZN tice site has. M(B,T ) = µ σ = µ N hσi = . B i B β ∂B The mean-field partition function is i N Y X βµB (B+Bint)σi 1D Ising Model ZN (B,T, hσi) = e . i=1 σi For a 1D, cyclic (meaning periodic boundary condi- tions), spin chain, we can write down the exact solu- For the case in which σ = ±1, we get tion. N N ZN (B,T, hσi) = 2 cosh (βµB(B + Bint)) . The partition function is We can calculate the magnetization h i X N N N ZN = hσ1|P|σ1i = Tr P = λ+ + λ− , 1 ∂ ln ZN σ1=±1 = µBN hσi , β ∂B T,hσi where " # to get the implicit relation eβ(J+µB B) e−βJ P = , hσi = tanh [β (µ B + qJ hσi)] . e−βJ eβ(J−µB B) B In the free field case (i.e. when B = 0) we have is called the transfer matrix, and hσi = tanh [βqJ hσi] . βJ −2βJ 2βJ 2 λ± = e cosh (βµBB)± e + e sinh (βµBB) , This has either one or three solutions depending on the are the two eigenvalues of P. temperature. If the slope of tanh [βqJ hσi] as a func- N N For large N, we know that λ+ >> λ− since λ+ tion of hσi is greater than one, then we have three solu- is always larger than λ−. So we can approximate tions. This condition means we can differentiate both ln ZN = N ln λ+. Then the magnetization is sides to find that the critical temperature (between one vs. three solutions) is ∂ ln Z M = N = µ B hσi , ∂B B Tc = qJ. 126 Phase Transitions In the B = 0 case and near the critical tempera- So C drops discontinuously to 0 at Tc indicating that ture Tc, we can expand in terms of small hσi. Then to this is a second order phase transition. leading order, The magnetic susceptibility of the mean-field Ising √ model is hσi ≈ −3t ∝ t1/2, ∂M Nµ2 1 − hσi2 χ = = B . Note, this is for T < Tc (i.e. for t < 0) since hσi = 0 2 ∂B T T 1 − (1 − hσi )Tc/T above Tc. Here we see the critical exponent β = 1/2. The entropy of the system is Just above and below the critical temperature, 2 S 1 + hσi 1 + hσi 1 − hσi 1 − hσi ( NµB = − ln + ln . if T > Tc T −Tc N 2 2 2 2 χ0 ≈ 2 NµB if T < Tc. 2(Tc−T ) The entropy is continuous, so there is apparently not a first order phase transition. Thus, we have the critical exponents γ = γ0 = 1. At B = 0, the energy is For the mean field Ising model, the magnetization can be written as qJ E(T ) = − hσi2 N. 2 1 ∂ ln ZN 1 M = = Tr [Mexp (−βH0 + βBM)] . β ∂B ZN Above Tc, we know hσi = 0, so for T > Tc, the specific heat is Then we can write the susceptibility in terms of the ∂E C = = 0. connected correlation function as ∂T ∂M h i V Note that C = CV = CP . For T < Tc, we can take χ = = β hM 2i − hMi2 = dDx Γ(~x). some implicit derivatives to find ∂B T ˆ 2 2 For the mean-field Ising model, we have the crit- ∂E Tc hσi 0 0 C = = N −1 . ical exponents β = 1/2, γ = γ = 1, α = α = 0, and ∂T T 2 1 − hσi − Tc/T δ = 3. Chapter 10 Landau-Ginzburg Field Theory 10.1 Introduction Suppose we have a general order parameter m that denotes the order present in a system. Above some critical temperature Tc, this order parameter is zero, and below Tc, it is finite. For example, in the Ising model, hσi is such an order parameter, and for liquid-gas phase transitions, the density difference between the two phases (i.e. ρL − ρG) at the transition is such an order parameter. Suppose also that we have a general field h that is conjugate to that order parameter. For example, in the Ising model, the conjugate field was the magnetic field, and for the liquid-gas phase transition, the conjugate field was the pressure. We want to study phase transitions in a more general way. Instead of averaging over the entire system as in the mean-field Ising model, we want to average over smaller regions, e.g., a few neighboring spins. That is, we want to study phase transitions at the mesoscopic scale, which is the scale between the macroscopic and microscopic scales. We will use the framework of Landau-Ginzburg field theory. Since we’re now averaging locally, our order parameter now depends on position m` → m`(~x), where the index ` indicates that we can have more than one order parameter. We could even have complex and vector order parameters. The partition function is −βHmicro ZN = Tr e . We will now define some new functions Z ≡ Dm (~x) W (m (~x); T, h) , N ˆ ` ` where the integration measure Dm`(~x) is the sum over all allowed field configurations, and W is the probability of a given configuration. Notice that W is an explicit function of m`, and an implicit function of T and h. We can also define an effective free energy βF (m`(~x); T, h) = − ln W. Now W looks like a Boltzmann factor in the integral above. We still don’t know how to calculate anything, but we can at least determine a few properties of these functions. • Locality and Uniformity: For a local theory, we expect the form βF = dDx Φ(m (~x), ~x; T, h) . ˆ ` 128 Landau-Ginzburg Field Theory For a uniform system, there is no dependence on position, then βF = dDx Φ(m (~x); T, h) . ˆ ` To include some non-local interactions, we can expand in terms of the spatial deriva- tives of m`, so βF = dDx Φ m , ∇m , ∇2m ,... ; T, h . ˆ ` ` ` This works for short-range interactions, but not for long-range ones like the Coulomb force. • Since Φ is mesoscopic, we are averaging over the microscopic scale, and thus, we expect it to be free of non-analytic behavior that might be present at the microscale. We also expect it to be free of the discontinuities that might be present at the macro- scopic scales. So it is reasonable to assume that Φ can be written as a polynomial in its parameters m, ∇m, and so on. • Next, we impose symmetries. The free energy βF should have the same symmetries as the system itself. For example, if the system is translationally invariant, then so should be the Landau free energy. We also expect it to be invariant under m → −m since we are averaging over the small scale. Applying these properties, we can write the Landau free energy in the general form h i F = F (T ) + N dDx q (T )m2 + q (T )m4 + κ(T )(∇m)2 + · · · − hm . 0 ˆ 2 4 Notice that this has the properties we need. It is a polynomial in m and its derivatives, and it only includes even powers of m and its derivatives. This free energy is the Landau- Ginzburg Hamiltonian. We still cannot easily calculate this integral, so we use saddle-point approxima- tion of the partition function. This means minimizing the free energy with respect to m(~x), and it assumes that the integral is dominated by the most probable field configu- ration. We assume any non-uniformity of the field takes us away from the minimum. I.e., we assume that uniformity is required for the lowest energy state. Thus, we calculate the variation of the free energy and set it equal to zero δF = N dDx 2q m + 4q m3 + · · · − h δm = 0. ˆ 2 4 We neglect the derivative terms for a uniform field. Thus, the condition for minimum is that 3 2q2m + 4q4m + · · · − h = 0. Integrating, we find the free energy per particle of 2 4 f(m, h, t) = q0(t) + q2(t)m + q4(t)m + · · · − mh, where T − T t = c , Tc and we minimize with respect to m to find the physical state. 10.2. Phase Transitions 129 Example 10.1.1 For example, how might we find a variational free energy for the Ising model? Earlier, we found that 1 + m 1 + m 1 − m 1 − m S = −N ln + ln . 