Added file with the solutions to the exercises of lecture 6: bonds and...

Added file with the solutions to the exercises of lecture 6: bonds and spectra. It contains a python code to return the images for question 3.

src/6_bonds_and_spectra_solutions.md

+# Solutions for bonds and spectra
+
+### Exercise 1: linear triatomic molecule 
+
+1. 
+
+In 1D, there are two normal modes and in 3D there are 4 normal modes
+
+2. 
+
+$$
+\begin{cases}
+    m \ddot{x}_1 & = -\kappa(x_1-x_2)\\
+    M \ddot{x}_2 & = -\kappa(2x_2-x_1-x_3)\\ 
+    m \ddot{x}_1 & = -\kappa(x_1-x_2)
+\end{cases}
+$$
+Where $m$ is the mass of the oxygen atoms and $M$ the mass of the carbon atom.
+
+3.
+
+$$
+\omega = \frac{\kappa}{m}
+$$
+
+4.
+
+$$
+\frac{x_1}{x_2} = \frac{M}{2m}
+$$
+
+5.
+
+$$
+\omega = \sqrt{\frac{\kappa(2m+M)}{mM}}
+$$
+
+### Exercise 2: Lennard-Jones potential
+
+1. 
+
+See lecture slides/internet
+
+2. 
+
+The equilibrium position is $r_0 = 2^{1/6}\sigma$. The energy at the inter atomic distance $r_0$ is given by:
+$$
+U(r_0) = -\epsilon
+$$
+
+3.
+
+$$
+U(r) = -\epsilon + \frac{\kappa}{2}(r-r_0)^2
+$$
+Where $\kappa = \frac{72\epsilon}{2^{1/3}\sigma^2}$
+
+4.
+
+The ground state energy is given by
+$$
+E_0 = -\epsilon+\frac{1}{2}\sqrt{\frac{2\kappa}{m}}
+$$
+And the breaking energy is given by
+$$
+E_{break} = \epsilon - \frac{1}{2}\sqrt{\frac{2\kappa}{m}}
+$$
+
+5.
+
+Distance from which $U(r)$ becomes anharmonic:
+$$
+r_{anharmonic} = \frac{62^{1/6}\sigma}{7}
+$$
+Number of phonons that fit in the potential before it becomes anharmonic
+$$
+n = \frac{36}{49}\frac{\epsilon}{\hbar\omega}-\frac{1}{2}
+$$
+
+### Exercise 3: Numerical simulation
+1.
+
+2.
+
+```python
+import numpy as np
+from numpy import linalg as LA
+import matplotlib.pyplot as plt
+
+# Creating function that initializes the spring matrix
+def initial_mat(N):
+    K = np.zeros((N,N), dtype = int)
+    b = np.ones(N-1)
+    np.fill_diagonal(K, 2); np.fill_diagonal(K[1:], -b); np.fill_diagonal(K[:, 1:], -b) 
+    K[0,0] = K[-1,-1] = 1
+    omega = np.sqrt(np.abs(LA.eigvalsh(K)))
+    
+    # outputting a histogram of the data
+    plt.figure()
+    plt.hist(omega, bins = 30)
+    plt.xlabel("$\omega$")
+    plt.ylabel("Number of levels per eigenfrequency")
+    plt.title('Number of levels per eigenfrequencies for N = %d' %N)
+
+# Running the code
+initial_mat(5)
+initial_mat(200)
+```
+
+3.
+
+```python
+import numpy as np
+from numpy import linalg as LA
+import matplotlib.pyplot as plt
+
+# Creating mass matrix
+def mass_matrix(N, m1, m2):
+    M = np.zeros((N,N), dtype = int)
+    j = np.linspace(0, N-1, N)
+    for j in range(N):
+        if j%2 ==0:
+            M[j, j] = m1
+        else:
+            M[j, j] = m2
+    return M
+
+# Creating function that initializes the spring matrix
+def initial_mat(N,M):
+    K = np.zeros((N,N), dtype = int)
+    b = np.ones(N-1)
+    np.fill_diagonal(K, 2); np.fill_diagonal(K[1:], -b); np.fill_diagonal(K[:, 1:], -b) 
+    K[0,0] = K[-1,-1] = 1
+    MK = np.dot(LA.inv(M), K)
+    omega = np.sqrt(np.abs(LA.eigvalsh(MK)))
+    
+    # outputting a histogram of the data
+    plt.figure()
+    plt.hist(omega, bins = 30)
+    plt.xlabel("$\omega$")
+    plt.ylabel("Number of levels per eigenfrequency")
+
+# Defining variables
+N = 200
+m1 = 1; m2 = 4
+
+# Running the code
+M = mass_matrix(N, m1, m2)
+initial_mat(N, M)
+```
+Where in this simulation, every even numbered atom in the chain has 4 times the mass of every odd numbered atom
+