Final: Do task 4.4, draft task 4.5

Final/Final - Tight-binding propagation method.ipynb

@@ -1530,7 +1530,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 19,
    "metadata": {
     "deletable": false,
     "nbgrader": {
@@ -1548,8 +1548,21 @@
    "outputs": [],
    "source": [
     "def getU_CN(tau, H):\n",
-    "    # YOUR CODE HERE\n",
+    "    \"\"\"\n",
-    "    raise NotImplementedError()\n",
+    "    Calculates the time-propagation matrix for time-step tau using\n",
+    "    the Crank-Nicolson algorithm.\n",
+    "    \n",
+    "    Args:\n",
+    "        tau: time-step numeric to calculate the time-propagation matrix at\n",
+    "        H:   n by n array representing the hamiltonian matrix\n",
+    "\n",
+    "    Returns:\n",
+    "        The time-propagation matrix U(tau).\n",
+    "    \"\"\"\n",
+    "    \n",
+    "    n = len(H)\n",
+    "    \n",
+    "    return (np.eye(n) - 1j*tau*H/2)@np.linalg.inv(np.eye(n) + 1j*tau*H/2)\n",
     "\n",
     "# Yann notes that the definition of $U_{CN}(\\tau)$ here is a little\n",
     "# different from what Malte used on the slides. He recommends using\n",
@@ -1602,7 +1615,80 @@
     }
    },
    "source": [
-    "YOUR ANSWER HERE"
+    "% BEGIN NOTES %\n",
+    "Koen: vergeet niet -1j*tau mee te nemen.\n",
+    "\n",
+    "H = H_1 + H_2\n",
+    "H_1 = [[0,t],[t,0]] bla bla blok\n",
+    "H_2 \n",
+    "\n",
+    "verschil n even en n oneven?\n",
+    "\n",
+    "Zie slide 14 en omgeving in 08_Partial_Diff.\n",
+    "\n",
+    "Vergeet niet 1 op de diagonalon\n",
+    "% END NOTES %\n",
+    "\n",
+    "In the Trotter-Suzuki decomposition, we seek for block matrix forms for $\\mathbf{H_1}$ and $\\mathbf{H_2}$ such that their sum equals the original matrix $\\mathbf{H}$. The exponential is than done for each block instead of the whole matrix, which is a lot less complicated. Afterwards the result is combined such that\n",
+    "$$\n",
+    "e^{-i\\tau\\mathbf{H}} = e^{-i \\tau\\left( \\mathbf{H_1} + \\mathbf{H_2} \\right)} \\approx e^{-i\\tau\\mathbf{H_1}} \\cdot e^{-i\\tau\\mathbf{H_2}}\n",
+    "$$\n",
+    "\n",
+    "Here are the forms for the matrices. For $\\mathbf{H}$ an $n \\times n$ matrix, the form for $n$ odd is written down below. For $n$ even, the last row and column for both matrices can be omitted to see the shape.\n",
+    "\n",
+    "$$\n",
+    "\\mathbf{H_1} =\n",
+    "\\begin{pmatrix}\n",
+    "0 & t &   &   &        &   &   &  \\\\\n",
+    "t & 0 &   &   &        &   &   &  \\\\\n",
+    "  &   & 0 & t &        &   &   &  \\\\\n",
+    "  &   & t & 0 &        &   &   &  \\\\\n",
+    "  &   &   &   & \\ddots &   &   &  \\\\\n",
+    "  &   &   &   &        & 0 & t &  \\\\\n",
+    "  &   &   &   &        & t & 0 &  \\\\\n",
+    "  &   &   &   &        &   &   & 0\n",
+    "\\end{pmatrix}\n",
+    "\\qquad\n",
+    "\\mathbf{H_2} =\n",
