From d3cb59361d16b5e1cbcb7959714d508837563c28 Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Sun, 11 Sep 2022 04:39:58 +0800 Subject: [PATCH 01/14] add lecture16 --- lecture15.ipynb | 156 ++++++++++++++++++++++++++++++++++++++ lecture16.ipynb | 146 +++++++++++++++++++++++++++++++++++ solutions/lecture16.ipynb | 81 ++++++++++++++++++++ 3 files changed, 383 insertions(+) create mode 100644 lecture15.ipynb create mode 100644 lecture16.ipynb create mode 100644 solutions/lecture16.ipynb diff --git a/lecture15.ipynb b/lecture15.ipynb new file mode 100644 index 0000000..63092df --- /dev/null +++ b/lecture15.ipynb @@ -0,0 +1,156 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# 量子线路的误差\n", + "\n", + "量子门永远都不可能是「完美」的，因为酉矩阵是连续的，我们只能使用连续变量来表示，可是现实中的计算总是有限精度的，这意味着误差是一定存在的。\n", + "\n", + "例如我们想要实现一个 $m$ 个门组成的量子线路，使用之前提到的通用量子门，记作 $U = U_m U_{m-1} \\cdots U_1$，但是因为某些未知的原因，我们实现的门序列是 $V = V_m V_{m-1} \\cdots V_1$，那么问题来了，它的表现有多差呢？这种「不完美」是否会毁掉我们的计算呢？\n", + "\n", + "首先让我们明确并定义**误差**的概念：\n", + "$$\n", + "E(U, V) \\equiv \\max_{|\\psi\\rangle} \\left\\Vert (U - V) |\\psi\\rangle \\right\\Vert\n", + "$$\n", + "其中范数使用欧几里得范数：\n", + "$$\n", + "\\left \\Vert |a\\rangle \\right \\Vert \\equiv \\sqrt{\\langle a | a \\rangle}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "接下来我们讨论这种**误差**的物理含义，我们将证明下面的定理。\n", + "\n", + "**定理：**\n", + "$$\n", + "\\left| P_U - P_V \\right| \\leq 2 E(U, V)\n", + "$$\n", + "其中 $P_U \\equiv \\langle \\psi | U^\\dagger M U | \\psi \\rangle$，$P_V \\equiv \\langle \\psi | V^\\dagger M V | \\psi \\rangle$ 分别代表某个测量POVM给出结果 $M$ 的概率。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "**Proof：**\n", + "\n", + "首先我们引入一个记号 $|\\Delta\\rangle \\equiv (U-V) |\\psi\\rangle$ ，这是一个未单位化的向量。\n", + "\n", + "首先看不等式的左端\n", + "$$\n", + "\\begin{align*}\n", + "|P_U - P_V| & = \\left| \\langle \\psi | U^\\dagger M U | \\psi\\rangle - \\langle \\psi | V^\\dagger M V | \\psi \\rangle \\right| \\\\\n", + "& = \\left| \\langle \\psi | U^\\dagger M (U-V) | \\psi\\rangle + \\langle \\psi | U^\\dagger M V | \\psi\\rangle + \\langle \\psi | (U - V)^\\dagger M V | \\psi \\rangle - \\langle \\psi | U^\\dagger M V | \\psi \\rangle \\right| \\\\\n", + "& = \\left| \\langle \\psi | U^\\dagger M | \\Delta \\rangle + \\langle \\Delta | M V | \\psi \\rangle \\right| \\\\\n", + "& \\leq \\left| \\langle \\psi | U^\\dagger M | \\Delta \\rangle \\right| + \\left|\\langle \\Delta | M V | \\psi \\rangle \\right|\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$M$ 是POVM中的一项，所以 $M$ 是正定矩阵，其特征值大于等于 $0$，由于完备性关系 $\\sum_i M_i = I$，所以 $M$ 的特征值小于等于 $1$，那么有：\n", + "$$\n", + "\\begin{align*}\n", + "& \\left| \\langle \\psi | U^\\dagger M | \\Delta \\rangle \\right| \\\\\n", + "= & \\left| \\langle \\psi_U | M | \\Delta \\rangle \\right| \\\\\n", + "\\leq & \\left \\Vert |\\psi_U\\rangle \\right \\Vert \\cdot \\left\\Vert M|\\Delta \\rangle \\right\\Vert \\\\\n", + "\\leq & \\left\\Vert |\\Delta\\rangle \\right\\Vert\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "因此，我们就得到了结论\n", + "$$\n", + "\\left| P_U - P_V \\right| \\leq 2 \\left\\Vert |\\Delta\\rangle \\right\\Vert \\leq 2 E(U, V)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "对于多个门，我们有下面的不等式：\n", + "$$\n", + "E(U, V) = E(U_m U_{m-1} \\cdots U_1, V_m, V_{m-1} \\cdots V_1) \\leq \\sum_{i=1}^m E(U_i, V_i)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "这里只证明两个门的情况（多于两个门可以用数学归纳法证明）。\n", + "\n", + "$$\n", + "\\begin{align*}\n", + "& \\max_{|\\psi\\rangle} \\left\\Vert (U_1U_2 - V_1V_2) |\\psi\\rangle \\right\\Vert \\\\\n", + "= & \\max_{|\\psi\\rangle} \\left\\Vert (U_1U_2 - U_1V_2 + U_1V_2 - V_1V_2) |\\psi\\rangle \\right\\Vert \\\\\n", + "= & \\max_{|\\psi\\rangle} \\left\\Vert U_1(U_2 - V_2) |\\psi\\rangle + (U_1 - V_1)V_2 |\\psi\\rangle \\right\\Vert \\\\\n", + "\\leq & \\max_{|\\psi\\rangle} \\left(\\left\\Vert U_1(U_2 - V_2) |\\psi\\rangle \\right\\Vert + \\left\\Vert (U_1 - V_1)V_2 |\\psi\\rangle \\right\\Vert \\right) \\\\\n", + "\\leq & \\max_{|\\psi\\rangle} \\left\\Vert U_1(U_2 - V_2) |\\psi\\rangle \\right\\Vert + \\max_{|\\psi\\rangle}\\left\\Vert (U_1 - V_1)V_2 |\\psi\\rangle \\right\\Vert \\\\\n", + "= & \\max_{|\\psi\\rangle} \\left\\Vert (U_2 - V_2) |\\psi\\rangle \\right\\Vert + \\max_{|\\psi\\rangle}\\left\\Vert (U_1 - V_1) |\\psi\\rangle \\right\\Vert \\\\\n", + "= & E(U_1, V_1) + E(U_2, V_2)\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "该定理表明，误差是线性叠加的，所以如果我们希望最终的测量结果误差不超过 $\\delta$，即 $|P_U - P_V|\\leq \\delta$，线路有 $m$ 个门，那么我们只需要控制每个门的误差\n", + "$$\n", + "E(U_i, V_i) \\leq \\frac{\\delta}{2m}\n", + "$$\n", + "\n", + "这表明，如果我们的量子线路越长，门的数目越多，单个门的保真度就要越高。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# 二量子门的标准分解\n", + "\n", + "一般的，任意二量子比特门都可以表示为「标准形式」（Kraus and Cirac 2001）：\n", + "$$\n", + "U = A_L\\otimes B_L e^{i(\\alpha_{xx}XX + \\alpha_{yy}YY + \\alpha_{zz}ZZ)} A_R\\otimes B_R\n", + "$$\n", + "其中 $A_L, A_R$ 和 $B_L, B_R$ 都是单量子比特门，$\\alpha\\in \\mathbb{R}$。\n", + "\n", + "为了证明这一点，下面将介绍几个事实。" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3.9.0 64-bit ('3.9.0')", + "language": "python", + "name": "python3" + }, + "language_info": { + "name": "python", + "version": "3.9.0" + }, + "orig_nbformat": 4, + "vscode": { + "interpreter": { + "hash": "0ed0b97269d4f102e0624f582249018cd3d8c77b83fc4d003b4aaace9dedf79d" + } + } + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/lecture16.ipynb b/lecture16.ipynb new file mode 100644 index 0000000..8112d1f --- /dev/null +++ b/lecture16.ipynb @@ -0,0 +1,146 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# 通用量子门\n", + "\n", + "在之前的章节中，我们使用单量子比特门的集合。因为一些技术原因，在允许误差的量子计算中，我们想要仅仅使用Hadamard门和T门来代替单量子比特门的集合。\n", + "$$\n", + "H = {1\\over\\sqrt{2}} \\begin{bmatrix} \n", + "1 & 1\\\\\n", + "1 & -1\n", + "\\end{bmatrix} \\quad , \\quad\n", + "T = \\begin{bmatrix}\n", + "1 & 0 \\\\\n", + "0 & e^{i\\pi / 4}\n", + "\\end{bmatrix}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "> 下面介绍的方法也可以拓展到其他可能的一对量子门。\n", + "\n", + "最终，我们可以使用 H、T和CNOT 来近似模拟任意量子线路，我们称这一组门的集合为通用量子门。\n", + "\n", + "首先，我们要说明如何使用H和T来近似任意单量子门。回忆之前的章节，任意单量子门都可以用一对旋转门来分解（拓展的Z-Y-Z分解）：\n", + "$$\n", + "U = e^{i\\alpha} R_n(\\beta_1) R_m(\\gamma_1) R_n(\\beta_2) R_m(\\gamma_2) \\cdots\n", + "$$\n", + "需要的门的数目取决于 $\\vec{n}$ 和 $\\vec{m}$ 之间的夹角。\n", + "\n", + "任意 $2\\times 2$ 的矩阵都可以用 Pauli 基分解：\n", + "$$\n", + "A = a_0 I + a_x X + a_y Y + a_z Z\n", + "$$\n", + "不难写出\n", + "$$\n", + "R_n(\\theta_{unit}) \\propto THTH = \\cos{\\theta_{unit}\\over 2} I - i(\\vec{n}\\cdot \\vec{\\sigma}) \\sin{\\theta_{unit}\\over 2}\n", + "$$\n", + "这确定了 $\\vec{n} = (\\cos{\\pi\\over 8}, \\sin{\\pi\\over 8}, \\cos{\\pi\\over 8})$ （未单位化） 和 $\\theta_{unit}$ 。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "现在我们需要担心能否产生连续的旋转角度。显然如果使用 $(THTH)^k$ 可以产生 $R_n(k \\theta_{unit})$，所以问题是我们能否使用 $R_n(\\theta_{unit})$ 对任意连续值进行逼近？