# I What is the physical significance of Bell's math?

Tags:
1. Feb 14, 2017

### N88

Bell (1964) http://cds.cern.ch/record/111654/files/vol1p195-200_001.pdf has 3 unnumbered equations following his equation (14). Let them be (14a)-(14c). Bell then uses his equation (1) to move from (14a)-(14b). It seems to me that he uses this:

[A(b,λ)]2 = 1. (X)

Now Bell (page 195) does not mind whether λ is continuous or discrete. So if we let λ be discrete, Bell needs this:

[A(bi)][A(bj)] = 1; (Y)

because each λ is drawn from a different run of the experiment (N times with λi from the test with the detectors set at a and b; N times with λj from the other test with the detectors set at a and c). So (with i = 1, 2, ..., N; j = N+1, N+2, …, 2N), is Bell assuming that he has a set of particles that he can test twice, and in the same order, in each test? So then λi = λj every time?

And is such an assumption in keeping with EPR and EPRB, the paper and the experiment that he is studying?

Because if λi ≠ λj:

[A(bi)][A(bj)] = ± 1; (Z)

and then his (14a) ≠ (14b).

Thank you.
Edited to fix brackets.

2. Feb 15, 2017

### Zafa Pi

You are getting at something important. However, once you accept (14a), (14b) follows whether you are using sums or integrals Y would be [A(bk)]2 = 1.

What his assumption of the hidden variables λ allows him to do in (14a) is say that for a particular instance, say, λ0 we get the same value A(a0) whether the B detectors were set at b or c. This is compatible with EPR, the reality facing A (and hence which λ) is unaffected by what B is doing. This sometimes goes by the name of realism.
Equivalently, if we made the measurement at instance λ0 and get values A(a0) and B(b0) there would still have been a value B(c0) even though it was never measured. This goes by the name counterfactual definite (CFD).

Now you want to say that we shouldn't be allowed to use the same λ0 for A(a0) when the B detectors are at b or c. Different runs you say. Well what you're doing is objecting to the consequences of hidden variables (or realism, or CFD). You're in good company - except for those that deny locality, i.e. A does affect B, some faster than light phenomena, spooky action at a distance.

Last edited: Feb 15, 2017
3. Feb 15, 2017

### N88

Thanks Zafa Pi; if I understand you correctly, I am happy to be in that good company. That is, in company with those who deny nonlocality, FTL-causality, spooky-action, etc. But I wonder:

1. Is CFD is being properly used here. To me, as a realist, the counter-fact would be this: IF we had tested B(c0) instead of B(b0), THEN the result would have been consistent with A(a0). But there would be no value B(b0); for what did not happen did not happen. In other words, the value B(b0) needs be produced via a decoherent interaction between λ0 and the detector-field represented by b -- and that did not happen.

2. However, putting such complexities aside for the moment: there appears to be a simpler resolution of my difficulty.

That is, we can rightly (by which I mean: without controversy) say that Bell's theorem applies to any setting that satisfies his assumption [A(b,λ)]2 = 1. (X) Full stop.

Then, since all classical situations known to me do just that -- they satisfy (X) -- Bell's theorem is a valid limit on all such classical situations. Full stop.

But EPRB, as studied in Bell (1964), is not such a situation. So I am not bound to accept the popular belief that Bell's theorem is relevant to EPRB, or to QM in general.

Hopefully: this allows me to remain a hopeful local-realist in good standing with that good company?

4. Feb 15, 2017

### DrChinese

If you don't share the definition of "realism" per EPR (their "elements of reality"), then naturally you disagree about Bell.

Not too many will be standing with you, but there are always a determined few.

5. Feb 15, 2017

### Strilanc

I'm not sure why you think there's any assumption about the product of two different measurement axes equaling to one. There's not. And it's very strange for you to say you're a realist when you used an argument talking about how measurements that happen to not happen aren't well defined.

Maybe it will help if I break down Bell's steps to reach 14(b) more than his paper does.

