叠加状态和混合状态有什么区别？

到目前为止，我的理解是：纯状态是系统的基本状态，混合状态表示系统的不确定性，即系统处于一组具有某种（经典）概率的状态中。但是，叠加似乎也是状态的一种混合，那么它们如何适合这种情况呢？

例如，考虑一个公平的硬币翻转。你可以把它表示的“头”的混合状态$\left|0\right>$和“尾” $\left|1\right>$：

$$\rho_1 = \sum_j \frac{1}{2} \left|\psi_j\right> \left<\psi_j\right| = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

但是，我们也可以使用“头”和“尾”的叠加：特定状态与密度$\psi = \frac{1}{\sqrt{2}}\left( \left|0\right> + \left|1\right> \right)$

$$\rho_2 = \left|\psi\right> \left<\psi\right| = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$

如果我们以计算为基础进行测量，我们将得到相同的结果。叠加状态和混合状态有什么区别？

没有，一个叠加两个不同的状态的是一个完全不同的野兽比混合物相同的状态。虽然它可能从你的例子看来，和 ρ 2产生相同的测量结果（这的确是这样），只要你在不同的基础上，他们会给可测量不同的结果衡量。$\rho_1$$\rho_2$

A“叠加”像是纯态。这意味着它是完全特征化的状态。换句话说，没有任何信息会增加其描述的“不确定性”。注意，每个纯状态都可以写为其他纯状态的叠加。写一个给定的状态 | 从某种意义上说，⟩⟩与其他状态的叠加实际上与写向量 v是一样的：您始终可以更改基础并找到 v的不同表示形式。$\newcommand{\up}{|\!\!\uparrow\rangle}\newcommand{\down}{|\!\!\downarrow\rangle}|\psi\rangle=\frac{1}{\sqrt2}(\up+\down)$$|\psi\rangle$$\boldsymbol v$$\boldsymbol v$

这是直接的对比一个混合的状态像你的问题。在的情况下，ρ 1，成果的概率性质取决于我们对国家本身的无知。这意味着，在原则上，有可能获得一些额外的信息，这将告诉我们是否ρ 2确实是在状态$\rho_1$$\rho_1$$\rho_2$或处于状态$\up$。$\down$

通常，混合状态不能写为纯状态。从上面的物理直觉中应该可以清楚地看出：混合状态代表我们对物理状态的无知，而纯状态是完全定义的状态，由于量子力学的工作方式，碰巧仍会产生概率结果。

确实，有一个简单的准则可以判断给定的（通常是混合的）状态可以写成| |。ψ ⟩ ⟨ ψ | 对于某些（纯）状态| ψ ⟩：计算其纯度。状态ρ的纯度定义为$\rho$$|\psi\rangle\langle\psi|$$|\psi\rangle$$\rho$，并且它是一个标准的结果状态的纯度为 1当且仅当该状态是纯的（和更小比 1以其他方式）。$\operatorname{Tr} \,(\rho^2)$$1$ $1$

简短的答案是，量子信息要比“不确定性”更多。这是因为测量状态的方法不止一种。而这是因为在其中，在原则上，可以存储和检索信息的一个以上的基础。叠加使您可以用不同于计算基础的基础来表达信息，但是无论您使用哪种基础查看状态，混合都可以描述概率元素的存在。

更长的答案如下：

如您所描述的，度量是在计算基础上的度量。为了简洁起见，这通常被称为“度量”，而社区的大部分子集认为这是度量事物的主要方式。但是在许多物理系统中，可以选择测量基础。

