Complete Statistical Mechanics

The Statistical Basis of Thermodynamics

Core ideas

Statistical mechanics explains thermodynamics from microscopic states. Entropy counts compatible microstates, equilibrium is overwhelmingly probable, and macroscopic laws emerge from averages over many degrees of freedom. Temperature and chemical potential are derivatives of entropy.

For review, be able to derive thermodynamic variables from entropy, distinguish microstate and macrostate, and explain why equilibrium is statistical. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$S=k_B\ln\Omega,\qquad \frac1T=\left(\frac{\partial S}{\partial E}\right)_{V,N},\qquad \frac{\mu}{T}=-\left(\frac{\partial S}{\partial N}\right)_{E,V}$$

Worked example

Consider a paramagnet of $N=4$ distinguishable spins-$1/2$ in a magnetic field $B=1\,\mathrm{T}$. Each spin has magnetic moment $\mu$ (Bohr magneton, $\mu\approx9.27\times10^{-24}\,\mathrm{J/T}$). The up state has energy $-\mu B$ and the down state $+\mu B$. Take the macrostate with total energy $E=-2\mu B$, which requires three spins up and one down.

1. Counting microstates: The number of ways to choose which spin is down is

   $$\Omega=\binom{4}{1}=4.$$

2. Entropy: Using $S=k_B\ln\Omega$,

   $$S=k_B\ln 4\approx1.386\,k_B\approx1.91\times10^{-23}\,\mathrm{J/K}.$$

3. Temperature: If the energy is raised to $E=0$ (two up, two down), $\Omega'=\binom{4}{2}=6$ and $S'=k_B\ln 6$. For the small energy step $\Delta E=2\mu B\approx1.85\times10^{-23}\,\mathrm{J}$,

   $$\frac1T\approx\frac{\Delta S}{\Delta E}=\frac{k_B(\ln6-\ln4)}{2\mu B}=\frac{k_B}{2\mu B}\ln\frac32.$$

   Thus

   $$T\approx\frac{2\mu B}{k_B\ln(3/2)}\approx\frac{1.85\times10^{-23}}{(1.38\times10^{-23})(0.405)}\approx3.3\,\mathrm{K}.$$

Problems with Solutions

1. Microstate counting for an Einstein solid.
   Problem: An Einstein solid consists of $N=3$ oscillators, each with energy spacing $\hbar\omega$. How many microstates correspond to total energy $E=2\hbar\omega$?
   Solution. This is the number of ways to distribute $q=2$ indistinguishable energy quanta among $N=3$ oscillators:

   $$\Omega=\binom{q+N-1}{q}=\binom{4}{2}=6.$$

2. Entropy and temperature from counting.
   Problem: A system has $\Omega(E)=C\,E^{3N/2}$ with $N=2$ and $C$ a constant. Find $S(E)$ and $T(E)$.
   Solution. $S=k_B\ln(C E^3)=k_B\ln C+3k_B\ln E$. Then

   $$\frac1T=\left(\frac{\partial S}{\partial E}\right)_{V,N}=\frac{3k_B}{E}\quad\Rightarrow\quad E=3k_BT.$$

   This is the equipartition result for $N=2$ structureless particles in 3D.

3. Chemical potential estimate.
   Problem: For a monatomic ideal gas, $\Omega(N,E,V)\propto V^N E^{3N/2}/N!$. Use Stirling’s approximation to estimate $\mu/T$ for large $N$.
   Solution. With $\ln N!\approx N\ln N-N$,

   $$S\approx Nk_B\ln\!\left(\frac{V}{N}\right)+\frac{3N}{2}k_B\ln E+\text{const}.$$

   Keeping $E$ fixed,

   $$\frac{\mu}{T}=-\left(\frac{\partial S}{\partial N}\right)_{E,V}\approx -k_B\ln\!\left(\frac{V}{N}\lambda_T^{-3}\right),$$

   which is the familiar Sackur—Tetrode form.

Section summary. Thermodynamics is the large-number limit of microscopic statistics.

Elements of Ensemble Theory

Core ideas

An ensemble is a probability distribution over microstates representing given macroscopic information. Microcanonical, canonical, and grand canonical ensembles differ by what is held fixed and what can be exchanged with a reservoir.

For review, be able to choose the right ensemble, compute averages, relate fluctuations to response, and understand equivalence of ensembles in the thermodynamic limit. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\langle A\rangle=\sum_s p_sA_s,\qquad p_s\ge0,\qquad \sum_sp_s=1$$

Worked example

Consider a 3-state system with energies $E_1=0$, $E_2=\varepsilon$, and $E_3=2\varepsilon$, where $\varepsilon=0.1\,\mathrm{eV}$. In the canonical ensemble at $T=300\,\mathrm{K}$ ($k_BT\approx0.02585\,\mathrm{eV}$), the partition function is

$$Z=\sum_{s=1}^3 e^{-\beta E_s}=1+e^{-\beta\varepsilon}+e^{-2\beta\varepsilon}.$$