也就是说，对于 $\\forall \\beta \\in [0, 2\\pi)$，找到某个 $k\\in N$，使得下面的关系成立\n", + "$$\n", + "R_n(\\beta) \\approx R_n(k \\theta_{unit}) = R_n(\\theta_{unit})^k = (THTH)^k\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "证明的关键在于，如果 $\\theta_{unit}$ 不是 $2\\pi$ 的有理数倍（$\\theta_{unit} / 2\\pi \\notin \\mathbb{Q}$），那么可以保证 $\\forall \\epsilon > 0$，$\\exists k \\in N$，使得\n", + "$$\n", + "E(R_n(\\beta), R_n(k\\theta_{unit})) < \\epsilon\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 圆上的稠密性\n", + "\n", + "### Fact 1\n", + "\n", + "如果 $\\alpha \\in [0, 1)$ 是无理数，那么点集 $n\\alpha \\mod 1$ 全部是不同的。\n", + "\n", + "**Proof:** \n", + "\n", + "假设存在两个点是相同的，即 $n \\alpha - m\\alpha = k$，其中 $k\\in \\mathbb{Z}$。那么我们可以将 $\\alpha$ 写成 $\\alpha = k / (n-m)$，这和 $\\alpha$ 是无理数矛盾。因此点集 $\\lbrace n\\alpha \\mod 1\\rbrace$ 中没有重复元素。\n", + "\n", + "### Fact 2\n", + "\n", + "对于 $\\epsilon > 0$，存在 $N\\in \\mathbb{N}$ 使得 $\\epsilon > 1/N$，现在把 $[0, 1]$ 区间分成 $N$ 份，每份区间长度为 $1/N$，根据抽屉原理（鸽笼原理），考虑 $\\alpha, 2\\alpha, \\ldots, N\\alpha, (N+1)\\alpha$ 模 $1$ 意义下，一定存在 $m\\alpha \\mod 1$ 和 $n\\alpha \\mod 1$ 落在同一个区间，即 $|m-n|\\alpha \\mod 1 < \\epsilon$ 。令 $r \\equiv |n-m|$，注意到 $r\\alpha \\mod 1$ 的倍数形成了单位环上等间距的点集，间距小于 $\\epsilon$ 。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "现在我们可以对角度进行任意逼近，下面说明旋转算子的误差满足：(证明留作习题)\n", + "$$\n", + "E(R_n(\\theta_1), R_n(\\theta_2)) \\leq \\left| e^{i\\theta_1} - e^{i\\theta_2} \\right| \\leq |\\theta_1 - \\theta_2|\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "同理，我们可以定义 $R_m(\\gamma)$\n", + "$$\n", + "R_m(\\gamma) \\equiv HR_n(\\gamma)H = HTHT\n", + "$$\n", + "\n", + "综上所述，我们可以构造\n", + "$$\n", + "E(U, R_n(k_1 \\theta_{unit}) R_m(k_2 \\theta_{unit}) R_n(k_3 \\theta_{unit}) R_m(k_4 \\theta_{unit}) \\cdots) < c\\epsilon\n", + "$$\n", + "其中 $c$ 代表旋转门的数目。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# 习题\n", + "\n", + "## Exercise 1\n", + "\n", + "证明：\n", + "$$\n", + "E(R_n(\\theta_1), R_n(\\theta_2)) \\leq \\left| e^{i\\theta_1} - e^{i\\theta_2} \\right| \\leq |\\theta_1 - \\theta_2|\n", + "$$" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3.9.0 64-bit ('3.9.0')", + "language": "python", + "name": "python3" + }, + "language_info": { + "name": "python", + "version": "3.9.0" + }, + "orig_nbformat": 4, + "vscode": { + "interpreter": { + "hash": "0ed0b97269d4f102e0624f582249018cd3d8c77b83fc4d003b4aaace9dedf79d" + } + } + }, + "nbformat": 4, + "nbformat_minor": 2 +} diff --git a/solutions/lecture16.ipynb b/solutions/lecture16.ipynb new file mode 100644 index 0000000..ccbd658 --- /dev/null +++ b/solutions/lecture16.ipynb @@ -0,0 +1,81 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Exercise 1\n", + "\n", + "证明：\n", + "$$\n", + "E(R_n(\\theta_1), R_n(\\theta_2)) \\leq \\left| e^{i\\theta_1} - e^{i\\theta_2} \\right| \\leq |\\theta_1 - \\theta_2|\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "**Proof:**\n", + "\n", + "我们知道对于任意旋转矩阵 $R_n(\\theta)$，一定可以写作\n", + "$$\n", + "R_n(\\theta) = |+\\rangle \\langle + | + e^{i\\theta} |-\\rangle \\langle -|\n", + "$$\n", + "其中 $|+\\rangle, |-\\rangle$ 分别表示两个特征向量，也是 $\\vec{n}$ 穿过Bloch球面上的两点。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "那么 \n", + "$$\n", + "\\begin{align*}\n", + "& \\left(R_n(\\theta_1) - R_n(\\theta_2)\\right) |\\psi\\rangle \\\\\n", + "= & (e^{i\\theta_1} - e^{i\\theta_2}) |-\\rangle \\langle -| (\\alpha|+\\rangle + \\beta |-\\rangle) \\\\\n", + "= & (e^{i\\theta_1} - e^{i\\theta_2}) \\beta |-\\rangle\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "计算 $E$\n", + "\n", + "$$\n", + "\\begin{align*}\n", + "& E(R_n(\\theta_1), R_n(\\theta_2)) \\\\\n", + "= & \\max_{|\\psi\\rangle} \\left\\Vert \\left( R_n(\\theta_1) - R_n(\\theta_2) \\right) |\\psi\\rangle \\right\\Vert \\\\\n", + "= & \\max_{|\\psi\\rangle} \\left\\Vert \\beta (e^{i\\theta_1} - e^{i\\theta_2}) |-\\rangle \\right\\Vert \\\\\n", + "= & \\left| e^{i\\theta_1} - e^{i\\theta_2} \\right| \\\\\n", + "= & |e^{i\\theta_1}| \\cdot |1 - e^{\\theta_2 - \\theta_1}| \\\\\n", + "= & 2 \\sin{|\\theta_2 - \\theta_1| \\over 2} \\\\\n", + "\\leq & |\\theta_2 - \\theta_1|\n", + "\\end{align*}\n", + "$$" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3.9.0 64-bit ('3.9.0')", + "language": "python", + "name": "python3" + }, + "language_info": { + "name": "python", + "version": "3.9.0" + }, + "orig_nbformat": 4, + "vscode": { + "interpreter": { + "hash": "0ed0b97269d4f102e0624f582249018cd3d8c77b83fc4d003b4aaace9dedf79d" + } + } + }, + "nbformat": 4, + "nbformat_minor": 2 +} -- Gitee From 0d2b744686351e40a38a9cb764b8650edab2a3ae Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Mon, 19 Sep 2022 22:56:59 +0800 Subject: [PATCH 02/14] add proof in lecture15 --- .gitignore | 3 +- lecture15.ipynb | 312 +++++++++++++++++++++++++++++++++++++- solutions/lecture15.ipynb | 234 ++++++++++++++++++++++++++++ 3 files changed, 547 insertions(+), 2 deletions(-) create mode 100644 solutions/lecture15.ipynb diff --git a/.gitignore b/.gitignore index 58461f2..aac22fa 100644 --- a/.gitignore +++ b/.gitignore @@ -1 +1,2 @@ -.ipynb_checkpoints \ No newline at end of file +.ipynb_checkpoints +.vscode \ No newline at end of file diff --git a/lecture15.ipynb b/lecture15.ipynb index 63092df..bceceac 100644 --- a/lecture15.ipynb +++ b/lecture15.ipynb @@ -130,7 +130,317 @@ "$$\n", "其中 $A_L, A_R$ 和 $B_L, B_R$ 都是单量子比特门，$\\alpha\\in \\mathbb{R}$。\n", "\n", - "为了证明这一点，下面将介绍几个事实。" + "下面将通过几个引理，一步步的证明这个结论。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## 纠缠\n", + "\n", + "首先介绍一个概念：纠缠熵。对于一个二量子比特纯态 $|\\psi\\rangle$，其纠缠熵定义如下：\n", + "$$\n", + "E(|\\psi\\rangle) \\equiv - tr \\left ( \\rho_A \\log_2 \\rho_A \\right ) = - tr \\left ( \\rho_B \\log_2 \\rho_B \\right )\n", + "$$\n", + "其中\n", + "$$\n", + "\\begin{align*}\n", + "\\rho & = |\\psi \\rangle \\langle \\psi | \\\\\n", + "\\rho_A & = tr_B \\rho \\\\\n", + "\\rho_B & = tr_A \\rho\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "我们知道，对于二量子比特纯态 $|\\psi\\rangle$，存在Schmidt分解：\n", + "$$\n", + "|\\psi\\rangle = \\sum_i \\sqrt{p_i} |i^A\\rangle |i^B\\rangle \n", + "$$\n", + "则可以计算A和B的密度矩阵：\n", + "$$\n", + "\\begin{align*}\n", + "\\rho_A & = \\sum_i p_i |i^A \\rangle \\langle i^A | \\\\\n", + "\\rho_B & = \\sum_i p_i |i^B \\rangle \\langle i^B |\n", + "\\end{align*}\n", + "$$\n", + "因此其纠缠熵为\n", + "$$\n", + "E(|\\psi\\rangle) = -tr \\left ( \\sum_i p_i \\log_2 p_i |i^A \\rangle \\langle i^A | \\right ) = - p_1 \\log_2 p_1 - p_2 \\log_2 p_2\n", + "$$\n", + "其中 $p_1, p_2$ 为 $\\rho_A, \\rho_B$ 的两个特征向量。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Concurrence\n", + "\n", + "为了方便计算，我们引入concurrence：\n", + "$$\n", + "C \\equiv \\left | \\sum_i \\alpha_i^2 \\right |\n", + "$$\n", + "其中 $\\alpha_i$ 是 $|\\psi\\rangle$ 在magic基下的振幅，magic基是由Bell基变换而来的：\n", + "$$\n", + "\\begin{align*}\n", + "m_{00} & \\equiv \\frac{|00\\rangle + |11\\rangle}{\\sqrt{2}} \\\\\n", + "m_{01} & \\equiv i \\frac{|00\\rangle - |11\\rangle}{\\sqrt{2}} \\\\\n", + "m_{10} & \\equiv i \\frac{|01\\rangle + |10\\rangle}{\\sqrt{2}} \\\\\n", + "m_{11} & \\equiv \\frac{|01\\rangle - |10\\rangle}{\\sqrt{2}}\n", + "\\end{align*}\n", + "$$\n", + "\n", + "> 注意，$C$ 的定义中是 $\\alpha_i^2$ 而不是 $|\\alpha|^2$。