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 depends only on their momenta. This means that only the three momentum degrees of freedom can store energy and thereby contribute to the heat capacity. In contrast, for atoms described by harmonic oscillators (e.g. masses on springs), both position and momentum determine the energy. This means that three momentum and three positional degrees of freedom can store energy and thereby contribute to the heat capacity. Finally, equipartition tells us that each degree of freedom contributes $k_BT/2$ to the energy.
Following the reasoning of the previous question, we get $C = 2k_B$ per atom.
See the plot with the slider in the lecture notes. For illustration, the Einstein distribution 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 of order $k_BT$, 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 as 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, we found $\langle E\rangle \approx k_BT$. Taking the derivative with respect to temperature therefore gives

\[ C = \frac{d\langle E\rangle}{dT} \approx k_B. \]

So in this limit the heat capacity of a 1D harmonic oscillator is $k_B$.

Question 4¶

At zero temperature, the harmonic oscillator has no phonon excitations, so $\langle n \rangle = 0$. Its energy is still nonzero because of the zero-point contribution $\hbar\omega_0/2$.
Phonons start getting excited at $T \sim \hbar\omega_0/k_B \equiv T_E$.
In 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 the Bose-Einstein distribution. For example, the Bose-Einstein distribution is an expected number of particles, whereas 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 $$ Z = \sum_{n = 0}^{\infty} e^{-\beta\hbar\omega_0(n + 1/2)} = e^{-\beta\hbar\omega_0/2} \frac{1}{1 - e^{-\beta\hbar\omega_0}} = \frac{1}{2\sinh(\beta\hbar\omega_0/2)} $$

where we used the geometric-series result $\sum_{n = 0}^{\infty}r^n = \frac{1}{1 - r}$.
We calculate the expectation value of the energy using $E = -d\ln(Z)/d\beta$. This yields $$ E = -\frac{1}{Z}\frac{\partial Z}{\partial\beta} = \frac{\hbar\omega_0}{2}\coth\frac{\beta\hbar\omega_0}{2} = \hbar\omega_0\left(\frac{1}{e^{\beta\hbar\omega_0} - 1} + \frac{1}{2}\right) = \hbar\omega_0\left(n_B(\beta\hbar\omega_0) + \frac{1}{2}\right). $$
We use $C = dE/dT$, yielding $$ C = \frac{\partial \langle E\rangle}{\partial T} = \frac{\partial\langle E\rangle}{\partial\beta}\frac{\partial\beta}{\partial T} = k_B(\beta\hbar\omega_0)^2\frac{e^{\beta\hbar\omega_0}}{(e^{\beta\hbar\omega_0} - 1)^2}. $$

In the high-temperature limit, $\beta \hbar \omega_0 \ll 1$. We then use

\[ e^{\beta\hbar\omega_0} \approx 1 + \beta\hbar\omega_0, \]

so the numerator becomes approximately $1 + \beta\hbar\omega_0 \approx 1$, while the denominator becomes

\[ (e^{\beta\hbar\omega_0} - 1)^2 \approx (\beta\hbar\omega_0)^2. \]

Therefore,

\[ C \approx k_B(\beta\hbar\omega_0)^2 \frac{1}{(\beta\hbar\omega_0)^2} = k_B, \]

which is the same result as in Exercise 1.3.
It is given by the Bose-Einstein distribution: $$ \langle n\rangle = \frac{1}{e^{\beta\hbar\omega_0} - 1}. $$

Exercise 3*: Total heat capacity of a two-isotope 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 in the lecture notes, apart from the factor 3 from the dimensionality.

A harmonic oscillator starts to occupy its ground state predominantly when the temperature is reduced below its Einstein temperature. Note that the Einstein temperature is the oscillator eigenfrequency expressed in units of kelvin. Therefore, upon lowering the temperature, the atoms with the higher eigenfrequency (or equivalently, higher Einstein temperature) go into their ground state first, followed by the atoms with the lower eigenfrequency.