{
"cells": [
{
"cell_type": "markdown",
"id": "32ff158f",
"metadata": {},
"source": [
"# 4.3 受控运算"
]
},
{
"cell_type": "markdown",
"id": "e45c7851",
"metadata": {},
"source": [
"本节的习题也可以参考 Qiskit 网站：\n",
"\n",
"> https://qiskit.org/textbook/ch-gates/more-circuit-identities.html"
]
},
{
"cell_type": "code",
"execution_count": 1,
"id": "fa4b576f",
"metadata": {},
"outputs": [],
"source": [
"import numpy as np\n",
"from opt_einsum import contract as einsum\n",
"from qiskit import QuantumCircuit\n",
"from qiskit.quantum_info import Operator\n",
"import warnings\n",
"\n",
"np.set_printoptions(4, suppress=True)\n",
"warnings.filterwarnings(\"ignore\")"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "9590e18f",
"metadata": {},
"outputs": [],
"source": [
"# Vector in computational basis\n",
"vec0 = np.array([1, 0])\n",
"vec1 = np.array([0, 1])\n",
"# Matrices in computational basis\n",
"I = np.eye(2)\n",
"X = np.array([[0, 1], [1, 0]])\n",
"Y = np.array([[0, -1j], [1j, 0]])\n",
"Z = np.array([[1, 0], [0, -1]])\n",
"H = 1/np.sqrt(2) * np.array([[1, 1], [1, -1]])\n",
"T = np.array([[1, 0], [0, np.exp(1j*np.pi/4)]])\n",
"S = T @ T\n",
"P0 = np.outer(vec0, vec0)\n",
"P1 = np.outer(vec1, vec1)"
]
},
{
"cell_type": "markdown",
"id": "a3eca18c",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 预练习：受控门的张量积表示"
]
},
{
"cell_type": "markdown",
"id": "5146e270",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 预练习\n",
"\n",
"请表明可以作用于 $|x_1x_2\\rangle$ 的、以 $x_1$ 为控制量子比特、以 $x_2$ 为目标量子比特的受控非门 $\\mathrm{CNOT}_{x_1, x_2}$ 可以表示为\n",
"\n",
"$$\n",
"\\mathrm{CNOT}_{x_1, x_2} = P_0 \\otimes I + P_1 \\otimes X\n",
"$$\n",
"\n",
"其中，$P_0, P_1$ 是计算基下的投影算符：\n",
"\n",
"$$\n",
"P_0 = | 0 \\rangle \\langle 0 |, \\quad P_1 = | 1 \\rangle \\langle 1 |\n",
"$$\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "ac8f24eb",
"metadata": {
"hidden": true
},
"source": [
"我们以后经常会使用这类受控门的张量积表示，来进行证明或计算。该控制门的图示是 ($x_1, x_2$ 对应到下图的 $q_0, q_1$)"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "190d451d",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"image/svg+xml": [
"\r\n",
"\r\n",
"\r\n",
"\r\n"
],
"text/plain": [
""
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"circ = QuantumCircuit(2)\n",
"circ.cx(0, 1)\n",
"circ.draw()"
]
},
{
"cell_type": "markdown",
"id": "4c134625",
"metadata": {
"hidden": true
},
"source": [
"**证明 1**\n",
"\n",
"我们令待证等式右边为算符 $A = P_0 \\otimes I + P_1 \\otimes X$ 。将 $A$ 作用于两个量子比特 $x_1, x_2$ 所构成的态 $|x_1 x_2\\rangle$ 上：\n",
"\n",
"$$\n",
"A |x_1 x_2\\rangle = P_0 |x_1\\rangle \\otimes I |x_2\\rangle + P_1 |x_1\\rangle \\otimes X |x_2\\rangle\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "f0c7ca03",
"metadata": {
"hidden": true
},
"source": [
"我们只需要考虑 $|x_1 x_2\\rangle$ 为计算基 $\\{ |00\\rangle, |01\\rangle, |10\\rangle, |11\\rangle \\}$ 的情况。\n",
"\n",
"- 若 $|x_1\\rangle = |0\\rangle$，那么 $P_0 |0\\rangle = |0\\rangle$ 且 $P_0 |0\\rangle = 0$。因此，\n",
" \n",
" $$\n",
" A |0 x_2\\rangle = |0\\rangle \\otimes I |x_2\\rangle = |0x_2\\rangle\n",
" $$\n",
" \n",
" 这意味着如果 $|x_1\\rangle = |0\\rangle$，那么经过算符 $A$ 的作用后，$|x_2\\rangle$ 仍然保持不变。\n",
"\n",
"- 若 $|x_1\\rangle = |1\\rangle$，那么 $P_0 |1\\rangle = 0$ 且 $P_1 |1\\rangle = |1\\rangle$。因此，\n",
" \n",
" $$\n",
" A |1 x_2\\rangle = |1\\rangle \\otimes X |x_2\\rangle\n",
" $$\n",
" \n",
" 这意味着如果 $|x_1\\rangle = |1\\rangle$，那么经过算符 $A$ 的作用后，$|x_2\\rangle$ 会变成 $X |x_2\\rangle$，即将非门作用于 $|x_2\\rangle$。\n",
"\n",
"因此，算符 $A$ 完全符合受控非门的定义。\n",
"\n",
"对于上述的计算过程，请不要搞混 $x_1, x_2$ 的下标与 $|0\\rangle, |1\\rangle$。"
]
},
{
"cell_type": "markdown",
"id": "476ef6c4",
"metadata": {
"hidden": true
},
"source": [
"**证明 2**"
]
},
{
"cell_type": "markdown",
"id": "8fa5ebc3",
"metadata": {
"hidden": true
},
"source": [
"直接计算等式右的 $P_0 \\otimes I + P_1 \\otimes Z$，并验证它是否是书中式 (4.23) 所表示的矩阵：\n",
"\n",
"\\begin{equation*}\n",
"\\mathrm{CNOT}_{x_1, x_2} = \\begin{bmatrix} \n",
"1 & 0 & 0 & 0 \\\\\n",
"0 & 1 & 0 & 0 \\\\\n",
"0 & 0 & 0 & 1 \\\\\n",
"0 & 0 & 1 & 0\n",
"\\end{bmatrix}\n",
"\\tag{4.23}\n",
"\\end{equation*}\n",
"\n",
"求取张量积 (Kronecker tensor product) 的过程实际上也可以通过 `np.kron` 函数实现。"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "b1618bf1",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"array([[1., 0., 0., 0.],\n",
" [0., 1., 0., 0.],\n",
" [0., 0., 0., 1.],\n",
" [0., 0., 1., 0.]])"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.kron(P0, I) + np.kron(P1, X)"
]
},
{
"cell_type": "markdown",
"id": "0f2563d1",
"metadata": {
"hidden": true
},
"source": [
"**补充**"
]
},
{
"cell_type": "markdown",
"id": "c1960b3c",
"metadata": {
"hidden": true
},
"source": [
"我们有时还需要用以 $x_2$ 为控制量子比特、以 $x_1$ 为目标量子比特的受控非门 $\\mathrm{CNOT}_{x_1, x_2}$。其表示是\n",
"\n",
"$$\n",
"\\mathrm{CNOT}_{x_2, x_1} = I \\otimes P_0 + X \\otimes P_1\n",
"$$\n",
"\n",
"请注意相对于 $\\mathrm{CNOT}_{x_1, x_2}$，上式的变化**不是** $P_0$ 与 $P_1$ 交换，而是张量积的顺序发生了变化。其图示是 ($x_1, x_2$ 对应到下图的 $q_0, q_1$)"
]
},
{
"cell_type": "code",
"execution_count": 5,
"id": "b71bf216",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"image/svg+xml": [
"\r\n",
"\r\n",
"\r\n",
"\r\n"
],
"text/plain": [
""
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"circ = QuantumCircuit(2)\n",
"circ.cx(1, 0)\n",
"circ.draw()"
]
},
{
"cell_type": "markdown",
"id": "57d64e84",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.16 (多量子比特门的矩阵表示)"
]
},
{
"cell_type": "markdown",
"id": "2c2ba9bd",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.16\n",
"\n",
"在计算基下，如下线路的 $4 \\times 4$ 酉矩阵是什么？\n",
"\n",
"\n",
"\n",
"如下线路在计算基下的酉矩阵是什么？\n",
"\n",
"\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "60a07144",
"metadata": {
"hidden": true
},
"source": [
"这里有意将原书中的 $x_1, x_2$ 颠倒过来。一般来说，不管是程序还是书中，排在上面的线路的量子比特序号都比较小。\n",
"\n",
"在这一小节中，我们会尽量同时使用 qiskit 来绘制线路图，以熟悉程序的编写。"
]
},
{
"cell_type": "markdown",
"id": "23c7614a",
"metadata": {
"hidden": true
},
"source": [
"第一个线路的酉矩阵是\n",
"\n",
"$$\n",
"H \\otimes I = \\frac{1}{\\sqrt{2}} \\begin{bmatrix}\n",
"1 & 0 & 1 & 0 \\\\\n",
"0 & 1 & 0 & 1 \\\\\n",
"1 & 0 & -1 & 0 \\\\\n",
"0 & 1 & 0 & -1 \\\\\n",
"\\end{bmatrix}\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 6,
"id": "72928d5b",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"array([[ 0.7071, 0. , 0.7071, 0. ],\n",
" [ 0. , 0.7071, 0. , 0.7071],\n",
" [ 0.7071, 0. , -0.7071, -0. ],\n",
" [ 0. , 0.7071, -0. , -0.7071]])"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.kron(H, I)"
]
},
{
"cell_type": "markdown",
"id": "804ac078",
"metadata": {
"hidden": true
},
"source": [
"该线路的绘制可以通过下述代码实现："
]
},
{
"cell_type": "code",
"execution_count": 7,
"id": "3ecd96ef",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"image/svg+xml": [
"\r\n",
"\r\n",
"\r\n",
"\r\n"
],
"text/plain": [
""
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"circ = QuantumCircuit(2)\n",
"circ.h(0)\n",
"circ.draw()"
]
},
{
"cell_type": "markdown",
"id": "76137e78",
"metadata": {
"hidden": true
},
"source": [
"第二个线路的酉矩阵是\n",
"\n",
"$$\n",
"I \\otimes H = \\frac{1}{\\sqrt{2}} \\begin{bmatrix}\n",
"1 & 1 & 0 & 0 \\\\\n",
"1 & -1 & 0 & 0 \\\\\n",
"0 & 0 & 1 & 1 \\\\\n",
"0 & 0 & 1 & -1 \\\\\n",
"\\end{bmatrix}\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 8,
"id": "1cfe0523",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"array([[ 0.7071, 0.7071, 0. , 0. ],\n",
" [ 0.7071, -0.7071, 0. , -0. ],\n",
" [ 0. , 0. , 0.7071, 0.7071],\n",
" [ 0. , -0. , 0.7071, -0.7071]])"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.kron(I, H)"
]
},
{
"cell_type": "markdown",
"id": "364d7ee1",
"metadata": {
"hidden": true
},
"source": [
"该线路的绘制可以通过下述代码实现："
]
},
{
"cell_type": "code",
"execution_count": 9,
"id": "7a61d6ac",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"image/svg+xml": [
"\r\n",
"\r\n",
"\r\n",
"\r\n"
],
"text/plain": [
""
]
},
"execution_count": 9,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"circ = QuantumCircuit(2)\n",
"circ.h(1)\n",
"circ.draw()"
]
},
{
"cell_type": "markdown",
"id": "317a813e",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.17 (由受控 $Z$ 门建造受控非门)"
]
},
{
"cell_type": "markdown",
"id": "57051928",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.17\n",
"\n",
"从一个受控 $Z$ 门 (即一个在计算基下的作用由如下酉矩阵表示)\n",
"\n",
"$$\n",
"\\text{ctrl-}Z = \\begin{bmatrix}\n",
"1 & 0 & 0 & 0 \\\\\n",
"0 & 1 & 0 & 0 \\\\\n",
"0 & 0 & 1 & 0 \\\\\n",
"0 & 0 & 0 & -1\n",
"\\end{bmatrix}\n",
"$$\n",
"\n",
"和两个 Hadamard 门构造一个受控非门，并指定控制量子比特和目标量子比特。\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "5aabe799",
"metadata": {
"hidden": true
},
"source": [
"该练习是多解题。我们将在练习 4.18 中证明到\n",
"\n",
"\n",
"\n",
"因此，对于受控 $Z$ 门而言，控制量子比特和目标量子比特是完全等价的。我们将受控 $Z$ 门的矩阵表示用程序变量 `ctrl_Z` 写出。"
]
},
{
"cell_type": "code",
"execution_count": 10,
"id": "2afeae63",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"array([[ 1., 0., 0., 0.],\n",
" [ 0., 1., 0., 0.],\n",
" [ 0., 0., 1., 0.],\n",
" [ 0., 0., 0., -1.]])"
