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

[stevendaryl]

I said that I was going to drop out, but after thinking about the model that @N88 was sketching, I realize that it is almost exactly the model Bell considered in "Speakable and Unspeakable in Quantum Mechanics". Bell didn't talk in terms of the spin vector rotating, but it amounts to the same thing:

• Assume that there is an intrinsic spin vector $\vec{\sigma}$ associated with each spin-1/2 particle.
• If you measure the particle's spin along axis $\hat{a}$ then you get +1, if the angle between $\hat{a}$ and $\vec{\sigma}$ is less than 90 degrees, and -1 otherwise.
• In correlated twin-pairs, if one particle has intrinsic spin $\vec{\sigma}$, then the other particle has spin $-\vec{\sigma}$.

So this is the same as the model of @N88, with the specific choice:

$\hat{a} \circ \vec{\sigma} = sign(\hat{a} \cdot \vec{\sigma})$

where $sign(x) = \pm 1$ depending on whether $x > 0$ or $x < 0$

This model gives the correlation $\langle (\hat{a} \circ \vec{\sigma})(\hat{b} \circ -\vec{\sigma}) \rangle = \frac{2 \phi}{\pi} - 1$

where $\phi$ is the angle between $\hat{a}$ and $\hat{b}$. This gives the same answer as the QM prediction for the special cases $\phi = 0$ and $\phi = \pi$, but gives the wrong answer for other values of $\phi$. (The quantum prediction is $E(\hat{a}, \hat{b}) = - cos(\phi)$)

You're not going to come up with a local realistic model that makes the same predictions as QM, because there provably are none (subject to known loopholes).

 

[DrChinese]

We seem to differ slightly here: I take Bell's (1) to be the local realism assumption.

(1) is part of it, where we have independence for a and b. So that's fine.

But the next part is adding unit vector c, which is done right after (14a). This is the assumption that EPR made, that there were elements of reality even to quantum attributes that could not be simultaneously observed. You could observe 2 out of a, b, c after all. But you can't observe all 3 simultaneously. (EPR: "No reasonable definition of reality could be expected" to require they also be simultaneously observable.) So (14b) is the mathematical expression of the EPR statement.

Now the bigger picture here is to understand that Bell did not spell this out, he knew his readers (the very few) would get this. And it doesn't matter whether everyone points this out with a big arrow, the fact is that every proof of Bell does the same thing one way or another. There is always a, b and c. It is realizing that the relationships between the 3 cannot be made to work out, even if you hand pick the outcomes yourself. If you haven't tried to do this, this is the time.

And as to your disagreeing with the move from (14a) to (14b): it is your right to reject the realism assumption and replace it with something else that represents realism to you. Just be aware that won't match EPR, and you shouldn't expect agreement from other scientists. Obviously, this has already been thoroughly considered by many.

 

[Jilang]

I said that I was going to drop out, but after thinking about the model that @N88 was sketching, I realize that it is almost exactly the model Bell considered in "Speakable and Unspeakable in Quantum Mechanics". Bell didn't talk in terms of the spin vector rotating, but it amounts to the same thing:

• Assume that there is an intrinsic spin vector $\vec{\sigma}$ associated with each spin-1/2 particle.
• If you measure the particle's spin along axis $\hat{a}$ then you get +1, if the angle between $\hat{a}$ and $\vec{\sigma}$ is less than 90 degrees, and -1 otherwise.
• In correlated twin-pairs, if one particle has intrinsic spin $\vec{\sigma}$, then the other particle has spin $-\vec{\sigma}$.

So this is the same as the model of @N88, with the specific choice:

$\hat{a} \circ \vec{\sigma} = sign(\hat{a} \cdot \vec{\sigma})$

where $sign(x) = \pm 1$ depending on whether $x > 0$ or $x < 0$

This model gives the correlation $\langle (\hat{a} \circ \vec{\sigma})(\hat{b} \circ -\vec{\sigma}) \rangle = \frac{2 \phi}{\pi} - 1$

where $\phi$ is the angle between $\hat{a}$ and $\hat{b}$. This gives the same answer as the QM prediction for the special cases $\phi = 0$ and $\phi = \pi$, but gives the wrong answer for other values of $\phi$. (The quantum prediction is $E(\hat{a}, \hat{b}) = - cos(\phi)$)

You're not going to come up with a local realistic model that makes the same predictions as QM, because there provably are none (subject to known loopholes).

Is there a readily available paper of that? I have struggled To get that outcome. Thanks!

 

[PeterDonis]

Is there a readily available paper of that?

Bell's 1964 paper is linked to in post #3 of this thread.

 

[Jilang]

Bell's 1964 paper is linked to in post #3 of this thread.

Thanks, I am struggling to get from equation 9 to 10. Is there a proof if this?

