Commit 24da1992 authored by Anton Akhmerov's avatar Anton Akhmerov
Browse files

delete duplicate files

parent e8ff4949
# Solutions for lecture 1 exercises
### Exercise 1: Heat capacity of a classical oscillator.
1.
$$
Z = \int_{-\infty}^{\infty}dp \int_{-\infty}^{\infty} dx e^{-\frac{\beta}{2m}p^2-\frac{\beta k}{2}x^2} = \sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi}{\beta k}} = \frac{2\pi}{\beta}\sqrt{\frac{m}{k}},
$$
where we used $\int_{-\infty}^{\infty}e^{-\alpha x^2} = \sqrt{\frac{\pi}{\alpha}}$.
2.
$$
\langle E \rangle = -\frac{1}{Z}\frac{\partial Z}{\partial \beta} = \frac{1}{\beta}
$$
3.
$$
C = \frac{\partial\langle E\rangle}{\partial T} = k_B
$$
4.
Since this is a 1D system, the prefactor is $1$ as expected.
### Exercise 2: Quantum harmonic oscillator.
1.
Take a look into the graph from the notes.
2.
$$
Z = \sum_{n = 0}^{\infty} e^{-\beta\hbar\omega(n + 1/2)} = e^{-\beta\hbar\omega/2}\frac{1}{1 - e^{-\beta\hbar\omega}} = \frac{1}{2\sinh(\beta\hbar\omega/2)},
$$
where we used $\sum_{n = 0}^{\infty}r^n = \frac{1}{1 - r}$.
3.
$$
\langle E\rangle = -\frac{1}{Z}\frac{\partial Z}{\partial\beta} = \frac{\hbar\omega}{2}\coth\frac{\beta\hbar\omega}{2} = \hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right) = \hbar\omega\left(n_B(\beta\hbar\omega) + \frac{1}{2}\right).
$$
4.
$$
C = \frac{\partial \langle E\rangle}{\partial T} = \frac{\partial\langle E\rangle}{\partial\beta}\frac{\partial\beta}{\partial T} = k_B(\beta\hbar\omega)^2\frac{e^{\beta\hbar\omega}}{(e^{\beta\hbar\omega} - 1)^2}.
$$
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 1.1.
5.
We can see that for $\beta\hbar\omega = 1$ we already have $C \approx 0.92k_B$, so we can define a characteristic temperature $T_E = \hbar\omega/k_B$ such that temperatures above this one can be considered high.
6.
$$
\langle n\rangle = \frac{1}{Z}\sum_{n = 0}^{\infty} ne^{-\beta\hbar\omega(n + 1/2)} = 2\frac{e^{\beta\hbar\omega/2} - e^{-\beta\hbar\omega/2}}{2}e^{-\beta\hbar\omega/2}\frac{e^{-\beta\hbar\omega}}{(1 - e^{-\beta\hbar\omega})^2} = \frac{1}{e^{\beta\hbar\omega} - 1},
$$
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.
$E = N_{^6Li}\hbar\omega_{^6Li}(2 + 1/2)+N_{^7Li}\hbar\omega_{^7Li}(4 + 1/2)$.
3.
$E = N_{^6Li}\hbar\omega_{^6Li}\left(n_B(\beta\hbar\omega_{^6Li}) + \frac{1}{2}\right) + N_{^7Li}\hbar\omega_{^7Li}\left(n_B(\beta\hbar\omega_{^7Li}) + \frac{1}{2}\right)$.
4.
$C = \frac{N_{^6Li}}{N}C_{^6Li} + \frac{N_{^7Li}}{N}C_{^7Li}$ where the heat capacities are calculated with the formula from Excercise 2.4.
---
jupyter:
jupytext:
text_representation:
extension: .md
format_name: markdown
format_version: '1.0'
jupytext_version: 1.0.2
---
# Solutions for lecture 2 exercises
### Exercise 1: Debye model: concepts.
1, 2, 3.
Look at the lecture notes.
4.
$$
g(\omega) = \frac{dN}{d\omega} = \frac{dN}{dk}\frac{dk}{d\omega} = \frac{1}{v}\frac{dN}{dk}.
$$
We assume that in $d$ dimensions there are $d$ polarizations.
