Statistical Mechanics

A microstate is a complete specification of the state of a system (positions and momenta of all particles). A macrostate is specified by macroscopic variables (energy, volume, particle number).

The fundamental postulate of statistical mechanics states that for an isolated system in equilibrium, every accessible microstate is equally probable.

Definition. The multiplicity $\Omega(E, V, N)$ is the number of microstates consistent with the macrostate $(E, V, N)$. The statistical entropy is

$S = k_B \ln \Omega$

Proposition 2.1. This definition of entropy agrees with the thermodynamic entropy: $\Delta S = k_B \ln(\Omega_f / \Omega_i)$.

Theorem 2.1 (Canonical Ensemble). For a system in thermal equilibrium with a heat bath at temperature $T$The probability of the system being in microstate $i$ with energy $E_i$ is

$P_i = \frac{1}{Z} e^{-E_i / (k_B T)}$

Where the partition function is

$Z = \sum_{i} e^{-E_i / (k_B T)}$

Proof. Consider the combined system (system + reservoir) with total energy $E_{\mathrm{tot}}$. The probability of the system being in state $i$ is proportional to the number of reservoir microstates compatible with it, which is $\Omega_R(E_{\mathrm{tot} - E_i)}$. Using $S_R = k_B \ln \Omega_R$:

P_i \propto \Omega_R(E_{\mathrm{tot} - E_i) = \exp\left(\frac{S_R(E_{\mathrm{tot} - E_i)}{k_B}\right)}}

Expanding $S_R$ around $E_{\mathrm{tot}}$: $S_R(E_{\mathrm{tot} - E_i) \approx S_R(E_{\mathrm{tot}) - E_i \left(\frac{\partial S_R}{\partial E}\right) = S_R(E_{\mathrm{tot}) - \frac{E_i}{T}}}}$

Since $(\partial S_R / \partial E) = 1/T$. Therefore $P_i \propto e^{-E_i / (k_BT)}$And normalising gives the result. $\blacksquare$

Theorem 2.2. The partition function determines all thermodynamic quantities:

$\langle E \rangle = -\frac{\partial \ln Z}{\partial \beta}, \quad F = -k_B T \ln Z, \quad S = \frac{\langle E \rangle - F}{T}$

Where $\beta = 1/(k_BT)$.

Proof. $\langle E \rangle = \sum_i E_i P_i = \frac{1}{Z}\sum_i E_i e^{-\beta E_i} = -\frac{1}{Z}\frac{\partial Z}{\partial \beta} = -\frac{\partial \ln Z}{\partial \beta}$.

$F = -k_BT \ln Z$ follows from $F = U - TS = \langle E \rangle - TS$ and the identification $Z = e^{-\beta F}$. $\blacksquare$

Theorem 2.3 (Partition Function of an Ideal Gas). For $N$ indistinguishable particles in a 3D box of volume $V$:

Z_N = \frac{1}{N!}\left(\frac{V}{\lambda_{\mathrm{th}^3}\right)^N, \quad \lambda_{\mathrm{th} = \frac{h}{\sqrt{2\pi m k_B T}}}}

Where $\lambda_{\mathrm{th}}$ is the thermal de Broglie wavelength.

Proof. The single-particle energy levels in a 3D box of side $L$ ($V = L^3$) are:

$\varepsilon_{n_x, n_y, n_z} = \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2)$

The single-particle partition function is:

$z = \sum_{n_x, n_y, n_z = 0}^{\infty} e^{-\beta h^2(n_x^2 + n_y^2 + n_z^2)/(8mL^2)} \approx \left(\int_0^{\infty} e^{-\beta h^2 n^2/(8mL^2)} dn\right)^3 = \left(\frac{L}{h}\sqrt{\frac{2\pi m}{\beta}}\right)^3 = \frac{V}{\lambda_{\mathrm{th}^3}}$

For $N$ indistinguishable particles (correct Boltzmann counting): $Z_N = z^N / N!$. $\blacksquare$

Corollary 2.4. From $Z_N$We recover the ideal gas law:

F = -k_BT \ln Z_N = -k_BT\left[N\ln\left(\frac{V}{\lambda_{\mathrm{th}^3}\right) - \ln N!\right]}

$P = -\left(\frac{\partial F}{\partial V}\right)_T = \frac{Nk_BT}{V}$

Giving $PV = Nk_BT$.

