 |
|
| Science Forum Index » Physics - Research Forum » Bell's Theorem?... |
|
Page 2 of 2 Goto page Previous 1, 2 |
|
| Author |
Message |
| Ah Ha... |
 Posted: Sun Sep 27, 2009 2:09 am |
|
|
|
Guest
|
On Sep 26, 1:59 am, harry <harald.vanlin... at (no spam) epfl.ch> wrote:
[quote:277e546f7b]
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."
[/quote:277e546f7b]
As a mathematician looking in on this, I wonder if the problem comes
down to the definition of the term "variable". It appears to me from
this discussion, that Bell defines variables to be elements of a
field, in this case real numbers, and Christian broadens this
definition to include the Clifford, or perhaps exterior, algebra (over
the same field) where non-commutativity of rotations in 3-space is
naturally represented.
Physicists are quite used to using separable Hilbert spaces of
differential forms to model physics. However, there are advantages
to representing physical structures by reversing the variance and
using locally convex topological vector spaces of "differential
chains" and their operator algebras instead of differential forms and
their operator algebras. Physical realism is one. Differential
chains can be thought of as "distributions without the test
functions". Indeed, there exists a powerful new "operator calculus"
of differential chains that offers mathematical support to
Christian's idea. The theory has been many years in the making, and I
plan to submit a formal announcement to a mathematics journal
shortly. The mathematics of Christian's idea will be seen as sound,
once the entire calculus is revealed. It builds upon what one might
call "hidden variables", namely discrete sections of the exterior
tangent bundle over a manifold, and higher order versions of these
sections, such as dipoles and quadrupoles. This provides a new way,
beyond fiber bundles, to move from the exterior algebra, which is
defined on the tangent space of each point in a manifold, to
structures globally defined on the manifold where these experiments
are taking place.
Jenny Harrison |
|
|
| Back to top |
|
|
|
| Jim Black... |
| Posted: Mon Sep 28, 2009 7:23 am |
|
|
|
Guest
|
On Thu, 24 Sep 2009 09:07:53 +0000 (UTC), Joy Christian wrote:
[quote:5d9bb9542b]I appreciate the effort that Jim Black has made to understand my work.
I would like to demonstrate however why his argument
is invalid. To begin with, it is not clear which quantum state he has
in mind. If it is the singlet state,
[/quote:5d9bb9542b]
It is.
[quote:5d9bb9542b]then quantum mechanics
does not predict what he claims. It is well known that quantum
mechanics predicts the expectation value of -- cos(theta) for his
experimental setup.
[/quote:5d9bb9542b]
But the expectation value of what? The point was that you can label
the observed states as +1 and -1, or +2 and -2, or with bivectors, or
whatever labels you choose, but when you do so, that changes the
expectation value of their product. For example, if the states are
labeled +/- 2, the expectation value of their product, according to
QM, is -4 cos(theta).
If you label them with unit bivectors, in a Clifford algebra, aligned
with the direction of measured spin, then QM predicts the funny
expectation value I derived. If you label them with bivectors in your
algebra, then I'm not sure what you get, but I'm hardly convinced it
will be -cos(theta).
[quote:5d9bb9542b]The corresponding calculation within my model is
done as follows: The observables, in his notation, are
mu.A = +/- xy and mu.B = +/- [ cos(theta) xy - sin(theta) yz ]
The product then is
(mu.A) (mu.B) = - cos(theta) -/+ xz sin(theta)
When the hidden variable mu (i.e., -/+ ) is summed over, we get -- cos
(theta), just as in quantum mechanics .
[/quote:5d9bb9542b]
It isn't very well hidden if you can figure out its value from the
product of two observables, is it?
All of this fuss could have been avoided if you simply calculated the
expected number of counts in each combination of output channels
(i.e., ++, +-, -+, and --) in your original paper, and from there
calculated the correlation. Since these counts are measured in
experiments, your model must be able to predict them if it is
complete. Have you done this calculation anywhere?
