FlorinM said:
Thank you very much. I appreciate that we are back from wooly vague words and into the realm of verifiable math. Please bear with me, this might take a while.
We are still talking about
http://arxiv.org/abs/1106.0748 as it appears to be far more detailed than the original paper that started this topic.
Start with eq(1). Here the author gives the results predicted by QM and observed in experiments:
\mathcal{A}(\alpha)=\pm 1, \mathcal{B}(\beta)=\pm 1
E(\alpha)=0, E(\beta)=0
E(\alpha,\beta)=-\cos^2(\alpha-\beta)
(eq 1)
Here E(\alpha,\beta) represents the expected value of simultaneously observing remote measurement results \mathcal{A}(\alpha) and \mathcal{B}(\beta) along the polarization angles \alpha and \beta, respectively.
First a small clarification, the text should read: "E(\alpha,\beta) represents the expected value
of the product \mathcal{A}(\alpha)\mathcal{B}(\beta) of simultaneously observing...". I added the words in bold because it is important exactly what kind of product we are dealing with here.
Now, the range of \mathcal{A}(\alpha) and \mathcal{B}(\beta) is a set of {-1, +1}. I stress that these are normal ordinary everyday integer +1 and -1, not some fancy Grassman +1 and -1 and the multiplication between \mathcal{A}(\alpha) and \mathcal{B}(\beta) for the purposes of computing E(\alpha,\beta) is normal everyday multiplication, not an inner product, not an outer product, not an wedge product, not a geometric product, not any other fancy kind of product.
Why is that so? Because Bell chose it to be so. The experiment itself can produce any kind of indication of the outcome, it could be 0 or 1, 'X' or 'O', up or down, red LED or green LED. Bell chose to associate these outcomes with numbers +1 and -1 for the purposes of deriving his inequality. And this is how the data is presented in real Bell-type experiments.
Obviously these numbers, be it theoretical results or real experimental data, are computed using normal everyday arithmetic, normal everyday definitions of expectation value, standard deviation, correlation etc, taken from the statistics 101.
Therefore if the author claims to disprove Bell and to demonstrate \cos^2 rule arising from locally realistic \mathcal{A}(\alpha) and \mathcal{B}(\beta), then he has to play by the rules. This means, internally \mathcal{A}(\alpha) and \mathcal{B}(\beta) can use whatever fancy math you want, but their outcomes should be counted the same way the outcomes of real experiments are counted.
To summarize: for the results to be relevant to Bell's theorem and to real-life experiments, functions \mathcal{A}(\alpha) and \mathcal{B}(\beta) should return either -1 or +1 which are to be treated as normal integer numbers using normal arithmetic and statistics. Do you agree with this statement?
Why do I have to explain is so painstakingly? Because I'm sick of people saying "this is not an ordinary multiplication/You won't understand/Go read a book" when in fact it is (should have been) ordinary multiplication.
Now, fast-forward to eq (16).
To this end, we have assumed that the complete state of the photons is given by \mu = \pm I, where I is the fundamental trivector defined in Eq. (2). The detections of photon polarizations observed by Alice and Bob along their respective axes \alpha and \beta, with the bivector basis fixed by the trivector \mu, can then be represented intrinsically as points of the physical space S^3, by the following two local variables:
S^3 \ni \mathcal{A} (\alpha,\mu)=(-I \cdot \tilde{a})(+\mu \cdot \tilde{a})= \begin{cases} +1 & \text{if } \mu = +I \\ -1 & \text{if } \mu = -I \end{cases}
(eq 16)
and
S^3 \ni \mathcal{B} (\beta,\mu)=(+I \cdot \tilde{b})(+\mu \cdot \tilde{b})= \begin{cases} -1 & \text{if } \mu = +I \\ +1 & \text{if } \mu = -I \end{cases}
(eq 17)
with equal probabilities for \mu being either +I or -I, and the rotating vectors \tilde{a} and \tilde{b} defined as
\tilde{a} = e_x \cos 2\alpha + e_y \sin 2\alpha, \tilde{b} = e_x \cos 2\beta + e_y \sin 2\beta
(eq 18)
[not sure it is meant to be \cos 2\alpha or \cos^2\alpha it won't matter much though]
and further down:
Putting these two results together, we arrive at the following standard scores corresponding to the raw scores (16) and (17):
A(\alpha,\mu) = \frac {\mathcal{A} (\alpha,\mu) - \overline{\mathcal{A} (\alpha,\mu)}} {\sigma(\mathcal{A})} = \frac {\mathcal{A} (\alpha,\mu) - 0} {(-I \cdot \tilde{a})} = (+\mu \cdot \tilde{a})
(eq 25)
B(\beta,\mu) = \frac {\mathcal{B} (\beta,\mu) - \overline{\mathcal{B} (\beta,\mu)}} {\sigma(\mathcal{B})} = \frac {\mathcal{B} (\beta,\mu) - 0} {(+I \cdot \tilde{b})} = (+\mu \cdot \tilde{b})
(eq 26)
The question is: which one of these should be identified with A(a,\lambda) and B(b,\lambda) from Bell's paper and with the outcomes collected in the actual experiments to compute E(a,b)? Should it be \mathcal{A} (\alpha,\mu) and \mathcal{B} (\beta,\mu) from eq 16-17, or "normalized" A(\alpha,\mu) and B(\beta,\mu) from eq 25-26? Please answer.
