{ "cells": [ { "cell_type": "markdown", "id": "e5e44fb3", "metadata": {}, "source": [ "## Ejemplo 8.2: Transmisión por entrelazamiento" ] }, { "cell_type": "markdown", "id": "71cd75b7", "metadata": {}, "source": [ "$\\newcommand{\\ket}[1]{|{#1}\\rangle}\\newcommand{\\bra}[1]{\\langle{#1}|}$La idea de la transmisión de información por entrelazamiento aparece representada en la siguiente figura:\n", "\n", "\"Medida\n", "\n", "Definimos el estado máximamente entrelazado\n", "\\begin{align*}\n", " \\delta^{AB} = \\frac{1}{2} \\left[\\begin{smallmatrix}\n", " 1 & 0 & 0 & 1\\\\\n", " 0 & 0 & 0 & 0\\\\\n", " 0 & 0 & 0 & 0\\\\\n", " 1 & 0 & 0 & 1 \\end{smallmatrix}\\right]\n", "\\end{align*}\n", "donde el superíndice $AB$ hace referencia a que la primera parte del estado pertenece a Alice y la segunda a Bob." ] }, { "cell_type": "code", "execution_count": 5, "id": "e51496e6", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "deltaAB =\n", " [[0.5 0. 0. 0.5]\n", " [0. 0. 0. 0. ]\n", " [0. 0. 0. 0. ]\n", " [0.5 0. 0. 0.5]]\n" ] } ], "source": [ "import numpy as np\n", "\n", "deltaAB = np.array([[1,0,0,1],[0,0,0,0],[0,0,0,0],[1,0,0,1]])/2\n", "print('deltaAB =\\n', deltaAB)" ] }, { "cell_type": "markdown", "id": "7625c52a", "metadata": {}, "source": [ "Consideramos los proyectores de medida:\n", "\\begin{align*}\n", "\\Pi_{0}^A=\\begin{bmatrix}1&0\\\\0&0\\end{bmatrix},\\quad\n", "\\Pi_{1}^A=\\begin{bmatrix}0&0\\\\0&1\\end{bmatrix},\\quad\n", "\\Pi_{+}^A=\\frac{1}{2}\\begin{bmatrix}+1&+1\\\\+1&+1\\end{bmatrix},\\quad\n", "\\Pi_{-}^A=\\frac{1}{2}\\begin{bmatrix}+1&-1\\\\-1&+1\\end{bmatrix}.\n", "\\end{align*}\n" ] }, { "cell_type": "code", "execution_count": 6, "id": "da91962a", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Pi0A =\n", " [[1 0]\n", " [0 0]] \n", "Pi1A =\n", " [[0 0]\n", " [0 1]] \n", "PipA =\n", " [[0.5 0.5]\n", " [0.5 0.5]] \n", "PimA =\n", " [[ 0.5 -0.5]\n", " [-0.5 0.5]]\n" ] } ], "source": [ "Pi0A = np.array([[1,0],[0,0]])\n", "Pi1A = np.array([[0,0],[0,1]])\n", "PipA = np.array([[+1,+1],[+1,+1]])/2\n", "PimA = np.array([[+1,-1],[-1,+1]])/2\n", "\n", "print('Pi0A =\\n', Pi0A, '\\nPi1A =\\n', Pi1A,'\\nPipA =\\n', PipA,'\\nPimA =\\n', PimA)" ] }, { "cell_type": "markdown", "id": "0a8faad0", "metadata": {}, "source": [ "Ahora, Alice aplica una medida a su parte del estado compartido con Bob:\n", "\n", "- Si $m=0$, Alice aplica la medida con respecto al POVM$_{01} = \\bigl\\{\\Pi^A_{0},\\Pi^A_{1}\\bigr\\}$.\n", "- Si $m=1$, Alice aplica la medida con respecto al POVM$_{+-} =\\bigl\\{\\Pi^A_{+},\\Pi^A_{-}\\bigr\\}$.\n", "\n", "Para aplicar la medida al estado conjunto, definimos las matrices de dimensión $4\\times 4$ dadas por $\\Pi_k^{AB} = \\Pi_k^A \\otimes \\boldsymbol{I}_2^B$, donde $k\\in\\{0,1,+,-\\}$. " ] }, { "cell_type": "code", "execution_count": 7, "id": "746f0b09", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Pi0AB =\n", " [[1 0 0 0]\n", " [0 1 0 0]\n", " [0 0 0 0]\n", " [0 0 0 0]] \n", "Pi1AB =\n", " [[0 0 0 0]\n", " [0 0 0 0]\n", " [0 0 1 0]\n", " [0 0 0 1]] \n", "PipAB =\n", " [[0.5 0. 0.5 0. ]\n", " [0. 0.5 0. 0.5]\n", " [0.5 0. 0.5 0. ]\n", " [0. 0.5 0. 0.5]] \n", "PimAB =\n", " [[ 0.5 0. -0.5 -0. ]\n", " [ 0. 0.5 -0. -0.5]\n", " [-0.5 -0. 0.5 0. ]\n", " [-0. -0.5 0. 