# Observers

Discussion in 'Physics & Math' started by arfa brane, Apr 18, 2017.

1. ### iceauraValued Senior Member

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It doesn't matter whether they are coins or not. The inequality holds for any collection of objects and any three such properties. In classical world.
You mean the quantum level formulation of it? Good thinking on his part.
Because it is proven to hold under very basic assumptions, that most people until then and many people today regard as self-evident. Proven, mathematically, formally, from those assumptions.

So if you have a violation of it, you have to discard or modify those assumptions. That's what a proof gives you - it shows you where you must focus your doubts.

And they are very basic, intuitively fundamental assumptions. A world in which they are invalid is a world in which human intuition works with great difficulty, one the human mind is not naturally built to handle.
? I have no idea what that question means.

3. ### arfa branecall me arfValued Senior Member

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Yes it does. It matters when you want to use coins as the collection (of countable things).
You say "none of the coins needs to have any of the properties", is that a general thing, you can have some countable collection with "none of the properties?".

But coins are objects that do have properties, hence my confusion.

5. ### iceauraValued Senior Member

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26,898
Your preferences regarding the objects you want to count are irrelevant to the Inequality.
The don't necessarily have any of the three properties being counted. Feathers, weight over five kilograms, flammable, for example.

7. ### arfa branecall me arfValued Senior Member

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If none of the coins have any of the properties being counted, what gets counted? Is it not the case that the collection (let's stay with coins) needs to have three properties, each true or false for any coin (is or isn't copper, is or isn't shiny, etc) ? Isn't that the point of the whole exercise?

The properties chosen are freely chosen, but you have to choose at least three, and, the three have to have some distribution over the collection. This is not irrelevant to the inequality because the inequality applies to any such collection.

8. ### iceauraValued Senior Member

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26,898
The inequality holds for any collection of objects and any three properties that fit the assumptions, which are very basic and general. In classical world.

That accounts for some of the power of Bell making explicit the fact that it does not always hold in QED, and then experiment agreeing with QED. That was a shock, for most people.

9. ### arfa branecall me arfValued Senior Member

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So Bell's inequalities are violated by quantum experiments, but why?

Is it because entanglement actually reduces the number of degrees of freedom, but a classical ensemble always has the same number?

I also note that entanglement is more striking when small number of particles are correlated, but there must be "bulk" entanglement out there in the real world otherwise Susskind, Maldacena and others are just wrong about it. But I don't think so; it must be all around me since, the electromagnetic field has complementary degrees of freedom (!)

(you don't say?)

10. ### iceauraValued Senior Member

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26,898
Dunno.

The theory of QED predicts these violations, and this prediction agrees with experiment perfectly - as does every other prediction of QED so far checked. But the question of what exactly these equations are modeling has as yet no generally satisfactory answer.

11. ### The GodValued Senior Member

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3,546
Observing Entanglement: Violations of
Bell Inequalities..
By Peter Heur University of Rochester.

This article gives the standard photon polarization entanglement experiment results and establishes the detector value 2.6 which is > 2 so the inequality is violated.

Cannot exactly figure out what is the correlation between the mathematical expression of inequality and the detected value 2.6? The simple point is 2.6 is too big a difference, what if it was 2.001 or 4? What exactly 2.6-2 = 0.6 or around 30% represent? Strength of entanglement? The zero entanglement would have given a value less than 2, what would be the value for 100% entanglement? Could they have created the kind of set up required with the help of TAs and students?

12. ### Confused2Registered Senior Member

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499
After looking at the link suggested by The God...
Now my thoughts.
Imagine we produce entangled pairs with horizontal/vertical entanglement.
Send both through a vertical polariser. If entanglement works then if one gets through V polarised then the other will be H polarised and won't. So if we counted coincidences of photon pairs getting through both polarisers then ideally there shouldn't be any. This seems to be too simple to be true.

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14. ### The GodValued Senior Member

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3,546
Thanks, I thought I had linked the PDF with the title in first line....but probably missed it.

The interesting part is that, they produced these photons here itself, so its a local experiment, and there is nothing remote which could influence their polarization...entangled photons locally created and locally trapped cannot violate Bells inequality...can they?

Last edited: Jul 9, 2017
15. ### iceauraValued Senior Member

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They have to, when measured, according to QED. And QED agrees with experiment.

16. ### arfa branecall me arfValued Senior Member

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In type-II SPDC, there are two places in the output where photons are maximally entangled.
What about the rest of the output?

17. ### arfa branecall me arfValued Senior Member

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The diagram shows type II SPDC.
The two green crosses are where it's impossible to tell which cone the photon is in, this is superposition.
It's a function which is position-dependent.
There are two cones because these are where the horizontally and vertically polarized photons emerge in pairs, for each pump photon.

