Bell's theorem and local realism

[atto]

It can be violated by a theory in which $P(\vec a,\vec b)$ isn't of the form as given by equation (2). This is the case for QM.

No. This means only the condition can be violated, not the inequality itself.

Triangle has 3 sides - yes?
This means: triangle with 4 sides is imposible.

And the QM reasoning - logics work in this way:
the square is a triangle with 4 sides; so, this fact breaks the reality;
and we are very naive beings, because we always believe it's impossible. :)

 

[atto]

I think you are confused about this topic. You seem to be expressing a view that is at odds with what everyone else has said about Bell's inequality. As I said recently in a different thread, the fact that something is an establishment view doesn't make it right, but it makes Physics Forums the wrong place for you to be arguing about it.

I understand, but this is a fact:
there are no series, which can break the inequalities, the same the inequality has never been broken, despite the many sensational reports of experimenters.

 

[stevendaryl]

OK. This is a formal proof of the inequality.
So, it's true - it can't be violated in any way.

Let me try one more time. QM gives us a function

P(\vec{a}, \vec{b}) = the probability of Alice measuring spin-up in direction \vec{a} and Bob measuring spin-up in direction \vec{b}. This function is given by (in the spin-1/2 EPR case):

P(\vec{a}, \vec{b}) = \frac{1}{2} cos^2(\frac{\theta}{2})

where \theta = the angle between \vec{a} and \vec{b}. What Bell proved is that it is impossible to write this function in the form:

P(\vec{a}, \vec{b}) = \sum P_1(\lambda) P_A(\lambda, \vec{a}) P_B(\lambda, \vec{b})

where the sum ranges over possible values of the hidden variable \lambda, and P_1 is the probability for each value.

So Bell showed that the joint probability distribution did not "factor" into local probability distributions. He did not prove that the original probability distribution is impossible. Of course, it's possible, and experiments bear out that it correctly describes the EPR results.

 

[rubi]

No. This means only the condition can be violated, not the inequality itself.

Bell's inequality is $1+P(\vec b,\vec c) \geq \left|P(\vec a,\vec b)-P(\vec a,\vec c)\right|$. This can be violated by a function $P(\vec a,\vec b)$ that is not of the form $P(\vec a,\vec b) = -\int\mathrm d\lambda\rho(\lambda)A(\vec a,\lambda) A(\vec b,\lambda)$. For example if $P(\vec a,\vec b) = -2$, then the inequality says $-1 \geq 0$.

 

[stevendaryl]

I understand, but this is a fact:
there are no series, which can break the inequalities, the same the inequality has never been broken, despite the many sensational reports of experimenters.

Yes, Bell's theorem is a theorem. There is no way to produce 4 lists of numbers that violate his inequality. Everybody agrees with that. Quantum mechanics is not in violation of Bell's theorem, because it's a theorem, and you can't come up with a counter-example to a theorem. Bell's theorem, together with the predictions of QM, can be used to prove that there is no "local realistic" implementation of the predictions of QM.

QM does not contradict Bell's theorem. QM plus Bell's theorem contradicts local realism.

 

[atto]

Bell's inequality is $1+P(\vec b,\vec c) \geq \left|P(\vec a,\vec b)-P(\vec a,\vec c)\right|$. This can be violated by a function $P(\vec a,\vec b)$ that is not of the form $P(\vec a,\vec b) = -\int\mathrm d\lambda\rho(\lambda)A(\vec a,\lambda) A(\vec b,\lambda)$. For example if $P(\vec a,\vec b) = -2$, then the inequality says $-1 \geq 0$.

Of course. For example the inequality is easily breakable:
1 + x >= |y-z|; where: x,y,z - free, independent - arbitrary parameters.

In the oryginal inequality the x,y,z are inter correlated - dependend.

 

[atto]

Let me try one more time. QM gives us a function

P(\vec{a}, \vec{b}) = the probability of Alice measuring spin-up in direction \vec{a} and Bob measuring spin-up in direction \vec{b}. This function is given by (in the spin-1/2 EPR case):

P(\vec{a}, \vec{b}) = \frac{1}{2} cos^2(\frac{\theta}{2})

This is just the fantastic scenario, i mentioned earlier, ie. we assume the knowledge about the outcome on other arm... or maybe the setting angle alone will be sufficient.

So Bell showed that the joint probability distribution did not "factor" into local probability distributions. He did not prove that the original probability distribution is impossible. Of course, it's possible, and experiments bear out that it correctly describes the EPR results.

It's impossible - the measured series of {1,-1} don't break the inequality - with probability 1 exactly, and certainly.

 

[rubi]

Of course. For example the inequality is easily breakable:
1 + x >= |y-z|; where: x,y,z - free, independent - arbitrary parameters.

In the oryginal inequality the x,y,z are inter correlated - dependend.

The correlation of the x, y, z is exactly defined by the form of $P(\vec a,\vec b)$ that is given by the integral that I quoted earlier. So if we experimentally find that the inequality is broken, we have automatically ruled out all theories that require $P(\vec a,\vec b)$ to be of that form. However, we haven't ruled out QM, since QM doesn't require $P(\vec a,\vec b)$ to be of that form.

 

[atto]

QM does not contradict Bell's theorem. QM plus Bell's theorem contradicts local realism.

No. The realism is just the math.

The results of experiments rather show there must be an error in the realisation of the experiments... maybe in the further data processing.

 

[stevendaryl]

No. The realism is just the math.

The results of experiments rather show there must be an error in the realisation of the experiments... maybe in the further data processing.

Really, I'm going to have to ask you to stop posting on this topic. If you believe that the standard results are all wrong, Physics Forums is not the place to argue about them.

Personally, I don't think that you know what you're talking about, but this forum is not the place to argue about it.

I am notifying a moderator.

 

[rubi]

The results of experiments rather show there must be an error in the realisation of the experiments.

This claim is equivalent to the claim that there can be no consistent theory that doesn't predict $P(\vec a,\vec b) = -\int\mathrm d\lambda \rho(\lambda) A(\vec a,\lambda) A(\vec b,\lambda)$. Do you have any evidence for this bold claim?

 

[stevendaryl]

This claim is equivalent to the claim that there can be no consistent theory that doesn't predict $P(\vec a,\vec b) = -\int\mathrm d\lambda \rho(\lambda) A(\vec a,\lambda) A(\vec b,\lambda)$. Do you have any evidence for this bold claim?

I really don't think that Physics Forums is the correct avenue for breaking new results. I don't think it's appropriate to discuss this here. Atto's opinion is contrary to just about all published papers about Bell's theorem. So if there is anything to it, it's new research, and this is not a forum for new research.

Like Jeopardy, he should have put it in the form of a question: "I don't understand...how is Bell's theorem compatible with the predictions of QM?" instead of declaring that it's not.

 

[rubi]

I really don't think that Physics Forums is the correct avenue for breaking new results. I don't think it's appropriate to discuss this here. Atto's opinion is contrary to just about all published papers about Bell's theorem. So if there is anything to it, it's new research, and this is not a forum for new research.

I agree. When I started writing my post, yours wasn't there yet.

 

[atto]

Really, I'm going to have to ask you to stop posting on this topic. If you believe that the standard results are all wrong, Physics Forums is not the place to argue about them.

Personally, I don't think that you know what you're talking about, but this forum is not the place to argue about it.

I am notifying a moderator.

OK.
By the way: do not forget to ask the moderator for these fantastic binary series, which breaks the Bell's-type tautology or eventually the whole mathematical world, at least.

 

[Doc Al]

Closed pending moderation.