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  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Kana\u0142y kwantowe\n",
      "===============\n",
      "\n",
      "Plan\n",
      "----\n",
      "1. Kana\u0142y kwantowe: definicja i zastosowania\n",
      "2. Reprezentacja Kraussa kana\u0142\u00f3w kwantowych\n",
      "3. Przyk\u0142ady kana\u0142\u00f3w kwantowych\n",
      "4. Uk\u0142ady z\u0142o\u017cone\n",
      "5. Wizualizacja wp\u0142ywu kana\u0142\u00f3w kwantowych na w\u0142aso\u015bci stan\u00f3w\n",
      "6. Zadania"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Kana\u0142y kwantowe, inaczej opracje kwantowe, to najbardziej podstawowa forma zapisu dla transformacji stanu kwantowego bez wnikania w wewn\u0119trzn\u0105 natur\u0119 procesu. Najpro\u015bciej m\u00f3wi\u0105c kana\u0142 kwantowy to operacja\n",
      "\n",
      "$\\rho\\rightarrow \\rho'=E(\\rho)$\n",
      "\n",
      "Opis taki ma szereg zastosowa\u0144. Mo\u017ce obrazowa\u0107 ewolucj\u0119 kwantow\u0105, dekoherencj\u0119, operacje kwantowe zwi\u0105zane z kwantowymi bramkami logicznymi. \n",
      "\n",
      "Tutaj ograniczymy si\u0119 do przypadku, gdy $\\rho'$ jest stanem, to znaczy zak\u0142adamy, \u017ce kana\u0142 kwantowy ma __reprezentacj\u0119 Kraussa__: \n",
      "\n",
      "$E(\\rho)=\\sum_i K_i \\rho K_i^\\dagger$\n",
      "\n",
      "ograniczenie to oznacza, \u017ce nie rozwa\u017camy problem\u00f3w niemarkowowskich (zale\u017cnych od czasu) oraz pomiar\u00f3w kwantowych. \n",
      "\n",
      "Operatory  $K_i$ spe\u0142nia\u0107 musz\u0105 warunek:\n",
      "\n",
      "$\\sum_i K_i^\\dagger K_i\\le\\mathbf{I}$\n",
      " \n",
      "__Uwaga__ warunek $\\sum_i K_i^\\dagger K_i=\\mathbf{I}$ gwarantuje, \u017ce $E(\\rho)$ jest stanem\n"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Dla zilustriowania uzyteczno\u015bci formalizmu Kraussa rozwa\u017cmy ewolucj\u0119 unitarn\u0105 uk\u0142adu $A$ i jego \"otoczenia\" $B$:\n",
      "\n",
      "$\\rho^{AB}=U^{AB} \\rho^A\\otimes |i_B\\rangle\\langle i_B| U^{AB\\dagger}$\n",
      "\n",
      "obliczaj\u0105c \u015blad cz\u0119\u015bciowy: \n",
      "\n",
      "$\\rho_A'=\\mbox{Tr}_B \\rho^{AB} =\\sum_n \\langle e_n^B|U^{AB}|i^B\\rangle \\rho^A \\langle i^B |U^{AB\\dagger} |e_n^B\\rangle$\n",
      "\n",
      "wprowadzaj\u0105c $K_n =\\langle e_n^B|U^{AB}|i^B\\rangle$\n",
      "\n",
      "otrzymujemy \n",
      "\n",
      "$E(\\rho^A)=\\sum_n K_n\\rho^A k_n^\\dagger$\n",
      "\n",
      "Zauwa\u017cmy, \u017ce \n",
      "\n",
      "$\\sum_n K_n^\\dagger K_n=\\mathbf{I}$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Dalej zbudujemy implementacj\u0119 numeryczn\u0105 najbardziej typowych kana\u0142\u00f3w kwantowych. Ograniczamy si\u0119 do kana\u0142\u00f3w kwantowych okre\u015blonych na stanach qubit\u00f3w:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from qutip import *\n",
      "from pylab import *\n",
      "%pylab inline"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Populating the interactive namespace from numpy and matplotlib\n"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stderr",
       "text": [
        "WARNING: pylab import has clobbered these variables: ['power', 'linalg', 'draw_if_interactive', 'random', 'save', 'load', 'info', 'fft']\n",
        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Wprowad\u017amy uproszczone oznaczenia dla wektor\u00f3w bazy:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "up=basis(2,0)\n",