\n", + "\n", + "通过计算可以证明下面这个定理（留作习题）：\n", + "$$\n", + "E(|\\psi\\rangle) = H\\left ( \\frac{1 + \\sqrt{1 - C^2}}{2} \\right )\n", + "$$\n", + "其中\n", + "$$\n", + "H(x) \\equiv -x \\log_2 x - (1-x) \\log_2 (1-x)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Lemma 1\n", + "\n", + "- 如果 $E(|\\psi\\rangle) = 0$，即 $C = 0$，那么 $|\\psi\\rangle = |a\\rangle|b\\rangle$ 是可分态；\n", + "- 如果 $E(|\\psi\\rangle) = 1$，即 $C = 1$，那么 $|\\psi\\rangle$ 是最大纠缠态，$\\rho_A = \\rho_B = I / 2$ 。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Lemma 2\n", + "\n", + "根据 Lemma 1，$|\\psi\\rangle$ 是最大纠缠态，当且仅当 $\\alpha_i$ 都是实数（忽略一个全局相位 $e^{i\\varphi}$）。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Lemma 3\n", + "\n", + "给定两个正交的态 $|\\psi\\rangle$ 和 $|\\psi^\\bot\\rangle$，如果这两个态在magic基下的表示都是实向量，那么 $\\frac{|\\psi\\rangle \\pm i |\\psi^\\bot\\rangle}{\\sqrt{2}}$ 是可分态。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "**Proof:**\n", + "\n", + "假设 $|\\psi\\rangle = \\sum_i a_i |m_i\\rangle$，$|\\psi^\\bot\\rangle = \\sum_i b_i |m_i\\rangle$，\n", + "其中 $a_i, b_i \\in \\mathbb{R}$，且 $\\sum_i a_i b_i = 0$，\n", + "\n", + "$$\n", + "\\begin{align*}\n", + "C \\left ( \\frac{|\\psi\\rangle \\pm |\\psi^\\bot\\rangle}{\\sqrt{2}} \\right )\n", + "& = \\left | \\sum_i \\left(\\frac{a_i \\pm i b_i}{\\sqrt{2}}\\right)^2 \\right | \\\\\n", + "& = \\frac{1}{2} \\left | \\sum_i a_i ^2 - b_i^2 \\pm 2a_ib_i i \\right | \\\\\n", + "& = 0\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "更进一步，我们记\n", + "$$\n", + "\\begin{align*}\n", + "|\\psi_+\\rangle & \\equiv \\frac{|\\psi\\rangle + i|\\psi^\\bot\\rangle}{\\sqrt{2}} = |A_+\\rangle|B_+\\rangle\\\\\n", + "|\\psi_-\\rangle & \\equiv \\frac{|\\psi\\rangle - i|\\psi^\\bot\\rangle}{\\sqrt{2}} = |A_-\\rangle|B_-\\rangle\n", + "\\end{align*}\n", + "$$\n", + "因为 $\\langle \\psi | \\psi^\\bot \\rangle = 0$，所以 $\\langle \\psi_+ | \\psi_- \\rangle = 0$，因此 $\\langle A_+ | A_- \\rangle \\cdot \\langle B_+ | B_- \\rangle = 0$，下面证明二者都为0.\n", + "\n", + "不妨设 $\\langle A_+| A_-\\rangle = 0$，我们可以用 $|A_+\\rangle|B_+\\rangle$ 和 $|A_-\\rangle|B_-\\rangle$ 来表示 $|\\psi\\rangle$ 和 $|\\psi^\\bot\\rangle$\n", + "$$\n", + "\\begin{align*}\n", + "|\\psi\\rangle & = \\frac{1}{\\sqrt{2}} \\left ( |A_+\\rangle |B_+\\rangle + |A_-\\rangle|B_-\\rangle\\right ) \\\\\n", + "|\\psi^\\bot\\rangle & = \\frac{-i}{\\sqrt{2}} \\left ( |A_+\\rangle |B_+\\rangle - |A_-\\rangle|B_-\\rangle\\right )\n", + "\\end{align*}\n", + "$$\n", + "\n", + "因为 $|\\psi\\rangle$ 是最大纠缠态，所以 $\\rho_B = I / 2$，故 $\\langle B_+ | B_-\\rangle = 0$ 。\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Lemma 4 \n", + "\n", + "对于任意酉矩阵 $U$，存在分解\n", + "$$\n", + "U = O_L D_\\phi O_R\n", + "$$\n", + "其中 $O_L, O_R$ 都是实正交矩阵，$D_\\phi$ 是对角矩阵，对角元都是模长为1的复数 $e^{i\\phi}$。\n", + "\n", + "> 该定理这里不予证明，参考论文 《Tucci, Robert R. 《An Introduction to Cartan’s KAK Decomposition for QC Programmers》. arXiv, 2005年7月18日. http://arxiv.org/abs/quant-ph/0507171. 》\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Proof\n", + "\n", + "下面给出定理的证明。\n", + "\n", + "首先定义 $M$ 矩阵：\n", + "$$\n", + "M \\equiv \\sum_k |m_k\\rangle \\langle k |\n", + "$$\n", + "\n", + "容易发现，对于实向量 $|\\psi_{real}\\rangle \\in \\mathbb{R}^4$ ，作用 $M$ 将得到一个最大纠缠态\n", + "$$\n", + "M |\\psi_{real}\\rangle = \\sum_k a_k |m_k\\rangle \\quad \\text{where } a_i \\in \\mathbb{R}\n", + "$$\n", + "\n", + "对于任意酉矩阵 $U \\in SU(4)$，令 $V \\equiv M^\\dagger U M$，根据 Lemma 4，$V = O_L D_\\phi O_R$，则\n", + "$$\n", + "U = MO_L D_\\phi O_R M^\\dagger\n", + "$$\n", + "\n", + "因为实正交矩阵 $O_L$ 的每一列 $|o_i\\rangle$ 都是实向量，所以\n", + "$$\n", + "M O_L = \\sum_k M |o_i \\rangle \\langle k | = \\sum_k |e_k \\rangle \\langle k |\n", + "$$\n", + "其中 $|e_k\\rangle$ 是最大纠缠态。\n", + "\n", + "根据Lemma 3，$|e_{00}\\rangle$ 和 $|e_{01}\\rangle$ 可以表示成\n", + "$$\n", + "\\begin{align*}\n", + "|e_{00}\\rangle & = \\frac{1}{\\sqrt{2}} \\left ( |A_+\\rangle |B_+\\rangle + |A_-\\rangle |B_-\\rangle \\right ) \\\\\n", + "|e_{01}\\rangle & = \\frac{-i}{\\sqrt{2}} \\left ( |A_+\\rangle |B_+\\rangle - |A_-\\rangle |B_-\\rangle \\right )\n", + "\\end{align*}\n", + "$$\n", + "\n", + "因为 $|e_{10}\\rangle, |e_{11}\\rangle$ 和 $|e_{00}\\rangle, |e_{01}\\rangle$ 正交，所以一定可以被 $|A_+\\rangle |B_-\\rangle, |A_-\\rangle|B_+\\rangle$ 线性表示，又因为是最大纠缠态，所以 $\\rho_A = \\rho_B = I /2$，通过计算可以表示为\n", + "$$\n", + "\\begin{align*}\n", + "|e_{10}\\rangle & = \\frac{e^{i\\lambda_3}}{\\sqrt{2}} \\left ( e^{-i\\delta} |A_+\\rangle |B_-\\rangle + e^{i\\delta} |A_-\\rangle |B_+\\rangle\\right ) \\\\\n", + "|e_{11}\\rangle & = \\frac{e^{i\\lambda_4}}{\\sqrt{2}} \\left ( e^{-i\\delta} |A_+\\rangle |B_-\\rangle - e^{i\\delta} |A_-\\rangle |B_+\\rangle\\right ) \\\\\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "令 \n", + "$$\n", + "\\begin{align*}\n", + "A & \\equiv |0 \\rangle \\langle A_+ | + e^{i\\delta} |1 \\rangle \\langle A_- | \\\\\n", + "B & \\equiv |0 \\rangle \\langle B_+ | + e^{-i\\delta} |1 \\rangle \\langle B_- |\n", + "\\end{align*}\n", + "$$\n", + "\n", + "则\n", + "$$\n", + "\\begin{align*}\n", + "A\\otimes B & = |00 \\rangle \\langle A_+ B_+ | \\\\\n", + " & + e^{i\\delta} |01 \\rangle \\langle A_+ B_- | \\\\\n", + " & + e^{-i\\delta} |10 \\rangle \\langle A_- B_+ | \\\\\n", + " & + |11 \\rangle \\langle A_- B_- |\n", + "\\end{align*}\n", + "$$\n", + "\n", + "计算\n", + "$$\n", + "\\begin{align*}\n", + "M O_L & = \\sum_k |e_k \\rangle \\langle k | \\\\\n", + "& = \\frac{1}{\\sqrt{2}} \\left ( |A_+\\rangle |B_+\\rangle + |A_-\\rangle |B_-\\rangle\\right ) \\\\\n", + "& + \\frac{-i}{\\sqrt{2}} \\left ( |A_+\\rangle |B_+\\rangle - |A_-\\rangle |B_-\\rangle\\right ) \\\\\n", + "& + \\frac{e^{i\\lambda_3}}{\\sqrt{2}} \\left ( e^{-i\\delta} |A_+\\rangle |B_-\\rangle + e^{i\\delta} |A_-\\rangle |B_+\\rangle\\right ) \\\\\n", + "& + \\frac{e^{i\\lambda_4}}{\\sqrt{2}} \\left ( e^{-i\\delta} |A_+\\rangle |B_-\\rangle - e^{i\\delta} |A_-\\rangle |B_+\\rangle\\right ) \\\\\n", + "\\end{align*}\n", + "$$\n", + "\n", + "则\n", + "$$\n", + "\\begin{align*}\n", + "(A\\otimes B) M O_L & = \n", + "\\frac{|00\\rangle + |11\\rangle}{\\sqrt{2}} \\langle 00| \\\\\n", + "& + (-i) \\frac{|00\\rangle - |11\\rangle}{\\sqrt{2}} \\langle 01| \\\\\n", + "& + e^{i\\lambda_3} \\frac{|01\\rangle + |10\\rangle}{\\sqrt{2}} \\langle 10| \\\\\n", + "& + e^{i\\lambda_4} \\frac{|01\\rangle - |10\\rangle}{\\sqrt{2}} \\langle 11| \\\\\n", + "& = \\sum_k e^{\\phi_k} |m_k \\rangle \\langle k |\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "同理，记\n", + "$$\n", + "\\begin{align*}\n", + "M O_L & = A_L^\\dagger \\otimes B_L^\\dagger \\sum_k e^{i\\phi_k} |m_k \\rangle \\langle k | \\\\\n", + "M O_R^\\dagger & = A_R^\\dagger \\otimes B_R^\\dagger \\sum_k e^{i\\phi'_k} |m_k \\rangle \\langle k |\n", + "\\end{align*}\n", + "$$\n", + "\n", + "则\n", + "$$\n", + "U = MO_L D_\\phi O_R M^\\dagger = A_L^\\dagger \\otimes B_L^\\dagger \\left(\\sum_k e^{i\\varphi_k} |m_k \\rangle \\langle m_k | \\right) A_R \\otimes B_R\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "因为 $|m_k\\rangle$ 是 $XX, YY, ZZ$ 的特征向量，如果忽略全局相位， $U \\in SU(4)$，则 $\\sum_k \\varphi_k = 0$，可以表示为\n", + "$$\n", + "\\sum_k e^{i \\varphi_k} |m_k \\rangle \\langle m_k | = e^{i a XX + b YY + c ZZ}\n", + "$$\n", + "其中 $a, b, c \\in \\mathbb{R}$，可以由 $\\varphi_k$ 解出来。