]
},
"execution_count": 10,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"ctrl_Z = np.kron(P0, I) + np.kron(P1, Z)\n",
"ctrl_Z"
]
},
{
"cell_type": "markdown",
"id": "20173e1a",
"metadata": {
"hidden": true
},
"source": [
"受控非门 CNOT 是受控 $X$ 门，依据控制量子比特是 $x_1$ 或 $x_2$、目标量子比特是 $x_2$ 或 $x_1$，我们可能写出两种 CNOT 的矩阵表示，分别用变量 $\\mathrm{CNOT}_{x_1, x_2}$ `ctrl_1_X` 与 $\\mathrm{CNOT}_{x_2, x_1}$ `ctrl_2_X` 写出。"
]
},
{
"cell_type": "code",
"execution_count": 11,
"id": "2a887061",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"(array([[1., 0., 0., 0.],\n",
" [0., 1., 0., 0.],\n",
" [0., 0., 0., 1.],\n",
" [0., 0., 1., 0.]]),\n",
" array([[1., 0., 0., 0.],\n",
" [0., 0., 0., 1.],\n",
" [0., 0., 1., 0.],\n",
" [0., 1., 0., 0.]]))"
]
},
"execution_count": 11,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"ctrl_1_X = np.kron(P0, I) + np.kron(P1, X)\n",
"ctrl_2_X = np.kron(I, P0) + np.kron(X, P1)\n",
"ctrl_1_X, ctrl_2_X"
]
},
{
"cell_type": "markdown",
"id": "dc6bdbbc",
"metadata": {
"hidden": true
},
"source": [
"我们很容易地可以凑出下述结果：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "2c23b365",
"metadata": {
"hidden": true
},
"source": [
"用数学的方式来表达上述过程，则是\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"(I \\otimes H) \\, \\text{ctrl-}Z \\, (I \\otimes H) &= \\mathrm{CNOT}_{x_1, x_2} \\\\\n",
"(I \\otimes H) (P_0 \\otimes I + P_1 \\otimes Z) (I \\otimes H) &= P_0 \\otimes I + P_1 \\otimes X\n",
"\\end{align*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "1a9405be",
"metadata": {
"hidden": true
},
"source": [
"有两种证明方法。第一种是给出所有矩阵表示并作验证："
]
},
{
"cell_type": "code",
"execution_count": 12,
"id": "57d4fc94",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"True"
]
},
"execution_count": 12,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.allclose(np.kron(I, H) @ ctrl_Z @ np.kron(I, H), ctrl_1_X)"
]
},
{
"cell_type": "markdown",
"id": "a5ab48c2",
"metadata": {
"hidden": true
},
"source": [
"第二种方法是作代数证明。回顾到 $HZH = X, H^2 = I$，因此\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"&\\quad (I \\otimes H) (P_0 \\otimes I + P_1 \\otimes Z) (I \\otimes H) \\\\\n",
"&= (I P_0 I) \\otimes (H I H) + (I P_1 I) \\otimes (H Z H) \\\\\n",
"&= P_0 \\otimes I + P_1 \\otimes X\n",
"\\end{align*}\n",
"$$\n",
"\n",
"证明完毕。"
]
},
{
"cell_type": "markdown",
"id": "49340837",
"metadata": {
"hidden": true
},
"source": [
"其中一个门路的图像可以绘制如下 ($x_1, x_2$ 对应到下图的 $q_0, q_1$)："
]
},
{
"cell_type": "code",
"execution_count": 13,
"id": "60ddd6f9",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"image/svg+xml": [
"\r\n",
"\r\n",
"\r\n",
"\r\n"
],
"text/plain": [
""
]
},
"execution_count": 13,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"circ = QuantumCircuit(2)\n",
"circ.h(1); circ.cz(0, 1); circ.h(1)\n",
"circ.draw()"
]
},
{
"cell_type": "markdown",
"id": "81c77b2c",
"metadata": {
"hidden": true
},
"source": [
"由于受控 $Z$ 门实际上是关于控制与目标量子比特 $x_1, x_2$ 对称的，因此也很容易地给出另一种情况：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "e54890b0",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.18"
]
},
{
"cell_type": "markdown",
"id": "b99a09b5",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.18\n",
"\n",
"证明下图：\n",
"\n",
"\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "12202e47",
"metadata": {
"hidden": true
},
"source": [
"回顾到预练习，用数学的语言表述，该题希望证明\n",
"\n",
"$$\n",
"P_0 \\otimes I + P_1 \\otimes Z = I \\otimes P_0 + Z \\otimes P_1\n",
"$$\n",
"\n",
"这个证明恐怕还是需要用矩阵表示做比较方便："
]
},
{
"cell_type": "code",
"execution_count": 14,
"id": "b8c83bb6",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"True"
]
},
"execution_count": 14,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.allclose(\n",
" np.kron(P0, I) + np.kron(P1, Z),\n",
" np.kron(I, P0) + np.kron(Z, P1))"
]
},
{
"cell_type": "markdown",
"id": "39da431c",
"metadata": {
"hidden": true
},
"source": [
"但也可以使用代数做法。我们可以利用下述结论：\n",
"\n",
"$$\n",
"P_0 = \\frac{I+Z}{2} = \\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix}, \\quad P_1 = \\frac{I-Z}{2} = \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "261c72d3",
"metadata": {
"hidden": true
},
"source": [
"因此，\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"&\\quad P_0 \\otimes I + P_1 \\otimes Z \\\\\n",
"&= \\frac{1}{2} \\big( (I + Z) \\otimes I + (I - Z) \\otimes Z \\big) \\\\\n",
"&= \\frac{1}{2} \\big( I \\otimes I + Z \\otimes I + I \\otimes Z - Z \\otimes Z \\big) \\\\\n",
"&= \\frac{1}{2} \\big( I \\otimes (I+Z) + Z \\otimes (I-Z) \\big) \\\\\n",
"&= I \\otimes P_0 + Z \\otimes P_1\n",
"\\end{align*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "733d3797",
"metadata": {
"hidden": true
},
"source": [
"因此，受控 $Z$ 门是关于两个量子比特对称的。一般来说，它的图像不使用方块圈起来的 $Z$ 给出，而是像下图一样 (也可以回顾练习 4.17 的解答)："
]
},
{
"cell_type": "code",
"execution_count": 15,
"id": "81c91594",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"image/svg+xml": [
"\r\n",
"\r\n",
"\r\n",
"\r\n"
],
"text/plain": [
""
]
},
"execution_count": 15,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"circ = QuantumCircuit(2)\n",
"circ.cz(0, 1)\n",
"circ.draw()"
]
},
{
"cell_type": "markdown",
"id": "190f518e",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.19 (受控非门在密度矩阵上的作用)"
]
},
{
"cell_type": "markdown",
"id": "260c7c70",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.19\n",
"\n",
"受控非门在密度矩阵 $\\rho$ 上的作用是重排矩阵元素的位置，故仅为一置换。在计算基下清晰地写出它的作用。\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "dde9a34c",
"metadata": {
"hidden": true
},
"source": [
"实际上，不仅对于密度矩阵，对于任意的 $4 \\times 4$ 矩阵 $A$ 都是置换作用。我们令计算基下的 $A$ 矩阵："
]
},
{
"cell_type": "code",
"execution_count": 16,
"id": "44ca71b8",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"array([[11, 12, 13, 14],\n",
" [21, 22, 23, 24],\n",
" [31, 32, 33, 34],\n",
" [41, 42, 43, 44]])"
]
},
"execution_count": 16,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A = np.array(np.zeros((4, 4)) + np.array([1, 2, 3, 4]) + np.array([10, 20, 30, 40])[:, None], dtype=int)\n",
"A"
]
},
{
"cell_type": "markdown",
"id": "70622419",
"metadata": {
"hidden": true
},
"source": [
"可以知道，$\\mathrm{CNOT}_{x_1, x_2} A$ 的效应是将矩阵的第 3、4 行进行对调："
]
},
{
"cell_type": "code",
"execution_count": 17,
"id": "38139943",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"array([[11, 12, 13, 14],\n",
" [21, 22, 23, 24],\n",
" [41, 42, 43, 44],\n",
" [31, 32, 33, 34]])"
]
},
"execution_count": 17,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.array(ctrl_1_X @ A, dtype=int)"
]
},
{
"cell_type": "markdown",
"id": "f1da1693",
"metadata": {
"hidden": true
},
"source": [
"可能更有意义的作用是 $\\mathrm{CNOT}_{x_1, x_2} \\, A \\, \\mathrm{CNOT}_{x_1, x_2}^\\dagger$ 的效应是将第 3、4 行与第 3、4 列同时对调。我们同时注意到 $\\mathrm{CNOT}_{x_1, x_2}$ 是 Hermitian 矩阵，即 $\\mathrm{CNOT}_{x_1, x_2} = \\mathrm{CNOT}_{x_1, x_2}^\\dagger$。"
]
},
{
"cell_type": "code",
"execution_count": 18,
"id": "0411bf7a",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"array([[11, 12, 14, 13],\n",
" [21, 22, 24, 23],\n",
" [41, 42, 44, 43],\n",
" [31, 32, 34, 33]])"
]
},
"execution_count": 18,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.array(ctrl_1_X @ A @ ctrl_1_X, dtype=int)"
]
},
{
"cell_type": "markdown",
"id": "13609446",
"metadata": {
"hidden": true
},
"source": [
"因此，操作 $\\mathrm{CNOT}_{x_1, x_2} \\, A \\, \\mathrm{CNOT}_{x_1, x_2}^\\dagger$ 实际上相当于将计算基 $\\{ |00\\rangle, |01\\rangle, |10\\rangle, |11\\rangle \\}$ 转换到 $\\{ |00\\rangle, |01\\rangle, |11\\rangle, |10\\rangle \\}$。"