      "dn=basis(2,1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "oraz macierzy bazowych"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "rpp=up*up.dag()\n",
      "rpm=up*dn.dag()\n",
      "rmp=rpm.trans()\n",
      "rmm=dn*dn.dag()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 3
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "nast\u0119pnie zdefiniujemy funkcj\u0119 depol() opisuj\u0105c\u0105 __kana\u0142 depolaryzacujmy__:\n",
      "\n",
      "$E(\\rho)=\\sum_{n=0}^4  K_n \\rho K_n$, \n",
      "\n",
      "gdzie $K_0=\\sqrt{1-p}\\mathbf{I}$ za\u015b $K_i=\\sqrt{p/3}\\sigma_i$, $i=x,y,x$"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "def depol(rho,p):\n",
      "    K0=sqrt(1.0-p)*qeye(2)\n",
      "    K1=sqrt(p/3.0)*sigmax()\n",
      "    K2=sqrt(p/3.0)*sigmay()\n",
      "    K3=sqrt(p/3.0)*sigmaz()\n",
      "    x=K0*rho*K0.dag()+K1*rho*K1.dag()+K2*rho*K2.dag()+K3*rho*K3.dag()\n",
      "    return x"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 4
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Mo\u017cemy sprawdzi\u0107 dla wybranych $p$ warunek zachowania \u015bladu:\n",
      "$\\sum_n K_n^\\dagger K_n=\\mathbf{I}$:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "p=0.8\n",
      "K0=sqrt(1.0-p)*qeye(2)\n",
      "K1=sqrt(p/3.0)*sigmax()\n",
      "K2=sqrt(p/3.0)*sigmay()\n",
      "K3=sqrt(p/3.0)*sigmaz()\n",
      "K0*K0.dag()+K1*K1.dag()+K2*K2.dag()+K3*K3.dag()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "\\begin{equation}\\text{Quantum object: dims = [[2], [2]], shape = [2, 2], type = oper, isHerm = True}\\\\[1em]\\begin{pmatrix}1.0 & 0.0\\\\0.0 & 1.0\\\\\\end{pmatrix}\\end{equation}"
       ],
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 5,
       "text": [
        "Quantum object: dims = [[2], [2]], shape = [2, 2], type = oper, isherm = True\n",
        "Qobj data =\n",
        "[[ 1.  0.]\n",
        " [ 0.  1.]]"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Kolejnym u\u017cytecznym kana\u0142em jest kana\u0142 __t\u0142umienia fazy__ (phase damping channel). \n",
      "\n",
      "$K_0=|1\\rangle\\langle 1| +\\sqrt{1-p}|0\\rangle\\langle 0|$, $K_1=\\sqrt{p} |1\\rangle\\langle 0|$\n",
      "\n",
      "dzia\u0142anie takiego kana\u0142u opisuje funkcja damp():"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "def damp(rho,p):\n",
      "    K0=rpp+sqrt(1.0-p)*rmm\n",
      "    K1=sqrt(p)*rpm\n",
      "    x=K0*rho*K0.dag()+K1*rho*K1.dag()\n",
      "    return x"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 6
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Ostanim z rozwa\u017canych typ\u00f3w kana\u0142\u00f3w jest kana\u0142 odwr\u00f3cenia fazu (phase flip channel) opisany funkcj\u0105 flip():"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "def flip(rho,p):\n",
      "    K0=sqrt(p)*sigmaz()\n",
      "    K1=sqrt(1.0-p)*qeye(2)\n",
      "    x=K0*rho*K0.dag()+K1*rho*K1.dag()\n",
      "    return x"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 7
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Teraz jeste\u015bmy gotowi do przygotowania wizualizacji wybranej charakterystyki kana\u0142u. Za\u0142o\u017cymy na wej\u015bciu stan czysty: "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "psi=(up+dn).unit()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 8
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "i jego macierz g\u0119sto\u015bci:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "rho=ket2dm(psi)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 9