\n", + "\n", + "至此，我们完成了定理的证明。" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# 习题\n", + "\n", + "## Exercise 1\n", + "\n", + "证明\n", + "$$\n", + "E\\left ( \\prod_{i=1}^n U_i, \\prod_{i=1}^n V_i \\right ) \\leq \\sum_{i=1}^n E(U_i, V_i)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exercise 2\n", + "\n", + "证明\n", + "\n", + "$$\n", + "E(|\\psi\\rangle) = H\\left ( \\frac{1 + \\sqrt{1 - C^2}}{2}\\right )\n", + "$$" ] } ], diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb new file mode 100644 index 0000000..eb1d9ea --- /dev/null +++ b/solutions/lecture15.ipynb @@ -0,0 +1,234 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Exercise 1\n", + "\n", + "证明\n", + "$$\n", + "E\\left ( \\prod_{i=1}^n U_i, \\prod_{i=1}^n V_i \\right ) \\leq \\sum_{i=1}^n E(U_i, V_i)\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "**Proof**\n", + "\n", + "$$\n", + "\\begin{align*}\n", + "E\\left ( \\prod_{i=1}^n U_i, \\prod_{i=1}^n V_i \\right ) \n", + "& = E \\left ( U_1 \\prod_{i=2}^n U_i, V_1 \\prod_{i=2}^n V_i \\right ) \\\\\n", + "& \\leq E(U_1, V_1) + E\\left ( \\prod_{i=2}^n U_i, \\prod_{i=2}^n V_i \\right ) \\\\\n", + "& \\vdots \\\\\n", + "& \\leq \\sum_{i=1}^n E(U_i, V_i)\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Exercise 2\n", + "\n", + "证明\n", + "\n", + "$$\n", + "E(|\\psi\\rangle) = H\\left ( \\frac{1 + \\sqrt{1 - C^2}}{2}\\right )\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "**Proof**\n", + "\n", + "设 $|\\psi\\rangle$ 的Schmidt分解为\n", + "$$|\\psi\\rangle = \\sum_k \\sqrt{p_k} |i^A_k\\rangle |i^B_k\\rangle$$\n", + "\n", + "则\n", + "$$\n", + "\\begin{align*}\n", + "\\rho_A & = \\sum_k p_k |i^A_k \\rangle \\langle i^A_k | \\\\\n", + "\\rho_B & = \\sum_k p_k |i^B_k \\rangle \\langle i^B_k |\n", + "\\end{align*}\n", + "$$\n", + "\n", + "因此纠缠熵计算结果为\n", + "$$\n", + "E = -p_1 \\log_2 p_1 - p_2 \\log_2 p_2\n", + "$$\n", + "其中 $p_1, p_2$ 是 $\\rho_A, \\rho_B$ 的两个特征值。\n", + "\n", + "对比等式右边\n", + "$$\n", + "\\begin{align*}\n", + "H\\left ( \\frac{1 + \\sqrt{1 - C^2}}{2}\\right )\n", + "& = - \\frac{1 + \\sqrt{1 - C^2}}{2} \\log_2 \\frac{1 + \\sqrt{1 - C^2}}{2} - \\frac{1 - \\sqrt{1 - C^2}}{2} \\log_2 \\frac{1 - \\sqrt{1 - C^2}}{2} \\\\\n", + "\\end{align*}\n", + "$$\n", + "只需要证明 $\\frac{1 \\pm \\sqrt{1 - C^2}}{2}$ 是$\\rho_A, \\rho_B$ 的两个特征值。\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "我们将 $|\\psi\\rangle$ 用Bell基表示：\n", + "$$\n", + "|\\psi\\rangle = \\sum_{xy} \\beta_{xy} |\\beta_{xy}\\rangle = \\sum_{xy} \\beta_{xy} \\frac{|0, x\\rangle + (-1)^y |1, \\bar{x}\\rangle}{\\sqrt{2}}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "计算密度矩阵\n", + "$$\n", + "\\begin{align*}\n", + "\\rho & = \\sum_{xy} \\sum_{ab} \\beta_{xy} \\bar{\\beta}_{ab} |\\beta_{xy} \\rangle \\langle \\beta_{ab} | \\\\\n", + "& = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \\Big [ |0, x \\rangle \\langle 0, a | + (-1)^b |0, x \\rangle \\langle 1, \\bar{a} | + (-1)^y |1, \\bar{x} \\rangle \\langle 0, a | + (-1)^{y+b} |1, \\bar{x} \\rangle \\langle 1, \\bar{a} | \\Big ] \\\\\n", + "\\rho_B & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \\Big[ |x \\rangle \\langle a | + (-1)^{y+b} |\\bar{x} \\rangle \\langle \\bar{a} | \\Big]\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "考虑计算第一部分\n", + "$$\n", + "\\begin{align*}\n", + "P_1 & = \\sum_{x,a,a,b} \\frac{\\beta_{xt} \\bar{\\beta}_{ab}}{2} |x \\rangle \\langle a | \\\\\n", + "& = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta_{xy}}{2} \\left ( \\bar{\\beta}_{aa} + \\bar{\\beta}_{a, \\bar{a}}\\right ) \\\\\n", + "& = \\sum_{x,a}|x \\rangle \\langle a | \\frac{\\beta_{xx} + \\beta_{x, \\bar{x}}}{\\sqrt{2}} \\left(\\frac{\\beta_{aa} + \\beta_{a, \n", + "bar{a}}}{\\sqrt{2}} \\right)^\\star \\\\\n", + "& = \\sum_{x, a} |x \\rangle \\langle a | A_x \\cdot \\bar{A}_a\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "同理可以计算出来第二部分\n", + "$$\n", + "P_2 = \\sum_{x,a} |x \\rangle \\langle a | (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B}_{\\bar{a}}\n", + "$$\n", + "其中\n", + "$$\n", + "\\begin{align*}\n", + "A_x & \\equiv \\frac{\\beta_{xx} + \\beta_{x, \\bar{x}}}{\\sqrt{2}} \\\\\n", + "B_x & \\equiv \\frac{\\beta_{xx} - \\beta_{x, \\bar{x}}}{\\sqrt{2}}\n", + "\\end{align*}\n", + "$$\n", + "\n", + "因此\n", + "$$\n", + "\\begin{align*}\n", + "\\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\Big [ A_x \\cdot \\bar{A}_{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B}_{\\bar{a}} \\Big] \\\\\n", + "& = \\begin{bmatrix}\n", + "A_0 \\bar{A}_0 + B_1 \\bar{B}_1 & A_0 \\bar{A}_1 - B_1 \\bar{B}_0 \\\\\n", + "A_1 \\bar{A}_0 - B_0 \\bar{B}_1 & A_1 \\bar{A}_1 + B_0 \\bar{B}_0\n", + "\\end{bmatrix} \\\\\n", + "& = \\begin{bmatrix}\n", + "A & B \\\\\n", + "C & D\n", + "\\end{bmatrix}\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "计算 $\\rho_B$ 的行列式\n", + "$$\n", + "\\begin{align*}\n", + "\\det \\rho_B & = AD - BC \\\\\n", + "& = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 \\Re(A_0 B_0 \\bar{A}_1 \\bar{B}_1) \\\\\n", + "& = |A_0B_0 + A_1 B_1|^2 \\\\\n", + "& = \\frac{1}{4} |\\beta_{00}^2 - \\beta_{01}^2 - \\beta_{10}^2 + \\beta_{11}^2|\n", + "\\end{align*} \n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "计算 $\\rho_B$ 的特征方程：\n", + "$$\n", + "\\begin{align*}\n", + "0 & = \\det (\\rho_B - \\lambda I) \\\\\n", + "& = (A - \\lambda)(D - \\lambda) - BC \\\\\n", + "& = \\lambda^2 - tr(\\rho_B) \\lambda + \\det \\rho_B\n", + "\\end{align*}\n", + "$$\n", + "\n", + "方程的解为\n", + "$$\n", + "\\lambda_{1,2} = \\frac{1 \\pm \\sqrt{1 - 4\\det \\rho_B}}{2}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "最后，我们把Bell基换成magic基就行了\n", + "\n", + "$$\n", + "\\begin{align*}\n", + "|\\psi\\rangle \n", + "& = \\beta_{00} \\frac{|00\\rangle + |11\\rangle}{\\sqrt{2}} \\quad & = \\alpha_{00} \\frac{|00\\rangle + |11\\rangle}{\\sqrt{2}} \\\\\n", + "& + \\beta_{01} \\frac{|00\\rangle - |11\\rangle}{\\sqrt{2}} \\quad & + i\\alpha_{01} \\frac{|00\\rangle - |11\\rangle}{\\sqrt{2}} \\\\\n", + "& + \\beta_{10} \\frac{|01\\rangle + |10\\rangle}{\\sqrt{2}} \\quad & + i\\alpha_{10} \\frac{|01\\rangle + |10\\rangle}{\\sqrt{2}} \\\\\n", + "& + \\beta_{11} \\frac{|01\\rangle - |10\\rangle}{\\sqrt{2}} \\quad & + \\alpha_{11} \\frac{|01\\rangle - |10\\rangle}{\\sqrt{2}} \\\\\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "因此\n", + "$$\n", + "\\det \\rho_B = \\frac{1}{4} \\left|\\sum_k \\alpha_k^2 \\right|^2 = \\frac{1}{4} C^2\n", + "$$\n", + "\n", + "证毕。" