量子状态博弈的最优策略

9

考虑以下游戏：

我掷一枚公平的硬币，根据结果（正面或反面），我将为您提供以下状态之一：

| 0 ⟩ or \cos (x) | 0 ⟩ + \sin (x) | 1 ⟩ .

$|0\rangle \text{ or } \cos(x)|0\rangle + \sin(x)|1\rangle.$

这里， $x$ 是已知的恒定角度。但是，我不告诉你我给你的状态。

我如何描述一个测量过程（即正交的量子比特基础），以猜测我处于哪个状态，同时最大化正确的机会？有没有最佳的解决方案？

我一直在研究量子计算，并且遇到了这个练习。我真的不知道该如何开始，我将不胜感激。

我认为一个好的策略是使用

[\begin{matrix} \cos (x) & - \sin (θ) \\ \sin (x) & \cos (θ) \end{matrix}] .

$\begin{bmatrix} \cos(x) & -\sin(\theta)\\ \sin(x) & \cos(\theta) \end{bmatrix}.$

不能取得太大进展...

quantum-state measurement games

— 吉比德
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直观上，答案是在计算基础上进行度量，因为我们可以限制

x

$x$ 至

[0, \frac{π}{2}]

$[0, \frac{\pi}{2}]$ 什么时候

x = 0

$x=0$ 状态是无法区分的，何时

x = \frac{π}{2}

$x=\frac{\pi}{2}$ 状态是正交的，但是我不确定如何证明这一点。

— ahelwer

8

我们只需将量子位测量的二进制结果转换为我们的猜测，即它是第一种状态还是第二种状态，计算出每种可能的量子位测量成功的概率，然后更多地找到两个变量的函数最大值（在两球）。

首先，我们不需要的东西是对状态的精确描述。可以同时依赖于叠加以及经典公平硬币的系统的完整状态可以被编码在密度矩阵中

ρ = \frac{1个}{2} （ \begin{matrix} 1个 & 0 \\ 0 & 0 \end{matrix} ） + \frac{1个}{2} （ \begin{matrix} \cos^{2} X & 罪 X \cos X \\ 罪 X \cos X & 罪^{2} X \end{matrix} ）

$\rho = \frac 12 \pmatrix{1&0\\0&0} + \frac 12 \pmatrix{\cos^2x &\sin x \cos x\\ \sin x\cos x & \sin^2 x}$ 其中左列和上一行对应于基本状态“零”，其余的对应于“一”。以4元素为基础重写密度矩阵很有帮助

2 \times 2

$2\times 2$ 矩阵

ρ = \frac{1个}{2} + \frac{罪 X \cos X}{2} σ_{X} + （ \frac{\cos^{2} X - 罪^{2} X}{4} + \frac{1个}{4} ） σ_{ž}

$\rho = \frac 12+ \frac{\sin x \cos x}{2} \sigma_x + \left(\frac{\cos^2 x - \sin^2 x}{4}+\frac 14\right) \sigma_z$ 那可以用角度来写

2 x

$2x$ ：

ρ = \frac{1个}{2} + \frac{罪 2 X}{4} σ_{X} + \frac{\cos 2 X + 1个}{4} σ_{ž}

$\rho = \frac 12 + \frac {\sin 2x}{4} \sigma_x + \frac{\cos 2x +1}{4} \sigma_z$ 现在，无论混合状态如何，这仍然是一个两级系统，并且二维希尔伯特空间上的所有度量都是微不足道的（

c

$c$ 数）或等效于沿轴的旋转量度，即

V = \vec{ñ} \cdot \vec{σ}

$V = \vec n \cdot \vec \sigma$ 这是一个单位3D向量乘以Pauli矩阵的向量。好吧，如果我们测量

V

$V$ ？的特征值

V

$V$ 是加一或减一。每个的概率可以从的期望值获得

V

$V$ 这是

⟨ V ⟩ = T r (V ρ)

$\langle V \rangle = {\rm Tr} (V \rho)$ 产品的痕迹只有在以下情况下才起作用

1

$1$ 遇见

1

$1$ （但我们假设在

V

$V$ ）要么

σ_{x}

$\sigma_x$ 遇见

σ_{x}

$\sigma_x$ 等等，在这种情况下，矩阵的迹线会产生2的额外因子。

⟨ V ⟩ = \frac{\sin 2 x}{2} n_{x} + \frac{\cos 2 x + 1}{2} n_{z}

$\langle V \rangle = \frac{\sin 2x}{2}n_x + \frac{\cos 2x +1}{2} n_z$ 我们得到特征值

\pm 1

$\pm 1$ 与概率

(1 \pm ⟨ V ⟩) / 2

$(1\pm\langle V \rangle) / 2$ ，分别。恰好在

\cos x = 0

$\cos x = 0$ ，则两个初始的“头尾”状态彼此正交（基本上

| 0 ⟩

$|0\rangle$ 和

| 1 ⟩

$|1\rangle$ ），我们可能会完全区分它们。做出概率

0, 1

$0,1$ ，我们必须选择

\vec{n} = (0, 0, \pm 1)

$\vec n=(0,0,\pm 1)$ ; 请注意，

\vec{n}

$\vec n$ 程序无关紧要。

现在，对于 $\cos x \neq 0$ ，状态在量子意义上是非正交的，即“不互斥”，我们无法直接测量硬币是正面还是反面，因为这些可能性混合在密度矩阵中。实际上，密度矩阵包含所有测量的所有概率，因此，如果通过抛硬币的可能状态的不同混合来获得相同的密度矩阵，则qubit的状态将完全无法区分。

如果成功，我们的成功概率将低于100％ $\cos x\neq 0$ 。但是使用经典位的唯一有意义的方法 $V=\pm 1$ 从测量中可以直接将其转化为我们对初始状态的猜测。不失一般性，我们的翻译可以选择为

(V = + 1) \to | i ⟩ = | 0 ⟩

$(V = +1) \to |i\rangle = |0\rangle \\$ 和

(V = - 1) \to | i ⟩ = \cos x | 0 ⟩ + \sin x | 1 ⟩ .

