gill1109 wrote:Yablon wrote:. . . Anybody not agree?Jay
I now understand your notation. "lambda" of a self-adjoint operator stands for "an eigenvalue" of that operator. So technically speaking, it is a many-valued function, there is a spectrum of distinct eigenvalues. On the right-hand side you have something you call "h" which stands for +/1.
I agree that if the one takes either of the two eigenvalues of both of those two operators and multiply them, the result is what is given by the right-hand side either with h = -1 or h = +1.
Good, we are on the same page.
The only fine point is that “lambda” here stands for the eigenvalues of the
non-adjoint i.e.,
non-Hermitian operators
![](http://latex.codecogs.com/png.latex?(\sigma\cdot\mathbf a)(\sigma\cdot\mathbf s))
and
![](http://latex.codecogs.com/png.latex?(\sigma\cdot (-\mathbf{s}))(\sigma\cdot \mathbf{b}))
. Because each of these is non-adjoint resulting from their Hermitian operator halves being non-commuting with one another, there is an uncertainty relation and a simultaneous measurement limitation associated with each of these, which when we get to the physics, I will argue causes h and also
s to become "hidden" variables. But that is by way of preview, no need to discuss at the moment.
gill1109 wrote:But you also say "with 50% -- 50% probability for each". If probabilities come out, them probabilities must have gone in. So you have implicitly put a joint probability distribution on the eigenvalues of the two operators you started with. If it is a uniform distribution, each of the four combinations has probability 25%, then the product equals +/-1 with probability half each.
Yes, and I am glad you caught that (as I would expect from a statistics expert).
![Smile :)](./images/smilies/icon_e_smile.gif)
I slightly mis-spoke when I said that the relation
![](http://latex.codecogs.com/png.latex?\lambda((\sigma\cdot\mathbf a)(\sigma\cdot\mathbf s))\lambda((\sigma\cdot (-\mathbf{s}))(\sigma\cdot \mathbf{b}))= (\mathbf{a}\cdot \mathbf{s}+ih\left\| \mathbf{a}\times \mathbf{s} \right\|)(-\left( \mathbf{s}\cdot \mathbf{b}+ih\left\| \mathbf{s}\times \mathbf{b} \right\| \right)))
![](http://latex.codecogs.com/png.latex?\,=-\left( \mathbf{a}\cdot \mathbf{b}+ih\left\| \mathbf{a}\times \mathbf{b} \right\| \right))
with,
![](http://latex.codecogs.com/png.latex?h=\pm1)
with 50% -- 50% probability for each
was derived “using standard eigen-mathematics.” You are correct that whether h = -1 or h = +1 has an
a priori 50%-50% probability for
![](http://latex.codecogs.com/png.latex?(\sigma\cdot\mathbf a)(\sigma\cdot\mathbf s))
, and likewise an
independent a priori 50%-50% probability for
![](http://latex.codecogs.com/png.latex?(\sigma\cdot (-\mathbf{s}))(\sigma\cdot \mathbf{b}))
. So you are correct that there is a 25% probability,
a priori, for each one of the four combinations. That is the result of the "standard eigen-mathematics." I do then introduce the
hypothesis that if h=+1 for
![](http://latex.codecogs.com/png.latex?(\sigma\cdot\mathbf a)(\sigma\cdot\mathbf s))
then,
jointly, h=+1 for
![](http://latex.codecogs.com/png.latex?(\sigma\cdot (-\mathbf{s}))(\sigma\cdot \mathbf{b}))
, and similarly for h=-1. That is how I get to the 50%-50% probability.
gill1109 wrote: So I can agree with your result but you need to further explain the notation. You are putting a joint probability distribution over the Cartesian product of the spectra of two non-commuting observables. They both depend on an unspecified variable "s". The right-hand side does not depend on "s"
You make a point which I want to stop to agree with and emphasize: "The right-hand side does not depend on
s." In physics language, I would say that means that this expression is
invariant with regard to
s, and depends only upon
a,
b and h. This will be become very important when we discuss the physics which builds upon this mathematics.
gill1109 wrote: Sorry to be pedantic about notation. As a mathematics professor, it's part of my job description. . .
You have done your job.
![Smile :)](./images/smilies/icon_e_smile.gif)
I think / hope my above clarifications remove any further ambiguity.
gill1109 wrote:So: you can define everything so that what you say is true but I don't think it has anything whatever to do with physics!
Well, you keep saying that you are a mathematician not a physicist, and I take you at your word. So, as a physicist myself, I a glad that an esteemed mathematics professor has agreed with the mathematics of my result, subject to the forgoing clarifications.
