Undergrad Problem with CHSH's version of Bell's inequalities

[Nicky665]

So what are now supposed to be the limits of bell inequation +-2 or +-2.82

 

[facenian]

So what are now supposed to be the limits of bell inequation +-2 or +-2.82

2.82 is limit for the quantum mechanical calculation and this value was not at discussion

 

[stevendaryl]

Here's another response to the original poster:

If, as he suggests, we consider different values of \lambda for the different terms, we would have the expression:

\mathcal{C}(a,b,a',b',\lambda, \lambda', \lambda'', \lambda''') = A(a,\lambda) B(b,\lambda) + A(a', \lambda') B(b, \lambda') + A(a, \lambda'') B(b', \lambda'') - A(a', \lambda''') B(b', \lambda''')

But we don't actually observe \lambda, (by definition, it is a "hidden variable"). So what we can observe is the average of \mathcal{C} over all values of \lambda, \lambda', \lambda'', \lambda''':

\mathcal{C}_{av}(a,b,a',b') = \int d\lambda d\lambda' d\lambda'' d\lambda''' P(\lambda) P(\lambda') P(\lambda'') P(\lambda''') \mathcal{C}(a,b,a',b',\lambda, \lambda', \lambda'', \lambda''')

It's elementary to prove that this is equal to:

\mathcal{C}_{av}(a,b,a',b') = \int d\lambda P(\lambda) \mathcal{C}(a,b,a',b',\lambda, \lambda, \lambda, \lambda)

with all four lambdas the same.

 

[facenian]

Here's another response to the original poster:It's elementary to prove that this is equal to:

\mathcal{C}_{av}(a,b,a',b') = \int d\lambda P(\lambda) \mathcal{C}(a,b,a',b',\lambda, \lambda, \lambda, \lambda)

with all four lambdas the same.

Sorry, but I don't know that elemtary proof

 

[stevendaryl]

Sorry, but I don't know that elemtary proof

Well, we have the definition:

\mathcal{C}(a,b,a',b',\lambda, \lambda', \lambda'', \lambda''')
= A(a,\lambda) B(b,\lambda) + A(a, \lambda') B(b', \lambda') + A(a', \lambda'') B(b, \lambda'') - A(a', \lambda''') B(b', \lambda''')

Now, integrate over \lambda''':

\int d\lambda''' P(\lambda''') \mathcal{C}(a,b,a',b',\lambda,\lambda', \lambda'', \lambda''')
= \int d\lambda''' P(\lambda''') A(a,\lambda) B(b,\lambda) + \int d\lambda''' P(\lambda''') A(a, \lambda') B(b', \lambda') + \int d\lambda''' P(\lambda''') A(a', \lambda'') B(b, \lambda'') - \int d\lambda''' P(\lambda''') A(a', \lambda''') B(b', \lambda''')
= A(a,\lambda) B(b,\lambda) + A(a, \lambda') B(b', \lambda') + A(a', \lambda'') B(b, \lambda'') - \int d\lambda''' P(\lambda''') A(a', \lambda''') B(b', \lambda''')

This just uses the fact that since A(a, \lambda) B(b, \lambda) doesn't depend on \lambda''', it can be pulled outside the integral: \int d\lambda''' P(\lambda''') A(a,\lambda) B(b,\lambda) = A(a,\lambda) B(b,\lambda) \int d\lambda''' P(\lambda'''). And \int d\lambda''' P(\lambda''') = 1. Similarly for all the other terms that do not involve \lambda'''.

