{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# 量子相位估计算法\n", "\n", "[![下载Notebook](https://mindspore-website.obs.cn-north-4.myhuaweicloud.com/website-images/master/resource/_static/logo_notebook.svg)](https://mindspore-website.obs.cn-north-4.myhuaweicloud.com/notebook/master/mindquantum/zh_cn/case_library/mindspore_quantum_phase_estimation.ipynb) \n", "[![下载样例代码](https://mindspore-website.obs.cn-north-4.myhuaweicloud.com/website-images/master/resource/_static/logo_download_code.svg)](https://mindspore-website.obs.cn-north-4.myhuaweicloud.com/notebook/master/mindquantum/zh_cn/case_library/mindspore_quantum_phase_estimation.py) \n", "[![查看源文件](https://mindspore-website.obs.cn-north-4.myhuaweicloud.com/website-images/master/resource/_static/logo_source.svg)](https://gitee.com/mindspore/docs/blob/master/docs/mindquantum/docs/source_zh_cn/case_library/quantum_phase_estimation.ipynb)\n", "\n", "## 概述\n", "\n", "量子相位估计算法(Quantum Phase Estimation Algorithm，简称QPE)，是很多量子算法的关键。假设一个幺正算符 $U$，这个幺正算符作用在其本征态 $|u\\rangle$ 上会出现一个相位 $e^{2\\pi i \\varphi}$，现在我们假设 $U$ 算符的本征值未知，也就是 $\\varphi$ 未知，但是 $U$ 算符和本征态 $|u\\rangle$ 已知，相位估计算法的作用就是对这个相位 $\\varphi$ 进行估计。" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "![quantum phase estimation](https://mindspore-website.obs.cn-north-4.myhuaweicloud.com/website-images/master/docs/mindquantum/docs/source_zh_cn/images/quantum_phase_estimation.png)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 算法解析\n", "\n", "量子相位估计算法的实现需要两个寄存器(register)，第一寄存器包含$t$个初始在 $|0\\rangle$ 的量子比特，比特数和最后相位估计的结果的精度和算法的成功概率相关；第二个寄存器初始化在幺正算符 $U$ 的本征态 $|u\\rangle$ 上。相位估计算法主要分为三步：\n", "\n", "### 步骤一\n", "\n", "对第一寄存器的所有量子比特进行 [Hadamard](https://www.mindspore.cn/mindquantum/docs/zh-CN/master/core/gates/mindquantum.core.gates.HGate.html) 门操作，对第二寄存器连续进行 `控制U` 门操作，其中 $U$ 门的幂次依次为 $2^0, 2^1,...,2^{t-1}$，控制比特依次为 $q_{t-1}, q_{t-2},..., q_{1}, q_{0}$。这时第一寄存器中的态就会变为\n", "\n", "$$\n", "|\\psi_1\\rangle=\\frac{1}{2^{t/2}}\\left(|0\\rangle+e^{i2\\pi 2^{t-1}\\varphi}|1\\rangle\\right)\\left(|0\\rangle+e^{i2\\pi2^{t-2}\\varphi}|1\\rangle\\right)...\\left(|0\\rangle+e^{i2\\pi 2^{0}\\varphi}|1\\rangle\\right) = \\frac{1}{2^{t/2}}\\sum_{k=0}^{2^t-1}e^{i2\\pi\\varphi k}|k\\rangle\n", "$$\n", "\n", "其中$k$为直积态的十进制表示，比如 $k=0$ 表示第一寄存器中t个比特全部在基态 $|00...00\\rangle$， $k=2$ 表示 $|00...10\\rangle$，以此类推。\n", "\n", "### 步骤二\n", "\n", "对第一寄存器的进行量子傅里叶变换的逆变换(Inverse Quantum Fourier Transform)，在线路中表示成 $QFT^\\dagger$, 对 $|\\psi_1\\rangle$ 进行逆量子傅里叶变换可得 $|\\psi_2\\rangle$\n", "\n", "$$\n", "|\\psi_2\\rangle=QFT^\\dagger|\\psi_1\\rangle =\\frac{1}{2^t}\\sum_{x=0}^{2^t-1}a_x|x\\rangle\n", "$$\n", "\n", "其中\n", "\n", "$$\n", "a_x=\\sum_{k=0}^{2^t-1}e^{2\\pi i k(\\varphi-x/2^t)}\n", "$$\n", "\n", "为本征基矢 $|x\\rangle$ ($x=0.1,...,2^t$) 对应的概率幅。由上式可得，当 $2^t\\varphi$ 为整数，且满足 $x=2^t\\varphi$ 时，概率幅取最大值1，此时第一寄存器的末态可以精确反映 $\\varphi$；当 $2^t\\varphi$ 不是整数时，$x$ 为 $\\varphi$ 的估计，且$t$越大，估计精度越高。\n", "\n", "### 步骤三\n", "\n", "对第一寄存器的量子比特进行测量，得到第一寄存器的末态 $f=\\sum_{x}^{2^t-1}a_x|x\\rangle$, $x=0,1,...,2^t$，从中找到最大的振幅 $a_{max}$，其对应的本征基矢 $|x\\rangle$ 中的 $x$ 再除以 $2^t$ 即为相位的估计值。