# I Thermal interpretation and Bell's inequality

#### N88

Summary
Concerning p.198 of Bell's famous 1964 paper http://cds.cern.ch/record/111654/files/vol1p195-200_001.pdf How does TI explain Bell's move from the first equation to the second equation?
Concerning p.198 of Bell's famous 1964 paper http://cds.cern.ch/record/111654/files/vol1p195-200_001.pdf

How does TI explain Bell's move from the first equation to the second equation?

Under TI, what is the physical significance of this move?

Thank you.

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#### PeterDonis

Mentor
How does TI explain Bell's move from the first equation to the second equation?
Why should TI have to explain it? The move from the first equation to the second equation involves an assumption which is violated in QM. Since TI is an interpretation of QM, it can't explain something which violates QM.

• vanhees71

#### N88

Why should TI have to explain it? The move from the first equation to the second equation involves an assumption which is violated in QM. Since TI is an interpretation of QM, it can't explain something which violates QM.
So what is that assumption via your explanation, thanks?

#### PeterDonis

Mentor
what is that assumption via your explanation, thanks?
Have you read the paper? What does it say in the sentence right after the first equation? That sentence even starts with the words "The vital assumption is..."

#### N88

Have you read the paper? What does it say in the sentence right after the first equation? That sentence even starts with the words "The vital assumption is..."
Yes, I've read the paper and that assumption. And I agree with that assumption. And I note Bell says that he uses that first equation to make the move that I am enquiring about.

But how does he use that qualifier of the first equation, and that equation itself, to make that move that I am enquiring about?

PS: You said that QM violates this assumption? Bell's eqn (3) seems not to?

#### PeterDonis

Mentor
But how does he use that qualifier of the first equation, and that equation itself, to make that move that I am enquiring about?
What move? The move from equation (1) to equation (2)? That's just the standard formula for an expectation value of a quantity, where the quantity in this case is the product of A and B.

You said that QM violates this assumption? Bell's eqn (3) seems not to?
Bell's equation (3) is the QM expectation value, which does indeed violate the assumption used in equations (1) and (2), because it violates the inequality that Bell derives in the rest of the paper based on that assumption. Which is the whole point of the paper. So, once again, have you read the paper? Because the questions you are asking don't seem like questions that someone who has read and understood the paper would be asking.

• vanhees71

#### N88

What move? The move from equation (1) to equation (2)? That's just the standard formula for an expectation value of a quantity, where the quantity in this case is the product of A and B.

Bell's equation (3) is the QM expectation value, which does indeed violate the assumption used in equations (1) and (2), because it violates the inequality that Bell derives in the rest of the paper based on that assumption. Which is the whole point of the paper. So, once again, have you read the paper? Because the questions you are asking don't seem like questions that someone who has read and understood the paper would be asking.
The move that I am enquiring about is at the top of p.198; from the first eqn there to the second eqn there.

#### PeterDonis

Mentor
The move that I am enquiring about is at the top of p.198; from the first eqn there to the second eqn there.
Ok. This is still irrelevant to the thermal interpretation, since it's part of the derivation of the Bell inequality, which has nothing to do with QM.

As far as I can tell, to go from the first equation at the top of p. 198 to the second, he is simply using equation (1), which says that both A and B only take the values +1 or -1.

#### N88

Ok. This is still irrelevant to the thermal interpretation, since it's part of the derivation of the Bell inequality, which has nothing to do with QM.

As far as I can tell, to go from the first equation at the top of p. 198 to the second, he is simply using equation (1), which says that both A and B only take the values +1 or -1.
So how do you see Bell using his eqn (1)?

I am trying to go further than as far as you can tell. That's why I thought the TI might provide me with an Aldi good/different insight.

#### PeterDonis

Mentor
how do you see Bell using his eqn (1)?
Um, exactly the way he says in the paper?

I am trying to go further than as far as you can tell.
I have no idea what this means.