Case 1: the answer is the former (\mathcal{A} (\alpha,\mu) and \mathcal{B} (\beta,\mu)):
We agreed (I hope) that individual outcomes of measurements are represented by (mapped onto) normal integer numbers -1 and 1. So the first order of business is to drop the notion of \mathcal{A} \in S^3 and replace it with simple \mathcal{A} \in \{-1, +1\} (by establishing 1:1 map if you wish).
The next thing we do is define \mu_+ = +I, \mu_- = -I. Once this is done we can rewrite eq 16-17, removing all traces of Grassman algebra from them:
\mathcal{A} (\alpha,\mu)=\begin{cases} +1 & \text{if } \mu = \mu_+ \\ -1 & \text{if } \mu = \mu_- \end{cases}, \mathcal{B} (\beta,\mu)=\begin{cases} -1 & \text{if } \mu = \mu_+ \\ +1 & \text{if } \mu = \mu_- \end{cases}
where \mu \in \{ \mu_+, \mu_- \} is some opaque random parameter taking up one of the two opaque values with equal probability.
From here we can immediately obtain:
\mathcal{A} (\alpha,\mu)\mathcal{B} (\beta,\mu)=-1, \forall \mu \in \{ \mu_+, \mu_- \}
and therefore
E(a,b)=-1, \forall a,b
and therefore
|E(a,b) + E(a',b) + E(a,b') -E(a',b')|= 2, \forall a,b,a',b'
So far the results agree with Bell and do not exhibit \cos^2 rule, which is exactly the opposite of what the author claimed.
Case 2: The answer is A(\alpha,\mu) and B(\beta,\mu) from eq 25-26. That appears to be author's intention because that's what he uses in eq 30 to calculate E(a,b). But what is the value of A(\alpha,\mu)? It is a whatsis bivector in whatever space.
Since the goal is to provide a working model explaining experimental results of \cos^2 rule (and thus disproves Bell) , we need to identify A(\alpha,\mu) unambiguously with the outcome of a measurement, such as either detector D+ or D- clicking in a typical two-channel Bell type experiment by mapping it into { -1, +1 }. The answer is that we cannot because A(\alpha,\mu) is not a two-valued function. It's value, whatever is it, cannot be obtained in the experiment, therefore it cannot be used to calculate E(a,b) (since E(a,b) is calculated from experimental data and we wish to provide a model for it).
As it is, A(\alpha,\mu) might refer to some internal state of the system, but an extra step is required to obtain the actual outcome of a measurement. This extra step ( which can be achieved by some sort of map M: A(\alpha,\mu) \mapsto \{-1, +1\} will encapsulate in itself the process of measurement. And to maintain connection with actual physical experiments, we would have to use the value of this M(\alpha,\mu) and not the unobservable A(\alpha,\mu). Well, guess what, doing this will bring us back to agreement with Bell and disagreement with reality.
So where it all went wrong? Well, when calculating standard deviation.
To begin with, the whole issue of standard deviation and "normalizing" is a red herring. If you bother to read Bell's original paper, you will see that there is no reference to mean or standard deviation. What's more, Bell's derivation works just fine for any A(a,\lambda) as long as A(a,\lambda) \in \{-1, +1 \} and A(a,\lambda)=-B(a,\lambda). The mean does not have to be 0 and sigma does not have to be 1 and there is no need to "normalize" anything.
Having said that, everyone knows that standard deviation of individual measurements in Bell type experiment is 1 (assuming ideal 100% efficient detector) . It is so bleedingly obvious that no-one needs to explain that. Still, there is nothing wrong with actually calculating one, as long as one's math is correct. The sigma would come out as 1, eq 25-26 would be exactly the same as 16-17 and we would be back to where we started.
But the math is not correct. Instead of directly calculating σ from the definition, which would be far easier but would not produce the desired effect, the author [STRIKE]averts his eyes and carefully walks along the wall pretending there is no elephant in the room[/STRIKE] starts mucking around with it with no clear purpose.
As I already pointed out, eq (23) is wrong. I said and you agreed that σ(aA)=aσ(A) is in general incorrect. You said:
yes, the correct formula is σ(aA)=norm(a)σ(A) when a is a constant (because the expectation value can be redefined with norms).
Well, I have news for you: σ(aA)=norm(a)σ(A) does not work either. I gave you the example already:
\sigma( \vec{a} \cdot \vec{b} ) \ne \vec{a} \cdot \sigma( \vec{b} ) \ne ||\vec{a}||\sigma( \vec{b} ) \ne \vec{a}\sigma( ||\vec{b})|| ) \ne ||\vec{a}||\sigma( ||\vec{b}|| )
in fact, 2d ad 3rd terms simply do not compute and 4th term gives a value of a vector where the original was a scalar. This is, by the way, exactly the case with eq. 23-24.
Alternatively σ(aA)=aσ(A) when σ(A) is (re)defined correctly.
Please enlighten us, what is the correct redefinition of σ(A) that allows σ(aA)=aσ(A). All I can see in eq (24) is the same old σ with the argument A (which is a vector) quietly replaced with its norm |A|, which bring us back to my previous point.
This is all so wrong and so crude I'm surprised anyone can fall for this trick. The whole thing reminds me of http://en.wikipedia.org/wiki/Technology_in_The_Hitchhiker%27s_Guide_to_the_Galaxy#Bistromathic_drive"
DK
So thanks for sharing this.
Hello-o-o?!
