{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# 量子相位估计算法\n",
"\n",
"[](https://mindspore-website.obs.cn-north-4.myhuaweicloud.com/notebook/r2.0.0-alpha/mindquantum/zh_cn/mindspore_quantum_phase_estimation.ipynb) \n",
"[](https://mindspore-website.obs.cn-north-4.myhuaweicloud.com/notebook/r2.0.0-alpha/mindquantum/zh_cn/mindspore_quantum_phase_estimation.py) \n",
"[](https://gitee.com/mindspore/docs/blob/r2.0.0-alpha/docs/mindquantum/docs/source_zh_cn/quantum_phase_estimation.ipynb) \n",
"[](https://authoring-modelarts-cnnorth4.huaweicloud.com/console/lab?share-url-b64=aHR0cHM6Ly9taW5kc3BvcmUtd2Vic2l0ZS5vYnMuY24tbm9ydGgtNC5teWh1YXdlaWNsb3VkLmNvbS9ub3RlYm9vay9yMi4wLjAtYWxwaGEvbWluZHF1YW50dW0vemhfY24vbWluZHNwb3JlX3F1YW50dW1fcGhhc2VfZXN0aW1hdGlvbi5pcHluYg%3D%3D&imageid=b711ac95-db2b-45b7-ab9b-98de275dd57e)\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": [
""
]
},
{
"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",
"1. 对第一寄存器的所有量子比特进行 `Hadamard` 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",
"2. 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",
"3. 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MindQuantum实现量子相位估计算法，选择 `T` 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` 门的相位信息，只知道幺正算符 $U$ 是 `T` 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` 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` 可以得到指定量子门的指数形式。因为我们已知 `T` 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` 对 $q_0, q_1, q_2, q_3$ 进行 `Hadamard` 门操作， 用 `X` 门对 $q_4$ 进行翻转，得到 `T` 门的本征态 $|1\\rangle$。"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [
{
"data": {
"image/svg+xml": "",
"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": "",
"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": "",
"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": "",
"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": "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": {
"interpreter": {
"hash": "6cd6e2203b621035efd3b4ac9716079b52ce7fc5622f6651a3ae71459e0d54ce"
},
"kernelspec": {
"display_name": "MindSpore",
"language": "python",
"name": "mindspore"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.7.5-final"
}
},
"nbformat": 4,
"nbformat_minor": 4
}