What is the physical significance of Bell's math?

[PeterDonis]

on the i-th run of the experiment, the former is the HV that Alice receives and the latter is the HV that Bob receives

Then you are certainly not making the same assumptions as Bell. In Bell's model, the HVs are the same for both Alice and Bob; that's the whole point. The only difference is how they choose to orient their detectors.

 

[N88]

Then you are certainly not making the same assumptions as Bell. In Bell's model, the HVs are the same for both Alice and Bob; that's the whole point. The only difference is how they choose to orient their detectors.

Are you sure? In EPRB, the case that he studies in Bell (1964), the particles are anti-correlated.

In some tests with photons, the HV are the same; under the same rule re conservation of angular momentum.

 

[PeterDonis]

In EPRB, the case that he studies in Bell (1964), the particles are anti-correlated.

The observed spins, given that both Alice and Bob choose the same orientation for their detectors, are anti-correlated. That is not the same as the hidden variables being anti-correlated. The hidden variables contain all of the things that could affect either of the measurements; there is no separation into "hidden variables that affect Alice's measurement" and "hidden variables that affect Bob's measurement". If Alice's and Bob's measurements are perfectly anti-correlated, then there will be one set of HVs $\lambda$ that effect that anti-correlation. In other words, as Bell writes, $A(a, \lambda) = - B(a, \lambda)$: the same $\lambda$, the same measurement direction $a$, but opposite results A and B.

 

[stevendaryl]

Are you sure? In EPRB, the case that he studies in Bell (1964), the particles are anti-correlated.

I think that you have been misunderstanding the role of $\lambda$. It is supposed to represent the information that is shared between the two particles. The assumption is that Alice's result is a function $A(\alpha, \lambda)$ that depends on the shared information, $\lambda$, and Alice's setting, $\alpha$. Bob's result is a function $B(\beta, \lambda)$ taht depends on $\lambda$ and Bob's setting, $\beta$. The anti-correlation is accomplished by the fact that if $\alpha = \beta$, then

$B(\alpha, \lambda) = - A(\alpha, \lambda)$

It's not that they have different values of $\lambda$---the whole point of $\lambda$ is to explain the correlations/anti-correlations in terms of shared state information.

 

[N88]

Once again, we have:

$g_n = a_n b_n + a_n b_n' + a_n' b_n - a_n' b_n'$

We rearrange it as follows:

$g_n = a_n (b_n + b_n') + a_n' (b_n - b_n')$

Since $b_n$ and $b_n'$ are each $\pm 1$, it follows that either

• Case A: $b_n + b_n' = 0$ (if they are opposite signs), or

• Case B: $b_n - b_n' = 0$ (if they are the same sign).

So in Case A,

$g_n = a_n' (b_n - b_n') = 2 a_n' b_n = \pm 2$

In Case B,

$g_n = a_n (b_n + b_n') = 2 a_n b_n = \pm 2$

So in either case, $g_n = \pm 2$.

If for every single $n$, $g_n = \pm 2$, then obviously the average value of $g_n$ must be in the range $-2 \leq \langle g \rangle \leq +2$.

We agreed earlier that

$\langle g \rangle = \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle$

So putting those two facts together, we get Bell's inequality (or actually, the CHSH inequality)

$-2 \leq \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle \leq +2$

Thank you for this detail. But please note: to achieve this latest analysis, you have NOT used the FOUR independent equations that I gave you.

That means they have been modified. It is the reasoning/justification behind these modifications that I am seeking to understand, as spelt out in the OP. Bell does it via a mathematical assumption. I am seeking to understand its physical significance.

Before we proceed, I'd better be clear about this point: Do you see that you cannot derive your result from my four valid equations? That is, they are defining equations that are valid, both classically and quantum mechanically.

 

[N88]

The observed spins, given that both Alice and Bob choose the same orientation for their detectors, are anti-correlated. That is not the same as the hidden variables being anti-correlated. The hidden variables contain all of the things that could affect either of the measurements; there is no separation into "hidden variables that affect Alice's measurement" and "hidden variables that affect Bob's measurement". If Alice's and Bob's measurements are perfectly anti-correlated, then there will be one set of HVs $\lambda$ that effect that anti-correlation. In other words, as Bell writes, $A(a, \lambda) = - B(a, \lambda)$: the same $\lambda$, the same measurement direction $a$, but opposite results A and B.

