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

N88
Messages
225
Reaction score
12
Admins: Please excuse my E and brackets in the title, and correct if possible.

My questions are these, please:

1. What is the physical significance of (\hat{a}\cdot\boldsymbol{\sigma}_{1})<br />
in <br /> \left\langle (\hat{a}\cdot\boldsymbol{\sigma}_{1})(-\boldsymbol{\sigma}_{2}\cdot\hat{b})\right\rangle <br />?

2. This QM formulation appears (to me) to be local and realistic: but I take it that that is not the mainstream view?

The questions arise in calculating the expectation for the famous EPRB experiment.

I understand the orientations of the detectors and representing the intrinsic spin of a spin-half particle by s = (hbar/2) \boldsymbol{\sigma} where \boldsymbol{\sigma} is the intrinsic spin vector. So the QM formulation looks (to me) to be local and realistic.

PS: I have a derivation of the correct result which looks to be local and realistic. But I have not found the same approach in any textbook, so my derivation may be non-mainstream. That's why I would like to check the physical significance of the mainstream result first (and in detail).

Thank you.
 
Physics news on Phys.org
N88 said:
What is the physical significance of ##(\hat{a}\cdot\boldsymbol{\sigma}_{1})##
in ##\left\langle (\hat{a}\cdot\boldsymbol{\sigma}_{1})(-\boldsymbol{\sigma}_{2}\cdot\hat{b})\right\rangle## ?

Where are you getting this equation from? What is the context?

N88 said:
I understand the orientations of the detectors and representing the intrinsic spin of a spin-half particle by s = (hbar/2) ##\boldsymbol{\sigma}## where ##\boldsymbol{\sigma}## is the intrinsic spin vector.

The symbol ##\boldsymbol{\sigma}## refers to a "vector" of matrices, the Pauli matrices: i.e., a "vector" with components ##\sigma_1##, ##\sigma_2##, ##\sigma_3##, which are the Pauli matrices:

$$
\sigma_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
$$
$$
\sigma_2 = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix}
$$
$$
\sigma_3 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}
$$

So the notation ##\hat{a} \cdot \boldsymbol{\sigma}## refers to the "dot product" of some vector ##\hat{a}## with the "vector" of matrices given above. If, as I suspect, ##\hat{a}## is meant to represent a unit vector in some particular direction (the direction of the axis along which the spin is to be measured), then ##\hat{a} \cdot \boldsymbol{\sigma}## would give an operator which would represent the proper "mixture" of the three Pauli spin matrices to denote measuring the spin of a spin-1/2 particle about the axis ##\hat{a}##. But an operator is not the same thing as a state, which is what "the intrinsic spin of a spin-1/2 particle" should be. So I'm confused about this notation and wondering where it comes from.
 
PeterDonis said:
Where are you getting this equation from? What is the context?
The symbol ##\boldsymbol{\sigma}## refers to a "vector" of matrices, the Pauli matrices: i.e., a "vector" with components ##\sigma_1##, ##\sigma_2##, ##\sigma_3##, which are the Pauli matrices:

$$
\sigma_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
$$
$$
\sigma_2 = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix}
$$
$$
\sigma_3 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}
$$

So the notation ##\hat{a} \cdot \boldsymbol{\sigma}## refers to the "dot product" of some vector ##\hat{a}## with the "vector" of matrices given above. If, as I suspect, ##\hat{a}## is meant to represent a unit vector in some particular direction (the direction of the axis along which the spin is to be measured), then ##\hat{a} \cdot \boldsymbol{\sigma}## would give an operator which would represent the proper "mixture" of the three Pauli spin matrices to denote measuring the spin of a spin-1/2 particle about the axis ##\hat{a}##. But an operator is not the same thing as a state, which is what "the intrinsic spin of a spin-1/2 particle" should be. So I'm confused about this notation and wondering where it comes from.

The expectation equation is from Bell (1964), eqn (3).
http://cds.cern.ch/record/111654/files/vol1p195-200_001.pdf

\hat{a} is a detector unit-vector; the orientation of the principal axis.

The intrinsic spin equation (in which I replaced his m with s),* is from p.149, eqn (48):
Dirac, P. A. M. (1982). The Principles of Quantum Mechanics (4th ed., rev.). Clarendon, Oxford.

* To have s related to intrinsic spin just better suits the way I think.

I will certainly benefit from discussion of the physical significance of the 'vector'**
\boldsymbol{\sigma} being defined by \boldsymbol{\sigma}\equiv(\sigma_{x}, \sigma_{y}, \sigma_{z}).

** I would like to call it by its correct name, more fully reflecting its physical significance (so that we don't need to put in "scare quotes" -- which makes it appear even scarier to a beginner.)

Thank you.
 
Last edited:
N88 said:
Admins: Please excuse my E and brackets in the title, and correct if possible.

My questions are these, please:

1. What is the physical significance of (\hat{a}\cdot\boldsymbol{\sigma}_{1})<br />
in <br /> \left\langle (\hat{a}\cdot\boldsymbol{\sigma}_{1})(-\boldsymbol{\sigma}_{2}\cdot\hat{b})\right\rangle<br />?

2. This QM formulation appears (to me) to be local and realistic: but I take it that that is not the mainstream view?

I don't understand in what sense you would call that "local".

