東京大学 新領域創成科学研究科 メディカル情報生命専攻 2024年1月実施 問題11

标签：

• Tokyo-University
• Quantum-Computing
• Complex-Numbers
• Hermitian-Matrix

Author

zephyr, 祭音Myyura (with the help of an anonymous contributor)

Description

A quantum state of a 1-qubit quantum computer can be represented by a $2 \times 2$ complex matrix $\rho = \frac{1}{2} \begin{pmatrix} 1 + a & b - ic \\ b + ic & 1 - a \end{pmatrix}$ called a density matrix. Here, $i = \sqrt{-1}$ is the imaginary unit that satisfies $i^2 = -1$ and $a, b, c$ are real numbers that satisfy $a^2 + b^2 + c^2 \leq 1$. By measuring the qubit represented by matrix $\rho$ in the computational basis (also called the Z-basis), we observe state 0 or state 1 with probabilities $p_0 = \frac{1 + a}{2}$ and $p_1 = \frac{1 - a}{2}$, which are the diagonal elements of matrix $\rho$, respectively. Answer the following questions with mathematical derivation.

1. Show that all the eigenvalues of matrix $\rho$ are non-negative real numbers.

2. Answer the probability that state 0 is observed by measurement in the computational basis after applying quantum gate operation $H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ to quantum state $\rho$.

3. Answer the probability that state 0 is observed by measurement in the computational basis after applying quantum gate operation $Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$ to quantum state $\rho$.

4. Let $\rho' = \frac{1}{2} \begin{pmatrix} 1 + a' & b' - ic' \\ b' + ic' & 1 - a' \end{pmatrix}$ be the quantum state after applying quantum gate operation $U = \begin{pmatrix} u_{00}' + iu_{00}'' & u_{01}' + iu_{01}'' \\ u_{10}' + iu_{10}'' & u_{11}' + iu_{11}'' \end{pmatrix}$ to quantum state $\rho$. Compute $a'^2 + b'^2 + c'^2$.

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1个量子比特量子计算机的量子态可以表示为一个 $2 \times 2$ 的复数矩阵 $\rho = \frac{1}{2} \begin{pmatrix} 1 + a & b - ic \\ b + ic & 1 - a \end{pmatrix}$，称为密度矩阵。这里，$i = \sqrt{-1}$ 是满足 $i^2 = -1$ 的虚数单位，$a, b, c$ 是满足 $a^2 + b^2 + c^2 \leq 1$ 的实数。通过在计算基（也称为Z基）中测量由矩阵 $\rho$ 表示的量子比特，我们以概率 $p_0 = \frac{1 + a}{2}$ 和 $p_1 = \frac{1 - a}{2}$ 观察到状态0或状态1，分别对应密度矩阵 $\rho$ 的对角元素。用数学推导回答以下问题。

1. 证明矩阵 $\rho$ 的所有特征值都是非负实数。

2. 在对量子态 $\rho$ 施加量子门操作 $H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ 后，在计算基中测量观察到状态0的概率是多少。

3. 在对量子态 $\rho$ 施加量子门操作 $Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$ 后，在计算基中测量观察到状态0的概率是多少。

4. 设 $\rho' = \frac{1}{2} \begin{pmatrix} 1 + a' & b' - ic' \\ b' + ic' & 1 - a' \end{pmatrix}$ 是施加量子门操作 $U = \begin{pmatrix} u_{00}' + iu_{00}'' & u_{01}' + iu_{01}'' \\ u_{10}' + iu_{10}'' & u_{11}' + iu_{11}'' \end{pmatrix}$ 后的量子态。计算 $a'^2 + b'^2 + c'^2$。

Kai

解题思路

这道题目涉及量子计算中的密度矩阵和量子门操作，需要运用线性代数和复数运算的知识。我们将逐步解答每个小问：

1. 求密度矩阵的特征值，证明它们是非负实数。
2. 计算 Hadamard 门（H 门）操作后的测量概率。
3. 计算 Y 门操作后的测量概率。
4. 计算一般量子门 U 操作后的密度矩阵参数。

