Lab 8 - SHO.ipynb

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  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Lab 8 - Simple Harmonic Oscillator states\n",
    "Problems from Chapter 12"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "from numpy import sqrt\n",
    "from qutip import *"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Define the standard operators"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [],
   "source": [
    "N = 10  # pick a size for our state-space\n",
    "a = destroy(N)\n",
    "n = a.dag()*a"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 12.1:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "Quantum object: dims = [[10], [10]], shape = (10, 10), type = oper, isherm = True\\begin{equation*}\\left(\\begin{array}{*{11}c}1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 1.000 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 1.000 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 1.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 1.000 & 0.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & -9.0\\\\\\end{array}\\right)\\end{equation*}"
      ],
      "text/plain": [
       "Quantum object: dims = [[10], [10]], shape = (10, 10), type = oper, isherm = True\n",
       "Qobj data =\n",
       "[[ 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.]\n",
       " [ 0.  1.  0.  0.  0.  0.  0.  0.  0.  0.]\n",
       " [ 0.  0.  1.  0.  0.  0.  0.  0.  0.  0.]\n",
       " [ 0.  0.  0.  1.  0.  0.  0.  0.  0.  0.]\n",
       " [ 0.  0.  0.  0.  1.  0.  0.  0.  0.  0.]\n",
       " [ 0.  0.  0.  0.  0.  1.  0.  0.  0.  0.]\n",
       " [ 0.  0.  0.  0.  0.  0.  1.  0.  0.  0.]\n",
       " [ 0.  0.  0.  0.  0.  0.  0.  1.  0.  0.]\n",
       " [ 0.  0.  0.  0.  0.  0.  0.  0.  1.  0.]\n",
       " [ 0.  0.  0.  0.  0.  0.  0.  0.  0. -9.]]"
      ]
     },
     "execution_count": 19,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a*a.dag() - a.dag()*a"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 12.2:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "Quantum object: dims = [[10], [10]], shape = (10, 10), type = oper, isherm = False\\begin{equation*}\\left(\\begin{array}{*{11}c}0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\\1.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 1.414 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 1.732 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 2.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 2.236 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.449 & 0.0 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.646 & 0.0 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 2.828 & 0.0 & 0.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 3.000 & 0.0\\\\\\end{array}\\right)\\end{equation*}"
      ],
      "text/plain": [
       "Quantum object: dims = [[10], [10]], shape = (10, 10), type = oper, isherm = False\n",
       "Qobj data =\n",
       "[[ 0.          0.          0.          0.          0.          0.          0.\n",
       "   0.          0.          0.        ]\n",
       " [ 1.          0.          0.          0.          0.          0.          0.\n",
       "   0.          0.          0.        ]\n",
       " [ 0.          1.41421356  0.          0.          0.          0.          0.\n",
       "   0.          0.          0.        ]\n",
       " [ 0.          0.          1.73205081  0.          0.          0.          0.\n",
       "   0.          0.          0.        ]\n",
       " [ 0.          0.          0.          2.          0.          0.          0.\n",
       "   0.          0.          0.        ]\n",
       " [ 0.          0.          0.          0.          2.23606798  0.          0.\n",
       "   0.          0.          0.        ]\n",
       " [ 0.          0.          0.          0.          0.          2.44948974\n",
       "   0.          0.          0.          0.        ]\n",
       " [ 0.          0.          0.          0.          0.          0.\n",
       "   2.64575131  0.          0.          0.        ]\n",
       " [ 0.          0.          0.          0.          0.          0.          0.\n",
       "   2.82842712  0.          0.        ]\n",
       " [ 0.          0.          0.          0.          0.          0.          0.\n",
       "   0.          3.          0.        ]]"
      ]
     },
     "execution_count": 20,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n*a.dag() - a.dag()*n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 21,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n*a.dag() - a.dag()*n == a.dag()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 12.3 (use n=2 as a test-case):\n",
    "To define $|2\\rangle$ use the `basis(N,n)` command where `N` is the dimension of the vector, and `n` is the quantum number."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {},
   "outputs": [],
   "source": [
    "psi = basis(N,2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 12.5 and 12.6:\n",
    "These are simple, just view the matrix representation of the operators"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 12.7:\n",
    "First, define $\\hat{X}$ and $\\hat{P}$ operators"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "X=\n",
    "P="
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 92,
   "metadata": {},
   "outputs": [],
   "source": [
    "psi = 1/sqrt(2)*(basis(N,1)+basis(N,2))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 12.8:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [],
   "source": [
    "psi = 1/sqrt(2)*(basis(N,2)+basis(N,4))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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