 

[PeterDonis]

I am struggling to get from equation 9 to 10

Equation 9 appears to have a typo; it should read

$$
A(\vec{a}, \vec{\lambda}) = \text{sign} \ \vec{a} \cdot \vec{\lambda}
$$

$$
B(\vec{b}, \vec{\lambda}) = - \text{sign} \ \vec{b} \cdot \vec{\lambda}
$$

 

[PeterDonis]

I am struggling to get from equation 9 to 10

If you plug equation 9 into equation 2, and use $\rho(\lambda) = 1$ (uniform distribution, as specified in the text just before equation 9), you get (I'm leaving off the vector symbols for ease of typing)

$$
P(a, b) = \int d\lambda A(a, \lambda) B(b, \lambda) = - \int d\lambda \ \text{sign} \ a \cdot \lambda \ \text{sign} \ b \cdot \lambda = - \langle \text{sign} \ a \cdot b \rangle
$$

In other words, P(a, b) is minus the expectation value of the sign of $a \cdot b$. But that expectation value is given by equation 5 in the paper (more precisely, the same logic that led from equation 4 to equation 5 in the paper leads to the above). This gives us equation 10.

 

[stevendaryl]

Bell's 1964 paper is linked to in post #3 of this thread.

I got it from "Speakable and Unspeakable in Quantum Mechanics", but I see that that chapter is taken from the 1964 paper. It's the section III called "Illustration"

 

[N88]

(1) is part of it, where we have independence for a and b. So that's fine.

But the next part is adding unit vector c, which is done right after (14a). This is the assumption that EPR made, that there were elements of reality even to quantum attributes that could not be simultaneously observed. You could observe 2 out of a, b, c after all. But you can't observe all 3 simultaneously. (EPR: "No reasonable definition of reality could be expected" to require they also be simultaneously observable.) So (14b) is the mathematical expression of the EPR statement.

Now the bigger picture here is to understand that Bell did not spell this out, he knew his readers (the very few) would get this. And it doesn't matter whether everyone points this out with a big arrow, the fact is that every proof of Bell does the same thing one way or another. There is always a, b and c. It is realizing that the relationships between the 3 cannot be made to work out, even if you hand pick the outcomes yourself. If you haven't tried to do this, this is the time.

And as to your disagreeing with the move from (14a) to (14b): it is your right to reject the realism assumption and replace it with something else that represents realism to you. Just be aware that won't match EPR, and you shouldn't expect agreement from other scientists. Obviously, this has already been thoroughly considered by many.

DrChinese, thank you, I like this very much; it's very helpful to me.

But, since you know (at my present stage of learning) there's a few 'buts' coming, here's the first: BUT I disagree with EPR's definition! So a disproof of their (IMHO) badly-worded definition of an element of physical reality is (for me) to be expected and accepted without argument! So if, as you seem to correctly and clearly insist, Bell refutes EPR under EPRB, then we agree! For I then understand this: Bell's move from (14a) to (14b) is based on EPR. And that basis is exactly AS EXPLAINED by d'Espagnat (1979) in SciAm.

So I trust I have this right: Every proof of Bell does the same thing one way or another. There's always a, b and c. Bell's bigger picture is the realisation that the relationships between the 3 cannot be made to work out under the EPR definition of an element of physical reality. Moreover, a large part of the mainstream world in physics thinks that all local-realistic definitions fail similarly.

But this leaves me with this continuing problem: Is there not a better expression of local-realism -- exactly like in d'Espagnat (1979) 3-part wording, which I fully accept (without the EPR-based inference) -- that voids Bell's move from (14a) to (14b)?

(Perhaps like Arthur and the Holy Grail), I think there is: but, as you say, a large part of the mainstream world in physics thinks otherwise.

No need to reply; I need to think further. With my thanks again for your clarity, N88.

 

[stevendaryl]

DrChinese, thank you, I like this very much; it's very helpful to me.

But, since you know (at my present stage of learning) there's a few 'buts' coming, here's the first: BUT I disagree with EPR's definition! So a disproof of their (IMHO) badly-worded definition of an element of physical reality is (for me) to be expected and accepted without argument! So if, as you seem to correctly and clearly insist, Bell refutes EPR under EPRB, then we agree! For I then understand this: Bell's move from (14a) to (14b) is based on EPR. And that basis is exactly AS EXPLAINED by d'Espagnat (1979) in SciAm.

So I trust I have this right: Every proof of Bell does the same thing one way or another. There's always a, b and c. Bell's bigger picture is the realisation that the relationships between the 3 cannot be made to work out under the EPR definition of an element of physical reality. Moreover, a large part of the mainstream world in physics thinks that all local-realistic definitions fail similarly.

