Why are Bell's inequalities violated?

bohm2

If one argues that something is local, realism is implied as above posts, I think. Analogously, if non-realism, then the issue of locality vs non-locality is kind of pointless since there's no ontological issues. I mean what ontological difference would there be between local vs non-local non-realism? Anyway, that's how I understood it. I think Gisin argues similarily here:
Is realism compatible with true randomness?
http://arxiv.org/pdf/1012.2536v1.pdf

harrylin

I certainly recall that point, which was the subject of many discussions... I just never interpreted it in the sense as expressed in that paper!

Their clarification ("assign a property to a system (e.g. the position of an electron)") plus your explanation allows me to make sense of Neumaier's claim (which he did not prove and with which you probably disagree). He holds that:
"all proofs of Bell type results [..] become invalid when particles have a temporal and spatial extension, with an internal structure that is modified when interacting in a beam splitter."
- http://www.mat.univie.ac.at/~neum/ms/lightslides.pdf

So, he suggests that that is why Bell's inequalities are violated.

DrChinese

He makes several statements like this that are difficult for me to place in a suitable context. Photons have both temporal and spatial extension in my view. Not sure if they do to a local realist though. And I do not follow his reasoning on how that connects to Bell. So I guess I would say that I disagree. If you are enforcing strict Einsteinian locality as is done in Weihs et al (1998), the spatial extent of a photon is already accounted for.

harrylin

Neither did I follow his reasoning until you explained it to me! Now I guess that I can follow it, for the first time.

Assuming that light is a wave and not a collection of photon particles, then the photon positions are undefined or non-existent until they are measured - and the same for some other properties. If so, then Bell's theorem may not apply because the existence of such unmeasured properties is required for that theorem - is that correct? It should be, following our posts #49 and #50 here above.

DrChinese

I keep saying that you have it backwards. QM does not assert photons have well-defined values for non-commuting operators - realists do!

EPR asserts that there is an element of reality IF Alice can predict Bob's outcome with certainty. Because this is experimentally demonstrated, you must accept EPR's key challenge. Which is that if QM is complete (read accurate in this instance), the reality of Bob's measurement is a function of Alice's choice of what to observe. Therefore if QM predicts correctly, we live in an observer dependent universe. This is directly from EPR. It does require the assumption of SIMULTANEOUS elements of reality (anything else is an unreasonable definition of reality, they say) and the assumption that there is no spooky action at a distance.

Bell simply takes those one step forward in his proof. So sure, Bell doesn't "apply" in the sense that one of the local realists' (and EPR's) 2 assumptions is invalid (much as you say: "the existence of such unmeasured properties"). But that is simply agreeing with Bell, disagreeing with EPR and denying local realism in one breath.

akhmeteli

I don't know, maybe A. Neumaier has revised his text since you looked at it, but I find a slightly different phrase there: "All proofs of Bell type results (including the present argument) become invalid when "particles" have a temporal and spatial extension over the whole experimental domain, with an internal structure that is modified when interacting in a beam splitter."

These extra words ("over the whole experimental domain") make me wonder if what he had in mind might be pretty much the same as the locality loophole.

DrChinese

I saw that too, couldn't figure out what he meant. Does he mean: non-local? Obviously the locality loophole itself was closed a while back.

akhmeteli

Only separately:-)

DrChinese

I would expect no less!

JK423

In a previous post i wrote down the CHSH inequality that any hidden variable model satisfies
I have understood quite well why every photon pair should satisfy this inequality (if local realism holds). Consequently, the mean value [itex]\left\langle S \right\rangle [/itex], over all photon pairs, should satisfy it as well. However, this quantity is unmeasurable in a real experiment, due to the fact that it involves unmeasurable quantities in each run, e.g. the covariances in different angles.
Now my problem is to express the CHSH inequality in an equivalent way, so that it involves only measurable quantities, like the number of pairs detected anti-correlated when measured in the angle [itex]\theta_1[/itex], and consequently will be applicable in a real experiment. But i want to do this completely equivalently! For example, the fact that in the quoted post i have written down the inequality satisfied by the mean value of S, i.e.
[itex] - 2 \le \,\left\langle S \right\rangle \le 2[/itex],
it doesn't mean that this inequality would also be satisfied by the mean value of the measurements! The mean value of S (including all unmeasurable quantities), that satisfies the above inequality, and the mean value of S which includes only experimental mean values (that don't include the unmeasurable quantities) are not equivalent! They are not equivalent because, in the first case the mean value of each quantity [itex]\left\langle {A\left( {{a_1}} \right)B\left( {{b_2}} \right)} \right\rangle [/itex] involves all photon pairs in the experiment, while in a real experiment the corresponding mean value would include only 1/4 of the total photon pairs (if we suppose that the four angles are chosen with the same probability), since with each photon pair we can measure only one of the 4 observables. The two quantities are totally different.