Now, integrate over \lambda'', \lambda' and \lambda. This gives:
\int d\lambda P(\lambda) A(a,\lambda) B(b,\lambda) + \int d\lambda' P(\lambda') A(a, \lambda') B(b', \lambda') + \int d\lambda'' P(\lambda'') A(a', \lambda'') B(b, \lambda'') - \int d\lambda''' P(\lambda''') A(a', \lambda''') B(b', \lambda''')

But \lambda, \lambda', \lambda'', \lambda''' are just dummy integration variables now. You can rename them all to \lambda, to get:
\int d\lambda P(\lambda) A(a,\lambda) B(b,\lambda) + \int d\lambda A(a, \lambda) B(b', \lambda) + \int d\lambda P(\lambda) A(a', \lambda) B(b, \lambda) - \int d\lambda P(\lambda) A(a', \lambda) B(b', \lambda)

By linearity of integration, this is the same as:
\int d\lambda P(\lambda) [A(a,\lambda) B(b,\lambda) + A(a, \lambda) B(b', \lambda) + A(a', \lambda) B(b, \lambda) - A(a', \lambda) B(b', \lambda)]

 

[facenian]

You can then compute a bit more complicatedly the other values of Chsh namely p (Chsh=-2) p (Chsh=0) and so on.

The average of course gives <Chsh>=-2 but all possible values are here computed not making the "same lambda" assumption.

I don't understand why you say that <Chsh>=-2 since this value is a function of the "classical" probabilities \theta_{ab},\theta_{ab&#039;},\theta_{a&#039;b},\theta_{a&#039;b&#039;}
and while these numbers can take "independandly" any value between 0 and 1, <Chsh> takes any value between -4 and +4. Obiously there is something here that I could not understand

 

[facenian]

Well, we have the definition:
.
.
.

By linearity of integration, this is the same as:
\int d\lambda P(\lambda) [A(a,\lambda) B(b,\lambda) + A(a, \lambda) B(b&#039;, \lambda) + A(a&#039;, \lambda) B(b, \lambda) - A(a&#039;, \lambda) B(b&#039;, \lambda)]

Yes, it is very interesting we end up with same expression you already wrote previously(more simply) and in both occasions your calculations are correct however
they don't solve the logical problem I'm striving with and which I'll explain again but in this occasion and after your explanations I find it even more subtle. I will state some facts I believe are correct:
1) Your expression

E(a,b) = \int d\lambda P(\lambda) A(a,\lambda) B(b, \lambda)

then

E(a,b) + E(a, b&#039;) + E(a&#039;, b) - E(a&#039;,b&#039;) = \int d\lambda P(\lambda) (A(a,\lambda) B(b, \lambda) + A(a,\lambda) B(b&#039;, \lambda) + A(a&#039;,\lambda) B(b, \lambda) - A(a&#039;,\lambda) B(b&#039;, \lambda))

That's just a fact.

is correct
2) I have problems not the with equations in 1) but with the bounding of it which is based on use of same lambda outside the integral, that is, the integrand is bounded before the integration.
3) As I see it we can not take the integrand outside the simbol of integration and interpret it as the outcome of a series of experimets(with the same lambda), by experimentes I mean emission of entangled pairs
4) By now I studied the original bibliography and found that such a "mistake"(from my viewpoint, of course) is not present. This is because the original authors did not bound the integrand before the integration and in doing so my objection dissapears.

 

[N88]

As rubi points out, this is the realism assumption at play. You don't have to accept it. But it makes little sense to reject it unless you have a better formulation. This is in fact considered an easy-to-swallow step.

Please explain in detail please why "this is in fact an easy-to-swallow step." I find it impossible to accept. In my head is the known fact (from Bohr) that a measurement disturbs the particle.

Let us run experiments at 4 different locations, S, T, U, V. Then we can derive the experimental correlation

C(a,b,a&#039;,b&#039;) \equiv E(a,b)_S + E(a,b&#039;)_T + E(a&#039;,b)_U - E(a&#039;, b&#039;)_V. There is no lambda involved here, just experimental outcomes.

So why is it "easy to swallow" that the theoretical calculation should be based on something like the same lambda at each location?

Are you maybe saying that there are only 4 particles involved at each site? Lambda_S = A⁺B⁺, A⁺B^-, A^-B⁺, A^-B^-; etc?

Is this what Bell accepted, based on d'Espagnat's inferences in his Scientific American article? But again, in my head is the known fact (from Bohr) that a measurement disturbs the particle.