上的向量空间有一个以上的基础（甚至超过一个正交的基础），并且在数学层面上，除了让数学家考虑方便的地方以外，没有什么比另一个更特殊的基础了。在量子力学中也是如此：除非您指定某些特定的动力学，否则没有比其他动力学更特别的基础。这意味着计算基础 | 0 ⟩ = [ $\mathbb C$ 是没有从根本上不同物理另一个基础如 | + ⟩ =

$$\lvert 0 \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad \lvert 1 \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$ 这也是一个标准正交基。这意味着应该有一种“测量”状态的方法。ψ⟩＆Element;C ^2以这样的方式，该结果的概率依赖于突起到这些状态| +⟩和| -⟩。 $$\lvert + \rangle = \tfrac{1}{\sqrt 2}\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \qquad \lvert - \rangle = \tfrac{1}{\sqrt 2}\begin{bmatrix} 1 \\ -1 \end{bmatrix},$$ $\lvert \psi \rangle \in \mathbb C^2$$\lvert + \rangle$$\lvert - \rangle$

$$\Pi_+ = \lvert + \rangle\!\langle + \rvert = \tfrac{1}{2}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, \qquad \Pi_- = \lvert - \rangle\!\langle - \rvert = \tfrac{1}{2}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$ $\lvert \varphi_+ \rangle := \Pi_+ \lvert \psi \rangle$ and $\lvert \varphi_- \rangle := \Pi_- \lvert \psi \rangle$. The norm-squared of $\lvert \varphi_\pm \rangle$ determines the probability of "measuring $\lvert + \rangle$" and of "measuring $\lvert - \rangle$"; and normalising $\lvert \varphi_+ \rangle$ or $\lvert \varphi_- \rangle$ to have a norm of 1 yields the post-measurement state. (For a state on a single qubit, this will just be $\lvert + \rangle$ or $\lvert - \rangle$. More interesting post-measurement states may result if we consider multi-qubit states, and consider the projector $\Pi_+$ or $\Pi_-$ acting on one of many qubits.)

For density operators, one takes the state $\rho$ which you want to perform a measurement on, and consider $\rho_+ := \Pi_+ \rho \Pi_+$ and $\rho_- := \Pi_- \rho \Pi_-$. These operators may be sub-normalised in the same way that the states $\lvert \varphi_\pm \rangle$ might be, in the sense that they may have trace less than 1. The value of the trace of $\rho_\pm$ is the probability of obtaining the outcome $\lvert + \rangle$ or $\lvert - \rangle$ of the measurement; to renormalise, simply scale the projected operator to have trace 1.

Consider your state $\rho_2$ above. If you measure it with respect to the $\lvert \pm \rangle$ basis, what you will find is that $\rho_2 = \rho_{2,+} := \Pi_+ \rho_2 \Pi_+$. This means that projecting the operator with $\Pi_+$ does change the state, and that the probability of obtaining the outcome $\lvert + \rangle$ to the measurement is 1. If you do this instead with $\rho_1$, you will find a 50/50 chance of obtaining either $\lvert + \rangle$ or $\lvert - \rangle$. So the state $\rho_1$ is a mixed state, while $\rho_2$ is not --- the difference being that $\rho_2$ has a definite outcome in a different measurement basis than the standard basis. You might say that $\rho_2$ stores a definite piece of information, albeit in a different basis than the computational basis.

More generally, a mixed state is one whose largest eigenvalue is less than 1, meaning that there is no basis in which you can measure it to get a definite outcome. Superpositions allow you to express information in a different basis than the computational basis; mixtures represent a degree of randomness about the state of the system you're considering, regardless of how you measure that system.

Along with glS' post:

A mixed state would be if you had a can of paint, but you weren't sure if it was blue or yellow. You know it is either one of the two, and once you pop the top and measure it, you'd know, but until you do it is in one of those two pure states. If you picked it up from a stack of cans where you knew there were equally many cans of blue paint as yellow, you would expect an equal chance of it being one or the other. 50% of the time it would be 100% yellow and 50% of the time it would be 100% blue.

A superposition is more like if you take half a can of blue and half a can of yellow and pour them together. You've now constructed a new pure state that is expressible as a combination of other pure states. If you test its 'blueness', it is about 50%. If you test its 'yellowness' it is about 50%. It is both yellow and blue at the same time. 100% of the time it is both 50% blue and 50% yellow.

If you measured the amount of blue and yellow in one stack of blue or yellow cans and then in another stack of green, you might be confused to see you have just as much blue and yellow in both stacks, but the difference is that the 'blueness' and 'yellowness' is in a mixed state in the later stack but is in a superposition in the latter.