from matplotlib import pyplot as plt
from mpl_toolkits.axes_grid1 import make_axes_locatable

import numpy as np
from scipy.optimize import curve_fit
from scipy.integrate import quad

from common import draw_classic_axes, configure_plotting

configure_plotting()
  

Solutions for lecture 1 exercises

Warm-up exercises

1. An ideal gas only contains 3 positional degrees of freedom.
2. $C=2k_B$.
3. See image below (with $T_1<T_2$)
4. Minus sign in the exponent. This would result in $n_B(T=0)=-1$, which is not physical.
5. See plot with slider

fig, ax = plt.subplots()
omega = np.linspace(0.1, 3)
T = [1,2]
ax.plot(omega, 1/(np.exp(omega/T[0]) - 1), label = r'$T_1$')
ax.plot(omega, 1/(np.exp(omega/T[1]) - 1), label = r'$T_2$')
ax.set_ylim([0,3])
ax.set_xlim([0,3])
ax.set_xlabel('$\hbar \omega$')
ax.set_xticks([0])
ax.set_xticklabels(['$0$'])
ax.set_ylabel('$n_B$')
ax.set_yticks([0,1, 2])
ax.set_yticklabels(['$0$','$1$', '$2$'])
ax.legend()
draw_classic_axes(ax, xlabeloffset=.2)

Exercise 1: Heat capacity of a classical oscillator.

1. The answer is

   $$Z={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dp{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx{e}^{-\frac{\beta }{2m}{p}^2-\frac{\beta k}{2}{x}^2}=\sqrt{\frac{2\pi m}{\beta }}\sqrt{\frac{2\pi }{\beta k}}=\frac{2\pi }{\beta }\sqrt{\frac{m}{k}},$$

   where we used ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\alpha x^2}=\sqrt{\frac{\pi }{\alpha }}$.

2.

$$
\langle E \rangle = -\frac{1}{Z}\frac{\partial Z}{\partial \beta} = \frac{1}{\beta}
$$

3. $$C=\frac{\partial ⟨E⟩}{\partial T}=k_B$$

The heat capacity is temperature independent.

Exercise 2: Quantum harmonic oscillator.

1.

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

where we used $\sum_{n = 0}^{\infty}r^n = \frac{1}{1 - r}$.

2.

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

3.

$$
C = \frac{\partial \langle E\rangle}{\partial T} = \frac{\partial\langle E\rangle}{\partial\beta}\frac{\partial\beta}{\partial T} = k_B(\beta\hbar\omega)^2\frac{e^{\beta\hbar\omega}}{(e^{\beta\hbar\omega} - 1)^2}.
$$

In the high temperature limit $\beta \to 0$ and $e^{\beta \hslash \omega }\approx 1+\beta \hslash \omega$, so $C\to k_B$ which is the same result as in Exercise 1.3.

1. Compare your result with the plot with the slider. Did you correctly indicate the where the Einstein temperature is?

5.

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

where we used $\sum_{n = 0}^{\infty}nr^n = \frac{r}{(1 - r)^2}$.

Exercise 3: Total heat capacity of a diatomic material.

1. Use the formula $\omega =\sqrt{\frac{k}{m}}$.

2. Energy per atom is given by

   $$E=\frac{N_{{}^{6}Li}}{N}\hslash {\omega }_{{}^{6}Li}(2+1/2)+\frac{N_{{}^{7}Li}}{N}\hslash {\omega }_{{}^{7}Li}(4+1/2).$$

3. Energy per atom is given by

   $$E=\frac{N_{{}^{6}Li}}{N}\hslash {\omega }_{{}^{6}Li}(n_B(\beta \hslash {\omega }_{{}^{6}Li})+\frac{1}{2})+\frac{N_{{}^{7}Li}}{N}\hslash {\omega }_{{}^{7}Li}(n_B(\beta \hslash {\omega }_{{}^{7}Li})+\frac{1}{2}).$$

4. Heat capacity per atom is given by

   $$C=\frac{N_{{}^{6}Li}}{N}{C}_{{}^{6}Li}+\frac{N_{{}^{7}Li}}{N}{C}_{{}^{7}Li},$$

where the heat capacities are calculated with the formula from Excercise 2.4.