Lab 8 - SHO Solutions.ipynb

{
 "cells": [
  {
   "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": 2,
   "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": 3,
   "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": 3,
     "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": 4,
   "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": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n*a.dag() - a.dag()*n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 5,
     "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": 6,
   "metadata": {},
   "outputs": [],
   "source": [
    "psi = basis(N,2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "inputHidden": false,
    "outputHidden": false
   },
   "outputs": [
    {
     "data": {
      "text/latex": [
       "Quantum object: dims = [[10], [1]], shape = (10, 1), type = ket\\begin{equation*}\\left(\\begin{array}{*{11}c}0.0\\\\0.0\\\\1.0\\\\0.0\\\\0.0\\\\0.0\\\\0.0\\\\0.0\\\\0.0\\\\0.0\\\\\\end{array}\\right)\\end{equation*}"
      ],
      "text/plain": [
       "Quantum object: dims = [[10], [1]], shape = (10, 1), type = ket\n",
       "Qobj data =\n",
       "[[ 0.]\n",
       " [ 0.]\n",
       " [ 1.]\n",
       " [ 0.]\n",
       " [ 0.]\n",
       " [ 0.]\n",
       " [ 0.]\n",
       " [ 0.]\n",
       " [ 0.]\n",
       " [ 0.]]"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "psi"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "Quantum object: dims = [[10], [1]], shape = (10, 1), type = ket\\begin{equation*}\\left(\\begin{array}{*{11}c}0.0\\\\0.0\\\\0.0\\\\1.732\\\\0.0\\\\0.0\\\\0.0\\\\0.0\\\\0.0\\\\0.0\\\\\\end{array}\\right)\\end{equation*}"
      ],
      "text/plain": [
       "Quantum object: dims = [[10], [1]], shape = (10, 1), type = ket\n",
       "Qobj data =\n",
       "[[ 0.        ]\n",
       " [ 0.        ]\n",
       " [ 0.        ]\n",
       " [ 1.73205081]\n",
       " [ 0.        ]\n",
       " [ 0.        ]\n",
       " [ 0.        ]\n",
       " [ 0.        ]\n",
       " [ 0.        ]\n",
       " [ 0.        ]]"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a.dag()*psi"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a.dag()*basis(N,2) == sqrt(3)*basis(N,3)"
   ]
  },
  {
   "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": 10,
   "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 & 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 & 3.0\\\\0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 & 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.          1.          0.          0.          0.          0.          0.\n",
       "   0.          0.          0.        ]\n",
       " [ 0.          0.          1.41421356  0.          0.          0.          0.\n",
       "   0.          0.          0.        ]\n",
       " [ 0.          0.          0.          1.73205081  0.          0.          0.\n",
       "   0.          0.          0.        ]\n",
       " [ 0.          0.          0.          0.          2.          0.          0.\n",
       "   0.          0.          0.        ]\n",
       " [ 0.          0.          0.          0.          0.          2.23606798\n",
       "   0.          0.          0.          0.        ]\n",
       " [ 0.          0.          0.          0.          0.          0.\n",
       "   2.44948974  0.          0.          0.        ]\n",
       " [ 0.          0.          0.          0.          0.          0.          0.\n",
       "   2.64575131  0.          0.        ]\n",
       " [ 0.          0.          0.          0.          0.          0.          0.\n",
       "   0.          2.82842712  0.        ]\n",
       " [ 0.          0.          0.          0.          0.          0.          0.\n",
       "   0.          0.          3.        ]\n",
       " [ 0.          0.          0.          0.          0.          0.          0.\n",
       "   0.          0.          0.        ]]"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "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 & 3.0 & 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": 11,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a.dag()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 12.7:\n",
    "First, define $\\hat{X}$ and $\\hat{P}$ operators"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [],
   "source": [
    "X = 1/2 * (a + a.dag())\n",
    "P = 1/2j * (a - a.dag())"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [],
   "source": [
    "psi = 1/sqrt(2)*(basis(N,1)+basis(N,2))\n",
    "ex = psi.dag()*X*psi\n",
    "exq = psi.dag()*X*X*psi\n",
    "ep = psi.dag()*P*psi\n",
    "epq = psi.dag()*P*P*psi"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [],
   "source": [
    "deltaX = sqrt(exq[0][0][0] - ex[0][0][0]**2)\n",
    "deltaP = sqrt(epq[0][0][0] - ep[0][0][0]**2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(1.4142135623730951+0j)"
      ]
     },
     "execution_count": 15,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "deltaX * deltaP * 2 # compare to uncertainty relation (ΔxΔp >= 1/2)\n",
    "# the factor of two is to convert from the unitless version of the operator"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Alternatively, we can find the indeterminacy bound for ΔX and ΔP (the unitless operators): $$\\Delta X \\Delta P \\geq \\frac{1}{2}\\left|\\left\\langle \\left[\\hat{X},\\hat{P}\\right] \\right\\rangle\\right|$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.24999999999999994"
      ]
     },
     "execution_count": 31,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "1/2*(psi.dag()*commutator(X,P)*psi).norm()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Which is also satisfied by the calculated value (1.41 > 0.25)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem 12.8:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [],
   "source": [
    "psi = 1/sqrt(2)*(basis(N,2)+basis(N,4))\n",
    "ex = psi.dag()*X*psi\n",
    "exq = psi.dag()*X*X*psi\n",
    "ep = psi.dag()*P*psi\n",
    "epq = psi.dag()*P*P*psi"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [],
   "source": [
    "deltaX = sqrt(exq[0][0][0] - ex[0][0][0]**2)\n",
    "deltaP = sqrt(epq[0][0][0] - ep[0][0][0]**2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(3.0413812651491092+0j)"
      ]
     },
     "execution_count": 19,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "deltaX * deltaP * 2 # to compare to book solution which uses the full x and p operators with units"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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