With $\beta\varepsilon\approx3.87$, we have $e^{-3.87}\approx0.0209$ and $e^{-7.74}\approx4.36\times10^{-4}$, giving

$$Z\approx1.0213.$$

The probabilities are $p_1\approx0.979$, $p_2\approx0.0205$, $p_3\approx4.3\times10^{-4}$. The average energy is

$$\langle E\rangle=\frac{0+\varepsilon e^{-\beta\varepsilon}+2\varepsilon e^{-2\beta\varepsilon}}{Z}\approx\frac{0.00209\,\mathrm{eV}+0.000087\,\mathrm{eV}}{1.0213}\approx2.1\times10^{-3}\,\mathrm{eV}\approx3.4\times10^{-22}\,\mathrm{J}.$$

Problems with Solutions

1. Two-level system energy and heat capacity.
   Problem: A system has states with energies $0$ and $\varepsilon$. Find $\langle E\rangle$ and the heat capacity $C=\partial\langle E\rangle/\partial T$.
   Solution. $Z=1+e^{-\beta\varepsilon}$. Then

   $$\langle E\rangle=\frac{\varepsilon e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}}=\frac{\varepsilon}{e^{\beta\varepsilon}+1}.$$

   Differentiating,

   $$C=\frac{\varepsilon^2}{k_BT^2}\frac{e^{\beta\varepsilon}}{(1+e^{\beta\varepsilon})^2}=k_B\left(\frac{\varepsilon}{2k_BT}\right)^2\operatorname{sech}^2\!\left(\frac{\varepsilon}{2k_BT}\right).$$

2. Gibbs entropy.
   Problem: For the 3-state example above, compute the Gibbs entropy $S=-k_B\sum_s p_s\ln p_s$.
   Solution. Using the probabilities found,

   $$S\approx-k_B\bigl[0.979\ln(0.979)+0.0205\ln(0.0205)+4.3\times10^{-4}\ln(4.3\times10^{-4})\bigr].$$

   Evaluating the logarithms gives $S\approx0.104\,k_B\approx1.4\times10^{-24}\,\mathrm{J/K}$.

3. Energy fluctuations and response.
   Problem: Show that $\langle(\Delta E)^2\rangle=-\partial_\beta\langle E\rangle$ and explain why this implies $\langle(\Delta E)^2\rangle=k_BT^2C_V$.
   Solution. Start from $\langle E\rangle=-\partial_\beta\ln Z$. Then

   $$-\partial_\beta\langle E\rangle=\partial_\beta^2\ln Z=\frac{\partial_\beta^2 Z}{Z}-\left(\frac{\partial_\beta Z}{Z}\right)^2=\langle E^2\rangle-\langle E\rangle^2.$$

   Since $\partial_\beta=-k_BT^2\partial_T$, we have $-\partial_\beta\langle E\rangle=k_BT^2(\partial\langle E\rangle/\partial T)_V=k_BT^2C_V$.

Section summary. Ensembles turn incomplete microscopic information into predictions.

The Canonical Ensemble

Core ideas

The canonical ensemble describes systems exchanging energy with a heat bath at fixed temperature. The partition function normalizes probabilities and generates energy, entropy, heat capacity, and free energy.

For review, be able to compute $Z$, derive $F$, $U$, $S$, and heat capacity, and use factorization for independent degrees of freedom. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$Z=\sum_s e^{-\beta E_s},\qquad F=-k_BT\ln Z,\qquad U=-\partial_\beta\ln Z$$

Worked example

A quantum harmonic oscillator of frequency $\omega=2\pi\times10^{13}\,\mathrm{s^{-1}}$ ($\hbar\omega\approx0.0414\,\mathrm{eV}$) is in thermal equilibrium at $T=300\,\mathrm{K}$ ($k_BT\approx0.02585\,\mathrm{eV}$). Its canonical partition function is

$$Z=\sum_{n=0}^\infty e^{-\beta\hbar\omega(n+1/2)}=\frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}.$$

With $\beta\hbar\omega\approx1.60$, $e^{-0.80}\approx0.449$ and $e^{-1.60}\approx0.202$, so

$$Z\approx\frac{0.449}{0.798}\approx0.563.$$

The average energy is

$$U=\frac{\hbar\omega}{2}+\frac{\hbar\omega}{e^{\beta\hbar\omega}-1}\approx0.0207\,\mathrm{eV}+\frac{0.0414}{3.95}\,\mathrm{eV}\approx0.0312\,\mathrm{eV}\approx5.0\times10^{-21}\,\mathrm{J}.$$

The heat capacity is

$$C=k_B(\beta\hbar\omega)^2\frac{e^{\beta\hbar\omega}}{(e^{\beta\hbar\omega}-1)^2}\approx(1.38\times10^{-23})(2.56)\frac{4.96}{15.6}\approx1.1\times10^{-23}\,\mathrm{J/K}.$$

Problems with Solutions

1. Classical limit of oscillator heat capacity.
   Problem: Show that for $k_BT\gg\hbar\omega$, the quantum oscillator heat capacity reduces to the classical equipartition value $k_B$.
   Solution. Let $x=\beta\hbar\omega\ll1$. Then $e^x-1\approx x+x^2/2$, so

   $$C\approx k_Bx^2\frac{1+x}{x^2(1+x/2)^2}\approx k_B.$$

2. $N$ independent spins.
   Problem: $N$ non-interacting spin-$1/2$ particles in a field $B$ have energies $\pm\mu B$. Find the total partition function $Z_N$ and the free energy $F$.
   Solution. The single-spin partition function is $Z_1=e^{\beta\mu B}+e^{-\beta\mu B}=2\cosh(\beta\mu B)$. Since the spins are independent,

   $$Z_N=[2\cosh(\beta\mu B)]^N,\qquad F=-Nk_BT\ln[2\cosh(\beta\mu B)].$$

3. Paramagnetic susceptibility.
   Problem: From the free energy of the $N$-spin system, derive the magnetization $M=-\partial F/\partial B$ and the zero-field susceptibility $\chi=\partial M/\partial B|_{B=0}$.
   Solution. $M=N\mu\tanh(\beta\mu B)$. For small $B$, $\tanh x\approx x$, so

   $$M\approx\frac{N\mu^2B}{k_BT},\qquad\chi=\frac{N\mu^2}{k_BT},$$

   which is Curie’s law.

