From 42ba863172d6845aa21fcc0c7d571230a933ef90 Mon Sep 17 00:00:00 2001
From: dsdsdshe <jedzhou99@163.com>
Date: Wed, 10 Aug 2022 21:39:54 +0800
Subject: [PATCH] lecture3 and lecture5

---
 images/commutative_relation.png | Bin 0 -> 35834 bytes
 lecture3.ipynb                  | 261 +++++++++++++++++++++++
 lecture5.ipynb                  | 358 ++++++++++++++++++++++++++++++++
 3 files changed, 619 insertions(+)
 create mode 100644 images/commutative_relation.png
 create mode 100644 lecture3.ipynb
 create mode 100644 lecture5.ipynb

diff --git a/images/commutative_relation.png b/images/commutative_relation.png
new file mode 100644
index 0000000000000000000000000000000000000000..39f0782c80f6bd6b3f81e0e6fca9eb3ca7504d67
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diff --git a/lecture3.ipynb b/lecture3.ipynb
new file mode 100644
index 0000000..813dd9e
--- /dev/null
+++ b/lecture3.ipynb
@@ -0,0 +1,261 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Lecture 3\n",
+    "## 贝尔态\n",
+    "- 两比特的最大纠缠态称为贝尔态（总共四种）：\n",
+    "$$\n",
+    "|\\psi_{Bell} \\rangle = \\frac{1}{\\sqrt{2}}(|00\\rangle + |11\\rangle)\n",
+    "$$\n",
+    "\n",
+    "- 剩下三种贝尔态为\n",
+    "$$\n",
+    "\\frac{1}{\\sqrt{2}}(|00\\rangle - |11\\rangle),\\enspace\\frac{1}{\\sqrt{2}}(|01\\rangle + |10\\rangle),\\enspace \\frac{1}{\\sqrt{2}}(|01\\rangle - |10\\rangle)\n",
+    "$$\n",
+    "\n",
+    "- 多比特量子态表示为：\n",
+    "$$\n",
+    "|\\psi_n \\rangle = \\sum_{x\\in[0,1]^n}c_{x_1 x_2 \\cdots x_n}|x_1 x_2 \\cdots x_n\\rangle = \\sum_{x\\in [0,1]^n} c_x|x\\rangle \\ ,\n",
+    "$$\n",
+    "其中 $\\sum_x |c_x|^2=1$。\n",
+    "\n",
+    "- 若想在MindQuantum中将模拟器的量子态设置为第一种贝尔态，则需要得知贝尔态的列向量：\n",
+    "$$\n",
+    "\\frac{1}{\\sqrt{2}}(|00\\rangle + |11\\rangle) = \\frac{1}{\\sqrt{2}}\\begin{pmatrix}\n",
+    "1 \\\\ 0 \\\\ 0 \\\\ 1\n",
+    "\\end{pmatrix}\n",
+    "$$\n",
+    "然后通过`set_qs`接口改变模拟器状态，该接口会自动将输入向量归一化，因此可以省略系数，如下所示："
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "√2/2¦00⟩\n",
+      "√2/2¦11⟩\n"
+     ]
+    }
+   ],
+   "source": [
+    "from mindquantum.simulator import Simulator\n",
+    "import numpy as np\n",
+    "\n",
+    "sim = Simulator('projectq', 2)\n",
+    "sim.set_qs(np.array([1, 0, 0, 1]))\n",
+    "bell_state = sim.get_qs(ket=True)\n",
+    "print(bell_state)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## 幺正变换\n",
+    "- 先前提到量子态是一个（列）向量$|\\psi\\rangle$（ket），并且其振幅的平方就是概率。与之对应的则是（行）向量$\\langle \\psi|$（bra），\n",
+    "\n",
+    "$$\n",
+    "|\\psi \\rangle = \\alpha |0\\rangle + \\beta |1 \\rangle \\to \\langle \\psi| \\equiv \\alpha^* \\langle 0 | + \\beta^* \\langle 1|\n",
+    "$$\n",
+    "\n",
+    "内积则称为 \"bracket\"\n",
+    "\n",
+    "$$\n",
+    "\\langle \\psi| \\psi \\rangle = |\\alpha|^2+|\\beta|^2=1\n",
+    "$$\n",
+    "\n",