]
},
{
"cell_type": "markdown",
"id": "18149aec",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.20 (受控非门的基变换)"
]
},
{
"cell_type": "markdown",
"id": "3b162d32",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.20\n",
"\n",
"不像理想的经典门，理想的量子门没有 (如电子工程师说的) 高阻抗输入。实际上，“控制”和“目标”的角色是随意的——它们的区分依赖于你认为该设备在哪组基下计算。我们已经描述了受控非门相对计算基的行为，并且在这个描述中控制量子比特是不变的。然而，如果考虑一组不同的基，则控制量子比特的确要变：我们将看到依赖于“目标”量子比特的状态，它的相位要翻转。证明下图：\n",
"\n",
"\n",
"\n",
"引入基状态组 $|\\pm\\rangle \\equiv (|0\\rangle \\pm |1\\rangle) / \\sqrt{2}$，利用这个线路恒等式关系证明一个第一个量子比特作为控制、第二个量子比特作为目标的受控非门的作用为\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"|+\\rangle |+\\rangle &\\rightarrow |+\\rangle |+\\rangle \\tag{4.24} \\\\\n",
"|-\\rangle |+\\rangle &\\rightarrow |-\\rangle |+\\rangle \\tag{4.25} \\\\\n",
"|+\\rangle |-\\rangle &\\rightarrow |-\\rangle |-\\rangle \\tag{4.26} \\\\\n",
"|-\\rangle |-\\rangle &\\rightarrow |+\\rangle |-\\rangle \\tag{4.27}\n",
"\\end{align*}\n",
"$$\n",
"\n",
"因此，相对这组新基，目标的量子比特的状态不变，而控制量子比特的状态在目标是 $|-\\rangle$ 的时候翻转，否则保持不变；即在这组基下，目标和控制的角色实际上互换了。\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "0a3ed7ab",
"metadata": {
"hidden": true
},
"source": [
"首先证明线路图等式。该线路图等式用数学语言表述，是\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"(H \\otimes H) \\mathrm{CNOT}_{x_1, x_2} (H \\otimes H) &= \\mathrm{CNOT}_{x_2, x_1} \\\\\n",
"(H \\otimes H) (P_0 \\otimes I + P_1 \\otimes X) (H \\otimes H) &= I \\otimes P_0 + X \\otimes P_1\n",
"\\end{align*}\n",
"$$\n",
"\n",
"证明如下。我们需要利用到的结论有\n",
"\n",
"$$\n",
"P_0 = \\frac{I+Z}{2}, \\quad P_1 = \\frac{I-Z}{2}, \\quad HXH=Z, \\quad H^2 = I\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "d2a6b4d9",
"metadata": {
"hidden": true
},
"source": [
"$$\n",
"\\begin{align*}\n",
"\\mathrm{LHS}\n",
"&= (H P_0 H) \\otimes (H I H) + (H P_1 H) \\otimes (H X H) \\\\\n",
"&= \\frac{1}{2} \\big( (HIH + HZH) \\otimes I + (HIH - HZH) \\otimes Z \\big) \\\\\n",
"&= \\frac{1}{2} \\big( I \\otimes I + X \\otimes I + I \\otimes Z - X \\otimes Z \\big) \\\\\n",
"&= \\frac{1}{2} \\big( I \\otimes (I+Z) + X \\otimes (I-Z) \\big) \\\\\n",
"&= I \\otimes P_0 + X \\otimes P_1 = \\mathrm{RHS}\n",
"\\end{align*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "74f0dfe4",
"metadata": {
"hidden": true
},
"source": [
"我们当然也可以利用矩阵乘法直接证明等式："
]
},
{
"cell_type": "code",
"execution_count": 19,
"id": "08b56bdd",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"True"
]
},
"execution_count": 19,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.allclose(\n",
" np.kron(H, H) @ ctrl_1_X @ np.kron(H, H),\n",
" ctrl_2_X)"
]
},
{
"cell_type": "markdown",
"id": "2fa2ae92",
"metadata": {
"hidden": true
},
"source": [
"---"
]
},
{
"cell_type": "markdown",
"id": "bf97e193",
"metadata": {
"hidden": true
},
"source": [
"第二个论断的另一种叙述方式是，若计算基下的态 $|x_1 x_2\\rangle$ 等价于 $\\{ |+\\rangle, |-\\rangle \\}$ 基下的态 $|y_1 y_2\\rangle$，那么 $\\mathrm{CNOT}_{x_1, x_2}$ 作为 $x_1$ 控制、$x_2$ 目标的受控非门，在基 $\\{ |+\\rangle, |-\\rangle \\}$ 下恰好等价于 $\\mathrm{CNOT}_{y_2, y_1}$ 即 $|y_2\\rangle$ 控制、$|y_1\\rangle$ 目标的受控非门。"
]
},
{
"cell_type": "markdown",
"id": "a6f46f3c",
"metadata": {
"hidden": true
},
"source": [
"实际上左乘 $H$，右乘 $H^\\dagger = H$ 就相当于将算符的表示矩阵从计算基 $\\{ |0\\rangle, |1\\rangle \\}$ 转换到基 $\\{ |+\\rangle, |-\\rangle \\}$。依照这个思路，我们作如下简单的说明。我们首先要回顾到\n",
"\n",
"$$\n",
"H |0\\rangle = \\frac{|0\\rangle + |1\\rangle}{\\sqrt{2}} = |+\\rangle, \\quad H |1\\rangle = \\frac{|0\\rangle - |1\\rangle}{\\sqrt{2}} = |-\\rangle\n",
"$$\n",
"\n",
"以及\n",
"\n",
"$$\n",
"H |+\\rangle = H^2 |0\\rangle = |0\\rangle, \\quad H |-\\rangle = H^2 |1\\rangle = |1\\rangle\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "edc34dda",
"metadata": {
"hidden": true
},
"source": [
"因此，\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"\\mathrm{CNOT}_{x_1, x_2} |+\\rangle |+\\rangle\n",
"&= \\mathrm{CNOT}_{x_1, x_2} \\big[ |+\\rangle \\otimes |+\\rangle \\big] \\\\\n",
"&= \\mathrm{CNOT}_{x_1, x_2} \\big[ \\big( H |0\\rangle \\big) \\otimes \\big( H |0\\rangle \\big) \\big] \\\\\n",
"&= \\mathrm{CNOT}_{x_1, x_2} (H \\otimes H) |0\\rangle |0\\rangle \\\\\n",
"&= (H \\otimes H)^2 \\mathrm{CNOT}_{x_1, x_2} (H \\otimes H) |0\\rangle |0\\rangle \\\\\n",
"&= (H \\otimes H) \\mathrm{CNOT}_{x_2, x_1} |0\\rangle |0\\rangle \\\\\n",
"&= (H \\otimes H) |0\\rangle |0\\rangle \\\\\n",
"&= \\big( H |0\\rangle \\big) \\otimes \\big( H |0\\rangle \\big) = |+\\rangle |+\\rangle\n",
"\\end{align*}\n",
"$$\n",
"\n",
"上式中用到的技巧是 $(H \\otimes H)^2 = I$。"
]
},
{
"cell_type": "markdown",
"id": "af674b9d",
"metadata": {
"hidden": true
},
"source": [
"类似地，\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"\\mathrm{CNOT}_{x_1, x_2} |+\\rangle |-\\rangle\n",
"&= \\mathrm{CNOT}_{x_1, x_2} (H \\otimes H) |0\\rangle |1\\rangle \\\\\n",
"&= (H \\otimes H) \\mathrm{CNOT}_{x_2, x_1} |0\\rangle |1\\rangle \\\\\n",
"&= (H \\otimes H) |1\\rangle |0\\rangle = |-\\rangle |+\\rangle\n",
"\\end{align*}\n",
"$$\n",
"\n",
"对于 $|-\\rangle |+\\rangle, |-\\rangle |-\\rangle$ 来说，证明过程是类似的。"
]
},
{
"cell_type": "markdown",
"id": "2994f40a",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.21"
]
},
{
"cell_type": "markdown",
"id": "5c7a2d57",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.21\n",
"\n",
"验证下图的 $C^2 (U)$ 的运算。\n",
"\n",
"\n",
"\n",
"其中，$V^2 = U$，且 $U, V$ 均为酉算符。\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "a3342408",
"metadata": {
"hidden": true
},
"source": [
"该练习尽管可以使用数学方法作证明，但过程非常繁琐。分类讨论是最快的处理方法。我们考察下图：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "21324774",
"metadata": {
"hidden": true
},
"source": [
"上图的过程是指，对于初始三粒子态 $|\\psi_0\\rangle = |x_1 x_2\\rangle \\otimes |x_3\\rangle$，经过该门路后得到 $|\\psi_5\\rangle$。我们对 $|x_1 x_2\\rangle = |00\\rangle, |01\\rangle, |10\\rangle, |11\\rangle$ 分别进行讨论。下面需要利用到酉矩阵的性质 $V V^\\dagger = V^\\dagger V = I$。"
]
},
{
"cell_type": "markdown",
"id": "1821daa0",
"metadata": {
"hidden": true
},
"source": [
"$$\n",
"\\begin{alignat*}{20}\n",
"& |\\psi_0\\rangle && |\\psi_1\\rangle && |\\psi_2\\rangle && |\\psi_3\\rangle && |\\psi_4\\rangle && |\\psi_5\\rangle &&\n",
"\\\\\n",
"& \\text{init}\n",
"&& \\text{ctrl-}V_{x_2, x_3}\n",
"&& \\mathrm{CNOT}_{x_1, x_2}\n",
"&& \\text{ctrl-}V^\\dagger_{x_2, x_3}\n",
"&& \\mathrm{CNOT}_{x_1, x_2}\n",
"&& \\text{ctrl-}V_{x_1, x_3}\n",
"&& \\quad\\text{equivalent}\n",
"\\\\\n",
"& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& = |00\\rangle \\otimes |x_3\\rangle\n",
"\\\\\n",
"& |01\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |01\\rangle \\otimes V |x_3\\rangle \\quad\n",
"&& |01\\rangle \\otimes V |x_3\\rangle \\quad\n",
"&& |01\\rangle \\otimes V V^\\dagger |x_3\\rangle \\quad\n",
"&& |01\\rangle \\otimes V V^\\dagger |x_3\\rangle \\quad\n",
"&& |01\\rangle \\otimes V V^\\dagger |x_3\\rangle \\quad\n",
"&& = |01\\rangle \\otimes |x_3\\rangle\n",
"\\\\\n",
"& |10\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |10\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |11\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |11\\rangle \\otimes V^\\dagger |x_3\\rangle \\quad\n",
"&& |10\\rangle \\otimes V^\\dagger |x_3\\rangle \\quad\n",
"&& |10\\rangle \\otimes V^\\dagger V |x_3\\rangle \\quad\n",
"&& = |10\\rangle \\otimes |x_3\\rangle\n",
"\\\\\n",
"& |11\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |11\\rangle \\otimes V |x_3\\rangle \\quad\n",
