 |
|
| Science Forum Index » Physics - Research Forum » Bell's Theorem?... |
|
Page 1 of 2 Goto page 1, 2 Next |
|
| Author |
Message |
| Lester Welch... |
 Posted: Sat Sep 05, 2009 11:59 pm |
|
|
|
Guest
|
Is Joy Christian disproof of Bell's theorem (arXiv:quant-ph/0703179)
accepted?
I believe the following simple example to be consistent with
Christian's disproof:
Let q= cos(a)sin(b) I +cos(a) cos(b) J + sin(a) K
where I,J,K are the basis of the quaternions. a,b are real and
arbitrary.
Then q^2 = -1 and q is isomorphic to i, the complex imaginary. Thus
in quantum mechanics, whereever we have i we can replace it with q -
with identical results and yet it has two hidden variables - a and b.
If two particles are entangled then a,b of one particle is related to
a,b of the second. If two particles are unentangled then a,b of one
particle is completely unrelated to a,b of the second. |
|
|
| Back to top |
|
|
|
| Jim Black... |
| Posted: Tue Sep 08, 2009 11:09 pm |
|
|
|
Guest
|
On Sep 6, 2:59 am, Lester Welch <lester.we... at (no spam) gmail.com> wrote:
[quote:d3c1ff3b81]Is Joy Christian disproof of Bell's theorem (arXiv:quant-ph/0703179)
accepted?
I believe the following simple example to be consistent with
Christian's disproof:
Let q= cos(a)sin(b) I +cos(a) cos(b) J + sin(a) K
where I,J,K are the basis of the quaternions. a,b are real and
arbitrary.
Then q^2 = -1 and q is isomorphic to i, the complex imaginary. Thus
in quantum mechanics, whereever we have i we can replace it with q -
with identical results and yet it has two hidden variables - a and b.
If two particles are entangled then a,b of one particle is related to
a,b of the second. If two particles are unentangled then a,b of one
particle is completely unrelated to a,b of the second.
[/quote:d3c1ff3b81]
Joy Christian's counterexample is invalid.
He considers an experiment where the measured result at each detector
is
either +1 or -1. But in his model of this experiment, he equates
these
*measured quantities* to bivectors. Not some intermediate
theoretical
construct, but the actual measured results.
(He also states that the bivectors have "values +/- 1," which seems
to
equate bivectors with scalars.)
He then computes the correlation of the measurements using a Clifford
product to get a correlation that matches experiment. But the
correlation
isn't something you directly measure; it's something you compute from
the
measured quantities, which are real numbers, using the ordinary real
number
product. What he has done, changing both the algebraic type of the
measured
quantities and the means by which the correlation is computed, amounts
to
altering the experiment.
--
Jim E. Black |
|
|
| Back to top |
|
|
|
| harry... |
| Posted: Thu Sep 10, 2009 6:38 am |
|
|
|
Guest
|
On Sep 9, 11:09 am, Jim Black <trams... at (no spam) yahoo.com> wrote:
[quote:4cd9048748]On Sep 6, 2:59 am, Lester Welch <lester.we... at (no spam) gmail.com> wrote:
Is Joy Christian disproof of Bell's theorem (arXiv:quant-ph/0703179)
accepted?
I believe the following simple example to be consistent with
Christian's disproof:
Let q= cos(a)sin(b) I +cos(a) cos(b) J + sin(a) K
where I,J,K are the basis of the quaternions. a,b are real and
arbitrary.
Then q^2 = -1 and q is isomorphic to i, the complex imaginary. Thus
in quantum mechanics, whereever we have i we can replace it with q -
with identical results and yet it has two hidden variables - a and b.
If two particles are entangled then a,b of one particle is related to
a,b of the second. If two particles are unentangled then a,b of one
particle is completely unrelated to a,b of the second.
Joy Christian's counterexample is invalid.
He considers an experiment where the measured result at each detector
is
either +1 or -1. But in his model of this experiment, he equates
these *measured quantities* to bivectors. Not some intermediate
theoretical construct, but the actual measured results.
[/quote:4cd9048748]
Yes, that's as much as I understood as well.
[quote:4cd9048748](He also states that the bivectors have "values +/- 1," which seems
to equate bivectors with scalars.)