I am using the alternative technique that Bell specifically approved: see the second paragraph following Bell (1964), eqn (3). This technique is more realistic and physically significant to me. And it changes nothing in the analyses.

 

[stevendaryl]

Thank you for this detail. But please note: to achieve this latest analysis, you have NOT used the FOUR independent equations that I gave you.

I don't know what you're talking about. We agreed that:

• $\langle g \rangle = \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle$

I proved that

• $-2 \leq \langle g \rangle \leq +2$

It follows that

• $-2 \leq \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle \leq +2$

I really have no idea what your point is. I guess I should give up, because I'm getting very frustrated.

Before we proceed, I'd better be clear about this point: Do you see that you cannot derive your result from my four valid equations?

No, I do not agree with that. I really have no idea what you are talking about. I'm using the exact same mathematics that I thought we both had agreed with.

 

[PeterDonis]

I am using the alternative technique that Bell specifically approved: see the second paragraph following Bell (1964), eqn (3).

No, you're not, because Bell explicitly says that that technique uses the same equations: "this possibility is contained in the above, since $\lambda$ stands for any number of variables and the dependencies thereon of $A$ and $B$ are unrestricted". So using this "alternative technique" doesn't change any of the equations, yet you are changing them.

 

[N88]

...

No, I do not agree with that. I really have no idea what you are talking about. I'm using the exact same mathematics that I thought we both had agreed with.

I am sorry for your frustration. If you do not agree with my statement, then please do it. That is, please: start with four equations that we agree with and derive a conclusion that we disagree with. NB: These are not the same equations that you have been using.

The four equations, valid classically and under QM are (quoting my earlier post #133): Without any assumptions, what is actually derived from measurements is four separate expectations:

• (1) $\langle a b \rangle \equiv \frac{1}{N_i} \sum^{N_i}_1 a_i b_i$
• (2) $\langle a b' \rangle \equiv \frac{1}{N_j} \sum^{N_j}_1 a_j b'_j$
• (3) $\langle a' b \rangle \equiv \frac{1}{N_k} \sum^{N_k}_1 a'_k b_k$
• (4) $\langle a' b' \rangle \equiv \frac{1}{N_l} \sum^{N_l}_1 a'_l b'_l$

To be clear, please show how these 4 equations, in the CHSH format (which I assume is what you're using), cannot exceed 2. Alternatively, what is the physical significance of the changes that you require in (1)-(4) to justify that limit of 2?

Thanks.

 

[N88]

No, you're not, because Bell explicitly says that that technique uses the same equations: "this possibility is contained in the above, since $\lambda$ stands for any number of variables and the dependencies thereon of $A$ and $B$ are unrestricted". So using this "alternative technique" doesn't change any of the equations, yet you are changing them.

Please see the next sentence re initial values. The consequence is that I make no functional change to Bell's equations; I simply use those initial values to arrive at his identical equations, or their equivalents. Specifically, I use:
$$A(a, \lambda) = - B(a, \lambda).$$

 

[PeterDonis]

my equations (1)-(4) are independent.

I don't know what your equations (1)-(4) mean, because they bring in extra indexes that I don't understand. Stevendaryl's equations (1)-(4), as given in post #130, are straightforward. Do you agree with his equations (1)-(4) in post #130? If you don't, then there's where the disagreement starts.

If you do agree with his equations (1)-(4) in post #130, then he showed in post #139 how they lead to the CHSH inequality, i.e., $|g| \le 2$.

 

[PeterDonis]

Specifically, I use:

A(a,λ)=−B(a,λ)​

But that's not what you wrote earlier; in post #138, you wrote:

Using your example

A(a,λ)=−B(a,λ)=B(a,−λ′).​

The second equality is the problem; it's not right, as both I and stevendaryl have explained.

 

[stevendaryl]

The four equations, valid classically and under QM are (quoting my earlier post #133): Without any assumptions, what is actually derived from measurements is four separate expectations:

• (1) $\langle a b \rangle \equiv \frac{1}{N_i} \sum^{N_i}_1 a_i b_i$
• (2) $\langle a b' \rangle \equiv \frac{1}{N_j} \sum^{N_j}_1 a_j b'_j$
• (3) $\langle a' b \rangle \equiv \frac{1}{N_k} \sum^{N_k}_1 a'_k b_k$
• (4) $\langle a' b' \rangle \equiv \frac{1}{N_l} \sum^{N_l}_1 a'_l b'_l$

To be clear, please show how these 4 equations, in the CHSH format (which I assume is what you're using), cannot exceed 2.