When Alice measures the spin of her particle along axis \vec{a}, she is measuring the observable
\vec{a} \cdot \vec{\sigma_1}. When Bob measures the spin of his particle along axis \vec{b}, he is measuring the observable \vec{b} \cdot \vec{\sigma_2}. Quantum-mechanically, the expectation value for the product (\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2}) is given by:

\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle

where |\Psi\rangle is the two-particle spin-state, which in the case of anti-correlated entangled spins is the state |\Psi\rangle = \frac{1}{\sqrt{2}} (|u\rangle |d\rangle - |d\rangle |u\rangle). Since the first operator only acts on the first particle, and the second operator only acts on the second particle, we can calculate:

\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle = - \vec{a} \cdot \vec{b}

That's not the explanation, that's the fact that needs to be explained. What Alice actually gets from her measurement of spin is not \vec{a} \cdot \vec{\sigma_1}, but one of the eigenvalues of that operator, which are \pm 1. Apparently, it's completely random which value (plus or minus 1) she actually gets. It's also apparently random which of \pm 1 Bob gets. But despite their being random, they are correlated. (If they weren't correlated then the expectation value of the product of Bob's result and Alice's result would be zero.) So the question is how to explain the correlation through local mechanisms. Suppose that they both decide to measure spins along axis \vec{a}. Suppose further that Alice gets her result first, and it's +1. Then at that point, she knows for certain that Bob will measure -1. That seems to be a fact about Bob and his particle that is not reflected by any variable local to Bob. So what guarantees that Bob will get -1?

Pointing out that \langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle = -1 in this case (the case in which \vec{b} = \vec{a}) doesn't seem like an explanation for the fact, it's just a mathematical way of stating the fact.

The derivation of the expectation value

Since |\Psi\rangle = \frac{1}{\sqrt{2}} (|u\rangle |d\rangle - |d\rangle |u\rangle) and \vec{a} \cdot \vec{\sigma_1} only affects the first component of the state, and \vec{b} \cdot \vec{\sigma_2} only affects the second component of the state, we can write:

\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle
= \frac{1}{2} [\langle u |(\vec{a} \cdot\vec{\sigma_1}) |u\rangle \langle d|(\vec{b} \cdot \vec{\sigma_2})|d\rangle
- \langle u |(\vec{a} \cdot\vec{\sigma_1}) |d\rangle \langle d|(\vec{b} \cdot \vec{\sigma_2})|u\rangle
- \langle d |(\vec{a} \cdot\vec{\sigma_1}) |u\rangle \langle u|(\vec{b} \cdot \vec{\sigma_2})|d\rangle
+ \langle d |(\vec{a} \cdot\vec{\sigma_1}) |d\rangle \langle u|(\vec{b} \cdot \vec{\sigma_2})|u\rangle]

So now we can use the representation of the Pauli matrices to derive:

\langle u | \vec{a} \cdot \vec{\sigma} |u\rangle = a_z
\langle d | \vec{a} \cdot \vec{\sigma} |u\rangle = (a_x + i a_y)
\langle u | \vec{a} \cdot \vec{\sigma} |d\rangle = (a_x - i a_y)
\langle d | \vec{a} \cdot \vec{\sigma} |d\rangle = -a_z
\langle u | \vec{b} \cdot \vec{\sigma} |u\rangle = b_z
\langle d | \vec{b} \cdot \vec{\sigma} |u\rangle = (b_x + i b_y)
\langle u | \vec{b} \cdot \vec{\sigma} |d\rangle = (b_x - i b_y)
\langle d | \vec{b} \cdot \vec{\sigma} |d\rangle = -b_z

So:
\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle
= \frac{1}{2} [- a_z b_z - (a_x - i a_y) (b_x + i b_y) - (a_x + i a_y) (b_x - i b_y) - a_z b_z]
= - a_z b_z - a_x b_x - a_y b_y
 
  • Like
Likes N88
N88 said:
I will certainly benefit from discussion of the physical significance of the 'vector'**
##\boldsymbol{\sigma}## being defined by σ≡(σxyz).

The physical significance is that the dot product of this vector with a vector ##\hat{a}## that points along some axis, gives you an operator (i.e., a matrix) that represents measuring the spin about that axis. So, as stevendaryl said, ##\hat{a} \cdot \boldsymbol{\sigma_1}## is the operator that represents Alice measuring the spin of her particle (hence the ##1## subscript on the ##\boldsymbol{\sigma}##) about axis ##\hat{a}##. And similarly for Bob, with the subscript ##2## and the axis ##\hat{b}##.
 
PeterDonis said:
The physical significance is that the dot product of this vector with a vector ##\hat{a}## that points along some axis, gives you an operator (i.e., a matrix) that represents measuring the spin about that axis. So, as stevendaryl said, ##\hat{a} \cdot \boldsymbol{\sigma_1}## is the operator that represents Alice measuring the spin of her particle (hence the ##1## subscript on the ##\boldsymbol{\sigma}##) about axis ##\hat{a}##. And similarly for Bob, with the subscript ##2## and the axis ##\hat{b}##.

Thanks for this. Please, what name do you give to "this vector" -- "the dot product of this vector with a vector ##\hat{a}##".