Section summary. The canonical partition function is the engine of fixed-temperature statistical mechanics.

The Grand Canonical Ensemble

Core ideas

The grand canonical ensemble allows both energy and particles to fluctuate. It is natural for gases, quantum many-particle systems, adsorption, and reactions. The grand potential controls pressure and particle number.

For review, be able to compute $\mathcal Z$, obtain $\langle N\rangle$, relate $\Omega$ to pressure, and use fugacity. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\mathcal Z=\sum_{N,s}e^{-\beta(E_{Ns}-\mu N)},\qquad \Omega=-k_BT\ln\mathcal Z,\qquad \langle N\rangle=k_BT\,\partial_\mu\ln\mathcal Z$$

Worked example

Consider a single lattice site that can be either empty (energy $0$, particle number $0$) or occupied by one particle (energy $-\varepsilon$, particle number $1$). The site is in contact with a reservoir at $T=500\,\mathrm{K}$ and chemical potential $\mu=0.1\,\mathrm{eV}$, with binding energy $\varepsilon=0.3\,\mathrm{eV}$.

The grand partition function is

$$\mathcal Z=\sum_{N=0,1}e^{-\beta(E_N-\mu N)}=1+e^{\beta(\mu+\varepsilon)}.$$

Here $\beta(\mu+\varepsilon)=0.4\,\mathrm{eV}/(8.617\times10^{-5}\,\mathrm{eV/K}\times500\,\mathrm{K})\approx9.28$. Thus $e^{9.28}\approx1.07\times10^4$ and

$$\mathcal Z\approx1.07\times10^4,\qquad\langle N\rangle=\frac{e^{\beta(\mu+\varepsilon)}}{\mathcal Z}\approx0.9999.$$

The average energy is $\langle E\rangle=-\varepsilon\langle N\rangle\approx-0.3\,\mathrm{eV}$.

Problems with Solutions

1. Ideal gas in the grand canonical ensemble.
   Problem: For an ideal gas, show that $\mathcal Z=\exp(zV/\lambda_T^3)$ where $z=e^{\beta\mu}$ is the fugacity and $\lambda_T$ the thermal wavelength.
   Solution. The canonical partition function for $N$ indistinguishable particles is $Z_N=z_1^N/N!$ with $z_1=V/\lambda_T^3$. Summing over $N$,

   $$\mathcal Z=\sum_{N=0}^\infty\frac{(zz_1)^N}{N!}=e^{zz_1}=\exp\!\left(\frac{zV}{\lambda_T^3}\right).$$

2. Particle number and pressure.
   Problem: From the ideal-gas grand potential $\Omega=-k_BT\ln\mathcal Z=-k_BT\,zV/\lambda_T^3$, derive $\langle N\rangle$ and the equation of state.
   Solution. Using $\langle N\rangle=k_BT\,\partial_\mu\ln\mathcal Z=z\partial_z\ln\mathcal Z$,

   $$\langle N\rangle=\frac{zV}{\lambda_T^3}.$$

   Since $\Omega=-PV$, we have $PV=k_BT\,zV/\lambda_T^3=\langle N\rangle k_BT$, the ideal gas law.

3. Lattice-gas variance.
   Problem: For the single-site example above, calculate the variance $\langle(\Delta N)^2\rangle$ and show it satisfies the fluctuation-response relation.
   Solution. For a two-state system, $N^2=N$, so $\langle N^2\rangle=\langle N\rangle$. Thus

   $$\langle(\Delta N)^2\rangle=\langle N\rangle(1-\langle N\rangle).$$

   Using $\langle N\rangle=\mathcal Z^{-1}e^{\beta(\mu+\varepsilon)}$, we find $\partial\langle N\rangle/\partial\mu=\beta\langle N\rangle(1-\langle N\rangle)$. The fluctuation-response relation $\langle(\Delta N)^2\rangle=k_BT(\partial\langle N\rangle/\partial\mu)_{T,V}$ is therefore satisfied.

Section summary. Grand canonical methods handle particle exchange cleanly.

Formulation of Quantum Statistics

Core ideas

Quantum statistics accounts for indistinguishable particles. Bosons occupy symmetric states and can pile up; fermions occupy antisymmetric states and obey Pauli exclusion. Occupation-number language replaces classical labeling of particles.