But this leaves me with this continuing problem: Is there not a better expression of local-realism -- exactly like in d'Espagnat (1979) 3-part wording, which I fully accept (without the EPR-based inference) -- that voids Bell's move from (14a) to (14b)?

I think it's a mistake to replace a precise, completely clear definition be replaced by a fuzzy definition that is too vague to reason about. The fact is that there is no model that anyone has proposed that makes the same predictions as quantum mechanics for EPR that is clearly intuitively local, by any definition. The fact that you tried to sketch such a model and ended up with exactly the model that Bell used to prove his point is, I think, telling.

 

[PeterDonis]

Bell's bigger picture is the realisation that the relationships between the 3 cannot be made to work out under the EPR definition of an element of physical reality.

No, Bell's bigger picture is the proof that the relationships between the 3 cannot be made to work out under his mathematical definition of "local realism". Whether that mathematical definition properly captures the EPR definition, which was in ordinary language, is a different question.

Is there not a better expression of local-realism -- exactly like in d'Espagnat (1979) 3-part wording, which I fully accept (without the EPR-based inference) -- that voids Bell's move from (14a) to (14b)?

Nobody has found one. If you think there is, the only way to find out is to try and find one.

 

[N88]

No, Bell's bigger picture is the proof that the relationships between the 3 cannot be made to work out under his mathematical definition of "local realism". Whether that mathematical definition properly captures the EPR definition, which was in ordinary language, is a different question.

Nobody has found one. If you think there is, the only way to find out is to try and find one.

Thanks Peter, this is clear.

My problem is this: As I read it in Bell's annotated move from 1964:(14a) to (14b), Bell relies on his (1); which we both find acceptable. But I do not find his manipulation of his (1) convincing, except under EPR (as DrChinese notes).

That is, under Bell's broad specification of λ, I do not see why λ cannot vary from run to run (in such a way that it is hardly ever the same). Thus the mathematical support for (A(a, λ))²= 1 continues to escape me, except as DrChinese notes (wherein different runs are enforced/unavoidable).

 

[PeterDonis]

under Bell's broad specification of λ, I do not see why λ cannot vary from run to run

It can.

the mathematical support for (A(a, λ))²= 1 continues to escape me

$A(a, \lambda)$ represents the result of Alice's measurement given a measuring device setting $a$ and some particular value for $\lambda$. Since Alice will always measure either spin up or spin down, i.e, +1 or -1, it follows that the square of whatever result she gets, for any $a$ and any $\lambda$, must be 1.

 

[N88]

I said that I was going to drop out, but after thinking about the model that @N88 was sketching, I realize that it is almost exactly the model Bell considered in "Speakable and Unspeakable in Quantum Mechanics". Bell didn't talk in terms of the spin vector rotating, but it amounts to the same thing:

• Assume that there is an intrinsic spin vector $\vec{\sigma}$ associated with each spin-1/2 particle.
• If you measure the particle's spin along axis $\hat{a}$ then you get +1, if the angle between $\hat{a}$ and $\vec{\sigma}$ is less than 90 degrees, and -1 otherwise.
• In correlated twin-pairs, if one particle has intrinsic spin $\vec{\sigma}$, then the other particle has spin $-\vec{\sigma}$.

So this is the same as the model of @N88, with the specific choice:

$\hat{a} \circ \vec{\sigma} = sign(\hat{a} \cdot \vec{\sigma})$

where $sign(x) = \pm 1$ depending on whether $x > 0$ or $x < 0$

This model gives the correlation $\langle (\hat{a} \circ \vec{\sigma})(\hat{b} \circ -\vec{\sigma}) \rangle = \frac{2 \phi}{\pi} - 1$

where $\phi$ is the angle between $\hat{a}$ and $\hat{b}$. This gives the same answer as the QM prediction for the special cases $\phi = 0$ and $\phi = \pi$, but gives the wrong answer for other values of $\phi$. (The quantum prediction is $E(\hat{a}, \hat{b}) = - cos(\phi)$)

You're not going to come up with a local realistic model that makes the same predictions as QM, because there provably are none (subject to known loopholes).

Thanks for this. Again, as always, I appreciate your detail. However, with respect: The sign in my model is determined by the "spin-flip" in that (to use your example), it allows* for opposite signs to yours whether $\phi > 0$ or $\phi < 0$. So the constraint you propose does not apply.

* The explanation for this allowance is that the interaction of the spin-vector $\vec{\sigma}$ with the field-orientation/gradient $\hat{a}$ involves more complex dynamics than Bell's model permits.

 

[N88]

It can.

$A(a, \lambda)$ represents the result of Alice's measurement given a measuring device setting $a$ and some particular value for $\lambda$. Since Alice will always measure either spin up or spin down, i.e, +1 or -1, it follows that the square of whatever result she gets, for any $a$ and any $\lambda$, must be 1.

