
“Spookiness” Confirmed by the First Loophole-Free Quantum Test - flurpitude
http://fqxi.org/community/forum/topic/2581
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semi-extrinsic
From TFA (or I guess that should be TFP):

"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."

It's a bit strong to call it loophole-free then, isn't it? But "free-of-the-
two-most-common-loopholes" is a lot less sexy. Nevertheless, cool work.

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deepnet
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.

That’s not what I showed. [http://www.bbc.co.uk/iplayer/episode/b04tr9x9/the-
secrets-of...](http://www.bbc.co.uk/iplayer/episode/b04tr9x9/the-secrets-of-
quantum-physics-1-einsteins-nightmare)

In fact, the subtlety is that the third term, the one that had a negative
value, was already negative. The inequality read:

P(a,b) + P(a,b’) – P(a’,b’) + P(a’,b) ≤ 2,

So, plugging all the numbers, this looks like:

0.56 + 0.82 – (–0.59) + 0.56 = 0.56 + 0.82 + 0.59 + 0.56 = 2.53

So, sorry Einstein, victory goes to Bohr instead.

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oofabz
If you want to know more about quantum entanglement, here is an excellent
lecture on the subject:

[https://www.youtube.com/watch?v=dEaecUuEqfc](https://www.youtube.com/watch?v=dEaecUuEqfc)

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qrybam
Slightly tangential question: I know that entanglement doesn'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'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'd need very accurate timing at both locations to know
when to attempt to read the message at the destination.

Or am I missing something along the lines of "its not possible to measure the
entangled counterparts without affecting something else, therefore making the
whole thing impossible"? I'm sure I am but hope someone could explain.

~~~
SAI_Peregrinus
The easiest way to think of it is as follows: When you entangle the particles
they have opposite states, and each has their state hidden in a box. When you
open one box, you know what the state of the other particle must be.

Only there is no box, and the state isn't well defined until you measure it.

Or there is no box, and the state isn't well defined for anything that hasn't
interacted with the particles, but upon interacting whatever did the
interaction becomes entangled too and sees the defined state of the particles
but the rest of the universe wouldn't know the state of whatever did the
interacting.

Or there is a box, but it's the size of the whole universe, and every time
anything interacts with it (looks inside the box) the box splits up into all
the possible things that could have been seen by that interaction and you get
lots of universes.

Or a number of other ways to think about the issue depending on your preferred
interpretation of quantum mechanics, but no matter how you look at it it's
nothing like a normal box.

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thisrod
Congratulations! The preparation scheme sounds very clever.

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effie
> 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.

Why should a local realist theory accept that? In determinism, there are no
random events.

~~~
n4r9
This has been studied. The kind of determinism that a local realistic theory
requires in order to explain Bell violations is generally regarded as absurd:
[https://en.wikipedia.org/wiki/Superdeterminism](https://en.wikipedia.org/wiki/Superdeterminism)

~~~
effie
But that is just regular determinism; everything is determined by the prior
state of the system, including the operation of the number generator and the
experimenter's movements. As far as physics models go, I cannot see anything
absurd about it.

~~~
n4r9
Sorry, I didn't word that very well.

"Regular" determinism doesn't necessarily affect Bell's Theorem. All you need
is that the measurement choices at each end are "effectively free for the
purpose at hand". The kind of determinism you'd need to actually affect Bell's
Theorem is a pathological, conspiratorial one in which the universe is well
aware of which measurement choices you are going to make and just sets things
up so that you'll get the same outcomes that quantum theory predicts.

This is discussed in Section D of the following paper: [http://www-e.uni-
magdeburg.de/mertens/teaching/seminar/theme...](http://www-e.uni-
magdeburg.de/mertens/teaching/seminar/themen/AJP001261.pdf)

~~~
effie
That's a very interesting paper, quite well written too. Thanks for the link.

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stefantalpalaru
> 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.

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'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't add epicycles upon epicycles.

BTW, there are many more quantum mechanics interpretations besides the most
popular one, and some of them don't violate locality:
[https://en.wikipedia.org/wiki/Interpretations_of_quantum_mec...](https://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics#Tabular_comparison)

~~~
semi-extrinsic
> how can you prove that both particles were not having the same state from
> the start?

