Undergrad Is Bell's Theorem Valid Despite Fundamental Theoretical Loopholes?

[stevendaryl]

Seeking clarity, where does this response fail?

Property Y: $P(AB|abλ) = P(A|aλ)P(B|bλ).$
Y means their is no logical connection between $A$ and $B$.
But there is a logical connection between $A$ and $B$ because of common condition $λ$.

I just want to remind you that nobody is proposing Property Y as a general property about all correlations. Only in the specific case in which

1. A and B are assumed to be separated so that it is impossible for A to influence B or vice-versa.
2. \lambda is the complete set of variables in the common causal past of both A and B.

As to Property Y saying that there is no logical connection between A and B; no, that's not true. Take the simple example of a pair of shoes: You take a pair of shoes, and put each into a box and mix up the boxes. Then you give one box to Alice and another box to Bob. There is definitely a "logical" relationship between what Alice and Bob discover in their respective boxes: If Alice sees a left shoe, then Bob sees a right shoe.

That example has correlations but does not violate Property Y. In that case, the "hidden variables" would consist of

1. The information about which box the left shoe was put in, and which box the right shoe was put in.
2. The information about how it was decided which box to send to Alice and which box to send to Bob.

If Alice knew 1&2, then she would know what shoe she was going to find even before she opened her box. No additional information about what Bob discovered would make any difference.

Therefore QM and all the related experiments have these three properties:
P: $P(AB|abλ) = P(A|aλ)P(B|abλA) = P(B|bλ)P(A|abλB).$
Q: $P(A|aλ) = P(B|bλ) = \frac{1}{2}.$

We don't have any information about whether Q is true. To establish Q, you would have to repeat the experiment many times in which you control for the value of \lambda. But we don't know what \lambda might be, so we can't control for it.

R: $P(AB|abλ) = P(A|aλ)P(B|abλA) = P(B|bλ)P(A|abλB) \neq P(A|aλ)P(B|bλ).$

S: So Bell's property Y is false.

I would put it as: QM does not satisfy property Y. That's what Bell proved.

T: So theories of type X are those in which there is no logical connection between $A$ and $B$.

I don't know what you mean by that. You can have locally realistic theories with perfect correlation between A and B. My two-shoe example is one.

U: Type X theories are therefore irrelevant here.

Definitely QM is not a theory of type X. That was what Bell proved.

 

[stevendaryl]

Please explain.

For me: Y is the product rule for the conjunction of $AB$ under EPRB.

So, if you have full knowledge of $λ$, then Y becomes Y*:

Y*: $P(AB|abλ) = P(A|aλ)P(B|bλ) = 1\cdot1 =1.$

But, under EPRB, we cannot have full knowledge of $λ$, so P holds:

P: $P(AB|abλ) = P(A|aλ)P(B|abλA) = P(B|bλ)P(A|abλB).$

So, I conclude: Bell's Y is false under EPRB.

I don't know why you want to put it that way. Bell proved that QM does not satisfy property Y. It's like a proof that \sqrt{2} is irrational. You assume that it is rational, and show that it leads to a contradiction.

 

[stevendaryl]

@N88, I'm having trouble understanding what point you are making. Saying that the correlation between A and B can be accounted for by a third variable is not saying that there is no relationship between A and B. Just the opposite: it's saying quite precisely what the relationship is.

 

[stevendaryl]

@N88, I'm having trouble understanding what point you are making. Saying that the correlation between A and B can be accounted for by a third variable is not saying that there is no relationship between A and B. Just the opposite: it's saying quite precisely what the relationship is.

Let me give an example of a hidden-variables theory. Suppose that Alice and Bob are married. Suppose that Alice occasionally gets sick--one day out of 57. Suppose that Bob also gets sick one day out of 57. But one day out of 94, they are both sick.

So letting B be: Bob is sick, and letting A be: Alice is sick, we have:

P(A) = 0.0175
P(B) = 0.0175
P(A) P(B) = 0.00031
P(A \wedge B) = 0.0106

So we conclude:
P(A \wedge B) \neq P(A) P(B)

So their illnesses are correlated. The local epidemiologist investigates and finds out that:

1. There is a 0.01 chance each day that the couple eats bad food for dinner.
2. If they eat bad food, there is a 0.75 chance that they will get sick.
3. There is a 0.02 chance each day that there will be flu viruses around their home
4. If they are exposed to flu viruses, there is a 0.50 chance they will get sick.
5. (There is a negligible chance of having both flu and bad food)

Then these two factors completely explain the correlation. Let \lambda_{BF} mean that there is bad food and let \lambda_F mean that there is flu around.