每个小问都需要详细的数学推导。

1. Show that all the eigenvalues of matrix $\rho$ are non-negative real numbers

To find the eigenvalues of $\rho$, we need to solve the characteristic equation:

$$\det(\rho - \lambda I) = 0$$

$$\det\left(\begin{pmatrix} \frac{1+a}{2} - \lambda & \frac{b-ic}{2} \\ \frac{b+ic}{2} & \frac{1-a}{2} - \lambda \end{pmatrix}\right) = 0$$

$$(\frac{1+a}{2} - \lambda)(\frac{1-a}{2} - \lambda) - (\frac{b-ic}{2})(\frac{b+ic}{2}) = 0$$

$$\lambda^2 - \lambda + \frac{1-a^2-b^2-c^2}{4} = 0$$

The solutions to this quadratic equation are:

$$\lambda = \frac{1 \pm \sqrt{1-(1-a^2-b^2-c^2)}}{2} = \frac{1 \pm \sqrt{a^2+b^2+c^2}}{2}$$

Since $a^2 + b^2 + c^2 \leq 1$, we have $0 \leq \sqrt{a^2+b^2+c^2} \leq 1$. Therefore, both eigenvalues are real and non-negative.

2. Probability of observing state 0 after applying H gate

The Hadamard gate operation transforms $\rho$ to $H\rho H^\dagger$. Let's calculate this:

$$\begin{aligned} H\rho H^\dagger &= \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1+a & b-ic \\ b+ic & 1-a \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \\ &= \frac{1}{4}\begin{pmatrix} (1+a)+(b+ic)+(b-ic)+(1-a) & (1+a)-(b+ic)+(b-ic)-(1-a) \\ (1+a)+(b+ic)-(b-ic)-(1-a) & (1+a)-(b+ic)-(b-ic)+(1-a) \end{pmatrix} \\ &= \frac{1}{2}\begin{pmatrix} 1+b & a-ic \\ a+ic & 1-b \end{pmatrix} \end{aligned}$$

The probability of observing state 0 is the top-left element of this matrix:

$$p_0 = \frac{1+b}{2}$$

3. Probability of observing state 0 after applying Y gate

Similarly, for the Y gate:

$$\begin{aligned} Y\rho Y^\dagger &= \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 1+a & b-ic \\ b+ic & 1-a \end{pmatrix} \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix} \\ &= \begin{pmatrix} 1-a & -b-ic \\ -b+ic & 1+a \end{pmatrix} \end{aligned}$$

The probability of observing state 0 is:

$$p_0 = \frac{1-a}{2}$$

4. Compute $a'^2 + b'^2 + c'^2$ after applying U gate (by zephyr)

Let $U = \begin{pmatrix} u_{00}' + iu_{00}'' & u_{01}' + iu_{01}'' \\ u_{10}' + iu_{10}'' & u_{11}' + iu_{11}'' \end{pmatrix}$ and $U^\dagger = \begin{pmatrix} u_{00}' - iu_{00}'' & u_{10}' - iu_{10}'' \\ u_{01}' - iu_{01}'' & u_{11}' - iu_{11}'' \end{pmatrix}$

We need to calculate $U\rho U^\dagger$. This is a complex calculation, so let's break it down:

$$U\rho = \begin{pmatrix} (u_{00}'+iu_{00}'')(1+a)+(u_{01}'+iu_{01}'')(b+ic) & (u_{00}'+iu_{00}'')(b-ic)+(u_{01}'+iu_{01}'')(1-a) \\ (u_{10}'+iu_{10}'')(1+a)+(u_{11}'+iu_{11}'')(b+ic) & (u_{10}'+iu_{10}'')(b-ic)+(u_{11}'+iu_{11}'')(1-a) \end{pmatrix}$$

Now, multiplying this by $U^\dagger$ and focusing on the diagonal elements:

$$\rho' = U\rho U^\dagger = \begin{pmatrix} \frac{1+a'}{2} & \frac{b'-ic'}{2} \\ \frac{b'+ic'}{2} & \frac{1-a'}{2} \end{pmatrix}$$