That's why I thought the TI might provide me with an Aldi good/different insight.
I don't understand what you need insight about or why you think any QM interpretation might provide it. Bell's reasoning in the paper is perfectly clear, and, as I've already said, has nothing whatever to do with any interpretation of QM, since his derivation of his inequality is not based on QM. The only use of QM in the paper at all is to point out that QM's expectation value violates the inequality.

#### N88

Um, exactly the way he says in the paper? [emph. added]

I don't understand what you need insight about. Bell's reasoning in the paper is perfectly clear, and, as I've already said, has nothing whatever to do with any interpretation of QM, since his derivation of his inequality is not based on QM. The only use of QM in the paper at all is to point out that QM's expectation value violates the inequality.
But, as far as I can see: Bell does not say exactly what way he uses his eqn (1). So it might help to interpret my seeking as: Why does Bell used eqn (1) in the way that he does in his move atop p.198? What is the physical significance of his way? Thanks.

#### PeterDonis

Mentor
Bell does not say exactly what way he uses his eqn (1).
Sure he does; he says "using (1)" right after the two equations at the top of p. 198. Yes, he leaves it to you to see exactly how using (1) gets you from the first to the second; are you unable to see how that works?

Why does Bell used eqn (1) in the way that he does in his move atop p.198?
Um, because that's a necessary step in the logic to get the inequality he wants?

What is the physical significance of his way?
What is the physical significance of equation (1)? Look at it. It's just saying that A and B are observables that can only produce the results +1 and -1 when measured, and that the things that affect which result gets produced are the corresponding setting of the measurement device ($\vec{a}$ or $\vec{b}$) and the "hidden variables" $\lambda$.

#### N88

Sure he does; he says "using (1)" right after the two equations at the top of p. 198. Yes, he leaves it to you to see exactly how using (1) gets you from the first to the second; are you unable to see how that works?
Could you tell me how you see that it works, please?

Um, because that's a necessary step in the logic to get the inequality he wants?
This seems back-to-front to me.

What is the physical significance of equation (1)? Look at it. It's just saying that A and B are observables that can only produce the results +1 and -1 when measured, and that the things that affect which result gets produced are the corresponding setting of the measurement device ($\vec{a}$ or $\vec{b}$) and the "hidden variables" $\lambda$.
No, that is not my question here. I've been asking for the physical significance of the way that Bell uses his (1) to make the move atop p.198.

#### PeterDonis

Mentor
Could you tell me how you see that it works, please?
We have the first equation at the top of p. 198:

$$P(\vec{a}, \vec{b}) - P(\vec{a}, \vec{b}) = - \int d \lambda \rho(\lambda) \left[ A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a}, \lambda) A(\vec{c}, \lambda) \right]$$

Since by equation (1), each $A$ can only take on the values $\pm 1$, then for any vector, such as $\vec{b}$, we must have $A(\vec{b}, \lambda) A(\vec{b}, \lambda) = 1$, so we can insert it anywhere we like as a factor. So we can insert it into the last term on the RHS of the above:

$$P(\vec{a}, \vec{b}) - P(\vec{a}, \vec{b}) = - \int d \lambda \rho(\lambda) \left[ A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a}, \lambda) A(\vec{b}, \lambda) A(\vec{b}, \lambda) A(\vec{c}, \lambda) \right]$$

Then we just flip the sign by reversing the order of the terms, and factor out $A(\vec{a}, \lambda) A(\vec{b}, \lambda)$ in the obvious way, to get the second equation at the top of p. 198:

$$P(\vec{a}, \vec{b}) - P(\vec{a}, \vec{b}) = \int d \lambda \rho(\lambda) A(\vec{a}, \lambda) A(\vec{b}, \lambda) \left[ A(\vec{b}, \lambda) A(\vec{c}, \lambda) - 1 \right]$$