+    "这也被称为归一化条件。\n",
+    "\n",
+    "- 因此若量子态是一个向量，则改变量子态的操作必须是一个矩阵（基于线性代数）。\n",
+    "$$\n",
+    "|\\psi\\rangle \\to U|\\psi\\rangle\n",
+    "$$\n",
+    "在对偶空间则表示为\n",
+    "$$\n",
+    "\\langle\\psi|\\to \\langle\\psi|U^\\dagger\n",
+    "$$\n",
+    "其中$A^\\dagger \\equiv (A^T)^*$。\n",
+    "\n",
+    "- 为了保证概率守恒，我们需要\n",
+    "$$\n",
+    "\\langle \\psi|\\psi\\rangle = \\langle \\psi|U^\\dagger U|\\psi\\rangle = 1\n",
+    "$$\n",
+    "对于所有可能的$|\\psi\\rangle$都成立。\n",
+    "\n",
+    "- 这就意味着\n",
+    "$$\n",
+    "U^\\dagger U = I \\Rightarrow U^\\dagger = U^{-1}\n",
+    "$$\n",
+    "这正是幺正矩阵的定义。\n",
+    "\n",
+    "### 总结性回顾\n",
+    "- 任何对量子态的改变都可以用一个矩阵来表示。\n",
+    "\n",
+    "- 由于概率守恒，该矩阵必须保证幺正性。\n",
+    "\n",
+    "- 换句话说，任何量子计算的量子动力学行为都必须是一种幺正变换（的子集）。\n",
+    "\n",
+    "### 作为幺正变换的量子计算？\n",
+    "- 接下来，我们可以想象量子计算是由一类幺正变换来表示，例如\n",
+    "$$\n",
+    "U|\\psi_{in}\\rangle=|\\psi_{out}\\rangle\n",
+    "$$\n",
+    "- 传统上，物理学家以这种方式解决问题：已知初态和哈密顿量，寻找末态。\n",
+    "- 但对于量子计算，通常我们已知初态与末态，而目标是找到量子线路$U$（量子算法）。\n",
+    "- 与经典计算相比较：\n",
+    "$$\n",
+    "f(x_{in}）=y_{out}\n",
+    "$$\n",
+    "- 我们稍后会表明量子计算是更加普适的，这意味着我们也可以用量子计算机完成经典计算任务。\n",
+    "- 问题的关键是一旦定义好输入与输出之间的逻辑关系，我们需要知道如何在物理上实现其对应的幺正变换（称为量子电路）。\n",
+    "- 回想一下，经典电路由基本的逻辑运算组成。（AND, OR, NOT, NAND等）\n",
+    "- 量子电路同样应该由基本逻辑运算组成，它们是什么？（我们会在之后发现）\n",
+    "\n",
+    "\n",
+    "## 术语\n",
+    "- 量子线路：实现量子算法的一种方式（更大的幺正矩阵）\n",
+    "\n",
+    "- 量子门：量子电路的元素（较小的幺正矩阵）\n",
+    "\n",
+    "- 算符：矩阵（不一定幺正）\n",
+    "\n",
+    "## 泡利`X`门\n",
+    "**单量子比特门$\\equiv$单量子比特幺正矩阵（变换）**\n",
+    "\n",
+    "- 让我们从经典的单比特门入手：\n",
+    "$$\n",
+    "Logical \\enspace NOT \\enspace gate: \\enspace 1 \\to 0 \\enspace and \\enspace 0 \\to 1\n",
+    "$$\n",
+    "- 而现在我们想要如下效果\n",
+    "$$\n",
+    "Logical \\enspace NOT \\enspace gate: \\enspace |1\\rangle \\to |0\\rangle \\enspace and \\enspace |0\\rangle \\to |1\\rangle\n",
+    "$$\n",
+    "这也被称为**比特翻转**。\n",
+    "\n",
+    "- 使用泡利`X`门就可以做到上述操作\n",
+    "$$\n",
+    "\\sigma_x \\equiv U_X \\equiv X = |1\\rangle\\langle 1| = \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}\n",
+    "$$\n",
+    "\n",
+    "- 可以验证泡利`X`门是幺正矩阵：\n",
+    "$$\n",
+    "X^\\dagger X = \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} = I\n",
+    "$$\n",
+    "\n",
+    "- 用狄拉克符号可以容易的验证：\n",
+    "\n",
+    "$$\n",
+    "X|1\\rangle = (|1\\rangle\\langle 0|+|0\\rangle\\langle 1|)|1\\rangle = |1\\rangle\\langle 0|1\\rangle + |0\\rangle\\langle 1|1\\rangle = |0\\rangle\n",
+    "$$\n",
+    "\n",
+    "$$\n",
+    "X|0\\rangle = (|1\\rangle\\langle 0|+|0\\rangle\\langle 1|)|0\\rangle = |1\\rangle\\langle 0|0\\rangle + |0\\rangle\\langle 1|0\\rangle = |1\\rangle\n",
+    "$$\n",
+    "\n",
+    "- 泡利`X`门作用在任意量子态的效果如下：\n",
+    "$$\n",