2 2 2 2 A guess for the variational free energy per particle is E TS f (m) = − var N N q TS = −J m2 − mh − 2 N q 1 + m 1 + m 1 − m 1 − m = −J m2 − mh + T ln + ln , 2 2 2 2 2 q 2 where h = µB, and the energy E = −JN 2 m − mhN comes from the mean-field Ising model. Minimizing fvar(m) with respect to m gives us q m = tanh J m + hm . 2 This is indeed the same mean-field self-consistency equation that we found previ- ously. However, fvar(m) is still not exactly in the form we want. If we Taylor expand it, we get q T m2 T m4 f (m) = −J m2 − mh − T ln 2 + + + ··· var 2 2 12 This implies that the first several coefficients are q0(T ) = −T ln 2 1 1 q (T ) = (T − qJ) = (T − T ) 2 2 2 c T q (T ) = . 4 3 Notice that q2(T ) changes sign at the critical point. This is useful for analyzing the phase transition. 10.2 Phase Transitions Previously, we found that the uniform free energy has the form 2 4 f(m, h, t) = q0(t) + q2(t)m + q4(t)m + · · · − mh, where T − T t = c . Tc Notice that if h = 0 and all the coefficients qi are positive, then the minimum free energy occurs when m = 0. On the other hand, if any of the qi changes sign, the minimum may shift away from m = 0—signalling a phase transition. 130 Landau-Ginzburg Field Theory Near the critical point, m << 1 and t << 1, and we can expand each coefficient as a Taylor series in t. For example, q2(t) ≈ q2,0 + q2,1t. Suppose q2,0 = 0, and q2,1, q4,0 > 0 then 2 4 f(m, h, t) = q2,1tm + q4,0m − mh. Minimizing with respect to m gives us 3 2q2,1t + 4q4,0m − h = 0. In the zero-field case, i.e., h = 0, the minima include m = 0, but also the two minima s −q2,1 mh=0 = ± t. 2q4,0 Above Tc, we have only the minimum at m = 0, but below Tc, we have all three minima. (0 for all t m = q 0 ± −q2,1 t for t < 0. 2q4,0 This is a second order phase transition. The susceptibility is ( −1 1 for t > 0 ∂h 1 2q2,1t χ = = 2 = 1 . ∂m 2q2,1t + 12q4,0m for t < 0 t,h=0 4q2,1|t| On the other hand, suppose q2,1 = 0, and q4,0 < 0, then 4 f(t, m, h) = q4,0m − mh. This is a problem because it implies thermodynamic instability. 10.3. Fluctuations and Correlations 131 Then we have to include the O(m6) term. Finally, suppose, 2 4 6 f(t, m, h) = tm − |q4,0|m + |g6,0|m . Here we have a discontinuous evolution of the order parameter implying that we have a first-order transition. The point is, depending on the relative values of the coefficients, we could have either a first or a second order transition. 10.3 Fluctuations and Correlations We are interested in the spatial size of the fluctuations in a liquid-gas system, since these will determine the wavelength of particles scattered when we probe the system. At finite momentum transfer ~q, we expect scattering intensity to be S(~q) ∝ h|ρ(~q)|i . This is a quantity that we can calculate from the fluctuations in our system. The probability of a particular field configuration in space is κ 1 W (m(~x)) ∝ exp −β dDx (∇m)2 + tm2 + um4 . ˆ 2 2 Previously, we found that the Landau free energy is minimized for a uniform field m. We can write m(~x) = m + φ(~x), where φ(~x) is the fluctuation away from the equilibrium field. The saddle-point approxi- mation of W implies that (0 for t > 0 m = q t . − 4u for t < 0 If we assume the fluctuations are small, then (∇m)2 = (∇φ)2 ← since ∇m = 0 m2 = m2 + 2mφ + φ2 m4 = m4 + 4m3φ + 6m2φ2 + O(φ3). Then κ t f = (∇φ)2 + tmφ + 4um3φ + φ2 + um4 + 6um2φ2 + O(φ3). 2 2 But since m = 0 or m = p−t/4u, the first-order φ terms cancel. The free energy is then given by t κ h i βF = − ln W = βV m2 + um4 + β dDx (∇φ)2 + ξ−2φ2 + O(φ3), 2 ˆ 2 132 Landau-Ginzburg Field Theory where the length scale is ( t + 12um2 t for t > 0 ξ−2 = = κ . κ 2t − κ for t < 0 Now we move to the Fourier transform of φ 1 X φ(~x) = φ ei~q·~x. V ~q ~q In the infinite volume limit, this can be written in the integral form dDq φ(~x) = φ ei~q·~x. ˆ (2π)D ~q This analysis is at the mesoscopic scale. There’s some cutoff scale Λ on q that we’ve imposed. Using the infinite volume limit, we find that Λ D D 0 D 2 d q d q 0 D i(~q+~q 0)·~x d x (∇φ) = (−~q · ~q ) φ φ 0 d x e . ˆ ˆ (2π)D (2π)D ~q ~q ˆ But 0 dDx ei(~q+~q )·~x = (2π)Dδ(~q + ~q 0), ˆ ∗ and since φ is real, φ~q = φ−~q, we end up with Λ D 2 d q 1 X dDx (∇φ) = q2φ φ ≈ q2|φ |2. ˆ ˆ (2π)D ~q −~q V ~q ~q So the probability of finding a given fluctuation is Y h κ 2i W ({φ }) ∝ exp −β q2 + ξ−2 |φ | . ~q 2 ~q ~q Thus, each mode follows a Gaussian distribution with mean 0 and standard deviation D D 2E T T (2π) 0 |φ | = hφ φ 0 i = δ 0 = δ (~q + ~q ) . ~q ~q ~q κ (q2 + ξ−2) ~q,−~q κ (q2 + ξ−2) We want to find the real space connected correlation function Γ(~r) = h(m(~x + ~r) − m)(m(~x) − m)i = hφ(~x + ~r)φ(~x)i . Via the infinite volume Fourier transform, we find D D 0 d q d q i~q·(~r+~x)+i~q 0·~x hφ(~x + ~r)φ(~x)i = e hφ φ 0 i ˆ (2π)D (2π)D ~q ~q D D 0 D d q d q 0 T (2π) = ei~q·(~r+~x)+i~q ·~x δ (~q + ~q 0) ˆ (2π)D (2π)D κ (q2 + ξ−2) T dDq ei~q·~r = . κ ˆ (2π)D q2 + ξ−2 In 3D, this integral becomes 1 ∞ iqr cos θ T 2 e T sin(kr) Γ(r) = 2 d(cos θ) dq q 2 −2 = 2 dq q 2 −2 . κ(2π) ˆ−1 ˆ0 q + ξ κ2π ˆ q + ξ 10.4. Symmetry Breaking 133 We can evaluate this integral using contour integration to get T Γ(r) = e−r/ξ. πκr We can also do this in other dimensions. In D-dimensions, 1 (D−2)/2 r Γ(r) = K , rξ (D−2)/2 ξ where Ka is the modified Bessel function. Asymptotically, the Bessel function goes as ( x−(D−2)/2 for x << 1 K (x) ∼ . (D−2)/2 x−1/2e−x for x >> 1 Then asymptotically, the correlation function goes as ( rD−2 for r << ξ Γ(r) ∼ . r−(D−1)/2ξ−(D−3)/2e−r/ξ for r >> ξ This is called the Ornstein-Zernike form of the correlation function. The parameter ξ is the correlation length. It goes as −2 T − Tc ξ ∝ = |t|, Tc so ξ ∝ |t|−1/2. The correlation length is the spatial scale of relevance. Notice that it diverges at the critical point. This means at the critical temperature, we’ll have correlations up to the macroscopic scale. Phenomena such as critical opalescence occur when there are cor- relations across all length scales like this—causing light of all frequencies to scatter and giving the system a milky appearance. 10.4 Symmetry Breaking There are two kinds of symmetry breaking that we will see—discrete and continuous symmetry breaking. For a system in which the free energy is invariant under m → −m and in zero field, both possible values of the order parameter have the same energy, however, it’s not possible to go continuously from one value to the other. Thus, in the absence of an external field forcing the value, the system has to spontaneously choose the value of the order parameter. This is an example of spontaneous symmetry breaking. Discrete Symmetry Breaking Suppose we have a large system, then it’s possible for some regions of the system to take one value of the order parameter and other regions to take the other value. These regions are called domains and they are separated by domain walls. For example, in the Ising model of a ferromagnet, some regions may have spins aligned in the up direction, while adjacent regions have spins aligned in the down direction. 