For review, be able to derive Bose-Einstein and Fermi-Dirac distributions, use density of states, and identify classical limits. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\bar n_i=\frac{1}{e^{\beta(\epsilon_i-\mu)}\mp1},\qquad (-)\ \mathrm{bosons},\quad (+)\ \mathrm{fermions}$$

Worked example

Consider a photon mode of angular frequency $\omega=3.0\times10^{15}\,\mathrm{s^{-1}}$ ($\hbar\omega\approx1.98\,\mathrm{eV}$) in a blackbody cavity at $T=5000\,\mathrm{K}$ ($k_BT\approx0.431\,\mathrm{eV}$). Photons are bosons with $\mu=0$. The average occupation number is

$$\bar n=\frac{1}{e^{\beta\hbar\omega}-1}=\frac{1}{e^{1.98/0.431}-1}\approx\frac{1}{e^{4.59}-1}\approx\frac{1}{98.5-1}\approx0.0103.$$

For a fermionic state at the same energy and temperature, with $\mu=0.2\,\mathrm{eV}$,

$$\bar n_F=\frac{1}{e^{\beta(\hbar\omega-\mu)}+1}=\frac{1}{e^{1.78/0.431}+1}\approx\frac{1}{e^{4.13}+1}\approx\frac{1}{62.1+1}\approx0.0158.$$

Problems with Solutions

1. Classical limit.
   Problem: Show that when $\bar n\ll1$, both the Bose—Einstein and Fermi—Dirac distributions reduce to the Maxwell—Boltzmann form $\bar n\approx e^{-\beta(\epsilon-\mu)}$.
   Solution. If $e^{\beta(\epsilon-\mu)}\gg1$, the $\mp1$ in the denominator is negligible:

   $$\bar n_{BE}=\frac{1}{e^{\beta(\epsilon-\mu)}-1}\approx e^{-\beta(\epsilon-\mu)},\qquad \bar n_{FD}=\frac{1}{e^{\beta(\epsilon-\mu)}+1}\approx e^{-\beta(\epsilon-\mu)}.$$

2. Degenerate Fermi gas at $T=0$.
   Problem: At $T=0$, all states with $\epsilon<\mu$ are occupied and all with $\epsilon>\mu$ are empty. Show that the total number of electrons in volume $V$ with spin degeneracy $g=2$ is $N=gV(2m\mu)^{3/2}/(6\pi^2\hbar^3)$.
   Solution. Integrate the zero-temperature distribution:

   $$N=gV\int_0^\mu\frac{4\pi p^2\,dp}{(2\pi\hbar)^3}=\frac{gV}{2\pi^2\hbar^3}\int_0^{p_F}p^2\,dp=\frac{gVp_F^3}{6\pi^2\hbar^3},$$

   where $p_F=\sqrt{2m\mu}$. Substituting gives the result.

3. Bose—Einstein condensation condition.
   Problem: For a 3D gas of bosons with fixed $N$, write the condition that the excited-state population equals $N$ at $T=T_c$.
   Solution. Using the density of states $g(\epsilon)\propto\epsilon^{1/2}$,

   $$N=\int_0^\infty\frac{g(\epsilon)\,d\epsilon}{e^{\epsilon/k_BT_c}-1}=V\left(\frac{mk_BT_c}{2\pi\hbar^2}\right)^{3/2}\zeta(3/2),$$

   which defines the critical temperature $T_c$.

Section summary. Indistinguishability changes the counting of many-particle states.

The Theory of Simple Gases

Core ideas

Simple gases show how partition functions become equations of state. The ideal gas follows from translational states; corrections come from interactions, internal modes, and quantum degeneracy. Equipartition works only for quadratic classical modes.

For review, be able to derive ideal gas law, thermal wavelength, Sackur-Tetrode entropy, and limits of equipartition. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$PV=Nk_BT,\qquad \lambda_T=\frac{h}{\sqrt{2\pi mk_BT}},\qquad Z_N=\frac{1}{N!}\left(\frac{V}{\lambda_T^3}\right)^N$$

Worked example

Calculate the Sackur—Tetrode entropy of $1\,\mathrm{mol}$ of neon gas at standard temperature and pressure ($T=273.15\,\mathrm{K}$, $P=1\,\mathrm{atm}=1.013\times10^5\,\mathrm{Pa}$). The mass of a neon atom is $m\approx20.18\,\mathrm{u}=3.35\times10^{-26}\,\mathrm{kg}$.

First, the thermal de Broglie wavelength:

$$\lambda_T=\frac{h}{\sqrt{2\pi mk_BT}}=\frac{6.626\times10^{-34}}{\sqrt{2\pi(3.35\times10^{-26})(1.38\times10^{-23})(273.15)}}\approx2.35\times10^{-11}\,\mathrm{m}.$$

The molar volume is $V_m=RT/P\approx0.0224\,\mathrm{m^3}$. The entropy per particle is

$$\frac{S}{Nk_B}=\ln\!\left(\frac{V_m}{N\lambda_T^3}\right)+\frac52.$$

With $N\lambda_T^3\approx6.022\times10^{23}(2.35\times10^{-11})^3\approx7.8\times10^{-9}\,\mathrm{m^3}$,

$$\frac{V_m}{N\lambda_T^3}\approx\frac{0.0224}{7.8\times10^{-9}}\approx2.9\times10^6.$$

Thus

$$S\approx Nk_B[\ln(2.9\times10^6)+2.5]\approx Nk_B(14.9+2.5)=17.4\,R\approx145\,\mathrm{J/(mol\cdot K)},$$

close to the experimental value ($\approx146\,\mathrm{J/(mol\cdot K)}$).