+    "X|\\psi\\rangle=X(\\alpha|0\\rangle+\\beta|1\\rangle)=\\alpha X|0\\rangle+\\beta X|1\\rangle = \\alpha |1\\rangle+\\beta |0\\rangle\n",
+    "$$\n",
+    "\n",
+    "$$\n",
+    "X\\begin{pmatrix}\\alpha \\\\ \\beta \\end{pmatrix} = \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix} \\begin{pmatrix}\\alpha \\\\ \\beta \\end{pmatrix} = \\begin{pmatrix}\\beta \\\\ \\alpha \\end{pmatrix}\n",
+    "$$\n",
+    "\n",
+    "<br>\n",
+    "\n",
+    "- 现在我们可以尝试在MindQuantum中搭建一个简单的量子电路——对一个量子比特作用泡利`X`门。首先创建一个量子电路："
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1¦0⟩\n"
+     ]
+    }
+   ],
+   "source": [
+    "from mindquantum.core.circuit import Circuit\n",
+    "\n",
+    "circ = Circuit()\n",
+    "quantum_state = circ.get_qs(ket=True)\n",
+    "print(quantum_state)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "可以看到电路的初始量子态处于$|0\\rangle$态，现在我们作用泡利`X`门，然后打印电路量子态："
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "q0: ──X──\n",
+      "1¦1⟩\n"
+     ]
+    }
+   ],
+   "source": [
+    "from mindquantum.core.gates import X\n",
+    "\n",
+    "circ += X.on(0) #对第0个比特作用X门\n",
+    "print(circ)     #打印线路\n",
+    "new_state = circ.get_qs(ket=True)\n",
+    "print(new_state)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "可以看到，对$q0$作用泡利`X`门后，量子态变为$|1\\rangle$态。"
+   ]
+  }
+ ],
+ "metadata": {
+  "interpreter": {
+   "hash": "128f52b0cdc33bc7d2583e313ce49af13392215631bffc3f029664446222cd2a"
+  },
+  "kernelspec": {
+   "display_name": "Python 3.9.12 ('base')",
+   "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": "3.9.9"
+  },
+  "orig_nbformat": 4
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
diff --git a/lecture5.ipynb b/lecture5.ipynb
new file mode 100644
index 0000000..c7e4fdb
--- /dev/null
+++ b/lecture5.ipynb
@@ -0,0 +1,358 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Lecture 5\n",
+    "\n",
+    "**Lecture 4 课后练习解答**\n",
+    "- 考虑$R_x(\\theta)=e^{-i\\theta X/2}$，让我们回顾一下$e$指数上矩阵的定义——级数展开，现在我们考虑如下简单情形：\n",
+    "$$\n",
+    "Q_\\theta = e^{\\theta X} = I+\\theta X + \\frac{\\theta^2}{2!}X^2+\\frac{\\theta^3}{3!}X^3+\\cdots\n",
+    "$$\n",
+    "先前提到$X^2 = I$，因此\n",
+    "$$\n",
+    "Q_{\\theta}=e^{\\theta X}=I+\\theta X+\\frac{\\theta^{2}}{2 !}+\\frac{\\theta^{3}}{3 !} X+\\ldots=\\left(1+\\frac{\\theta^{2}}{2 !}+\\ldots\\right) I+\\left(1+\\frac{\\theta^{4}}{3 !}+\\ldots\\right) X\n",
+    "$$\n",
+    "又有$e^x=\\cosh{x}+\\sinh{x}$，因此我们可以将上式总结为$Q_\\theta=I\\cosh{\\theta}+X\\sinh{\\theta}$。\n",
+    "\n",
+    "- 将$\\theta$替换为$-i\\theta/2$，即得到\n",
+    "\n",
+    "$$\n",
+    "R_{x}(\\theta)=I \\cos \\frac{\\theta}{2}-i X \\sin \\frac{\\theta}{2}=\\left(\\begin{array}{cc}\n",
+    "c_{\\theta} & 0 \\\\\n",
+    "0 & c_{\\theta}\n",
+    "\\end{array}\\right)-i\\left(\\begin{array}{cc}\n",
+    "0 & s_{\\theta} \\\\\n",
+    "s_{\\theta} & 0\n",