But this is exactly what these experiments aim to prove. That kind of
predeterminism is precisely a form of local hidden variable theory. Such a
theory is disproved if we can prove the violation of a Bell inequality. The
standard reference for understanding the connection is this laymans-terms
paper by Mermin:

[http://web.pdx.edu/~pmoeck/pdf/Mermin%20short.pdf](http://web.pdx.edu/~pmoeck/pdf/Mermin%20short.pdf)

If we accept that this experiment is indeed a standard-loophole-free violation
of Bell's inequality, we then conclude that we must give up one of three
things: locality, realism, or all free will. This rules out any local hidden-
variable theory. The mainstream view is then to keep locality and free will.
Some alternative theories (notably Bohmian ones) give up locality and keep
realism, but this is strongly at odds with what we know from special
relativity, and indeed locality is assumed in all work on quantum field theory
(through Lorenz invariance) which has been verified to match experiments to a
precision above that of any other physical theory. The third option, which
essentially no-one supports, is that free will is impossible in our universe
and that everything everywhere is predetermined.

~~~
tene
Is there any chance that you can expand on the precise meaning of "free will"
in a quantum physics context? My understand of the usual common-sense
interpretation is roughly "The actions that I take are supernatural, and are
not a consequence of physical laws", which seems quite absurd to me, and I'd
expect that to have very little support in the physics community. If not this,
then what does "realism" and "free will" mean here?

~~~
api
Free will is typically defined to mean you[n+1] != classical_function(you[n]),
at least sometimes or with some probability. That does not necessarily imply
anything supernatural, though if the supernatural (or any kind of dualism)
exists that would explain it. It could also arise due to quantum noise or any
other process that violates classical determinism.

There are those who define free will a bit differently though. It's not a
precise term. Another definition is that you[n+1] cannot be computed from any
function other than you or something isomorphic with you -- in other words you
are not coarse-grainable or predictable using any subset of your state.
Someone would have to literally make a copy of you to predict your behavior,
possibly down to the atomic or quantum level.

I've heard functions/processes with this property called computationally
irreducible:

[https://en.wikipedia.org/wiki/Computational_irreducibility](https://en.wikipedia.org/wiki/Computational_irreducibility)

Basically the computational irreducibility definition of free will just means
nothing outside of you can predict what you're going to do unless it has an
exact copy of you or something functionally equivalent (uploaded mind, etc.).

Another variation on the same idea is the "arrow of time" view put forward by
Ilya Prigogine:

[http://www.amazon.com/The-End-Certainty-Ilya-
Prigogine/dp/06...](http://www.amazon.com/The-End-Certainty-Ilya-
Prigogine/dp/0684837056)

This is IMHO very close to if not identical to Wolfram's computational
irreducibility, but framed a bit differently.

Obviously humans are somewhat predictable, but somewhat predictable doesn't
imply deterministic. My personal opinion is that the second theory
(irreducibility/arrow of time) is almost certainly true, and the first is also
probably true. So we are probably both irreducible and indeterminate. I'd say
the same is likely true of any living thing and possibly other complex natural
processes.

~~~
daemeh
My opinion is that we are predictable and deterministic, but we choose to
cling to the idea of us being more than that because we can't deal with the
other option. The other option is quite simple, from what I see: we'll never
manage to completely read, simulate and predict a complex system like our
body, so in reality nobody will be able to predict what we'll do - and for me
that's enough to feel comfortable. From what I see free will is defined by
others as some kind of magic process through which our decisions are based on
some random factor which is unpredictable. I'd say the randomness doesn't need
to exist as long as the unpredictability holds.

~~~
stefantalpalaru
If we cannot really choose, we are not really responsible of anything we do.
Are you comfortable with that?

~~~
dllthomas
This appears to be an appeal to consequences.

That aside, however, I would say that this depends on what you mean by
"responsible".