$(V = -1) \to |i\rangle = \cos x |0\rangle + \sin x |1 \rangle.$ 如果我们想要相反的头尾交叉识别和

V

$V$ ，我们可以简单地通过翻转

\vec{n} \to - \vec{n}

$\vec n \to -\vec n$ 。

我们将第一个简单的初始状态称为“ heads”（零），将第二个较简单的初始状态称为“ tails”（余弦-正弦叠加）。鉴于我们的翻译是 $+1$ 头和 $-1$ 到尾巴，

P_{s u c c e s s} = P (H) P (+ 1 | H) + P (T) P (- 1 | T) .

$P_{\rm success} = P(H) P(+1|H) + P(T) P(-1|T).$ 因为这是一个公平的硬币，所以上面包括的两个因素是

P (H) = P (T) = 1 / 2

$P(H)=P(T)=1/2$ 。四个概率中最困难的计算是

P (- 1 | T)

$P(-1|T)$ 。但是上面我们已经进行了更困难的计算，这是

(1 - ⟨ V ⟩) / 2

$(1-\langle V\rangle) / 2$ 。在这里，我们只是忽略与

n_{z}

$n_z$ 并乘以二：

P (- 1 | T) = \frac{1}{2} - \sin 2 x \frac{n_{x}}{2} - \cos 2 x \frac{n_{z}}{2}

$P(-1|T ) = \frac 12 - \sin 2x \frac{ n_x}2 - \cos 2x \frac{ n_z}2$ The result for "heads" is simply obtained by setting

x = 0

$x=0$ because the "heads" state equals "tails" states with

x = 0

$x=0$ substituted. So

P (- 1 | H) = \frac{1 - n_{z}}{2}

$P(-1|H) = \frac{1-n_z}{2}$ and the complementary

1 - P

$1-P$ probability is

P (+ 1 | H) = \frac{1 + n_{z}}{2}

$P(+1|H) = \frac{1+n_z}{2}$ Substitute those results to our "success probability" to get

P_{s u c c e s s} = \frac{1 + n_{z} + 1 - (\sin 2 x) n_{x} - (\cos 2 x) n_{z}}{4}

$P_{\rm success} = \frac{1+n_z +1 - (\sin 2x)n_x - (\cos 2x)n_z}{4}$ or

P_{s u c c e s s} = \frac{1}{2} - \frac{n_{x}}{4} \sin 2 x + \frac{n_{z}}{4} (1 - \cos 2 x)

$P_{\rm success} = \frac 12 - \frac{n_x}4 \sin 2x + \frac{n_z}{4} (1-\cos 2x )$ If we define

(n_{x}, n_{z}) = (- \cos α, - \sin α)

$(n_x,n_z)=(-\cos \alpha,-\sin\alpha)$ , we may also write it as

P_{s u c c e s s} = \frac{1}{2} + \frac{\sin (2 x + α) - \sin α}{4} = \frac{1}{2} + \frac{\sin x \cos (x + α)}{2}

$P_{\rm success} = \frac 12 +\frac{\sin(2x+\alpha)-\sin \alpha}{4} =\frac 12+\frac{\sin x \cos(x+\alpha)}{2}$ We want to maximize that over

α

$\alpha$ . Clearly, the maximum is for

\cos (x + α) = \pm 1

$\cos(x+\alpha)=\pm 1$ where the sign agrees with that of

\sin x

$\sin x$ i.e.

α = - x

$\alpha=-x$ or

α = π - x

$\alpha=\pi -x$ and the value at this maximum is

P_{s u c c e s s} = \frac{1 + | \sin x |}{2}

$P_{\rm success} = \frac{1+|\sin x|}{2}$ which sits in the interval 50% and 100%.

That's a nice measurement which is really quantum mechanical. We use a different measurement than that of $\sigma_z$ , i.e. the classical measurement of the bit. Instead, we measure the spin along the axis in the $xz$ -plane that is defined by the same nonzero angle as the angle $x$ at the beginning, with some correct signs and shifts by multiples of $\pi/2$ . Note that if you measured simply $\sigma_z$ , the classical bit, the success rate would be just $(3-\cos 2x)/4$ , also between 50% and 100%, but smaller than our result. In particular, for a small $x=0+\epsilon$ , our optimal result would be Taylor-expanded as $1/2+|x|/2$ while the non-optimum result using the classical measurement would increase above $1/2$ more slowly, as $1/2+x^2/2$ .

For many hours, a wrong answer (a mistake in the final portions) was posted here, despite the fact that I had previously fixed many wrong factors of two. I posted a slightly edited version of this answer on my weblog where some discussion may take place:

The Reference Frame: A fun simple problem in quantum computing

On that page, I also write the eigenstates of the measured operator in the appendix. The arguments in the angles may be surprising for some folks who think that this problem is obvious in terms of the wave functions or that the wave functions after the measurement have to be simple.

— Luboš Motl
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I might need to read the question and answer more carefully, but isn't this a special case of the problem solved in arxiv.org/abs/1805.03477?

— glS

Maybe, I am not familiar with the paper and I can't see it is the generalization of this problem, at least not within minutes. But I am not claiming to have solved any cutting-edge paper-style problem. This question is probably an exercise in some textbooks which is expected to be solved by students.

— Luboš Motl

1

The key is in the optimal strategy for distinguishing two non-orthogonal states. This is something called the Helstrom measurement, which I described here.

— DaftWullie
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