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3.9.0 64-bit ('3.9.0')", + "language": "python", + "name": "python3" + }, + "language_info": { + "name": "python", + "version": "3.9.0" + }, + "orig_nbformat": 4, + "vscode": { + "interpreter": { + "hash": "0ed0b97269d4f102e0624f582249018cd3d8c77b83fc4d003b4aaace9dedf79d" + } + } + }, + "nbformat": 4, + "nbformat_minor": 2 +} -- Gitee From fe4dce4e4afefa96756f4bb71109c19978d7de24 Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Mon, 19 Sep 2022 23:07:12 +0800 Subject: [PATCH 03/14] try --- solutions/lecture15.ipynb | 1 + 1 file changed, 1 insertion(+) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index eb1d9ea..9238989 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -124,6 +124,7 @@ "$$\n", "P_2 = \\sum_{x,a} |x \\rangle \\langle a | (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B}_{\\bar{a}}\n", "$$\n", + "\n", "其中\n", "$$\n", "\\begin{align*}\n", -- Gitee From 653cb57c44362d22915cb296212cdffe74d754ca Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Mon, 19 Sep 2022 23:11:45 +0800 Subject: [PATCH 04/14] test --- solutions/lecture15.ipynb | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index 9238989..f44d1d6 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -94,8 +94,8 @@ "$$\n", "\\begin{align*}\n", "\\rho & = \\sum_{xy} \\sum_{ab} \\beta_{xy} \\bar{\\beta}_{ab} |\\beta_{xy} \\rangle \\langle \\beta_{ab} | \\\\\n", - "& = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \\Big [ |0, x \\rangle \\langle 0, a | + (-1)^b |0, x \\rangle \\langle 1, \\bar{a} | + (-1)^y |1, \\bar{x} \\rangle \\langle 0, a | + (-1)^{y+b} |1, \\bar{x} \\rangle \\langle 1, \\bar{a} | \\Big ] \\\\\n", - "\\rho_B & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \\Big[ |x \\rangle \\langle a | + (-1)^{y+b} |\\bar{x} \\rangle \\langle \\bar{a} | \\Big]\n", + "& = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \\big [ |0, x \\rangle \\langle 0, a | + (-1)^b |0, x \\rangle \\langle 1, \\bar{a} | + (-1)^y |1, \\bar{x} \\rangle \\langle 0, a | + (-1)^{y+b} |1, \\bar{x} \\rangle \\langle 1, \\bar{a} | \\big ] \\\\\n", + "\\rho_B & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \\big[ |x \\rangle \\langle a | + (-1)^{y+b} |\\bar{x} \\rangle \\langle \\bar{a} | \\big]\n", "\\end{align*}\n", "$$" ] @@ -136,7 +136,7 @@ "因此\n", "$$\n", "\\begin{align*}\n", - "\\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\Big [ A_x \\cdot \\bar{A}_{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B}_{\\bar{a}} \\Big] \\\\\n", + "\\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\big [ A_x \\cdot \\bar{A}_{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B}_{\\bar{a}} \\big] \\\\\n", "& = \\begin{bmatrix}\n", "A_0 \\bar{A}_0 + B_1 \\bar{B}_1 & A_0 \\bar{A}_1 - B_1 \\bar{B}_0 \\\\\n", "A_1 \\bar{A}_0 - B_0 \\bar{B}_1 & A_1 \\bar{A}_1 + B_0 \\bar{B}_0\n", @@ -157,7 +157,7 @@ "$$\n", "\\begin{align*}\n", "\\det \\rho_B & = AD - BC \\\\\n", - "& = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 \\Re(A_0 B_0 \\bar{A}_1 \\bar{B}_1) \\\\\n", + "& = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 Re(A_0 B_0 \\bar{A}_1 \\bar{B}_1) \\\\\n", "& = |A_0B_0 + A_1 B_1|^2 \\\\\n", "& = \\frac{1}{4} |\\beta_{00}^2 - \\beta_{01}^2 - \\beta_{10}^2 + \\beta_{11}^2|\n", "\\end{align*} \n", -- Gitee From 115e5c18ec70b1ec1832f9a0e92170ae7dcd2a66 Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Mon, 19 Sep 2022 23:16:08 +0800 Subject: [PATCH 05/14] test --- solutions/lecture15.ipynb | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index f44d1d6..58bac68 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -136,10 +136,10 @@ "因此\n", "$$\n", "\\begin{align*}\n", - "\\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\big [ A_x \\cdot \\bar{A}_{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B}_{\\bar{a}} \\big] \\\\\n", + "\\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\big [ A_x \\cdot \\bar{A} _{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B} _{\\bar{a}} \\big] \\\\\n", "& = \\begin{bmatrix}\n", - "A_0 \\bar{A}_0 + B_1 \\bar{B}_1 & A_0 \\bar{A}_1 - B_1 \\bar{B}_0 \\\\\n", - "A_1 \\bar{A}_0 - B_0 \\bar{B}_1 & A_1 \\bar{A}_1 + B_0 \\bar{B}_0\n", + "A_0 \\bar{A} _0 + B_1 \\bar{B} _1 & A_0 \\bar{A} _1 - B_1 \\bar{B} _0 \\\\\n", + "A_1 \\bar{A} _0 - B_0 \\bar{B} _1 & A_1 \\bar{A} _1 + B_0 \\bar{B} _0\n", "\\end{bmatrix} \\\\\n", "& = \\begin{bmatrix}\n", "A & B \\\\\n", @@ -157,8 +157,8 @@ "$$\n", "\\begin{align*}\n", "\\det \\rho_B & = AD - BC \\\\\n", - "& = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 Re(A_0 B_0 \\bar{A}_1 \\bar{B}_1) \\\\\n", - "& = |A_0B_0 + A_1 B_1|^2 \\\\\n", + "& = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 \\Re(A_0 B_0 \\bar{A} _1 \\bar{B} _1) \\\\\n", + "& = |A_0 B_0 + A_1 B_1|^2 \\\\\n", "& = \\frac{1}{4} |\\beta_{00}^2 - \\beta_{01}^2 - \\beta_{10}^2 + \\beta_{11}^2|\n", "\\end{align*} \n", "$$" -- Gitee From 36beee461c68781876cce09f9e4bdc11ead2952a Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Mon, 19 Sep 2022 23:20:26 +0800 Subject: [PATCH 06/14] test --- solutions/lecture15.ipynb | 9 ++++++++- 1 file changed, 8 insertions(+), 1 deletion(-) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index 58bac68..c3704e1 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -91,12 +91,19 @@ "metadata": {}, "source": [ "计算密度矩阵\n", - "$$\n", + "\n", + "$$\n", + "\\begin{aligned}\n", + "\\rho &=\\sum_{x y} \\sum_{a b} \\beta_{x y} \\bar{\\beta}_{a b}\\left|\\beta_{x y}\\right\\rangle\\left\\langle\\beta_{a b}\\right| \\\\\n", + "&=\\sum_{x, y, a, b} \\frac{\\beta_{x y} \\bar{\\beta}_{a b}}{2}\\left[|0, x\\rangle\\left\\langle 0, a\\left|+(-1)^{b}\\right| 0, x\\right\\rangle\\left\\langle 1, \\bar{a}\\left|+(-1)^{y}\\right| 1, \\bar{x}\\right\\rangle\\left\\langle 0, a\\left|+(-1)^{y+b}\\right| 1, \\bar{x}\\right\\rangle\\langle 1, \\bar{a}|\\right] \\\\\n", + "\\rho_{B} &=\\sum_{x, y, a, b} \\frac{\\beta_{x y} \\bar{\\beta}_{a b}}{2}\\left[|x\\rangle\\left\\langle a\\left|+(-1)^{y+b}\\right| \\bar{x}\\right\\rangle\\langle\\bar{a}|\\right]\n", + "\\end{aligned}\n", "$$" ] }, -- Gitee From 15802c0aa376f0d55781f9caf3086b82ed1d1d54 Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Mon, 19 Sep 2022 23:22:31 +0800 Subject: [PATCH 07/14] test --- solutions/lecture15.ipynb | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index c3704e1..0d4c790 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -100,9 +100,9 @@ "$$ -->\n", "$$\n", "\\begin{aligned}\n", - "\\rho &=\\sum_{x y} \\sum_{a b} \\beta_{x y} \\bar{\\beta}_{a b}\\left|\\beta_{x y}\\right\\rangle\\left\\langle\\beta_{a b}\\right| \\\\\n", - "&=\\sum_{x, y, a, b} \\frac{\\beta_{x y} \\bar{\\beta}_{a b}}{2}\\left[|0, x\\rangle\\left\\langle 0, a\\left|+(-1)^{b}\\right| 0, x\\right\\rangle\\left\\langle 1, \\bar{a}\\left|+(-1)^{y}\\right| 1, \\bar{x}\\right\\rangle\\left\\langle 0, a\\left|+(-1)^{y+b}\\right| 1, \\bar{x}\\right\\rangle\\langle 