I want to be clear about the terminology that I use in replying to stevendaryl.
 
N88 said:
Thanks for this. Please, what name do you give to "this vector" -- "the dot product of this vector with a vector ##\hat{a}##".

I want to be clear about the terminology that I use in replying to stevendaryl.

You can call it "the component of spin in the \hat{a} direction".
 
stevendaryl said:
You can call it "the component of spin in the \hat{a} direction".

I'd like to be clear about this. With respect to the vector that I asked about, PeterDonis (Post #2) wrote: The symbol σ refers to a "vector" of matrices, the Pauli matrices.

So I thought it was the dot-product of σ with a unit-vector \hat{a} that gave "the component of spin in the \hat{a} direction.

So my new question is this: Is σ called "the Pauli-vector"?

So when it is described as "a vector of matrices" with a dot-product, this is a slightly misleading for beginners unless they understand that it is a special element of a vector space (maybe)?

Because when I searched for "Pauli vector" I found this: https://www.physicsforums.com/threads/pauli-vector.635079/

So, hoping I have not misunderstood, I am thinking that dextercioby's answer is helpful: we are working with "a shorthand notation which is useful, but can be misleading".

PS: I'm happy to work with the shorthand, but I want to be clear for my response to your nicely expanded work above. Thanks
 
N88 said:
I'd like to be clear about this. With respect to the vector that I asked about, PeterDonis (Post #2) wrote: The symbol σ refers to a "vector" of matrices, the Pauli matrices.

Yes. It's a vector with components \sigma_x, \sigma_y, \sigma_z, where each of those is a 2x2 matrix.

So I thought it was the dot-product of σ with a unit-vector \hat{a} that gave "the component of spin in the \hat{a} direction.

Yes, except it's an operator, not a value. In the same way that p_x = -i \hbar \frac{\partial}{\partial x} is the component of momentum in the x-direction, but it's an operator, not a value.

So my new question is this: Is σ called "the Pauli-vector"?

So when it is described as "a vector of matrices" with a dot-product, this is a slightly misleading for beginners unless they understand that it is a special element of a vector space (maybe)?

I can't see how it's misleading to call it a vector of matrices. It's not an element of a vector space, it is a vector of matrices in the sense of a 3-vector with an x-component, a y-component and a z-component. But those components are matrices, instead of numbers.
 
  • #10
stevendaryl said:
I don't understand in what sense you would call that "local".

When Alice measures the spin of her particle along axis \vec{a}, she is measuring the observable
\vec{a} \cdot \vec{\sigma_1}. When Bob measures the spin of his particle along axis \vec{b}, he is measuring the observable \vec{b} \cdot \vec{\sigma_2}. Quantum-mechanically, the expectation value for the product (\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2}) is given by:

\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle

where |\Psi\rangle is the two-particle spin-state, which in the case of anti-correlated entangled spins is the state |\Psi\rangle = \frac{1}{\sqrt{2}} (|u\rangle |d\rangle - |d\rangle |u\rangle). Since the first operator only acts on the first particle, and the second operator only acts on the second particle, we can calculate:

\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle = - \vec{a} \cdot \vec{b}

That's not the explanation, that's the fact that needs to be explained. What Alice actually gets from her measurement of spin is not \vec{a} \cdot \vec{\sigma_1}, but one of the eigenvalues of that operator, which are \pm 1. Apparently, it's completely random which value (plus or minus 1) she actually gets. It's also apparently random which of \pm 1 Bob gets. But despite their being random, they are correlated. (If they weren't correlated then the expectation value of the product of Bob's result and Alice's result would be zero.) So the question is how to explain the correlation through local mechanisms. Suppose that they both decide to measure spins along axis \vec{a}. Suppose further that Alice gets her result first, and it's +1. Then at that point, she knows for certain that Bob will measure -1. That seems to be a fact about Bob and his particle that is not reflected by any variable local to Bob. So what guarantees that Bob will get -1?

Pointing out that \langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle = -1 in this case (the case in which \vec{b} = \vec{a}) doesn't seem like an explanation for the fact, it's just a mathematical way of stating the fact.

The derivation of the expectation value

Since |\Psi\rangle = \frac{1}{\sqrt{2}} (|u\rangle |d\rangle - |d\rangle |u\rangle) and \vec{a} \cdot \vec{\sigma_1} only affects the first component of the state, and \vec{b} \cdot \vec{\sigma_2} only affects the second component of the state, we can write:

\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle
= \frac{1}{2} [\langle u |(\vec{a} \cdot\vec{\sigma_1}) |u\rangle \langle d|(\vec{b} \cdot \vec{\sigma_2})|d\rangle
- \langle u |(\vec{a} \cdot\vec{\sigma_1}) |d\rangle \langle d|(\vec{b} \cdot \vec{\sigma_2})|u\rangle
- \langle d |(\vec{a} \cdot\vec{\sigma_1}) |u\rangle \langle u|(\vec{b} \cdot \vec{\sigma_2})|d\rangle
+ \langle d |(\vec{a} \cdot\vec{\sigma_1}) |d\rangle \langle u|(\vec{b} \cdot \vec{\sigma_2})|u\rangle]