"&& |10\\rangle \\otimes V |x_3\\rangle \\quad\n",
"&& |10\\rangle \\otimes V |x_3\\rangle \\quad\n",
"&& |11\\rangle \\otimes V |x_3\\rangle \\quad\n",
"&& |11\\rangle \\otimes V^2 |x_3\\rangle \\quad\n",
"&& = |11\\rangle \\otimes U |x_3\\rangle\n",
"\\end{alignat*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "4f2cc276",
"metadata": {
"hidden": true
},
"source": [
"这其中，只有 $|11\\rangle \\otimes |x_3\\rangle$ 给出了 $|11\\rangle \\otimes U |x_3\\rangle$，其它情况都保持初态不变。因此，该门路从定义上确实符合 $x_1, x_2$ 为控制量子比特、$x_3$ 为目标量子比特的受控 $U$ 门路。"
]
},
{
"cell_type": "markdown",
"id": "faca13c0",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.22"
]
},
{
"cell_type": "markdown",
"id": "2423f905",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.22\n",
"\n",
"证明一个 $C^2 (U)$ 门 (对任意单量子比特酉算符 $U$) 可至多用 8 个单量子比特门和 6 个受控非门构造出来。\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "09641be7",
"metadata": {
"hidden": true
},
"source": [
"该证明实际上还是比较精妙的。这里先给出大体的证明过程；但对于其中一部分细节，我们在后续的引理证明部分给出。"
]
},
{
"cell_type": "markdown",
"id": "81d7588e",
"metadata": {
"hidden": true
},
"source": [
"该题的参考资料是\n",
"\n",
"> https://quantumcomputing.stackexchange.com/a/6951/14843"
]
},
{
"cell_type": "markdown",
"id": "2e2130b2",
"metadata": {
"hidden": true
},
"source": [
"### 第一步：用 CNOT 与单量子比特酉算符表示 $C^2(U)$ 门"
]
},
{
"cell_type": "markdown",
"id": "fedc8a2e",
"metadata": {
"hidden": true
},
"source": [
"证明的第一步很容易。我们回顾到图 4.6，任意的酉矩阵 $V$ 总可以写为酉矩阵 $A, B, C$ 与实数相位参数 $\\alpha$ 的表达式：\n",
"\n",
"$$\n",
"\\begin{gather*}\n",
"V = e^{i \\alpha} A X B X C \\\\\n",
"ABC = I\n",
"\\end{gather*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "80d14511",
"metadata": {
"hidden": true
},
"source": [
"为了方便，我们定义下述相位算符：\n",
"\n",
"$$\n",
"P_\\alpha = \\begin{bmatrix} 1 & 0 \\\\ 0 & e^{i \\alpha} \\end{bmatrix}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "4bb31949",
"metadata": {
"hidden": true
},
"source": [
"因此，书中的图 4.6 可以表示如下 (注意为了这道例题的记号方便，我们有意用 $V$ 替代 $U$ 表示任意酉矩阵)：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "47e6e4c9",
"metadata": {
"hidden": true
},
"source": [
"其转置 $V = C^\\dagger X B^\\dagger X (e^{-i \\alpha} A^\\dagger)$ 也是容易给出的。我们只要将转换为自伴算符，并对调线路顺序即可：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "5a136c80",
"metadata": {
"hidden": true
},
"source": [
"我们将这两个线路代入图 4.8 即可 (红色线条隔开受控 $V$ 或 $V^\\dagger$ 算符的代入区域)：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "0c2ff2c3",
"metadata": {
"hidden": true
},
"source": [
"我们发现，上述的图像中，有两个可以简化的线路。所有蓝色虚线框中标出的部分实际上是恒等作用 $A A^\\dagger = C^\\dagger C = I$ (回顾到 $A, B, C$ 都是酉矩阵)，因此进一步简化为\n",
"\n",
"\n",
"\n",
"到现在为止，我们简化到 8 个单量子比特门路、与 8 个 CNOT 门路。我们还需要减掉两个 CNOT 门路。下一步中，我们会先对上图中的蓝色区域所包含的 CNOT 作简化。"
]
},
{
"cell_type": "markdown",
"id": "c9457f42",
"metadata": {
"hidden": true
},
"source": [
"### 第二步：简化 CNOT 门路"
]
},
{
"cell_type": "markdown",
"id": "319d783f",
"metadata": {
"hidden": true
},
"source": [
"这一步需要用到下述**引理一**：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "8a2f92fa",
"metadata": {
"hidden": true
},
"source": [
"我们可以立即简化得到 (这里仍然有 7 个 CNOT 门路，还没有达到要求)\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "4e54045e",
"metadata": {
"hidden": true
},
"source": [
"注意到第三行的 $B^\\dagger$ 门路与第一、二行的 CNOT 门路相互不受影响，因此我们交换一下这两个门路：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "d486f9cb",
"metadata": {
"hidden": true
},
"source": [
"对上述蓝色区域使用**引理一推论**：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "4b3bddcb",
"metadata": {
"hidden": true
},
"source": [
"可以得到\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "d06a149f",
"metadata": {
"hidden": true
},
"source": [
"### 第三步：相位转移"
]
},
{
"cell_type": "markdown",
"id": "0ebea8ad",
"metadata": {
"hidden": true
},
"source": [
"看起来问题又复杂化了，重新回到了 8 个 CNOT 门路。但实际上，依据**引理二**与**引理三**，上述蓝色虚线框中的部分可以比较随意地移动。迁移之后，得到\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "4753a90f",
"metadata": {
"hidden": true
},
"source": [
"橙色的区域是两个相同的 CNOT 算符。由于 CNOT 算符在双量子比特空间下是实对称酉算符，因此可以相互抵消。最终，我们得到\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "d26c81e6",
"metadata": {
"hidden": true
},
"source": [
"### 引理一：CNOT 门路简化"
]
},
{
"cell_type": "markdown",
"id": "26d3fa89",
"metadata": {
"hidden": true
},
"source": [
"我们需要证明下述路线图：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "3528d1ef",
"metadata": {
"hidden": true
},
"source": [
"证明方法与练习 4.21 相同。我们对等式左作考察："
]
},
{
"cell_type": "markdown",
"id": "892e2f05",
"metadata": {
"hidden": true
},
"source": [
"$$\n",
"\\begin{alignat*}{20}\n",
"& |\\psi_0\\rangle && |\\psi_1\\rangle && |\\psi_2\\rangle && |\\psi_3\\rangle &&\n",
"\\\\\n",
"& \\text{init}\n",
"&& \\mathrm{CNOT}_{x_2, x_3}\n",
"&& \\mathrm{CNOT}_{x_1, x_2}\n",
"&& \\mathrm{CNOT}_{x_1, x_3}\n",
"&& \\quad\\text{equivalent}\n",
"\\\\\n",
"& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& = |00\\rangle \\otimes |x_3\\rangle\n",
"\\\\\n",
"& |01\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |01\\rangle \\otimes X |x_3\\rangle \\quad\n",
"&& |01\\rangle \\otimes X |x_3\\rangle \\quad\n",
"&& |01\\rangle \\otimes X |x_3\\rangle \\quad\n",
"&& = |01\\rangle \\otimes X |x_3\\rangle\n",
"\\\\\n",
"& |10\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |10\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |11\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |11\\rangle \\otimes X |x_3\\rangle \\quad\n",
"&& = |11\\rangle \\otimes X |x_3\\rangle\n",
"\\\\\n",
"& |11\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |11\\rangle \\otimes X |x_3\\rangle \\quad\n",
"&& |10\\rangle \\otimes X |x_3\\rangle \\quad\n",
"&& |10\\rangle \\otimes X^2 |x_3\\rangle \\quad\n",
"&& = |10\\rangle \\otimes |x_3\\rangle\n",
"\\end{alignat*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "74a0148f",
"metadata": {
"hidden": true
},
"source": [
"再对等式右作考察："
]
},
{
"cell_type": "markdown",
"id": "841df2d0",
"metadata": {
"hidden": true
},
"source": [
"$$\n",
"\\begin{alignat*}{20}\n",
"& |\\psi_0\\rangle && |\\psi_4\\rangle && |\\psi_5\\rangle\n",
"\\\\\n",
"& \\text{init}\n",
"&& \\mathrm{CNOT}_{x_1, x_2}\n",
"&& \\mathrm{CNOT}_{x_2, x_3}\n",
"\\\\\n",
"& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |00\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |00\\rangle \\otimes |x_3\\rangle\n",
"\\\\\n",
"& |01\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |01\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |01\\rangle \\otimes X |x_3\\rangle \\quad\n",
"\\\\\n",
"& |10\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |11\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |11\\rangle \\otimes X |x_3\\rangle\n",
"\\\\\n",
"& |11\\rangle \\otimes |x_3\\rangle \\quad\n",
"&& |10\\rangle \\otimes X |x_3\\rangle \\quad\n",
"&& |10\\rangle \\otimes |x_3\\rangle\n",
"\\end{alignat*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "3313fbfb",
"metadata": {
"hidden": true
},
"source": [
"我们就发现 $|\\psi_3\\rangle = |\\psi_5\\rangle$，即等式两边相等。"
]
},
{
"cell_type": "markdown",
"id": "3a3b7c7d",
"metadata": {
"hidden": true
},
"source": [
"### 引理一推论"
]
},
{
"cell_type": "markdown",
"id": "d25bae1a",
"metadata": {
"hidden": true
},
"source": [
"我们给出下述等式关系：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "e7293abc",
"metadata": {
"hidden": true
},
"source": [
"我们对**引理一**的等式左右取自伴算符，随后对调等式左右，就得到上式的第一个等号。\n",
"\n",
"第二个等号相当于一种对易关系：\n",
"\n",
"$$\n",