For 1D we have that $N = \frac{L}{2\pi}\int dk$, hence $g(\omega) = \frac{L}{2\pi v}$.
For 2D we have that $N = 2\left(\frac{L}{2\pi}\right)^2\int d^2k = 2\left(\frac{L}{2\pi}\right)^2\int 2\pi kdk$, hence $g(\omega) = \frac{L^2\omega}{\pi v^2}$.
For 3D we have that $N = 3\left(\frac{L}{2\pi}\right)^3\int d^3k = 3\left(\frac{L}{2\pi}\right)^3\int 4\pi kdk$, hence $g(\omega) = \frac{3L^3\omega^2}{2\pi^2v^3}$.
### Exercise 2: Debye model in 2D.
1.
Look at the lecture notes.
2.
$$
E = \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega = \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\beta\hbar\omega_D}\frac{x^2}{e^{x} - 1}dx + T \text{ independent constant}.
$$
3.
High temperature implies $\beta \rightarrow 0$, hence $E = \frac{L^2}{\pi v^2\hbar^2\beta^3}\frac{(\beta\hbar\omega_D)^2}{2} + T \text{ independent constant}$, and then $C = \frac{k_BL^2\omega^2_D}{2\pi v^2} = 2Nk_B$. We've used the value for $\omega_D$ calculated from $2N = \int_{0}^{\omega_D}g(\omega)d\omega$.
4.
In the low temperature limit we have that $\beta \rightarrow \infty$, hence $E \approx \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx + T \text{ independent constant} = \frac{2\zeta(3)L^2}{\pi v^2\hbar^2\beta^3} + T \text{ independent constant}$. Finally $C = \frac{6\zeta(3)k^3_BL^2}{\pi v^2\hbar^2}T^2$. We used the fact that $\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx = 2\zeta(3)$ where $\zeta$ is the Riemann zeta function.
### Exercise 3: Different phonon modes.
1.
$$
g(\omega) = \sum_{\text{polarizations}}\frac{dN}{dk}\frac{dk}{d\omega} = \left(\frac{L}{2\pi}\right)^3\sum_{\text{polarizations}}4\pi k^2\frac{dk}{d\omega} = \frac{L^3}{2\pi^2}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\omega^2
$$
$$
E = \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega = \frac{L^3}{2\pi^2\hbar^3\beta^4}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\int_{0}^{\beta\hbar\omega_D}\frac{x^3}{e^{x} - 1}dx + T \text{ independent constant}.
$$
2.
Note that we can get $\omega_D$ from $3N = \int_{0}^{\omega_D}g(\omega)$ so everything cancels as usual and we are left with the Dulong-Petit law $C = 3Nk_B$.
3.
In the low temperature limit we have that $C \sim \frac{2\pi^2k_B^4L^3}{15\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)T^3$. We used that $\int_{0}^{\infty}\frac{x^3}{e^{x} - 1}dx = \frac{\pi^4}{15}$.
### Exercise 4: Anisotropic sound velocities.
$$
E = 3\left(\frac{L}{2\pi}\right)^3\int d^3k\hbar\omega(\mathbf{k})\left(n_B(\beta\hbar\omega(\mathbf{k})) + \frac{1}{2}\right) = 3\left(\frac{L}{2\pi}\right)^3\frac{1}{v_xv_yv_z}\int d^3\kappa\frac{\hbar\kappa}{e^{\beta\hbar\kappa} - 1} + T \text{ independent part},
$$
where we used the substitutions $\kappa_x = k_xv_x,\kappa_y = k_yv_y, \kappa_z = k_zv_z$. Finally
$$
E = \frac{3\hbar L^3}{2\pi^2}\frac{1}{v_xv_yv_z}\int_0^{\kappa_D} d\kappa\frac{\kappa^3}{e^{\beta\hbar\kappa} - 1} + T \text{ independent part} = \frac{3L^3}{2\pi^2\hbar^3\beta^4}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} dx\frac{x^3}{e^{x} - 1} + T \text{ independent part},
$$
hence $C = \frac{\partial E}{\partial T} = \frac{6k_B^4L^3T^3}{\pi^2\hbar^3}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} dx\frac{x^3}{e^{x} - 1}$. We see that the result is similar to the one with the linear dispersion, the only difference is the factor $1/v_xv_yv_y$ instead of $1/v^3$.
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment