一个量子比特的测量如何影响其他量子比特？

为了表示量子计算机的状态，所有量子位都贡献一个状态向量（据我所知，这是量子计算与经典计算之间的主要区别之一）。我的理解是，可以在一个多量子位系统中仅测量一个量子位。测量一个量子比特如何影响整个系统（具体来说，它如何影响状态向量）？

查看量子位的方法有很多，状态向量形式主义只是其中之一。在一般的线性代数意义上，度量是投影到基础上的。在这里，我将从Pauli可观察的角度提供一个示例，这是QC的常用电路模型。

首先，感兴趣的是提供状态向量的基础-每个测量运算符都带有一组本征态，以及您要查看的任何测量值（例如$X,Y,Z, XX, XZ$等）。确定最适合写状态向量的基础。回答问题的最简单方法是，您是否知道哪个基础对您感兴趣，更重要的是，该基础是否与您刚刚进行的测量相减。

因此，为简单起见，假设您以两个$Z$位为基础，以Z为基础，以处于任意状态的两个耦合的qubit开始：

$$| \psi \rangle = a | 0_{Z} \rangle \otimes | 0_{Z} \rangle +b | 0_{Z} \rangle \otimes | 1_{Z} \rangle + c | 1_{Z} \rangle \otimes | 0_{Z} \rangle + d | 1_{Z} \rangle \otimes | 1_{Z} \rangle$$

您可能做的最简单的测量就是$Z_{1}$，即第一个量子位上的$Z$运算符，然后是$Z_{2}$，第二个量子位上的$Z$运算符。测量做什么？它将状态投影到本征状态之一。您可以将其视为消除了所有可能的答案，这些答案与我们刚刚测得的答案不一致。例如，假设我们测量$Z_{1}$并获得结果$1$，那么我们将得到的结果状态将是：

$$| \psi \rangle = \frac{1}{\sqrt{|c|^{2} +|d|^{2}}} \left(c | 1_{Z} \rangle \otimes | 0_{Z} \rangle + d | 1_{Z} \rangle \otimes | 1_{Z} \rangle \right)$$

请注意，前面的系数仅用于重新规格化。因此，我们测量概率$Z_{2}=0$为。请注意，这不同于我们在初始状态下的概率| 一个| 2+| c| 2。$\frac{1}{|c|^{2} +|d|^{2}} |c^{2}|$$|a|^{2}+|c|^{2}$

但是，假设您进行的下一次测量与上一次测量不同步。这比较棘手，因为您必须根据状态向量实现基础的更改才能理解概率。但是，通过Pauli测量，由于本征基以一种很好的方式关联，所以它往往很容易，即：

$$| 0_{Z} \rangle = \frac{1}{\sqrt{2}} (|0_{X}\rangle + |1_{X} \rangle )$$

$$| 1_{Z} \rangle = \frac{1}{\sqrt{2}} (|0_{X}\rangle - |1_{X} \rangle )$$

A good way to check your understanding: What is the probability of measuring $X= +1$ after the $Z_{1}=1$ measurement above? What is the probability if we have not made the $Z_{1}$ measurement? Then a more complicated question is to look at product operators that act on both qubits at once, for instance, how does a measurement of $Z_{1}Z_{2}=+1$ affect the initial state? Here $Z_{1}Z_{2}$ measures the product of the two operators.

Suppose that, prior to measurement, your $n$-qubit system is in some state $\lvert \psi \rangle \in \mathcal H_2^{\otimes n}$, where $\mathcal H_2 \cong \mathbb C^2$ is the Hilbert space of a single qubit. Write

$$\lvert \psi \rangle = \sum_{x \in \{0,1\}^n} u_x \lvert x \rangle$$ for some coefficients $u_x \in \mathbb C$ such that $\sum_x \lvert u_x \rvert^2 = 1$.

• $$\begin{aligned} \lvert \varphi_0 \rangle &= \!\!\!\!\!\sum_{x' \in \{0,1\}^{n-1}}\!\!\!\!\!\! u_{0x'} \,\lvert0\rangle \lvert x' \rangle, \\ \lvert \varphi_1 \rangle &= \!\!\!\!\!\sum_{x' \in \{0,1\}^{n-1}}\!\!\!\!\!\! u_{1x'} \,\lvert1\rangle \lvert x' \rangle,\end{aligned}$$ and let $\lvert \psi_0 \rangle = \lvert \varphi_0 \rangle \big/\! \sqrt{\langle \varphi_0 \vert \varphi_0 \rangle}\,$ and $\,\lvert \psi_1 \rangle = \lvert \varphi_1 \rangle \big/\! \sqrt{\langle \varphi_1 \vert \varphi_1 \rangle}\,$. It is not too difficult to show that, if you measure the first qubit and obtain the state $\lvert 0 \rangle$, the state of the entire system "collapses" to $\lvert \psi_0 \rangle$, and if you obtain $\lvert 1 \rangle$ what you obtain is $\lvert \psi_1 \rangle$.