[/quote:4cd9048748]
Hmmm... giving him the benefit of the doubt (rather little doubt, as
he's an expert in the field), I guess him to mean that these are unit
vectors that have +1 or -1 as possible values. Then they correspond to
either "spin up" or "spin down, just as the typical measurement
results of QM.
[quote:4cd9048748]He then computes the correlation of the measurements using a Clifford
product to get a correlation that matches experiment. But the
correlation
isn't something you directly measure; it's something you compute from
the
measured quantities, which are real numbers, using the ordinary real
number product.
[/quote:4cd9048748]
Please explain why you claim that instead of for example "spin up" and
spin down" (which correspond to vectors), real numbers are measured.
[quote:4cd9048748]What he has done, changing both the algebraic type of the
measured
quantities and the means by which the correlation is computed, amounts
to altering the experiment.
[/quote:4cd9048748]
It appears to me that you are mistaken; but if not, please elaborate!
Note: this was helpful fo me, for now I start to understand a little
of Joy Christian's first* paper on this. :-)
Thanks,
Harald
*For later clarifications see his overview on:
http://www.perimeterinstitute.ca/index.php?option=com_content&task=view&id=30&Itemid=72&pi=Joy_Christian |
|
|
| Back to top |
|
|
|
| Lester Welch... |
| Posted: Sun Sep 13, 2009 9:14 am |
|
|
|
Guest
|
On Sep 6, 5:59 am, Lester Welch <lester.we... at (no spam) gmail.com> wrote:
[quote:c641ffca74]Is Joy Christian disproof of Bell's theorem (arXiv:quant-ph/0703179)
accepted?
I believe the following simple example to be consistent with
Christian's disproof:
Let q= cos(a)sin(b) I +cos(a) cos(b) J + sin(a) K
where I,J,K are the basis of the quaternions. a,b are real and
arbitrary.
Then q^2 = -1 and q is isomorphic to i, the complex imaginary. Thus
in quantum mechanics, whereever we have i we can replace it with q -
with identical results and yet it has two hidden variables - a and b.
If two particles are entangled then a,b of one particle is related to
a,b of the second. If two particles are unentangled then a,b of one
particle is completely unrelated to a,b of the second.
[/quote:c641ffca74]
Thanks. From that irrefutable source, Wikipedia [  ]
"Later, Bell's theorem would prove (in the opinion of most physicists
and contrary to Einstein's assertion) that local hidden variables are
impossible."
So, regardless of whether Joy Christian is right or not, does Bell's
theorem address the question of hidden variables? |
|
|
| Back to top |
|
|
|
| Ilja... |
| Posted: Sun Sep 13, 2009 11:47 pm |
|
|
|
Guest
|
On 10 Sep., 21:38, harry <harald.vanlin... at (no spam) epfl.ch> wrote:
[quote:ecb565cc79]Please explain why you claim that instead of for example "spin up" and
spin down" (which correspond to vectors), real numbers are measured.
[/quote:ecb565cc79]
"Spin up" and "spin down" correspond to two different labels, thus, a
quite simple set {u,d} of two elements.
To identify them with anything more complicated than a set of two
elements, like {-1,1}, seems a waste of allowed values. But, of
course,
you can feel free to identify them with any subset of any vector
space,
"up" as one element, and "down" with -"up".
But the more important point is that the description of the experiment
completely specifies what has to be done with the labels "u" and
"d" to obtain E(a,b). Namely, you have to look if the values
of A and B are equal (above "u" or above "d"), and in this case
memorize a +1 for this run of the experiment, or if they are unequal,
then you have to memorize a -1. And then you have to compute
the average over all runs.
This is the _definition_ of the E(a,b) used in Bell's inequality,
thus, Christian is not free to change it. But he changes it.
If you compute, instead of this E(a,b), something else, you are
not explaining the violation of the BI.
[quote:ecb565cc79]It appears to me that you are mistaken; but if not, please elaborate!