\n", "\n", "## QPE代码实现\n", "\n", "下面用一个实例来演示如何在MindSpore Quantum实现量子相位估计算法，选择 [T](https://www.mindspore.cn/mindquantum/docs/zh-CN/master/core/gates/mindquantum.core.gates.TGate.html) 门作为进行估计的幺正算符，由定义\n", "\n", "$$\n", "T|1\\rangle=e^{i\\pi/4}|1\\rangle\n", "$$\n", "\n", "可知需要估计的相位角为 $\\varphi=\\frac{1}{8}$。\n", "\n", "现在假设我们不知道 [T](https://www.mindspore.cn/mindquantum/docs/zh-CN/master/core/gates/mindquantum.core.gates.TGate.html) 门的相位信息，只知道幺正算符 $U$ 是 [T](https://www.mindspore.cn/mindquantum/docs/zh-CN/master/core/gates/mindquantum.core.gates.TGate.html) 门且本征态为 $|1\\rangle$ ，接下来我们需要用量子相位估计算法求出其对应的本征值，即需要估计本征值指数上的相位角。\n", "\n", "首先导入相关依赖。" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "from mindquantum.core.gates import T, H, X, Power, BARRIER\n", "from mindquantum.core.circuit import Circuit, UN\n", "from mindquantum.simulator import Simulator\n", "from mindquantum.algorithm.library import qft\n", "import numpy as np" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "[UN](https://www.mindspore.cn/mindquantum/docs/zh-CN/master/core/circuit/mindquantum.core.circuit.UN.html) 可以指定量子门，目标比特和控制比特，从而在线路中搭建门操作； [Power](https://www.mindspore.cn/mindquantum/docs/zh-CN/master/core/gates/mindquantum.core.gates.Power.html) 可以得到指定量子门的指数形式。因为我们已知 [T](https://www.mindspore.cn/mindquantum/docs/zh-CN/master/core/gates/mindquantum.core.gates.TGate.html) 门的本征态为 $|1\\rangle$，所以第二寄存器只需1个比特，而在第一寄存器中的比特数越多，得到的结果就越准确，在这里我们使用4个比特。\n", "\n", "因此我们需要搭建5比特线路， $q_0, q_1, q_2, q_3$ 比特用于估计，属于第一寄存器， $q_4$ 属于第二寄存器用于传入 $T$ 算符的本征态。\n", "\n", "利用 [UN](https://www.mindspore.cn/mindquantum/docs/zh-CN/master/core/circuit/mindquantum.core.circuit.UN.html) 对 $q_0, q_1, q_2, q_3$ 进行 [Hadamard](https://www.mindspore.cn/mindquantum/docs/zh-CN/master/core/gates/mindquantum.core.gates.HGate.html) 门操作，用 [X](https://www.mindspore.cn/mindquantum/docs/zh-CN/master/core/gates/mindquantum.core.gates.XGate.html) 门对 $q_4$ 进行翻转，得到 [T](https://www.mindspore.cn/mindquantum/docs/zh-CN/master/core/gates/mindquantum.core.gates.TGate.html) 门的本征态 $|1\\rangle$。" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [ { "data": { "image/svg+xml": [ "q0: q1: q2: q3: q4: H H H H X " ], "text/plain": [ "" ] }, "execution_count": 2, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# pylint: disable=W0104\n", "n = 4\n", "circ = Circuit()\n", "circ += UN(H, n) # 对前4个比特作用力H门\n", "circ += X.on(n) # 对q4作用X门\n", "circ.svg()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "以 $q_4$ 为目标比特，添加控制$T^{2^i}$门。" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "data": { "image/svg+xml": [ "q0: q1: q2: q3: q4: H H H H X T^1 T^2 T^4 T^8 " ], "text/plain": [ "" ] }, "execution_count": 3, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# pylint: disable=W0104\n", "for i in range(n):\n", " circ += Power(T, 2**i).on(n, n - i - 1) # 添加T^2^i门，其中q4为目标比特，n-i-1为控制比特\n", "circ.svg()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "对第一寄存器中的比特进行逆量子傅里叶变换。" ] }, { "cell_type": "code", "execution_count": 4, "metadata": {}, "outputs": [ { "data": { "image/svg+xml": [ "q0: q1: q2: q3: q4: H H H H X T^1 T^2 T^4 T^8 H PS -π/2 H PS -π/4 PS -π/2 H PS -π/8 PS -π/4 PS -π/2 H " ], "text/plain": [ "" ] }, "execution_count": 4, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# pylint: disable=W0104\n", "circ += BARRIER\n", "circ += qft(range(n)).hermitian() # 对前4个比特作用量子傅立叶变换的逆变换\n", "circ.svg()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "选择后端、传入总比特数创建模拟器，对量子线路进行演化，得到末态。" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "tags": [] }, "outputs": [ { "data": { "image/svg+xml": [ "Shots:\n", " 100 Keys: q3 q2 q1 q0 0.0 0.2 0.4 0.6 0.8 1.0 0100 100 probability " ], "text/plain": [ "" ] }, "execution_count": 5, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# pylint: disable=W0104\n", "from mindquantum.core.gates import Measure\n", "sim = Simulator('mqvector', circ.n_qubits) # 创建模拟器\n", "sim.apply_circuit(circ) # 用模拟器演化线路\n", "qs = sim.get_qs() # 获得演化得到的量子态\n", "res = sim.sampling(UN(Measure(), circ.n_qubits - 1), shots=100) # 在寄存器1中加入测量门并对线路进行100次采样，获得统计结果\n", "res.svg()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "需要注意的是，测量结果作为二进制串的读取顺序应为$|q_0q_1q_2q_3\\rangle$，因此我们得到寄存器1的测量结果为`0010`，概率幅为1，该末态可以精准地反映相位$\\varphi$。但`0010`是二进制结果，因此我们将它转回十进制后再除以$2^n$，就得到了我们最终的估计值：$\\varphi=\\frac{2}{2^4}=\\frac{1}{8}$。\n", "\n", "我们也可以通过线路演化得到的量子态 `qs` 找出第一寄存器中振幅最大值 $a_{max}$ 的位置，进而得到其对应的本征基矢 $|x\\rangle$ ，其中的 $x$ 再除以 $2^t$ 即为相位的估计值。" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false, "pycharm": { "name": "#%%\n" }, "tags": [] }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "10100\n" ] } ], "source": [ "index = np.argmax(np.abs(qs))\n", "print(bin(index)[2:])" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false, "pycharm": { "name": "#%% md\n" } }, "source": [ "需要注意的是，`qs` 对应的是整个量子线路的末态，因此得到的 ``index`` 也包含第二寄存器中的比特，不能直接得到第一寄存器末态中 $a_{max}$ 对应的 $|x\\rangle$ ，需要将 ``index`` 转成二进制后将 $q4$ 对应的比特位剔除，然后得到的才是第一寄存器的 $|x\\rangle$ 。" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "tags": [] }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "0010\n" ] } ], "source": [ "bit_string = bin(index)[2:].zfill(circ.n_qubits)[1:] # 将index转换成01串并剔除q4\n", "bit_string = bit_string[::-1] # 将比特串顺序调整为q0q1q2q3\n", "print(bit_string)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "再将二进制转回十进制，得到我们最终的估计值。" ] }, { "cell_type": "code", "execution_count": 8, "metadata": {}, "outputs": [ { "data": { "text/plain": [ "0.125" ] }, "execution_count": 8, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# pylint: disable=W0104\n", "theta_exp = int(bit_string, 2) / 2**n\n", "theta_exp" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "可见得到的估计相位和 $\\varphi$ 近似相等。" ] }, { "cell_type": "code", "execution_count": 9, "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", " \n", " \n", " \n", " \n", "\n", "\n", "\n", "\n", " \n", " \n", "\n", "\n", "
SoftwareVersion
mindquantum0.9.11
scipy1.10.1
numpy1.23.5
SystemInfo
Python3.9.16
OSLinux x86_64
Memory8.3 GB
CPU Max Thread8
DateSun Dec 31 23:59:31 2023
\n" ], "text/plain": [ "" ] }, "execution_count": 9, "metadata": {}, "output_type": "execute_result" } ], "source": [ "from mindquantum.utils.show_info import InfoTable\n", "\n", "InfoTable('mindquantum', 'scipy', 'numpy')" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 参考文献\n", "\n", "[1] Michael A. Nielsen and Isaac L. Chuang. [Quantum computation and quantum information](www.cambridge.org/9781107002173)" ] } ], "metadata": { "kernelspec": { "display_name": "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": "3.9.16" } }, "nbformat": 4, "nbformat_minor": 4 }