--
Jim E. Black |
|
|
| Back to top |
|
|
|
| Joy Christian... |
| Posted: Mon Sep 28, 2009 7:24 am |
|
|
|
Guest
|
On 25 Sep, 15:40, Jack <jackmal... at (no spam) yahoo.com> wrote:
[quote:b8644b0e84]On Sep 24, 5:07Â am, Joy Christian <joy.christ... at (no spam) wolfson.ox.ac.uk
wrote:
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). Â 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).
The end results of the experiment are just binary values. Â You could
label them +/-, or cat/dog, or whatever. Â They are just macroscopic
results and are in no way describable the way you are saying. Â 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).
[/quote:b8644b0e84]
Your assertion suggests that you have not read my papers. You will
find the answer to your puzzle here: http://arxiv.org/abs/0904.4259
Joy Christian |
|
|
| Back to top |
|
|
|
| Ilja... |
| Posted: Mon Sep 28, 2009 8:08 am |
|
|
|
Guest
|
On 27 Sep., 17:09, Ah Ha <profharri... at (no spam) gmail.com> wrote:
[quote:e8aa29722e]As a mathematician looking in on this, I wonder if the problem comes
down to the definition of the term "variable". It appears to me from
this discussion, that Bell defines variables to be elements of a
field, in this case real numbers, and Christian broadens this
definition to include the Clifford, or perhaps exterior, algebra (over
the same field) where non-commutativity of rotations in 3-space is
naturally represented.
[/quote:e8aa29722e]
No, it has nothing to do with it.
In Bell's approach, the hidden variables (beables) may be anything,
you can use an arbitrary mathematical space Lambda to specify
the states of reality lambda in Lambda. If there is some product
structure
in Lambda or not, if it is commutative or not, is not specified.
But a realistic model of something of course needs some connection
to the something it is supposed to explain. In this case, what
one has to explain are the observable measurement results
(detector clicked or not - two possible values at the two
places A,B in {-1,1}) in dependence
of the parameters the experimenters can fix by their free
decision (the orientation of the measurement devices a,b).
One is not free to replace the observations to be explained
by something else.
And if the experimenters decide to compute, from their
data (probability distributions p(A,B|a,b))
expectation values of the product
E(AB|a,b) = sum AB p(A,B|a,b),
and these expectation values have strange properties,
we cannot consider some other expectation values of
other functions
E(f|a,b) = sum f(A,B) p(A,B|a,b)
and name this a realistic model. |
|
|
| Back to top |
|
|
|
| Joy Christian... |
| Posted: Sun Oct 04, 2009 6:40 am |
|
|
|
Guest
|
On Oct 4, 4:34Â am, Jim Black <fmla... at (no spam) organization.edu> wrote:
[quote:7d1a3eab45]
Okay, so how about the expectation value of A^2, i.e. the square of the
outcome at one detector. Â Since the possible outcomes are +/- 1, we don't
even need to do the experiment to see that the answer should be +1. Â But if
we work it out according to your model, following the example of Eq. 19 in
quant-ph/0703179 with another a in place of the b, we get -1. Â So even if
all we care about are expectation values, your use of bivectors where one
would normally use the numbers +/- 1 can yield wrong answer.
[/quote:7d1a3eab45]
Once again you are comparing apples with
oranges and concluding that bananas are blue.
For any given quantum mechanical observable one
can always find its dispersion-free counterpart
(a multi-vector in the case of my model) such that
the corresponding expectation value is reproduced.
I am sure you will figure out what the correct
multi-vector is for the scenario you have described.
(Hint: bivectors represent binary rotations and hence
two successive bivectors amount to a sign-flip, but
being multi-vectors bivectors are also invertible).
Joy Christian |
|
|
| Back to top |
|
|
|
| Jim Black... |
| Posted: Wed Oct 07, 2009 5:43 am |
|
|
|
Guest
|
On Sun, 04 Oct 2009 12:40:45 EDT, Joy Christian wrote:
[quote:1da1402b47]On Oct 4, 4:34Â am, Jim Black <fmla... at (no spam) organization.edu> wrote:
Okay, so how about the expectation value of A^2, i.e. the square of the
outcome at one detector. Â Since the possible outcomes are +/- 1, we don't
even need to do the experiment to see that the answer should be +1. Â But if
we work it out according to your model, following the example of Eq. 19 in
quant-ph/0703179 with another a in place of the b, we get -1. Â So even if
all we care about are expectation values, your use of bivectors where one
would normally use the numbers +/- 1 can yield wrong answer.