Problems with Solutions

1. Pressure of an ideal gas.
   Problem: What is the pressure of $2\,\mathrm{mol}$ of $\mathrm{H_2}$ in a $10\,\mathrm{L}$ container at $T=300\,\mathrm{K}$?
   Solution. Using $PV=nRT$,

   $$P=\frac{(2\,\mathrm{mol})(8.314\,\mathrm{J/(mol\cdot K)})(300\,\mathrm{K})}{0.010\,\mathrm{m^3}}\approx4.99\times10^5\,\mathrm{Pa}\approx4.9\,\mathrm{atm}.$$

2. Thermal wavelength of an electron.
   Problem: Calculate the thermal de Broglie wavelength for a free electron at $T=300\,\mathrm{K}$.
   Solution. Using $m_e=9.11\times10^{-31}\,\mathrm{kg}$,

   $$\lambda_T=\frac{h}{\sqrt{2\pi m_ek_BT}}=\frac{6.626\times10^{-34}}{\sqrt{2\pi(9.11\times10^{-31})(1.38\times10^{-23})(300)}}\approx6.2\times10^{-9}\,\mathrm{m}=6.2\,\mathrm{nm}.$$

3. Quantum—classical crossover.
   Problem: The classical ideal-gas partition function $Z_N=(V/\lambda_T^3)^N/N!$ is valid when $n\lambda_T^3\ll1$. Evaluate this quantity for helium gas at $T=5\,\mathrm{K}$ and $P=10^5\,\mathrm{Pa}$.
   Solution. Number density $n=P/(k_BT)\approx10^5/(1.38\times10^{-23}\times5)\approx1.45\times10^{27}\,\mathrm{m^{-3}}$. For $^4\mathrm{He}$, $m\approx6.64\times10^{-27}\,\mathrm{kg}$:

   $$\lambda_T=\frac{h}{\sqrt{2\pi mk_BT}}\approx\frac{6.626\times10^{-34}}{\sqrt{2\pi(6.64\times10^{-27})(1.38\times10^{-23})(5)}}\approx2.8\times10^{-10}\,\mathrm{m}.$$

   Then $n\lambda_T^3\approx1.45\times10^{27}(2.8\times10^{-10})^3\approx3.2\times10^{-2}\ll1$. Even at $5\,\mathrm{K}$, helium is still classical at atmospheric pressure.

Section summary. Ideal gases are the baseline for statistical mechanics.

Ideal Bose Systems

Core ideas

Ideal bosons can macroscopically occupy the ground state, producing Bose-Einstein condensation. Photons and phonons are bosonic gases with variable particle number and zero chemical potential in equilibrium.

For review, be able to compute critical temperature qualitatively, distinguish condensate and thermal cloud, and use Planck distribution. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$N_0/N=1-\left(T/T_c\right)^{3/2},\qquad u(\omega)d\omega=\frac{\hbar\omega\,g(\omega)d\omega}{e^{\beta\hbar\omega}-1}$$

Worked example

Bose—Einstein condensation of $^{87}\mathrm{Rb}$ atoms. Consider $N=10^4$ atoms in a 3D harmonic trap with geometric mean frequency $\bar\omega=2\pi\times10^4\,\mathrm{s^{-1}}$. The atomic mass is $m=1.44\times10^{-25}\,\mathrm{kg}$.

For a trapped gas, the critical temperature is

$$k_BT_c=\hbar\bar\omega\left(\frac{N}{\zeta(3)}\right)^{1/3},\qquad\zeta(3)\approx1.202.$$

With $\hbar\bar\omega\approx6.63\times10^{-30}\,\mathrm{J}\approx k_B\times0.48\,\mu\mathrm{K}$, and $(10^4/1.202)^{1/3}\approx20.3$, we obtain

$$T_c\approx0.48\,\mu\mathrm{K}\times20.3\approx9.7\,\mu\mathrm{K}\approx10\,\mu\mathrm{K}.$$

At $T=0.5T_c$, the condensate fraction is

$$\frac{N_0}{N}=1-\left(\frac{T}{T_c}\right)^3=1-(0.5)^3=0.875.$$

So about 87.5% of the atoms are in the condensate.

Problems with Solutions

1. Critical temperature in a box.
   Problem: Show that for a uniform 3D gas of bosons, $T_c=\frac{2\pi\hbar^2}{mk_B}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}$.
   Solution. The number of particles in excited states is $N_{ex}=V\lambda_T^{-3}g_{3/2}(z)$. At $T_c$, $z\to1$ and $N_{ex}=N$. Using $g_{3/2}(1)=\zeta(3/2)$ and $\lambda_T=h/\sqrt{2\pi mk_BT}$, solving for $T_c$ gives the result.

2. Photon gas energy density.
   Problem: Use the Planck distribution to write the energy density $u(\omega)$ of blackbody radiation at temperature $T$.
   Solution. With two polarization states, the density of photon states is $g(\omega)\,d\omega=V\omega^2/(\pi^2c^3)\,d\omega$. Thus

   $$u(\omega)=\frac{\hbar\omega^3}{\pi^2c^3}\frac{1}{e^{\beta\hbar\omega}-1}.$$

3. Condensate fraction in a trap.
   Problem: For the trapped gas example, what fraction of atoms remains in the thermal cloud at $T=0.8T_c$?
   Solution. For a harmonic trap, $N_{ex}/N=(T/T_c)^3$. At $T/T_c=0.8$,

   $$\frac{N_{ex}}{N}=(0.8)^3=0.512.$$

   Thus 51.2% are still thermally distributed and 48.8% are in the condensate.