+    "\\end{array}\\right)=\\left(\\begin{array}{cc}\n",
+    "c_{\\theta} & -i s_{\\theta} \\\\\n",
+    "-i s_{\\theta} & c_{\\theta}\n",
+    "\\end{array}\\right) .\n",
+    "$$\n",
+    "\n",
+    "- 同样的，由于$Y|0\\rangle=i|1\\rangle$，并且\n",
+    "$$\n",
+    "R_y(\\theta)=Icos\\frac{\\theta}{2}-iYsin\\frac{\\theta}{2}\n",
+    "$$\n",
+    "这意味着$R_y(\\theta)|0\\rangle=cos\\frac{\\theta}{2}|0\\rangle+sin\\frac{\\theta}{2}|1\\rangle$。\n",
+    "\n",
+    "---\n",
+    "\n",
+    "**下面的特性你需要了解：**\n",
+    "$$\n",
+    "X R_{z}(\\theta) X=R_{z}(-\\theta)=R_{z}(\\theta)^{-1}=R_{z}(\\theta)^{\\dagger}\n",
+    "$$\n",
+    "最后的关系式来源于$R_z(\\theta)$的幺正性。验证一下，$(Z^\\dagger=Z)$\n",
+    "$$\n",
+    "R_{z}(\\theta)^{\\dagger}=I \\cos \\frac{\\theta}{2}+i Z \\sin \\frac{\\theta}{2}=e^{i \\theta Z / 2}\n",
+    "$$\n",
+    "第一个关系式也是正确的，有如下两种方式验证：\n",
+    "1. 直接的办法，已知$R_{z}(\\theta)(\\alpha|0\\rangle+\\beta|1\\rangle)=\\alpha e^{-i \\theta / 2}|0\\rangle+\\beta e^{i \\theta / 2}|1\\rangle$，则有\n",
+    "$$\n",
+    "\\begin{aligned}\n",
+    "X R_{z}(\\theta) X(\\alpha|0\\rangle+\\beta|1\\rangle) &=X R_{z}(\\theta)(\\alpha|1\\rangle+\\beta|0\\rangle) \\\\\n",
+    "&=X\\left(\\alpha e^{i \\theta / 2}|1\\rangle+\\beta e^{-i \\theta / 2}|0\\rangle\\right) \\\\\n",
+    "&=\\alpha e^{i \\theta / 2}|0\\rangle+\\beta e^{-i \\theta / 2}|1\\rangle\n",
+    "\\end{aligned}\n",
+    "$$\n",
+    "然后与$R_{z}(-\\theta)(\\alpha|0\\rangle+\\beta|1\\rangle)=\\alpha e^{i \\theta / 2}|0\\rangle+\\beta e^{-i \\theta / 2}|1\\rangle$比较。\n",
+    "\n",
+    "2. “更好”的方法可能是：\n",
+    "$$\n",
+    "X R_{z}(\\theta) X=X I X \\cos \\frac{\\theta}{2}-i X Z X \\sin \\frac{\\theta}{2}=I \\cos \\frac{\\theta}{2}+i Z \\sin \\frac{\\theta}{2}\n",
+    "$$\n",
+    "因为我们知道$X Z X=-Z $，或 $ Z X+X Z=0 $ 亦或 $ X Z=-X Z$（对于任何泡利矩阵的组合都成立）\n",
+    "\n",
+    "**Figure from wikipedia**\n",
+    "\n",
+    "![commutative_relation](images/commutative_relation.png)\n",
+    "\n",
+    "- 注意到$XY-YX=2iZ$，这意味着$XY=iZ$。两个不同的泡利矩阵相乘与第三个泡利矩阵成正比。\n",
+    "\n",
+    "<br>\n",
+    "\n",
+    "- 现在让我们在 MindQuantum 中验证上述性质，我们随便选取一个$\\theta$的值，然后观察$X R_{z}(\\theta) X$、$R_{z}(-\\theta)$、$R_{z}(\\theta)^{-1}$和$R_{z}(\\theta)^{\\dagger}$作用在$|0\\rangle$态上的结果是否相同。\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "q0: ──X────RZ(1)────X──\n",
+      "(0.8775825618903728+0.479425538604203j)¦0⟩\n"
+     ]
+    }
+   ],
+   "source": [
+    "from mindquantum.core.gates import RZ, X\n",
+    "from mindquantum.core.circuit import Circuit\n",
+    "\n",
+    "theta = 1\n",
+    "\n",
+    "# X Rz(𝜃) X\n",
+    "circ = Circuit()\n",
+    "circ += X.on(0)\n",
+    "circ += RZ(theta).on(0)\n",
+    "circ += X.on(0)\n",
+    "print(circ)\n",
+    "print(circ.get_qs(ket=True))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "q0: ──RZ(-1)──\n",