So now we can use the representation of the Pauli matrices to derive:

\langle u | \vec{a} \cdot \vec{\sigma} |u\rangle = a_z
\langle d | \vec{a} \cdot \vec{\sigma} |u\rangle = (a_x + i a_y)
\langle u | \vec{a} \cdot \vec{\sigma} |d\rangle = (a_x - i a_y)
\langle d | \vec{a} \cdot \vec{\sigma} |d\rangle = -a_z
\langle u | \vec{b} \cdot \vec{\sigma} |u\rangle = b_z
\langle d | \vec{b} \cdot \vec{\sigma} |u\rangle = (b_x + i b_y)
\langle u | \vec{b} \cdot \vec{\sigma} |d\rangle = (b_x - i b_y)
\langle d | \vec{b} \cdot \vec{\sigma} |d\rangle = -b_z

So:
\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle
= \frac{1}{2} [- a_z b_z - (a_x - i a_y) (b_x + i b_y) - (a_x + i a_y) (b_x - i b_y) - a_z b_z]
= - a_z b_z - a_x b_x - a_y b_y

Elsewhere, as I recall, you asked about my local-realistic views. Above you wondered about the sense of "local" that I use.

I'm attaching a 1-page PDF extract from an essay that I'm working on. Equations marked (X.) are not relevant to our current discussion. The QM results are mainstream: whether I have derived them properly, under local-realism, remains the question. Also: Do they answer the correlation difficulties that you mention above?

With my thanks again for the detail that you provide, I look forward to your comments.
 

Attachments

  • #11
N88 said:
Elsewhere, as I recall, you asked about my local-realistic views. Above you wondered about the sense of "local" that I use.

I'm attaching a 1-page PDF extract from an essay that I'm working on. Equations marked (X.) are not relevant to our current discussion. The QM results are mainstream: whether I have derived them properly, under local-realism, remains the question. Also: Do they answer the correlation difficulties that you mention above?

With my thanks again for the detail that you provide, I look forward to your comments.

No, I don't see how it answers the problems. You seem to be giving a local realistic model of the type that Bell proved impossible. As I said, the fact about Pauli matrices, that \langle \Psi |(\vec{a} \cdot \vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2}) \Psi \rangle = - \vec{a} \cdot \vec{b} does not explain EPR correlations, it is simply a mathematically succinct statement of those correlations.

You say in that paper that

\vec{s} is then spin-up or spin-down with respect to \hat{a} after q(\vec{\sigma}) interacts with a polarizer-field oriented \hat{a}

That is the starting point of a possible local-realistic model of the type that Bell proved is impossible. What you're saying is that
  • Alice's particle has an intrinsic spin \vec{\sigma} pointing in some direction.
  • The particle interacts with the electromagnetic field of Alice's detector (which has orientation \hat{a}).
  • As a result of the interaction, the particle's spin afterward is either spin-up or spin-down relative to \hat{a}.
Before we move on to Bob, let's ask a follow-up question about this: Given \vec{\sigma} and \hat{a}, is it nondeterministic, or deterministic whether the particle ends up being measured to have spin-up or spin-down relative to \hat{a}?
  • If it's deterministic, then you're proposing a deterministic local hidden-variable theory, and Bell proved that no such theory produces the EPR correlations.
  • If it's nondeterministic, it's provable that no nondeterministic theory can reproduce the perfect correlations of EPR.
The fact about products of Pauli spin matrices seems completely irrelevant. Or at least, I don't see the relevance. Once again, let's consider the case in which Alice has measured spin along axis \hat{a} and gotten result +1 (or +\frac{\hbar}{2}). Bob has not yet measured his particle, but he also plans to measure spin along the same axis, \hat{a}. Alice can confidently predict that Bob will get result -1. That's a fact to be explained.

If you want to say that it's explained by saying that Bob's particle has an intrinsic spin vector \vec{\sigma_2}, fine. How does that explain that Bob will definitely get result -1? In keeping with what you wrote (quoted above), you want to say that the interaction of Bob's particle with the electromagnetic field of Bob's detector results in the outcome -1. And, since Alice's prediction is certain, that means that it is certain that the combination of Bob's detector setting, \vec{a}, and Bob's particle's intrinsic spin vector, \vec{\sigma_2} results in -1. How is that different from saying that there is a function, B(\hat{b}, \vec{\sigma}) giving \pm 1 depending on Bob's setting \hat{b} and the particle's spin \vec{\sigma}?

It seems to me that you don't have a different idea of a local realistic theory. You have the same idea that Bell had, and which Bell proved is impossible.
 
  • Like
Likes N88
  • #12
stevendaryl said:
No, I don't see how it answers the problems. You seem to be giving a local realistic model of the type that Bell proved impossible. As I said, the fact about Pauli matrices, that \langle \Psi |(\vec{a} \cdot \vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2}) \Psi \rangle = - \vec{a} \cdot \vec{b} does not explain EPR correlations, it is simply a mathematically succinct statement of those correlations.

You say in that paper that
N88 said:
Elsewhere, as I recall, you asked about my local-realistic views. Above you wondered about the sense of "local" that I use.