Section summary. Bosons reveal collective occupation of single quantum states.

Ideal Fermi Systems

Core ideas

Fermions fill states up to the Fermi energy at low temperature. Degeneracy pressure, heat capacity linear in $T$, and Fermi surfaces explain electrons in metals, white dwarfs, and many quantum fluids.

For review, be able to compute Fermi energy, use density of states near the Fermi surface, and explain Pauli pressure. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$E_F=\frac{\hbar^2}{2m}(3\pi^2n)^{2/3},\qquad f(\epsilon)=\frac{1}{e^{\beta(\epsilon-\mu)}+1}$$

Worked example

Conduction electrons in copper can be modeled as a free-electron gas with density $n=8.5\times10^{28}\,\mathrm{m^{-3}}$. The electron mass is $m_e=9.11\times10^{-31}\,\mathrm{kg}$.

1. Fermi wavevector:

   $$k_F=(3\pi^2 n)^{1/3}=(3\pi^2\times8.5\times10^{28})^{1/3}\approx(2.52\times10^{30})^{1/3}\approx1.36\times10^{10}\,\mathrm{m^{-1}}.$$

2. Fermi energy:

   $$E_F=\frac{\hbar^2k_F^2}{2m_e}=\frac{(1.055\times10^{-34})^2(1.36\times10^{10})^2}{2(9.11\times10^{-31})}\approx\frac{2.06\times10^{-48}}{1.822\times10^{-30}}\approx1.13\times10^{-18}\,\mathrm{J}\approx7.1\,\mathrm{eV}.$$

3. Fermi temperature:

   $$T_F=\frac{E_F}{k_B}\approx\frac{1.13\times10^{-18}}{1.38\times10^{-23}}\approx8.2\times10^4\,\mathrm{K}.$$

4. Degeneracy pressure at $T=0$:

   $$P=\frac{2}{5}nE_F=0.4(8.5\times10^{28})(1.13\times10^{-18})\approx3.8\times10^{10}\,\mathrm{Pa}\approx3.8\times10^5\,\mathrm{atm}.$$

Problems with Solutions

1. Fermi surface of silver.
   Problem: Silver has one conduction electron per atom and number density $n=5.86\times10^{28}\,\mathrm{m^{-3}}$. Calculate $k_F$ and $E_F$.
   Solution.

   $$k_F=(3\pi^2\times5.86\times10^{28})^{1/3}\approx(1.74\times10^{30})^{1/3}\approx1.20\times10^{10}\,\mathrm{m^{-1}}.$$ $$E_F=\frac{\hbar^2k_F^2}{2m_e}\approx\frac{(1.055\times10^{-34})^2(1.20\times10^{10})^2}{1.822\times10^{-30}}\approx8.3\times10^{-19}\,\mathrm{J}\approx5.2\,\mathrm{eV}.$$

2. Low-temperature heat capacity.
   Problem: Show that for a degenerate 3D Fermi gas, the electronic heat capacity is $C_V=\frac{\pi^2}{2}Nk_B\frac{T}{T_F}$ for $T\ll T_F$.
   Solution. At low $T$, only electrons within $\sim k_BT$ of $E_F$ are thermally excited. The Sommerfeld expansion gives

   $$U\approx U_0+\frac{\pi^2}{4}Nk_B\frac{T^2}{T_F},$$

   so $C_V=\partial U/\partial T=\frac{\pi^2}{2}Nk_B(T/T_F)$.

3. White-dwarf estimate.
   Problem: A white dwarf has $N\sim10^{57}$ electrons and radius $R\sim10^7\,\mathrm{m}$. Estimate the electron degeneracy pressure.
   Solution. Number density $n\sim3N/(4\pi R^3)\sim2.4\times10^{35}\,\mathrm{m^{-3}}$. For non-relativistic electrons,

   $$E_F\sim\frac{\hbar^2}{2m_e}(3\pi^2n)^{2/3}\sim3\times10^{-14}\,\mathrm{J}\sim0.2\,\mathrm{MeV},$$

   so relativistic corrections are significant. The pressure is of order $10^{22}\,\mathrm{Pa}$, balancing gravitational collapse.

Section summary. Fermi systems are controlled by states near the Fermi surface.

Cluster Expansion and Pseudopotential Methods

Core ideas

Interactions correct ideal-gas behavior. Virial and cluster expansions organize dilute-gas corrections by density. Effective pseudopotentials replace complicated short-range interactions with low-energy scattering parameters.