Once again you are comparing apples with
oranges and concluding that bananas are blue.
For any given quantum mechanical observable one
can always find its dispersion-free counterpart
(a multi-vector in the case of my model) such that
the corresponding expectation value is reproduced.
I am sure you will figure out what the correct
multi-vector is for the scenario you have described.
(Hint: bivectors represent binary rotations and hence
two successive bivectors amount to a sign-flip, but
being multi-vectors bivectors are also invertible).
[/quote:1da1402b47]
No, I'm not going to try to guess how you would have me do this
computation. As the model-builder, it is your job to provide an objective
prescription for how to calculate any observable quantity, without having
to use intuition or make guesses. If following the same procedure as in
quant-ph/0703179 is not correct, feel free to refer me to where you
describe the correct procedure.
--
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 |
|
|
|
| Joy Christian... |
| Posted: Tue Oct 13, 2009 8:48 am |
|
|
|
Guest
|
On Oct 7, 4:43 pm, Jim Black <fmla... at (no spam) organization.edu> wrote:
[quote:0d2508d92c]No, I'm not going to try to guess how you would have me do this
computation. As the model-builder, it is your job to provide an objective
prescription for how to calculate any observable quantity, without having
to use intuition or make guesses. If following the same procedure as in
quant-ph/0703179 is not correct, feel free to refer me to where you
describe the correct procedure.
[/quote:0d2508d92c]
Fine.
The model discussed in http://arxiv.org/abs/quant-ph/0703179
is for the standard EPR correlations for two different
observers. You, on the other hand, are considering an entirely
different physical scenario. You want to know what would be
the outcome when two (or a sequence of) spin measurements
are performed at one and the same detector. Bell himself
considered such a scenario within his own local model for
spin. He constructed a different observable for it from the
one he constructed for the EPR correlations, as these present
a different physical scenario. Subsequently, the scenario
was also discussed at greater length by Clauser, within his
own extended hidden variable model. I too have discussed
the scenario at some length within my own hidden variable
framework. You can find my treatment of it in Section V,
page 10, of this paper: http://arxiv.org/abs/0707.1333
Joy Christian |
|
|
| Back to top |
|
|
|
| Jim Black... |
| Posted: Wed Oct 14, 2009 12:39 am |
|
|
|
Guest
|
On Tue, 13 Oct 2009 19:48:00 +0100 (BST), Joy Christian wrote:
[quote:e60395a60b]The model discussed in http://arxiv.org/abs/quant-ph/0703179
is for the standard EPR correlations for two different
observers. You, on the other hand, are considering an entirely
different physical scenario. You want to know what would be
the outcome when two (or a sequence of) spin measurements
are performed at one and the same detector.
[/quote:e60395a60b]
No, this is the exact same experimental setup. (And thus discussion of
alternate experiments is irrelevant.) I just want to take the one
measurement, square it, and compute the expectation value of the square
according to your model. How do I do that?
I'm trying to be open-minded here. It's already clear that in your model,
you have to pretend that the +/- 1 values the experimenter is recording,
multiplying, and averaging are actually bivectors. This is flat-out wrong,
but let's bear with it and see if it gives us the right answers. And it
seems that in your view, "gives us the right answers" means "gives us the
right expectation values." That's acceptable, since if you can compute
arbitrary expectation values, you can compute arbitrary probabilities. A
probability is just the expectation value of a projection operator.
So the question is, can your model accurately predict the expectation
values of arbitrary functions of measurable quantities? As a basic
consistency check, I tried to calculate the expectation value of the square
of one measurement result. The correct answer is trivially +1. Following
your method, I obtained -1. If my attempt to follow your method was wrong,
then how would you do the calculation?
