Physical significance of a.σ in expectation -E(a.σ b.σ)?

[stevendaryl]

You also make this point in post #3, "So what guarantees that Bob will get -1?". One way is to only count the pairs that match within a specific short time-frame and assume any that do not match are noise or were not entangled in the first place.

Yes, that's a different type of loophole. Bell's analysis of EPR with twin pairs is about an idealized experiment in which (1) every twin-pair that is produced is measured, and none are missed and (2) no electrons/positrons are erroneously detected that are not from a twin pair, and (3) the experimenters can unambiguously determine which particles at one detector correspond to which particles at the other detector. In a real experiment, none of these is guaranteed to be true. I don't know much about efforts to analyze and close these loopholes.

 

[Jilang]

The idea behind the following picture is this:

Alice picks axis $\vec{a}$ and Bob picks axis $\vec{b}$. You can always choose a coordinate system such that $\vec{a}$ is in the x-direction and $\vec{b}$ is at an angle $\theta$ away from $\vec{a}$ in the x-y plane.

The intrinsic spin vector, $\vec{\lambda}$, can be in any direction, but we can write it as $\vec{\lambda} = \vec{\lambda_z} + \vec{\lambda_{xy}}$, where $\lambda_z$ is the component of $\lambda$ in the z-direction, and $\vec{\lambda_{xy}}$ is the component in the x-y plane. For the purposes of determining whether Alice and Bob get spin-up or spin-down, only $\lambda_{xy}$ is relevant, so in the diagram, $\lambda$ just refers to this component in the x-y plane.

So if $\lambda$ is in the x-y plane, we assume that it has equal likelihood of pointing anywhere in the x-y plane.

So let $A$ be Alice's result and let $B$ be Bob's result. The first picture shows how Alice's result depends on $\lambda$: If $\lambda$ lies anywhere in the yellow region, then Alice gets +1. Otherwise, she gets -1. The second picture shows how Bob's result depends on $\lambda$: If $\lambda$ is in the red region, Bob gets +1, and otherwise, he gets -1.

The third picture shows the joint probabilities: Alice and Bob get the same result if $\lambda$ is in the orange and white regions, which occur with probability $\theta/\pi$. Alice and Bob get opposite results if $\lambda$ is in the yellow or red regions, which occur with probability $(1 - \theta/\pi)$. So the product $A B$ is +1 with probability $\theta/\pi$ and $-1$ with probability $1-\theta/\pi$. So the expectation value of $A B$ is $(+1)(\theta/\pi) + (-1) (1-\theta/\pi) = -1 + 2\theta/\pi$

View attachment 114570

Thanks for this. I can't see though how you can factor out the orthogonal component in the way you are suggesting. Angles in 2 dimensions are closer than in 3 CosA=CosB CosC

 

[N88]

First of all, you do realize that the model you were sketching, in which you assume that there is a function $\hat{a} \circ \vec{\lambda}$ that yields Alice's result, OBEYS CFD? So your model has nothing to do with exploiting the CFD loophole.

stevendaryl, Is it OK to continue discussing my model as physics? I am concerned that it is not. It came into this thread via the subject of local realism. And it aims for the same results as QM. But it is personal research. And I can't see where it obeys CFD. Thanks, N88.

 

[PeterDonis]

Is it OK to continue discussing my model as physics?

At this point I would say no. We should stick to the model(s) described in Bell's paper. Are there any open questions concerning that paper at this point?

 

[N88]

At this point I would say no. We should stick to the model(s) described in Bell's paper. Are there any open questions concerning that paper at this point?

In terms of the OP, I understand the following general points:

$(\hat{a}\cdot\boldsymbol{\sigma}_{1})$ represents the 'dot-product' of an operator $\hat{a}$ with $\boldsymbol{\sigma}_{1}$, a Pauli 'vector', a vector of Pauli matrices.

Outcomes include the expectations: $\left\langle \hat{a}\cdot\boldsymbol{\sigma}_{1}\right\rangle =0;<br /> <br /> \left\langle (\hat{a}\cdot\boldsymbol{\sigma}_{1})^{2}\right\rangle =1$;

$\left\langle{\sigma}_{x1}^{2}\right\rangle=\left\langle{\sigma}_{y1}^{2}\right\rangle=\left\langle{\sigma}_{z1}^{2}\right\rangle=\left\langle{\sigma}_{1}^{2}\right\rangle/3=1$.

So the question arises: do I understand aright? Thanks.

 

[PeterDonis]

$(\hat{a}\cdot\boldsymbol{\sigma}_{1})$ represents the 'dot-product' of an operator $\hat{a}$ with $\boldsymbol{\sigma}_{1}$, a Pauli 'vector', a vector of Pauli matrices.

No, $\hat{a}$ is a unit vector pointing in a particular direction. The result of this dot product is an operator that represents measuring spin about the axis $\hat{a}$.

Outcomes include the expectations: $\left\langle \hat{a}\cdot\boldsymbol{\sigma}_{1}\right\rangle =0$; $\left\langle (\hat{a}\cdot\boldsymbol{\sigma}_{1})^{2}\right\rangle =1$ ;

These look ok to me.

$\left\langle{\sigma}_{x1}^{2}\right\rangle=\left\langle{\sigma}_{y1}^{2}\right\rangle=\left\langle{\sigma}_{z1}^{2}\right\rangle=\left\langle{\sigma}_{1}^{2}\right\rangle/3=1$.

I don't know about these, since no unit vector $\hat{a}$ representing the direction of the spin axis to be measured is included.

 

[stevendaryl]

stevendaryl, Is it OK to continue discussing my model as physics? I am concerned that it is not. It came into this thread via the subject of local realism. And it aims for the same results as QM. But it is personal research. And I can't see where it obeys CFD. Thanks, N88.

If $\circ$ is a function (that is, $\vec{a} \circ \vec{\lambda}$ always gives the same result for the same values of $\vec{a}$ and $\vec{\lambda}$), then your model satisfies CFD: If Alice had chosen $\vec{a'}$ instead of $\vec{a}$, her result would have definitely been $\vec{a'} \circ \vec{\lambda}$.

If $\circ$ is not a function, if the result is not uniquely determined by $\vec{a}$ and $\vec{\lambda}$, then as I already said, the fact that Alice and Bob always get opposite values is not explained by the use of $\circ$.

 

[stevendaryl]

I don't know about these, since no unit vector $\hat{a}$ representing the direction of the spin axis to be measured is included.

Well, I interpreted $\sigma_{x1}$ as the operator $\hat{a} \cdot \vec{\sigma_1}$ in the case $\hat{a} = \hat{x}$.

 

[stevendaryl]

Thanks for this. I can't see though how you can factor out the orthogonal component in the way you are suggesting. Angles in 2 dimensions are closer than in 3 CosA=CosB CosC

The three-dimensional way of seeing it is by considering a sphere of radius 1, which you can think of as a globe. Every unit vector in space corresponds to a unique point on the surface of the sphere. For any point on the sphere, there is a corresponding hemisphere consisting of the half of the sphere that is closest to that point. For example, on the globe, the hemisphere corresponding to the North Pole is the northern hemisphere.

So in terms of the sphere, the rules for computing Alice's and Bob's results are:

• If $\lambda$ lies in the hemisphere corresponding to Alice's setting $\hat{a}$, then she gets +1.
• If it is in the other hemisphere, she gets -1.
• If $\lambda$ lies in the hemisphere corresponding to Bob's setting $\hat{b}$, then he gets -1 (because his particle is anti-correlated with Alice's, meaning it has the opposite sign for results).
• Otherwise, Bob gets +1.

So we have two hemispheres: the one corresponding to Bob's setting, and the one corresponding to Alice's setting. There are 4 possible cases for $\lambda$:

1. $\lambda$ is in the intersection of Alice's hemisphere and Bob's hemisphere. In this case, Alice gets +1, and Bob gets -1, so the product of their results is -1.
2. $\lambda$ is in Alice's hemisphere, but not Bob's. In this case, both get +1. So the product is +1
3. $\lambda$ is in Bob's hemisphere, but not Alice's. In this case, both get -1. So the product is +1.
4. $\lambda$ is in neither hemisphere. In this case, Alice gets -1, and Bob gets +1. So the product is -1.

So this divides the sphere up into 4 regions. Let $\mathcal{A}$ be the area of the intersection of Alice's and Bob's hemispheres. Alice and Bob's hemispheres each have area $2 \pi$ (a full sphere of radius 1 has area 4 $\pi$, so a hemisphere has area $2 \pi$) Then

• Region 1 has area $\mathcal{A}$.
  