+      "(0.8775825618903728+0.479425538604203j)¦0⟩\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Rz(-𝜃)\n",
+    "circ2 = Circuit()\n",
+    "circ2 += RZ(-theta).on(0)\n",
+    "print(circ2)\n",
+    "print(circ2.get_qs(ket=True))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "q0: ──RZ(1)^-1──\n",
+      "(0.877582561890373+0.4794255386042032j)¦0⟩\n"
+     ]
+    }
+   ],
+   "source": [
+    "# inverse of Rz(𝜃)\n",
+    "from mindquantum.core.gates import Power\n",
+    "circ3 = Circuit()\n",
+    "circ3 += Power(RZ(theta), -1).on(0)\n",
+    "print(circ3)\n",
+    "print(circ3.get_qs(ket=True))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "q0: ──RZ(-1)──\n",
+      "(0.8775825618903728+0.479425538604203j)¦0⟩\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Rz(𝜃) with dagger\n",
+    "circ4 = Circuit()\n",
+    "circ4 += RZ(theta).hermitian().on(0)\n",
+    "print(circ4)\n",
+    "print(circ4.get_qs(ket=True))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "可以看到四种方式得到的线路末态完全一致。"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "\n",
+    "- 最后，为何一定是$X, Y, Z$？我们也可以考虑一个任意的旋转。这需要我们定义\n",
+    "$$\n",
+    "R_{n}(\\theta) \\equiv e^{-i(\\theta / 2) n \\cdot \\sigma}=c_{\\theta} I-i s_{\\theta}(n \\cdot \\sigma)\n",
+    "$$\n",
+    "其中$n\\cdot\\sigma\\equiv n_x X+n_y Y+n_z Z$，且$n_x^2+n_y^2+n_z^2=1$。（第二个等式是一个好的练习/作业）\n",
+    "\n",
+    "- 该旋转算子可以生成任意的单量子比特旋转$U$（相差一个全局相位）。从某种意义上说，给定任何的$U$，人们总是可以选择角度$\\theta$和旋转轴$\\hat{n}$，使得（**想想看**）\n",
+    "$$\n",
+    "U=e^{i\\alpha}R_n(\\theta)\n",
+    "$$\n",
+    "\n",
+    "---\n",
+    "\n",
+    "- 部分证明：首先，任何 $2\\times 2$ 矩阵都可以用泡利矩阵展开（完全基）。回想一下，任何状态都可以用 0，1 基分解，即$|\\psi\\rangle=a|0\\rangle+b|1\\rangle$。一开始你不需要知道 a 和 b。 为了确定它们，我们可以使用内积，$\\langle 0|\\psi\\rangle=a\\langle 0|0\\rangle + b\\langle 0|1\\rangle=a$。\n",
+    "\n",
+    "- 类似的，我们可以对泡利矩阵做同样的事。即对任意$2\\times 2$矩阵$U$，我们总是可以写出\n",
+    "$$\n",
+    "U=aI+bX+cY+dZ\n",
+    "$$\n",
+    "\n",
+    "- 第一步，$X, Y, Z$泡利矩阵是无迹的。（迹 $\\operatorname{Tr}$ 含义是对矩阵对角线上的值求和；$\\operatorname{Tr}$ 是线性运算）\n",
+    "$$\n",
+    "\\operatorname{Tr}(U)=a \\operatorname{Tr}(I)+b \\operatorname{Tr}(X)+c \\operatorname{Tr}(Y)+d \\operatorname{Tr}(Z)=2 a\n",
+    "$$\n",
+    "这意味着$a=\\operatorname{Tr}(U)/2$。\n",
+    "\n",
+    "- 接下来我们将$X$作用于$U$，得到\n",
+    "$$\n",
+    "XU=aX+bI+cXY+dXZ\n",
+    "$$\n",
+    "\n",
+    "- 然后对上式求迹，\n",
+    "$$\n",
+    "\\operatorname{Tr}(X U)=b \\operatorname{Tr}(I)+c \\operatorname{Tr}(X Y)+d \\operatorname{Tr}(X Z)=2 b\n",
+    "$$\n",
+    "这意味着 $b=\\operatorname{Tr}(XU)/2$ 。同理有 $c=\\operatorname{Tr}(YU)/2$ 和 $d=\\operatorname{Tr}(ZU)/2$ 。\n",
+    "\n",
+    "- 至此，如果我们使$a=|a|e^{i\\alpha}$，离证明完成已经非常接近了。"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "\n",
+    "---\n",
+    "\n",