I just did that! You tell me which step in the following you don't agree with:

1. For each j, $-2 \leq a_j b_j + a_j b_j' + a_j' b_j - a_j' b_j' \leq +2$
2. If it is true for each j, then it is true for the average: $-2 \leq \langle a b + a b' + a' b - a' b'\rangle \leq +2$
3. $\langle a b + a b' + a' b - a' b'\rangle = \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle$

Just to be super-clear, in the above, the notion of average is $\langle Q \rangle \equiv \frac{1}{N} \sum_{n=1}^N Q_n$. So there is an additional assumption that:

$\frac{1}{N_1} \sum_{1,n} a_n b_n = \frac{1}{N} \sum_n a_n b_n$
(where $N_1$ is the number of times that Alice chose to measure $a$ and Bob chose to measure $b$, and $\sum_{1,n}$ means the sum over just those values of $n$)

and similarly for the other averages. The possibility that that's not the case is one of the loopholes that I mentioned in a previous post. For the average of the quantity $a_n b_n$ to depend on choices made by Alice and Bob would be a very strange situation. It's comparable to the following:

I hide 100 balls in 5000 boxes, all of which look identical on the outside. So on the average, 1 out of 50 boxes contains a ball. Then I ask you to choose 1000 boxes. You would expect that 1 in 50 of those would contain a ball, that the ratios are the same as for the full set. That isn't a logical necessity, but it is what people normally assume when they use sampling to get an idea about the likelihood of something.

 

[Mentz114]

@N88, I think I understand your problem with this.

$-2 \leq a_j b_j + a_j b_j' + a_j' b_j - a_j' b_j' \leq +2$

This is only a mathematical identity if $a, a', b, b'$ is the exactly the same thing in each term. This is not possible in practical terms so one must assume that the identical preparation and many repetitions is enough for this to hold within a small margin.

We could write
$-2 \leq a_j b_j + \bar{a}_j b_j' + a_j' \bar{b}_j - \bar{a}_j' \bar{b}_j' \leq +2$

is true iff all barred things belong to the same equivalence class as the unbarred things.

 

[rubi]

For each j, $-2 \leq a_j b_j + a_j b_j' + a_j' b_j - a_j' b_j' \leq +2$

This step doesn't work in N88's case, since you don't know that the $a_j$ in the first term is the same as the $a_j$ in the second term. These $a_j$'s come from different sequences of measurements. What you really measure is a huge list $(a_j,b_j,\alpha_j,\beta_j)$ (the experimenter has to memorize the angle settings) and the individual correlations are given by $C_{\alpha\beta}=\frac 1 {N(\alpha,\beta)} \sum _{\alpha_j=\alpha,\beta_j=\beta} a_j b_j$, i.e. you perform a sum over subsequences. For example the first $j$ such that $\alpha_j=\alpha$ and $\beta_j=\beta$ might be $j=3$ and the first $j$ such that $\alpha_j=\alpha$ and $\beta_j=\beta'$ might be $j=5$. In order to apply your inequality to the sum $C_{\alpha\beta} + C_{\alpha\beta'} + \cdots$, you need that $a_3 = a_5$, because this was used in the proof of the bound of your inequality ($a_j b_j + a_j b_j' + \cdots = a_j(b_j+b_j') + \cdots$). If you don't make such an assumption, the bound will be $4$ instead of $2$.

Your proof applies to the situation, where one measures a list $(a_j,b_j,a_j',b_j')$ of 4 spins. But in a Bell test experiment, one measures 2 spins and 2 angles instead. In order to map your proof onto the situation of a real Bell test experiment, you need assumptions on the sequence $(a_j,b_j,\alpha_j,\beta_j)$.

 

[Mentz114]

..
..
In order to map your proof onto the situation of a real Bell test experiment, you need assumptions on the sequence $(a_j,b_j,\alpha_j,\beta_j)$.

One assumption that is required is the the sequence $a$ and $\bar{a}$ etc ( see my post above) must contain the same number of 1's. Sounds unlikely but this property alone sets the correlation limits between any sequences.