"\\big[ \\mathrm{CNOT}_{x_1, x_3}, \\mathrm{CNOT}_{x_1, x_2} \\mathrm{CNOT}_{x_2, x_3} \\big] = 0\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "a62e81c8",
"metadata": {
"hidden": true
},
"source": [
"对于这个对易，它是比较容易看出来的，我们不作仔细证明。\n",
"\n",
"- 我们说，$\\mathrm{CNOT}_{x_1, x_3}$ 对于对目标量子比特 $|x_3\\rangle$ 的作用是 $X$ 或 $I$。而其它可能对 $|x_3\\rangle$ 产生作用的算符是 $\\mathrm{CNOT}_{x_2, x_3}$，它也是 $X$ 或 $I$。因此，可能作用于 $|x_3\\rangle$ 上的所有算符都相互对易。\n",
"- 同时，$|x_1\\rangle$ 的作用仅仅是控制，因此 $|x_1\\rangle$ 的值始终不变。\n",
"- 综合上面的两点考虑，$\\mathrm{CNOT}_{x_1, x_3}$ 的位置在这个门路中是可以任意交换。"
]
},
{
"cell_type": "markdown",
"id": "9ddb423e",
"metadata": {
"hidden": true
},
"source": [
"但我们也必须要说明，$\\mathrm{CNOT}_{x_1, x_2}$ 与 $\\mathrm{CNOT}_{x_2, x_3}$ 在门路中不可以交换。这个由读者自行验证吧。一般来说，如果一个量子比特同时可能作为控制比特出现，也被单量子比特算符作用或作为目标比特作用，那么这些算符之间不可交换。"
]
},
{
"cell_type": "markdown",
"id": "63f395a6",
"metadata": {
"hidden": true
},
"source": [
"### 引理二：对角算符在受控门路下的平移"
]
},
{
"cell_type": "markdown",
"id": "5fbb953b",
"metadata": {
"hidden": true
},
"source": [
"我们之前说到，一般来说，不是所有算符都可以与 CNOT 的控制部分作对换。一个简单的例子会是\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "38eb6868",
"metadata": {
"hidden": true
},
"source": [
"但我们马上会证明下述结论。对于任意的对角矩阵 $\\Lambda$ 与受控 $A$ 门路 (未必要求是酉矩阵)，其门路满足下式\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "06716d18",
"metadata": {
"hidden": true
},
"source": [
"既然 $\\Lambda$ 是对角矩阵，那么它一定能通过参数 $\\lambda_0, \\lambda_1 \\in \\mathbb{C}^2$ 表示：\n",
"\n",
"$$\n",
"\\Lambda = \\begin{bmatrix} \\lambda_0 & 0 \\\\ 0 & \\lambda_1 \\end{bmatrix} = \\lambda_0 P_0 + \\lambda_1 P_1\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "35b7bee6",
"metadata": {
"hidden": true
},
"source": [
"那么\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"\\mathrm{RHS} &= (\\Lambda \\otimes I) (P_0 \\otimes I + P_1 \\otimes A) = (\\lambda_0 P_0 \\otimes I + \\lambda_1 P_1 \\otimes I) (P_0 \\otimes I + P_1 \\otimes A) \\\\\n",
"&= \\lambda_0 P_0^2 \\otimes I^2 + \\lambda_1 P_1^2 \\otimes IA + \\big( \\text{terms involves } P_0 P_1 = P_1 P_0 = 0 \\big) \\\\\n",
"&= (P_0 \\otimes I + P_1 \\otimes A) (\\lambda_0 P_0 \\otimes I + \\lambda_1 P_1 \\otimes I) \\\\\n",
"&= (P_0 \\otimes I + P_1 \\otimes A) (\\Lambda \\otimes I) = \\mathrm{LHS}\n",
"\\end{align*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "eceadf23",
"metadata": {
"hidden": true
},
"source": [
"### 引理三"
]
},
{
"cell_type": "markdown",
"id": "6ab1ece6",
"metadata": {
"hidden": true
},
"source": [
""
]
},
{
"cell_type": "markdown",
"id": "31393df5",
"metadata": {
"hidden": true
},
"source": [
"该引理不作证明。直接代入矩阵定义很容易给予证明。引理三是要表明，该符合算符在双量子比特空间下为对角矩阵，因此可以利用**引理二**，该符合算符可以与原问题的第一行或第二行控制节点进行交换。"
]
},
{
"cell_type": "markdown",
"id": "ba358eec",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.23"
]
},
{
"cell_type": "markdown",
"id": "1d28d929",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.23\n",
"\n",
"仅用受控非门和单量子比特门构造一个 $U = R_x(\\theta)$ 和 $U = R_y(\\theta)$ 的 $C^1(U)$ 门。你能够把所需的单量子比特门的数量减到三个或两个吗？\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "8d03d22a",
"metadata": {
"hidden": true
},
"source": [
"我们先构造较为简单的受控 $R_y(\\theta)$。回顾练习 4.7，我们有\n",
"\n",
"$$\n",
"R_y(\\theta) = X R_y(-\\theta) X\n",
"$$\n",
"\n",
"因此，我们构造下述矩阵拆分：\n",
"\n",
"$$\n",
"R_y(\\theta) = e^{i \\alpha} AXBXC = X R_y(-\\theta/2) X R_y(\\theta/2)\n",
"$$\n",
"\n",
"其中，$\\alpha = 0$、$A = I$、$B = R_y(-\\theta/2)$、$C = R_y(\\theta/2)$。容易证明 $ABC = I$。因此，\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "0621d5f4",
"metadata": {
"hidden": true
},
"source": [
"类似地，我们也能得到 $R_z (\\theta) = X R_z (-\\theta) X$。但我们还能同时得到另一个结论：\n",
"\n",
"$$\n",
"R_x(\\theta) = H R_z(\\theta) H\n",
"$$\n",
"\n",
"证明过程很简单。套用式 (4.4, 4.6, 4.18)，可以得到\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"R_x(\\theta) &= \\cos \\frac{\\theta}{2} I - i \\sin \\frac{\\theta}{2} X \\\\\n",
"&= \\cos \\frac{\\theta}{2} H^2 - i \\sin \\frac{\\theta}{2} HZH \\\\\n",
"&= H \\left( \\cos \\frac{\\theta}{2} I - i \\sin \\frac{\\theta}{2} Z \\right) \\\\\n",
"&= H R_z(\\theta) H\n",
"\\end{align*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "7372af53",
"metadata": {
"hidden": true
},
"source": [
"基于上述结论，以及仿照 $R_y$ 的矩阵拆解\n",
"\n",
"$$\n",
"R_z(\\theta) = X R_z(-\\theta/2) X R_z(\\theta/2)\n",
"$$\n",
"\n",
"我们立即可以得到\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"R_x(\\theta) &= e^{i \\alpha} AXBXC \\\\\n",
"&= H R_z(\\theta) H = H X R_z(-\\theta/2) X R_z(\\theta/2) H\n",
"\\end{align*}\n",
"$$\n",
"\n",
"其中，$\\alpha = 0$、$A = H$、$B = R_z(-\\theta/2)$、$C = R_z(\\theta/2) H$。同样容易证明 $ABC = I$。因此，\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "576b3769",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.24"
]
},
{
"cell_type": "markdown",
"id": "ef593245",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.24\n",
"\n",
"验证图 4.9 所实现的 Toffoli 门。\n",
"\n",
"\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "deda323e",
"metadata": {
"hidden": true
},
"source": [
"我们这里采用与原来书中稍有不同的记号。我们令\n",
"\n",
"$$\n",
"V = \\frac{(1 + i) (I - i X)}{2}\n",
"$$\n",
"\n",
"它具有性质 $V^2 = X$。"
]
},
{
"cell_type": "code",
"execution_count": 20,
"id": "2bfe03bb",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"array([[0.5+0.5j, 0.5-0.5j],\n",
" [0.5-0.5j, 0.5+0.5j]])"
]
},
"execution_count": 20,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"V = 0.5 * (1 + 1j) * (I - 1j * X)\n",
"V"
]
},
{
"cell_type": "code",
"execution_count": 21,
"id": "017c771b",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"True"
]
},
"execution_count": 21,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.allclose(V.conj() @ V.conj(), X)"
]
},
{
"cell_type": "markdown",
"id": "9dbb2fe0",
"metadata": {
"hidden": true
},
"source": [
"可以验证 $V$ 具有如下分解：\n",
"\n",
"$$\n",
"\\begin{gather*}\n",
"A = H T, \\quad B = T^\\dagger, \\quad C = H, \\quad \\alpha = \\pi / 4 \\\\\n",
"V = e^{i \\alpha} AXBXC\n",
"\\end{gather*}\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 22,
"id": "e389f08e",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"True"
]
},
"execution_count": 22,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"A = H @ T\n",
"B = T.conj().T\n",
"C = H\n",
"α = np.pi / 4\n",
"np.allclose(np.exp(1j * α) * A @ X @ B @ X @ C, V)"
]
},
{
"cell_type": "markdown",
"id": "07dd250d",
"metadata": {
"hidden": true
},
"source": [
"那么依据练习 4.22 的结论，我们立即能得到\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "e37efb5c",
"metadata": {
"hidden": true
},
"source": [
"---"
]
},
{
"cell_type": "markdown",
"id": "ca23c35b",
"metadata": {
"hidden": true
},
"source": [
"上图与待证的图 4.9 在线路图的第一、二行上稍有差别。第一行依据练习 4.22 的引理二，容易知道其实是等价的，即第一行的 $T$ 门位置并不重要。对于第二行，我们需要表明 (下图中的 $S = T^2$)\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "895c8608",
"metadata": {
"hidden": true
},
"source": [
"第一个量子比特是控制比特，等式左右当然是相等的。我们要分别考察第二个目标量子比特在第一个比特为 $|0\\rangle, |1\\rangle$ 时的行为。对于控制量子比特为 $|0\\rangle$ 时，上图的等式证明问题就化为\n",
"\n",
"$$\n",
"T^\\dagger T = S T^\\dagger T^\\dagger\n",
"$$\n",
"\n",
"这个结论是比较显然的，等式左右为恒等算符。"
]
},
{
"cell_type": "code",
"execution_count": 23,
"id": "bbaafcf8",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"True"
]
},
"execution_count": 23,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.allclose(\n",
" T.conj().T @ T,\n",
" S @ T.conj().T @ T.conj().T)"
]
},
{
"cell_type": "markdown",
"id": "7fb9348e",
"metadata": {
"hidden": true
},
"source": [
"控制量子比特为 $|1\\rangle$ 时，上图的等式证明就化为\n",
"\n",