Where:

$$\begin{aligned} a' &= (|u_{00}|^2-|u_{01}|^2)(1+a) + (u_{00}'u_{01}'+u_{00}''u_{01}'')2b \\ &+ (u_{00}'u_{01}''-u_{00}''u_{01}')2c + (|u_{10}|^2-|u_{11}|^2)(1-a) \end{aligned}$$

$$\begin{aligned} b' &= (u_{00}'u_{10}'+u_{00}''u_{10}'')(1+a) + (u_{01}'u_{11}'+u_{01}''u_{11}'')(1-a) \\ &+ (u_{00}'u_{11}'+u_{00}''u_{11}''+u_{01}'u_{10}'+u_{01}''u_{10}'')b - (u_{00}'u_{11}''-u_{00}''u_{11}'-u_{01}'u_{10}''+u_{01}''u_{10}')c \end{aligned}$$

$$\begin{aligned} c' &= (u_{00}'u_{10}''-u_{00}''u_{10}')(1+a) + (u_{01}'u_{11}''-u_{01}''u_{11}')(1-a) \\ &+ (u_{00}'u_{11}''-u_{00}''u_{11}'-u_{01}'u_{10}''+u_{01}''u_{10}')b + (u_{00}'u_{11}'+u_{00}''u_{11}''-u_{01}'u_{10}'-u_{01}''u_{10}'')c \end{aligned}$$

Now, we need to compute $a'^2 + b'^2 + c'^2$. This is a very long and complex calculation. However, we can use a property of unitary matrices: $UU^\dagger = I$. This implies that the transformation $\rho \to U\rho U^\dagger$ preserves the trace and the purity of the density matrix.

The purity of a density matrix is defined as $\text{Tr}(\rho^2) = \frac{1+a^2+b^2+c^2}{2}$.

Since this quantity is preserved under unitary transformations, we have:

$$\frac{1+a^2+b^2+c^2}{2} = \frac{1+a'^2+b'^2+c'^2}{2}$$

Therefore:

$$a'^2 + b'^2 + c'^2 = a^2 + b^2 + c^2$$

This result shows that the sum of squares of the parameters in the density matrix is invariant under unitary transformations.

4. Compute $a'^2 + b'^2 + c'^2$ after applying U gate (by an anonymous contributor)

Note that

$$\det P = \frac{1 - a^2 - b^2 - c^2}{4}$$

$$\det P' = \det(UPU^*) = \det(PU^*U) = \det P = \frac{1 - a^2 - b^2 - c^2}{4}$$

and

$$\det P' = \frac{1 - a'^2 - b'^2 - c'^2}{4}$$

Therefore,

$$a'^2 + b'^2 + c'^2 = a^2 + b^2 + c^2$$

Knowledge

难点思路

第 4 小问的计算过程非常复杂，直接计算会非常繁琐。关键是要认识到酉变换的性质，即它保持密度矩阵的纯度不变。这样可以大大简化计算。

解题技巧和信息

1. 在处理密度矩阵时，要注意其特殊性质：Hermitian（自伴）、半正定、迹为 1。
2. 量子门操作可以表示为 $U\rho U^\dagger$，其中 $U$ 是酉矩阵。
3. 酉变换保持密度矩阵的迹和纯度不变，意味着新态的 $a'^2 + b'^2 + c'^2$ 保持不变。这是解决复杂问题的关键。
4. 在计算复杂的矩阵乘法时，可以先关注最终需要的元素，而不必计算整个矩阵。
5. Hadamard 门 $H$ 将计算基的状态均匀地混合到对角线基。测量概率可以通过变换后的密度矩阵来计算。
6. Pauli-Y 门 $Y$ 交换计算基的状态并引入相位因子。

重点词汇

• density matrix 密度矩阵
• eigenvalue 特征值
• quantum gate 量子门
• Hadamard gate H 门
• unitary transformation 酉变换
• purity 纯度
• trace 迹

参考资料

1. Nielsen, M. A., & Chuang, I. L. (2010). Quantum Computation and Quantum Information: 10th Anniversary Edition. Cambridge University Press. Chapter 2 and 4.
2. Wilde, M. M. (2017). Quantum Information Theory. Cambridge University Press. Chapter 3.