So, what i am looking for is an equivalent but measurable expression. Does anyone have any idea of how to do this?
Note: The above CHSH inequality holds for every initial preparation of the photons, so in the proposed derivation we should not include any condition on the initial state in order to have a more general result.

DrChinese

This is a point that is so confusing, I would say most folks reason do not understand at all. S is a derived formula, and to a certain extent, an arbitrary expression. There is absolutely NO need to be able to measure this in a single experiment.

What you are really attempting to do is to verify the QM prediction of cos^2(theta) for matches. That is experimentally verifiable. If QM predicts accurately (with cos^2), then Bell tells us that LR is violated. There is then no need to have the CHSH inequality. And if you look at a lot of the Bell tests, they graph the QM predicted value against the experimental results to show this. (The LR function would need to be a straight line, in contrast.)

However, it is possible to look at the graphed results and say, "hmmm, maybe a straight line as close to the observed results as the QM prediction". This is where CHSH comes in. It is a concrete way to determine that LR is rejected while QM is confirmed.

So specifically: the coincidence prediction for 60 degrees for QM is .25 and for LR is .33 or higher. All you have to do is measure that, it is directly measurable exactly as you hope! Then you see that the measured value is quite close to .25 and far away from .33 (by perhaps 30+ standard deviations). And you rule out LR because its prediction is flat out wrong.

JK423

Thank you both, DrChinese and Gordon Watson, for your feedback.
Let me restate where my doubts are specifically located, so that my point will become more clear and you will be able to give me more "targeted" help. First, let me repeat the necessary formulas:
Now, the reason why the CHSH inequality holds is because [itex]S_j[/itex] is always [itex]S_j=±2[/itex] for every photon pair separately. And this is due to the locality assumption, i.e. that the outcome in Alice's side does not depend on what Bob measures, etc. Mathematically this assumption is expressed by the fact that [itex]{A_j}\left( {{a_1}} \right)[/itex] is the same in both [itex]{A_j}\left( {{a_1}} \right) \cdot {B_j}\left( {{\beta _1}} \right)[/itex] and [itex]{A_j}\left( {{a_1}} \right) \cdot {B_j}\left( {{\beta _2}} \right)[/itex], i.e. whether Bob measures [itex]\beta_1[/itex] or [itex]\beta_2[/itex] is irrelevant, the outcome [itex]{A_j}\left( {{a_1}} \right)[/itex] will be the same. The same reasoning applies to [itex]{A_j}\left( {{a_2}} \right)[/itex].
The important thing here is that this is always true because [itex]{A_j}\left( {{a_1}} \right)[/itex] corresponds to the same photon in these two expressions [itex]{A_j}\left( {{a_1}} \right) \cdot {B_j}\left( {{\beta _1}} \right)[/itex] and [itex]{A_j}\left( {{a_1}} \right) \cdot {B_j}\left( {{\beta _2}} \right)[/itex], so it cannot be different if locality is assumed.

Now, look what happens if you consider the measureable edition of [itex]\left\langle S \right\rangle[/itex], where each of the quantities [itex]\left\langle {A\left( {{a_i}} \right)B\left( {{b_j}} \right)} \right\rangle[/itex] are mean values over the measurements. For simplicity let me consider only one of the measured values (instead of the whole mean value) in order to make my point clear. So assume just one run:
[itex]\left\langle {A\left( {{a_1}} \right)B\left( {{b_1}} \right)} \right\rangle = {A_1}\left( {{a_1}} \right){B_1}\left( {{b_1}} \right)[/itex] , corresponding to the measured value of the photon pair "1",
[itex]\left\langle {A\left( {{a_1}} \right)B\left( {{b_2}} \right)} \right\rangle = {A_2}\left( {{a_1}} \right){B_2}\left( {{b_2}} \right)[/itex], corresponding to the measured value of the photon pair "2".
Now take the sum of these in order to form the first half part of the quantity S:
[itex]\left\langle {A\left( {{a_1}} \right)B\left( {{b_1}} \right)} \right\rangle + \left\langle {A\left( {{a_1}} \right)B\left( {{b_2}} \right)} \right\rangle = {A_1}\left( {{a_1}} \right){B_1}\left( {{b_1}} \right) + {A_2}\left( {{a_1}} \right){B_2}\left( {{b_2}} \right)[/itex]. (1)

