Update docs/2_debye_model_solutions.md, docs/2_debye_model.md

docs/2_debye_model.md

@@ -419,7 +419,7 @@ What is the material-dependent parameter that plays a similar role in the Debye
 Hint: assume that the number of atoms is given by $N=V/a^3$. Discuss if the answer is reasonable.
 
 ### Exercise 1*: Deriving the density of states for the linear dispersion relation of the Debye model.
-In this lecture, we found that the linear dispersion considered in the Debye model yields a density of states $g(\oemga)\propto \omega^2$ (in three dimensions). In this exercise, we will practice this important derivation again and extend it to 1D and 2D. 
+In this lecture, we found that the linear dispersion considered in the Debye model yields a density of states $g(\omega)\propto \omega^2$ (in three dimensions). In this exercise, we will practice this important derivation again and extend it to 1D and 2D. 
 
 1. Write down the dispersion relation of the vibrational modes in the Debye model.
 2. Assume periodic boundary conditions. What is the distance between nearest-neighbour points in $\mathbf{k}$-space? What is the density of $\mathbf{k}$-points in 1, 2, and 3 dimensions?
@@ -429,9 +429,10 @@ In this lecture, we found that the linear dispersion considered in the Debye mod
 
 ### Exercise 2*: Debye model in 2D
 Here, we analyze the phonon energy and heat capacity of a two-dimensional Debye solid. 
+
 1. Formulate an integral expression for the energy stored in the vibrational modes of a two-dimensional Debye solid as a function of $T$.
-3. Calculate the heat capacity in the high $T$ limit.
-4. At low $T$, show that $C_V=KT^{n}$. Find $n$. Express $K$ as an indefinite integral (similarly to what was done during the lecture)[^3].
+2. Calculate the heat capacity in the high $T$ limit.
+3. At low $T$, show that $C_V=KT^{n}$. Find $n$. Express $K$ as an indefinite integral (similarly to what was done during the lecture)[^3].
 
 ### Exercise 3: Longitudinal and transverse vibrations with different sounds velocities 
 *(adapted from ex 2.6a of "The Oxford Solid State Basics" by S.Simon)*

docs/2_debye_model_solutions.md

@@ -67,44 +67,49 @@ ax.set_yticklabels(['$0$'])
 ax.legend();
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-### Exercise 1: Debye model: concepts.
-
-#### Question 1.
-$k = \frac{4\pi}{L}$ and $k = -\frac{4\pi}{L}$.  
-#### Question 2.
+### Exercise 1*: Deriving the density of states for the linear dispersion relation of the Debye model.
+1. $\omega = v_s|\mathbf{k}|$
+2. The distance between nearest-neighbour points in $\mathbf{k}$-space is $2\pi/L$. The density of $\mathbf{k}$-points in 1, 2, and 3 dimensions is $L/(2\pi)$, $L^2/(2\pi)^2$, and $L^3/(2\pi)^3$ respectively.
+3. Express the number of states between frequencies $0<\omega<\omega_0$ as an integral over k-space. Do so for 1D, 2D and 3D. Do not forget the possible polarizations. We assume that in $d$ dimensions there are $d_p$ polarizations.
+    \begin{align}
+    N_\text{states, 1D} & = 1_p \frac{L}{2\pi} \int_0^{k_0} 2 dk \\
+    N_\text{states, 2D} & = 2_p \left(\frac{L}{2\pi}\right)^2 \int_0^{k_0} 2\pi k dk \\
+    N_\text{states, 3D} & = 3_p  \left(\frac{L}{2\pi}\right)^3 \int_0^{k_0} 4\pi k^2 dk 
+    \end{align}
+4.   We use $k=\mathbf{k}| = \omega/v_s$ and $dk = d\omega/v_s$ to get 
+  \begin{align}
+  N_\text{states, 1D} & = 1_p \frac{L}{2\pi} \int_0^{\omega_0} 2 \frac{1}{v_s} d\omega  := \int_0^{\omega_0} 2 g_{1D}(\omega) d\omega\\
+  N_\text{states, 2D} & = 2_p \left(\frac{L}{2\pi}\right)^2 \int_0^{omega_0} 2\pi \frac{\omega}{v_s^2} d\omega := \int_0^{\omega_0} 2 g_{2D}(\omega) d\omega\\
+  N_\text{states, 3D} & = 3_p  \left(\frac{L}{2\pi}\right)^3 \int_0^{omega_0} \frac{\omega^2}{v_s^3} k^2 d\omega := \int_0^{\omega_0} 2 g_{3D}(\omega) d\omega
+  \end{align}
+  The integral boundaries set the frequency region in which you calculate the density of states.
+5. The density of states is the number of states per unit frequency. It has units of 1 over frequency
 