 

[N88]

I just did that! You tell me which step in the following you don't agree with:

1. For each j, $-2 \leq a_j b_j + a_j b_j' + a_j' b_j - a_j' b_j' \leq +2$
2. If it is true for each j, then it is true for the average: $-2 \leq \langle a b + a b' + a' b - a' b'\rangle \leq +2$
3. $\langle a b + a b' + a' b - a' b'\rangle = \langle a b \rangle + \langle a b' \rangle + \langle a' b \rangle - \langle a' b' \rangle$

Just to be super-clear, in the above, the notion of average is $\langle Q \rangle \equiv \frac{1}{N} \sum_{n=1}^N Q_n$. So there is an additional assumption that:

$\frac{1}{N_1} \sum_{1,n} a_n b_n = \frac{1}{N} \sum_n a_n b_n$
(where $N_1$ is the number of times that Alice chose to measure $a$ and Bob chose to measure $b$, and $\sum_{1,n}$ means the sum over just those values of $n$)

and similarly for the other averages. The possibility that that's not the case is one of the loopholes that I mentioned in a previous post. For the average of the quantity $a_n b_n$ to depend on choices made by Alice and Bob would be a very strange situation. It's comparable to the following:

I hide 100 balls in 5000 boxes, all of which look identical on the outside. So on the average, 1 out of 50 boxes contains a ball. Then I ask you to choose 1000 boxes. You would expect that 1 in 50 of those would contain a ball, that the ratios are the same as for the full set. That isn't a logical necessity, but it is what people normally assume when they use sampling to get an idea about the likelihood of something.

You write: For each j, $-2 \leq a_j b_j + a_j b_j' + a_j' b_j - a_j' b_j' \leq +2.$ (SD-1)

But for the 4 equations that I gave you, the fundamental expression is:

|$a_i b_i + a_j b_j' + a_k' b_k - a_l' b_l'$| $\leq +4.$ (N88-1)

Please note, term by term, that you did not do what was requested. To be clear, my four equations, valid classically and quantum mechanically, follow. Without any assumptions, they represent what is actually derived from measurements; four separate expectations:

• (1) $\langle a b \rangle \equiv \frac{1}{N_i} \sum^{N_i}_1 a_i b_i$
• (2) $\langle a b' \rangle \equiv \frac{1}{N_j} \sum^{N_j}_1 a_j b'_j$
• (3) $\langle a' b \rangle \equiv \frac{1}{N_k} \sum^{N_k}_1 a'_k b_k$
• (4) $\langle a' b' \rangle \equiv \frac{1}{N_l} \sum^{N_l}_1 a'_l b'_l$

Please compare -- term by term -- your requirement in the heart of (SD-1) to my (N88-1); which is valid valid classically and quantum mechanically. In my terms, your equations would be valid for an ordered ensemble of classical objects that could be repeatedly tested in the same order. But (it seems to me) such objects would have no place under EPRB (Bell 1964).

So my question remains: What is the physical significance of the changes that you require for (SD-1) to be valid, please?

 

[N88]

@N88, I think I understand your problem with this.

$-2 \leq a_j b_j + a_j b_j' + a_j' b_j - a_j' b_j' \leq +2$

This is only a mathematical identity if $a, a', b, b'$ is the exactly the same thing in each term. This is not possible in practical terms so one must assume that the identical preparation and many repetitions is enough for this to hold within a small margin.

We could write
$-2 \leq a_j b_j + \bar{a}_j b_j' + a_j' \bar{b}_j - \bar{a}_j' \bar{b}_j' \leq +2$

is true iff all barred things belong to the same equivalence class as the unbarred things.

Thanks for this; I believe you do understand my problem. I trust the reply just posted (for stevendaryl) shows how my equations make your point. Critical comments on such a comparison would be welcome: but I suspect the equivalence class would need to be be the smallest such: an equality relation.

 

[Mentz114]

Thanks for this; I believe you do understand my problem. I trust the reply just posted (for stevendaryl) shows how my equations make your point. Critical comments on such a comparison would be welcome: but I suspect the equivalence class would need to be be the smallest such: an equality relation.

Thanks. [ I deleted some nonsense here].

The point is that this is a classical theorem and QM does seem to break it.

 

[N88]

Thanks. But it is possible that an experiment will ( very nearly) satisfy the conditions that the limit theorem requires. It is a loophole that ( to my knowledge) is not tested. One assumption is that all subsequences appear random - but I'm not sure what random means in this context.

I'm not at all sure re the possibility that you raise; and I'm not sure that rubi's careful checking could do the job, except with classical objects. I believe the facts go this way:

(SD-1) will be satisfied by classical objects, as suggested earlier. (N88-1) will provide a limit of 2 for classical objects and limit of 2√2 for quantum-entangled objects (eg, EPRB).

