如果所有量子门必须是单一的，那么测量呢？

所有量子操作必须统一才能允许可逆性，但是测量又如何呢？可以将测量表示为一个矩阵，并将该矩阵应用于量子位，因此这似乎等效于量子门的操作。这绝对是不可逆的。有没有可能允许非单一门的情况？

ary运算只是量子运算的一种特殊情况，量子运算是将密度算符映射到密度算符的线性，完全正图（“通道”）。这在信道的克劳斯的表示变得明显，其中所谓的克劳斯运营ķ 我满足Σ Ñ 我= 1 ķ † 我 ķ 我 ≤ 我（符号

但是，量子门是单一的，因为它们是通过哈密顿量在特定时间内完成的，这根据Schrödinger方程给出了单一的时间演化。

简短答案

量子操作并不需要是单一的。实际上，许多量子算法和协议都利用非单一性。

长答案

测量，可以说是非统一转变为算法的基本组成部分（在这个意义上，一个“测量”是等同于从退相干操作后的概率分布采样的最明显的例子）。$\sum_k c_k\lvert k\rangle\mapsto\sum_k |c_k|^2\lvert k\rangle\langle k\rvert$

More generally, any quantum algorithm that involves probabilistic steps requires non-unitary operations. A notable example that comes to mind is HHL09's algorithm to solve linear systems of equations (see 0811.3171). A crucial step in this algorithm is the mapping $|\lambda_j\rangle\mapsto C\lambda_j^{-1}|\lambda_j\rangle$, where $|\lambda_j\rangle$ are eigenvectors of some operator. This mapping is necessarily probabilistic and therefore non-unitary.

任何利用（经典）前馈的算法或协议也都在利用非单一运算。这就是单向量子计算协议的全部（顾名思义，它们要求不可逆的操作）。

使用单个光子进行光学量子计算的最著名方案还需要进行测量，有时还需要进行后选择，以纠缠不同光子的状态。例如，KLM协议会产生概率门，因此概率门至少部分不可逆。关于该主题的一个不错的评论是quant-ph / 0512071。

耗散诱导的量子态工程（例如1402.0529或srep10656）提供了不太直观的示例。在这些协议中，人们使用开放式地图耗散动态，并设计状态与环境的交互关系，以使系统的长期静态状态成为理想状态。

冒着脱离量子计算进入物理学的风险，我将回答我认为与该主题相关的子问题，并用它来指导量子计算中gate门的讨论。

The question here is: Why do we want unitarity in quantum gates?

The less specific answer is as above, it gives us 'reversibility', or as physicists often talk about it, a type of symmetry for the system. I'm taking a course in quantum mechanics right now, and the way unitary gates cropped up in that course was motivated by the desire to have physical transformations $\hat{U}$: that act as symmetries. This imposed two conditions on the transformation $\hat{U}$:

1. The transformations should act linearly on the state (this is what gives us a matrix representation).
2. The transformations should preserve probability, or more specifically inner product. This means that if we define:

的内积来保存该。从第二个规范中可以得出统一性（有关完整详细信息，请参见van Raamsdonk博士的注释，此处$\langle \phi | | \psi \rangle= \langle \phi' | | \psi'\rangle$）。

因此，这回答了为什么保持事物“可逆”的操作必须统一的问题。

为什么测量本身不统一的问题更多地与量子计算有关。度量是对基准的投影；本质上，它必须以一个或多个基本状态作为状态本身来“回答”。它还以与度量的“答案”一致的方式离开状态，而不与状态开始时的潜在概率一致。因此，该运算满足变换规格1. ，但绝对不满足规格2。并非所有矩阵都相等！$U$

为了使事情回到量子计算上，测量是破坏性的和射影的（即，我们只能通过重复测量相同状态来重建叠加，而每个测量只能给出0/1的答案），这是使得量子计算和常规计算之间的区分非常微妙（这也是很难确定这一点的部分原因）。人们可能会认为，由于希尔伯特空间的大小，量子计算的功能更加强大，所有这些状态叠加都对我们可用。但是我们提取这些信息的能力受到很大限制。

As far as I understand it this shows that for information storage purposes, a qubit is only as good as a regular bit, and no better. But we can be clever in quantum computation with the way that information is traded around, because of the underlying linear-algebraic structure.

There are several misconceptions here, most of them originate from exposure to only the pure state formalism of quantum mechanics, so let's address them one by one:

1. All quantum operations must be unitary to allow reversibility, but what about measurement?

This is false. In general, the states of a quantum system are not just vectors in a Hilbert space $\mathcal{H}$ but density matrices $-$ unit-trace, positive semidefinite operators acting on the Hilbert space $\mathcal{H}$ i.e., $\rho: \mathcal{H} \rightarrow \mathcal{H}$, $Tr(\rho) = 1$, and $\rho \geq 0$ (Note that the pure state vectors are not vectors in the Hilbert space but rays in a complex projective space; for a qubit this amounts to the Hilbert space being $\mathbb{C}P^1$ and not $\mathbb{C}^2$). Density matrices are used to describe a statistical ensemble of quantum states.