I told you previously that CHSH holds because [itex]{A_j}\left( {{a_1}} \right)[/itex] has the same value in these two quantities, since it corresponds to the same photon. But now that we have considered the mean values over measurements, [itex]{A_1}\left( {{a_1}} \right)[/itex] and [itex]{A_2}\left( {{a_1}} \right)[/itex] are, generally, different since they correspond to different photons "1" and "2", and that way they could be mimicking non-locality, since it looks as if [itex]{A}\left( {{a_1}} \right)[/itex] depends on what Bob measures.
You can generalize (1) for N photon pairs and take a more appropriate mean value. The moral in this story is that the photons are different in each quantity, so there is no obvious reason why CHSH would not be violated.

I hope that i made my point clear..
I'm looking forward to your feedback!

Giannis

nanosiborg

Thanks for helping me wade through this bohm2.
What about, eg., QFT?

We're (as Bell was) concerned with lhv formalism as it relates to qm and quantum entanglement experiments ... and not with how lhv formalism relates to the ontology of the world.
Can we agree that realism, for our purposes, means the formal expression of hidden variables?

We see by dBB that hidden variables and therefore hv formalisms aren't ruled out. But lhv (at least Bell lhv) formalisms are. So, it seems to come down to something to do with the formal expression of locality in terms of hidden variables ... and I'm reminded again of Jarrett's (and similar) treatment(s) of this which say that BI violation might be ruling out Bell type lhv models without also ruling out the possibility that nature is exclusively local.

harrylin

As I just discovered and clarified, it's exactly the meaning of those "simultaneous elements of reality" that appears to be an unreasonable requirement if that means "counterfactual" in the sense as cited in post #49. So, I guess that we now discovered in two parallel threads (and for different reasons) that it may be useful to focus more on Bell's "realist" criteria.
Once more: thanks for pointing out that Bell-realism is only a particular form of realism, different from that of Neumaier and myself. I will have to read again EPR to verify if their formulation of "realism" was as limited as Bell's.
I did a copy-paste so that's puzzling... I forgot what is meant with "locality loophole", but I'm pretty sure that he refers to his interpretation of QFT [EDIT:"Photons are intrinsically nonlocal objects"] which I think differs from the kind of spatial extension that DrChinese has in mind.

morrobay

In this very basic table of the 2³ possible cases of a spin 1/2 system with 3
axis settings, x,y,z ;

A _________ B
x y z ______ x y z
+++ _______ ---
++- _______ --+
+-+ _______ -+-
+-- _______ -++
-++ _______ +--
-+- _______ +-+
--+ _______ ++-
--- _______ +++

P(x⁺y⁺) < P(x⁺z⁺)+P(z⁺y⁺)
In the above inequality what are the exact counts that violate it ? And if the magnetic
field in the detector not only alters the spin on y axis when detecting spin on x but also
alters the spin on the axis being measured, x , then how is this violation valid ?

DrChinese

The cases you show assume a realistic/hidden variable perspective. OK, that is fine for a starting point for a Bell Inequality. Are you thinking that x, y and z are 3 perpendicular spatial axes (not clear to me from the example) ? Because if so you can't get a Bell Inequality from those. Instead, you need something like:

x=0
y=135
z=90

Assuming this is fine with you, and these angles are on the same plane:

xz= theta of 90 degrees
yz= theta of 45 degrees
xy= theta of 135 degrees

The quantum mechanical prediction for Matches (M) when anti-correlated is: 1 - cos^2(theta/2) or simply sin^2(theta/2). Your realistic requirement is M(xy) < M(xz) + M(yz) which is the same as saying: 0 < M(xz) + M(yz) - M(xy). Substituting the right hand side, you get something like:

M(xz)* + M(yz) - M(xy) =

sin^2(90/2) + sin^2(45/2) - sin^2(135/2) =

.5 + .1465 - .8535 =

-.2070**

Oops, this was supposed to be greater than zero per your realism requirement! So the realism requirement is flat out inconsistent with the predictions of QM. So here are the specific values that lead to a violation.

*There was a minor issue in your formula that became immaterial because of the angle settings I selected.
** And this is reduced by half to -.1035 if we only look at the ++ match cases, not that it really matters. The QM prediction is still less than zero and realism requires it be non-negative.

morrobay

The question on the above ( as an outsider to QM ) I have is that you are applying QM predictions to negate the realism requirement. That would be like using realism predictions
to negate the QM requirement. That is why I asked what the actual data is that violates
the inequality. Having said that, I could give up realism in all these discussions for
saving locality. Especially when Bell/EPR experiments are done with photons.