{ "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": 2, "id": "e51496e6", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "deltaAB =\n", "\n", " 0.5000 0 0 0.5000\n", " 0 0 0 0\n", " 0 0 0 0\n", " 0.5000 0 0 0.5000\n", "\n" ] } ], "source": [ "deltaAB = 1/2*[[1,0,0,1];[0,0,0,0];[0,0,0,0];[1,0,0,1]]" ] }, { "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": 4, "id": "da91962a", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Pi0A =\n", "\n", " 1 0\n", " 0 0\n", "\n", "Pi1A =\n", "\n", " 0 0\n", " 0 1\n", "\n", "PipA =\n", "\n", " 0.5000 0.5000\n", " 0.5000 0.5000\n", "\n", "PimA =\n", "\n", " 0.5000 -0.5000\n", " -0.5000 0.5000\n", "\n" ] } ], "source": [ "Pi0A = [[1,0];[0,0]]\n", "Pi1A = [[0,0];[0,1]]\n", "PipA = [[+1,+1];[+1,+1]]/2\n", "PimA = [[+1,-1];[-1,+1]]/2" ] }, { "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": 5, "id": "746f0b09", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "I2B =\n", "\n", "Diagonal Matrix\n", "\n", " 1 0\n", " 0 1\n", "\n", "Pi0AB =\n", "\n", " 1 0 0 0\n", " 0 1 0 0\n", " 0 0 0 0\n", " 0 0 0 0\n", "\n", "Pi1AB =\n", "\n", " 0 0 0 0\n", " 0 0 0 0\n", " 0 0 1 0\n", " 0 0 0 1\n", "\n", "PipAB =\n", "\n", " 0.5000 0 0.5000 0\n", " 0 0.5000 0 0.5000\n", " 0.5000 0 0.5000 0\n", " 0 0.5000 0 0.5000\n", "\n", "PimAB =\n", "\n", " 0.5000 0 -0.5000 0\n", " 0 0.5000 0 -0.5000\n", " -0.5000 0 0.5000 0\n", " 0 -0.5000 0 0.5000\n", "\n" ] } ], "source": [ "I2B = eye(2) % Bob no hace nada en su parte\n", "% Definimos las medidas conjuntas:\n", "Pi0AB = kron(Pi0A,I2B)\n", "Pi1AB = kron(Pi1A,I2B)\n", "PipAB = kron(PipA,I2B)\n", "PimAB = kron(PimA,I2B)" ] }, { "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": 7, "id": "dfd352b5", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "theta01AB =\n", "\n", " 0.5000 0 0 0\n", " 0 0 0 0\n", " 0 0 0 0\n", " 0 0 0 0.5000\n", "\n" ] } ], "source": [ "theta01AB = Pi0AB*deltaAB*Pi0AB + Pi1AB*deltaAB*Pi1AB" ] }, { "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": 8, "id": "cbb0598a", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "thetapmAB =\n", "\n", " 0.2500 0 0 0.2500\n", " 0 0.2500 0.2500 0\n", " 0 0.2500 0.2500 0\n", " 0.2500 0 0 0.2500\n", "\n" ] } ], "source": [ "thetapmAB = PipAB*deltaAB*PipAB + PimAB*deltaAB*PimAB" ] }, { "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": 10, "id": "df045697", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "sigma_01B =\n", "\n", " 0.5000 0\n", " 0 0.5000\n", "\n", "sigma_pmB =\n", "\n", " 0.5000 0\n", " 0 0.5000\n", "\n" ] } ], "source": [ "function [sigma] = traceA(delta) % traces out part A of a bipartite system\n", " sigma = [delta(1,1)+delta(3,3) delta(1,2)+delta(3,4); ...\n", " delta(2,1)+delta(4,3) delta(2,2)+delta(4,4)];\n", "end\n", "\n", "sigma_01B = traceA(theta01AB)\n", "sigma_pmB = traceA(thetapmAB)" ] }, { "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": "Octave", "language": "octave", "name": "octave" }, "language_info": { "file_extension": ".m", "help_links": [ { "text": "GNU Octave", "url": "https://www.gnu.org/software/octave/support.html" }, { "text": "Octave Kernel", "url": "https://github.com/Calysto/octave_kernel" }, { "text": "MetaKernel Magics", "url": "https://metakernel.readthedocs.io/en/latest/source/README.html" } ], "mimetype": "text/x-octave", "name": "octave", "version": "8.2.0" } }, "nbformat": 4, "nbformat_minor": 5 }