The density matrix is called pure if $\rho^2 = \rho$ and mixed if $\rho^2 < \rho$. Once we are dealing with a pure state density matrix (that is, there's no statistical uncertainty involved), since $\rho^2 = \rho$, the density matrix is actually a projection operator and one can find a $|\psi\rangle \in \mathcal{H}$ such that $\rho = |\psi\rangle \langle\psi|$.

The most general quantum operation is a CP-map (completely positive map), i.e., $\Phi: L(\mathcal{H}) \rightarrow L(\mathcal{H})$ such that

Now, coming to the OP's claim that all quantum operations are unitary to allow reversibility -- this is just not true. The unitarity of time evolution operator ($e^{-iHt/\hbar}$) in quantum mechanics (for closed system quantum evolution) is simply a consequence of the Schrödinger equation.

However, when we consider density matrices, the most general evolution is a CP-map (or CPTP for a closed system to preserve the trace and hence the probability).

2. Are there any situations where non-unitary gates might be allowed?

Yes. An important example that comes to mind is open quantum systems where Kraus operators (which are not unitary) are the "gates" with which the system evolves.

Note that if there is only a single Kraus operator then, $\sum_i K_i^\dagger K_i = \mathbb{I}$. But there's only one $i$, therefore, we have, $K^\dagger K = \mathbb{I}$ or, $K$ is unitary. So the system evolves as $\rho \rightarrow U \rho U^\dagger$ (which is the standard evolution that you may have seen before). However, in general, there are several Kraus operators and therefore the evolution is non-unitary.

Coming to the final point:

3. Measurement can be represented as a matrix, and that matrix is applied to qubits, so that seems equivalent to the operation of a quantum gate. That's definitively not reversible.

In standard quantum mechanics (with wavefunctions etc.), the system's evolution is composed of two parts $-$ a smooth unitary evolution under the system's Hamiltonian and then a sudden quantum jump when a measurement is made $-$ also known as wavefunction collapse. Wavefunction collapses are described as some projection operator say $|\phi\rangle \langle\phi|$ acting on the quantum state $|\psi\rangle$ and the $|\langle\phi|\psi\rangle|^2$ gives us the probability of finding the system in the state $|\phi\rangle$ after the measurement. Since the measurement operator is after all a projector (or as the OP suggests, a matrix), shouldn't it be linear and physically similar to the unitary evolution (also happening via a matrix). This is an interesting question and in my opinion, difficult to answer physically. However, I can shed some light on this mathematically.

If we are working in the modern formalism, then measurements are given by POVM elements; Hermitian positive semidefinite operators, $\{M_{i}\}$ on a Hilbert space $\mathcal{H}$ that sum to the identity operator (on the Hilbert space) $\sum _{{i=1}}^{n}M_{i}=\mathbb{I}$. Therefore, a measurement takes the form

The $\text{Tr}(E_i \rho E_i^\dagger) =: p_i$ is the probability of the measurement outcome being $M_i$ and is used to renormalize the state to unit trace. Note that the numerator, $\rho \rightarrow E_i \rho E_i^\dagger$ is a linear operation, but the probabilistic dependence on $p_i$ is what brings in the non-linearity or irreversibility.

Edit 1: You might also be interested Stinespring dilation theorem which gives you an isomorphism between a CPTP map and a unitary operation on a larger Hilbert space followed by partial tracing the (tensored) Hilbert space (see 1, 2).

I'll add a small bit complementing the other answers, just about the idea of measurement.

Measurement is usually taken as a postulate of quantum mechanics. There's usually some preceding postulates about hilbert spaces, but following that

• Every measurable physical quantity $A$ is described by an operator $\hat{A}$ acting on a Hilbert space $\mathcal{H}$. This operator is called an observable, and it's eigenvalues are the possibly outcomes of a measurement.
• If a measurement is made of the observable $A$, in the state of the system $\psi$, and the outcome is $a_n$, then the state of the system immediately after measurement is $$\frac{\hat{P}_n|\psi\rangle}{\|\hat{P}_n|\psi\rangle\|},$$ where $\hat{P}_n$ is the projector onto the eigen-subspace of the eigenvalue $a_n$.

Normally the projection operators themselves should satisfy $\hat{P}^\dagger=\hat{P}$ and $\hat{P}^2=\hat{P}$, which means they themselves are observables by the above postulates, and their eigenvalues $1$ or $0$. Supposing we take one of the $\hat{P}_n$ above, we can interpret the $1,0$ eigenvalues as a binary yes/no answer to whether the observable quantity $a_n$ is available as an outcome of measurement of the state $|\psi\rangle$.

Measurements are unitary operations, too, you just don't see it: A measurement is equivalent to some complicated (quantum) operation that acts not just on the system but also on its environment. If one were to model everything as a quantum system (including the environment), one would have unitary operations all the way.

However, usually there is little point in this because we usually don't know the exact action on the environment and typically don't care. If we consider only the system, then the result is the well-known collapse of the wave function, which is indeed a non-unitary operation.

Quantum states can change in two ways: 1. quantumly, 2. classically.

1. All the state changes taking place quantumly, are unitary. All the quantum gates, quantum errors, etc., are quantum changes.

2. There is no obligation on classical changes to be unitary, e.g. measurement is a classical change.

All the more reason, why it is said that the quantum state is 'disturbed' once it's measured.