  This is broadly analogous to the idea of conditional probability distributions: you might think of $\lvert \psi_0 \rangle$ as the state of the system conditioned on the first qubit being $\lvert 0 \rangle$, and $\lvert \psi_1 \rangle$ as the state of the system conditioned on the first qubit being $\lvert 1 \rangle$ (except of course that the story is a bit more complicated, on account of the fact that the first qubit is not "secretly" in either the state $0$ or $1$).

• The above is not strongly dependent on measuring the first qubit: we can define $\lvert \varphi_0 \rangle$ and $\lvert \varphi_1 \rangle$ in terms of fixing any particular bit in the bit string $x$ to either $0$ or $1$, summing over only those components which are consistent with either the choice $0$ or $1$, and proceeding as above.

• The above is also not strongly dependent on measuring in the standard basis, as Emily indicates. If we wish to consider measuring the first qubit in the basis $\lvert \alpha \rangle, \lvert \beta \rangle$, where $\lvert \alpha \rangle = \alpha_0 \lvert 0 \rangle + \alpha_1 \lvert 1 \rangle$ and $\lvert \beta \rangle = \beta_0 \lvert 0 \rangle + \beta_1 \lvert 1 \rangle$, we define

  $$\begin{aligned} \lvert \varphi_0 \rangle &= \Bigl(\lvert \alpha \rangle\!\langle \alpha \lvert \otimes I^{\otimes n-1}\Bigr)\lvert \psi\rangle = \!\!\!\!\!\sum_{x' \in \{0,1\}^{n-1}}\!\!\!\!\!\! \bigl(\alpha_0^\ast u_{0x'} + \alpha_1^\ast u_{1x'}\bigr) \,\lvert\alpha\rangle \lvert x' \rangle\,, \\ \lvert \varphi_1 \rangle &= \Bigl(\lvert \beta\rangle\!\langle \beta \lvert \otimes I^{\otimes n-1}\Bigr)\lvert \psi\rangle = \!\!\!\!\!\sum_{x' \in \{0,1\}^{n-1}}\!\!\!\!\!\! \bigl(\beta_0^\ast u_{0x'} + \beta_1^\ast u_{1x'}\bigr) \,\lvert\beta\rangle \lvert x' \rangle\,, \end{aligned}$$ and then proceeding as above.

Less formally-stated than the other answers, but for beginners I like the intuitive method outlined by Prof. Vazirani in this video.

Suppose you have a general two-qbit state:

$|\psi\rangle = \begin{bmatrix} \alpha_{00} \\ \alpha_{01} \\ \alpha_{10} \\ \alpha_{11} \end{bmatrix} = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle$

Now suppose you measure the most-significant (leftmost) qbit in the computational basis (as in, collapse it to either $|0\rangle$ or $|1\rangle$). There are two questions we might ask:

1. What is the probability that the measured qbit collapses to $|0\rangle$? What about $|1\rangle$?
2. What is the state of the 2-qbit system after measurement?

For the first question, the intuitive answer is this: take the sum of squares of all amplitudes associated with the value for which you want to find the probability of collapse. So, if you want to know the probability of the measured qbit collapsing to $|0\rangle$, you'd look at the amplitudes associated with cases $|00\rangle$ and $|01\rangle$, because those are the cases where the measured qbit is $|0\rangle$. Thus:

$P[|0\rangle] = |\alpha_{00}|^2 + |\alpha_{01}|^2$

Similarly, for $|1\rangle$ you look at the amplitudes associated with cases $|10\rangle$ and $|11\rangle$, so:

$P[|1\rangle] = |\alpha_{10}|^2 + |\alpha_{11}|^2$

As for the state of the 2-qbit system after measurement, what you do is cross out all the components of the superposition which are inconsistent with the answer you got. So, if you measured $|0\rangle$, then the state after measurement is:

$\require{cancel} |\psi\rangle = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \cancel{\alpha_{10}|10\rangle} + \cancel{\alpha_{11}|11\rangle} = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle$

However, this state is not normalized - the sum of squares does not add up to 1, and so you have to normalize it:

$|\psi\rangle = \frac{\alpha_{00}|00\rangle + \alpha_{01}|01\rangle}{\sqrt{|\alpha_{00}|^2 + |\alpha_{01}|^2}}$

Similarly, if you measured $|1\rangle$ then you'd get:

$\require{cancel} |\psi\rangle = \cancel{\alpha_{00}|00\rangle} + \cancel{\alpha_{01}|01\rangle} + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle = \alpha_{10}|10\rangle + \alpha_{11}|11\rangle$

Normalized:

$|\psi\rangle = \frac{\alpha_{10}|10\rangle + \alpha_{11}|11\rangle}{\sqrt{|\alpha_{10}|^2 + |\alpha_{11}|^2}}$

And that's how you calculate the action of measuring one qbit in a multi-qbit state, in the simplest case!