So, if you could place both the cones in a spatial superposition, all pairs would be (maximally) entangled. This is what you get with type I SPDC and two crystals of BBO, one rotated by $\pi/2$. But the two types each generate half the four Bell states.

Last edited: Jul 12, 2017
18. ### iceauraValued Senior Member

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26,898
According to your sources (for the pictures) and the information on the pictures themselves the horizontal and vertical polarizations of the photons separated in the cones are correlated - not entangled. You can clearly see the labels - upper cone photons are measured, their polarization is known, it is vertical, for example.

19. ### arfa branecall me arfValued Senior Member

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Right. But the places where photons are entangled is where the correlated states are in superposition.

Is there a way to entangle the photons in the separated cones? Sure, just bring the cones together so they aren't separated (like in the second diagram). In practice this might be difficult to do with the setup as is.

20. ### arfa branecall me arfValued Senior Member

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Although it's quite math-heavy, I wanted to look at how this paper, the source of the second diagram, treats the whole entanglement-Bell's theorem connection.

After all, if you ever want to design a quantum algorithm, it seems to be something you'll need to get your head around, like you need to get your head around conditional branching and looping in classical algorithms.
It seems to me to be entirely reasonable to think I can map the gate-paradigm to type I SPDC, at least partially. Gate-wise you have two inputs, in SPDC you have a pump beam as (one) input. This input beam has to be considered then, as being equivalent to two inputs.

--https://quantumexperience.ng.bluemi...=a3ac6cc67b61b902cf009293bc2aed73&pageIndex=0

One of the inputs is rotated into the Hadamard basis, something that's not too hard to visualise with a Bloch sphere, then the two inputs "go through" a CNOT gate. The result is two entangled photons because, well at this point I can think it's because the CNOT gate does something when its control bit is neither 0 nor 1--it's in a superposition of both--that doesn't resolve in classical logic, I need some other kind at this point.

On the other hand, with type I and a pair of crystals, as the second diagram shows, rotating one of the two crystals does the same thing, and each pump photon can 'entangle' with one crystal or the other, the two outputs are the cones. Classically most of the pump photons don't entangle, but stay in a coherent beam.
It would be helpful if the physics of BBO crystals was a bit more specific, and a clearer explanation maybe in terms of a field theory, as to why a certain percentage of the input radiation gets spin-correlated (then entangled depending on certain spatial parameters), but most doesn't.

Last edited: Jul 13, 2017
21. ### arfa branecall me arfValued Senior Member

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This is the output for the above setup, again from the bluemix site:

It isn't exactly a 50% spread, but the number of measurements they've made is large enough to be statistically significant.
You should get the same results if you 'invert' both inputs with the X gate, which rotates |0> to |1> and conversely. So the outputs will be indistiguishable, and you can't tell if the inputs are both |0> or both |1> (except, you know what the inputs are, this isn't a trivial thing). You get two Bell states, and a violation of the inequality.

Bell's theorem, and it's attendant inequalities are about the commutativity of information, I think.
That's because any derivation of his theorem means there is a nonempty subset of probabilities which is the intersection where property A, property B, and property C all have the same binary probability of being true (resp. false!). This means the probabilities commute, and so does the information.

On the other hand, because quantum measurements violate Bell's inequality there must be noncommutative information, with it's attendant probability and entropy. By entropy I mean what you don't know about what the input information is; you can easily introduce some entropy by assuming you only have the output.

22. ### iceauraValued Senior Member

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26,898
The derivation of Bell's Inequality makes no assumptions about any actual probabilities involved, and in fact does not involve probability at all. It holds for all probability distributions of the properties measured, in classical world. What are you referring to, exactly?
They are not correlated when entangled, when in superposition. That is not a quibble - it's important. If they were correlated, they would obey Bell's Inequality.

23. ### arfa branecall me arfValued Senior Member

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Isn't Bell's inequality really about commutativity? Isn't it really about the least number of "properties" that can distribute over what we normally understand are "measurements", from which we "gain information about the system". And isn't measurement already predicated on the existence of general "systems", "physicality" etc?
After all, a set of countable objects could be n physically distinct positions, the physical distinctness is a property. Is probability a property of distinct "things"?

Right, the "correlation information" doesn't commute with "entanglement".
Nonetheless, we say the entangled states are correlated states in superposition--because it's something we "know" . . .

Here's a heuristic I've been using. You have a coin in your closed hand. You know if you open your hand you can see if the coin is heads or tails, and that it will be one or the other.

Suppose you open your hand and without looking at the coin, turn it over and close up your hand again. Now you know that if you look at the coin you will gain information about the current state and the state before it, of the coin.
And you don't need to look at the coin to know this (!). You know all that because you've measured something.
Something that was a completely determined sequence, no probability (yay!), but it changed the state of your knowledge about the coin.

Last edited: Jul 16, 2017