[/quote:ecb565cc79]
I also think Christian is mistaken, and for the same reasons. |
|
|
| Back to top |
|
|
|
| Ilja... |
| Posted: Mon Sep 14, 2009 2:29 am |
|
|
|
Guest
|
On 14 Sep., 00:14, Lester Welch <lester.we... at (no spam) gmail.com> wrote:
[quote:1aebddd08a]Thanks. From that irrefutable source, Wikipedia [ ]
"Later, Bell's theorem would prove (in the opinion of most physicists
and contrary to Einstein's assertion) that local hidden variables are
impossible."
So, regardless of whether Joy Christian is right or not, does Bell's
theorem address the question of hidden variables?
[/quote:1aebddd08a]
It does address the problem of local hidden variables.
The problem of hidden variables in general (possibly nonlocal,
that means with a hidden preferred frame in the relativistic domain)
was addressed by Bohm with an explicit example:
de Broglie-Bohm pilot wave theory. |
|
|
| Back to top |
|
|
|
| Jim Black... |
| Posted: Mon Sep 14, 2009 2:29 am |
|
|
|
Guest
|
On Thu, 10 Sep 2009 16:38:01 +0000 (UTC), harry wrote:
[quote:f2b9e6fec9]On Sep 9, 11:09 am, Jim Black <trams... at (no spam) yahoo.com> wrote:
[/quote:f2b9e6fec9]
[regarding content at
http://www.perimeterinstitute.ca/index.php?option=com_content&task=view&id=30&Itemid=72&pi=Joy_Christian]
[quote:f2b9e6fec9]Joy Christian's counterexample is invalid.
He considers an experiment where the measured result at each detector is
either +1 or -1. But in his model of this experiment, he equates
these *measured quantities* to bivectors. Not some intermediate
theoretical construct, but the actual measured results.
Yes, that's as much as I understood as well.
(He also states that the bivectors have "values +/- 1," which seems
to equate bivectors with scalars.)
Hmmm... giving him the benefit of the doubt (rather little doubt, as
he's an expert in the field), I guess him to mean that these are unit
vectors that have +1 or -1 as possible values. Then they correspond to
either "spin up" or "spin down, just as the typical measurement
results of QM.
[/quote:f2b9e6fec9]
To be clear, unit bivectors. Also known as two-forms or antisymmetric
tensors of rank two in three dimensions. They can be represented as
pseudovectors -- like vectors but they don't flip sign under a parity
transformation.
[quote:f2b9e6fec9]He then computes the correlation of the measurements using a Clifford
product to get a correlation that matches experiment. But the correlation
isn't something you directly measure; it's something you compute from the
measured quantities, which are real numbers, using the ordinary real
number product.
Please explain why you claim that instead of for example "spin up" and
spin down" (which correspond to vectors), real numbers are measured.
[/quote:f2b9e6fec9]
You can express the measured spin values as bivectors if you want to.
What you cannot do is compute the expectation value of the Clifford
product of those bivectors and pretend it's the same as the
expectation value of real numbers which Bell's theorem is concerned
with.
So let us express the spin values as bivectors and ask: Does quantum
mechanics agree with Joy Christian's result?
In what follows, x, y, and z mean the unit vectors in the x, y, and z
directions. In this notation, the rules for the Clifford algebra Cl_
{3,0} can be summarized as:
x^2 = y^2 = z^2 = 1
xy = -yx, xz = -zx, yz = -zy
The usual associative and distributive axioms apply. Addition is
commutative, and it's perfectly okay to add any combination of
scalars, vectors, and all the possible products of vectors. We define
I = xyz, as in Joy Christian's paper.
Let detector A measure spin along the z-axis. We can represent "spin
up" as the bivector Iz = xy, and "spin down" as the bivector -xy. In
the real number representation, these measurements are represented as
+1 for spin up and -1 for spin down.
Let detector B measure spin along the axis
cos(theta) z + sin(theta) x.
Spin aligned with the axis (+1 in the numeric rep.) is then
represented by
I [cos(theta) z + sin(theta) x] = cos(theta) xy + sin(theta) yz,
and the opposite spin (-1) is
- cos(theta) xy - sin(theta) yz.
According to quantum mechanics, the probabilities of the various
measurement outcomes are:
++ : (1/2) sin^2(theta/2)
+- : (1/2) cos^2(theta/2)
-+ : (1/2) cos^2(theta/2)
-- : (1/2) sin^2(theta/2)
where "+-" indicates +1 at A and -1 at B.