 

[stevendaryl]

Let us run experiments at 4 different locations, S, T, U, V. Then we can derive the experimental correlation

C(a,b,a&#039;,b&#039;) \equiv E(a,b)_S + E(a,b&#039;)_T + E(a&#039;,b)_U - E(a&#039;, b&#039;)_V. There is no lambda involved here, just experimental outcomes.

So why is it "easy to swallow" that the theoretical calculation should be based on something like the same lambda at each location?

I don't know what you mean by "the same lambda at each location". The assumption that Bell made was that for each twin pair produced, there was an associated value of lambda. It could very well be that the same value of lambda never appears twice. But all the values of lambda come from the same probability distribution.

Bell assumed that in the twin pair EPR experiment (with the usual experimenters, Alice and Bob), Alice's measurement result A(a,\lambda) for each pair is a deterministic function of the value of \lambda for that pair and her setting, a. Bob's result B(b,\lambda) is a function of \lambda and his setting, b. The lambdas might never repeat (and probably won't, if there is a continuum of possible lambda values). But Bell assumed that the lambdas were produced according to a probability distribution P(\lambda). The expression that the original poster was complaining about was \mathcal{C}(a,b,a&#039;,b&#039;,\lambda) \equiv A(a,\lambda) B(b,\lambda) + A(a,\lambda) B(b&#039;,\lambda) + A(a&#039;,\lambda) B(b, \lambda) - A(a&#039;,\lambda) B(b&#039;,\lambda). There is no assumption that this expression corresponds to anything directly measurable. The only reason for introducing it is the following two facts: (which are purely mathematical facts, independent of any experimental confirmation)

1. For each value of \lambda, |\mathcal{C}(a,b,a&#039;,b&#039;,\lambda)| \leq 2, under the assumption that A(a,\lambda) = \pm 1 and B(a,\lambda) = \pm 1.
   
2. \int d\lambda P(\lambda) \mathcal{C}(a,b,a&#039;,b&#039;,\lambda) = E_{th}(a,b) + E_{th}(a,b&#039;) + E_{th}(a&#039;,b) - E_{th}(a&#039;,b&#039;), where E_{th}(a,b) is the theoretical prediction for the correlation between Alice's result and Bob's result, E_{th}(a,b) = \int d\lambda P(\lambda) A(a,\lambda) B(b,\lambda)

These are facts about the mathematical model, not about experimental results. The connection between the two is that IF the model is a correct description of the physics, then E_{th}(a,b) = E(a,b), where E(a,b) is the experimentally determined correlation.

So Bell is not assuming anything about "the same value of lambda for different locations". The constraint on E_{th} is just a fact about the model. On the other side of the equation is the experimentally determined E(a,b). No lambdas are involved in that, at all; it's just an experimentally determined value.

But again, in my head is the known fact (from Bohr) that a measurement disturbs the particle.

Yes, but in the case of twin pairs, Einstein and Podolsky and Rosen argued that it is unreasonable to believe that the measurement of one particle could disturb the OTHER particle, which might be far away when it is measured. So they argued that if there is perfect correlation between the distant particles, it must be because the two particles share some property, which is the idea behind the hidden variable \lambda. Bell's argument shows that EPR's assumptions are incorrect---the correlations cannot be explained in terms of shared properties and purely local interactions.

 

[DrChinese]

Please explain in detail please why "this is in fact an easy-to-swallow step." I find it impossible to accept. In my head is the known fact (from Bohr) that a measurement disturbs the particle...

Does a measurement "here" also affect a particle "there"? Because that is what is appears to happen if you assume preexisting variables that determine the outcome.

The "easy-to-swallow step" is simply trying to come up with some definition - whatever it may look like to you - that expresses "realism" or "pre-existing variables" or whatever. That's lambda. They do not need to be the same at each measurement location, but consider this: you can predict the outcomes of any identical measurements with certainty. So you would certainly expect those variables to be the same (or at least leading to same outcomes).