![Smile :-)](./images/smilies/icon_e_smile.gif)
As I said in the first sentence of
https://jayryablon.files.wordpress.com/ ... lrhvqm.pdf which I used to start this thread, it is my present goal to lay out “the mathematical skeleton upon which quantum mechanics as represented through Pauli linear algebra, can, in my humble opinion, be made local and realistic using hidden variables.” When I get to the physics, as I will, I want to make sure that we all have gotten beyond any disagreement regarding the correctness of the underlying mathematics, and have weeded out any ambiguity or miscommunication. Your agreement with my result subject to the foregoing clarifications is encouraging in that direction.
So now we have gone through Q1 through Q3 from my post at
viewtopic.php?f=6&t=412&sid=462f446b1d57aeb59ecc271a8b39fd99#p10335, and I believe Richard and I have substantial agreement regarding the mathematics.
----------------------------------
Therefore, I now want to return now to my Q4, which is for you Richard, and for anybody and everybody else:
“Q4: Expressions such as
![](http://latex.codecogs.com/png.latex?sign (\mathbf a \cdot \mathbf s) = \pm 1)
appear regularly in the EPRB literature. When you see such an expression, what does it mean to you, physically?” For example, this expression is in equation (9) of Bell’s original paper at, e.g.,
https://cds.cern.ch/record/111654/files ... 00_001.pdf, with
![](http://latex.codecogs.com/png.latex?\lambda)
there being what I call
s here. And specifically, don't just make the conclusory statement that it leads to Bell's equation (10) for the classical correlation. What I am asking is this:
When
![](http://latex.codecogs.com/png.latex?sign (\mathbf a \cdot \mathbf s) = +1)
, what is that telling you about the
physics of
![](http://latex.codecogs.com/png.latex?\mathbf s)
in relation to
![](http://latex.codecogs.com/png.latex?\mathbf a)
? And when
![](http://latex.codecogs.com/png.latex?sign (\mathbf a \cdot \mathbf s) = -1)
, what is that alternatively telling you about the
physics of
![](http://latex.codecogs.com/png.latex?\mathbf s)
in relation to
![](http://latex.codecogs.com/png.latex?\mathbf a)
? What I am looking for is
physics understandings along the lines of "before an observation occurs . . ." and "when an observation occurs . . ." and "after an observation occurs . . ." What is the
physics that comes to mind when you see this expression, on its own terms?
Jay
[quote="gill1109"][quote="Yablon"]. . . Anybody not agree?Jay[/quote]I now understand your notation. "lambda" of a self-adjoint operator stands for "an eigenvalue" of that operator. So technically speaking, it is a many-valued function, there is a spectrum of distinct eigenvalues. On the right-hand side you have something you call "h" which stands for +/1.
I agree that if the one takes either of the two eigenvalues of both of those two operators and multiply them, the result is what is given by the right-hand side either with h = -1 or h = +1.[/quote]
Good, we are on the same page.
The only fine point is that “lambda” here stands for the eigenvalues of the [i]non-adjoint[/i] i.e., [i]non-Hermitian[/i] operators [tex](\sigma\cdot\mathbf a)(\sigma\cdot\mathbf s)[/tex] and [tex](\sigma\cdot (-\mathbf{s}))(\sigma\cdot \mathbf{b})[/tex]. Because each of these is non-adjoint resulting from their Hermitian operator halves being non-commuting with one another, there is an uncertainty relation and a simultaneous measurement limitation associated with each of these, which when we get to the physics, I will argue causes h and also [b]s[/b] to become "hidden" variables. But that is by way of preview, no need to discuss at the moment.