0.5]]\n" ] } ], "source": [ "I2B = np.array([[1,0],[0,1]]) # Bob no hace nada en su parte\n", "# Definimos las medidas conjuntas:\n", "Pi0AB = np.kron(Pi0A,I2B)\n", "Pi1AB = np.kron(Pi1A,I2B)\n", "PipAB = np.kron(PipA,I2B)\n", "PimAB = np.kron(PimA,I2B)\n", "\n", "print('Pi0AB =\\n', Pi0AB, '\\nPi1AB =\\n', Pi1AB,'\\nPipAB =\\n', PipAB,'\\nPimAB =\\n', PimAB)" ] }, { "cell_type": "markdown", "id": "94c1b5e6", "metadata": {}, "source": [ "Si Alice aplica el POVM$_{01}= \\bigl\\{\\Pi_{0}^A,\\Pi_{1}^A\\bigr\\}$, y teniendo en cuenta que Bob no puede conocer el resultado de la medida, obtenemos el nuevo estado conjunto:\n", "\\begin{align*} \n", " \\theta_{01}^{AB} = \\Pi_{0}^{AB} \\delta^{AB} \\Pi_{0}^{AB} + \n", " \\Pi_{1}^{AB} \\delta^{AB} \\Pi_{1}^{AB} = \\frac{1}{2} \\left[\\begin{smallmatrix}\n", " 1 & 0 & 0 & 0\\\\\n", " 0 & 0 & 0 & 0\\\\\n", " 0 & 0 & 0 & 0\\\\\n", " 0 & 0 & 0 & 1 \\end{smallmatrix}\\right].\n", "\\end{align*}\n", "\n" ] }, { "cell_type": "code", "execution_count": 10, "id": "dfd352b5", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "theta01AB =\n", " [[0.5 0. 0. 0. ]\n", " [0. 0. 0. 0. ]\n", " [0. 0. 0. 0. ]\n", " [0. 0. 0. 0.5]]\n" ] } ], "source": [ "theta01AB = Pi0AB@deltaAB@Pi0AB + Pi1AB@deltaAB@Pi1AB\n", "\n", "print('theta01AB =\\n', theta01AB)" ] }, { "cell_type": "markdown", "id": "fd44255a", "metadata": {}, "source": [ "Si Alice aplica el POVM$_{+-} =\\bigl\\{\\Pi_{+}^A,\\Pi_{-}^A\\bigr\\}$, como Bob no conoce el resultado de la medida, obtenemos:\n", "\\begin{align*} \n", " \\theta_{+-}^{AB} = \\Pi_{+}^{AB} \\delta^{AB} \\Pi_{+}^{AB} + \n", " \\Pi_{-}^{AB} \\delta^{AB} \\Pi_{-}^{AB} = \\frac{1}{4} \\left[\\begin{smallmatrix}\n", " 1 & 0 & 0 & 1\\\\\n", " 0 & 1 & 1 & 0\\\\\n", " 0 & 1 & 1 & 0\\\\\n", " 1 & 0 & 0 & 1 \\end{smallmatrix}\\right].\n", "\\end{align*}" ] }, { "cell_type": "code", "execution_count": 11, "id": "cbb0598a", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "thetapmAB =\n", " [[0.25 0. 0. 0.25]\n", " [0. 0.25 0.25 0. ]\n", " [0. 0.25 0.25 0. ]\n", " [0.25 0. 0. 0.25]]\n" ] } ], "source": [ "thetapmAB = PipAB@deltaAB@PipAB + PimAB@deltaAB@PimAB\n", "\n", "print('thetapmAB =\\n', thetapmAB)" ] }, { "cell_type": "markdown", "id": "d54c0c2f", "metadata": {}, "source": [ "El estado conjunto tras aplicar la medida es diferente en los dos casos, $\\theta_{01}^{AB} \\neq \\theta_{+-}^{AB}$.\n", "\n", "Sin embargo, Bob no tiene acceso a este estado conjunto, sino solo a una parte del mismo. Para conocer el estado desde el punto de vista de Bob, debemos marginalizar $\\theta^{AB}$ eliminando la parte de Alice. Se tiene que\n", "\\begin{align*} \n", " \\sigma^B_{01} = \\text{Tr}_A\\bigl[\\theta_{01}^{AB}\\bigr] = \\frac{1}{2} \\left[\\begin{matrix} 1 & 0 \\\\ 0 & 1 \\end{matrix}\\right],\\qquad\n", " \\sigma^B_{+-} = \\text{Tr}_A\\bigl[\\theta_{+-}^{AB}\\bigr] = \\frac{1}{2} \\left[\\begin{matrix} 1 & 0 \\\\ 0 & 1 \\end{matrix}\\right].\n", "\\end{align*}" ] }, { "cell_type": "code", "execution_count": 13, "id": "df045697", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "sigma_01B =\n", " [[0.5 0. ]\n", " [0. 0.5]] \n", "sigma_pmB =\n", " [[0.5 0. ]\n", " [0. 0.5]]\n" ] } ], "source": [ "def traceA(delta): # traces out part A of a bipartite system\n", " return np.array([[delta[0, 0] + delta[2, 2], delta[0, 1] + delta[2, 3]],\n", " [delta[1, 0] + delta[3, 2], delta[1, 1] + delta[3, 3]]])\n", "\n", "sigma_01B = traceA(theta01AB)\n", "sigma_pmB = traceA(thetapmAB)\n", "\n", "print('sigma_01B =\\n', sigma_01B, '\\nsigma_pmB =\\n', sigma_pmB)" ] }, { "cell_type": "markdown", "id": "8e225f01", "metadata": {}, "source": [ "Podemos ver que $\\sigma^B_0 = \\sigma^B_1$. Así, Bob no puede distinguir ambos casos teniendo aceso únicamente a su parte del estado. Por tanto, la comunicación no es posible con este esquema." ] } ], "metadata": { "celltoolbar": "Raw Cell Format", "kernelspec": { "display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.10.11" } }, "nbformat": 4, "nbformat_minor": 5 }