1. P(\lambda_{BF}) = 0.01
2. P(A | \lambda_{BF}) = 0.75
3. P(B | \lambda_{BF}) = 0.75
4. P(\lambda_F) = 0.02
5. P(A | \lambda_F) = 0.5
6. P(B | \lambda_F) = 0.5
7. P(A \wedge B) = P(\lambda_F) P(A | \lambda_F) P(B | \lambda_F) + P(\lambda_{BF}) P(A | \lambda_{BF}) P(B | \lambda_{BF}) = 0.02 \cdot 0.5 \cdot 0.5 + 0.01 \cdot 0.75 \cdot 0.75 = 0.0106

So for this model:

P(A \wedge B) = \sum_{\lambda} P(\lambda) P(A | \lambda) P(B | \lambda)

where the sum is over the two possibilities: \lambda_F and \lambda_{BF}

The fact that Alice's and Bob's illnesses are independent, given that we have accounted for bad food and viruses doesn't mean that there is no relation between their illnesses; it just means that the correlation is explained by the shared food and/or viruses.

 

[DrChinese]

Do you agree that if Alice had picked a setting different from the one she actually picked she would have obtained a result? That result, whatever it is, is the counterfactual result. "

I am not the one asserting there are counterfactual cases. Everything I see points to an observer dependent universe, one that lacks local realism. So when someone asks about a particular local realistic theory, one which is excluded by Bell but that the author claims is not, I always ask: Please describe those counterfactual cases.

 

[PeterDonis]

if you have full knowledge of $λ$, then Y becomes Y*

No it doesn't. Full knowledge of $\lambda$ does not mean all probabilities are 1.

 

[Nicky665]

I am not the one asserting there are counterfactual cases. Everything I see points to an observer dependent universe, one that lacks local realism. So when someone asks about a particular local realistic theory, one which is excluded by Bell but that the author claims is not, I always ask: Please describe those counterfactual cases.

" Let us break down the situation fully:

Monday:
If Alice picks setting "a" on Tuesday she will obtain result "A"
If Alice picks setting "b" on Tuesday she will obtain result "B"
If Alice picks setting "c" on Tuesday she will obtain result "C"

Tuesday:
Alice picks setting "a" and obtains result "A"

Wednesday:
Alice picked setting "a" on Tuesday and obtained result "A"
If Alice had picked setting "b" on Tuesday she would have obtained result "B"
If Alice had picked setting "c" on Tuesday she would have obtained result "C"

The original question then becomes clear: Do you agree that if Alice had picked a different setting on Tuesday than the one she picked, she would have obtained a result ?

Note that I don't care what the value of the actual result is, just the fact that there is a result. Also note that A, B, C are not actual values, but just labels for whatever result it is that Alice obtains for the corresponding setting. "

 

[stevendaryl]

" The original question then becomes clear: Do you agree that if Alice had picked a different setting on Tuesday than the one she picked, she would have obtained a result ?

In discussions of Bell's inequality, "counterfactual" is mentioned in the context of "counterfactual definiteness". CFD says that there is a definite answer to questions such as "What result would Alice have gotten if she had measured X instead of Y?"

 

[Nicky665]

It's not Bell's problem. Bell proved a theorem along the lines of: "All theories of type X have property Y. QM does not have property Y. Therefore, QM is not a theory of type X."

What are you disagreeing about?

"Perhaps if you spell out exactly what X and Y mean mathematically, I we will show you what we are disagreeing about.

For example, please describe to me a physical situation for which the relationship P(AB|λ) = P(A|λ)P(B|λ) is true
But the relationship P(AB|λ) = P(A|λ)P(B|Aλ) is false."

In discussions of Bell's inequality, "counterfactual" is mentioned in the context of "counterfactual definiteness". CFD says that there is a definite answer to questions such as "What result would Alice have gotten if she had measured X instead of Y?"

"And you think, QM disagrees with that?

Bell in 1964 Paper wrote:Measurements can be made, say by Stern-Gerlach magnets, on selected components of the spins

and

. If measurement of the component

, where

is some unit vector, yields the value +1 then, according to quantum mechanics, measurement of

must yield the value -1 and vice versa.Bell says according to QM, Alice would have gotten a definite answer if she had measured at the same angle as Bob. Is that CFD or not?

BTW, you didn't answer the first question. Would Alice have definitely obtained a result (any result at all) had she measured at a different setting?"

 

[DrChinese]

Would Alice have definitely obtained a result (any result at all) had she measured at a different setting?"

Obviously Alice literally obtains a result when she measures at any setting. But the relevant point is whether there is any dependency on what Bob does. It always comes back to the same place: The answer is interpretation dependent.