 

[jk22]

I don't understand why you say that <Chsh>=-2 since this value is a function of the "classical" probabilities \theta_{ab},\theta_{ab&#039;},\theta_{a&#039;b},\theta_{a&#039;b&#039;}
and while these numbers can take "independandly" any value between 0 and 1, <Chsh> takes any value between -4 and +4. Obiously there is something here that I could not understand

<> means average of chsh. The values involved are -4,-2,0,2,4 but the average on a large number of samples is smaller classically

 

[facenian]

<> means average of chsh. The values involved are -4,-2,0,2,4 but the average on a large number of samples is smaller classically

Yes, I did get that, however I don't see why you said that <Chsh>=-2 since Chsh is a function of the four numbers \theta_{ab},\theta_{ab&#039;},\theta_{a&#039;b},\theta_{a&#039;b&#039;} and they belong to the range 0\leq | \theta_{ik}|\leq1 so -4\leq \,&lt;Chsh&gt; \,\leq+4

 

[jk22]

That's just the probability of 4 that dépends on the four angles. To be formal : $$CHSH\in \{-4,-2,-0,2,4\}$$ and $$|\langle CHSH\rangle|\leq 2$$ since $$p(-4)\leq\approx 0.32$$ and $$p(-2)\leq\approx .42$$ this implies $$|\langle CHSH\rangle|\leq \approx 2.$$

 

[Mentz114]

Yes, I did get that, however I don't see why you said that <Chsh>=-2 since Chsh is a function of the four numbers \theta_{ab},\theta_{ab&#039;},\theta_{a&#039;b},\theta_{a&#039;b&#039;} and they belong to the range 0\leq | \theta_{ik}|\leq1 so -4\leq \,&lt;Chsh&gt; \,\leq+4

This is not true because $a,b,a',b'$ each are used twice which introduces a dependency and reduces the limit to $\pm2$.

 

[jk22]

The latter fact rests on the following assumption : in the two pairs $$A (a,\lambda)B (b,\lambda) $$ and $$A (a,\lambda)B (b',\lambda) $$ A is fully determined by lambda and a.

However one could imagine that A corresponds to a measurement two different times and is not stored in a variable.

In fact lambda could be simply taken as the angle of polarization of the photon, its more physical. Then its clear from the onset that the measurement angle and the polarization are not sufficient to determine the result.

One could imagine strange functions like +1 for rational numbers and -1 for algebraic non rational numbers but they need to reproduce the probabilities which already becomes quite complicated.

Then since those numbers are dense any small modification could change the result and Bell theorem applies.

At this point it becomes rather a philosophical question but in experiments the accuracy is finite and this can be probably better studied with computers.

It is then easy to write a code who gives different results for the same angle and polarization, it just needs a random generator in the function A.

 

[facenian]

This is not true because $a,b,a',b'$ each are used twice which introduces a dependency and reduces the limit to $\pm2$.

Knowing that the correct answer is $\pm2$ I think that you might be right

 

[facenian]

Hello everybody! We had this discussion some time ago and I decided to write a paper about the error I'm claiming to exist in the way the CHSH inequality is derived in many references. Though I think the mistake is trivial it seems to be very common to the point that the paper was unanimously rejected by a Journal claiming the the free will principle enables one to take the same value of lambda in the equation. Finally the paper was published by Foundations of Physics-Springer.
If somebody is interested in it I can upload it.

 

[Mentz114]

Hello everybody! We had this discussion some time ago and I decided to write a paper about the error I'm claiming to exist in the way the CHSH inequality is derived in many references. Though I think the mistake is trivial it seems to be very common to the point that the paper was unanimously rejected by a Journal claiming the the free will principle enables one to take the same value of lambda in the equation. Finally the paper was published by Foundations of Physics-Springer.
If somebody is interested in it I can upload it.

I'm interested.

 

[facenian]

I'm interested.

http://rdcu.be/oFPR

 

[jk22]

If I take independent lhv I get : $$|<CHSH>|=2\leq <|CHSH|>=\frac {71}{32}\approx 2.22$$

Hence there is a discrepancy with Bell's theorem which obtains the modulus of the average equals to the average of the modulus, because it makes the assumption of dependency.

 

[facenian]

Sorry but I really don't know what you are talking about. I' m referring to the usual CHSH inequality where there is no such thing as the average of the modulus.