[quote="gill1109"]
But you also say "with 50% -- 50% probability for each". If probabilities come out, them probabilities must have gone in. So you have implicitly put a joint probability distribution on the eigenvalues of the two operators you started with. If it is a uniform distribution, each of the four combinations has probability 25%, then the product equals +/-1 with probability half each. [/quote]
Yes, and I am glad you caught that (as I would expect from a statistics expert). :) I slightly mis-spoke when I said that the relation
[tex]\lambda((\sigma\cdot\mathbf a)(\sigma\cdot\mathbf s))\lambda((\sigma\cdot (-\mathbf{s}))(\sigma\cdot \mathbf{b}))= (\mathbf{a}\cdot \mathbf{s}+ih\left\| \mathbf{a}\times \mathbf{s} \right\|)(-\left( \mathbf{s}\cdot \mathbf{b}+ih\left\| \mathbf{s}\times \mathbf{b} \right\| \right))[/tex][tex]\,=-\left( \mathbf{a}\cdot \mathbf{b}+ih\left\| \mathbf{a}\times \mathbf{b} \right\| \right)[/tex] with, [tex]h=\pm1[/tex] with 50% -- 50% probability for each
was derived “using standard eigen-mathematics.” You are correct that whether h = -1 or h = +1 has an [i]a priori[/i] 50%-50% probability for [tex](\sigma\cdot\mathbf a)(\sigma\cdot\mathbf s)[/tex], and likewise an [i][u]independent[/u] a priori[/i] 50%-50% probability for [tex](\sigma\cdot (-\mathbf{s}))(\sigma\cdot \mathbf{b})[/tex]. So you are correct that there is a 25% probability, [i]a priori[/i], for each one of the four combinations. That is the result of the "standard eigen-mathematics." I do then introduce the [i]hypothesis[/i] that if h=+1 for [tex](\sigma\cdot\mathbf a)(\sigma\cdot\mathbf s)[/tex] then, [i]jointly[/i], h=+1 for [tex](\sigma\cdot (-\mathbf{s}))(\sigma\cdot \mathbf{b})[/tex], and similarly for h=-1. That is how I get to the 50%-50% probability.
[quote="gill1109"] So I can agree with your result but you need to further explain the notation. You are putting a joint probability distribution over the Cartesian product of the spectra of two non-commuting observables. They both depend on an unspecified variable "s". The right-hand side does not depend on "s"[/quote]
You make a point which I want to stop to agree with and emphasize: "The right-hand side does not depend on [b]s[/b]." In physics language, I would say that means that this expression is [i]invariant[/i] with regard to [b]s[/b], and depends only upon [b]a[/b], [b]b[/b] and h. This will be become very important when we discuss the physics which builds upon this mathematics.
[quote="gill1109"] Sorry to be pedantic about notation. As a mathematics professor, it's part of my job description. . . [/quote]
You have done your job. :) I think / hope my above clarifications remove any further ambiguity.
[quote="gill1109"]So: you can define everything so that what you say is true but I don't think it has anything whatever to do with physics![/quote]
Well, you keep saying that you are a mathematician not a physicist, and I take you at your word. So, as a physicist myself, I a glad that an esteemed mathematics professor has agreed with the mathematics of my result, subject to the forgoing clarifications. :-) As I said in the first sentence of https://jayryablon.files.wordpress.com/2019/10/lrhvqm.pdf which I used to start this thread, it is my present goal to lay out “the mathematical skeleton upon which quantum mechanics as represented through Pauli linear algebra, can, in my humble opinion, be made local and realistic using hidden variables.” When I get to the physics, as I will, I want to make sure that we all have gotten beyond any disagreement regarding the correctness of the underlying mathematics, and have weeded out any ambiguity or miscommunication. Your agreement with my result subject to the foregoing clarifications is encouraging in that direction.
So now we have gone through Q1 through Q3 from my post at http://www.sciphysicsforums.com/spfbb1/viewtopic.php?f=6&t=412&sid=462f446b1d57aeb59ecc271a8b39fd99#p10335, and I believe Richard and I have substantial agreement regarding the mathematics.
----------------------------------
Therefore, I now want to return now to my Q4, which is for you Richard, and for anybody and everybody else:
“Q4: Expressions such as [tex]sign (\mathbf a \cdot \mathbf s) = \pm 1[/tex] appear regularly in the EPRB literature. When you see such an expression, what does it mean to you, physically?” For example, this expression is in equation (9) of Bell’s original paper at, e.g., https://cds.cern.ch/record/111654/files/vol1p195-200_001.pdf, with [tex]\lambda[/tex] there being what I call [b]s[/b] here. And specifically, don't just make the conclusory statement that it leads to Bell's equation (10) for the classical correlation. What I am asking is this:
When [tex]sign (\mathbf a \cdot \mathbf s) = +1[/tex], what is that telling you about the [i]physics[/i] of [tex]\mathbf s[/tex] in relation to [tex]\mathbf a[/tex]? And when [tex]sign (\mathbf a \cdot \mathbf s) = -1[/tex], what is that alternatively telling you about the [i]physics[/i] of [tex]\mathbf s[/tex] in relation to [tex]\mathbf a[/tex]? What I am looking for is [i]physics[/i] understandings along the lines of "before an observation occurs . . ." and "when an observation occurs . . ." and "after an observation occurs . . ." What is the [i]physics[/i] that comes to mind when you see this expression, on its own terms?
Jay