-#### Question 3.
-The number of states per $k$ or per frequency.
-#### Question 4.
+###  Exercise 2: Debye model in 2D.
+Here, we analyze the phonon energy and heat capacity of a two-dimensional Debye solid. 
+1. Formulate an integral expression for the energy stored in the vibrational modes of a two-dimensional Debye solid as a function of $T$.
+    \begin{align}
+    E & = \int_0^{\omega_D}(n_B(\omega(\mathbf{k}))+\frac{1}{2})\hbar\omega(\mathbf{k}) d\mathbf{k} \\
+    & = \int_0^{\omega_D}(n_B(\omega(\mathbf{k}))+\frac{1}{2})\hbar\omega(\mathbf{k}) d\mathbf{k} +E_0\\
+    & = \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\beta\hbar\omega_D}\frac{x^2}{e^{x} - 1}dx + E_0
+    \end{align}
 
   
+2. The high-$T$ limit implies $\beta \rightarrow 0$. Therefore, $n_B \approx = k_BT/\hbar\omega$, and the integral becomes particularly illuminating:
     $$
-g(\omega) = \frac{dN}{d\omega} = \frac{dN}{dk}\frac{dk}{d\omega} = \frac{1}{v}\frac{dN}{dk}.
+    E = \int_0^\omega_D \hbar\omega  n_B(\omega) g(\omega) d\omega \approx \int_0^\omega_D k_BT g(\omega) d\omega = N_\text{modes} k_B T
     $$
+    where we neglected the zero-point energy. In 3D, we have $N_\text{modes} = 3_p N_\text{atoms}$
 
-We assume that in $d$ dimensions there are $d$ polarizations.
-
-For 1D we have that $N = \frac{L}{2\pi}\int_{-k}^{k} dk$, hence $g(\omega) = \frac{L}{\pi v}$.
-
-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 $N = 3\left(\frac{L}{2\pi}\right)^3\int d^3k = 3\left(\frac{L}{2\pi}\right)^3\int 4\pi k^2dk$, hence $g(\omega) = \frac{3L^3\omega^2}{2\pi^2v^3}$.
-
-###  Exercise 2: Debye model in 2D.
-
-#### Question 1.
-See lecture notes.
-#### Question 2.
-
-\begin{align}
-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 + C.
-\end{align}
+3.  In the low temperature limit, the high-energy modes are not excited so we can safely let the upper boundary of the integral go to infinity. For convenience, we write $g(\omega) = alpha \omega$, with $\alpha = \frac{L^2}{\pi v_s^2}$. We get 
+    $$ 
+    E = \int_0^\omega_D\frac{\hbar\omega g(\omega)}{e^{\hbar\omega/k_BT}- 1}d\omega = \alpha\frac{k^3T^3}{\hbar^2}\int_)^\infty\frac{x^2}{e^x-1}dx 
+    $$
+    From which we find  $C_v = dE/dT =K T^2$ with 
     
-#### Question 3.
-High temperature implies $\beta \rightarrow 0$, hence $E = \frac{L^2}{\pi v^2\hbar^2\beta^3}\frac{(\beta\hbar\omega_D)^2}{2} + C$, 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 $2N = \int_{0}^{\omega_D}g(\omega)d\omega$.
-#### Question 4.
-In the low temperature limit we have that $\beta \rightarrow \infty$, hence $E \approx \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx + C = \frac{2\zeta(3)L^2}{\pi v^2\hbar^2\beta^3} + C$. 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.
+    $$
+    K = 3\alpha\frac{k^3}{\hbar^2}\int_)^\infty\frac{x^2}{e^x-1}dx 
+    $$
 
 ###  Exercise 3: Different phonon modes.
 #### Question 1.