134 Landau-Ginzburg Field Theory What is the energety cost of having these domain walls? Consider a system with a scalar order parameter m and a Hamiltonian density t κ H(~x) = m2 + um4 + (∇m)2 − hm. 2 2 In the zero field case (i.e. h = 0), we clearly have symmetry under m → −m. Both possible values of the order parameter give the same energy, but we cannot continuously deform the system from one value to the other. Consider a 1D slice of the graphic above (the dotted line), which crosses a domain wall. On one side of the wall, the order parameter is m(~x) = m, and on the other side, it is m(~x) = −m. This is plotted below, illustrating the transition. We know that at ±∞, m(~x) = ±m. What structure of the domain wall minimizes the free energy subject to these constraints? The variation of H with respect to m(~x) is ∞ ∂H δH = dx δm ˆ−∞ ∂m ∞ ∂m ∂(δm) = dx tm δm + 4um3 δm + κ . ˆ−∞ ∂x ∂x We can integrate the last term by parts to get ∞ 2 3 ∂ m δH = dx tm + 4um − κ 2 δm. ˆ−∞ ∂x We want the variation to be zero, so ∂2m tm + 4um3 − κ = 0. ∂x2 Thus, the structure of the domain wall is specified by the differential equation ∂2m κ = tm + 4um3, ∂x2 subject to our boundary conditions at ±∞. The solution is x − x m = m tanh 0 , w where x0 is the middle of the transition region, and r−t r 2κ m = , w = . 4u −t The quantity w defines the width of the transition region. 10.4. Symmetry Breaking 135 Now we can calculate the free energy cost of having a domain wall " # ∞ t κ ∂m2 βF (m(x)) − βF (m) = dx m2(x) − m2 + u m4(x) − m4 + . ˆ−∞ 2 2 ∂x Note that F (m) is the free energy if there were no domain wall. This integral evaluates to 2 F (m(x)) − F (m) = − tm2wA ∝ (−t)3/2, 3 where A ∝ LD−1 is the area of the domain wall. Notice that this is positive for t < 0. The area of the domain wall is proportional to some macroscopic length scale; A ∼ LD−1. In the 1D case, the energy difference is microscopic at all t, and one can have fluctuations creating and destroying domain walls. This destroys all long-range order, implying that there can be no phase transitions in 1D systems. This is consistent with what we found for the exact 1D Ising model. At higher dimensions, macroscopic energy is required to create domain walls. Thus domain walls are suppressed and long-range order is a stable state of the system. We say that the lower critical dimension is 1. Above this, we expect phase transitions. Continuous Symmetry Breaking Now, consider a system with the same Hamiltonian density as before t κ H(~x) = |ψ(~x)|2 + u |ψ(~x)|4 + |∇ψ|2 , 2 2 but now with a complex order parameter (or ordering field) ψ(~x) = |ψ(~x)| eiθ(~x). We want the system to have free energy that is invariant under change of phase. The saddle point minimization gives us r t |ψ| = − , 4u for t < 0, and ψ = 0 for t > 0. The phase θ can have any value, so this gives us the Mexican hat potential. That θ can have any value is the source of the continuous symmetry in this example. If we assume that only the phase varies slowly with position, then we can write ψ(~x) = ψeiθ(~x). Then we find that 2 κψ βF = βF + dDx (∇θ)2 , 0 2 ˆ where F0 is the saddle point free energy which does not depend on the gradient, and the integral adds any contribution we might get from the phase of the order parameter. Now we look at the Fourier transform of the phase 1 X θ(~x) = θ ei~q·~x. V ~q ~q Then, κψ X 2 βF = βF + q2 | θ | . 0 2 ~q 136 Landau-Ginzburg Field Theory Any spontaneously broken continuous symmetry gives rise to long-range modes that can be excited. These are called Goldstone modes. An example is the phonons from the Debye model. I.e. these are new low energy excitations of the system which appear below the critical point. The correlation function for these θ modes is δ~q,~q 0 hθ~q θ~q 0 i = 2 . κψ q2 What we want to know is the correlation function of our entire order parameter ψ. There are a lot of steps. The final result is (for large r) ( 2 ψ if D > 2 lim hψ(~r)ψ∗(0)i ∼ r→∞ 0 if D ≤ 2 2 where D is the dimension. The finite value ψ for D > 2 implies long-range order in the system. In general, a finite (nonzero) value of the correlation function over the critical point implies long-range order. In this case, for D ≤ 2, there is no long-range order, which implies there is no phase transition. This is an example of the Mermin-Wagner theorem, which tells us that no spon- taneous symmetry breaking occurs for continuous symmetries for D ≤ 2. So there is a lower critical dimension of 2. Remember that for discrete symmetries, the lower critical dimension is 1. Example 10.4.1 A string, confined in a 2D plane, is stretched between two points and has Hamiltonian H = σ|S − L| where S is the distance between the two endpoints, L is the unstretched length of the string, and σ is the tension. Find the Landau- Ginzburg partition function and identify the Goldstone modes of the system. Consider a very small segment of the string when it is displaced from equi- librium. We can model this displaced piece of string as a triangle with width ∆x, height ∆h, and hypotenuse ∆S. Then s ∂h2 ∆S2 = ∆x2 + ∆h2 → dS = 1 + dx. ∂x Then the length of the string can be written in terms of displacement from equi- librium as s L ∂h2 S = 1 + dx. ˆ0 ∂x Then the Landau-Ginzburg Hamiltonian in the limit of small displacements is s L ∂h2 σ L ∂h2 F = H = σ|S − L| = σ dx 1 + − 1 ' dx . ˆ0 ∂x 2 ˆ0 ∂x 10.5. The Ginzburg Criterion 137 In general, the Landau-Ginzburg partition function is Z = Dm W(m), with βF = − ln W. In our case, βF = − ln W implies that ´ " # σ L ∂h2 W = e−βF = exp −β dx , 2 ˆ0 ∂x and so the Landau-Ginzburg partition function is " # σ L ∂h2 Z = Dh(x) W(h) = Dh exp −β dx . ˆ ˆ 2 ˆ0 ∂x Note, in our case, the order parameter is the height; m`(~x) ≡ h(x). Next, we look at the Fourier transform of the fluctuations, 1 X dq h(x) = √ h eiqx → h eiqx. q ˆ 2π q L q Then ∂h dq = iqh eiqx ∂x ˆ 2π q 2 0 ∂h dq dq 0 i(q+q0)x = (−qq )h h 0 e ∂x ˆ 2π ˆ 2π q q L 2 L 0 ∂h dq dq 0 i(q+q0)x dx = dx (−qq )hqhq0 e ˆ0 ∂x ˆ0 ˆ 2π ˆ 2π 0 L dq dq 0 i(q+q0)x = (−qq )hqhq0 dx e ˆ 2π ˆ 2π ˆ0 0 dq dq 0 = (−qq )h h 0 Lδ 0 ˆ 2π ˆ 2π q q q,−q 1 X 0 = (−qq )h h 0 Lδ 0 L q q q,−q q,q0 X 2 = q hqh−q. q So the Goldstone modes have energy 2 σ L ∂h σ X F = dx = q2h h . 2 ˆ ∂x 2 q −q 0 q 10.5 The Ginzburg Criterion Why is mean-field theory wrong for D < 4? We now go back to a scalar order parameter of the form m(~x) = m + φ(~x). The partition function is the path integral t κ Z ' exp −βV m2 + um4 Dφ[~x] exp −β dDx (∇φ)2 + ξ−2φ2 . 