"$$\n",
"X T^\\dagger X T = S X T^\\dagger X T^\\dagger\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 24,
"id": "f845ae27",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"True"
]
},
"execution_count": 24,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.allclose(\n",
" X @ T.conj().T @ X @ T,\n",
" S @ X @ T.conj().T @ X @ T.conj().T)"
]
},
{
"cell_type": "markdown",
"id": "b1e6a15a",
"metadata": {
"hidden": true
},
"source": [
"因此，上图的等式成立。到这里，我们就借着练习 4.22 的结论，完成了 Toffoli 门的证明。"
]
},
{
"cell_type": "markdown",
"id": "54066b0c",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.25 (Fredkin 门的构造)"
]
},
{
"cell_type": "markdown",
"id": "9026337f",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.25\n",
"\n",
"回忆 Fredkin (受控交换，controlled-swap) 门进行的运算：\n",
"\n",
"$$\n",
"\\begin{bmatrix}\n",
"1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\\n",
"0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\\n",
"0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\\n",
"0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\\n",
"0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\\n",
"0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\\n",
"0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\\\n",
"0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\n",
"\\end{bmatrix}\n",
"$$\n",
"\n",
"1. 给出使用三个 Toffoli 门构造的 Fredkin 门 (提示：想想这个交换门的结构——可以每次控制一个门)。\n",
"2. 证明第一个和最后一个 Toffoli 门可以用受控非门代替。\n",
"3. 用图 4.8 所示线路代替中间的 Toffoli 门，来得到仅用 6 个双量子比特门所构造的 Fredkin 门。\n",
"4. 有更简单的仅用 5 个双量子比特门的结构吗？\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "7a40b6cd",
"metadata": {
"hidden": true
},
"source": [
"**第一问**"
]
},
{
"cell_type": "markdown",
"id": "e4b26028",
"metadata": {
"hidden": true
},
"source": [
"首先我们要验证交换门路：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "99bdb210",
"metadata": {
"hidden": true
},
"source": [
"我们假设态处于 $|x_1\\rangle |x_2\\rangle$，其中 $x_1, x_2 \\in \\{0, 1\\}$。那么\n",
"\n",
"- 经过第一个 CNOT，得到 $|x_1\\rangle |x_2 \\oplus x_1\\rangle$；\n",
"- 经过第二个 CNOT，得到 $|x_1 \\oplus (x_2 \\oplus x_1)\\rangle |x_2 \\oplus x_1\\rangle = |x_2\\rangle |x_2 \\oplus x_1\\rangle$；\n",
"- 经过第三个 CNOT，得到 $|x_2\\rangle |(x_2 \\oplus x_1) \\oplus x_2\\rangle = |x_2\\rangle |x_1\\rangle$；\n",
"\n",
"这就将 $|x_1\\rangle |x_2\\rangle$ 转换到了 $|x_2\\rangle |x_1\\rangle$。"
]
},
{
"cell_type": "markdown",
"id": "f3a6e350",
"metadata": {
"hidden": true
},
"source": [
"因此，受控交换门只是再加上了一个控制节点而已：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "d4ef376e",
"metadata": {
"hidden": true
},
"source": [
"---"
]
},
{
"cell_type": "markdown",
"id": "588d83e5",
"metadata": {
"hidden": true
},
"source": [
"**第二问**"
]
},
{
"cell_type": "markdown",
"id": "b091a82f",
"metadata": {
"hidden": true
},
"source": [
"这一问答案是\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "65c65434",
"metadata": {
"hidden": true
},
"source": [
"比较正常的解法会是对该算符考察其真值表 (原书中图 3.15)。当然，我们也可以用数学的语言来证明。上图的三量子比特下的 CNOT 门 (等式右的左右两个门路) 可以写为 $I \\otimes \\mathrm{CNOT}$，而等式右的中间 Toffoli 门表示为 $I^{\\otimes 3} + P_1 \\otimes (X-I) \\otimes P_1$。我们需要证明的是\n",
"\n",
"$$\n",
"\\text{ctrl-swap} = \\big( I \\otimes \\mathrm{CNOT} \\big) \\big( I^{\\otimes 3} + P_1 \\otimes (X-I) \\otimes P_1 \\big) \\big( I \\otimes \\mathrm{CNOT} \\big)\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": 25,
"id": "69005351",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"array([[1., 0., 0., 0., 0., 0., 0., 0.],\n",
" [0., 1., 0., 0., 0., 0., 0., 0.],\n",
" [0., 0., 1., 0., 0., 0., 0., 0.],\n",
" [0., 0., 0., 1., 0., 0., 0., 0.],\n",
" [0., 0., 0., 0., 1., 0., 0., 0.],\n",
" [0., 0., 0., 0., 0., 0., 1., 0.],\n",
" [0., 0., 0., 0., 0., 1., 0., 0.],\n",
" [0., 0., 0., 0., 0., 0., 0., 1.]])"
]
},
"execution_count": 25,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"ctrl_12_X = np.kron(I, ctrl_1_X)\n",
"ctrl_13_toff = np.eye(8) + np.kron(np.kron(P1, X-I), P1)\n",
"ctrl_12_X @ ctrl_13_toff @ ctrl_12_X"
]
},
{
"cell_type": "markdown",
"id": "25388965",
"metadata": {
"hidden": true
},
"source": [
"上述的计算结果与题目所给出的 Fredkin 门矩阵表示完全一致。"
]
},
{
"cell_type": "markdown",
"id": "4cd2cd8f",
"metadata": {
"hidden": true
},
"source": [
"---"
]
},
{
"cell_type": "markdown",
"id": "790ff1e3",
"metadata": {
"hidden": true
},
"source": [
"**第三问**"
]
},
{
"cell_type": "markdown",
"id": "a5075dbc",
"metadata": {
"hidden": true
},
"source": [
"由于交换算符的对称性，门路的第二、三行可以互相交换。为了方便起见，我们交换这两行，并代入图 4.8：\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "58788be2",
"metadata": {
"hidden": true
},
"source": [
"---"
]
},
{
"cell_type": "markdown",
"id": "5c25900e",
"metadata": {
"hidden": true
},
"source": [
"**第四问**"
]
},
{
"cell_type": "markdown",
"id": "277238d1",
"metadata": {
"hidden": true
},
"source": [
"这里需要利用到一些对易关系。首先，利用到 $V V^\\dagger = V^\\dagger V = I$ 的对易关系，可以表明左数第 6 个门路与左数 3, 4, 5 门路可以交换。因此，\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "df0c7e8d",
"metadata": {
"hidden": true
},
"source": [
"随后利用到 $[X, X] = 0$ 的关系，右数第 1, 2 个门路可以对易。因此，\n",
"\n",
"\n",
"\n",
"上图中的五个蓝色虚线框，就是题目所要求的 5 个双量子比特门路。"
]
},
{
"cell_type": "markdown",
"id": "77ff4cc1",
"metadata": {
"hidden": true
},
"source": [
"本题的参考资料是\n",
"\n",
"> https://journals.aps.org/pra/abstract/10.1103/PhysRevA.53.2855"
]
},
{
"cell_type": "markdown",
"id": "8841e43c",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.26"
]
},
{
"cell_type": "markdown",
"id": "1de3c758",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.26\n",
"\n",
"证明如下线路\n",
"\n",
"\n",
"\n",
"与 Toffoli 门的差别仅是相对相位。即该线路把 $|c_1, c_2, t\\rangle$ 变为 $e^{i \\theta(c_1, c_2, t)} |c_1, c_2, t \\oplus c_1 c_2\\rangle$，其中 $e^{i \\theta(c_1, c_2, t)}$ 是某个相对相位因子。这类门在实验实现中有时有用，它可以实现除了一个相对相位外与 Toffoli 门相同的门，但比直接实现 Toffoli 门容易得多。\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "b0d4663e",
"metadata": {
"hidden": true
},
"source": [
"本题使用练习 4.21 的解法会比较方便。我们同时回顾到练习 4.7 的结论 $X R_y(\\theta) X = R_y(-\\theta)$ 以及 $X^2 = 1$。那么，我们考察不同的 $c_1, c_2$ 下，第三行算符的实际作用 (注意算符图上线路的顺序与矩阵乘法顺序是反过来的)：\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"(c_0, c_1) = (0, 0)\n",
"&\\rightarrow R_y\\left(- \\frac{\\pi}{4}\\right) R_y\\left(- \\frac{\\pi}{4}\\right) R_y\\left(\\frac{\\pi}{4}\\right) R_y\\left(\\frac{\\pi}{4}\\right) = I \\\\\n",
"(c_0, c_1) = (0, 1)\n",
"&\\rightarrow R_y\\left(- \\frac{\\pi}{4}\\right) X \\underline{R_y\\left(- \\frac{\\pi}{4}\\right) R_y\\left(\\frac{\\pi}{4}\\right)} X R_y\\left(\\frac{\\pi}{4}\\right) \\\\\n",
"&= R_y\\left(- \\frac{\\pi}{4}\\right) X I X R_y\\left(\\frac{\\pi}{4}\\right) = I \\\\\n",
"(c_0, c_1) = (1, 0)\n",
"&\\rightarrow \\underline{R_y\\left(- \\frac{\\pi}{4}\\right) R_y\\left(- \\frac{\\pi}{4}\\right)} X \\underline{R_y\\left(\\frac{\\pi}{4}\\right) R_y\\left(\\frac{\\pi}{4}\\right)} \\\\\n",
"&= R_y\\left(- \\frac{\\pi}{2}\\right) \\underline{X R_y\\left(\\frac{\\pi}{2}\\right) X} X \\\\\n",
"&= \\underline{R_y\\left(- \\frac{\\pi}{2}\\right) R_y\\left(- \\frac{\\pi}{2}\\right)} X \\\\\n",
"&= R_y\\left(- \\pi\\right) X = Z \\\\\n",
"(c_0, c_1) = (1, 1)\n",
"&\\rightarrow R_y\\left(- \\frac{\\pi}{4}\\right) X R_y\\left(- \\frac{\\pi}{4}\\right) \\underline{X R_y\\left(\\frac{\\pi}{4}\\right) X} R_y\\left(\\frac{\\pi}{4}\\right) \\\\\n",
"&= R_y\\left(- \\frac{\\pi}{4}\\right) X \\underline{R_y\\left(- \\frac{\\pi}{4}\\right) R_y\\left(- \\frac{\\pi}{4}\\right) X R_y\\left(\\frac{\\pi}{4}\\right)} \\\\\n",