First, we assume there is some hidden variable $\lambda$ that determines the measurement outcomes for both Alice and Bob no matter what direction $v$ they each measure. By experiment, we know the measurement result must always be +1 or -1:

$$\forall v: A_\lambda(v) \in \{-1, +1\}$$ $$\forall v: B_\lambda(v) \in \{-1, +1\}$$
Furthermore, by experiment, we know that when Alice and Bob measure in the same direction then the measurement outcomes must be opposite:

$$\forall v: A_\lambda(v) = -B_\lambda(v)$$
If Alice measures along $a$ and Bob measures along $b$, and then they multiply their results together, they get the parity measurement result $A_\lambda(a) \cdot B_\lambda(b)$ which will also either be -1 or +1. We assume the observed probability distribution $P$ of this parity measurement result is determined by some hidden, but consistent across experiments, probability distribution $p$ of $\lambda$:

$$\forall a, b: P(a, b) = \sum_\lambda p(\lambda) \cdot A_\lambda(a) \cdot B_\lambda(b)$$
**NO MORE ASSUMPTIONS ARE INTRODUCED BEYOND THIS POINT. JUST THE MATH OF SUMS.**

Using the fact that $A$ is opposite to $B$, we can rewrite the above equation in terms of just $A$:

$$\forall a, b: P(a, b) = -\sum_\lambda p(\lambda) \cdot A_\lambda(a) \cdot A_\lambda(b)$$
For compactness, I'm going to shorten $A_\lambda(x)$ into just $x_\lambda$ for various symbols $x$. The compact version of the above equation is:

$$\forall a, b: P(a, b) = -\sum_\lambda p(\lambda) a_\lambda b_\lambda$$
Now consider what happens when we compute the difference in predicted probabilities between two possible observations:

$$P(a, b) - P(a, c)$$
We expand the definition inline:

$$\forall a, b, c: P(a, b) - P(a, c) = \left(-\sum_\lambda p(\lambda) a_\lambda b_\lambda\right) - \left(-\sum_\lambda p(\lambda) a_\lambda c_\lambda\right)$$
Because the two sums are over the same set, and addition is associative and commutative, we can merge the sums:

$$\forall a, b, c: P(a, b) - P(a, c) = -\sum_\lambda \big(p(\lambda) a_\lambda b_\lambda - p(\lambda) a_\lambda c_\lambda\big)$$
We factor out $p(\lambda) a_\lambda$ and flip the subtraction to cancel out the leading negation:

$$\forall a, b, c: P(a, b) - P(a, c) = \sum_\lambda p(\lambda) a_\lambda \left(c_\lambda - b_\lambda\right)$$
Now, because $b_\lambda$ is either -1 or +1, we can multiply by $b_\lambda^2=1$ without changing the computed result:

$$\forall a, b, c: P(a, b) - P(a, c) = \sum_\lambda p(\lambda) a_\lambda b_\lambda^2 \left(c_\lambda - b_\lambda\right)$$
We keep one $b_\lambda$ outside, and distribute the other one over the subtraction:

$$\forall a, b, c: P(a, b) - P(a, c) = \sum_\lambda p(\lambda) a_\lambda b_\lambda \left(b_\lambda c_\lambda - b_\lambda b_\lambda\right)$$
Again, we know that $b_\lambda b_\lambda = 1$, so we can simplify:

$$\forall a, b, c: P(a, b) - P(a, c) = \sum_\lambda p(\lambda) a_\lambda b_\lambda \left(b_\lambda c_\lambda - 1\right)$$
This last equation is the one you were saying we couldn't reach without assuming that $A_\lambda(x) \cdot A_\lambda(y) = 1$ for $x \neq y$. But notice that I never made that assumption. I only ever assumed that $A_\lambda(x)^2 = 1$.

It's true that, in practice, you will experimentally measure the difference in predicted probabilities by doing many runs of an experiment measuring each part. But that doesn't change the fact that the math should still give the right answer. If the system was really like a probability distribution over a hidden variable, we'd be able to sample the difference in probabilities by sampling each probability and then subtracting.

6. Feb 15, 2017

### DrChinese

I couldn't figure that one out either. By my reading, N88 rejects CFD and is not a realist. Nothing wrong with that position, but usually they don't call themselves local realists.

But hey, people can label themselves however they like.

7. Feb 15, 2017

### N88

With Bell, I share d'Espagnat's definition of realism: regularities in observed phenomena are caused by some physical reality whose existence is independent of human observers.

8. Feb 15, 2017

### N88

Thanks Strilanc, much appreciated. 2 points:

1. I AM not sure why you (Strilanc) think there's any assumption about the product of two different measurement axes equaling to one!

2. Perhaps I'm confused with this next? You provide 13 equations. Numbering them (1)-(13), I'm not sure how to interpret (5). You appear to be specifying a probability P that can take negative values?

Are you using the same notation as Bell? Because his P denotes an expectation, not a probability. Then, for a less confusing more compact notation, you could use $$\left\langle AB\right\rangle$$ to denote an expectation.

9. Feb 15, 2017

### DrChinese

That isn't useful in discussions of Bell, and makes any attempt to associate things with generally accepted science difficult, if not impossible. But you are certainly welcome to your opinion.

