Chinese Researchers Achieve Quantum Entanglement Record
Crash course in entanglement:<p>The state of each qbit is represented by a state vector of two complex numbers [a, b] where |a|^2 + |b|^2 = 1. There are two special qbit values called the <i>classical basis</i>: [1, 0] which is the classical bit 0, and [0, 1] which is the classical bit 1. If a qbit is not in one of the two classical states, we say it is in <i>superposition</i>. When a qbit is in superposition, we can measure it[0] and it will collapse probabilistically to 0 or 1; for a qbit [a, b], the probability that it collapses to 0 is |a|^2 and the probability that it collapses to 1 is |b|^2.<p>Things get more interesting when we have multiple qbits. If we have two qbits [a, b] and [c, d], we define their <i>product state</i> as their tensor product [ac, ad, bc, bd]. For example, if we have two qbits both in state [1&#x2F;sqrt(2), 1&#x2F;sqrt(2)], their product state would be [1&#x2F;2, 1&#x2F;2, 1&#x2F;2, 1&#x2F;2]. We use the product state to calculate the action of a quantum logic gate that operates on multiple qbits - for a gate which operates on two qbits, we can always represent its action as a 4x4 matrix.<p>Usually we can move back and forth between the product state representation and writing out the individual qbits states. However, in certain scenarios something very special happens: we cannot factor the product state back into the individual state representation! Consider the product state [1&#x2F;sqrt(2), 0, 0, 1&#x2F;sqrt(2)]. If you try to write this as a tensor product of two states [a, b] and [c, d], you cannot! It cannot be factored; the qbits have no individual value, and we say they are <i>entangled</i>.<p>Well, what does this mean? It means when you measure one qbit, even if the qbits are very far apart, you instantly know the value of the other qbit. So if I entangled two qbits in the state [1&#x2F;sqrt(2), 0, 0, 1&#x2F;sqrt(2)], give you one, then we go to opposite ends of the universe, if I measure my qbit and see a 0 I&#x27;ll know your qbit instantly also collapsed to 0 (or collapsed to 1 if I measured 1). This phenomenon has been experimentally-verified to occur faster than light. It is instantaneous, as far as we can tell. So, local realism is wrong! Spooky action at a distance is real!<p>There is an important caveat: while the qbits seem to coordinate in some faster-than-light way, you cannot use this to <i>communicate</i> in a faster-than-light way. All we have is a shared random number generator. I can&#x27;t send some chosen bit from my reference frame to yours. This is called the no-communication theorem.<p>If you found this interesting, I have a full video on quantum computing for computer scientists here: <a href="https:&#x2F;&#x2F;youtu.be&#x2F;F_Riqjdh2oM" rel="nofollow">https:&#x2F;&#x2F;youtu.be&#x2F;F_Riqjdh2oM</a><p>[0] IDGAF about your chosen quantum mechanics interpretation, don&#x27;t @ me