"&= X \\underline{X R_y\\left(- \\frac{\\pi}{4}\\right) X R_y\\left(- \\frac{\\pi}{4}\\right)} = X\n",
"\\end{align*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "0b6dd13a",
"metadata": {
"hidden": true
},
"source": [
"这里比较特殊的是 $(c_0, c_1) = (1, 0)$ 的情况。该控制门路的作用是将 $|t\\rangle$ 变为 $Z |t\\rangle$，而不是恒等变换。但我们也注意到，$Z$ 的作用等同于相位门 $P_\\pi = S^2 = T^4$。因此，对于这种特殊情况，确实可以认为是相差一个相位；但这个相位是 $|t\\rangle$ 中 $|0\\rangle$ 与 $|1\\rangle$ 的贡献分量相位之区别，而不是全局相位 $e^{i \\theta(c_1, c_2, t)}$。因此我倾向于认为，题目的问法是错的。"
]
},
{
"cell_type": "markdown",
"id": "e04627a2",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.27"
]
},
{
"cell_type": "markdown",
"id": "84b25c66",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.27\n",
"\n",
"仅用受控非门和 Toffoli 门，来构造一个执行如下变换的量子线路。\n",
"\n",
"$$\n",
"\\begin{bmatrix}\n",
"1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\\n",
"0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\\\\n",
"0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\\n",
"0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\\n",
"0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\\n",
"0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\\n",
"0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\\\n",
"0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\\n",
"\\end{bmatrix}\n",
"$$\n",
"\n",
"这类部分循环置换 (cyclic permutation) 运算将在后面第 7 章用到。\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "0c35031a",
"metadata": {
"hidden": true
},
"source": [
"很遗憾的是，这道题的答案目前还是凑出来的。但是我们至少可以验证下述结果。\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "00088236",
"metadata": {
"hidden": true
},
"source": [
"仍然按照练习 4.21 的做法，对每个可能的波函数进行讨论：\n",
"\n",
"$$\n",
"\\begin{alignat*}{20}\n",
"& \\mathrm{init}\\; && \\mathrm{Toffoli}_{x_2 x_3, x_1}\\; && \\mathrm{CNOT}_{x_3, x_2}\\; && \\mathrm{CNOT}_{x_2, x_3}\\; && \\mathrm{CNOT}_{x_1, x_2}\\; && \\mathrm{Toffoli}_{x_1 x_2, x_3}\\; && \\mathrm{CNOT}_{x_1, x_2} \\\\\n",
"& |000\\rangle\\; && |000\\rangle && |000\\rangle && |000\\rangle && |000\\rangle && |000\\rangle && |000\\rangle && \\\\\n",
"& |001\\rangle\\; && |001\\rangle && |011\\rangle && |010\\rangle && |010\\rangle && |010\\rangle && |010\\rangle && \\\\\n",
"& |010\\rangle\\; && |010\\rangle && |010\\rangle && |011\\rangle && |011\\rangle && |011\\rangle && |011\\rangle && \\\\\n",
"& |011\\rangle\\; && |111\\rangle && |101\\rangle && |101\\rangle && |111\\rangle && |110\\rangle && |100\\rangle && \\\\\n",
"& |100\\rangle\\; && |100\\rangle && |100\\rangle && |100\\rangle && |110\\rangle && |111\\rangle && |101\\rangle && \\\\\n",
"& |101\\rangle\\; && |101\\rangle && |111\\rangle && |110\\rangle && |100\\rangle && |100\\rangle && |110\\rangle && \\\\\n",
"& |110\\rangle\\; && |110\\rangle && |110\\rangle && |111\\rangle && |101\\rangle && |101\\rangle && |111\\rangle && \\\\\n",
"& |111\\rangle\\; && |011\\rangle && |001\\rangle && |001\\rangle && |001\\rangle && |001\\rangle && |001\\rangle && \\\\\n",
"\\end{alignat*}\n",
"$$\n",
"\n",
"上表满足题目所要求的循环置换的矩阵表示结果。"
]
},
{
"cell_type": "markdown",
"id": "6471d657",
"metadata": {
"hidden": true
},
"source": [
"---"
]
},
{
"cell_type": "markdown",
"id": "7ab15f93",
"metadata": {
"hidden": true
},
"source": [
"我们也可以利用 Python 代码结合 Qiskit 实现。但需要注意，由于原书中的量子比特记号约定是 $|x_1 x_2 x_3\\rangle$，但在 Qiskit 中则是 $|x_3 x_2 x_1\\rangle$，因此为了得到与书中相同的矩阵表示，我们需要额外增加两个 SWAP 门路。作为答案的线路，在下图中是被灰色虚线所围起来的线路。"
]
},
{
"cell_type": "code",
"execution_count": 26,
"id": "e413a85a",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"image/svg+xml": [
"\r\n",
"\r\n",
"\r\n",
"\r\n"
],
"text/plain": [
""
]
},
"execution_count": 26,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"circ = QuantumCircuit(3)\n",
"circ.swap(0, 2)\n",
"circ.barrier()\n",
"circ.toffoli(1, 2, 0)\n",
"circ.cnot(2, 1)\n",
"circ.cnot(1, 2)\n",
"circ.cnot(0, 1)\n",
"circ.toffoli(0, 1, 2)\n",
"circ.cnot(0, 1)\n",
"circ.barrier()\n",
"circ.swap(0, 2)\n",
"circ.draw()"
]
},
{
"cell_type": "markdown",
"id": "e2cc2c2e",
"metadata": {
"hidden": true
},
"source": [
"我们可以用下述代码，打出其矩阵表示。可以发现，该矩阵表示结果与题目要求完全相同。"
]
},
{
"cell_type": "code",
"execution_count": 27,
"id": "966ec526",
"metadata": {
"hidden": true
},
"outputs": [
{
"data": {
"text/plain": [
"array([[1, 0, 0, 0, 0, 0, 0, 0],\n",
" [0, 0, 0, 0, 0, 0, 0, 1],\n",
" [0, 1, 0, 0, 0, 0, 0, 0],\n",
" [0, 0, 1, 0, 0, 0, 0, 0],\n",
" [0, 0, 0, 1, 0, 0, 0, 0],\n",
" [0, 0, 0, 0, 1, 0, 0, 0],\n",
" [0, 0, 0, 0, 0, 1, 0, 0],\n",
" [0, 0, 0, 0, 0, 0, 1, 0]])"
]
},
"execution_count": 27,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"np.asarray(Operator(circ).data, dtype=int)"
]
},
{
"cell_type": "markdown",
"id": "144f0d85",
"metadata": {
"hidden": true
},
"source": [
"---"
]
},
{
"cell_type": "markdown",
"id": "adfa0243",
"metadata": {
"hidden": true
},
"source": [
"但本题最为直观的构造方式是，先对量子比特作加法，随后若 $|x_1 x_2, x_3\\rangle = |00, x_3\\rangle$，那么将其变换为 $|00, x_3\\oplus1\\rangle$。其实际操作过程非常直观：\n",
"\n",
"$$\n",
"\\begin{alignat*}{20}\n",
"& \\mathrm{init}\\; && \\mathrm{Addition}\\; && \\mathrm{Control} \\\\\n",
"& |000\\rangle\\; && |001\\rangle && |000\\rangle && \\\\\n",
"& |001\\rangle\\; && |010\\rangle && |010\\rangle && \\\\\n",
"& |010\\rangle\\; && |011\\rangle && |011\\rangle && \\\\\n",
"& |011\\rangle\\; && |100\\rangle && |100\\rangle && \\\\\n",
"& |100\\rangle\\; && |101\\rangle && |101\\rangle && \\\\\n",
"& |101\\rangle\\; && |110\\rangle && |110\\rangle && \\\\\n",
"& |110\\rangle\\; && |111\\rangle && |111\\rangle && \\\\\n",
"& |111\\rangle\\; && |000\\rangle && |001\\rangle && \\\\\n",
"\\end{alignat*}\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "d67ec702",
"metadata": {
"hidden": true
},
"source": [
"- 加法的线路构造思路是，需要在 $|x_3\\rangle$ 引入 $X$ 算符以实现 $|x_3 \\oplus 1\\rangle$，随后小心地通过引入控制门路以作合适的进位。\n",
"- 控制部分 $|x_1 x_2, x_3\\rangle = |00, x_3\\rangle$ 线路，实际上就是以 $|x_1 x_2\\rangle = |00\\rangle$ 来控制的 Toffoli 线路；我们回顾到先前定义的所有 Toffoli 线路都是以 $|x_1 x_2\\rangle = |11\\rangle$ 控制的；因此，依照书中的图 4.11 或图 4.12，可以给出该控制线路。\n",
"\n",
"下图的蓝色区域即是加法线路，橙色区域即是控制线路。该线路也是本题的一个解，且构造思路清晰；但由于包含了单量子比特 $X$ 门路，因此并不满足题目的要求。\n",
"\n",
""
]
},
{
"cell_type": "markdown",
"id": "26552bb4",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.28"
]
},
{
"cell_type": "markdown",
"id": "94adb808",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.28\n",
"\n",
"对于 $U = V^2$，其中 $V$ 是酉的，使用类似于图 4.10 的方式构造 $C^5 (U)$ 门，但不用工作量子比特，可以使用受控 $V$ 门与受控 $V^\\dagger$ 门。\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "38d5c8a4",
"metadata": {
"hidden": true
},
"source": [
":::{danger}\n",
"\n",
"本题无解。无解性的证明思路是作行列式值分析。详细的理由请参考\n",
"\n",
"> https://cs.stackexchange.com/a/138853/141343\n",
"\n",
"如果我们允许多控制节点的非门或 $V$ 门，那么 $C^5 (U)$ 门是可以构造出来的：\n",
"\n",
"\n",
"\n",
"同时，若定义 $U = V^{16}$，那么该练习也是可以完成的，但线路恐怕会非常复杂。\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "02012126",
"metadata": {
"heading_collapsed": true
},
"source": [
"## 练习 4.29, 4.30"
]
},
{
"cell_type": "markdown",
"id": "9eb128f4",
"metadata": {
"hidden": true
},
"source": [
":::{admonition} 练习 4.29\n",
"\n",
"不使用工作量子比特，求出一个包含 $O(n^2)$ Toffoli 门、受控非门和单量子比特门的可实现 $C^n(X)$ 门 ($n > 3$) 的线路。\n",
"\n",
":::"
]
},
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"source": [
":::{admonition} 练习 4.30\n",
"\n",
"设 $U$ 是一个单量子比特酉运算，不使用工作量子比特，求一个包含 $O(n^2)$ Toffoli 门、受控非门和单量子比特门，实现 $C^n (U)$ 门 ($n>3$) 的线路。\n",
"\n",
":::"
]
},
{