2 ˆ ˆ 2 138 Landau-Ginzburg Field Theory Treating this in continuous space is hard. For discontinuous space ~xi, Y Dφ = N 0 dφ . ˆ ˆ i i That’s what we do in real space. We can also do it in Fourier space. Then Y Dφ(~x) = N dφ dφ∗ . ˆ ˆ ~q ~q ~q ∗ Note that φ~q = φ−~q if φ is real. The fluctuating part of the partition function is κ Z = Dφ[~x] exp −β dDx (∇φ)2 + ξ−2φ2 F ˆ ˆ 2 Y κ X 2 = N dφ dφ∗ exp −β q2 + ξ−2 |φ | . ˆ ~q ~q 2V ~q ~q ~q This is now a series of Gaussian integrals, so we get s Y 2πT V Z = N . F κ(q2 + ξ−2) ~q The thermodynamic free energy (not to be mistaken with the Landau free energy) is T X 2πT V N 2 F = − ln . f 2 κ(q2 + ξ−2) ~q √ The contribution of the fluctuations to heat capacity (remember ξ = ξ0/ t), is 2 2 ∂ Ff 1 X 2T T Cvf = −T 2 = 1 − 2 2 −2 + 2 4 2 −2 2 . ∂T 2 Tcξ (q + ξ ) T ξ (q + ξ ) ~q 0 c 0 The first term is what we expect from the equipartition theorem—each mode contributes 1/2 to the heat capacity. The leading order piece of the second term is ΛD−2 if D > 2 Λ qD−1 dq ∼ ln Λ if D = 2 . ˆ q2 + ξ−2 0 ξ if D = 1 where Λ is the cutoff. For example, for the Ising model, Λ = π/a where a is the lattice spacing. The leading order piece of the third term is ( Λ qD−1 ΛD−4 if D ≥ 4 dq 2 −2 2 ∼ 4−D . ˆ0 (q + ξ ) ξ if D < 4 So for D < 4, fluctuations impact the divergence of CV . The cutoff Λ ∼ π/a does not diverge at the critical point, however, ξ does diverge at the critical point. So we get very different behavior depending on the dimension. For D < 4, 4−D −ν(4−D) −α CVf ∼ ξ ∼ |t| ∼ |t| , and ξ ∼ |t|−ν . 10.6. Scaling Hypothesis 139 Recall that in the mean field case, α = 0. Here, we have α = ν(4 − D) for D < 4. This is not consistent with the mean field case except for D = 4. Thus, the mean field or saddle point approach is exact for D ≥ 4. This defines the upper critical dimension to be D = 4. The Ginzburg criterion is related to how important fluctuations on the scale of the correlation length (for example) are. D E Γ(~x) = (m(~x) − m)(m(~0) − m) . The ratio to look at is Γ(ξ) ∼ |t|(D−2+η)ν−2β, m2 for Γ(r) ∼ r2−D−η. We want (D − 2 + η) ν ≥ 2β, for fluctuations to be unimportant. This is the Ginzburg criterion. This condition becomes D ≥ 4 for mean field critical exponents. 10.6 Scaling Hypothesis In Landau field theory we can write h = tm + 4um3, where r t m = − , 0 4u for t < 0. So we define a scale r |t| m = , s 2u which is valid on both sides of Tc. Then 3 3 m 3 m h = 2msu sign(t) + 4ums . ms ms Next, we define a scale for h 2−3/2 h = um3 = |t|3/2. s s u1/2 Then we have the non-dimensional relation 3 h m m −1 m = 2 sign(t) + 4 ≡ W± , hs ms ms ms where the subscript ± refers to the sign of t. Inverting this, we find h m(t, h) = msW± . hs The mean-field free energy can be written 2 " 2 4 # 2 t 2 4 |t| m m h m |t| h f(t, h) = m +um −hm = sign(t) + − ≡ Y± . 2 4u ms ms hs ms 4u hs This is all for mean-field theory. The Kadanoff scaling hypothesis asserts that we can apply this to general theories. 140 Landau-Ginzburg Field Theory The Kadanoff scaling hypothesis tells us that in general, we can write the scaling field in the form −1 ∆ hs = G |t| , where the exponent ∆ is called the gap exponent. Then we can write the free energy in the form h f = F |t|2−α± Y G . ± |t|∆ Continuity across t = 0 implies that α+ = α− = α, so we can write the free energy as h f = F |t|2−αY G . ± |t|∆ What does this predict for the value of our order parameter? ∂f 2−α−∆ 0 m = − = −FG|t| Y±(0). ∂h t,h=0 Recall that m ∼ |t|β, so by comparison, β = 2 − α − ∆. The susceptibility is ∂m 2 2−α−2∆ 00 χ ∼ = −FG |t| Y± (0), ∂h t,h=0 which implies that γ = γ0 = 2 − α − 2∆. Repeating a similar analysis and comparing with m ∼ h1/δ tells us that ∆ δ = . 2 − α − ∆ Similarly, 2 ∂ f −α CV ∼ 2 ∝ (2 − α)(1 − α)F |t| Y±(0). ∂t h=0 −α Comparing with CV ∼ |t| tells us that this new α is the same as the old α we were using. To summarize, we now have two critical exponents ∆ and α from which we can derive the previous four critical exponends γ, α, δ, and β. The critical exponents ∆ and α are really the only two that matter now. Can we get the rest of the critical exponents via ∆ and α? We now that ξ ∼ |t|−ν and Γ(r) ∼ r2−η−D near the critical point, so h ξ(t, h) ∼ |t|−ν χ , |t|∆ where χ is some homogeneous function of the same form as Y±. Since it’s extensive, we know that ln Z ∼ LD where L is the length scale of our system. But it should be dimensionless, so LD ln Z ∼ . ξD Then our free energy has the form ln Z h f(t, h) ∼ ∼ ξ−D ∼ |t|νDχ−D . LD |t|∆ 10.6. Scaling Hypothesis 141 This implies that 2 − α = νD or 2 − α ν = . D This is called the hyperscaling relation. This implies that the mean-field approach is only correct for D = 4. We can write ξ D−1 D r 2−η −ν(2−η) χ ∼ d x Γ(r) ∼ D−2+η dr ∼ ξ ∼ |t| , ˆ ˆ0 r where we use the upper limit ξ on the integral since correlations taper off above ξ. We also know that χ ∼ |t|−γ , so γ = ν(2 − η), This implies that 2D∆ η = 2 − D + . 2 − α So now we can express all six of our original critical exponents in terms of two—α and ∆. The critical exponents at a second order phase transition depend only on: • the dimension of the system • the symmetries of the order parameter • the range of the interaction Systems belong to the same universality class if they have the same critical exponents. Example 10.6.1 We will now analyze a 2D system with a Landau-Ginzburg free energy 2 F = d2x tψ~ · ψ~ + u ψ~ · ψ~ − 2u0ψ2ψ2 ˆ x y 3 κ κ +g ψ~ · ψ~ + (∇ψ )2 + (∇ψ )2 − ~h · ψ~ , 2 x 2 y where ψ~ = (ψx, ψy) is a two-component order parameter, t = (T − Tc)/Tc, u, g, and κ are constants, and ~h is an external ordering field. We start by finding the most probable values of the order parameter ψ~ when the system is uniform with u = u0 = 0, ~h = 0, and g > 0. For a uniform system, we can also ignore the derivative terms, so now, 3 F = d2x tψ~ · ψ~ + g ψ~ · ψ~ . ˆ Remember that to find the most probable value of the order parameter, we min- imize the free energy with respect to the order parameter. In this case, we can think of the vector order parameter as a pair of order parameters ψx and ψy, so we minimize F with respect to both. We can write h 3i F = d2x t(ψ2 + ψ2) + g ψ2 + ψ2 . ˆ x y x y Minimizing this means setting ∂F = 0 and ∂F = 0, which implies ∂ψx ∂ψy 2 2 2 2tψx + 6gψx(ψx + ψy) = 0 2 2 2 2tψy + 6gψy(ψx + ψy) = 0. 142 Landau-Ginzburg Field Theory This can be written as 4 ψx 2t + 6gψ = 0 4 ψy 2t + 6gψ = 0. Both of these imply that 4 t + 3gψ = 0. We know that the order parameter must be zero above the critical temperature, so we have that our most probable value above and below the critical temperature is (0 if t > 0 ψ = q . 4 −t 3g if t < 0 Going back to the original F , suppose u = u0 > 0 and g = 0. What is the critical exponent γx, where ∂ψx ∝ |t|−γx ? ∂hx Now the free energy is (we can again neglect the derivatives) 2 F = d2x tψ~ · ψ~ + u ψ~ · ψ~ − 2u0ψ2ψ2 − ~h · ψ~ ˆ x y = d2x t(ψ2 + ψ2) + u(ψ4 + ψ4) − h ψ − h ψ ˆ x y x y x x y y Minimizing F with respect to ψx gives us 3 2tψx + 4uψx − hx = 0, which gives us ∂hx 2 = 2t + 12uψx, ∂ψx or ∂ψx 1 = 2 . ∂hx 2t + 12uψx We know that ψx = 0 for t > 0. For t < 0, we assume hx = 0, then from the 3 2 equation 2tψx + 4uψx − hx = 0, we get ψx = −t/2u. Then ( 1 ∂ψx 2t if t > 0 = 1 . ∂hx 4|t| if t < 0 Thus, γ = γ0 = 1. 10.7. Summary: Landau-Ginzburg Field Theory 143 10.7 Summary: Landau-Ginzburg Field Theory Skills to Master • Write down the Landau-Ginzburg Hamiltonian for a system • Apply the saddle-point approximation • Identify symmetries and their effect on the system • Calculate critical exponents • Calculate the contribution of Goldstone modes to the energy In Landau-Ginzburg field theory, instead of av- where the integration measure Dm`(~x) is over all al- eraging over the entire system as in the mean-field lowed field configurations, and W is the probability of Ising model, we average over smaller regions, e.g., a a given configuration. We also define an effective free few neighboring spins. We study phase transitions at energy the mesoscopic scale. Since we’re now averaging lo- cally, our order parameters now depend on position βF (m`(~x); T, h) = − ln W. m` → m`(~x). We define the partition function to be of the form After applying assumptions of locality and unifor- Z ≡ Dm (~x) W (m (~x); T, h) , N ˆ ` ` mity, we come to a free energy of the form h i F = F (T ) + N dDx q (T )m2 + q (T )m4 + κ(T )(∇m)2 + · · · − hm . 