更改

想象一下，这两个纠缠的量子位被分开，每个给爱丽丝和鲍勃一个。爱丽丝对她的量子位进行测量，以相等的概率获得ーー要么是 < math > | 0 rangle </math > ，要么是 < math > | 1 rangle </math > ，也就是说，她现在可以判断她的量子位是值“0”还是“1”。由于量子比特的纠缠，鲍勃现在必须得到与爱丽丝完全相同的测量结果。例如，如果她测量一个 < math > | 0 rangle </math > ，那么 Bob 必须测量相同的值，因为 < math > | 00 rangle </math > 是 Alice 的量子位唯一的状态，其中的量子位是 < math > | 0 rangle </math > 。简而言之，对于这两个纠缠的量子比特，不管 Alice 测量什么，Bob 也会测量，不管它们之间的距离有多远，在任何基础上都有完美的相关性，即使它们都无法判断它们的量子比特值是“0”还是“1”——这是经典物理学无法解释的最令人惊讶的情况。

想象一下，这两个纠缠的量子位被分开，每个给爱丽丝和鲍勃一个。爱丽丝对她的量子位进行测量，以相等的概率获得ーー要么是 < math > | 0 rangle </math > ，要么是 < math > | 1 rangle </math > ，也就是说，她现在可以判断她的量子位是值“0”还是“1”。由于量子比特的纠缠，鲍勃现在必须得到与爱丽丝完全相同的测量结果。例如，如果她测量一个 < math > | 0 rangle </math > ，那么 Bob 必须测量相同的值，因为 < math > | 00 rangle </math > 是 Alice 的量子位唯一的状态，其中的量子位是 < math > | 0 rangle </math > 。简而言之，对于这两个纠缠的量子比特，不管 Alice 测量什么，Bob 也会测量，不管它们之间的距离有多远，在任何基础上都有完美的相关性，即使它们都无法判断它们的量子比特值是“0”还是“1”——这是经典物理学无法解释的最令人惊讶的情况。

==Quantum entanglement==

==Quantum entanglement量子纠缠==

An important distinguishing feature between qubits and classical bits is that multiple qubits can exhibit [[quantum entanglement]]. Quantum entanglement is a [[quantum nonlocality|nonlocal]] property of two or more qubits that allows a set of qubits to express higher correlation than is possible in classical systems.

An important distinguishing feature between qubits and classical bits is that multiple qubits can exhibit [[quantum entanglement]]. Quantum entanglement is a [[quantum nonlocality|nonlocal]] property of two or more qubits that allows a set of qubits to express higher correlation than is possible in classical systems.

The simplest system to display quantum entanglement is the system of two qubits.  Consider, for example, two entangled qubits in the <math>|\Phi^+\rangle</math> [[Bell state]]:

显示量子纠缠的最简单的系统是两个量子比特的系统。例如，考虑<math>|\Phi^+\rangle</math>[[贝尔态]]中的两个纠缠量子比特：

Controlled gates act on 2 or more qubits, where one or more qubits act as a control for some specified operation. In particular, the controlled NOT gate (or CNOT or cX) acts on 2 qubits, and performs the NOT operation on the second qubit only when the first qubit is <math>|1\rangle</math>, and otherwise leaves it unchanged. With respect to the unentangled product basis <math>\{|00\rangle</math>, <math>|01\rangle</math>, <math>|10\rangle</math>, <math>|11\rangle\}</math>, it maps the basis states as follows:

Controlled gates act on 2 or more qubits, where one or more qubits act as a control for some specified operation. In particular, the controlled NOT gate (or CNOT or cX) acts on 2 qubits, and performs the NOT operation on the second qubit only when the first qubit is <math>|1\rangle</math>, and otherwise leaves it unchanged. With respect to the unentangled product basis <math>\{|00\rangle</math>, <math>|01\rangle</math>, <math>|10\rangle</math>, <math>|11\rangle\}</math>, it maps the basis states as follows:

In this state, called an ''equal superposition'', there are equal probabilities of measuring either product state <math>|00\rangle</math> or <math>|11\rangle</math>, as <math>|1/\sqrt{2}|^2 = 1/2</math>.  In other words, there is no way to tell if the first qubit has value “0” or “1” and likewise for the second qubit.

In this state, called an ''equal superposition'', there are equal probabilities of measuring either product state <math>|00\rangle</math> or <math>|11\rangle</math>, as <math>|1/\sqrt{2}|^2 = 1/2</math>.  In other words, there is no way to tell if the first qubit has value “0” or “1” and likewise for the second qubit.

<math> | 1 1 \rangle \mapsto | 1 0 \rangle </math>.

<math> | 1 1 \rangle \mapsto | 1 0 \rangle </math>.

Imagine that these two entangled qubits are separated, with one each given to Alice and Bob.  Alice makes a measurement of her qubit, obtaining—with equal probabilities—either <math>|0\rangle</math> or <math>|1\rangle</math>, i.e., she can now tell if her qubit has value “0” or “1”.  Because of the qubits' entanglement, Bob must now get exactly the same measurement as Alice.  For example, if she measures a <math>|0\rangle</math>, Bob must measure the same, as <math>|00\rangle</math> is the only state where Alice's qubit is a <math>|0\rangle</math>.  In short, for these two entangled qubits, whatever Alice measures, so would Bob, with ''perfect'' correlation, in any basis, however far apart they may be and even though both can not tell if their qubit has value “0” or “1” — a most surprising circumstance that can ''not'' be explained by classical physics.

Imagine that these two entangled qubits are separated, with one each given to Alice and Bob.  Alice makes a measurement of her qubit, obtaining—with equal probabilities—either <math>|0\rangle</math> or <math>|1\rangle</math>, i.e., she can now tell if her qubit has value “0” or “1”.  Because of the qubits' entanglement, Bob must now get exactly the same measurement as Alice.  For example, if she measures a <math>|0\rangle</math>, Bob must measure the same, as <math>|00\rangle</math> is the only state where Alice's qubit is a <math>|0\rangle</math>.  In short, for these two entangled qubits, whatever Alice measures, so would Bob, with ''perfect'' correlation, in any basis, however far apart they may be and even though both can not tell if their qubit has value “0” or “1” — a most surprising circumstance that can ''not'' be explained by classical physics.

A common application of the C_NOT gate is to maximally entangle two qubits into the <math>|\Phi^+\rangle</math> Bell state.  To construct <math>|\Phi^+\rangle</math>, the inputs A (control) and B (target) to the C_NOT gate are:

A common application of the C_NOT gate is to maximally entangle two qubits into the <math>|\Phi^+\rangle</math> Bell state.  To construct <math>|\Phi^+\rangle</math>, the inputs A (control) and B (target) to the C_NOT gate are:

C < sub > NOT </sub > 门的一个常见应用是将两个量子位最大限度地纠缠到 < math > | Phi ^ + rangle </math > Bell 状态。为了构造 < math > | Phi ^ + rangle </math > ，输入 a (control)和 b (target)到 c < sub > NOT </sub > gate 是:

C_NOT门的一个常见应用是最大限度地将两个量子位纠缠成<math>|\Phi^+\rangle</math>贝尔态。要构造<math>|\Phi^+\rangle</math>，到C_NOT门的输入A（控制）和B（目标）是：