• Region 2 has area $2 \pi - \mathcal{A}$ (the area of Alice's hemisphere, minus the intersection with Bob's)
  
• Region 3 has area $2 \pi - \mathcal{A}$ (the area of Bob's hemisphere, minus the intersection with Alice's)
  
• Region 4 has area $\mathcal{A}$ (This might not be obvious, but it's true.)

So if we assume that $\lambda$ is equally likely to point to any spot on the sphere, then the probability of lambda falling in each region is given by $\frac{area(region)}{4 \pi}$. So the probability that the product of Alice's result and Bob's result will be +1 is just the sum of the areas for regions 2 and 3 divided by 4 $\pi$. So that's $\frac{4 \pi - 2 \mathcal{A}}{4\pi} = 1 - \frac{\mathcal{A}}{2 \pi}$. The probability that the product of their results will be -1 is the just the sum of the areas for regions 1 and 4 divided by 4 $\pi$. So that's $\frac{2 \mathcal{A}}{4 \pi} = \frac{\mathcal{A}}{2 \pi}$. So the expectation value of the product is:

$\langle A B \rangle = (+1)(1 - \frac{\mathcal{A}}{2 \pi}) + (-1)\frac{\mathcal{A}}{2\pi} = 1 - \frac{\mathcal{A}}{\pi}$

So the final step is to calculate $\mathcal{A}$. This is easiest to calculate if you choose a coordinate system so that $\hat{a}$ and $\hat{b}$ both lie on the equator, at a distance of $\theta$ apart. It should be obvious (I hope) that $\mathcal{A}$ is proportional to $\theta$. If you work it out, it turns out that $\mathcal{A} = 2 \theta$. So once again, you get:

$\langle A B \rangle = 1 - \frac{2 \theta}{\pi}$

It's the same result as I derived for the 2D case.

 

[stevendaryl]

At this point I would say no. We should stick to the model(s) described in Bell's paper. Are there any open questions concerning that paper at this point?

The relevant point about the model of @N88 is that it is exactly the type of model that Bell was proving his theorem about. So his details are completely unimportant, since Bell proved a fact about all such models.

 

[PeterDonis]

The relevant point about the model of @N88 is that it is exactly the type of model that Bell was proving his theorem about.

I agree, but I'm not sure @N88 does. But whether he does or not, the discussion here is about Bell's paper, so models that are not the type that Bell was talking about in his paper are off topic.

 

[stevendaryl]

I agree, but I'm not sure @N88 does. But whether he does or not, the discussion here is about Bell's paper, so models that are not the type that Bell was talking about in his paper are off topic.

Right.

 

[Jilang]

Thank you Steven Daryl for post #69. I have been drawing out the orange segments and I have it now!

 

[N88]

No, $\hat{a}$ is a unit vector pointing in a particular direction. The result of this dot product is an operator that represents measuring spin about the axis $\hat{a}$.

These look ok to me.

I don't know about these, since no unit vector $\hat{a}$ representing the direction of the spin axis to be measured is included.

Thank you. In terms of the OP, how's this:

$(\hat{a}\cdot\boldsymbol{\sigma}_{1})$ represents the 'dot-product' of a unit-vector $\hat{a}$ with $\boldsymbol{\sigma}_{1}$, the Pauli 'vector', a vector of Pauli matrices.

$\boldsymbol{\sigma}_{1}=\{{\sigma}_{x1},{\sigma}_{y1},{\sigma}_{z1}\}$. (1)

To be clear to beginners (like myself), I call it a 'dot-product' because, although $\boldsymbol{\sigma}_{1}$ is not a normal vector, the product yields:

$\hat{a}\cdot\boldsymbol{\sigma}_{1}$ = ${a}_x{\sigma}_{x1}+{a}_y{\sigma}_{y1}+{a}_z{\sigma}_{z1}$. (2)

[Peter, re this from you: "The result of this 'dot-product' is an operator that represents measuring spin about the axis $\hat{a}$." Is that still the meaning of (2)?]