#### N88

We have the first equation at the top of p. 198:

$$P(\vec{a}, \vec{b}) - P(\vec{a}, \vec{b}) = - \int d \lambda \rho(\lambda) \left[ A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a}, \lambda) A(\vec{c}, \lambda) \right]. (X)$$

Since by equation (1), each $A$ can only take on the values $\pm 1$, then for any vector, such as $\vec{b}$, we must have $A(\vec{b}, \lambda) A(\vec{b}, \lambda) = 1$, so we can insert it anywhere we like as a factor. So we can insert it into the last term on the RHS of the above:

$$P(\vec{a}, \vec{b}) - P(\vec{a}, \vec{b}) = - \int d \lambda \rho(\lambda) \left[ A(\vec{a}, \lambda) A(\vec{b}, \lambda) - A(\vec{a}, \lambda) A(\vec{b}, \lambda) A(\vec{b}, \lambda) A(\vec{c}, \lambda) \right]. (Y)$$

Then we just flip the sign by reversing the order of the terms, and factor out $A(\vec{a}, \lambda) A(\vec{b}, \lambda)$ in the obvious way, to get the second equation at the top of p. 198:

$$P(\vec{a}, \vec{b}) - P(\vec{a}, \vec{b}) = \int d \lambda \rho(\lambda) A(\vec{a}, \lambda) A(\vec{b}, \lambda) \left[ A(\vec{b}, \lambda) A(\vec{c}, \lambda) - 1 \right]. (Z)$$
OK, many thanks. So the overall effect is that Bell has made an assumption that leads to $$- A(\vec{a}, \lambda) A(\vec{c}, \lambda)$$ in (X) being equal to $$- A(\vec{a}, \lambda) A(\vec{b}, \lambda) A(\vec{b}, \lambda) A(\vec{c}, \lambda$$ in (Y).

Now can you explain please what QM (or TI) has to say about this? Rather than being about locality (as you suggested earlier) isn't it more to do with (something like) non-commuting observables not permitting such an equality? In other words: since I know Bell fairly well, I'd like to know what QM says of the situation.

Thanks again.

#### DarMM

Gold Member
Bell's inquality was known to mathematicians as the Triangle inequality as early as the 1920s. It's just a statement of the relation between the expectation values of variables defined on a common sample space.

Ultimately in the CHSH presentation for example the inequality basically assumes that there is a distribution $p\left(a,b,c,d\right)$ giving the probability of outcomes for each variable. In a given round we only measure two variables say $a,c$ and we have probabilities for their outcomes $p\left(a,c\right)$. That these are marginals of the overall distribution is what gives the inequality.

How QM violates the inequality is that there is no $p\left(a,b,c,d\right)$, the outcomes in a given trial are not the marginals of a four outcome case.

However this has nothing to do with the Thermal Interpretation in particular.

#### N88

Bell's inquality was known to mathematicians as the Triangle inequality as early as the 1920s. It's just a statement of the relation between the expectation values of variables defined on a common sample space.

Ultimately in the CHSH presentation for example the inequality basically assumes that there is a distribution $p\left(a,b,c,d\right)$ giving the probability of outcomes for each variable. In a given round we only measure two variables say $a,c$ and we have probabilities for their outcomes $p\left(a,c\right)$. That these are marginals of the overall distribution is what gives the inequality.

How QM violates the inequality is that there is no $p\left(a,b,c,d\right)$, the outcomes in a given trial are not the marginals of a four outcome case.

However this has nothing to do with the Thermal Interpretation in particular.
Many thanks for your reply. I agree re TI. My interest in TI was in seeking alternative explanations as to where Bell goes wrong physically; TI (as I understand it) being another theory that accept's Bell's inequality mathematically.

I am seeking to understand the physical objections to Bell's inequality. To understand the reasons given for its violation by experiments.

This comparison might make my interest clearer. Here are 3 inequalities with a common LHS.