--
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 |
|
|
|
| Joy Christian... |
| Posted: Wed Oct 14, 2009 2:44 am |
|
|
|
Guest
|
On Oct 14, 11:39 am, Jim Black <fmla... at (no spam) organization.edu> wrote:
[quote:c1db8ebdb3]
No, this is the exact same experimental setup.
[/quote:c1db8ebdb3]
I disagree. What you are considering is a different
experiment, and hence requires a different beable.
Read Bohr (1949) and Bell (1965) again. My model,
just like Bell's own, is a *contextual* local hidden
variable model. Different physical question means
different *context*, and that in turn means that
one has to consider different beables in general.
[quote:c1db8ebdb3]I just want to take the one
measurement, square it, and compute the expectation value of the square
according to your model. How do I do that?
As a basic
consistency check, I tried to calculate the expectation value of the square
of one measurement result. The correct answer is trivially +1. Following
your method, I obtained -1. If my attempt to follow your method was wrong,
then how would you do the calculation?
[/quote:c1db8ebdb3]
This is how I would do the calculation informally:
< A^2 > = < A* A > = < (- mu.a) (+ mu.a) > = +1 ,
where * represents the "reverse" in the language of
geometric algebra. This is correct as far as it
goes, but for a full understanding of how things
work in my framework I again urge you to look at
Section V, page 10, of http://arxiv.org/abs/0707.1333
See especially the derivation of eq.(40), which
is for a more general case than what you have been
considering. Put p = a in eq.(40), and you will have
your answer --- namely, +1.
Joy Christian |
|
|
| Back to top |
|
|
|
| Jim Black... |
| Posted: Wed Oct 14, 2009 1:53 pm |
|
|
|
Guest
|
On Wed, 14 Oct 2009 14:44:43 +0200 (CEST), Joy Christian wrote:
[quote:56bf09377b]On Oct 14, 11:39 am, Jim Black <fmla... at (no spam) organization.edu> wrote:
No, this is the exact same experimental setup.
I disagree. What you are considering is a different
experiment, and hence requires a different beable.
Read Bohr (1949) and Bell (1965) again. My model,
just like Bell's own, is a *contextual* local hidden
variable model. Different physical question means
different *context*, and that in turn means that
one has to consider different beables in general.
[/quote:56bf09377b]
It is not yours to disagree. I was posing the problem, and in the problem
I was posing, the experimental setup and the measurements performed are
specified to be the same as in the Bell's inequality experiment. Only the
calculations done by the experimenter to arrive at an expectation value are
different. If you disagree, then there has been a failure in
communication.
Either that, or you are considering the calculations done by the
experimenter to be part of the context, which I find conceivable but quite
bizarre. But more importantly, it would disprove the claim that the model
is local, since the calculations might be done in the distant future.
[quote:56bf09377b]This is how I would do the calculation informally:
A^2 > = < A* A > = < (- mu.a) (+ mu.a) > = +1 ,
where * represents the "reverse" in the language of
geometric algebra.
[/quote:56bf09377b]
Given that the experimental setup is the same, what is the reason for
taking the reverse in this calculation but not in the calculation of the
expectation value of AB?
[quote:56bf09377b]This is correct as far as it
goes, but for a full understanding of how things
work in my framework I again urge you to look at
Section V, page 10, of http://arxiv.org/abs/0707.1333
See especially the derivation of eq.(40), which
is for a more general case than what you have been
considering. Put p = a in eq.(40), and you will have
your answer --- namely, +1.
[/quote:56bf09377b]
I did look at it, but it appears to apply to a different experimental
setup, in which two measurements are taken, and then multiplied. This is
not what I am asking about.
--
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 |
|
|
|
| Joy Christian... |
| Posted: Thu Oct 15, 2009 12:26 am |
|
|
|
Guest
|
On Oct 15, 12:53 am, Jim Black <fmla... at (no spam) organization.edu> wrote:
[quote:6403dfff73]
It is not yours to disagree. I was posing the problem, and in the problem
I was posing, the experimental setup and the measurements performed are
specified to be the same as in the Bell's inequality experiment. Only the
calculations done by the experimenter to arrive at an expectation value are
different. If you disagree, then there has been a failure in
communication.