For review, be able to interpret virial coefficients, know when dilute expansions work, and connect scattering length to effective interactions. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\frac{P}{k_BT}=n+B_2(T)n^2+B_3(T)n^3+\cdots,\qquad g=\frac{4\pi\hbar^2a_s}{m}$$

Worked example

The second virial coefficient for a gas of hard spheres of diameter $\sigma$ is derived from the Mayer $f$-function. For hard spheres, the pair potential is $u(r)=\infty$ for $r<\sigma$ and $0$ for $r>\sigma$. Then

$$B_2(T)=-\frac12\int d^3r\,f(r)=-\frac12\int_0^\sigma4\pi r^2(-1)\,dr=\frac{2\pi\sigma^3}{3}.$$

For argon-like atoms with effective hard-sphere diameter $\sigma\approx3.4\,\text{\AA}=3.4\times10^{-10}\,\mathrm{m}$,

$$B_2=\frac{2\pi}{3}(3.4\times10^{-10})^3\approx8.2\times10^{-29}\,\mathrm{m^3}.$$

At $T=300\,\mathrm{K}$ and number density $n=2.5\times10^{25}\,\mathrm{m^{-3}}$ (roughly 1 atm), the first correction to pressure is

$$\frac{\Delta P}{P_{\text{ideal}}}=B_2n\approx(8.2\times10^{-29})(2.5\times10^{25})\approx2.1\times10^{-3}.$$

So the hard-core correction is only about 0.2% at STP.

Problems with Solutions

1. Square-well potential.
   Problem: A potential has $u(r)=\infty$ for $r<\sigma$, $u(r)=-\varepsilon$ for $\sigma<r<\lambda\sigma$, and $u(r)=0$ for $r>\lambda\sigma$. Find $B_2(T)$.
   Solution.

   $$B_2(T)=\frac{2\pi\sigma^3}{3}\Bigl[1-(\lambda^3-1)(e^{\beta\varepsilon}-1)\Bigr].$$

   The first term is the hard-sphere repulsion; the second is the attractive contribution.

2. Pseudopotential strength.
   Problem: For $^4\mathrm{He}$ atoms with scattering length $a_s\approx7.5\,\mathrm{nm}$ and mass $m\approx6.64\times10^{-27}\,\mathrm{kg}$, compute the effective interaction strength $g=4\pi\hbar^2a_s/m$.
   Solution.

   $$g=\frac{4\pi(1.055\times10^{-34})^2(7.5\times10^{-9})}{6.64\times10^{-27}}\approx\frac{1.05\times10^{-75}}{6.64\times10^{-27}}\approx1.6\times10^{-49}\,\mathrm{J\cdot m^3}.$$

3. Validity of virial expansion.
   Problem: Using the hard-sphere $B_2$ above, estimate the density at which the second-order term $B_2n$ is 10% of the ideal pressure.
   Solution. Set $B_2n=0.1$. Then

   $$n_{10\%}=\frac{0.1}{B_2}=\frac{0.1}{8.2\times10^{-29}}\approx1.2\times10^{27}\,\mathrm{m^{-3}}.$$

   This corresponds to roughly $P\approx nk_BT\sim5\times10^7\,\mathrm{Pa}\sim500\,\mathrm{atm}$ for argon at room temperature.

Section summary. Interaction effects can be organized systematically at low density.

Phase Transitions and Critical Phenomena

Core ideas

Phase transitions occur when free energies become nonanalytic in the thermodynamic limit. First-order transitions have latent heat; continuous transitions have diverging correlation length and universal critical exponents.

For review, be able to use order parameters, Landau free energy, susceptibility, correlation length, and scaling ideas. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$F[m]=F_0+am^2+bm^4-hm,\qquad \xi\sim |T-T_c|^{-\nu},\qquad \chi\sim |T-T_c|^{-\gamma}$$

Worked example

Consider the Landau free energy for a scalar order parameter $m$ near a continuous transition:

$$F[m]=F_0+a(T-T_c)m^2+bm^4,$$

with $a=1\,\mathrm{J/(m^3\,K)}$, $b=0.5\,\mathrm{J/m^3}$, and $T_c=100\,\mathrm{K}$. Minimizing with respect to $m$:

$$\frac{\partial F}{\partial m}=2a(T-T_c)m+4bm^3=0.$$

For $T>T_c$, the only real solution is $m=0$ (disordered phase). For $T<T_c$, there are two nonzero minima:

$$m=\pm\sqrt{\frac{a(T_c-T)}{2b}}.$$

At $T=90\,\mathrm{K}$,

$$m=\pm\sqrt{\frac{1\times(100-90)}{2\times0.5}}=\pm\sqrt{10}\approx\pm3.16\;(\text{in appropriate units}).$$

The susceptibility is found from $\chi^{-1}=\partial^2F/\partial m^2$ evaluated at the minimum. For $T>T_c$,

$$\chi^{-1}=2a(T-T_c)\quad\Rightarrow\quad\chi=\frac{1}{2a(T-T_c)}.$$

At $T=110\,\mathrm{K}$,

$$\chi=\frac{1}{2(1)(10)}=0.05\,\mathrm{m^3/J}.$$

Problems with Solutions

1. Critical exponent $\beta$.
   Problem: From the Landau result $m\sim(T_c-T)^{1/2}$, identify the critical exponent $\beta$ and explain how mean-field theory gets it.
   Solution. By definition $m\sim|T-T_c|^\beta$, so $\beta=1/2$. Mean-field theory neglects fluctuations and assumes the free energy can be expanded analytically in $m$, which yields the square-root law. In reality, fluctuations change $\beta$ (e.g., Ising 3D: $\beta\approx0.326$).

