Unghostly entanglement

[Lazarus]

 

If you have a hidden variable, i.e. the spin is already determined before it hits the detector, then half of your choices are not possible for detector A. You cannot detect a particle as spin down if it really is spin up.

 

That makes any further critique moot. Did you even watch the whole video?

 

I watched it several times.

 

As you want to end this thread, I will make my closing arguments that Bell’s Theorem does not prove that classic calculations are self contradicting or that Spooky Entanglement exists.

 

1. The video from ScienceForms.net, “Why Hidden Variables don’t Work” calculates an incorrect probability of 5/9 while a correct inclusion of all the probabilities give a correct result of ½.

 

1. The David Mermin “proof” from Physics Today, April 1985, “Is the moon really there when nobody looks?” makes the unproven assumption that turning both detectors 1 degree is the same as turning one detector 2 degrees. Also, it concludes that P(A,C) – P(B,A) – P(BC) <= 1 is equivalent to Bell’s Theorem. All that means is that P(A,C) is between 0 and 1 so is worthless.

 

1. Lorenzo’s proof by his interpretation of Bell’s Theorem is seriously flawed. He contends that P(A,B)+P(A,C)+P(B,C) has to be equal to or greater than 1. His 2 coins example shows his error. The coins can be Gold or Copper, Shinny or Dull, Large or Tiny. All the possible combinations with A as G or C, B as S or D, C as L or T are: GSL, GSD, GDL, GDT, CSL, CST, CDL, CDT. That makes 8 possibilities so the probabilities are, P(A,C)=1/4, P(B,A)=1/4, P(BC)=1/4. The total is ¾ which agrees with experimental and Quantum results. Note: ¾ is less than 1.

 

1. Further, the rectangular hole bowling ball detector goes through the same steps and gets similar results to the CH74 experiments and Quantum predictions and isn’t undetermined in flight.

 

1. Bell’s inserting lambda into the equation is unnecessary if Quantum Theory and Classic Theory get the same results or if lambda = 1. Almost like using an assumption to prove the assumption.

 

If none of this raises any suspicion about Bell’s proof or Spooky entanglement then there are more “True Believers” in science than there are in religion.

 

Thanks again for the enlightening and interesting posts.

[swansont]

I watched it several times.

 

As you want to end this thread, I will make my closing arguments that Bell’s Theorem does not prove that classic calculations are self contradicting or that Spooky Entanglement exists

The video from ScienceForms.net, “Why Hidden Variables don’t Work” calculates an incorrect probability of 5/9 while a correct inclusion of all the probabilities give a correct result of ½.

Until you go through and do a detailed (and correct) accounting of this probability, this is merely an assertion, seemingly based on misunderstanding the scenario. The calculation in the video is correct.

 

How about taking the 9 cases in the video and showing which calculations are wrong. Any other orientation of the detector, or hidden variable scenario, is mathematically equivalent.

 

 

• Lorenzo’s proof by his interpretation of Bell’s Theorem is seriously flawed. He contends that P(A,B)+P(A,C)+P(B,C) has to be equal to or greater than 1. His 2 coins example shows his error. The coins can be Gold or Copper, Shinny or Dull, Large or Tiny. All the possible combinations with A as G or C, B as S or D, C as L or T are: GSL, GSD, GDL, GDT, CSL, CST, CDL, CDT. That makes 8 possibilities so the probabilities are, P(A,C)=1/4, P(B,A)=1/4, P(BC)=1/4. The total is ¾ which agrees with experimental and Quantum results. Note: ¾ is less than 1.

 

This is the first time you have mentioned Lorenzo, and as there is no link, I have no idea what proof this is.

 

 

• Further, the rectangular hole bowling ball detector goes through the same steps and gets similar results to the CH74 experiments and Quantum predictions and isn’t undetermined in flight.

 

Second time for CH74, but again, no link.

 

You should probably focus on one example in order to identify and correct your misconception. Looking at multiple examples just leads to confusion, since the setups differ and so do the resulting explanations.

 

If none of this raises any suspicion about Bell’s proof or Spooky entanglement then there are more “True Believers” in science than there are in religion.