The value of the Clifford product of the spins is:
++ : xy * [cos(theta) xy + sin(theta) yz]
= - cos(theta) + sin(theta) xz
+- : + cos(theta) - sin(theta) xz
-+ : + cos(theta) - sin(theta) xz
-- : - cos(theta) + sin(theta) xz
Therefore quantum mechanics predicts an expectation value of:
cos^2(theta) - sin(theta) cos(theta) xz
Whereas Joy Christian's model predicts:
-cos(theta)
--
Jim E. Black (domain in headers)
How to filter out stupid arguments in 40tude Dialog:
!markread,ignore From "Name" +"<email address>"
[X] Watch/Ignore works on subthreads |
|
|
| Back to top |
|
|
|
| harry... |
| Posted: Mon Sep 14, 2009 8:36 pm |
|
|
|
Guest
|
On Sep 14, 2:29 pm, Jim Black <trams... at (no spam) yahoo.com> wrote:
[quote:8f1938c493]On Thu, 10 Sep 2009 16:38:01 +0000 (UTC), harry wrote:
On Sep 9, 11:09 am, Jim Black <trams... at (no spam) yahoo.com> wrote:
[regarding content athttp://www.perimeterinstitute.ca/index.php?option=com_content&task=vi...]
Joy Christian's counterexample is invalid.
He considers an experiment where the measured result at each detector is
either +1 or -1. But in his model of this experiment, he equates
these *measured quantities* to bivectors. Not some intermediate
theoretical construct, but the actual measured results.
Yes, that's as much as I understood as well.
(He also states that the bivectors have "values +/- 1," which seems
to equate bivectors with scalars.)
Hmmm... giving him the benefit of the doubt (rather little doubt, as
he's an expert in the field), I guess him to mean that these are unit
vectors that have +1 or -1 as possible values. Then they correspond to
either "spin up" or "spin down, just as the typical measurement
results of QM.
To be clear, unit bivectors. Also known as two-forms or antisymmetric
tensors of rank two in three dimensions. They can be represented as
pseudovectors -- like vectors but they don't flip sign under a parity
transformation.
He then computes the correlation of the measurements using a Clifford
product to get a correlation that matches experiment. But the correlation
isn't something you directly measure; it's something you compute from the
measured quantities, which are real numbers, using the ordinary real
number product.
Please explain why you claim that instead of for example "spin up" and
spin down" (which correspond to vectors), real numbers are measured.
You can express the measured spin values as bivectors if you want to.
What you cannot do is compute the expectation value of the Clifford
product of those bivectors and pretend it's the same as the
expectation value of real numbers which Bell's theorem is concerned
with.
So let us express the spin values as bivectors and ask: Does quantum
mechanics agree with Joy Christian's result?
In what follows, x, y, and z mean the unit vectors in the x, y, and z
directions. In this notation, the rules for the Clifford algebra Cl_
{3,0} can be summarized as:
x^2 = y^2 = z^2 = 1
xy = -yx, xz = -zx, yz = -zy
The usual associative and distributive axioms apply. Addition is
commutative, and it's perfectly okay to add any combination of
scalars, vectors, and all the possible products of vectors. We define
I = xyz, as in Joy Christian's paper.
Let detector A measure spin along the z-axis. We can represent "spin
up" as the bivector Iz = xy, and "spin down" as the bivector -xy. In
the real number representation, these measurements are represented as
+1 for spin up and -1 for spin down.
Let detector B measure spin along the axis
cos(theta) z + sin(theta) x.
Spin aligned with the axis (+1 in the numeric rep.) is then
represented by
I [cos(theta) z + sin(theta) x] = cos(theta) xy + sin(theta) yz,
and the opposite spin (-1) is
- cos(theta) xy - sin(theta) yz.
According to quantum mechanics, the probabilities of the various
measurement outcomes are:
++ : (1/2) sin^2(theta/2)
+- : (1/2) cos^2(theta/2)
-+ : (1/2) cos^2(theta/2)
-- : (1/2) sin^2(theta/2)
where "+-" indicates +1 at A and -1 at B.