 

[Mentz114]

I'm not at all sure re the possibility that you raise; and I'm not sure that rubi's careful checking could do the job, except with classical objects. I believe the facts go this way:

(SD-1) will be satisfied by classical objects, as suggested earlier. (N88-1) will provide a limit of 2 for classical objects and limit of 2√2 for quantum-entangled objects (eg, EPRB).

I deleted the offending stuff before you posted - sorry about the confusion.

 

[Jilang]

Isn't the difference just down to the fact that identically prepared classical entities will share the same $\lambda$ but identically prepared quantum mechanical entities won't?

 

[stevendaryl]

Thanks for this; I believe you do understand my problem. I trust the reply just posted (for stevendaryl) shows how my equations make your point. Critical comments on such a comparison would be welcome: but I suspect the equivalence class would need to be be the smallest such: an equality relation.

Let me try one more time, and make it super concrete. Suppose that we repeatedly do the following 4 measurements:

1. We produce a correlated pair. Alice measures the spin of her particle along axis $a$. Bob measures along axis $b$
2. We produce a correlated pair. Alice measures along $a$, Bob measures along $b'$
3. We produce a correlated pair. Alice measures $a'$. Bob measures $b$
4. We produce a correlated pair. Alice measures $a'$. Bob measures $b'$

We do these four things over and over, N times. (So we actually produce 4N correlated pairs)

Then we compute:
$C(a,b) = \frac{1}{N} \sum_n a_n b_n$ (where $n$ ranges over 1, 5, 9, etc.)
$C(a,b') = \frac{1}{N} \sum_n a_n b_n'$ (where $n$ ranges over 2, 6, 10, etc.)
$C(a',b) = \frac{1}{N} \sum_n a_n' b_n$ (where $n$ ranges over 3, 7, 11, etc.)
$C(a', b') = \frac{1}{N} \sum_n a_n' b_n'$ (where $n$ ranges over 4, 8, 12, etc.)

Now, the hidden-variable assumption is this: Although

• nobody measured $a_n b_n$ when $n=2, 3, 4, 6, 7, 8, 10, 11, 12...$
• nobody measured $a_n b_n'$ when $n=1, 3, 4, 5, 7, 8, 9, 11, 12...$
• nobody measured $a_n' b_n$ when $n=1, 2, 4, 5, 6, 8, 9, 10, 12...$
• nobody measured $a_n' b_n'$ when $n=1, 2, 3, 5, 6, 7, 9, 10, 11...$

Those variables had definite values. So even though we don't know what the values were for some variables on some rounds, it makes sense to talk about the following averages:

1. $D(a,b) = \frac{1}{4N} \sum_n a_n b_n$
2. $D(a,b') = \frac{1}{4N} \sum_n a_n b_n'$
3. $D(a',b) = \frac{1}{4N} \sum_n a_n' b_n$
4. $D(a',b') = \frac{1}{4N} \sum_n a_n' b_n'$

where this time, all sums extend over all values of $n$ from $1$ to $4N$.

The assumption is that

1. $D(a,b) \approx C(a,b)$
2. $D(a, b') \approx C(a, b')$
3. $D(a', b) \approx C(a', b)$
4. $D(a', b') \approx C(a', b')$

That is, the assumption is that the unmeasured quantities have the same statistical properties as the measured quantities. In the limit as $N \rightarrow \infty$, it is assumed that the averages $C$ approach the averages $D$.

 

[stevendaryl]

Isn't the difference just down to the fact that identically prepared classical entities will share the same $\lambda$ but identically prepared quantum mechanical entities won't?

No, there is no assumption classically that identically prepared systems will share the same $\lambda$, only that identically prepared sequences of systems will share the same distribution on possible values of $\lambda$.

 

[rubi]

it's certainly possible to fail to recognize that the stricter inequality is true.

The stricter inequality might be true, but it can't be applied in the case N88 is talking about. If you are in the situation of a Bell test experiment, where you recorded the sequence $(a_j,b_j,\alpha_j,\beta_j)$ of 2 spins and 2 angles, you just can't apply the stricter inequality to the sum $C_{\alpha\beta}+C_{\alpha\beta'}+C_{\alpha'\beta}-C_{\alpha'\beta'}$ and hence, you won't obtain the a bound of $2$. In fact, in this situation, the bound $2$ is false and can be violated easily. Here's a list that violates it: $(1,1,\alpha,\beta)$, $(1,1,\alpha,\beta')$, $(1,1,\alpha',\beta)$, $(1,-1,\alpha',\beta')$. In that case, the sum will be equal to $4$.