I'm attaching a 1-page PDF extract from an essay that I'm working on. Equations marked (X.) are not relevant to our current discussion. The QM results are mainstream: whether I have derived them properly, under local-realism, remains the question. Also: Do they answer the correlation difficulties that you mention above?

With my thanks again for the detail that you provide, I look forward to your comments.

stevendaryl said:
That is the starting point of a possible local-realistic model of the type that Bell proved is impossible. What you're saying is that
  • Alice's particle has an intrinsic spin \vec{\sigma} pointing in some direction.
  • The particle interacts with the electromagnetic field of Alice's detector (which has orientation \hat{a}).
  • As a result of the interaction, the particle's spin afterward is either spin-up or spin-down relative to \hat{a}.
Before we move on to Bob, let's ask a follow-up question about this: Given \vec{\sigma} and \hat{a}, is it nondeterministic, or deterministic whether the particle ends up being measured to have spin-up or spin-down relative to \hat{a}?
  • If it's deterministic, then you're proposing a deterministic local hidden-variable theory, and Bell proved that no such theory produces the EPR correlations.
  • If it's nondeterministic, it's provable that no nondeterministic theory can reproduce the perfect correlations of EPR.
The fact about products of Pauli spin matrices seems completely irrelevant. Or at least, I don't see the relevance. Once again, let's consider the case in which Alice has measured spin along axis \hat{a} and gotten result +1 (or +\frac{\hbar}{2}). Bob has not yet measured his particle, but he also plans to measure spin along the same axis, \hat{a}. Alice can confidently predict that Bob will get result -1. That's a fact to be explained.

If you want to say that it's explained by saying that Bob's particle has an intrinsic spin vector \vec{\sigma_2}, fine. How does that explain that Bob will definitely get result -1? In keeping with what you wrote (quoted above), you want to say that the interaction of Bob's particle with the electromagnetic field of Bob's detector results in the outcome -1. And, since Alice's prediction is certain, that means that it is certain that the combination of Bob's detector setting, \vec{a}, and Bob's particle's intrinsic spin vector, \vec{\sigma_2} results in -1. How is that different from saying that there is a function, B(\hat{b}, \vec{\sigma}) giving \pm 1 depending on Bob's setting \hat{b} and the particle's spin \vec{\sigma}?

It seems to me that you don't have a different idea of a local realistic theory. You have the same idea that Bell had, and which Bell proved is impossible. (Emphasis added.)

Thanks again for the detail. I have highlighted four key points but I start my reply with this (which I find very helpful):
stevendaryl said:
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&#039;) + C(a&#039;, b) - C(a&#039;, b&#039;) \leq 4
Continuing:
stevendaryl said:
[1] Before we move on to Bob, let's ask a follow-up question about this: Given \vec{\sigma} and \hat{a}, is it nondeterministic, or deterministic whether the particle ends up being measured to have spin-up or spin-down relative to \hat{a}?

To embrace both options (because that's the way I think here), my preferred answer is to say that it is "law-like": specifically, under EPRB, like Malus' Law.

By law-like I mean this: if a fair robot is tossing a fair Head/Tail coin under experiment Z, I can call: P(H|Z) = P(T|Z) = \tfrac{1}{2}. So my answer boils down to: "It doesn't matter, so I don't care: for I can handle both, as we'll see." (But I suspect -- if push comes to shove -- most (thinking like me), would see the case of EPRB being under determinism.)

Thus, in responding to the dilemma that you offer (if push comes to shove), I choose this option: "If it's deterministic, then you're proposing a deterministic local hidden-variable theory, and Bell proved that no such theory produces the EPR correlations."

BUT Bell 1964:(14a) ≠ Bell 1964:(14b) under my theory. So I join Peres (textbook, p. 162) and many others in saying: "Bell's theorem is not a property of quantum theory." [EDIT: in my opinion, EPR correlations are not at stake here. Some EPR ideas might be. But, as I see it, those are not relevant to our discussion.]

stevendaryl said:
[2] That's a fact to be explained.

As in that 1-page extract, I invoke my generalisation of Malus' Law to quantum systems under EPRB. (NB: In conventional quantum theory, some (eg, Aspect) discuss collapse and Malus' Law in the same context. I have no collapse -- so I interpret it in QT as good old Bayesian updating.

stevendaryl said:
[3] How is that different from saying that there is a function, B(\hat{b}, \vec{\sigma}) giving \pm 1 depending on Bob's setting \hat{b} and the particle's spin \vec{\sigma}?

Given Malus' Law, am I missing something? At the moment, I see no difference. However, to be clear re my view: Neither Alice's nor Bob's pristine particles need have the measured spin-direction prior to each particle-detector interaction. (Many senior physicists support this; eg, Kochen (2015).)

stevendaryl said:
[4 ] You have the same idea that Bell had, and which Bell proved is impossible.

As I see it, the difference arises in this way: We start with identical equations -- Bell 1964:(1)-(3) -- but proceed with a very different assumption. For, under Bell's assumption (as identified after his (14b) Bell has (14a) = (14b). I make no such assumption.

HTH, with my thanks again -- N88
 
Last edited:
  • #13
N88 said:
Given Malus' Law, am I missing something? At the moment, I see no difference. However, to be clear re my view: Neither Alice's nor Bob's pristine particles need have the measured spin-direction prior to each particle-detector interaction. (Many senior physicists support this; eg, Kochen (2015).)