The value of the Clifford product of the spins is:
++ : xy * [cos(theta) xy + sin(theta) yz]
= - cos(theta) + sin(theta) xz
+- : + cos(theta) - sin(theta) xz
-+ : + cos(theta) - sin(theta) xz
-- : - cos(theta) + sin(theta) xz
Therefore quantum mechanics predicts an expectation value of:
cos^2(theta) - sin(theta) cos(theta) xz
Whereas Joy Christian's model predicts:
-cos(theta)
--
[/quote:8f1938c493]
Jim thanks very much for the clarification; it is reassuring to see
that you now don't suggest anymore that he confuses bivectors with
scalars, and went along with him in the use of bivectors. But
regretfully (or happily, depending on one's bias), also your new
criticism is wrong IMHO. You specified that "addition is commutative",
and apparently you put that statement to use in your reproduction of
Christian's model.
To the contrary, Joy Christian elaborated on the fact that the
addition of his observables is NOT commutative; this is a main
discussion point in his follow-up papers. As he puts it in
http://arxiv.org/pdf/0707.1333v2 :
"To be sure, the non-commutativity of observables plays a central role
in our model".
Therefore I think to have reason to suspect that you have not
correctly reproduced his model, despite my lack of understanding of
this topic. Indeed, it would be quite surprising if such an expert
would make the kind of error that you suggest without noticing it over
the course of two years while in consultation with leaders in the
field. Thus I simply assume his math to be correct; my doubt is about
the physical meaning of his "beables".
Regards,
Harald |
|
|
| Back to top |
|
|
|
| Jim Black... |
| Posted: Tue Sep 15, 2009 6:53 pm |
|
|
|
Guest
|
On Sep 14, 11:36 pm, harry <harald.vanlin... at (no spam) epfl.ch> wrote:
[quote:f494843569]Jim thanks very much for the clarification; it is reassuring to see
that you now don't suggest anymore that he confuses bivectors with
scalars, and went along with him in the use of bivectors. But
regretfully (or happily, depending on one's bias), also your new
criticism is wrong IMHO. You specified that "addition is commutative",
and apparently you put that statement to use in your reproduction of
Christian's model.
To the contrary, Joy Christian elaborated on the fact that the
addition of his observables is NOT commutative; this is a main
discussion point in his follow-up papers. As he puts it inhttp://arxiv.or=
g/pdf/0707.1333v2:
"To be sure, the non-commutativity of observables plays a central role
in our model".
Therefore I think to have reason to suspect that you have not
correctly reproduced his model, despite my lack of understanding of
this topic. Indeed, it would be quite surprising if such an expert
would make the kind of error that you suggest without noticing it over
the course of two years while in consultation with leaders in the
field. Thus I simply assume his math to be correct; my doubt is about
the physical meaning of his "beables".
[/quote:f494843569]
It's multiplication, not addition, that's non-commutative here.
Nevertheless, the example calculation probably does not correspond to
the algebra(s?) Christian has in mind. The math in Christian's paper
is quite unclear, and I'm not sure I want to muddle through it. But
the calculation was not the main point. The main point was this:
[quote:f494843569]You can express the measured spin values as bivectors if you want to.
What you cannot do is compute the expectation value of the Clifford
product of those bivectors and pretend it's the same as the
expectation value of
[/quote:f494843569]
[the product of]
[quote:f494843569]real numbers which Bell's theorem is concerned
with.
[/quote:f494843569]
This is what Christian has done, and it's a conceptual error, not a
mathematical one. I don't know whether Christian's math is
consistent, but with a critical conceptual mistake, it doesn't matter.
--
Jim E. Black |
|
|
| Back to top |
|
|
|
| harry... |
| Posted: Fri Sep 18, 2009 9:55 am |
|
|
|
Guest
|
[Apparently my earlier message got lost]
"Jim Black" <tramspap at (no spam) yahoo.com> wrote in message
news:20ef93c6-5d47-4c62-bf46-399a333cb734 at (no spam) v23g2000pro.googlegroups.com...