+    "\\begin{pmatrix}\n",
+    "0 &   &   &   &   &        &   &  \\\\\n",
+    "  & 0 & t &   &   &        &   &  \\\\\n",
+    "  & t & 0 &   &   &        &   &  \\\\\n",
+    "  &   &   & 0 & t &        &   &  \\\\\n",
+    "  &   &   & t & 0 &        &   &  \\\\\n",
+    "  &   &   &   &   & \\ddots &   &  \\\\\n",
+    "  &   &   &   &   &        & 0 & t\\\\\n",
+    "  &   &   &   &   &        & t & 0\n",
+    "\\end{pmatrix}\n",
+    "$$\n",
+    "\n",
+    "For the exponents, we find the following for $n$ odd, again with the remark that the result for $n$ even can be reached by omitting the last row and column for both matrices and calculating the exponents for them.\n",
+    "\\begin{align}\n",
+    "\\exp{\\mathbf{H_1}} &=\n",
+    "\\begin{pmatrix}\n",
+    "\\exp{\\begin{pmatrix}0 & -i\\tau{}t \\\\ -i\\tau{}t & 0\\end{pmatrix}} & & & &\\\\\n",
+    " & \\exp{\\begin{pmatrix}0 & -i\\tau{}t \\\\ -i\\tau{}t & 0\\end{pmatrix}} & & &\\\\\n",
+    " & & \\ddots & &\\\\\n",
+    " & & & \\exp{\\begin{pmatrix}0 & -i\\tau{}t \\\\ -i\\tau{}t & 0\\end{pmatrix}} &\\\\\n",
+    " & & & & \\exp{0}\n",
+    "\\end{pmatrix}\n",
+    "\\\\\n",
+    "\\exp{\\mathbf{H_2}} &=\n",
+    "\\begin{pmatrix}\n",
+    "\\exp{0} & & & &\\\\\n",
+    " & \\exp{\\begin{pmatrix}0 & -i\\tau{}t \\\\ -i\\tau{}t & 0\\end{pmatrix}} & & &\\\\\n",
+    " & & \\exp{\\begin{pmatrix}0 & -i\\tau{}t \\\\ -i\\tau{}t & 0\\end{pmatrix}} & &\\\\\n",
+    " & & & \\ddots &\\\\\n",
+    " & & & & \\exp{\\begin{pmatrix}0 & -i\\tau{}t \\\\ -i\\tau{}t & 0\\end{pmatrix}}\\\\\n",
+    " & & & &\n",
+    "\\end{pmatrix}\n",
+    "\\end{align}\n",
+    "\n",
+    "As $\\exp{0} = 1$, we just calculate\n",
+    "$$\n",
+    "\\exp{\\begin{pmatrix}0 & -i\\tau{}t \\\\ -i\\tau{}t & 0\\end{pmatrix}}\n",
+    "= \\begin{pmatrix}\\cos{\\tau{}t} & -i\\sin{\\tau{}t} \\\\ -i\\sin{\\tau{}t} & \\cos{\\tau{}t}\\end{pmatrix}.\n",
+    "$$\n"
    ]
   },
   {
@@ -1625,6 +1711,26 @@
    "outputs": [],
    "source": [
     "def getU_TZ(tau, H):\n",
+    "    # TODO: Add docstring.\n",
+    "    \n",
+    "    # First we calculate the exponent of the block.\n",
+    "    exp_block = np.array([\n",
+    "        [    np.cos(tau*t),    -1j*np.sin(tau*t)],\n",
+    "        [-1j*np.sin(tau*t),        np.cos(tau*t)]\n",
+    "    ])\n",
+    "    \n",
+    "    # Now we use the block to calculate the exponent of matrices\n",
+    "    # H_1 and H_2.\n",
+    "    n = len(H)\n",
+    "    exp_H_1 = np.block([\n",
+    "        [1, 0],\n",
+    "        [0, np.kron(np.eye(n//2, dtype=int), exp_block)]\n",
+    "    ])\n",
+    "    exp_H_2 = np.block([\n",
+    "        [1,               0],\n",
+    "        [0, np.kron(np.eye(n//2, dtype=int), exp_block)]\n",
+    "    ])\n",
+    "    \n",
     "    # YOUR CODE HERE\n",
     "    raise NotImplementedError()\n",
     "\n",