$|E(a,b)-E(a,c)| ≤ 1-E(a,b)E(a,c)$ = True under Bell (1964) and its experiment.
$|E(a,b)-E(a,c)| ≤ 1+E(b,c)$ = False under QM and experiment = Bell's famous inequality.
$|E(a,b)-E(a,c)| ≤ 3/2+E(b,c)$ = True under Bell (1964) and its experiment.

So Bell is the odd one out. And, as I see it, the difference is this: as in Post #15 above, Bell does not maintain the requirement that the correlations relevant to expectations must be derived from test results obtained in the same instance. (As stated in the line before Bell's first equation.)

So in this sense, it seems to me that Bell involves a common mathematical/physical error. Mathematically and physically, it seems to me: Bell breaks the requirement about working with results obtained in the same instance.

In other words; see Post #15 above: he appears to bust one instance and makes it two and gets an invalid result. Whereas, when instances are not busted, valid results twice follow mathematically and physically and locally.

#### A. Neumaier

TI (as I understand it) being another theory that accept's Bell's inequality mathematically.
TI is consistent with standard quantum mechanics, which violates the Bell inequalities in the appropriate contexts.
I am seeking to understand the physical objections to Bell's inequality. To understand the reasons given for its violation by experiments.
Bell inequalities are theorems about classical probabilistic models, whose assumptions already contradict both quantum theory and experiment (since the conclusions do). Hence they say nothing at all about quantum systems.

To understand the reasons for their violation by experiments it is no help to analyze the proof of the inequalities (which you try to do) since the problem is already in the assumptions made.

• mattt and N88

#### N88

TI is consistent with standard quantum mechanics, which violates the Bell inequalities in the appropriate contexts.

Bell inequalities are theorems about classical probabilistic models, whose assumptions already contradict both quantum theory and experiment (since the conclusions do). Hence they say nothing at all about quantum systems.

To understand the reasons for their violation by experiments it is no help to analyze the proof of the inequalities (which you try to do) since the problem is already in the assumptions made.
Thank you; this is very helpful. But I question whether Bell is about classical probability models: since the two valid inequalities above are derived under locality and standard probability theory?

Also. It seems to me that the widespread belief -- that Bell-inequalities show that nature is nonlocal -- needs to be challenged by showing what can be derived locally by means of Bell's approach and ordinary probability theory. For, also, Bell did expect that "dilemmas" (associated with his work ) would be overcome.

May I take it that TI is fully Einstein-local?

#### DarMM

Gold Member
Bell assumes no retrocausality, no nonlocal effects, all variables come from a common sample space and that there is a single world.

QM violates the resulting equalities, thus one of these is wrong.

• N88

#### PeterDonis

Mentor
Now can you explain please what QM (or TI) has to say about this?
Absolutely nothing. As I've already said, this whole derivation in Bell's paper has nothing to do with QM. It is a derivation regarding a class of local hidden variable models that is intended to show that no such model can match the predictions of QM. That's the whole point of the paper. Obviously QM and interpretations of QM can't have anything to say about a class of models that are not QM and can't match the predictions of QM.

#### PeterDonis

Mentor
TI (as I understand it) being another theory that accept's Bell's inequality mathematically.
Huh? QM, and all interpretations of QM, cannot possibly "accept Bell's inequality mathematically", since they violate it.

#### PeterDonis

Mentor
Here are 3 inequalities with a common LHS.
Where are you getting the other two (the ones that are not Bell's inequality) from?

the two valid inequalities above are derived under locality and standard probability theory?
You didn't derive them; you just pulled them out of thin air.

#### PeterDonis

Mentor
May I take it that TI is fully Einstein-local?
What does "Einstein-local" mean?

#### PeterDonis

Mentor
the widespread belief -- that Bell-inequalities show that nature is nonlocal
No, the fact that QM violates the Bell inequalities shows that nature is either nonlocal, or retrocausal, or many-worlds, or whatever the short version of "not all variables come from a common sample space" is.

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