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

[N88]

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 $(\hat{a}\cdot\boldsymbol{\sigma}_{1})$
in $\left\langle (\hat{a}\cdot\boldsymbol{\sigma}_{1})(-\boldsymbol{\sigma}_{2}\cdot\hat{b})\right\rangle$?

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.

 

[PeterDonis]

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?

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.

 

[N88]

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.

 

[stevendaryl]

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 $(\hat{a}\cdot\boldsymbol{\sigma}_{1})$
in $\left\langle (\hat{a}\cdot\boldsymbol{\sigma}_{1})(-\boldsymbol{\sigma}_{2}\cdot\hat{b})\right\rangle$?

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

 

[PeterDonis]

I will certainly benefit from discussion of the physical significance of the 'vector'**
$\boldsymbol{\sigma}$ being defined by σ≡(σ_x,σ_y,σ_z).

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 $\hat{b}$.

 

[N88]

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.

 

[stevendaryl]

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".

 

[N88]

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

 

[stevendaryl]

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.

 

[N88]

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.

 

[stevendaryl]

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.

 

[N88]

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

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.

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):

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

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

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

Continuing:

[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.]

[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.

[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).)

[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

 

[stevendaryl]

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.

 

[N88]

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.

 

[stevendaryl]

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?

 

[N88]

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.)

 

[N88]

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.

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?

 

[stevendaryl]

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.

 

[stevendaryl]

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 $\hat{a}$,

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

 

[N88]

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.

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).

 

[DrChinese]

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.

 

[stevendaryl]

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.

 

[PeterDonis]

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.

 

[N88]

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.

 

[N88]

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.

 

[PeterDonis]

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.

 

[N88]

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.

 

[PeterDonis]

How does Bell present his definition of "local realism" mathematically?

Haven't you read his paper?

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.

 

[PeterDonis]

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.

 

[N88]

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?

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).

"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.