Theorem 2.5 (Equipartition). For a classical system in thermal equilibrium, each quadratic degree of freedom in the Hamiltonian contributes $\frac{1}{2}k_BT$ to the average energy.

Proof. Consider a single degree of freedom with Hamiltonian $H = ap^2$ (or $bq^2$). The average energy is:

$\langle ap^2 \rangle = \frac{\int_{-\infty}^{\infty} ap^2 e^{-\beta ap^2}\, dp}{\int_{-\infty}^{\infty} e^{-\beta ap^2}\, dp} = -\frac{\partial}{\partial \beta}\ln\left(\int_{-\infty}^{\infty} e^{-\beta ap^2}\, dp\right) = -\frac{\partial}{\partial \beta}\ln\left(\sqrt{\frac{\pi}{a\beta}}\right) = \frac{1}{2\beta} = \frac{k_BT}{2}$

The same calculation for $bq^2$ gives another $k_BT/2$. $\blacksquare$

Application. A monatomic ideal gas has 3 translational degrees of freedom: $U = \frac{3}{2}Nk_BT$ and $C_V = \frac{3}{2}Nk_B$. A diatomic gas also has 2 rotational degrees of freedom: $U = \frac{5}{2}Nk_BT$ and $C_V = \frac{5}{2}Nk_B$ (at temperatures where vibration is frozen out).

Fermi—Dirac Statistics (for fermions, particles with half-integer spin):

$\langle n_i \rangle = \frac{1}{e^{(E_i - \mu)/(k_BT)} + 1}$

Where $\mu$ is the chemical potential.

Bose—Einstein Statistics (for bosons, particles with integer spin):

$\langle n_i \rangle = \frac{1}{e^{(E_i - \mu)/(k_BT)} - 1}$

Maxwell—Boltzmann Statistics (classical limit, $\mu$ very negative):

$\langle n_i \rangle = e^{-(E_i - \mu)/(k_BT)}$

The classical limit applies when the thermal de Broglie wavelength is much smaller than the inter-particle spacing: $\lambda_{\mathrm{th}^3 \ll V/N}$.

Definition. The Fermi energy $\varepsilon_F$ is the chemical potential at $T = 0$:

$\varepsilon_F = \frac{\hbar^2}{2m}\left(3\pi^2 n\right)^{2/3}$

Where $n = N/V$ is the particle number density.

Proposition 2.6. At $T = 0$All states with $E \leq \varepsilon_F$ are occupied and all states with $E > \varepsilon_F$ are empty. The ground-state energy of a 3D Fermi gas is:

$U_0 = \frac{3}{5}N\varepsilon_F$

Proof. $U_0 = \int_0^{\varepsilon_F} E \cdot g(E)\, dE$ where $g(E) = \frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\sqrt{E}$ is the density of states. Evaluating: $U_0 = \frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2} \cdot \frac{2}{5}\varepsilon_F^{5/2} = \frac{3}{5}N\varepsilon_F$. $\blacksquare$

Planck”s Law gives the spectral energy density of blackbody radiation:

$u(\nu, T) = \frac{8\pi h \nu^3}{c^3} \cdot \frac{1}{e^{h\nu/(k_BT)} - 1}$

Stefan—Boltzmann Law: The total radiated power per unit area:

$j = \sigma T^4, \quad \sigma = \frac{\pi^2 k_B^4}{60 \hbar^3 c^2}$

Wien’s Displacement Law: The peak frequency satisfies $\nu_{\mathrm{max} / T = \mathrm{const}}$.

Problem. Calculate the Fermi energy and Fermi temperature for copper. Given: electron density $n \approx 8.5 \times 10^{28}\,\mathrm{m}^{-3}$, $m_e = 9.109 \times 10^{-31}$ kg.

• The classical limit does not always apply. When $\lambda_{\mathrm{th}^3 \gtrsim V/N}$Quantum …/4-statistics-and-probability/2_statistics (Fermi-Dirac or Bose-Einstein) must be used. This is critical for electrons in metals and for helium-4 at low temperatures.
• The Boltzmann distribution applies to systems in contact with a heat bath, not isolated systems. For isolated systems, use the microcanonical ensemble (all accessible microstates equally probable).
• The partition function must account for indistinguishability. The $1/N!$ factor in $Z_N$ is essential for obtaining the correct entropy (otherwise the entropy is not extensive and the Gibbs paradox arises).