And it certainly would be ridiculous to assert your opinion on that overlaps Bell's paper in any way. What Bell said at later times, in varying contexts, is not really relevant to Bell's Theorem.

10. Feb 15, 2017

### N88

Being endorsed by Bell and d'Espagnat, I thought I'd be on safe grounds. What definition do you prefer?

11. Feb 15, 2017

### DrChinese

The norm is to go back to EPR:

1. If a physical quantity of a system can be predicted with certainty without disturbing that system, there must be an element of reality associated with it.
2. It would be unreasonable to require that ALL such elements be able to be simultaneously predicted, to accord them status as elements of reality.

The combination of those leads to the hypothesis that quantum properties are counterfactually definite - what is often called hidden variables but certainly would be considered predetermined prior to measurement. This is what Bell went on to attack, and he expressed the hypothesis of CFD in the specific equations you are asking about at the start of this thread. Note the title of his paper: "On the Einstein Podolsky Rosen Paradox".

So no, your quote is quite a ways from the Bell paper.

12. Feb 15, 2017

### Strilanc

Blergh, right, it's an expected value not a probability. I should have used $E$ and not said "probability distribution" so much.

13. Feb 15, 2017

### Zafa Pi

I am glad you are happy. These are trying times.
I am using it as in Wikipedia: In quantum mechanics, counterfactual definiteness (CFD) is the ability to speak meaningfully of the definiteness of the results of measurements that have not been performed
There is no assumption here, [A(b,λ)] = + or - 1 so the square is 1.
I'm not following you here. (X) is true no matter what. You are bound to accept the popular belief, you've just failed to notice the chains around you.
BEWARE! There is no good company of local-realists. At worst they are evil, at best they are old-fashioned.

14. Feb 16, 2017

### stevendaryl

Staff Emeritus
I don't understand that at all. Bell proved that for a certain wide category of theories, correlations between distant (causally disconnected) measurements must obey a certain inequality. Quantum mechanics does not obey the inequality. Therefore, QM is not secretly one of those theories. QED

To say that Bell's theorem is not relevant to QM because QM isn't the type of theory to which Bell's proof applies seems weird. That conclusion is the whole point of his theorem.

15. Feb 16, 2017

### Zafa Pi

I totally agree with your statement. But Bell wrote his paper in 1964 before even a hint of testing. Since 1981 that has changed. Let me explain the relevance.
Essentially every time there is a discussion of the Bell Business in this Forum people get hung up in the details of QM, which misses the point.

The derivation of a Bell Theorem has nothing to do with QM. (and by the way the later theorems, e.g. CHSH, GHZ etc. are far easier to follow than Bell's own given in this thread, though of course he was the first to make the brilliant observation)

After a Bell Theorem has been presented it can be pointed out that lab tests refute the inequality, leaving QM out of it. (As a side note it could be pointed out that QM predicts the the lab results.) I have personally found that this approach keeps the focus on the essential question:
What are the assumptions (hypotheses) of Bell's Theorem that lead to a conflict with reality?

16. Feb 16, 2017

### stevendaryl

Staff Emeritus
Well, I agree with that.

17. Feb 16, 2017

### N88

I look forward to answers to Zafa Pi's question: What are the assumptions (hypotheses) of Bell's Theorem that lead to a conflict with reality?

DrChinese? Strilanc? stevendaryl? With my thanks in advance.

18. Feb 16, 2017

### Zafa Pi

19. Feb 16, 2017

### stevendaryl

Staff Emeritus
Bell's notion of a local realistic theory is pretty clear to me. What's hard for me to understand is what would count as a non-realistic theory. Presumably a completely relational model, where there is no objective answer to a question such as "What result did Bob get for his measurement", but there are only answers relative to a particular observer.

As for Bell's assumptions, he makes it all clearer in an essay called "The Theory of Local Beables".

Roughly speaking, a local realistic model means to me that
• there is a physical notion of the "state" of some little region of the universe at a particular time
• the future state of one little region depends only on the current state of that region and neighboring regions
• when you perform a measurement, the outcome reveals facts about the local state of the region where the measurement was performed (the region including both the measuring device and the system being measured)
The minimalist interpretation of quantum mechanics is not a local realistic model, because there is no notion of the state of a region. There is an overall state of the entire system under consideration, but this state is nonlocal.

20. Feb 17, 2017

### DrChinese

The assumption of counterfactual definiteness embodied in Bell's statement follows [14] in which Bell makes the assumption:

"It follows that c is another unit vector"

If it weren't assumed, the rest of the reasoning would not work.