0 ˆ 2 4 This is the Landau-Ginzburg Hamiltonian. Fluctuations & Correlations In the saddle-point approximation, we minimize The probability of a particular field configuration in the free energy with respect to m(~x), and assume that space is any non-uniformity takes us away from the minimum. Then we can neglect the derivative terms and set the κ 1 variation of the free energy to zero to obtain a free W (m(~x)) ∝ exp −β dDx (∇m)2 + tm2 + um4 . ˆ 2 2 energy per particle of the form The Landau free energy is minimized for a uniform field f(m, h, t) = q (t) + q (t)m2 + q (t)m4 + · · · − mh, 0 2 4 m. We can write which is minimized with respect to m to find the phys- ical state. m(~x) = m + φ(~x), Note: Near the critical point, m << 1 and t << 1. where φ(~x) is the fluctuation away from the equilib- Physical states correspond to minima of f with rium field. The saddle-point approximation of W im- respect to m. Different relative signs of the q (t) coef- i plies that ficients give us different behavior. If the minimum is ( at m = 0, then the system is in the disordered phase. 0 for t > 0 m = q . If the minimum could shift between m = 0 and m 6= 0 t − 4u for t < 0 depending on the qi(t), then there is the possibility of a phase transition occuring. If we assume fluctuations are small, then we find t κ h i βF = − ln W = βV m2 + um4 + β dDx (∇φ)2 + ξ−2φ2 + O(φ3). 2 ˆ 2 144 Landau-Ginzburg Field Theory Now we move to the Fourier transform of φ For a system with a scalar order parameter m and zero field, the Hamiltonian density is 1 X i~q·~x φ(~x) = √ φ~q e . t κ V H(~x) = m2 + um4 + (∇m)2 . ~q 2 2 In the infinite volume limit, Then we minimize the energy cost of the domain wall by calculating the variation of H with respect to m dDq φ(~x) = φ ei~q·~x. ∞ ∂H ˆ (2π)D ~q δH = dx δm . ˆ−∞ ∂m This leads us to This leads us to the differential equation D T (2π) 0 2 hφ φ 0 i = δ (~q + ~q ) . ∂ m 3 ~q ~q 2 −2 κ = tm + 4um , κ (q + ξ ) ∂x2 We want to find the real space connected correla- subject to our boundary conditions at ±∞. The solu- tion function tion is x − x m = m tanh 0 , Γ(~r) = h(m(~x + ~r) − m)(m(~x) − m)i = hφ(~x + ~r)φ(~x)i . w where x0 is the middle of the transition region, and Via the infinite volume Fourier transform, we find r−t r 2κ T dDq ei~q·~r m = , w = . hφ(~x + ~r)φ(~x)i = . 4u −t κ ˆ (2π)D q2 + ξ−2 The quantity w defines the width of the transition re- In 3D, we can do a contour integral to find gion. Then the free energy cost of having the domain wall turns out to be T Γ(r) = e−r/ξ. 3/2 πκr F (m(x)) − F (m) ∝ (−t) , In D dimensions, where F (m) is the free energy if there were no domain wall. 1 (D−2)/2 r Γ(r) = K rξ (D−2)/2 ξ Continuous Symmetry Breaking ( rD−2 for r << ξ Again, we have a system with the same Hamiltonian ∼ , r−(D−1)/2ξ−(D−3)/2e−r/ξ for r >> ξ density as before t 2 4 κ 2 where K is the modified Bessel function, and ξ, the H(~x) = |ψ(~x)| + u |ψ(~x)| + |∇ψ| , a 2 2 correlation length goes as but now with a complex order parameter (or ordering ξ ∝ |t|−1/2, field) ψ(~x) = |ψ(~x)| eiθ(~x). above and below the critical point. Since the relevant The phase angle θ can have any value, thus, this is an length scale of fluctuations ξ diverges at the critical example of a continuous symmetry. point, we have correlations at all length scales at the Again, saddle-point approximation gives us critical point. (0 if t > 0 |ψ| = q . Discrete Symmetry Breaking t − 4u if t < 0 In a large system, it is possible for some parts of the If we assume that only the phase varies slowly with system to have one value of the order parameter m and position, then we can write ψ(~x) = ψeiθ(~x). Then we other parts have the opposite value −m. Since there find that are only discrete possible values, this is an example of 2 a discrete symmetry that must be “broken” by the sys- κψ βF = βF + dDx (∇θ)2 . tem. These domains are separated by domain walls. 0 2 ˆ 10.7. Summary: Landau-Ginzburg Field Theory 145 1 X i~q·~x Problem Solving θ(~x) = √ θ~qe . V ~q In general, for Landau-Ginzburg field theory, the parti- Then, tion function is Z = Dm W(m), where βF = − ln W. The general procedure´ for solving a Landau-Ginzburg κψ X 2 2 βF = βF0 + q | θ~q| . field theory problem is as follows: 2 1. Write down the Landau-Ginzburg Hamiltonian, This leads us to the thermal expectation aka the free energy, which for a scalar order pa- rameter m has the form δ~q,~q 0 hθ θ 0 i = , ~q ~q 2 h i κψ q2 F = dDx q m2 + q m4 + κ (∇m)2 + · · · − hm , ˆ 2 4 for these long-range Goldstone modes. Going through the steps of finding the real space correlation functions, where the coefficients qi depend on temperature. we find the asymptotic result 2. Next, we perform a saddle point approximation, which means to minimize the free energy with ( 2 ψ if D > 2 respect to the order parameter, i.e., we set lim hψ(~r)ψ∗(0)i ∼ r→∞ 0 if D ≤ 2 ∂F = 0, ∂m So for order parameters with continuous symmetry, there is a lower critical dimension of two below which to get something like no spontaneous symmetry breaking occurs. 3 2q2m + 4q4m − h = 0. Ginzburg Criterion & Scaling Hypothesis When minimizing F , we can neglect the gradient For D < 4, the heat capacity depends on the corre- terms. We could also have a vector order param- lation length ξ which diverges at the critical point. eter like ~m = (mx, my), in which case you would use ∂F = ∂F = 0. Can also have a complex Thus, for D < 4, fluctuations impact the divergence of ∂mx ∂my the heat capacity, and the mean-field or saddle-point order parameter ψ = |ψ|eiθ. approximations are only correct for dimension four or 3. To find the most probable value of the order pa- higher. rameter, we let h = 0. Then we get something The Ginzburg criterion tells us that fluctuations like are not important if (0 if t > 0 m = q . −q2 if t < 0 (D − 2 + η) ν ≥ 2β. 2q4 Note that in general, the order parameter is zero This condition gives us D ≥ 4 for mean-field critical above the critical point, so m = 0 for t > 0. exponents. 4. To identify the order of the phase transition, de- The Kadanoff scaling hypothesis is that we can ∆ termine if the order parameter is continuous (sec- scale the field as hs ∝ |t| , where ∆ is called the gap ond order) at t = 0 or discontinuous (first order). exponent. Then we can write our critical exponents in 5. Spontaneous symmetry occurs when there are terms of only two critical exponents—α and ∆. different possible values of the order parame- ter that the system can choose. If the system β = 2 − α − ∆ can only choose discrete possible values such as 0 γ = γ = 2 − α − 2∆ m and −m, which occurs when you calculate a ∆ most probable value in terms of m2, then discrete δ = 2 − α − ∆ symmetry breaking occurs. If there are continu- 2 − α ous possible values, then continuous symmetry ν = D breaking occurs. 