+    "- 哈达玛门$H$ (Hadamard gate)：（我认为几乎所有量子电路都有哈达玛门）\n",
+    "$$\n",
+    "H \\equiv \\frac{1}{\\sqrt{2}}\\left(\\begin{array}{cc}\n",
+    "1 & 1 \\\\\n",
+    "1 & -1\n",
+    "\\end{array}\\right)=\\frac{1}{\\sqrt{2}}\\left(\\begin{array}{l}\n",
+    "1 \\\\\n",
+    "1\n",
+    "\\end{array}\\right)\\langle 0|+\\frac{1}{\\sqrt{2}}\\left(\\begin{array}{c}\n",
+    "1 \\\\\n",
+    "-1\n",
+    "\\end{array}\\right)\\langle 1|\n",
+    "$$\n",
+    "其中\n",
+    "$$\n",
+    "H|0\\rangle=\\frac{1}{\\sqrt{2}}(|0\\rangle+|1\\rangle) \\quad, H|1\\rangle=\\frac{1}{\\sqrt{2}}(|0\\rangle-|1\\rangle)\n",
+    "$$\n",
+    "\n",
+    "- 注意到$H$也是自逆的，即$H^2=I$，并且$H=(X+Z)/\\sqrt{2}$。\n",
+    "\n",
+    "- 以下是一些哈达玛门的性质会在之后用到：\n",
+    "$$\n",
+    "HXH=Z\\quad HZH=X\\quad HYH=-Y\n",
+    "$$\n",
+    "\n",
+    "- 例如，$HR_x(\\theta)H=Icos\\frac{\\theta}{2}-iHXHsin\\frac{\\theta}{2}=R_z(\\theta)$。\n",
+    "\n",
+    "- 最后我们还需要知道另外两个单比特门：\n",
+    "$$\n",
+    "S \\equiv\\left(\\begin{array}{ll}\n",
+    "1 & 0 \\\\\n",
+    "0 & i\n",
+    "\\end{array}\\right) \\quad, \\quad T \\equiv\\left(\\begin{array}{cc}\n",
+    "1 & 0 \\\\\n",
+    "0 & e^{i \\pi / 4}\n",
+    "\\end{array}\\right)\n",
+    "$$\n",
+    "并且有$S=T^2$。\n",
+    "\n",
+    "- 实际上，在实验室中，人们可能不具备从任意方向控制量子比特旋转的能力。\n",
+    "\n",
+    "- 比如说，如果我们只能实现$R_z$和$R_y$该怎么办？\n",
+    "\n",
+    "<br>\n",
+    "\n",
+    "在 MindQuantum 中也有 `H` , `S` 和 `T` 门，使用方法如下："
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "q0: ──H───────\n",
+      "\n",
+      "q1: ──S────T──\n"
+     ]
+    }
+   ],
+   "source": [
+    "from mindquantum.core.gates import H, S, T\n",
+    "from mindquantum.core.circuit import Circuit\n",
+    "\n",
+    "circ = Circuit()\n",
+    "circ += H.on(0)\n",
+    "circ += S.on(1)\n",
+    "circ += T.on(1)\n",
+    "print(circ)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### 作业\n",
+    "\n",
+    "1. 请用 MindQuantum 验证下式是否成立\n",
+    "$$\n",
+    "H|0\\rangle=\\frac{1}{\\sqrt{2}}(|0\\rangle+|1\\rangle) \\quad, H|1\\rangle=\\frac{1}{\\sqrt{2}}(|0\\rangle-|1\\rangle)\n",
+    "$$\n",
+    "\n",
+    "2. 请用 MindQuantum 验证下式是否成立\n",
+    "$$\n",
+    "HR_x(\\theta)H=R_z(\\theta)\n",
+    "$$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "interpreter": {
+   "hash": "aee8b7b246df8f9039afb4144a1f6fd8d2ca17a180786b69acc140d282b71a49"
+  },
+  "kernelspec": {
+   "display_name": "Python 3.9.9 64-bit",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
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+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
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+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.9.9"
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+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
-- 
Gitee