Commit 24da1992 by Anton Akhmerov

### 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\$.
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