You obtain the bound of $2$ only in the hypothetical situation in which CFD is not violated and you can assume that your recorded sequence $(a_j,b_j,a_j',b_j')$ consists of 4 spins in each run (even though you measure only 2 of them).

The assumption is that

1. $D(a,b) \approx C(a,b)$
2. $D(a, b') \approx C(a, b')$
3. $D(a', b) \approx C(a', b)$
4. $D(a', b') \approx C(a', b')$

That is, the assumption is that the unmeasured quantities have the same statistical properties as the measured quantities. In the limit as $N \rightarrow \infty$, it is assumed that the averages $C$ approach the averages $D$.

In the counterfactually definite situation, this can even be proved. However, if CFD is violated, then there is no reason to expect something like this to be true. The quantity $D$ can't even be meaningfully defined.

 

[stevendaryl]

Let me try one more time, and make it super concrete. Suppose that we repeatedly do the following 4 measurements:

1. We produce a correlated pair. Alice measures the spin of her particle along axis $a$. Bob measures along axis $b$
2. We produce a correlated pair. Alice measures along $a$, Bob measures along $b'$
3. We produce a correlated pair. Alice measures $a'$. Bob measures $b$
4. We produce a correlated pair. Alice measures $a'$. Bob measures $b'$

We do these four things over and over, N times. (So we actually produce 4N correlated pairs)

Then we compute:
$C(a,b) = \frac{1}{N} \sum_n a_n b_n$ (where $n$ ranges over 1, 5, 9, etc.)
$C(a,b') = \frac{1}{N} \sum_n a_n b_n'$ (where $n$ ranges over 2, 6, 10, etc.)
$C(a',b) = \frac{1}{N} \sum_n a_n' b_n$ (where $n$ ranges over 3, 7, 11, etc.)
$C(a', b') = \frac{1}{N} \sum_n a_n' b_n'$ (where $n$ ranges over 4, 8, 12, etc.)

Now, the hidden-variable assumption is this: Although

• nobody measured $a_n b_n$ when $n=2, 3, 4, 6, 7, 8, 10, 11, 12...$
• nobody measured $a_n b_n'$ when $n=1, 3, 4, 5, 7, 8, 9, 11, 12...$
• nobody measured $a_n' b_n$ when $n=1, 2, 4, 5, 6, 8, 9, 10, 12...$
• nobody measured $a_n' b_n'$ when $n=1, 2, 3, 5, 6, 7, 9, 10, 11...$

Those variables had definite values. So even though we don't know what the values were for some variables on some rounds, it makes sense to talk about the following averages:

1. $D(a,b) = \frac{1}{4N} \sum_n a_n b_n$
2. $D(a,b') = \frac{1}{4N} \sum_n a_n b_n'$
3. $D(a',b) = \frac{1}{4N} \sum_n a_n' b_n$
4. $D(a',b') = \frac{1}{4N} \sum_n a_n' b_n'$

where this time, all sums extend over all values of $n$ from $1$ to $4N$.

The assumption is that

1. $D(a,b) \approx C(a,b)$
2. $D(a, b') \approx C(a, b')$
3. $D(a', b) \approx C(a', b)$
4. $D(a', b') \approx C(a', b')$

That is, the assumption is that the unmeasured quantities have the same statistical properties as the measured quantities. In the limit as $N \rightarrow \infty$, it is assumed that the averages $C$ approach the averages $D$.

So the proof of Bell's inequality is a proof about the averages $D$. What we actually measure is a different kind of average, $C$. So that's one of the loopholes for Bell's theorem--maybe for some reason the averages $C$ are not equal to the averages $D$, and so the inequalities don't apply to the measured averages $C$.

@N88 is correct, that unless you assume that the averaging process $C$ gives approximately the same result as the theoretical averages $D$, then you can't prove Bell's inequality, and in fact, you have a much weaker inequality:

$-4 \leq C(a,b) + C(a, b') + C(a', b) - C(a', b') \leq 4$

 

[N88]

So the proof of Bell's inequality is a proof about the averages $D$. What we actually measure is a different kind of average, $C$. So that's one of the loopholes for Bell's theorem--maybe for some reason the averages $C$ are not equal to the averages $D$, and so the inequalities don't apply to the measured averages $C$.