Once again, if Alice and Bob both agree to perform measurements along axis \hat{a}, and Alice measures spin-up, then she knows for certain that Bob will measure spin-down. So the combination of Bob's setting and the intrinsic spin of Bob's particle is guaranteed to give the result -1. That's a deterministic model. It's provably false, according to Bell's theorem.
 
  • #14
stevendaryl said:
Once again, if Alice and Bob both agree to perform measurements along axis \hat{a}, and Alice measures spin-up, then she knows for certain that Bob will measure spin-down. So the combination of Bob's setting and the intrinsic spin of Bob's particle is guaranteed to give the result -1. That's a deterministic model. It's provably false, according to Bell's theorem.

1. Is my law-like model provably false? It agrees with QM. (I put that push-shove response in so that I could learn about my private passion to avoid it.) [EDIT: I understand that determinism does NOT mean that Alice's result determines Bob's result; and vice-vera.]

2. Since Bell and I start with the same equations -- but proceed with different assumptions -- can you help me by being more specific, please?

3. For example: where does the math change and what is the physical significance of that change, if it is not at our disagreement re (14a) = (14b)? Is it not (then) an experimental fact that: "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: …"?

That is, is this next not true? C delivers QM results but does not deliver D; QM delivers C results but does not deliver D.
 
  • #15
N88 said:
1. Is my law-like model provably false?

I don't see that you have a local model. Can you explain how it works in the situation I'm talking about?
 
  • #16
stevendaryl said:
I don't see that you have a local model. Can you explain how it works in the situation I'm talking about?

I take it that you are happy for me to discuss my model in the context of intrinsic spin only? It's easier; so that's what I'll do.

(For generality, I prefer to discuss it in the context of total-spin using Bell's λ.)
 
  • #17
stevendaryl said:
Once again, if Alice and Bob both agree to perform measurements along axis \hat{a}, and Alice measures spin-up, then she knows for certain that Bob will measure spin-down. So the combination of Bob's setting and the intrinsic spin of Bob's particle is guaranteed to give the result -1. That's a deterministic model. It's provably false, according to Bell's theorem.
stevendaryl said:
I don't see that you have a local model. Can you explain how it works in the situation I'm talking about?
As I see it:
Given conservation of intrinsic spin under EPRB: \boldsymbol{\sigma_1}+\boldsymbol{\sigma_2}=0. (1)
Alice's result is given by \hat{a}\circ\boldsymbol{\sigma_1}=+1. (2)
Bob's result is given by \hat{a}\circ\boldsymbol{\sigma_2}=-\hat{a}\circ\boldsymbol{\sigma_1}=-1. (3)
QED?
 
  • #18
N88 said:
As I see it:
Given conservation of intrinsic spin under EPRB: \boldsymbol{\sigma_1}+\boldsymbol{\sigma_2}=0. (1)
Alice's result is given by \hat{a}\circ\boldsymbol{\sigma_1}=+1. (2)
Bob's result is given by \hat{a}\circ\boldsymbol{\sigma_2}=-\hat{a}\circ\boldsymbol{\sigma_1}=-1. (3)
QED?

If \vec{\sigma} were an actual vector, yes. But if \vec{\sigma} were an actual vector, then there would be a nonzero probability of Alice choosing \hat{a} such that \hat{a} \cdot \vec{\sigma} = 0. So it's not an actual vector. It's an operator. Since it's an operator, it's not literally true that \hat{a} \cdot \vec{\sigma} = +1. The left-hand side is an operator, while the right-hand side is a number.
 
  • #19
stevendaryl said:
If \vec{\sigma} were an actual vector, yes. But if \vec{\sigma} were an actual vector, then there would be a nonzero probability of Alice choosing \hat{a} such that \hat{a} \cdot \vec{\sigma} = 0. So it's not an actual vector. It's an operator. Since it's an operator, it's not literally true that \hat{a} \cdot \vec{\sigma} = +1. The left-hand side is an operator, while the right-hand side is a number.

The issue is that there is no such thing as a vector \vec{\sigma} such that for any choice of a measurement axis \hat{a},

\vec{\sigma} \cdot \hat{a} = \pm 1
 
  • #20
stevendaryl said:
If \vec{\sigma} were an actual vector, yes. But if \vec{\sigma} were an actual vector, then there would be a nonzero probability of Alice choosing \hat{a} such that \hat{a} \cdot \vec{\sigma} = 0. So it's not an actual vector. It's an operator. Since it's an operator, it's not literally true that \hat{a} \cdot \vec{\sigma} = +1. The left-hand side is an operator, while the right-hand side is a number.

Not at all, in what I have written. As allowed by Bell under EPRB (via his λ), \hat{a} and \vec{\sigma} are both vectors. The operator is \circ; ie, the spin-product as introduced and defined in my 1-page PDF.

stevendaryl said:
The issue is that there is no such thing as a vector \vec{\sigma} such that for any choice of a measurement axis \hat{a},

\vec{\sigma} \cdot \hat{a} = \pm 1

I do not use the dot-product. I use the spin-product -- AKA the \circ-product and the \circ-product -- as defined above. Its output is definitely \pm 1 (and nothing else).
 