[quote:5e1cfe8aa5]On Sep 14, 11:36 pm, harry <harald.vanlin... at (no spam) epfl.ch> wrote:
Jim thanks very much for the clarification; it is reassuring to see
that you now don't suggest anymore that he confuses bivectors with
scalars, and went along with him in the use of bivectors. But
regretfully (or happily, depending on one's bias), also your new
criticism is wrong IMHO. You specified that "addition is commutative",
and apparently you put that statement to use in your reproduction of
Christian's model.
To the contrary, Joy Christian elaborated on the fact that the
addition of his observables is NOT commutative; this is a main
discussion point in his follow-up papers. As he puts it
inhttp://arxiv.or=
g/pdf/0707.1333v2:
"To be sure, the non-commutativity of observables plays a central role
in our model".
Therefore I think to have reason to suspect that you have not
correctly reproduced his model, despite my lack of understanding of
this topic. Indeed, it would be quite surprising if such an expert
would make the kind of error that you suggest without noticing it over
the course of two years while in consultation with leaders in the
field. Thus I simply assume his math to be correct; my doubt is about
the physical meaning of his "beables".
It's multiplication, not addition, that's non-commutative here.
[/quote:5e1cfe8aa5]
Sorry for the glitch.
[quote:5e1cfe8aa5]Nevertheless, the example calculation probably does not correspond to
the algebra(s?) Christian has in mind. The math in Christian's paper
is quite unclear, and I'm not sure I want to muddle through it. But
the calculation was not the main point. The main point was this:
You can express the measured spin values as bivectors if you want to.
What you cannot do is compute the expectation value of the Clifford
product of those bivectors and pretend it's the same as the
expectation value of
[the product of]
real numbers which Bell's theorem is concerned
with.
This is what Christian has done, and it's a conceptual error, not a
mathematical one. I don't know whether Christian's math is
consistent, but with a critical conceptual mistake, it doesn't matter.
[/quote:5e1cfe8aa5]
Thanks for the clarification. I now verified with Bell's original paper that
Bell's theorem according to Bell is not really about real numbers (which is
mathematics), but about the foundation of physics question if deterministic
local realism can be compatible with quantum mechanics. For sure that's also
the question that Joy Christian discusses.
Regretfully I can't (yet) picture Joy Christian's "beables", thus I'm not
convinced either.
Best regards,
Harald |
|
|
| Back to top |
|
|
|
| Joy Christian... |
| Posted: Thu Sep 24, 2009 7:32 am |
|
|
|
Guest
|
[quote:6261b688d0]Hello Joy,
I am so glad to see you have joined this discussion and enjoyed
reading your very clear response. I find your mathematics sound
and interesting, and physicists need to consider your proposed
application to physics carefully in an open discussion. SPR could
be just the right forum for this. I especially admire you for joining
the discussion after the derisive name calling. (Is someone
moderating these postings? It looks like graffiti scrawled in a back
alley. Is there no civility or decorum? I hope that you will delete
all occurrences immediately before further damage is done. )
Jenny Harrison
Professor of Mathematics
University of California, Berkeley
[/quote:6261b688d0]
Hello Jenny,
I am glad you liked my response. I would be happy to address any further
questions. There is also a talk I gave recently. It may provide further
clarification of some of the issues raised above. It can be found among
other talks posted at the FQXi website: http://fqxi.org/conference/talks
..
Joy Christian |
|
|
| Back to top |
|
|
|
| student... |
| Posted: Fri Sep 25, 2009 3:23 am |
|
|
|
Guest
|
On Sep 24, 7:07 pm, Joy Christian <joy.christ... at (no spam) wolfson.ox.ac.uk>
wrote:
[quote:5e264f5dff]Several people have asked me to respond to the comments made here
about my counterexample to Bell’s theorem,
so here is my response. In my view Bell’s theorem is based on a
serious topological error. The error lies in the very first
equation of Bell’s famous paper. He associates numbers +1 and -1 with
the end results of an EPR-type experiment, and
writes them as A ( a, L ) = +1 or -1. What could be wrong with such an
innocent assumption?
[/quote:5e264f5dff]
Nothing as such. An assumption is an assumption. Inequalities are
rigorously derived on the basis of this assumption.