@N88 is correct, that unless you assume that the averaging process $C$ gives approximately the same result as the theoretical averages $D$, then you can't prove Bell's inequality, and in fact, you have a much weaker inequality:

$-4 \leq C(a,b) + C(a, b') + C(a', b) - C(a', b') \leq 4$

Thanks for going "super-concrete". I trust I have it right: that $C$ denotes the measured correlations for EPRB. So, continuing to be super-concrete under EPRB (Bell 1964), and consistent with QM theory and QM calculations:

$C(a,b) = \frac{1}{N} \sum_n a_n b_n$ (where $n$ ranges over 1, 5, 9, etc.) $= -a\cdot b.$
$C(a,b') = \frac{1}{N} \sum_n a_n b_n'$ (where $n$ ranges over 2, 6, 10, etc.) $= -a\cdot b'.$
$C(a',b) = \frac{1}{N} \sum_n a_n' b_n$ (where $n$ ranges over 3, 7, 11, etc.) $= -a'\cdot b.$
$C(a', b') = \frac{1}{N} \sum_n a_n' b_n'$ (where $n$ ranges over 4, 8, 12, etc.) $= -a'\cdot b'.$

Re $D$, I'll have more to say soon.

Thanks again.

 

[Jilang]

So these are different sets? The Venn diagram that is often shown to explain the issue cannot apply then?
http://theory.physics.manchester.ac.uk/~judith/AQMI/PHYS30201se24.xhtml

 

[N88]

So these are different sets? The Venn diagram that is often shown to explain the issue cannot apply then?
http://theory.physics.manchester.ac.uk/~judith/AQMI/PHYS30201se24.xhtml

For me, the answers to your questions are Yes and Yes. The following article by d'Espagnat (1979) has Bell's (1980) endorsement. You will see the above Venn diagram developed there.

http://www.scientificamerican.com/media/pdf/197911_0158.pdf

General note: On p.158 we see these principles of local realism: (i) realism -- regularities in observed phenomena are caused by some physical reality whose existence is independent of human observers; (ii) locality -- no influence of any kind can propagate superluminally; (iii) induction -- legitimate conclusions can be drawn from consistent observations.

On p.166, in the last paragraph and continuing into p.167, we see d'Espagnat's subtle (but false) inference: that is, he ignores consistent observations re the validity of QM and the repeated verification of Bohr's well-known insight (that a measurement perturbs the measured system). Thus, in my view, the above principles of local realism remain valid.

So this is not a dispute about principles differing from the local realism of Bell and d'Espagnat. For me it's a lesson on the need to infer correctly to quantum-mechanically validated experimental results.

 

[N88]

Let me try one more time, and make it super concrete. Suppose that we repeatedly do the following 4 measurements:

1. We produce a correlated pair. Alice measures the spin of her particle along axis $a$. Bob measures along axis $b$
2. We produce a correlated pair. Alice measures along $a$, Bob measures along $b'$
3. We produce a correlated pair. Alice measures $a'$. Bob measures $b$
4. We produce a correlated pair. Alice measures $a'$. Bob measures $b'$

We do these four things over and over, N times. (So we actually produce 4N correlated pairs)

Then we compute:
$C(a,b) = \frac{1}{N} \sum_n a_n b_n$ (where $n$ ranges over 1, 5, 9, etc.)
$C(a,b') = \frac{1}{N} \sum_n a_n b_n'$ (where $n$ ranges over 2, 6, 10, etc.)
$C(a',b) = \frac{1}{N} \sum_n a_n' b_n$ (where $n$ ranges over 3, 7, 11, etc.)
$C(a', b') = \frac{1}{N} \sum_n a_n' b_n'$ (where $n$ ranges over 4, 8, 12, etc.)

Now, the hidden-variable assumption is this: Although

• nobody measured $a_n b_n$ when $n=2, 3, 4, 6, 7, 8, 10, 11, 12...$
• nobody measured $a_n b_n'$ when $n=1, 3, 4, 5, 7, 8, 9, 11, 12...$
• nobody measured $a_n' b_n$ when $n=1, 2, 4, 5, 6, 8, 9, 10, 12...$
• nobody measured $a_n' b_n'$ when $n=1, 2, 3, 5, 6, 7, 9, 10, 11...$

Those variables had definite values. So even though we don't know what the values were for some variables on some rounds, it makes sense to talk about the following averages:

1. $D(a,b) = \frac{1}{4N} \sum_n a_n b_n$
2. $D(a,b') = \frac{1}{4N} \sum_n a_n b_n'$
3. $D(a',b) = \frac{1}{4N} \sum_n a_n' b_n$
4. $D(a',b') = \frac{1}{4N} \sum_n a_n' b_n'$

where this time, all sums extend over all values of $n$ from $1$ to $4N$.