[/quote:6403dfff73]
The experimental setup is NOT the same in your question
and the question asked by EPR and Bell. Read Bell’s 1964
paper again, especially the paragraph around his equations
(4) to (7). He carefully---and separately---discusses the
case of repeated measurements made at a single detector.
Then read Bell’s “Reply to Critics” (1975). Let me quote
from his reply to make things clear. Concerning the EPR-Bell
setup he says this: “We are not at all concerned with
sequences of measurements on a given particle, or of pairs of
measurements on a given pair of particles. We are concerned
with experiments in which for each pair the “spin” of each
particle is measured *once only* (my emphasis).”
It should be clear then that what you are asking is a
different question from the question one is concerned with
in an EPR-Bell experiment. EPR and Bell are concerned with
the expectation value < AB >, but you are asking how
the expectation value < A^2 > is calculated within my
framework. These are two separate physical questions, and
they both have two separate answers within my framework,
which is a framework for a locally *contextual* hidden
variable theory. For the question you are asking, I have
already showed you how to perform the calculation
informally. Here it is again:
< A^2 > = < A* A > = < (- mu.a) (+ mu.a) > = +1 .
[quote:6403dfff73]
Given that the experimental setup is the same, what is the reason for
taking the reverse in this calculation but not in the calculation of the
expectation value of AB?
I did look at it, but it appears to apply to a different experimental
setup, in which two measurements are taken, and then multiplied. This is
not what I am asking about.
[/quote:6403dfff73]
The bottom line is not “why reverse” or “why not reverse”.
“Reverse” is simply how geometric algebra does the job
correctly. No; the bottom line is the expectation value
< AB > is equal to - a.b in one experimental question, and
the expectation value < A^2 > is equal to +1 in an entirely
different experimental question. Apples and Oranges.
Joy Christian |
|
|
| Back to top |
|
|
|
| Jim Black... |
| Posted: Thu Oct 15, 2009 10:22 am |
|
|
|
Guest
|
Joy Christian wrote:
[quote:6c001c98b0]On Oct 15, 12:53 am, Jim Black <fmla... at (no spam) organization.edu> wrote:
It is not yours to disagree. I was posing the problem, and in the problem
I was posing, the experimental setup and the measurements performed are
specified to be the same as in the Bell's inequality experiment. Only the
calculations done by the experimenter to arrive at an expectation value are
different. If you disagree, then there has been a failure in
communication.
The experimental setup is NOT the same in your question
and the question asked by EPR and Bell.
[/quote:6c001c98b0]
I'm sensing a pattern here. Given an experimental question, instead
of showing how to answer the question using your model, you attempt to
change the question. I specify that you take one measurement and
square it; you insist I take two measurements. Bell specifies that
you express the measured quantities as +/- 1 and then compute the
product of their expectation value; you insist that the measured
quantities be expressed as bivectors.
You will never understand Bell's theorem until you learn to follow an
operational description of an experiment. Bell's theorem is not about
the expectation value of some abstract quantum mechanical operator; it
is about the expectation value of the product of two measured
quantities, whose values can only be +1 or -1, never a bivector.
--
Jim E. Black |
|
|
| Back to top |
|
|
|
| Joy Christian... |
| Posted: Thu Oct 15, 2009 9:41 pm |
|
|
|
Guest
|
On Oct 15, 9:22 pm, Jim Black <trams... at (no spam) yahoo.com> wrote:
[quote:21b134cfae]
Bell's theorem is not about
the expectation value of some abstract quantum mechanical operator; it
is about the expectation value of the product of two measured
quantities, whose values can only be +1 or -1, never a bivector.
[/quote:21b134cfae]
You are absolutely right. Bell’s theorem is not about
two vectors---or even about one vector. It is not about
the expectation value of some abstract quantum mechanical
operator. It is about the expectation value of the product of
two measured quantities, whose values can only be +1 or -1.
Bell’s theorem is all about measurement results. Sadly for
Bell and his followers, however, physics is not about
measurement results. Measurement results are about physics.
Joy Christian |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Fri Nov 27, 2009 3:52 pm
|
|