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Posted by on Jul 2, 2011 in Quantum Crackpot RANDI Counter Challenge | 5 comments

Quantum Crackpot RANDI Challenge Taken: Part 2

In Part 1 the rules for individual particles were extracted from the  ”Hidden Variable Madness” blog in the Quantum Crackpot RANDI Challenge.  This part discusses the coincidence rule:

5.  “Thus, for every pair, we obtain a result Result = a,b(A,B), where a and b tell us the angles, and (A,B) are the values for the exits that Alice’s and Bob’s crystals put their ball through: (A,B) = (0,0), or (0,1), or (1,0), or (1,1)”.

I will use the outcomes: (A,B) = (+1,+1), or (-1,-1), or (+1,-1), or (-1,+1)

The Result represents two outcomes (2 clicks) from an EPR pair,  one click at Alice and the other at Bob.  The filter angles are chosen randomly after the pair has left its source. The experiment is repeated many times and the Results are stored in two computers. Coincidences are identified by common time stamps, ti:

Time stamps are unnecessary in my model because only one EPR pair is treated.  By definition, however, the LHV ( λ)  must be the same. Therefore for each EPR pair the time stamp is replaced by common LHV,

An unfiltered particle is called “Pristine” as defined in the website called Bell’s theorem refuted. The values of the LHV of a pristine particle are unknown. A pristine EPR pair is one before filtering.

A filter allows only those particles to pass which possess LHV consistent with giving a click of +1 or of -1.  By definition that is the role of a filter. There are no other outcomes but +1 and -1.

For a hands-on clarification of the role of filters, please see the Stern-Gerlach simulation at the Phet Physics site.

The above discussion of single coincidence events is illustrated in the image from the blog: Bell’s theorem refuted. We see the common source S; the two pristine particles that form an EPR pair before the filter;  the same LHV for each; and the four outcomes that result from filtering.

 

The filtered particles are indicated in blue with specific LHV which are consistent with outcomes of ±1 in the directions of the filter angles, a and b.

The question is then:

For each EPR pair, when the outcome is recorded at Alice (Bob), what is the probability for the outcome at Bob (Alice)?

If nobody disagrees, I will move on and present a model that shows a disentangled product state not only gives the quantum result, but predicts something new.

July 1st was Canada Day and July 4th is coming soon.  I think holidays will likely trump discussions on EPR.  So I will wait a bit for the next part.

5 Comments

  1. Dear Bryan,

    You write: “The question is then: For each EPR pair, when the outcome is recorded at Alice (Bob), what is the probability for the outcome at Bob (Alice)?”

    This question is easily answered: so I guess I’m disagreeing. For me, the more relevant question is: Can EPRB results be locally and realistically simulated on two classical computers? Specifically: E(a, b) = –a.b.

    So by all means: “… move on and present a model that shows a disentangled product state not only gives the quantum result, but predicts something new.”

    But I suspect my question will remain — as I’ll discuss off-line.

  2. Hi there Bryan thanks for your great blog post on Crackpot Science- Challenge RANDI Taken. the article was very helpful for a report I am putting together for class.

  3. Hi Sammie
    Please send me a copy of you report. Interested in your views.
    Bryan

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