2. Susceptibility divergence.
   Problem: Calculate the susceptibility exponent $\gamma$ in Landau theory for $T>T_c$ and $T<T_c$.
   Solution. For $T>T_c$, $\chi=[2a(T-T_c)]^{-1}\sim|T-T_c|^{-1}$, so $\gamma=1$. For $T<T_c$, at the minimum $m^2=a(T_c-T)/(2b)$, the curvature is

   $$\partial^2F/\partial m^2=2a(T-T_c)+12bm^2=2a(T-T_c)+6a(T_c-T)=4a(T_c-T),$$

   so $\chi=[4a(T_c-T)]^{-1}\sim|T-T_c|^{-1}$. Thus $\gamma=1$ on both sides in mean field.

3. First-order transition with cubic term.
   Problem: If $F[m]=F_0+\alpha(T-T_c)m^2-cm^3+bm^4$ with $\alpha,b,c>0$, show the transition becomes first order.
   Solution. The equation $\partial F/\partial m=2\alpha(T-T_c)m-3cm^2+4bm^3=0$ has solutions $m=0$ and

   $$2\alpha(T-T_c)-3cm+4bm^2=0.$$

   At a temperature $T_*>T_c$, the nonzero minimum appears with $F[m_*]=F[0]$, indicating a discontinuous jump in $m$. A latent heat is associated with this first-order transition.

Section summary. Critical behavior is governed by symmetry, dimension, and fluctuations.

Fluctuations

Core ideas

Fluctuations are not noise to ignore; they determine response and reveal microscopic physics. Variances of energy, particle number, and order parameters connect to heat capacity, compressibility, and susceptibility.

For review, be able to derive fluctuation-response relations, estimate relative fluctuations, and explain why fluctuations grow near criticality. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\langle(\Delta E)^2\rangle=k_BT^2C_V,\qquad \langle(\Delta N)^2\rangle=k_BT\left(\frac{\partial N}{\partial\mu}\right)_{T,V}$$

Worked example

For $1\,\mathrm{mol}$ of a monatomic ideal gas at $T=300\,\mathrm{K}$, the internal energy is $U=\frac32Nk_BT$ and the heat capacity at constant volume is $C_V=\frac32Nk_B$.

The rms energy fluctuation is

$$\Delta E_{\text{rms}}=\sqrt{\langle(\Delta E)^2\rangle}=\sqrt{k_BT^2C_V}.$$

With $N=N_A=6.022\times10^{23}$, $k_B=1.38\times10^{-23}\,\mathrm{J/K}$:

$$C_V=\frac32(6.022\times10^{23})(1.38\times10^{-23})\approx12.47\,\mathrm{J/K}.$$

Then

$$\Delta E_{\text{rms}}=\sqrt{(1.38\times10^{-23})(300)^2(12.47)}\approx\sqrt{1.55\times10^{-17}}\approx3.9\times10^{-9}\,\mathrm{J}.$$

The total internal energy is

$$U=\frac32Nk_BT\approx\frac32(8.314)(300)\approx3.74\times10^3\,\mathrm{J}.$$

The relative fluctuation is

$$\frac{\Delta E_{\text{rms}}}{U}\approx\frac{3.9\times10^{-9}}{3.74\times10^3}\approx1.0\times10^{-12}.$$

For a single macroscopic mole, energy fluctuations are utterly negligible; they become observable only in very small systems or near critical points.

Problems with Solutions

1. **1/\sqrt{N** scaling.}
   Problem: Show that the relative energy fluctuation in the canonical ensemble scales as $1/\sqrt{N}$ for an extensive system.
   Solution. Both $U$ and $C_V$ are extensive, scaling as $N$. Thus

   $$\frac{\Delta E_{\text{rms}}}{U}=\frac{\sqrt{k_BT^2C_V}}{U}\sim\frac{\sqrt{N}}{N}=\frac{1}{\sqrt{N}}.$$

2. Particle-number fluctuations in a gas.
   Problem: For an ideal gas in the grand canonical ensemble, calculate the variance $\langle(\Delta N)^2\rangle$ in a volume $V=1\,\mathrm{cm^3}$ at STP.
   Solution. For an ideal gas, $\langle(\Delta N)^2\rangle=\langle N\rangle$. At STP, $n=P/(k_BT)\approx2.69\times10^{25}\,\mathrm{m^{-3}}$. In $V=10^{-6}\,\mathrm{m^3}$,

   $$\langle N\rangle\approx2.69\times10^{19},\qquad\Delta N_{\text{rms}}=\sqrt{\langle N\rangle}\approx5.2\times10^9.$$

   The relative fluctuation is $\Delta N_{\text{rms}}/\langle N\rangle\approx1/\sqrt{\langle N\rangle}\approx1.9\times10^{-10}$.

3. Critical fluctuations.
   Problem: Near a critical point the correlation length diverges as $\xi\sim|T-T_c|^{-\nu}$ and the susceptibility as $\chi\sim|T-T_c|^{-\gamma}$. Use the fluctuation-dissipation relation $\chi\sim\xi^{2-\eta}$ to relate the critical exponents.
   Solution. Substituting the scaling forms,

   $$|T-T_c|^{-\gamma}\sim\left(|T-T_c|^{-\nu}\right)^{2-\eta}=|T-T_c|^{-\nu(2-\eta)}.$$

   Equating exponents gives the scaling law

   $$\gamma=\nu(2-\eta).$$

   In mean field, $\nu=1/2$ and $\eta=0$, so $\gamma=1$, consistent with Landau theory.

Section summary. Fluctuations quantify stability and response.