2D∆ 6. To analyze fluctuations, write the order parame- η = 2 − D + . 2 − α ter in terms of a uniform-field value plus a piece that can change Systems belong to the same universality class if they have the same critical exponents. m = m + φ, or ψ = ψeiθ. 146 Landau-Ginzburg Field Theory 7. Goldstone modes can occur when there is conti- Plug this back into the free energy F , now keep- nous symmetry breaking. To analyze them, take ing the gradient terms. You should find some- the Fourier transform of the order parameter (or thing like at least the changing part of it) X 2 2 F ∼ q |φ~q| , 1 X i~q·~x φ(x) = √ φ~qe . as the Goldstone mode contributions to the en- V ~q ergy. Chapter 11 Non-Equilibrium Statistical Mechanics 11.1 BBGKY Recall that the ensemble average of some quantity O is N Y hOi = d3x d3p O ({~x , ~p }) ρ ({~x , ~p }) , ˆ i i i i i i i=1 with ρ normalized such that h1i = 1. One quantity that we might want to know is the probability of finding any particular particle at a given position and momentum at time t. This is given by the single-particle distribution function * N + X f1(~x, ~p, t) = δ (~p − ~pi(t)) δ (~x − ~xi(t)) . i=1 Since the particles are identical, this becomes N Y f (~x, ~p, t) = N dz ρ (~x, ~p, z , . . . , z , t) , 1 ˆ i 2 N i=2 3 3 where dzi ≡ d xid pi. We could also define an m-particle distribution function " N # N! Y f (z , . . . , z ) = dz ρ, m 1 N (N − m)! ˆ i i=m+1 which gives the probability of finding one particle at z1, another at z2, and so on. To find the time evolution of these distribution functions, we differentiate with respect to time to get " N # ∂ N! Y ∂ρ f (z , . . . , z ) = dz . ∂t s 1 N (N − s)! ˆ i ∂t i=s+1 Recall Liouville’s equation N ∂ρ X ˙ ˙ = − ~xi · ∇ ρ + ~p · ∇ ρ ∂t ~xi i ~pi i=1 N X ~pi = − · ∇ ρ + F~ tot,i · ∇ ρ m ~xi ~pi i=1 ≡ −h1,N ρ, 148 Non-Equilibrium Statistical Mechanics where N N X ~pi ~ 1 X ~ hM,N = · ∇~xi + F ext,i · ∇~p + F ji · ∇~p − ∇~p , m i 2 i j i=M j=M where all forces depend only on position. We can split h1,N into two pieces: s N X X h = h + h + F~ · ∇ − ∇ . 1,N 1,s s+1,N ji ~pi ~pj i=1 j=s+1 Now we can write " N # Y dz h ρ = 0. ˆ i M,N i=M Thus, " N # ∂fs N! Y = − dz h ρ. ∂t (N − s)! ˆ i 1,N i=s+1 Plugging everything in, " N # s N ∂fs N! Y X X + h1,s fs = − dzi F~ ji · ∇ − ∇ ρ ∂t (N − s)! ˆ ~pi ~pj i=s+1 j=1 k=s+1 s X 3 3 = − d xs+1d ps+1F~ s+1,i · ∇ − ∇ fs+1. ˆ ~pi ~ps+1 i=1 This gives us the BBGKY equations, so named after the initials of those whom de- veloped it, for the evolution of the distribution functions. Notice that the left side is a partial derivative of fs and the right side is an integral over " N # N! Y f = dz ρ. s+1 (N − s − 1)! ˆ i i=s+2 Consider the time evolution equation for the 1-particle distribution function ∂ ~p1 3 3 + · ∇ + F~ ext · ∇ f1(~x, ~p, t = − d x2d p2 F~ 2,1 ∇ − ∇ f2(z1, z2). ∂t m ~x ~p ˆ ~p1 ~p2 Note that the quantity in brackets on the left is really just the total time derivative d/dt. The right side is often called the collision term and denoted ∂f 1 = − d3x d3p F~ ∇ − ∇ f (z , z ). 2 2 2,1 ~p1 ~p2 2 1 2 ∂t coll ˆ To get the solution, we need to see if we can approximate f2(z1, z2) in closed form. For that, we have ∂ ~p ~p + 1 · ∇ + F · ∇ + 2 · ∇ + F · ∇ f + F~ · (∇ − ∇ ) f = ∂t m x1 ext,1 p1 m x2 ext,2 p2 2 2,1 p1 p2 2 2 X = − d3x d3p F · ∇ f . ˆ 3 3 3,i pi 3 i=1 Relevant time scales are ~p · ∇ ∼ τ −1 ← streaming time scale m x s ~ −1 F ext · ∇p ∼ τext ← external time scale ~ −1 F ij · ∇p ∼ τc ← collision time scale. 11.1. BBGKY 149 −1 If in the low density limit, we expect τs to be large, so τs is small. We will assume that τc << τs << τext. Then F~ ext · ∇p ∼ 0. In the low density limit, the right-hand side of the time evolution equation for f2 is approximately zero. Using center-of-mass and relative coordinates, we write ~p − ~p ~r + ~r P~ = ~p + ~p , ~p = 2 1 , R~ = 1 2 , ~r = ~r − ~r . 1 2 2 2 2 1 Then we can write the time evolution equation for f2 as " # P~ ~p · ∇ + · ∇ + F~ · ∇ f (R~ , ~r, P~ , ~p) = 0. 2m R m r 2,1 p 2 After a time δt, ~p f (~r, ~p) = f ~r + δt, ~p + F~ δt . 2 m 2,1 We could also write this as ~p f ~r − δt, ~p = f ~r , ~p + F~ δt . 2 m 2 1 2,1 We can eventually show that ∂f1 1 3 3 ~ ~ ' d p2 d r (~p1 − ~p2) · ∇rf2 R, ~r, P , ~p . ∂t coll m ˆ ˆ In cylindrical coordinates, with the z-axis aligned with the relative momentum of the particles, ∂f1 1 3 ∂f2 = d p2 |~p1 − ~p2| dφ db dz , ∂t coll m ˆ ˆ ˆ ˆ ∂z where b is the coordinate distance from the axis of rotation. If the interaction scale is r0, then r0 ∂f2 ∂f2 0 dz = dz = f2 R~ , ~r , P~ , ~p − f2 R~ , ~r, P~ , ~p ˆ ∂z ˆ−r0 ∂z 0 = f2 R~ , ~r, P~ , ~p − f2 R~ , ~r, P~ , ~p . Outside of the interaction region, i.e. for r > r0, the particles are uncorrelated, so f2(~x1, ~p1, ~x2, ~p2) = f1(~x1, ~p1)f1(~x2, ~p2). Finally, ∂f1 1 3 h 0 0 i ' d p2 |~p1 − ~p2| dφ db f1(~x1, ~p1 )f1(~x1, ~p2 ) − f1(~x1, ~p1)f1(~x1, ~p2 ∂t coll m ˆ ˆ ˆ 1 dσ h i = d3p m dΩ |~v − ~v | f (~x , ~p 0)f (~x , ~p 0) − f (~x , ~p )f (~x , ~p . m ˆ 2 ˆ 2Ω 1 2 1 1 1 1 1 2 1 1 1 1 1 2 This is the Boltzmann equation, and it can be written in the form ∂ ~p + 1 · ∇ + F~ · ∇ f = d3p d3p0 d3p0 ∂t m x1 ext p1 1 ˆ 2 ˆ 1 ˆ 2 0 0 × r 0 0 δ(E + E − E − E ) p1p2→p1p2 1 2 1 2 (3) 0 0 h 0 0 i × δ (~p1 + ~p2 − ~p1 − ~p2 ) f1(p1)f1(p2) − f1(p1)f1(p2) . 150 Non-Equilibrium Statistical Mechanics The r 0 0 is the reaction rate in which two particles entering the scattering region p1p2→p1p2 0 0 with momenta p1 and p2 leave with momenta p1 and p2. The delta functions enforce energy and momentum conservation in the interaction. What if the distribution functions in the collision term are of the Maxwell-Boltzmann form −βE(p) f1(p) ∝ e ? Then −β(E1+E2) f1(p1)f1(p2) ∝ e . But energy conservation E1 + E2 = E3 + E4, implies then that 0 0 f1(p1)f1(p2) − f1(p1)f1(p2) = 0, then the whole collision term is zero. 11.2 Hydrodynamics We start by looking at the moments of the Boltzmann equation. These are the mass density, momentum density, and energy density, and they are given by (in order) ρ(~x) = d3p m f (~x, ~p, t) ˆ 1 ρ(~x)~u = d3p ~p f (~x, ~p, t) ˆ 1 p2 ρ(~x)E = d3p f (~x, ~p, t). ˆ 2m 1 To find their evolution, we integrate over the Boltzmann equation. For the collision term, 3 ∂f 3 3 3 3 (4) d p χ(p1) = d p1d p2d p3d p4 rp1p2→p3p4 δ (P1 + P2 − P3 − P4) ˆ ∂t coll ˆ h i × f1(p3)f1(p4) − f1(p1)f1(p2) χ(p1), 2 where χ(p1) can be either of m, ~p, or p /2m. We can relabel indices, which implies that 1h i χ(p ) = χ(p ) + χ(p ) − χ(p ) − χ(p ) , 1 4 1 2 3 4 so we can write 3 ∂f 1 3 3 3 3 (4) d p1 χ(p1) = d p1d p2d p3d p4 rp1p2→p3p4 δ (P1 + P2 − P3 − P4) ˆ ∂t coll 4 ˆ h ih i × f1(p3)f1(p4) − f1(p1)f1(p2) χ(p1) + χ(p2) − χ(p3) − χ(p4) . Notice that if χ is a quantity which is conserved in the scattering process, then the above is zero. 3 ∂f d p1 χ(p1) = 0. ˆ ∂t coll For the whole equation then 3 3 ∂f d p1 χ∂t + χ~v · ∇x + χF~ ext · ∇p f1(p1) = d p1 χ = 0, ˆ ˆ ∂t coll where ~v = ~p/m. If we define a new average 1 hQi ≡ Q(p)f d3p, n ˆ 1 11.2. Hydrodynamics 151 where n = d3p f , ˆ 1 then we can write the above result as ∂ (n hχi) + ∇ (n h~vχi) = nF~ · h∇ χi . ∂t x ext p For the mass density (i.e. χ = m), this implies ∂ρ + ∇ · (ρ~u) = 0, ∂t where ~u is the average velocity. Notice that this is a continuity equation. This is the statement of mass conservation. For the momentum density (i.e. χ = ~p), this implies ∂ (ρ~u) + ∇ · (ρ h~v~vi) = ρF~ . ∂t ext In component form, we would write this as ∂ ∂ (ρui) + (ρ hvivji) = ρFext,i. ∂t ∂xj We can also define ~v = ~u + ~w, such that h~vi = ~u and h ~wi = 0. Then hvivji = uiuj + hwiwji . We can then define ρ hwiwji ≡ P δij + πij, where ρ P = h ~w 2i , 3 and πij are the components of the viscous stress tensor. Then ∂ ∂ ∂ ∂ (ρui) + (ρuiuj) = − P + Fext,i + πij. ∂t ∂xi ∂xi ∂xj For the energy, we perform a similar sequeence of steps using the internal energy density per mass 1 ε = h ~w 2i . 