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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

Author

zephyr

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 \(a'^2 + b'^2 + c'^2\).

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

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:

\[ a' = (|u_{00}|^2-|u_{01}|^2)(1+a) + (u_{00}'u_{01}'+u_{00}''u_{01}'')2b + (u_{00}'u_{01}''-u_{00}''u_{01}')2c + (|u_{10}|^2-|u_{11}|^2)(1-a) \]
\[ b' = (u_{00}'u_{10}'+u_{00}''u_{10}'')(1+a) + (u_{01}'u_{11}'+u_{01}''u_{11}'')(1-a) + (u_{00}'u_{11}'+u_{00}''u_{11}''+u_{01}'u_{10}'+u_{01}''u_{10}'')b - (u_{00}'u_{11}''-u_{00}''u_{11}'-u_{01}'u_{10}''+u_{01}''u_{10}')c \]
\[ c' = (u_{00}'u_{10}''-u_{00}''u_{10}')(1+a) + (u_{01}'u_{11}''-u_{01}''u_{11}')(1-a) + (u_{00}'u_{11}''-u_{00}''u_{11}'-u_{01}'u_{10}''+u_{01}''u_{10}')b + (u_{00}'u_{11}'+u_{00}''u_{11}''-u_{01}'u_{10}'-u_{01}''u_{10}'')c \]

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.

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.