Editing Quantum entanglement (section)

=== Pure states ===
Consider two arbitrary quantum systems {{mvar|A}} and {{mvar|B}}, with respective [[Hilbert space]]s {{mvar|H<sub>A</sub>}} and {{mvar|H<sub>B</sub>}}. The Hilbert space of the composite system is the [[tensor product]]
: <math> H_A \otimes H_B.</math>

If the first system is in state <math>| \psi \rangle_A</math> and the second in state <math>| \phi \rangle_B</math>, the state of the composite system is
: <math>|\psi\rangle_A \otimes |\phi\rangle_B.</math>

States of the composite system that can be represented in this form are called separable states, or [[product state]]s. However, not all states of the composite system are separable. Fix a [[basis (linear algebra)|basis]] <math> \{|i \rangle_A\}</math> for {{mvar|H<sub>A</sub>}} and a basis <math> \{|j \rangle_B\}</math> for {{mvar|H<sub>B</sub>}}. The most general state in {{math|''H<sub>A</sub>'' ⊗ ''H<sub>B</sub>''}} is of the form
: <math>|\psi\rangle_{AB} = \sum_{i,j} c_{ij} |i\rangle_A \otimes |j\rangle_B</math>.

This state is separable if there exist vectors <math> [c^A_i], [c^B_j]</math> so that <math> c_{ij}= c^A_i c^B_j,</math> yielding <math display="inline"> |\psi\rangle_A = \sum_{i} c^A_{i} |i\rangle_A</math> and <math display="inline"> |\phi\rangle_B = \sum_{j} c^B_{j} |j\rangle_B.</math> It is inseparable if for any vectors <math> [c^A_i],[c^B_j]</math> at least for one pair of coordinates <math> c^A_i,c^B_j</math> we have <math> c_{ij} \neq c^A_ic^B_j.</math> If a state is inseparable, it is called an 'entangled state'.<ref name="Rieffel2011"/>{{rp|218}}<ref name="Zwiebach2022"/>{{rp|§1.5}}

For example, given two basis vectors <math> \{|0\rangle_A, |1\rangle_A\}</math> of {{mvar|H<sub>A</sub>}} and two basis vectors <math> \{|0\rangle_B, |1\rangle_B\}</math> of {{mvar|H<sub>B</sub>}}, the following is an entangled state:
: <math>\tfrac{1}{\sqrt{2}} \left ( |0\rangle_A \otimes |1\rangle_B - |1\rangle_A \otimes |0\rangle_B \right ).</math>

If the composite system is in this state, it is impossible to attribute to either system {{mvar|A}} or system {{mvar|B}} a definite [[pure state]]. Another way to say this is that while the [[von Neumann entropy]] of the whole state is zero (as it is for any pure state), the entropy of the subsystems is greater than zero. In this sense, the systems are "entangled". The above example is one of four [[Bell states]], which are (maximally) entangled pure states (pure states of the {{math|''H<sub>A</sub>'' ⊗ ''H<sub>B</sub>''}} space, but which cannot be separated into pure states of each {{mvar|H<sub>A</sub>}} and {{mvar|H<sub>B</sub>}}).<ref name="Zwiebach2022"/>{{rp|§18.6}}

Now suppose Alice is an observer for system {{mvar|A}}, and Bob is an observer for system {{mvar|B}}. If in the entangled state given above Alice makes a measurement in the <math> \{|0\rangle, |1\rangle\}</math> eigenbasis of {{mvar|A}}, there are two possible outcomes, occurring with equal probability: Alice can obtain the outcome 0, or she can obtain the outcome 1. If she obtains the outcome 0, then she can predict with certainty that Bob's result will be 1. Likewise, if she obtains the outcome 1, then she can predict with certainty that Bob's result will be 0. In other words, the results of measurements on the two qubits will be perfectly anti-correlated. This remains true even if the systems {{mvar|A}} and {{mvar|B}} are spatially separated. This is the foundation of the EPR paradox.<ref name="Nielsen-2010" />{{Rp|pages=113–114}}

The outcome of Alice's measurement is random. Alice cannot decide which state to collapse the composite system into, and therefore cannot transmit information to Bob by acting on her system. Causality is thus preserved, in this particular scheme. For the general argument, see [[no-communication theorem]].