2 We find that ∂ (ρε) + ∇ · (ρε~u) = −p∇ · ~u + Ψ − ∇ · F , ∂t k where 1 F = ρ w w2 , k 2 k is the conduction heat flux (which turns out to be zero), and ∂ui Ψ = πij , ∂xj 152 Non-Equilibrium Statistical Mechanics is the rate of viscous dissipation. In component form, ∂ ∂ ∂uk ∂Fk (ρε) + (ρεui) = −p − + Ψ. ∂t ∂xi ∂xk ∂xk We now have five evolution equations (mass, three components of momentum, and internal energy) in 13 independent variables. Thus, this is not a closed set. To lowest order, we can assume we have enough collisions to be in local thermody- namic equilibrium. This implies that our distribution function is a Maxwell-Boltzmann distribution. I.e., we assume that 3/2 2 1 −β (~p−m~u) f = f (~x, ~p, t) = n e 2m , MB 2πmT and ~w = ~p−~um. Now, the stress tensor and heat flux are both zero (πij = 0 and F~ = 0). Then we are left with Euler’s equations of hydrodynamics: ∂ρ + ∇ · (ρ~u) = 0 ∂t ∂~u F~ 1 + (~u · ∇) ~u = ext − ∇P ∂t m ρ ∂ε P + (~u · ∇) ε = − ∇ · ~u. ∂t ρ This is a closed system of equations assuming that P = P (ρ, ε). These are the lowest order equations of hydrodynamics. The next order approximation would be the Navier-Stokes equations which include heat diffusion and viscosity. Note that ~u is the velocity flow vector, and that in general, ~u = ~u(~x, t) and ρ = ρ(~x, t). We also find that ρ ρ 3T P = h ~w 2i = = nT, 3 3 m so P is indeed the thermal pressure. 11.3. Summary: Non-Equilibrium Statistical Mechanics 153 11.3 Summary: Non-Equilibrium Statistical Mechanics Skills to Master • Chapter 12 Review 12.1 Entropy We started with the Boltzmann entropy S = k ln Ω(E,N,V ), where Ω is the number of microstates with given E, N, and V . If there are Ni in the state i, then N! Ω = Q , i Ni! with the condition that X N = Ni. i Later, we derived the more useful Gibbs entropy X S = − pi ln pi, i where the sum is over the microstates of the ensemble, and pi is the probability of picking state i from the ensemble. In the thermodynamic limit, these two definitions of the entropy are the same. 12.2 Ensembles Next, we consider whole ensembles of systems, an ensemble being a set of systems sharing some property such as fixed energy. In general, to find the probability pi, we maximize the Gibbs entropy. In the microcanonical ensemble, we have fixed E, N, and V . We can maximize the Gibbs entropy by using the method of Lagrange multipliers. X L = λ − (pi ln pi + λpi) , i which implies ! X λ pi − 1 = 0, i which gives us pi = const. 12.3. Classical Limit 155 In the canonical ensemble, we have fixed N and V and we want some average energy hEi. Now the Lagrange multiplier approach gives us X L = λ + β hEi − (pi ln pi + λpi + βEipi) , i which implies that −βEi pi ∝ e . Normalizing gives us e−βEi pi = , ZN where X −βEi ZN = e , i is the canonical partition function. The ensemble average of a quantity Q is calculated as P −βEi X i Qie hQi = piQi = . P e−βEi i i In the grand canonical ensemble, we have fixed V but average values of energy and particle number hEi and hNi. The Lagrange multiplier approach gives us X L = λ + β hEi + α hNi − (pi ln pi + (λ + βEi + αNi)pi) , i which implies −β(Ei−µNi) pi ∝ e , −µβ = α. The normalization factor is now X Z = e−β(Ei−µNi). i Useful properties of the partition functions include ∂ 1 X ln Z = (−E )e−βEi = − hEi ∂β N Z i N i ∂2 ln Z = hE2i − hEi2 = h(E − hEi)2i = h∆E2i ∂β2 N 2 ∂ 2 2 ln Z = h∆N i . ∂α β Note that ∆E is the energy fluctuation about the mean. 12.3 Classical Limit The above is all for discrete states. What if we have continuous states as we do in the classical limit? Now the ensemble average can be calculated as hQN 3 3 i i d xid pi Q({~xi, ~pi})ρ hQi = ´ h i , QN 3 3 i d xid pi ρ ´ where ρ ∝ e−βH({~xi,~pi}), 156 Review is the phase space density. For classical, continuous phase space, the canonical partition function is " N D D # 1 Y d xid pi Z = e−βH, N N! ˆ (2π )D i=1 ~ where the factor of 1/N! out front is there only if the particles are indistinguishable, and ~ is found by comparison with the quantum case. Assuming there is no momentum dependence in the potential, we can write this as D N " N # N 1 d p 2 Y β X Z = e−p /2m dDx exp − u(~r − ~r ) . N N! ˆ (2π )D ˆ i 2 i j ~ i=1 i,j The first integral is just a Gaussian, so we get " N # N 1 1 Y D β X ZN = d xi exp − u(~ri − ~rj) , N! `ND ˆ 2 Q i=1 i,j where r 2π 2 ` = ~ , Q mT is the de Broglie wavelength. For a non-interacting gas, the second integral just gives the volume LND = V N . For an ideal gas, 1 L ND ZN,ideal = . N! `Q We can also have internal states for the N components of a system. Then the internal partition function is " #N V N X dDp Z = e−β(ε(p)+εα) , N,internal N! ˆ (2π )D α ~ where the sum is over the internal states. Equipartition Theorem The generalized equipartition theorem is ∂H dω ξi ρ ∂H ∂ξj ξi = ´ = T δij, ∂ξj dω ρ ´ where ξi and ξj are any pair of phase space coordinates. This generalized equipartition theorem is only true for classical systems. Chemical Equilibrium If we have multiple species, we can have reactions like A + B ↔ C + D. Then the equilibrium condition is that µA + µB = µC + µD. If 2A + B ↔ C + D, then we would have the condition 2µA + µB = µC + µD. 12.4. Thermodynamic Relations 157 12.4 Thermodynamic Relations Next, we try to connect partition functions to thermodynamic quantities. If we have two systems in thermal contact, and they are able to exchange energy, and E1 + E2 = Et, then Ω = Ω1(E1)Ω2(E2). Maximizing to identify the most probable state, we can derive the relation ∂ ln Ω ∂ ln Ω 1 1 = 2 = . ∂E1 ∂E2 T Then the fundamental thermodynamic relation 1 P µ dS = dE + dV − dN, T T T implies ∂S 1 ∂S P = , = . ∂E V,N T ∂V E,N T That’s for entropy. We can do the same for other quantities. For an extensive system λS(N,V,E) = S(λN, λV, λE). For example, if you double N, double V , and double E for a system, then its entropy will be doubled. Extensity of a system implies that E = ST − PV + µN. Plugging the canonical probability into the Gibbs entropy gives us X X 1 X S = − p ln p = − p e−βEi = p (ln Z + βE ) = ln Z + β hEi . i i i Z i N i N i N This implies that −T ln ZN = E − TS = F. Since E = E(S, N, V ), we have that dE = T