fixed latex rendering for nbviewer in 2.2 solution

2fe41b12 · Georg Wolfgang Winkler · 8e03403b · 2fe41b12
Commit 2fe41b12 authored 9 years ago by Georg Wolfgang Winkler
--- a/exercises/ex02_solution/anharmonic.ipynb
+++ b/exercises/ex02_solution/anharmonic.ipynb
@@ -26,22 +26,22 @@
    "\\end{equation}\n",
    "\n",
    "Let us now expand the $(a+a^{\\dagger})^4$ using the commutator relation $[a,a^{\\dagger}] = a a^{\\dagger} - a^{\\dagger} a = 1$ and the identity $a^{\\dagger} a = \\hat n$:\n",
-    "\\begin{align}\n",
+    "$$\\begin{align}\n",
    " ( a + a^{\\dagger})^4  &=  (a a + a a^{\\dagger} + a^{\\dagger} a + a^{\\dagger} a^{\\dagger})^2 \\\\\n",
    "  &=  ( a a + 2 a^{\\dagger} a + 1 + a^{\\dagger} a^{\\dagger})^2 \\\\\n",
    "  &=  a^4 + 2 a a \\hat n + 2 a a + 2 \\hat n a a + 6 \\hat n^2 + 6 \\hat n + 3 + 2 \\hat n a^{\\dagger} a^{\\dagger} \\\\\n",
    "  &\\quad + 2 a^{\\dagger} a^{\\dagger} \\hat n + 2 a^{\\dagger} a^{\\dagger} + (a^{\\dagger})^4\n",
-    "\\end{align}\n",
+    "\\end{align}$$\n",
    "\n",
    "Using $a | n \\rangle = \\sqrt{n} | n -1 \\rangle$, $ a^{\\dagger} | n \\rangle = \\sqrt{n+1} | n+1 \\rangle$ and $ \\langle m | n \\rangle = \\delta_{m,n}$\n",
    "we obtain following non-zero matrix elements:\n",
-    "\\begin{align}\n",
+    "$$\\begin{align}\n",
    "    \\langle n | 6 \\hat n^2 + 6 \\hat n + 3 | n \\rangle &= 6 n^2 + 6 n + 3 \\\\\n",
    "    \\langle n + 2 | 2 \\hat n a^{\\dagger} a^{\\dagger} + 2 a^{\\dagger} a^{\\dagger} + 2 a^{\\dagger} a^{\\dagger} \\hat n | n \\rangle & = \\langle n | 2 a a \\hat n + 2 a a + 2 \\hat a a | n + 2\\rangle \\\\\n",
    "    & =(4 n + 6) \\sqrt{ (n+1) (n+2) } \\\\\n",
    "    \\langle n + 4 | (a^{\\dagger})^4 | n \\rangle & = \\langle n | a^4 | n+4 \\rangle \\\\\n",
    "    & = \\sqrt{(n+1) \\cdot (n+2) \\cdot (n+3) \\cdot (n+4)}\n",
-    "\\end{align}\n",
+    "\\end{align}$$\n",
    "\n",
    "A solution code setting up and diagonalizing a matrix for a given cutoff $N$ is given below.\n",
    "The dependence of the  energy spectrum on the anharmonicity $K$ is shown in figure at the bottom.\n",
@@ -188,7 +188,7 @@
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
-   "version": "3.4.3"
+   "version": "3.4.4"
  }
 },
 "nbformat": 4,

 %% Cell type:markdown id: tags:

 # Anharmonic Oscillator

 %% Cell type:markdown id: tags:

 First we set $\hbar = m = 1$ to simplify the terms
 \begin{eqnarray}
 a=\sqrt{\frac{\omega}{2}} x + \frac{ip}{\sqrt{2\omega}}  \\
 a^{\dagger}=\sqrt{\frac{\omega}{2}} x - \frac{ip}{\sqrt{2\omega}} \, .
 \end{eqnarray}
 Adding the two operators results in
 \begin{equation}
 a + a^{\dagger} = \sqrt{2\omega} x \, ,
 \end{equation}
 hence the anharmonic part of the Hamiltonian is
 \begin{equation}
 H_{\text{anharmonic}} = \frac{K}{4\omega^2} (a + a^{\dagger})^4 \, .
 \end{equation}

 Let us now expand the $(a+a^{\dagger})^4$ using the commutator relation $[a,a^{\dagger}] = a a^{\dagger} - a^{\dagger} a = 1$ and the identity $a^{\dagger} a = \hat n$:
-\begin{align}
+$$\begin{align}
 ( a + a^{\dagger})^4  &=  (a a + a a^{\dagger} + a^{\dagger} a + a^{\dagger} a^{\dagger})^2 \\
  &=  ( a a + 2 a^{\dagger} a + 1 + a^{\dagger} a^{\dagger})^2 \\
  &=  a^4 + 2 a a \hat n + 2 a a + 2 \hat n a a + 6 \hat n^2 + 6 \hat n + 3 + 2 \hat n a^{\dagger} a^{\dagger} \\
  &\quad + 2 a^{\dagger} a^{\dagger} \hat n + 2 a^{\dagger} a^{\dagger} + (a^{\dagger})^4
-\end{align}
+\end{align}$$