Under DrChinese view, as I understand it, the product that Bell uses derives from different runs of the experiment (due the vector c that DrChinese refers to). So, as I understand it:

1. It is the EPR assumption that allows Bell to assume the same λ is available. Without EPR, the product (over different runs, and not now a squaring) might involve λ_i differing from λ_j and a possible result of -1.

2. In this way, with EPR setting the widely-accepted standard for "local realism", local realism fails.

3. I therefore interpret Bell's result as the failure of CFD and the survival of locality.

I hope this is now OK, and an acceptable view?

 

[Jilang]

I think it's a mistake to replace a precise, completely clear definition be replaced by a fuzzy definition that is too vague to reason about. The fact is that there is no model that anyone has proposed that makes the same predictions as quantum mechanics for EPR that is clearly intuitively local, by any definition. The fact that you tried to sketch such a model and ended up with exactly the model that Bell used to prove his point is, I think, telling.

The diagram in post 45 illustrates why I am struggling I think. If b is at a given angle from a should it be drawn as a cone around a rather than as a line?

 

[Jilang]

If you plug equation 9 into equation 2, and use $\rho(\lambda) = 1$ (uniform distribution, as specified in the text just before equation 9), you get (I'm leaving off the vector symbols for ease of typing)

$$
P(a, b) = \int d\lambda A(a, \lambda) B(b, \lambda) = - \int d\lambda \ \text{sign} \ a \cdot \lambda \ \text{sign} \ b \cdot \lambda = - \langle \text{sign} \ a \cdot b \rangle
$$

In other words, P(a, b) is minus the expectation value of the sign of $a \cdot b$. But that expectation value is given by equation 5 in the paper (more precisely, the same logic that led from equation 4 to equation 5 in the paper leads to the above). This gives us equation 10.

Thanks Peter, but I still don't see how the a.b gets in there.

 

[stevendaryl]

The diagram in post 45 illustrates why I am struggling I think. If b is at a given angle from a should it be drawn as a cone around a rather than as a line?

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 $(+1)(\theta/\pi) + (-1) (1-\theta/\pi) = -1 + 2\theta/\pi$

 

[stevendaryl]

Thanks for this. Again, as always, I appreciate your detail. However, with respect: The sign in my model is determined by the "spin-flip" in that (to use your example), it allows* for opposite signs to yours whether $\phi > 0$ or $\phi < 0$. So the constraint you propose does not apply.

Yes, Bell's model gave a specific value for the $\circ$ operator. His theorem, though, proves that there is no choice for the $\circ$ operator that gives the same predictions as EPR.

* The explanation for this allowance is that the interaction of the spin-vector $\vec{\sigma}$ with the field-orientation/gradient $\hat{a}$ involves more complex dynamics than Bell's model permits.

This is like schemes to devise a perpetual motion machine--it doesn't matter how complicated the interaction is, it's impossible. That's the beauty of mathematics---you don't have to try all possibilities in order to prove a universal fact.

 

[stevendaryl]

Under DrChinese view, as I understand it, the product that Bell uses derives from different runs of the experiment (due the vector c that DrChinese refers to). So, as I understand it:

1. It is the EPR assumption that allows Bell to assume the same λ is available. Without EPR, the product (over different runs, and not now a squaring) might involve λ_i differing from λ_j and a possible result of -1.

2. In this way, with EPR setting the widely-accepted standard for "local realism", local realism fails.

3. I therefore interpret Bell's result as the failure of CFD and the survival of locality.

I hope this is now OK, and an acceptable view?

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]

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.

You can allow that the operator $\circ$ is nondeterministic, but then we're back to the question: If it's nondeterministic, then how can you guarantee that Alice and Bob will get the same value for $\hat{a} \circ \vec{\lambda}$?

 

[stevendaryl]

For the sake of having consistent, on-topic threads, I think we should have a new thread about CFD and local realism, because this thread has not been about "the physical significance of a.σ" for a while now.

 

[PeterDonis]

I still don't see how the a.b gets in there.

See stevendaryl's post #53. What he explains there is equivalent to explaining that

$$
\int d\lambda \ \text{sign} \ a \cdot \lambda \ \text{sign} \ b \cdot \lambda = \langle \text{sign} \ a \cdot b \rangle
$$

 

[PeterDonis]

the product that Bell uses derives from different runs of the experiment (due the vector c that DrChinese refers to).

The vector c doesn't appear anywhere in the formula $(A(a, \lambda))^2 = 1$, which is what you asked about.

 

[edguy99]

You can allow that the operator $\circ$ is nondeterministic, but then we're back to the question: If it's nondeterministic, then how can you guarantee that Alice and Bob will get the same value for $\hat{a} \circ \vec{\lambda}$?

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.

 

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

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

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.