[quote:5e264f5dff]Well, the problem is that A
and B are supposed to represent values of the EPR elements of reality
(or spin components). But EPR-Bohm elements
of reality have a very specific topological structure---they live on a
unit 2-sphere (i.e., on the surface of a unit ball). This
topological structure differs from the topological structure presumed
by Bell in the functions A ( a, L ) = +1 or -1, which live
on a unit 0-sphere, not 2-sphere. Thus Bell’s theorem simply does not
apply to the EPR argument, unless one modifies
his main assumption by writing his function as A ( a, L ) = +1 or -1
about a. After all, no one has ever observed a “click” in
an experiment other than about some experimental direction a.
[/quote:5e264f5dff]
One observes results, for a given direction a, which can be
labelled as +1 or -1. Bell's theorem is a rigorous statement
about such results, based effectively on the assumption that
there is a hidden variable L which predetermines them. If your
model does not have such a prediction (i.e., functions A(a,L)
and B(b,L), then it does not use this assumption. However,
this does not constitute a 'disproof' of Bell theorem, it simply
means your model is not a 'local realistic' model as defined
by Bell.
[quote:5e264f5dff]With
this simple change the function A now takes on
values in a topological 2-sphere, not the real line, thereby correctly
representing the EPR elements of reality. The values
of the spin components are still +1 or -1, but they now reside on the
surface of a unit ball. This, in essence, is the only
change I have made in any of my papers. But once this change is made,
no contradiction with quantum mechanics arises.
In fact I have been able to reproduced many complicated quantum
mechanical results by implementing this corrected
assumption. And I have done this in a manifestly local and realistic
manner. Hence the title “disproof of Bell’s theorem.”
[/quote:5e264f5dff]
As you are perfectly well aware, most experts in the field do not
by any means consider that the above constitutes a 'local realistic'
model in the sense of Bell. In fact, you are explicitly making a
different definition of 'local realistic' to Bell (and everyone else).
I could redefine 2 to be different from 1+1, but this does not
disprove that 2=1+1 in the standard defintion!
Let me use the same logic to 'generalise' your model to give a
'local realistic' model of all QM (not just spin):
I represent
every measurement by a Hermitian matrix A having the
measurement results as eigenvalues relative to some basis.
After all, no one has ever observed a measurement result
without choosing a basis. With this simple change the
function A now takes values on a suitable space of matrices,
thereby correctly representing the EPR elements of reality.
The values of the outcomes are still the eigenvalues of A, but
they now reside in a Hermitian matrix. This is, in essence,
the only change I have made to your paper (one also replaces
your strange measure theory, that you gloss over above,
by density operators). Once this change is made no
contradiction with quantum mechanics arises. Indeed,
surprise, surprise, it IS quantum mechanics!! And I have
done this in a manifestly local and realistic way. Hence
the title "This spoof of Bell's theorem".
As I mention in an earlier comment, there is something
of value in your papers, in that they yield a
Grassmannian model for two spin-1/2 systems which in
some sense is 'simpler' that the model based on spin
operators. However, claims that either model is
'realistic' is misleading, and inconsistent with the
rest of the literature on Bell's theorem. Maybe you
should think of a better term (eg, 'nonrealistic' would
be quite appropriate). |
|
|
| Back to top |
|
|
|
| Jack... |
| Posted: Fri Sep 25, 2009 4:40 am |
|
|
|
Guest
|
On Sep 24, 5:07 am, Joy Christian <joy.christ... at (no spam) wolfson.ox.ac.uk>
wrote:
[quote:0225241bee]He associates numbers +1 and -1 with the end results of an EPR-type experiment, and writes them as A ( a, L ) = +1 or -1. What could be wrong with such an innocent assumption? Well, the problem is that A and B are supposed to represent values of the EPR elements of reality (or spin components). But EPR-Bohm elements of reality have a very specific topological structure---they live on a unit 2-sphere (i.e., on the surface of a unit ball).
[/quote:0225241bee]
The end results of the experiment are just binary values. You could
label them +/-, or cat/dog, or whatever. They are just macroscopic
results and are in no way describable the way you are saying. The
only way things could be more complicated is if there is more than one
actual outcome, or in other words, in a Many-Worlds Interpretation
(which can indeed be local). |
|
|
| Back to top |
|
|
|
| Joy Christian... |
| Posted: Fri Sep 25, 2009 1:21 pm |
|
|
|
Guest
|
On 25 Sep, 14:23, student <of_1001_nig... at (no spam) hotmail.com> wrote:
[quote:b9221b47e2]
Nothing as such. An assumption is an assumption. Inequalities are
rigorously derived on the basis of this assumption.