 Using $a | n \rangle = \sqrt{n} | n -1 \rangle$, $ a^{\dagger} | n \rangle = \sqrt{n+1} | n+1 \rangle$ and $ \langle m | n \rangle = \delta_{m,n}$
 we obtain following non-zero matrix elements:
-\begin{align}
+$$\begin{align}
    \langle n | 6 \hat n^2 + 6 \hat n + 3 | n \rangle &= 6 n^2 + 6 n + 3 \\
    \langle n + 2 | 2 \hat n a^{\dagger} a^{\dagger} + 2 a^{\dagger} a^{\dagger} + 2 a^{\dagger} a^{\dagger} \hat n | n \rangle & = \langle n | 2 a a \hat n + 2 a a + 2 \hat a a | n + 2\rangle \\
    & =(4 n + 6) \sqrt{ (n+1) (n+2) } \\
    \langle n + 4 | (a^{\dagger})^4 | n \rangle & = \langle n | a^4 | n+4 \rangle \\
    & = \sqrt{(n+1) \cdot (n+2) \cdot (n+3) \cdot (n+4)}
-\end{align}
+\end{align}$$

 A solution code setting up and diagonalizing a matrix for a given cutoff $N$ is given below.
 The dependence of the  energy spectrum on the anharmonicity $K$ is shown in figure at the bottom.
 We observe the expected equally spaced energy spectrum of the harmonic oscillator at $K=0$.
 For the anharmonic oscillator ($K > 0$) the difference between two consecutive energy levels grows with increasing anharmonicity and increasing energy.

 %% Cell type:code id: tags:

 ``` python
 import numpy as np
 import numpy.linalg as la
 from numpy import sqrt

 # interactive plots
 #%matplotlib notebook
 # nice inline plots
 %matplotlib inline

 import matplotlib.pyplot as plt
 plt.rcParams['figure.figsize'] = 16, 9
 ```

 %% Cell type:code id: tags:

 ``` python
 def anharm_mtx(N,K):
    # set m = hbar = 1
    w = 1
    f = K/(4*w*w)
    mtx = np.zeros((N,N))

    for row in range(N):
        for col in range(N):
            n = min(row,col)
            if col == row:
                mtx[row,col] += w*(row+0.5)
                mtx[row,col] += f*3*(row**2+(row+1)**2)
            if col == row + 2 or col + 2 == row:
                mtx[row,col] += f*(4*n+6)*sqrt((n+1)*(n+2))
            if col == row + 4 or col + 4 == row:
                 mtx[row,col] += f*sqrt((n+1)*(n+2)*(n+3)*(n+4))
    return mtx

 def anharm(N,K,verbose=False):
    # set m = hbar = 1
    mtx = anharm_mtx(N,K)
    if verbose:
        print('=== Hamiltonian for N=%s, K=%s ===' % (N, K))
        print(np.around(mtx, decimals=1))
    vals,vecs = la.eig(mtx)
    return np.sort(vals)
 ```

 %% Cell type:code id: tags:

 ``` python
 N = 100
 keeps = 5
 Ks = np.arange(0,5,0.1)

 markers = ['+', 'x', '-', 'o']
 colors = ['b', 'r', 'g', 'k', 'y']
 plt.figure()
 imark = 0
 for N in [20,60,100]:
    vals = np.zeros((len(Ks), keeps))
    for k,K in enumerate(Ks):
        eigs = anharm(N=N, K=K)
        for i in range(keeps):
            vals[k,i] = eigs[i]
    icolor = 0
    for i in range(keeps):
        plt.plot(Ks, vals[:,i], markers[imark % len(markers)] + colors[icolor % len(colors)],
                 label='$E_%d, N=%s$' % (i, N))
        icolor += 1
    imark += 1
 plt.legend(loc='best',ncol=3)
 plt.ylim(0,30)
 plt.xlabel('K')
 plt.ylabel('E')
 plt.grid()
 plt.show()
 ```

 %% Output



 %% Cell type:markdown id: tags:

 While for three lowest energy states ($E_0, E_1, E_2$) a cutoff of $N = 20$ eigen-states of the harmonic oscillator yields good approximations for $K < 5$, we need to use at least $N=60$ eigen-states to approximate the 3rd and 4th excited state properly.

 %% Cell type:code id: tags:

 ``` python
 ```