Problem Set

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

$= \frac{(1.055 \times 10^{-34})^2}{2 \times 9.109 \times 10^{-31}}(3\pi^2 \times 2.65 \times 10^{28})^{2/3}$

$(3\pi^2 \times 2.65 \times 10^{28})^{1/3} = (7.85 \times 10^{29})^{1/3} \approx 9.23 \times 10^9$

$(3\pi^2 n)^{2/3} = (9.23 \times 10^9)^2 = 8.52 \times 10^{19}$

$\varepsilon_F = \frac{1.113 \times 10^{-68}}{1.822 \times 10^{-30}} \times 8.52 \times 10^{19} \approx 5.20 \times 10^{-19}\,\mathrm{J} \approx 3.25\,\mathrm{eV}$

$T_F = \varepsilon_F/k_B = 5.20 \times 10^{-19}/1.381 \times 10^{-23} \approx 37700\,\mathrm{K}$

Above $T_c$ (classical regime): $C_V = \frac{3}{2}Nk_B$.

Below $T_c$: $C_V = \frac{15}{4}Nk_B\,\zeta(5/2)/\zeta(3/2) \cdot (T/T_c)^{3/2}$.

At $T = T_c^-$:

$C_V(T_c^-) = \frac{15}{4}Nk_B \cdot \frac{\zeta(5/2)}{\zeta(3/2)}$

$\zeta(5/2) \approx 1.341$, $\zeta(3/2) \approx 2.612$:

$C_V(T_c^-) = \frac{15}{4} \times \frac{1.341}{2.612}\,Nk_B \approx 1.926\,Nk_B$

At $T = T_c^+$: $C_V = \frac{3}{2}Nk_B = 1.5\,Nk_B$.

The jump is $\Delta C_V = C_V(T_c^-) - C_V(T_c^+) \approx 0.426\,Nk_B$.

The pressure of a real gas is expanded as $PV/(Nk_BT) = 1 + B_2(T)\,(N/V) + B_3(T)\,(N/V)^2 + \cdots$.

For a classical gas with pairwise interaction $V(r_{12})$:

$B_2(T) = -\frac{1}{2V}\int d^3\mathbf{r}_1\,d^3\mathbf{r}_2\,\left[e^{-\beta V(r_{12})} - 1\right]$

$= -2\pi \int_0^\infty \left[e^{-\beta V(r)} - 1\right] r^2\, dr$

For a hard-sphere gas ($V(r) = \infty$ for $r < d$, $V(r) = 0$ for $r > d$):

$B_2 = -2\pi\int_0^d (-1)\,r^2\,dr = \frac{2\pi d^3}{3} = \frac{2\pi}{3}\left(\frac{d}{2}\right)^3 \cdot 8 = 4v_0$

Where $v_0 = \pi d^3/6$ is the volume of one sphere. The van der Waals excluded volume parameter is $b = 4Nv_0 = N B_2$.

The Fermi-Dirac distribution is:

$f_{\mathrm{FD}(\varepsilon) = \frac{1}{e^{\beta(\varepsilon - \mu)} + 1}}$

The total number of particles is:

$N = \int_0^\infty \frac{g(\varepsilon)}{e^{\beta(\varepsilon - \mu)} + 1}\, d\varepsilon$

In the classical limit $e^{\beta(\varepsilon - \mu)} \gg 1$The $+1$ is negligible:

$N \approx \int_0^\infty g(\varepsilon)\, e^{-\beta(\varepsilon - \mu)}\, d\varepsilon = e^{\beta\mu} \int_0^\infty g(\varepsilon)\, e^{-\beta\varepsilon}\, d\varepsilon$

e^{\beta\mu} = \frac{N}{\int_0^\infty g(\varepsilon)\, e^{-\beta\varepsilon}\, d\varepsilon} = \frac{N\lambda_{\mathrm{th}^3}{V}}

The classical limit requires $e^{\beta\mu} \ll 1$I.e., $N\lambda_{\mathrm{th}^3/V \ll 1}$Or equivalently, the average inter-particle spacing $(V/N)^{1/3}$ must be much larger than $\lambda_{\mathrm{th}}$.