And we have the inevitable appeal to personal incredulity fallacy resulting in labeling this as religion. It raises no suspicion because the critique is so obviously flawed.

[uncool]

 

1. Lorenzo’s proof by his interpretation of Bell’s Theorem is seriously flawed. He contends that P(A,B)+P(A,C)+P(B,C) has to be equal to or greater than 1. His 2 coins example shows his error. The coins can be Gold or Copper, Shinny or Dull, Large or Tiny. All the possible combinations with A as G or C, B as S or D, C as L or T are: GSL, GSD, GDL, GDT, CSL, CST, CDL, CDT. That makes 8 possibilities so the probabilities are, P(A,C)=1/4, P(B,A)=1/4, P(BC)=1/4. The total is ¾ which agrees with experimental and Quantum results. Note: ¾ is less than 1.

 

Why do you think P(A, C) = 1/4 for this?

 

P(A, C) is the probability that A and C give the same result. That would be 1/2, not 1/4.

[imatfaal]

!

Moderator Note

 

Lazarus

 

Can you begin to back up your arguments with both citations for the works you quote and a bit more rigor. Your interpretation of Lorenzo Maconne's very simple explanation of the mathematical basis for Bell's Inequality (which I knew from here) is wrong in all important respects.

 

I am close to believing you are trolling the community with deliberately misinterpreted ideas. I really hope that you are not - and that is why I have made the request for links and rigor; you will note that so far in the this thread that whenever you have provided chapter and verse that Members have very quickly been able to show the flaws in your argument. This will continue to be the case when you present Lorenzo Maccone's proof (or more an explanation for non-mathematicians) and the reasons you think it is incorrect.

 

This thread will remain open and you will suffer no opprobrium whilst you are honestly challenging mainstream mathematics (even such a widely accepted idea) - but if we believe you are deliberately mis-stating your position to elicit a response from the forum then we may call troll and sanctions could follow.

 

Do not respond to this moderation - please report the post if you feel it is unfair.

 

[Lazarus]

Swansont said:

Until you go through and do a detailed (and correct) accounting of this probability, this is merely an assertion, seemingly based on misunderstanding the scenario. The calculation in the video is correct.

 

How about taking the 9 cases in the video and showing which calculations are wrong. Any other orientation of the detector, or hidden variable scenario, is mathematically equivalent.

Lazarus said:

I appreciate your letting me point out what I feel is a problem with the video.

 

The video is not very clear on how they choose the probabilities but here is one way it could get 5 out of 9 matches.

 

Alice’s detector gets heads up guys in all 3 angles and Bob’s detector gets heads down guys in all 3 angles. Since the number of combinations is the product of the two, the number of combinations comes out to be 9 and can produce the 5 of 9 result.

 

A problem with this approach is that it is ignoring the other 3 conditions, Alice’s detector gets heads down guys in all 3 angles and Bob’s detector gets heads up guys in all 3 angles.

 

The rules set up by this video are that there are 2 detectors at various settings at 0, 120 and 240 degrees, entangled pairs of guys with one head up and the other head down (spin) and the only restriction is that when both detectors are at the same angle, the one guy is head up and the other head down.

 

Here is another way to do the probability calculation.

 

There are 6 results that Alice’s detector can have and Bob’s detector can have the same 6 results making 36 combinations between the two detectors as follows.

 

18 Matched

 

A 1 up/B 1 down, A 1 up/B 2 down, A 1 up/B 3 down

A 2 up/B 1 down, A 2 up/B 2 down, A 2 up/B 3 down

A 3 up/B 1 down, A 3 up/B 2 down, A 3 up/B 3 down

 

B 1 up/A 1 down, B 1 up/A 2 down. B 1 up/A 3 down

B 2 up/A 1 down, B 2 up/A 2 down, B 2 up/A 3 down

B 3 up/A 1 down, B 3 up/A 2 down, B 3 up/A 3 down

 

18 Unmatched

 

A 1 up/B 1 up, A 1 up/B 2 up, A 1 up/B 3 up

A 2 up/B 1 up, A 2 up/B 2 up, A 2 up/B 3 up

A 3 up/B 1 up, A 3 up/B 2 up, A 3 up/B 3 up

 