  • #21
N88 said:
As I see it:
Given conservation of intrinsic spin under EPRB: \boldsymbol{\sigma_1}+\boldsymbol{\sigma_2}=0. (1)
Alice's result is given by \hat{a}\circ\boldsymbol{\sigma_1}=+1. (2)
Bob's result is given by \hat{a}\circ\boldsymbol{\sigma_2}=-\hat{a}\circ\boldsymbol{\sigma_1}=-1. (3)
QED?

You are missing the bigger picture. There can't be any important effect added by the measurement apparati themselves past what cancels out for them - and they are separated and independent. Else you would not get perfect correlations when the settings are the same. And yet you get that at any angle setting, which completely goes against the idea of local realistic resolution of the spin. This is not an assumption, this is a prediction of QM.

So the conservation rule - as you are trying to interpret it - doesn't explain how you get a +1 on 1 side and -1 on the other. It would need to be fully quantized (always yielding 1 or -1 and nothing in between) on both sides. Because whatever rule you supply to "solve" that runs afoul of Bell logic. If you over-focus on Bell's math (the presentation, not the idea), you miss this. You can't construct specific examples that fit your hypothetical rules and matches QM.
 
  • Like
Likes N88
  • #22
N88 said:
I do not use the dot-product. I use the spin-product -- AKA the \circ-product and the \circ-product -- as defined above. Its output is definitely \pm 1.

We've actually discussed this longer than is appropriate for Physics Forums. This isn't the forum to discuss your personal unrefereed theories. But it sounds like to me that you are trying to construct a model of the type that Bell proved impossible. You are saying that Alice's output is a deterministic function of the variables \hat{a} and the variable \vec{\sigma_1}. Similarly, Bob's output is a deterministic function of the variables \hat{b} and \vec{\sigma_2}. So that's exactly the sort of model that Bell was considering, and which he proved is impossible.
 
  • #23
N88 said:
Given conservation of intrinsic spin under EPRB: ##\boldsymbol{\sigma_1}+\boldsymbol{\sigma_2}=0##. (1)

This is not correct. ##\boldsymbol{\sigma}## is a vector whose components are the three Pauli matrices. Those matrices are the same for Alice and Bob, and for everybody else.

If you are trying to model mathematically whatever-it-is that causes Alice and Bob's results to be opposite, it would be the state ##\vert \Psi \rangle## of the entangled particles. Stevendaryl showed you how to write down such a description in post #4.
 
  • #24
stevendaryl said:
We've actually discussed this longer than is appropriate for Physics Forums. This isn't the forum to discuss your personal unrefereed theories. But it sounds like to me that you are trying to construct a model of the type that Bell proved impossible. You are saying that Alice's output is a deterministic function of the variables \hat{a} and the variable \vec{\sigma_1}. Similarly, Bob's output is a deterministic function of the variables \hat{b} and \vec{\sigma_2}. So that's exactly the sort of model that Bell was considering, and which he proved is impossible.

I understand the PF rules. The OP is my focus. My response was to your questioning the "local-realism" of my ideas. The Fröhner paper, refereed and published, has similar ideas. That was all.
 
  • Like
Likes Jilang
  • #25
DrChinese said:
You are missing the bigger picture. There can't be any important effect added by the measurement apparati themselves past what cancels out for them - and they are separated and independent. Else you would not get perfect correlations when the settings are the same. And yet you get that at any angle setting, which completely goes against the idea of local realistic resolution of the spin. This is not an assumption, this is a prediction of QM.

So the conservation rule - as you are trying to interpret it - doesn't explain how you get a +1 on 1 side and -1 on the other. It would need to be fully quantized (always yielding 1 or -1 and nothing in between) on both sides. Because whatever rule you supply to "solve" that runs afoul of Bell logic. If you over-focus on Bell's math (the presentation, not the idea), you miss this. You can't construct specific examples that fit your hypothetical rules and matches QM.

I'm interested in the bigger picture that you mention. Is this the correct thread to discuss it?

PS: In my reply to stevendaryl above, and in the piece that you quote, the results are \pm1 only: with nothing in between. So It is fully quantized (always yielding 1 or -1 and nothing in between) on both sides. There is nothing in my conclusions that goes against mainline QM.
 
  • #26
N88 said:
I'm interested in the bigger picture that you mention. Is this the correct thread to discuss it?

The bigger picture is just Bell's theorem. I'm not sure whether you are trying to understand how Bell's theorem works, trying to understand how models that Bell called "local realistic" (and which must satisfy his inequalities, which QM violates) work, or trying to dispute the theorem by constructing a "local realistic" model that violates the inequalities.

If it's #1, that's certainly on topic here (and there are already plenty of threads on it, as a search will show you).

If it's #2, that's on topic, but you have to be clear about the definition of "local realistic". You can't make up your own; you have to use the one Bell used.

If it's #3, that's off topic since it would be a personal theory.
 
  • Like
Likes N88
  • #27
PeterDonis said:
The bigger picture is just Bell's theorem. I'm not sure whether you are trying to understand how Bell's theorem works, trying to understand how models that Bell called "local realistic" (and which must satisfy his inequalities, which QM violates) work, or trying to dispute the theorem by constructing a "local realistic" model that violates the inequalities.