The assumption is that

1. $D(a,b) \approx C(a,b)$
2. $D(a, b') \approx C(a, b')$
3. $D(a', b) \approx C(a', b)$
4. $D(a', b') \approx C(a', b')$

That is, the assumption is that the unmeasured quantities have the same statistical properties as the measured quantities. In the limit as $N \rightarrow \infty$, it is assumed that the averages $C$ approach the averages $D$.

Continuing your appreciated "super-concrete" theme: Two posts above, I agreed with your analysis re $C$ and gave what I believe to be results agreed by us jointly. When it comes to $D$, imho we are dealing with counterfactuals; ie, counterfactual statements of the "IF this … THEN this" kind. Note that counterfactuals are NOT contrary (fake) facts. They are facts about what would have happened IF we had done something different.

So, IF we a dealing with quantum objects (where a decoherent interaction occurs), THEN only one of your four $D$ examples can be valid. Thus, in comparison with the $C$ factors:

1. IF we had conducted $D-1$ under EPRB, THEN the result would have been: $D(a,b) = C(a,b)$.
2. IF we had conducted $D-2$ under EPRB, THEN the result would have been: $D(a,b') = C(a,b')$.
3. IF we had conducted $D-3$ under EPRB, THEN the result would have been: $D(a',b) = C(a',b)$.
4. IF we had conducted $D-4$ under EPRB, THEN the result would have been: $D(a',b') = C(a',b')$.

However, IF you wish to insist that all four $D$ examples are jointly permissible: well then (in my view) you are (in fact) invoking ordered and unperturbed classical objects. Thus, whereas the preceding EPRB results would have delivered a maximised CHSH result of 2√2, your "jointly valid" and therefore classical $D$ examples would deliver a maximised CHSH result of 2.

My above reply to Jilang explains, for me, the departure of Bell's analysis from local-realism in a quantum setting; ie, from the EPRB setting that is the focus of Bell (1964). For me, the above (i)-(iii) Bell/d'Espagnat principles of local-realism "should" be acceptable to all local realists.

 

[stevendaryl]

Continuing your appreciated "super-concrete" theme: Two posts above, I agreed with your analysis re $C$ and gave what I believe to be results agreed by us jointly. When it comes to $D$, imho we are dealing with counterfactuals; ie, counterfactual statements of the "IF this … THEN this" kind. Note that counterfactuals are NOT contrary (fake) facts. They are facts about what would have happened IF we had done something different.

The realist assumption is that the variables $a_n, a_n', b_n, b_n'$ exist whether we measure them or not. So under this assumption, the averages D are not counterfactual---they are actual averages of unmeasured quantities.

 

[stevendaryl]

In the counterfactually definite situation, this can even be proved. However, if CFD is violated, then there is no reason to expect something like this to be true. The quantity $D$ can't even be meaningfully defined.

We're talking in the context of Bell's inequality, which was derived under the assumption that we had a local realistic theory. With such a theory, the variables $a_n, a_n', b_n, b_n'$ exist independently of whether anybody measures them, and so it makes sense to talk about averages of them.

You can certainly reject that assumption, which means rejecting the assumption that Bell proved false, anyway.

 

[Jilang]

The realist assumption is that the variables $a_n, a_n', b_n, b_n'$ exist whether we measure them or not. So under this assumption, the averages D are not counterfactual---they are actual averages of unmeasured quantities.

Are we able to assume that realistic properties do exist independent of measurement, but that the act of measurement changes their distribution?

 

[DrChinese]

Are we able to assume that realistic properties do exist independent of measurement, but that the act of measurement changes their distribution?

That's not possible if there are to be perfect correlations. In effect, the measurement apparatus itself cannot be more than a static participant (i.e. precisely the same impact on both sides). Else there would be some variability introduced at Alice or Bob's setups. And the outcomes would not match, as they actually do.

 

[stevendaryl]

Are we able to assume that realistic properties do exist independent of measurement, but that the act of measurement changes their distribution?

That's certainly a logical possibility, but there is no way to explain the perfect anti-correlations achieved in EPR that way.