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"id": "eb0bd116",
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"source": [
":::{warning}\n",
"\n",
"这两个问题不在此作证明。\n",
"\n",
"可以参考下述文献的 Corollary 7.6。尽管证明过程还算比较直观，但需要引入许多符号，证明过程也比较长。\n",
"\n",
"> https://arxiv.org/abs/quant-ph/9503016\n",
"\n",
"同时需要指出，对于任意 $n > 5$，如果允许一个工作量子比特，那么依据 Corollary 7.4，多受控非门的门路复杂度可以从 $O(n^2)$ 降为 $O(n)$。从这个角度来说，工作量子比特其实还是非常重要的。工作量子比特在这个问题中，或可能在其它问题中，可以对线路大大作简化。\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "12a8c599",
"metadata": {
"heading_collapsed": true
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"source": [
"## 练习 4.31 (更多的线路恒等式)"
]
},
{
"cell_type": "markdown",
"id": "686c511f",
"metadata": {
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"source": [
":::{admonition} 练习 4.31\n",
"\n",
"令下标表示算符作用的量子比特、并令 $C$ 是量子比特 1 为控制、量子比特 2 为目标的受控非门。证明如下恒等式：\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"C X_1 C &= X_1 X_2 \\tag{4.32} \\\\\n",
"C Y_1 C &= Y_1 X_2 \\tag{4.33} \\\\\n",
"C Z_1 C &= Z_1 \\tag{4.34} \\\\\n",
"C X_2 C &= X_2 \\tag{4.35} \\\\\n",
"C Y_2 C &= Z_1 Y_2 \\tag{4.36} \\\\\n",
"C Z_2 C &= Z_1 Z_2 \\tag{4.37} \\\\\n",
"R_{z,1} (\\theta) C &= C R_{z,1} (\\theta) \\tag{4.38} \\\\\n",
"R_{x,2} (\\theta) C &= C R_{x,2} (\\theta) \\tag{4.39}\n",
"\\end{align*}\n",
"$$\n",
"\n",
":::"
]
},
{
"cell_type": "markdown",
"id": "ad59c3f0",
"metadata": {
"heading_collapsed": true,
"hidden": true
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"source": [
"### 记号的细化定义"
]
},
{
"cell_type": "markdown",
"id": "e7ae7efe",
"metadata": {
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"source": [
"我们回顾到 $C$ 即 CNOT 门，因此\n",
"\n",
"$$\n",
"C = P_0 \\otimes I + P_1 \\otimes X\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "bf25c997",
"metadata": {
"hidden": true
},
"source": [
"$X_1$ 是只作用于第一个量子比特的门路，因此\n",
"\n",
"$$\n",
"X_1 = X \\otimes I\n",
"$$"
]
},
{
"cell_type": "markdown",
"id": "411e1fe6",
"metadata": {
"hidden": true
},
"source": [
"仿上，其它的符号也可以相应地给予定义。"
]
},
{
"cell_type": "markdown",
"id": "53195fb2",
"metadata": {
"heading_collapsed": true,
"hidden": true
},
"source": [
"### 式 (4.32)"
]
},
{
"cell_type": "markdown",
"id": "05a999eb",
"metadata": {
"hidden": true
},
"source": [
"\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"C X_1 C &= (P_0 \\otimes I + P_1 \\otimes X) (X \\otimes I) (P_0 \\otimes I + P_1 \\otimes X) \\\\\n",
"&= P_0 X P_0 \\otimes I I I + P_0 X P_1 \\otimes I I X + P_1 X P_0 \\otimes X I I + P_1 X P_1 \\otimes X I X \\\\\n",
"&= 0 \\otimes I + (P_0 X P_1 + P_1 X P_0) \\otimes X + 0 \\otimes I \\\\\n",
"&= X \\otimes X = (X \\otimes I) (I \\otimes X) = X_1 X_2\n",
"\\end{align*}\n",
"$$\n",
"\n",
"上式中，我们利用了 $P_0 X P_1 + P_1 X P_0 = I$ 以及 $P_0 X P_0 = P_1 X P_1 = 0$。这些可以通过矩阵乘法简单地验证，当然也可以用 $Z, I$ 的表达式展开 $P_0, P_1$ 来求解。"
]
},
{
"cell_type": "markdown",
"id": "d1929814",
"metadata": {
"heading_collapsed": true,
"hidden": true
},
"source": [
"### 式 (4.33)"
]
},
{
"cell_type": "markdown",
"id": "f112f5d1",
"metadata": {
"hidden": true
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"source": [
"\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"C Y_1 C &= (P_0 \\otimes I + P_1 \\otimes X) (Y \\otimes I) (P_0 \\otimes I + P_1 \\otimes X) \\\\\n",
"&= P_0 Y P_0 \\otimes I I I + P_0 Y P_1 \\otimes I I X + P_1 Y P_0 \\otimes X I I + P_1 Y P_1 \\otimes X I X \\\\\n",
"&= 0 \\otimes I + (P_0 Y P_1 + P_1 Y P_0) \\otimes X + 0 \\otimes I \\\\\n",
"&= Y \\otimes X = (Y \\otimes I) (I \\otimes X) = Y_1 X_2\n",
"\\end{align*}\n",
"$$\n",
"\n",
"上式中，我们利用了 $P_0 Y P_1 + P_1 Y P_0 = Y$ 以及 $P_0 Y P_0 = P_1 Y P_1 = 0$。"
]
},
{
"cell_type": "markdown",
"id": "1868f804",
"metadata": {
"heading_collapsed": true,
"hidden": true
},
"source": [
"### 式 (4.34)"
]
},
{
"cell_type": "markdown",
"id": "f5d15c82",
"metadata": {
"hidden": true
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"source": [
"\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"C Z_1 C &= (P_0 \\otimes I + P_1 \\otimes X) (Z \\otimes I) (P_0 \\otimes I + P_1 \\otimes X) \\\\\n",
"&= P_0 Z P_0 \\otimes I I I + P_0 Z P_1 \\otimes I I X + P_1 Z P_0 \\otimes X I I + P_1 Z P_1 \\otimes X I X \\\\\n",
"&= P_0 Z P_0 \\otimes I + 0 \\otimes X + 0 \\otimes X + P_1 Z P_1 \\otimes I \\\\\n",
"&= Z \\otimes I = Z_1\n",
"\\end{align*}\n",
"$$\n",
"\n",
"上式中，我们利用了 $P_0 Z P_0 + P_1 Z P_1 = Z$ 以及 $P_0 Z P_1 = P_1 Z P_0 = 0$。"
]
},
{
"cell_type": "markdown",
"id": "7d499755",
"metadata": {
"heading_collapsed": true,
"hidden": true
},
"source": [
"### 式 (4.35)"
]
},
{
"cell_type": "markdown",
"id": "511a85cb",
"metadata": {
"hidden": true
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"source": [
"\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"C X_2 C &= (P_0 \\otimes I + P_1 \\otimes X) (I \\otimes X) (P_0 \\otimes I + P_1 \\otimes X) \\\\\n",
"&= P_0 I P_0 \\otimes I X I + P_0 I P_1 \\otimes I X X + P_1 I P_0 \\otimes X X I + P_1 I P_1 \\otimes X X X \\\\\n",
"&= P_0 \\otimes X + 0 \\otimes I + 0 \\otimes I + P_1 \\otimes X \\\\\n",
"&= I \\otimes X = X_2\n",
"\\end{align*}\n",
"$$\n",
"\n",
"上式中，我们利用了 Pauli 矩阵所具有的性质 $X^2 = Y^2 = Z^2 = I$、投影算符的幂等性 $P_0^2 = P_0$ 及 $P_1^2 = P_1$，和投影算符性质 $P_0 + P_1 = I$。"
]
},
{
"cell_type": "markdown",
"id": "78c69816",
"metadata": {
"heading_collapsed": true,
"hidden": true
},
"source": [
"### 式 (4.36)"
]
},
{
"cell_type": "markdown",
"id": "c71affac",
"metadata": {
"hidden": true
},
"source": [
"\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"C Y_2 C &= (P_0 \\otimes I + P_1 \\otimes X) (I \\otimes Y) (P_0 \\otimes I + P_1 \\otimes X) \\\\\n",
"&= P_0 I P_0 \\otimes I Y I + P_0 I P_1 \\otimes I Y X + P_1 I P_0 \\otimes X Y I + P_1 I P_1 \\otimes X Y X \\\\\n",
"&= P_0 \\otimes Y - P_1 \\otimes Y \\\\\n",
"&= Z \\otimes Y = (Z \\otimes I) (I \\otimes Y) = Z_1 Y_2\n",
"\\end{align*}\n",
"$$\n",
"\n",
"上式中，我们利用了 $Z$ 在投影算符下的展开 $Z = P_0 - P_1$，以及一个恒等式 $X Y X = iZX = i^2 Y = -Y$。"
]
},
{
"cell_type": "markdown",
"id": "2f549707",
"metadata": {
"heading_collapsed": true,
"hidden": true
},
"source": [
"### 式 (4.37)"
]
},
{
"cell_type": "markdown",
"id": "221682f5",
"metadata": {
"hidden": true
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"source": [
"\n",
"\n",
"$$\n",
"\\begin{align*}\n",
"C Z_2 C &= (P_0 \\otimes I + P_1 \\otimes X) (I \\otimes Z) (P_0 \\otimes I + P_1 \\otimes X) \\\\\n",
"&= P_0 I P_0 \\otimes I Z I + P_0 I P_1 \\otimes I Z X + P_1 I P_0 \\otimes X Z I + P_1 I P_1 \\otimes X Z X \\\\\n",
"&= P_0 \\otimes Z - P_1 \\otimes Z \\\\\n",
"&= Z \\otimes Z = (Z \\otimes I) (I \\otimes Z) = Z_1 Z_2\n",
"\\end{align*}\n",
"$$\n",
"\n",
"上式中，我们利用了一个恒等式 $X Z X = iXY = i^2 Z = -Z$。"
]
},
{
"cell_type": "markdown",
"id": "cacd819f",
"metadata": {
"heading_collapsed": true,
"hidden": true
},
"source": [
"### 式 (4.38)"
]
},
{
"cell_type": "markdown",
"id": "cc100286",
"metadata": {
"hidden": true
},
"source": [
"\n",
"\n",
"我们知道 $R_z (\\theta)$ 是对角矩阵，因此可以与控制节点交换。这是练习 4.22 证明过程中的引理二所告诉我们的。"
]
},
{
"cell_type": "markdown",
"id": "3c8fc66e",
"metadata": {
"heading_collapsed": true,
"hidden": true
},
"source": [
"### 式 (4.39)"
]
},
{
"cell_type": "markdown",
"id": "5abfd8fe",
"metadata": {
"hidden": true
},
"source": [
"\n",
"\n",
"我们知道 $R_x (\\theta)$ 在目标量子比特上，与控制非门的非门 $X$ 是可以交换的。因此，上式成立。"
]
}
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