B 1 down/A 1 down, B 1 down/A 2 down, B 1 down/A 3 down

B 2 down/A 1 down. B 2 down/A 2 down, B 2 down/A 3 down

B 3 down/A 1 down, B 3 down/A 2 down, B 3 down/A 3 down

 

Which would give 50 % matched. However, there are 6 unmatched cases that are not allowed by the video rules. They are:

 

A 1 up/B 1 up, A 2 up/B 2 up, A 3 up/B 3 up

A 1 down/B 1 down, A 2 down/B 2 down, A 3 down/B 3 down

 

That leaves 30 counted cases, 18 are matched and 12 are unmatched for a result of 60 % matched, which does not match the video or the experimental results.

 

The test is biased by removing 6 unmatched cases.

 

The huge problem with the video proof is that the rules do not define what happens when the particle is not aligned with the detector.

 

If you assume that particles are only detected within 1 degree of the detector the there are only 6 matched and lots of unmatched. If you assume that particles are only detected within 10 degrees of the detector you get a different result. If you assume that particles are randomly detected as a function of the cosine of the angle you get a different result, etc.

 

--------------------------------------------

--------------------------------------------

 

Uncool said:

 

Why do you think P(A, C) = 1/4 for this?

P(A, C) is the probability that A and C give the same result. That would be 1/2, not 1/4.

 

 

Lazarus said:

 

Each coin with 3 conditions has 8 possible combinations. There are 64 possibilities with the 2 coins. By assigning A to Gold/Copper, B to Shinny/Dull and C to Large/Tiny we can show all possibilities with both coins. For A and C there are 16 combinations that are matched out of the 64 possible which makes the probability ¼. I cheated and used my little old computer.

 

Here are all the combinations:

Bob Alice

1 GSL GSL matched 1

2 GSL GST unmatched

3 GSL GDL matched 2

4 GSL GDT unmatched

5 GSL CSL unmatched

6 GSL CST unmatched

7 GSL CDL unmatched

8 GSL CDT unmatched

9 GST GSL unmatched

10 GST GST matched 3

11 GST GDL unmatched

12 GST GDT matched 4

13 GST CSL unmatched

14 GST CST unmatched

15 GST CDL unmatched

16 GST CDT unmatched

17 GDL GSL matched 5

18 GDL GST unmatched

19 GDL GDL matched 6

20 GDL GDT unmatched

21 GDL CSL unmatched

22 GDL CST unmatched

23 GDL CDL unmatched

24 GDL CDT unmatched

25 GDT GSL unmatched

26 GDT GST matched 7

27 GDT GDL unmatched

28 GDT GDT matched 8

29 GDT CSL unmatched

30 GDT CST unmatched

31 GDT CDL unmatched

32 GDT CDT unmatched

33 CSL GSL unmatched

34 CSL GST unmatched

35 CSL GDL unmatched

36 CSL GDT unmatched

37 CSL CSL matched 9

38 CSL CST unmatched

39 CSL CDL matched 10

40 CSL CDT unmatched

41 CST GSL unmatched

42 CST GST unmatched

43 CST GDL unmatched

44 CST GDT unmatched

45 CST CSL unmatched

46 CST CST matched 11

47 CST CDL unmatched

48 CST CDT matched 12

49 CDL GSL unmatched

50 CDL GST unmatched

51 CDL GDL unmatched

52 CDL GDT unmatched

53 CDL CSL matched 13

54 CDL CST unmatched

55 CDL CDL matched 14

56 CDL CDT unmatched

57 CDT GSL unmatched

58 CDT GST unmatched

59 CDT GDL unmatched

60 CDT GDT unmatched

61 CDT CSL unmatched

62 CDT CST matched 15

63 CDT CDL unmatched

64 CDT CDT matched 16

---------------------------------------------------

---------------------------------------------------

[swansont]

Swansont said:

Until you go through and do a detailed (and correct) accounting of this probability, this is merely an assertion, seemingly based on misunderstanding the scenario. The calculation in the video is correct.

 

How about taking the 9 cases in the video and showing which calculations are wrong. Any other orientation of the detector, or hidden variable scenario, is mathematically equivalent.