$Z_{\mathrm{HO} = \sum_{n=0}^{\infty} e^{-\beta\hbar\omega(n+1/2)} = e^{-\beta\hbar\omega/2}\sum_{n=0}^{\infty}\left(e^{-\beta\hbar\omega}\right)^n = \frac{e^{-\beta\hbar\omega/2}}{1 - e^{-\beta\hbar\omega}}}$

$\langle E \rangle = -\frac{\partial \ln Z}{\partial \beta} = \frac{\hbar\omega}{2} + \frac{\hbar\omega\,e^{-\beta\hbar\omega}}{1 - e^{-\beta\hbar\omega}} = \frac{\hbar\omega}{2} + \frac{\hbar\omega}{e^{\beta\hbar\omega} - 1} = \hbar\omega\left(\frac{1}{2} + n_B\right)$

At high $T$ ($\beta \to 0$): $\langle E \rangle \to k_BT$ (equipartition). At low $T$: $\langle E \rangle \to \hbar\omega/2$ (zero-point energy).

For a single spin-1/2 particle, the energy levels are $E_\uparrow = -\mu B$ and $E_\downarrow = +\mu B$. The single-particle partition function is:

$Z_1 = e^{\beta\mu B} + e^{-\beta\mu B} = 2\cosh(\beta\mu B)$

For $N$ non-interacting spins, $Z = Z_1^N = [2\cosh(\beta\mu B)]^N$.

The magnetisation is:

$M = \frac{1}{\beta}\frac{\partial \ln Z}{\partial B} = N\mu\tanh(\beta\mu B)$

The susceptibility at small $B$ (or high $T$, where $\beta\mu B \ll 1$):

$\chi = N\mu^2\beta\,\mathrm{sech}^2(\beta\mu B) \approx \frac{N\mu^2}{k_B T}$

This is the Curie law $\chi = C/T$ with $C = N\mu^2/k_B$, showing that paramagnetic susceptibility obeys a $1/T$ dependence at high temperatures. $\blacksquare$

The Planck distribution gives the spectral energy density:

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

Integrating over all frequencies:

$U(T) = \int_0^\infty u(\omega, T)\,d\omega = \frac{\hbar}{\pi^2 c^3}\int_0^\infty \frac{\omega^3}{e^{\beta\hbar\omega} - 1}\,d\omega$

Let $x = \beta\hbar\omega$:

$U(T) = \frac{(k_B T)^4}{\pi^2 c^3 \hbar^3}\int_0^\infty \frac{x^3}{e^x - 1}\,dx = \frac{\pi^2 k_B^4}{15 c^3 \hbar^3} T^4 = \sigma T^4$

where $\sigma = \pi^2 k_B^4/(60\hbar^3 c^2)$ is the Stefan-Boltzmann constant.

The photon number density is:

$n = \int_0^\infty \frac{\omega^2}{\pi^2 c^3}\frac{1}{e^{\beta\hbar\omega} - 1}\,d\omega = \frac{2\zeta(3)}{\pi^2}\left(\frac{k_B T}{\hbar c}\right)^3$

The average energy per photon is $\langle E \rangle = U/n = (\pi^4/30\zeta(3))k_B T \approx 2.701\,k_B T$.

$\blacksquare$

In mean-field theory, each spin experiences an effective field $B_{\text{eff}} = zJ\langle s \rangle$. The self-consistency equation is $\langle s \rangle = \tanh(\beta z J \langle s \rangle)$. For $T > T_c^{\text{MF}} = zJ/k_B$, only the trivial solution $\langle s \rangle = 0$ exists; below $T_c^{\text{MF}}$, a non-zero magnetisation appears.

For the square lattice, $z = 4$, so $T_c^{\text{MF}} = 4J/k_B$. The exact Onsager solution gives $T_c \approx 2.269 J/k_B$. The discrepancy arises because mean-field theory neglects fluctuations, which are significant in low dimensions. In 1D, mean-field theory incorrectly predicts a phase transition at finite $T$, while the exact solution shows no spontaneous magnetisation at any $T > 0$. The lower critical dimension for the Ising model is $d = 1$, below which fluctuations destroy long-range order. $\blacksquare$

(a) $Z_1 = 1 + e^{-\beta\varepsilon} + e^{-2\beta\varepsilon}$.

(b) For $N$ distinguishable particles, $Z = Z_1^N$. The average energy is: