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“Spookiness” Confirmed by the First Loophole-Free Quantum Test

96 点作者 flurpitude超过 9 年前

7 条评论

semi-extrinsic超过 9 年前
From TFA (or I guess that should be TFP):<p>&quot;Strictly speaking, no Bell experiment can exclude the infinite number of conceivable local realist theories, because it is fundamentally impossible to prove when and where free random input bits and output values came into existence. Even so, our loophole-free Bell test opens the possibility to progressively bound such less conventional theories.&quot;<p>It&#x27;s a bit strong to call it loophole-free then, isn&#x27;t it? But &quot;free-of-the-two-most-common-loopholes&quot; is a lot less sexy. Nevertheless, cool work.
deepnet超过 9 年前
The difference is that the whiteboard formula is actually called the Clauser-Horne-Shimony-Holt inequality, not the Bell inequality, and it is a slightly more sophisticated version, concocted by the four physicists it is named after, five years The first two numbers were 0.56 and 0.82. The third was –0.59, so it seems I would have to take this way from the running total. The fourth number, another 0.56, should then have left me with a total of 1.35 and victory for Einstein.<p>That’s not what I showed. <a href="http:&#x2F;&#x2F;www.bbc.co.uk&#x2F;iplayer&#x2F;episode&#x2F;b04tr9x9&#x2F;the-secrets-of-quantum-physics-1-einsteins-nightmare" rel="nofollow">http:&#x2F;&#x2F;www.bbc.co.uk&#x2F;iplayer&#x2F;episode&#x2F;b04tr9x9&#x2F;the-secrets-of...</a><p>In fact, the subtlety is that the third term, the one that had a negative value, was already negative. The inequality read:<p>P(a,b) + P(a,b’) – P(a’,b’) + P(a’,b) ≤ 2,<p>So, plugging all the numbers, this looks like:<p>0.56 + 0.82 – (–0.59) + 0.56 = 0.56 + 0.82 + 0.59 + 0.56 = 2.53<p>So, sorry Einstein, victory goes to Bohr instead.
oofabz超过 9 年前
If you want to know more about quantum entanglement, here is an excellent lecture on the subject:<p><a href="https:&#x2F;&#x2F;www.youtube.com&#x2F;watch?v=dEaecUuEqfc" rel="nofollow">https:&#x2F;&#x2F;www.youtube.com&#x2F;watch?v=dEaecUuEqfc</a>
qrybam超过 9 年前
Slightly tangential question: I know that entanglement doesn&#x27;t violate information transfer greater than the speed of light; c. But given that knowing the state of the measured particle means we also know the state of the entangled particle and measuring again could produce a different result, couldn&#x27;t you conceivably construct a machine which repeatedly measured a number of particles until you reach a desired state for each, therefore constructing a message via the entangled counterpart particles for someone to consume? I suppose you&#x27;d need very accurate timing at both locations to know when to attempt to read the message at the destination.<p>Or am I missing something along the lines of &quot;its not possible to measure the entangled counterparts without affecting something else, therefore making the whole thing impossible&quot;? I&#x27;m sure I am but hope someone could explain.
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thisrod超过 9 年前
Congratulations! The preparation scheme sounds very clever.
effie超过 9 年前
&gt; Our observation of a loophole-free Bell inequality violation thus rules out all local realist theories that accept that the number generators timely produce a free random bit and that the outputs are final once recorded in the electronics.<p>Why should a local realist theory accept that? In determinism, there are no random events.
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stefantalpalaru超过 9 年前
&gt; And quantum theory allows two entangled particles to become linked in such a way that when a measurement is performed on one (breaking it out of superposition, and clicking it into a well-defined state), the properties of its entangled partner will likewise become defined, instantaneously — no matter how far apart they are separated.<p>Besides the fact that measurement at that level is actually interaction, how can you prove that both particles were not having the same state from the start? If it&#x27;s a consequence of the wave function being placed in the configuration space by the Copenhagen interpretation, we need to be certain that we don&#x27;t add epicycles upon epicycles.<p>BTW, there are many more quantum mechanics interpretations besides the most popular one, and some of them don&#x27;t violate locality: <a href="https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Interpretations_of_quantum_mechanics#Tabular_comparison" rel="nofollow">https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Interpretations_of_quantum_mec...</a>
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