[/quote:b9221b47e2]
Bell’s theorem has foundational significance only within the context
of the EPR argument. Bell’s assumption is incompatible with the EPR
criteria of locality, reality, and completeness. This becomes evident
when one considers these criteria collectively within the coherence of
the EPR argument. An inequality derived using a faulty assumption
cannot have any relevance for the question of local realism. Just as
von Neumann’s theorem could not rule out all hidden variable theories
because of its faulty assumption, Bell’s theorem cannot---and does
not---rule out a local-realistic theory. These issues are discussed in
more detail on the pages 3 to 7 of http://arxiv.org/abs/0904.4259
[quote:b9221b47e2]
One observes results, for a given direction a, which can be
labelled as +1 or -1. Bell's theorem is a rigorous statement
about such results, based effectively on the assumption that
there is a hidden variable L which predetermines them. If your
model does not have such a prediction (i.e., functions A(a,L)
and B(b,L), then it does not use this assumption. However,
this does not constitute a 'disproof' of Bell theorem, it simply
means your model is not a 'local realistic' model as defined
by Bell.
[/quote:b9221b47e2]
Given L and a, my model does predict a definite binary outcome --- +1
or -1.
The only difference is that my beables take their values on a unit 2-
sphere,
thereby correctly representing the EPR elements of reality---whereas
Bell’s
beables take their values on a subset of the real line---thereby
incorrectly
representing the EPR elements of reality. This is better explained in
the
paper: http://arxiv.org/abs/0904.4259 . I fail to see why the numbers
+1 and
-1 should be regarded as less real when they are taken to have the
topology
of a 2-sphere, and more real when they are taken to have the topology
of the
real line. After all, EPR elements of reality are not lined up as a
real line.
They are organized as points of a unit 2-sphere. My models are
completely
consistent with the EPR criteria of locality, reality, and
completeness. They
also rigorously respect Bell’s own criterion of locality, or
factorizability.
Hence I do not recognize my model in some of your assertions.
[quote:b9221b47e2]
your strange measure theory, that you gloss over above,
[/quote:b9221b47e2]
I have only used the measure theory that was used by von Neumann in
his pioneering book, and by Bell himself in the first two of his
papers. I do
not find anything strange about it. Further discussion of this theory
can be
found on the page 8 of http://arxiv.org/abs/0904.4259
Joy Christian |
|
|
| Back to top |
|
|
|
| harry... |
| Posted: Fri Sep 25, 2009 10:59 pm |
|
|
|
Guest
|
On Sep 25, 3:23 pm, student <of_1001_nig... at (no spam) hotmail.com> wrote:
[quote:451bacc0af]On Sep 24, 7:07 pm, Joy Christian <joy.christ... at (no spam) wolfson.ox.ac.uk
wrote:
Several people have asked me to respond to the comments made here
about my counterexample to Bell’s theorem,
so here is my response. In my view Bell’s theorem is based on a
serious topological error. The error lies in the very first
equation of Bell’s famous paper. He associates numbers +1 and -1 with
the end results of an EPR-type experiment, and
writes them as A ( a, L ) = +1 or -1. What could be wrong with such an
innocent assumption?
Nothing as such. An assumption is an assumption. Inequalities are
rigorously derived on the basis of this assumption.
[/quote:451bacc0af]
Assumptions can be faulty; if a physics problem is wrongly translated
into math, then in principle the solution is wrong as well.
As I stressed earlier, Bell's theorem concerns the question if
deterministic local realism can be compatible with quantum mechanics.
Bell in his 1964 introduction:
"In this note that idea [of EPR that QM should be supplemented by
additional variables] will be formulated mathematically and shown to
be incompatible with the statistical predictions of quantum
mecanics."
and his generalization in that same note:
"for at least one quantum mechanical state [..], the statistical
predictions of quantum mechanics are incompatible with separable
predetermination."
[..]
Harald |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Sun Nov 29, 2009 4:25 am
|
|