Solutions for lecture 1 exercises¶

    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()

Warm-up exercises¶

By definition, the particles in an ideal gas do not interact. As such, their energy is independent of their position, and only depends on the momenta of the particles. This means that only the 3 momentum degrees of freedom can store energy and thereby contribute to the heat capacity. In contrast, both the position and the momentum of atoms described by harmonic oscillators (e.g. masses on springs) determine the energy. This means that 3 momentum + 3 positional degrees of freedom can store energy and thereby contribute to the heat capacity. Finally, equipartition tells us that each degree of freedom contributes $1/2k_BT$ to the energy.
Following last question's reasoning, we get $C = 2k_B$ per atom.
See the plot with the slider in the lecture notes. For fun, the Einstein dist. for two different values of $T$ is plotted below

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)

svg

Exercise 1: The harmonic oscillator and the Bose-Einstein distribution¶

Question 1¶

The number of phonons in the oscillator is given by the Bose-Einstein distribution $n_B(\beta\hbar\omega_0) = 1/(e^{\beta\hbar\omega_0} - 1)$.
In the high temperature limit, $\beta\hbar\omega_0 \ll 1$, so $\exp(\beta\hbar\omega_0) \approx 1 + \beta\hbar\omega_0$, and $n_B(\beta\hbar\omega_0) \approx 1/(\beta\hbar\omega_0)$.
This result has a natural interpretation: the thermal energy is $k_BT$, which requires $n = k_BT/\hbar\omega_0$ phonons to store it.

Question 2¶

The expectation value of the energy is given by $\langle E\rangle = \hbar\omega_0(n_B(\beta\hbar\omega_0) + 1/2)$.
In the high temperature limit, if we apply the first order Taylor expansion like above, we get an almost correct answer $\langle E\rangle \approx k_BT + \hbar\omega_0/2$. To get a more accurate result (not required for this problem), we need to use the second order expansion, $\exp(\beta\hbar\omega_0) \approx 1 + \beta\hbar\omega_0 + (\beta\hbar\omega_0)^2/2$, which gives $\langle E\rangle \approx k_BT$.
This result is consistent with the expected number of phonons in the high temperature limit.

Question 3¶

In the high temperature limit the heat capacity is the derivative of the above expression with temperature, and it is $C = k_B$.

Question 4¶

At zero temperature, the harmonic oscillator contains no phonons.
Phonons start getting excited at $T \sim \hbar\omega_0/k_B \equiv T_E$.
At the low temperature limit the number of phonons is exponentially small because of the exponential in the Bose-Einstein distribution.
This means that both the energy and the heat capacity are exponentially small at low temperatures.

Question 5¶

There are many ways to argue that this function cannot be Bose-Einstein distribution. For example, the Bose-Einstein distribution is an expected number of particles, and the function in question is negative.

Exercise 2: The quantum harmonic oscillator - connection with statistical physics¶

The energy spectrum is $E_n = (n+1/2)\hbar\omega_0$ with $n=0,1,2,...$
We calculate the partition function using $$ Z = \sum_{n = 0}^{\infty} e^{-\beta\hbar\omega(n + 1/2)} = e^{-\beta\hbar\omega/2} \frac{1}{1 - e^{-\beta\hbar\omega}} = \frac{1}{2\sinh(\beta\hbar\omega/2)} $$

,where we used the result of a geometric series $\sum_{n = 0}^{\infty}r^n = \frac{1}{1 - r}$.
We calculate the expectation value of the energy using $E = d\ln(Z)/d\beta)$. This yields $$ E = -\frac{1}{Z}\frac{\partial Z}{\partial\beta} = \frac{\hbar\omega}{2}\coth\frac{\beta\hbar\omega}{2} = \hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right) = \hbar\omega\left(n_B(\beta\hbar\omega) + \frac{1}{2}\right). $$
We use $C = dE/dT$, yielding $$ C = \frac{\partial \langle E\rangle}{\partial T} = \frac{\partial\langle E\rangle}{\partial\beta}\frac{\partial\beta}{\partial T} = k_B(\beta\hbar\omega)^2\frac{e^{\beta\hbar\omega}}{(e^{\beta\hbar\omega} - 1)^2}. $$

In the high temperature limit $\beta \rightarrow 0$ and $e^{\beta\hbar\omega} \approx 1 + \beta\hbar\omega$, so $C \rightarrow k_B$ which is the same result as in Exercise 1.3.
It is given by the Bose-Einstein distribution. This also follows from the partition function using $$ \langle n\rangle = \frac{1}{Z}\sum_{n = 0}^{\infty} ne^{-\beta\hbar\omega(n + 1/2) } = 2\frac{e^{\beta\hbar\omega/2} - e^{-\beta\hbar\omega/2}}{2}e^{-\beta\hbar\omega/2}\frac{e^{-\beta\hbar\omega}}{(1 - e^{-\beta\hbar\omega})^2} = \frac{1}{e^{\beta\hbar\omega} - 1} $$

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

Exercise 3*: Total heat capacity of a diatomic material.¶

We use $\omega_{1,2} = \sqrt{\frac{k}{m_{1,2}}}$.
The expectation value of the vibrational energy stored in a single atom modeled as a 3D harmonic oscillator with eigenfrequency $\omega_{1,2}$ is given by $$ E_{1,2} = 3 (n(\omega_{1,2})+1/2)\hbar\omega_{1,2} $$ , where $n(\omega)$ is given by the Bose-Einstein distribution function. The total energy $E$ of the crystal is obtained by summing over all the atoms: $E = N_1E_1 + N_2E_2$, where $N_1$ is the number of $^6$Li atoms and $N_2$ is the number of $^7$Li atoms.
The heat capacity is given by $dE/dT$, with $E$ given in the answer to the previous subquestion. The calculation is the same as that in the lecture notes (apart from the factor 3 for the dimensionality).

A harmonic oscillator starts to go mostly into its ground state when the temperature is reduced below its Einstein temperature (Note that the Einstein temperature is the eigenfrequency of the oscillator expressed in units of Kelvin). Therefore, upon reducing temperature, first the atoms with the higher eigenfrequency (or equivalently, higher Einstein temperature) will go into their ground state, and then the atoms with the lower eigenfrequency.