Outcomes include the expectations: $\left\langle \hat{a}\cdot\boldsymbol{\sigma}_{1}\right\rangle =0; (3)<br /> <br /> \left\langle (\hat{a}\cdot\boldsymbol{\sigma}_{1})^{2}\right\rangle =1$. (4)

$\left\langle{\sigma}_{x1}^{2}\right\rangle=\left\langle{\sigma}_{y1}^{2}\right\rangle=\left\langle{\sigma}_{z1}^{2}\right\rangle=\left\langle{\sigma}_{1}^{2}\right\rangle/3=1$. (5)

Thanks.

Well, I interpreted $\sigma_{x1}$ as the operator $\hat{a} \cdot \vec{\sigma_1}$ in the case $\hat{a} = \hat{x}$.

Does this fit with what I've summarised above? Thanks

 

[PeterDonis]

Peter, re this from you: "The result of this 'dot-product' is an operator that represents measuring spin about the axis $\hat{a}$." Is that still the meaning of (2)?

Yes.

 

[N88]

If $\circ$ is a function (that is, $\vec{a} \circ \vec{\lambda}$ always gives the same result for the same values of $\vec{a}$ and $\vec{\lambda}$), then your model satisfies CFD: If Alice had chosen $\vec{a'}$ instead of $\vec{a}$, her result would have definitely been $\vec{a'} \circ \vec{\lambda}$.

$\circ$ is a function: it always gives the same result for the same $\vec{\lambda}$. IF Alice had measured under the detector-orientation $\vec{a'}$ instead of $\vec{a}$, THEN her result would have definitely been $\vec{a'} \circ \vec{\lambda}$.

Please detail your definition of CFD in this situation. Thanks.

 

[PeterDonis]

Please detail your definition of CFD in this situation.

He already did, in the very statement you quoted.

As I've already said, further discussion of your personal model is off topic.

 

[N88]

He already did, in the very statement you quoted.

As I've already said, further discussion of your personal model is off topic.

Can I discuss CFD here?

 

[PeterDonis]

Can I discuss CFD here?

In connection with the model described in Bell's paper, yes.

 

[stevendaryl]

$\circ$ is a function: it always gives the same result for the same $\vec{\lambda}$. IF Alice had measured under the detector-orientation $\vec{a'}$ instead of $\vec{a}$, THEN her result would have definitely been $\vec{a'} \circ \vec{\lambda}$.

Please detail your definition of CFD in this situation. Thanks.

That is the definition of CFD. A model satisfies CFD if it gives definite answers to questions of the type: "If the experimenter had done A instead of B, her result would have been X instead of Y."

 

[N88]

That is the definition of CFD. A model satisfies CFD if it gives definite answers to questions of the type: "If the experimenter had done A instead of B, her result would have been X instead of Y."

This is rushed but it should get us to a point that we can progress from.

OK; thank you. But that is a different form of CFD to the one Bell proceeds under.*

Note that my CFD is licensed under EPRB by Bell (1964), eqns (1) and (13). Which means this: I can convert my CFD setting to an experimental test and it passes adequately. So mine is that special form of CFD that can be adequately proven; in this case consistent with Bell's model.

*A second variant of CFD can be adequately disproven. Bell demonstrates that when he derives his inequalities under EPRB and they are experimentally refuted. I do not use this variant and I am not conflicted by AAD and nonlocality: whereas, in his final year (1990), Bell talked about his dilemma on this front. Further, he was confident that someone would come up with answer; even that it might show that he had been rather silly.**
** http://www.quantumphil.org./Bell-indeterminism-and-nonlocality.pdf

Further, I share his motivation: "It is this possibility, of a homogeneous account of the world, which is for me the chief motivation of the study of the so-called ‘hidden variable' possibility,” Bell (Speakable and Unspeakable, 2004:28-29). So, all in all, I'm a keen student of Bell and keen to learn the more about the QM aspects of his search.

Therefore: I have no wish to breach PF policies about private studies; etc. But in answering past questions I had need to refer to some components of it; all based on Bell's model but not widely accepted.

So, as to the way ahead: I suggest that I only respond to specific questions and do not move beyond them; ie, I let comments like this slide by: So [N88's] details are completely unimportant, since Bell proved a fact about all such models.

Excuse rush here; and I do appreciate and welcome our exchanges; I learn much from such. Hope this is OK.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

my CFD

Is off topic here. The term "counterfactual definiteness" does not appear in Bell's paper at all, and Bell's paper is the topic of discussion. At this point we have wandered very far from the OP of this thread, and the question in the OP has been thoroughly discussed. The thread will remain closed.