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "W nast\u0119pnym kroku zdefiniujemy listy z warto\u015bciami $p$ oraz wynikami:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "plist=linspace(0,1.0,10.0)\n",
      "slist=zeros(len(plist))"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 10
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Niech intersuj\u0105c\u0105 charakterystyk\u0105 stanu wyj\u015bciowego b\u0119dzie entropia von Neumanna:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "for i1 in range(len(plist)):\n",
      "    p=plist[i1]\n",
      "    rho1=depol(rho,p)\n",
      "    slist[i1]=entropy_vn(rho1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 11
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Uzyskany wynik mo\u017cemy wykre\u015bli\u0107:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "plot(plist,slist,'-',label='', linewidth=4)\n",
      "xlabel('p',fontsize=20)\n",
      "ylabel('SvN',fontsize=20)\n",
      "show()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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AM8+4/WOJiFzC+pfo9PR0p99LsaDQ6XQwmUy1z00mE/R6vU2/EydOYO7cuTAY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\nKIiIqEEMCiIiahCDgoiIGvT/+y1xhix0fgEAAAAASUVORK5CYII=\n",
       "text": [
        "<matplotlib.figure.Figure at 0x2b459d0>"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Teraz rozwa\u017cmy problem bardziej z\u0142o\u017cony. Za\u0142\u00f3\u017cmy, \u017ce w parze qubit\u00f3w przygotowanych w stanie  Bella:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "BB0=(tensor(rpp,rmm)+tensor(rpm,rmp)+tensor(rmp,rpm)+tensor(rmm,rpp))/2.0"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 13
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "jeden z qubit\u00f3w podlega kana\u0142owi depolaryzacyjnemu z $p=0.4$. Sp\u00f3bujmy okre\u015bli\u0107 wp\u0142yw tego kana\u0142u na warto\u015b\u0107 spl\u0105tania stanu pary qubit\u00f3w. Rozpocznujmy od obliczenia dzia\u0142ania depolaryzacji na stany bazowe:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "p=0.4\n",
      "rpp1=depol(rpp,p)\n",
      "rpm1=depol(rpm,p)\n",
      "rmp1=depol(rmp,p)\n",
      "rmm1=depol(rmm,p)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 14
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "nast\u0119pnie przekszta\u0142\u0107my stan Bella:"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "BB0_E=(tensor(rpp1,rmm)+tensor(rpm1,rmp)+tensor(rmp1,rpm)+tensor(rmm1,rpp))/2.0 "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 15
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "i ostatecznie obliczmy jego spl\u0105tanie wykorzystuj\u0105c funkcj\u0119 concurrence():"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "concurrence(BB0_E)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 16,
       "text": [
        "0.20000000000000012"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Zadania\n",
      "-------\n",
      "\n",
      "1. Wykre\u015bli\u0107 entropi\u0119 w  funkcji $p$ dla pozosta\u0142ych kana\u0142\u00f3w omawianych powy\u017cej\n",
      "2. Wykre\u015bli\u0107 spl\u0105tanie zdepolaryzowanego stanu Bella  w funkcji p. \n",
      "3. Wykre\u015bli\u0107 spl\u0105tanie  stanu Bella zmodyfikowanego ze pomoc\u0105 innych omawianych kana\u0142\u00f3w w funkcji p."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 16
    }
   ],
   "metadata": {}
  }
 ]
}