dS − P dV − µ dN. which implies dF = −S dT − P dV + µ dN. This implies quantities like ∂F = µ. ∂N V,T Following the same procedure with the grand canonical partition function, we find the grand potential Ω = E − TS − µN = −T ln Z = −PV. We can also derive other potentials like the Gibbs free energy G(T,P,N) and the enthalpy H(S, N, P ). We can also make up new potentials. Using derivative identities, we can derive Maxwell relations like ∂2F ∂S ∂µ = − = . ∂T ∂N ∂N ∂T 158 Review We can also calculate response functions like the heat capacity δQ ∂E ∂S ∂2F CV = = = T = −T 2 . dT ∂T V,N ∂T V,N ∂T V,N We can relate the response functions to fluctuations of our system. For example, via the connection between F and ZN , we find that 1 ∂2 h∆E2i C = ln Z = . V T 2 ∂β2 N T 2 The probability of a fluctuation away from equilibrium (pf ) versus the most probable value is p Ω 1 ∂S ∂P f = f = e∆S = exp − ∆T 2 − ∆V 2 . peq Ωeq 2T ∂T ∂V Thus, we have two independent Gaussians. If we treat our system as a large reservoir r and a subsystem a, then the equilibrium state satisfies dStot = dSa + dSr = 0. T,V,N But by energy conservation, dE 1 dF dS + dS = dS − a = d(TS − E ) = − a . a r a T T a a T Thus, dFa dStot = − = 0, T,V,N T implying that we need to minimize our free energy. 12.5 Quantum Statistics Here, we only consider noninteracting quantum gases. For quantum gases, the spin nature of the particles becomes important. For quantum ensembles, we have the density matrix N 1 X ρˆ = |θi hθ| , N i where N is the number of ensemble members. For ρ to be an equilibrium ensemble, i.e. time-independent, then X ρˆ = ωj |Eji hEj| , j where |Eji are the energy eigenstates. To get a proper normalization, the partition function is h i Z = Tr e−β(Hˆ −µNˆ) . If we have n1 particles in state 1, n2 particles in state 2, and so on, then X En1n2··· = niεi, i 12.5. Quantum Statistics 159 where εi is the energy of the single-particle state i. The total number of particles is X Nn1n2··· = ni. i We can write the partition function as h i P −β(Hˆ −µNˆ) X X − niβ(εi−µ) Y Z = Tr e = ··· e = zi, n1 n2 i where X −niβ(εi−µ) zi = e . ni For fermions, −β(εi−µ) zi,F = 1 + e . For bosons, −1 −β(εi−µ) zi,B = 1 − e . The log of the partition function becomes X ∞ ln Z = ∓ ln 1 ∓ e−β(εj −µ) ' ∓ dε g(ε) ln 1 ∓ e−β(εj −µ) , ˆ j 0 where the sum over α is a sum over any internal states (i.e. it is a degeneracy factor). More specifically, it is a sum over any quantum numbers that do not depend on the momentum. The density of states is D D −1 L X L X ∂ε g(ε) = dDp δ(ε − ε (~p)) = pD−1 dΩ , 2π ˆ α 2π ∂p ˆ D ~ α ~ α where dΩD is the integral over the angles. For example, dΩD = 4π in 3D. It’s important´ to know how to calculate g(ε) for free (i.e. non-interacting)´ gases in various dimensions and with various dispersion relations. Useful thermodyanmic quantities include ∞ N = dε g(ε) f∓(ε) ˆ0 Ω P = − V ∞ E = dε g(ε) ε f∓(ε), ˆ0 where 1 f (ε) = . ∓ eβ(ε−µ) ∓ 1 Bosons For bosons, we are limited to µ < 0. Bosons exhibit the interesting property of Bose- Einstein condensation. For some dimensions, there’s some maximum number ∞ g(ε) Nmax = βε dε. ˆ0 e − 1 This gives us our first example of a phase transition. Phonons/Debye Model 160 Review Fermions An interesting property of fermions is that they are limited to one particle per state. This results in a degeneracy pressure even at zero temperature. The Fermi energy is calculated via the relation εF N = dε g(ε). ˆ0 In the low temperature limit, some fermionic quantities can be expanded using Som- merfeld expansion. In this expansion, integrals of the form ∞ µ 2 2 π T 0 4 I = dε φ(ε) f+(ε) ' dε φ(ε) + φ (µ) + O(T ), ˆ0 ˆ0 6 are expressed as a temperature independent piece plus a temperature dependent pertur- bation. For example, d ln g π2T 2 µ ' εF − dε ε=εF 6 εF π2T 2 E ' g(ε)dε + g(εF ). ˆ0 6 12.6 Interacting Gases Virial expansion 12.7 Phase Transitions 12.8 Landau-Ginzburg Field Theory Index Black-body radiation, Degeneracy, 88 Kadanoff scaling hy- 80 Density matrix, 52 pothesis, 140 Boltzmann constant,6 Digamma function,4 Kinetic theory, 65 Boltzmann democracy Discrete symmetry hypothesis, 24, breaking, 133 Lagrange multipliers, 25 27 Domain walls, 133 Landau free energy, 21 Boltzmann entropy,1 Domains, 133 Landau-Ginzburg Field Boltzmann equation, Theory, 127 149 Ehrenfest classification Landau-Ginzburg Boltzmann gas, 60 scheme, 109 Hamiltonian, Boltzmann gases, 64 Einstein model, 82 128 Boltzmann H-theorem, Ensemble, 12 Latent heat, 110 25 Enthalpy, 36 Legendre transform, 35 Bose gas, 60 Entropy,1 Liouville’s theorem, 13, Bose-Einstein condensa- Equipartition theorem, 14 tion, 75, 76 67 Liquid phase, 108 Boson, 55, 56 Extensive parameters, Locality, 127 Boson gas, 75 7, 10 Lower critical dimen- sion, 136 Canonical ensemble, 19, Fermi energy, 89 Magnetic susceptibility, 54 Fermi gas, 60, 87 90 Chemical equilibrium, Fermi-Dirac function, 87 Magnetization, 90 71 Fermion, 55, 56 Master equation, 14 Chemical potential,6, Fiducial element, 16 Maxwell construction, 60 Fluctuations, 42, 131 113 Classical gases, 64 Fugacity, 30 Maxwell relations, 39 Clausius-Clapeyron re- Gamma function,4 Maxwell-Boltzmann dis- lation, 111 Gap exponent, 140 tribution, 65 Compressibility, 46 Gas phase, 108 Mean field approxima- Connected correlation Gibb’s paradox, 19 tion, 115 function, 119 Gibbs free energy, 108 Mesoscopic, 127 Continuous symmetry Gibbs-Duhem relation, Mexican hat potential, breaking, 133, 7 135 135 Ginzburg criterion, 137, Meyer cluster expan- Correlation function, 139 sion, 101 119 Goldstone modes, 136 Microcanonical ensem- Correlation functions, Grand canonical ensem- ble, 15, 54 96 ble, 28, 54, 56 Correlation length, 120, Grand potential, 29 Order parameter, 109, 133 119 Correlations, 131 Heat capacity, 46 Ornstein-Zernike, 133 Critical dimension, 136 Helmholtz free energy, Critical exponents, 119 21 Partition function, 20, Critical opalescence, 133 Hydrodynamics, 150 22 Critical temperature, 77 Hyperscaling relation, Path integral, 137 Curie susceptibility, 90 141 Pauli spin susceptibility, 89 Debye function, 83 Intensive parameters,7, Pauli susceptibility, 90 Debye model, 82 10 Phase space density, 12 Debye temperature, 83 Interacting gases, 96 Phase transition, 129 Debye wavelength, 82 Ising model, 114 Phase Transitions, 108 162 INDEX Phase transitions, 109 Sommerfeld expansion, Thermodynanic iden- Phonons, 81 91 tity, 35 Photons, 80 Spin excess,2 Trace, 52 Poisson bracket, 12 Spinodal instability, 112 Transfer matrix, 118 Polylogarithm, 87 Spontaneous symmetry Pressure,6 breaking, 133 Uniformity, 127 Pure state, 53 Stirling approximation, Universality class, 120, 3 141 Rayleigh-Jeans law, 80 Susceptibility, 89, 90 Upper critical dimen- Reservoir, 19 Symmetry breaking, 133 sion, 139 Response function, 46 Thermal de Broglie Van der Waals Equa- Sackur-Tetrode equa- wavelength, 23 tion, 101 tion, 18 Thermal expansion, 46 van der Waals gas, 112 Saddle-point approxi- Thermodynamic poten- Virial Expansion, 99 mation, 128 tials, 35, 47 Virial theorem, 67 Scaling hypothesis, 139 Thermodynamic square, Solid phase, 108 38 Wien’s formula, 80