1, \\bar{a}|\\right] \\\\\n", - "\\rho_{B} &=\\sum_{x, y, a, b} \\frac{\\beta_{x y} \\bar{\\beta}_{a b}}{2}\\left[|x\\rangle\\left\\langle a\\left|+(-1)^{y+b}\\right| \\bar{x}\\right\\rangle\\langle\\bar{a}|\\right]\n", + "\\rho &=\\sum_{x y} \\sum_{a b} \\beta_{x y} \\bar{\\beta} _{a b}\\left|\\beta_{x y}\\right\\rangle\\left\\langle\\beta_{a b}\\right| \\\\\n", + "&=\\sum_{x, y, a, b} \\frac{\\beta_{x y} \\bar{\\beta} _{a b}}{2}\\left[|0, x\\rangle\\left\\langle 0, a\\left|+(-1)^{b}\\right| 0, x\\right\\rangle\\left\\langle 1, \\bar{a}\\left|+(-1)^{y}\\right| 1, \\bar{x}\\right\\rangle\\left\\langle 0, a\\left|+(-1)^{y+b}\\right| 1, \\bar{x}\\right\\rangle\\langle 1, \\bar{a}|\\right] \\\\\n", + "\\rho_{B} &=\\sum_{x, y, a, b} \\frac{\\beta_{x y} \\bar{\\beta} _{a b}}{2}\\left[|x\\rangle\\left\\langle a\\left|+(-1)^{y+b}\\right| \\bar{x}\\right\\rangle\\langle\\bar{a}|\\right]\n", "\\end{aligned}\n", "$$" ] -- Gitee From ae2f6b4fa4aa2f7dd75dc05d324aba11d4f1d72c Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Tue, 20 Sep 2022 23:50:31 +0800 Subject: [PATCH 08/14] test --- solutions/lecture15.ipynb | 48 ++++----------------------------------- 1 file changed, 4 insertions(+), 44 deletions(-) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index 0d4c790..a36b7c2 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -91,20 +91,7 @@ "metadata": {}, "source": [ "计算密度矩阵\n", - "\n", - "$$\n", - "\\begin{aligned}\n", - "\\rho &=\\sum_{x y} \\sum_{a b} \\beta_{x y} \\bar{\\beta} _{a b}\\left|\\beta_{x y}\\right\\rangle\\left\\langle\\beta_{a b}\\right| \\\\\n", - "&=\\sum_{x, y, a, b} \\frac{\\beta_{x y} \\bar{\\beta} _{a b}}{2}\\left[|0, x\\rangle\\left\\langle 0, a\\left|+(-1)^{b}\\right| 0, x\\right\\rangle\\left\\langle 1, \\bar{a}\\left|+(-1)^{y}\\right| 1, \\bar{x}\\right\\rangle\\left\\langle 0, a\\left|+(-1)^{y+b}\\right| 1, \\bar{x}\\right\\rangle\\langle 1, \\bar{a}|\\right] \\\\\n", - "\\rho_{B} &=\\sum_{x, y, a, b} \\frac{\\beta_{x y} \\bar{\\beta} _{a b}}{2}\\left[|x\\rangle\\left\\langle a\\left|+(-1)^{y+b}\\right| \\bar{x}\\right\\rangle\\langle\\bar{a}|\\right]\n", - "\\end{aligned}\n", - "$$" + "$$ \\begin{align*} \\rho & = \\sum_{xy} \\sum_{ab} \\beta_{xy} \\bar{\\beta}_{ab} |\\beta_{xy} \\rangle \\langle \\beta_{ab} | \\\\ & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \\big [ |0, x \\rangle \\langle 0, a | + (-1)^b |0, x \\rangle \\langle 1, \\bar{a} | + (-1)^y |1, \\bar{x} \\rangle \\langle 0, a | + (-1)^{y+b} |1, \\bar{x} \\rangle \\langle 1, \\bar{a} | \\big ] \\\\ \\rho_B & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \\big[ |x \\rangle \\langle a | + (-1)^{y+b} |\\bar{x} \\rangle \\langle \\bar{a} | \\big] \\end{align*}$$" ] }, { @@ -112,15 +99,7 @@ "metadata": {}, "source": [ "考虑计算第一部分\n", - "$$\n", - "\\begin{align*}\n", - "P_1 & = \\sum_{x,a,a,b} \\frac{\\beta_{xt} \\bar{\\beta}_{ab}}{2} |x \\rangle \\langle a | \\\\\n", - "& = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta_{xy}}{2} \\left ( \\bar{\\beta}_{aa} + \\bar{\\beta}_{a, \\bar{a}}\\right ) \\\\\n", - "& = \\sum_{x,a}|x \\rangle \\langle a | \\frac{\\beta_{xx} + \\beta_{x, \\bar{x}}}{\\sqrt{2}} \\left(\\frac{\\beta_{aa} + \\beta_{a, \n", - "bar{a}}}{\\sqrt{2}} \\right)^\\star \\\\\n", - "& = \\sum_{x, a} |x \\rangle \\langle a | A_x \\cdot \\bar{A}_a\n", - "\\end{align*}\n", - "$$" + "$$ \\begin{align*} P_1 & = \\sum_{x,a,a,b} \\frac{\\beta_{xt} \\bar{\\beta}_{ab}}{2} |x \\rangle \\langle a | \\\\ & = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta_{xy}}{2} \\left ( \\bar{\\beta}_{aa} + \\bar{\\beta}_{a, \\bar{a}}\\right ) \\\\ & = \\sum_{x,a}|x \\rangle \\langle a | \\frac{\\beta_{xx} + \\beta_{x, \\bar{x}}}{\\sqrt{2}} \\left(\\frac{\\beta_{aa} + \\beta_{a, \\bar{a}}}{\\sqrt{2}} \\right)^\\star \\\\ & = \\sum_{x, a} |x \\rangle \\langle a | A_x \\cdot \\bar{A} \\end{align*} $$" ] }, { @@ -141,19 +120,7 @@ "$$\n", "\n", "因此\n", - "$$\n", - "\\begin{align*}\n", - "\\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\big [ A_x \\cdot \\bar{A} _{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B} _{\\bar{a}} \\big] \\\\\n", - "& = \\begin{bmatrix}\n", - "A_0 \\bar{A} _0 + B_1 \\bar{B} _1 & A_0 \\bar{A} _1 - B_1 \\bar{B} _0 \\\\\n", - "A_1 \\bar{A} _0 - B_0 \\bar{B} _1 & A_1 \\bar{A} _1 + B_0 \\bar{B} _0\n", - "\\end{bmatrix} \\\\\n", - "& = \\begin{bmatrix}\n", - "A & B \\\\\n", - "C & D\n", - "\\end{bmatrix}\n", - "\\end{align*}\n", - "$$" + "$$ \\begin{align*} \\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\big [ A_x \\cdot \\bar{A} _{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B} _{\\bar{a}} \\big] \\\\ & = \\begin{bmatrix} A_0 \\bar{A} _0 + B_1 \\bar{B} _1 & A_0 \\bar{A} _1 - B_1 \\bar{B} _0 \\\\ A_1 \\bar{A} _0 - B_0 \\bar{B} _1 & A_1 \\bar{A} _1 + B_0 \\bar{B} _0 \\end{bmatrix} \\\\ & = \\begin{bmatrix} A & B \\\\ C & D \\end{bmatrix} \\end{align*} $$" ] }, { @@ -161,14 +128,7 @@ "metadata": {}, "source": [ "计算 $\\rho_B$ 的行列式\n", - "$$\n", - "\\begin{align*}\n", - "\\det \\rho_B & = AD - BC \\\\\n", - "& = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 \\Re(A_0 B_0 \\bar{A} _1 \\bar{B} _1) \\\\\n", - "& = |A_0 B_0 + A_1 B_1|^2 \\\\\n", - "& = \\frac{1}{4} |\\beta_{00}^2 - \\beta_{01}^2 - \\beta_{10}^2 + \\beta_{11}^2|\n", - "\\end{align*} \n", - "$$" + "$$ \\begin{align*} \\det \\rho_B & = AD - BC \\\\ & = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 \\Re(A_0 B_0 \\bar{A} _1 \\bar{B} _1) \\\\ & = |A_0 B_0 + A_1 B_1|^2 \\\\ & = \\frac{1}{4} |\\beta_{00}^2 - \\beta_{01}^2 - \\beta_{10}^2 + \\beta_{11}^2| \\end{align*} $$" ] }, { -- Gitee From f40f9119405ba277742b1c6bd2f014c8cf5daa60 Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Wed, 21 Sep 2022 00:02:22 +0800 Subject: [PATCH 09/14] test --- solutions/lecture15.ipynb | 37 ++++++++++++++++++++++++++++++++++--- 1 file changed, 34 insertions(+), 3 deletions(-) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index a36b7c2..130ca77 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -91,7 +91,24 @@ "metadata": {}, "source": [ "计算密度矩阵\n", - "$$ \\begin{align*} \\rho & = \\sum_{xy} \\sum_{ab} \\beta_{xy} \\bar{\\beta}_{ab} |\\beta_{xy} \\rangle \\langle \\beta_{ab} | \\\\ & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \\big [ |0, x \\rangle \\langle 0, a | + (-1)^b |0, x \\rangle \\langle 1, \\bar{a} | + (-1)^y |1, \\bar{x} \\rangle \\langle 0, a | + (-1)^{y+b} |1, \\bar{x} \\rangle \\langle 1, \\bar{a} | \\big ] \\\\ \\rho_B & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \\big[ |x \\rangle \\langle a | + (-1)^{y+b} |\\bar{x} \\rangle \\langle \\bar{a} | \\big] \\end{align*}$$" + "$$\n", + "\\begin{align*} \n", + "\\rho & = \\sum_{xy} \\sum_{ab} \\beta_{xy} \\bar{\\beta}_{ab} |\\beta_{xy} \\rangle \\langle \\beta_{ab} | \\\\ \n", + "& = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \n", + "\\big (\n", + "|0, x \\rangle \\langle 0, a | + \n", + "(-1)^b |0, x \\rangle \\langle 1, \\bar{a} | + \n", + "(-1)^y |1, \\bar{x} \\rangle \\langle 0, a | + \n", + "(-1)^{y+b} |1, \\bar{x} \\rangle \\langle 1, \\bar{a} | \n", + "\\big ) \\\\ \n", + "\\end{align*}\n", + "$$\n", + "\n", + "则\n", + "$$\n", + "\\rho_B = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \n", + "\\big( |x \\rangle \\langle a | + (-1)^{y+b} |\\bar{x} \\rangle \\langle \\bar{a} | \\big) \n", + "$$" ] }, { @@ -120,7 +137,16 @@ "$$\n", "\n", "因此\n", - "$$ \\begin{align*} \\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\big [ A_x \\cdot \\bar{A} _{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B} _{\\bar{a}} \\big] \\\\ & = \\begin{bmatrix} A_0 \\bar{A} _0 + B_1 \\bar{B} _1 & A_0 \\bar{A} _1 - B_1 \\bar{B} _0 \\\\ A_1 \\bar{A} _0 - B_0 \\bar{B} _1 & A_1 \\bar{A} _1 + B_0 \\bar{B} _0 \\end{bmatrix} \\\\ & = \\begin{bmatrix} A & B \\\\ C & D \\end{bmatrix} \\end{align*} $$" + "$$\n", + "\\begin{align*} \n", + "\\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\big [ A_x \\cdot \\bar{A} _{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B} _{\\bar{a}} \\big] \\\\ \n", + "& = \\begin{bmatrix} \n", + "A_0 \\bar{A} _0 + B_1 \\bar{B} _1 & A_0 \\bar{A} _1 - B_1 \\bar{B} _0 \\\\ \n", + "A_1 \\bar{A} _0 - B_0 \\bar{B} _1 & A_1 \\bar{A} _1 + B_0 \\bar{B} _0 \n", + "\\end{bmatrix} \\\\ \n", + "& = \\begin{bmatrix} A & B \\\\ C & D \\end{bmatrix} \n", + "\\end{align*} \n", + "$$" ] }, { @@ -128,7 +154,12 @@ "metadata": {}, "source": [ "计算 $\\rho_B$ 的行列式\n", - "$$ \\begin{align*} \\det \\rho_B & = AD - BC \\\\ & = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 \\Re(A_0 B_0 \\bar{A} _1 \\bar{B} _1) \\\\ & = |A_0 B_0 + A_1 B_1|^2 \\\\ & = \\frac{1}{4} |\\beta_{00}^2 - \\beta_{01}^2 - \\beta_{10}^2 + \\beta_{11}^2| \\end{align*} $$" + "$$\\begin{align*} \n", + "\\det \\rho_B & = AD - BC \\\\ \n", + "& = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 \\Re(A_0 B_0 \\bar{A} _1 \\bar{B} _1) \\\\ \n", + "& = |A_0 B_0 + A_1 B_1|^2 \\\\ \n", + "& = \\frac{1}{4} |\\beta_{00}^2 - \\beta_{01}^2 - \\beta_{10}^2 + \\beta_{11}^2|^2 \n", + "\\end{align*}$$" ] }, { -- Gitee From 9c48c3671b23292f19c2d182a1969a917f9a829d Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Wed, 21 Sep 2022 00:12:48 +0800 Subject: [PATCH 10/14] test --- solutions/lecture15.ipynb | 34 +++++++++++++++++++--------------- 1 file changed, 19 insertions(+), 15 deletions(-) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index 130ca77..b2d445c 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -91,17 +91,14 @@ "metadata": {}, "source": [ "计算密度矩阵\n", - "$$\n", - "\\begin{align*} \n", - "\\rho & = \\sum_{xy} \\sum_{ab} \\beta_{xy} \\bar{\\beta}_{ab} |\\beta_{xy} \\rangle \\langle \\beta_{ab} | \\\\ \n", - "& = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \n", - "\\big (\n", + "$$ \n", + "\\rho = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \n", + "\\left(\n", "|0, x \\rangle \\langle 0, a | + \n", "(-1)^b |0, x \\rangle \\langle 1, \\bar{a} | + \n", "(-1)^y |1, \\bar{x} \\rangle \\langle 0, a | + \n", "(-1)^{y+b} |1, \\bar{x} \\rangle \\langle 1, \\bar{a} | \n", - "\\big ) \\\\ \n", - "\\end{align*}\n", + "\\right)\n", "$$\n", "\n", "则\n", @@ -116,7 +113,14 @@ "metadata": {}, "source": [ "考虑计算第一部分\n", - "$$ \\begin{align*} P_1 & = \\sum_{x,a,a,b} \\frac{\\beta_{xt} \\bar{\\beta}_{ab}}{2} |x \\rangle \\langle a | \\\\ & = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta_{xy}}{2} \\left ( \\bar{\\beta}_{aa} + \\bar{\\beta}_{a, \\bar{a}}\\right ) \\\\ & = \\sum_{x,a}|x \\rangle \\langle a | \\frac{\\beta_{xx} + \\beta_{x, \\bar{x}}}{\\sqrt{2}} \\left(\\frac{\\beta_{aa} + \\beta_{a, \\bar{a}}}{\\sqrt{2}} \\right)^\\star \\\\ & = \\sum_{x, a} |x \\rangle \\langle a | A_x \\cdot \\bar{A} \\end{align*} $$" + "$$\n", + "\\begin{aligned}\n", + "P_1 & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} |x \\rangle \\langle a | \\\\ \n", + "& = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta_{xy}}{2} \\left ( \\bar{\\beta}_{aa} + \\bar{\\beta}_{a\\bar{a}}\\right ) \\\\ \n", + "& = \\sum_{x,a} |x \\rangle \\langle a | \\frac{\\beta_{xx} + \\beta_{x\\bar{x}}}{\\sqrt{2}} \\left(\\frac{\\beta_{aa} + \\beta_{a\\bar{a}}}{\\sqrt{2}} \\right)^\\star \\\\ \n", + "& = \\sum_{x,a} |x \\rangle \\langle a | A_x \\cdot \\bar{A} \\\\ \n", + "\\end{aligned}\n", + "$$" ] }, { @@ -130,22 +134,22 @@ "\n", "其中\n", "$$\n", - "\\begin{align*}\n", + "\\begin{aligned}\n", "A_x & \\equiv \\frac{\\beta_{xx} + \\beta_{x, \\bar{x}}}{\\sqrt{2}} \\\\\n", "B_x & \\equiv \\frac{\\beta_{xx} - \\beta_{x, \\bar{x}}}{\\sqrt{2}}\n", - "\\end{align*}\n", + "\\end{aligned}\n", "$$\n", "\n", "因此\n", "$$\n", - "\\begin{align*} \n", - "\\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\big [ A_x \\cdot \\bar{A} _{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B} _{\\bar{a}} \\big] \\\\ \n", + "\\begin{aligned} \n", + "\\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\left( A_x \\cdot \\bar{A} _{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B} _{\\bar{a}} \\right) \\\\ \n", "& = \\begin{bmatrix} \n", "A_0 \\bar{A} _0 + B_1 \\bar{B} _1 & A_0 \\bar{A} _1 - B_1 \\bar{B} _0 \\\\ \n", "A_1 \\bar{A} _0 - B_0 \\bar{B} _1 & A_1 \\bar{A} _1 + B_0 \\bar{B} _0 \n", "\\end{bmatrix} \\\\ \n", "& = \\begin{bmatrix} A & B \\\\ C & D \\end{bmatrix} \n", - "\\end{align*} \n", + "\\end{aligned} \n", "$$" ] }, @@ -154,12 +158,12 @@ "metadata": {}, "source": [ "计算 $\\rho_B$ 的行列式\n", - "$$\\begin{align*} \n", + "$$\\begin{aligned} \n", "\\det \\rho_B & = AD - BC \\\\ \n", "& = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 \\Re(A_0 B_0 \\bar{A} _1 \\bar{B} _1) \\\\ \n", "& = |A_0 B_0 + A_1 B_1|^2 \\\\ \n", "& = \\frac{1}{4} |\\beta_{00}^2 - \\beta_{01}^2 - \\beta_{10}^2 + \\beta_{11}^2|^2 \n", - "\\end{align*}$$" + "\\end{aligned}$$" ] }, { -- Gitee From d0bf7d3a234eb25219874affb17ae46e17958ce1 Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Wed, 21 Sep 2022 00:18:27 +0800 Subject: [PATCH 11/14] test --- solutions/lecture15.ipynb | 11 ++++++----- 1 file changed, 6 insertions(+), 5 deletions(-) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index b2d445c..af812f8 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -92,7 +92,7 @@ "source": [ "计算密度矩阵\n", "$$ \n", - "\\rho = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \n", + "\\rho = \\sum_{x,y,a,b} \\frac{\\beta _{xy} \\bar{\\beta}_{ab}}{2} \n", "\\left(\n", "|0, x \\rangle \\langle 0, a | + \n", "(-1)^b |0, x \\rangle \\langle 1, \\bar{a} | + \n", @@ -113,12 +113,13 @@ "metadata": {}, "source": [ "考虑计算第一部分\n", + "\n", "$$\n", "\\begin{aligned}\n", - "P_1 & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} |x \\rangle \\langle a | \\\\ \n", - "& = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta_{xy}}{2} \\left ( \\bar{\\beta}_{aa} + \\bar{\\beta}_{a\\bar{a}}\\right ) \\\\ \n", - "& = \\sum_{x,a} |x \\rangle \\langle a | \\frac{\\beta_{xx} + \\beta_{x\\bar{x}}}{\\sqrt{2}} \\left(\\frac{\\beta_{aa} + \\beta_{a\\bar{a}}}{\\sqrt{2}} \\right)^\\star \\\\ \n", - "& = \\sum_{x,a} |x \\rangle \\langle a | A_x \\cdot \\bar{A} \\\\ \n", + "P_1 & = \\sum_{x,y,a,b} \\frac{\\beta _{xy} \\bar{\\beta} _{ab}}{2} |x \\rangle \\langle a | \\\\ \n", + "& = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta _{xy}}{2} \\left ( \\bar{\\beta} _{aa} + \\bar{\\beta} _{a \\bar{a}}\\right ) \\\\ \n", + "& = \\sum_{x,a} |x \\rangle \\langle a | \\frac{\\beta _{xx} + \\beta _{x \\bar{x}}}{\\sqrt{2}} \\left(\\frac{\\beta _{aa} + \\beta _{a\\bar{a}}}{\\sqrt{2}} \\right)^* \\\\ \n", + "& = \\sum_{x,a} |x \\rangle \\langle a | A_x \\cdot \\bar{A} _a \\\\ \n", "\\end{aligned}\n", "$$" ] -- Gitee From d075355d234ffe8dce5a469ef2c28fd98bd00bdb Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Wed, 21 Sep 2022 00:20:48 +0800 Subject: [PATCH 12/14] test A_a --- solutions/lecture15.ipynb | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index af812f8..dc423f0 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -92,7 +92,7 @@ "source": [ "计算密度矩阵\n", "$$ \n", - "\\rho = \\sum_{x,y,a,b} \\frac{\\beta _{xy} \\bar{\\beta}_{ab}}{2} \n", + "\\rho = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \n", "\\left(\n", "|0, x \\rangle \\langle 0, a | + \n", "(-1)^b |0, x \\rangle \\langle 1, \\bar{a} | + \n", @@ -116,10 +116,10 @@ "\n", "$$\n", "\\begin{aligned}\n", - "P_1 & = \\sum_{x,y,a,b} \\frac{\\beta _{xy} \\bar{\\beta} _{ab}}{2} |x \\rangle \\langle a | \\\\ \n", - "& = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta _{xy}}{2} \\left ( \\bar{\\beta} _{aa} + \\bar{\\beta} _{a \\bar{a}}\\right ) \\\\ \n", - "& = \\sum_{x,a} |x \\rangle \\langle a | \\frac{\\beta _{xx} + \\beta _{x \\bar{x}}}{\\sqrt{2}} \\left(\\frac{\\beta _{aa} + \\beta _{a\\bar{a}}}{\\sqrt{2}} \\right)^* \\\\ \n", - "& = \\sum_{x,a} |x \\rangle \\langle a | A_x \\cdot \\bar{A} _a \\\\ \n", + "P_1 & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} |x \\rangle \\langle a | \\\\ \n", + "& = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta_{xy}}{2} \\left ( \\bar{\\beta}_{aa} + \\bar{\\beta}_{a \\bar{a}}\\right ) \\\\ \n", + "& = \\sum_{x,a} |x \\rangle \\langle a | \\frac{\\beta_{xx} + \\beta_{x \\bar{x}}}{\\sqrt{2}} \\left(\\frac{\\beta_{aa} + \\beta_{a\\bar{a}}}{\\sqrt{2}} \\right)^* \\\\ \n", + "& = \\sum_{x,a} |x \\rangle \\langle a | A_x \\cdot \\bar{A}_a \\\\ \n", "\\end{aligned}\n", "$$" ] -- Gitee From f957e8f19a942403a93165433dc2d633f066873d Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Wed, 21 Sep 2022 00:23:34 +0800 Subject: [PATCH 13/14] test for star --- solutions/lecture15.ipynb | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index dc423f0..aa71edf 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -116,8 +116,8 @@ "\n", "$$\n", "\\begin{aligned}\n", - "P_1 & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} |x \\rangle \\langle a | \\\\ \n", - "& = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta_{xy}}{2} \\left ( \\bar{\\beta}_{aa} + \\bar{\\beta}_{a \\bar{a}}\\right ) \\\\ \n", + "P_1 & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\beta_{ab}^\\star }{2} |x \\rangle \\langle a | \\\\ \n", + "& = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta_{xy}}{2} \\left ( \\beta_{aa}^\\star + \\beta_{a \\bar{a}}^\\star \\right ) \\\\ \n", "& = \\sum_{x,a} |x \\rangle \\langle a | \\frac{\\beta_{xx} + \\beta_{x \\bar{x}}}{\\sqrt{2}} \\left(\\frac{\\beta_{aa} + \\beta_{a\\bar{a}}}{\\sqrt{2}} \\right)^* \\\\ \n", "& = \\sum_{x,a} |x \\rangle \\langle a | A_x \\cdot \\bar{A}_a \\\\ \n", "\\end{aligned}\n", -- Gitee From e2464ecb636c78e22f30dced29f4b97795b895c1 Mon Sep 17 00:00:00 2001 From: zhuyk6 Date: Wed, 21 Sep 2022 00:29:14 +0800 Subject: [PATCH 14/14] test --- solutions/lecture15.ipynb | 16 ++++++++-------- 1 file changed, 8 insertions(+), 8 deletions(-) diff --git a/solutions/lecture15.ipynb b/solutions/lecture15.ipynb index aa71edf..c088dab 100644 --- a/solutions/lecture15.ipynb +++ b/solutions/lecture15.ipynb @@ -92,7 +92,7 @@ "source": [ "计算密度矩阵\n", "$$ \n", - "\\rho = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \n", + "\\rho = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\beta_{ab}^\\star }{2} \n", "\\left(\n", "|0, x \\rangle \\langle 0, a | + \n", "(-1)^b |0, x \\rangle \\langle 1, \\bar{a} | + \n", @@ -103,7 +103,7 @@ "\n", "则\n", "$$\n", - "\\rho_B = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\bar{\\beta}_{ab}}{2} \n", + "\\rho_B = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\beta_{ab}^\\star }{2} \n", "\\big( |x \\rangle \\langle a | + (-1)^{y+b} |\\bar{x} \\rangle \\langle \\bar{a} | \\big) \n", "$$" ] @@ -119,7 +119,7 @@ "P_1 & = \\sum_{x,y,a,b} \\frac{\\beta_{xy} \\beta_{ab}^\\star }{2} |x \\rangle \\langle a | \\\\ \n", "& = \\sum_{x,a} |x \\rangle \\langle a | \\sum_y \\frac{\\beta_{xy}}{2} \\left ( \\beta_{aa}^\\star + \\beta_{a \\bar{a}}^\\star \\right ) \\\\ \n", "& = \\sum_{x,a} |x \\rangle \\langle a | \\frac{\\beta_{xx} + \\beta_{x \\bar{x}}}{\\sqrt{2}} \\left(\\frac{\\beta_{aa} + \\beta_{a\\bar{a}}}{\\sqrt{2}} \\right)^* \\\\ \n", - "& = \\sum_{x,a} |x \\rangle \\langle a | A_x \\cdot \\bar{A}_a \\\\ \n", + "& = \\sum_{x,a} |x \\rangle \\langle a | A_x \\cdot A_a^\\star \\\\ \n", "\\end{aligned}\n", "$$" ] @@ -130,7 +130,7 @@ "source": [ "同理可以计算出来第二部分\n", "$$\n", - "P_2 = \\sum_{x,a} |x \\rangle \\langle a | (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B}_{\\bar{a}}\n", + "P_2 = \\sum_{x,a} |x \\rangle \\langle a | (-1)^{x+a} B_{\\bar{x}} \\cdot B_{\\bar{a}}^\\star\n", "$$\n", "\n", "其中\n", @@ -144,10 +144,10 @@ "因此\n", "$$\n", "\\begin{aligned} \n", - "\\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\left( A_x \\cdot \\bar{A} _{a} + (-1)^{x+a} B_{\\bar{x}} \\cdot \\bar{B} _{\\bar{a}} \\right) \\\\ \n", + "\\rho_B & = \\sum_{x,a} |x \\rangle \\langle a | \\left( A_x \\cdot A_{a}^\\star + (-1)^{x+a} B_{\\bar{x}} \\cdot B_{\\bar{a}}^\\star \\right) \\\\ \n", "& = \\begin{bmatrix} \n", - "A_0 \\bar{A} _0 + B_1 \\bar{B} _1 & A_0 \\bar{A} _1 - B_1 \\bar{B} _0 \\\\ \n", - "A_1 \\bar{A} _0 - B_0 \\bar{B} _1 & A_1 \\bar{A} _1 + B_0 \\bar{B} _0 \n", + "A_0 A_0^\\star + B_1 B_1^\\star & A_0 A_1^\\star - B_1 B_0^\\star \\\\ \n", + "A_1 A_0^\\star - B_0 B_1^\\star & A_1 A_1^\\star + B_0 B_0^\\star \n", "\\end{bmatrix} \\\\ \n", "& = \\begin{bmatrix} A & B \\\\ C & D \\end{bmatrix} \n", "\\end{aligned} \n", @@ -161,7 +161,7 @@ "计算 $\\rho_B$ 的行列式\n", "$$\\begin{aligned} \n", "\\det \\rho_B & = AD - BC \\\\ \n", - "& = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 \\Re(A_0 B_0 \\bar{A} _1 \\bar{B} _1) \\\\ \n", + "& = |A_0 B_0|^2 + |A_1 B_1|^2 + 2 \\Re(A_0 B_0 A_1^\\star B_1^\\star) \\\\ \n", "& = |A_0 B_0 + A_1 B_1|^2 \\\\ \n", "& = \\frac{1}{4} |\\beta_{00}^2 - \\beta_{01}^2 - \\beta_{10}^2 + \\beta_{11}^2|^2 \n", "\\end{aligned}$$" -- Gitee