If it's #1, that's certainly on topic here (and there are already plenty of threads on it, as a search will show you).

If it's #2, that's on topic, but you have to be clear about the definition of "local realistic". You can't make up your own; you have to use the one Bell used.

If it's #3, that's off topic since it would be a personal theory.

Thanks for this. #1 I use readily and happily here. #3 I have and suppress willingly here. Helping to focus my interests, and given recent clarifications to recent questions, my next question is thus under #2: How does Bell present his definition of "local realism" mathematically?

I ask because you seem to imply that Bell's "local realistic" definition satisfies his inequalities (which QM violates). I can accept that "Bell's local realistic" definition meets that criterion. But how does the related assumption that Bell (1964) makes -- ie, specifically, Bell's specified move from his (14a) to his (14b) -- relate mathematically "to local realism" in general?

In other words: It's not clear to me how it embraces all possible varieties of local realism.

To be clear: in words, and without question, I prefer the version of local-realism that d'Espagnat gave in his Scientific American article (which Bell endorsed). I do not understand how Bell's specified move relates to that definition:

(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.

I wonder now if I am missing some subtlety in this definition?

Thanks again.
 
  • #28
N88 said:
How does Bell present his definition of "local realism" mathematically?

Haven't you read his paper?

N88 said:
It's not clear to me how it embraces all possible varieties of local realism.

Can you give a mathematical definition of "all possible varieties of local realism"? If not, what makes you think it's a meaningful concept?

"Local realism" is just words. Bell's contribution was to give a precise mathematical property that, if any theory has it, leads to inequalities that QM violates. Therefore no theory that reproduces the predictions of QM can have that property. For anyone who doesn't think the property he defined captures "local realism", the challenge is to construct a better mathematical property. I haven't seen anyone do that yet.
 
  • #29
N88 said:
How does Bell present his definition of "local realism" mathematically?

Just in case it's not clear from the paper, Bell's equation (2) already contains the local realism assumption.
 
  • #30
PeterDonis said:
Haven't you read his paper?

I have read it. I disagree with Bell's move from his (14a) to his (14b). Could you explain please why you accept it?

PeterDonis said:
Can you give a mathematical definition of "all possible varieties of local realism"? If not, what makes you think it's a meaningful concept?

I think that Bell (1964), equation (1), is adequate. That's why I think it is a meaningful concept. But I can't relate it to Bell's move from his (14a) to his (14b).

PeterDonis said:
"Local realism" is just words. Bell's contribution was to give a precise mathematical property that, if any theory has it, leads to inequalities that QM violates. Therefore no theory that reproduces the predictions of QM can have that property. For anyone who doesn't think the property he defined captures "local realism", the challenge is to construct a better mathematical property. I haven't seen anyone do that yet.

In my opinion, local realism survives until Bell's (14b): So, to me, the critical property must be in Bell's move from his (14a) to his (14b). Hence my question above.
 
  • #31
N88 said:
I think that Bell (1964), equation (1), is adequate

What about his equation (2)?
 
  • #32
N88 said:
I disagree with Bell's move from his (14a) to his (14b).

Didn't we already discuss this ad nauseam in a previous thread?
 
  • #33
PeterDonis said:
Didn't we already discuss this ad nauseam in a previous thread?

Yes, we did. So I think I'll bow out of this discussion.
 
  • #34
PeterDonis said:
Just in case it's not clear from the paper, Bell's equation (2) already contains the local realism assumption.

We seem to differ slightly here: I take Bell's (1) to be the local realism assumption. To me, (2) defines the expectation of the product of the outcomes given in (1).
 
  • #35
N88 said:
I take Bell's (1) to be the local realism assumption

If you interpret it as limiting the functional dependence of A and B, then yes, I would agree.
 
  • #36
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 &gt; 0 or x &lt; 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).
 
  • #37
N88 said:
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.
 
  • #38
stevendaryl said:
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 &gt; 0 or x &lt; 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!
 
  • #39
Jilang said:
Is there a readily available paper of that?

Bell's 1964 paper is linked to in post #3 of this thread.
 
  • #40
PeterDonis said:
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?
 
  • #41
Jilang said:
I am struggling to get from equation 9 to 10

Equation 9 appears to have a typo; it should read

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

$$
B(\vec{b}, \vec{\lambda}) = - \text{sign} \ \vec{b} \cdot \vec{\lambda}
$$
 
  • #42
Jilang said:
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.
 
  • #43
PeterDonis said:
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"
 
  • #44
DrChinese said:
(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.
 
  • #45
N88 said:
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.
 

Attachments

  • bell-toy-model.jpg
    bell-toy-model.jpg
    23.8 KB · Views: 379
  • #46
N88 said:
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.

N88 said:
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.
 
  • Like
Likes N88
  • #47
PeterDonis said:
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).
 
  • #48
N88 said:
under Bell's broad specification of λ, I do not see why λ cannot vary from run to run

It can.

N88 said:
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.
 
  • #49
stevendaryl said:
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 &gt; 0 or x &lt; 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 &gt; 0 or \phi &lt; 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.
 
  • #50
PeterDonis said:
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 λi differing from λj 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?
 
Back
Top