 

Lazarus said:

I appreciate your letting me point out what I feel is a problem with the video.

 

The video is not very clear on how they choose the probabilities but here is one way it could get 5 out of 9 matches.

 

Alice’s detector gets heads up guys in all 3 angles and Bob’s detector gets heads down guys in all 3 angles. Since the number of combinations is the product of the two, the number of combinations comes out to be 9 and can produce the 5 of 9 result.

 

A problem with this approach is that it is ignoring the other 3 conditions, Alice’s detector gets heads down guys in all 3 angles and Bob’s detector gets heads up guys in all 3 angles.

 

 

 

 

If Alice gets heads up in all 3 orientations and Bob gets heads down, then that's 100%. One is always up, another is always down. That's not 5/9, that's 9/9. If you flip the directions, you once again get 100%. That's not ignoring the case, it's recognizing that it's the same condition with a - sign out in front, and so one need not waste time on it. (It's a video, and he's a fairly polished video maker. You omit irrelevant details that would make the video unnecessarily long) Including those new conditions don't change the answer

 

Since you botched this part of the analysis, your final answer will likely be wrong. Especially since it doesn't look like you defined the hidden variable condition. It looks like you are freely moving back and forth between the QM analysis and the hidden variable analysis. No wonder it's all jumbled up.

 

The huge problem with the video proof is that the rules do not define what happens when the particle is not aligned with the detector.

 

 

OK, you are completely missing the point. There is no such thing in QM. The orientation of the detector doesn't matter. If the state is undetermined, there is no such thing as being aligned or not with the detector. If it's a hidden variable, that result is baked in for each possible orientation.

[uncool]

Uncool said:

 

Why do you think P(A, C) = 1/4 for this?

P(A, C) is the probability that A and C give the same result. That would be 1/2, not 1/4.

 

 

Lazarus said:

 

Each coin with 3 conditions has 8 possible combinations. There are 64 possibilities with the 2 coins. By assigning A to Gold/Copper, B to Shinny/Dull and C to Large/Tiny we can show all possibilities with both coins. For A and C there are 16 combinations that are matched out of the 64 possible which makes the probability ¼. I cheated and used my little old computer.

 

Here are all the combinations:

Bob Alice

1 GSL GSL matched 1

2 GSL GST unmatched

3 GSL GDL matched 2

4 GSL GDT unmatched

5 GSL CSL unmatched

6 GSL CST unmatched

7 GSL CDL unmatched

8 GSL CDT unmatched

9 GST GSL unmatched

10 GST GST matched 3

11 GST GDL unmatched

12 GST GDT matched 4

13 GST CSL unmatched

14 GST CST unmatched

15 GST CDL unmatched

16 GST CDT unmatched

17 GDL GSL matched 5

18 GDL GST unmatched

19 GDL GDL matched 6

20 GDL GDT unmatched

21 GDL CSL unmatched

22 GDL CST unmatched

23 GDL CDL unmatched

24 GDL CDT unmatched

25 GDT GSL unmatched

26 GDT GST matched 7

27 GDT GDL unmatched

28 GDT GDT matched 8

29 GDT CSL unmatched

30 GDT CST unmatched

31 GDT CDL unmatched

32 GDT CDT unmatched

33 CSL GSL unmatched

34 CSL GST unmatched

35 CSL GDL unmatched

36 CSL GDT unmatched

37 CSL CSL matched 9

38 CSL CST unmatched

39 CSL CDL matched 10

40 CSL CDT unmatched

41 CST GSL unmatched

42 CST GST unmatched

43 CST GDL unmatched

44 CST GDT unmatched

45 CST CSL unmatched

46 CST CST matched 11

47 CST CDL unmatched

48 CST CDT matched 12

49 CDL GSL unmatched

50 CDL GST unmatched

51 CDL GDL unmatched

52 CDL GDT unmatched

53 CDL CSL matched 13

54 CDL CST unmatched

55 CDL CDL matched 14

56 CDL CDT unmatched

57 CDT GSL unmatched

58 CDT GST unmatched

59 CDT GDL unmatched

60 CDT GDT unmatched

61 CDT CSL unmatched

62 CDT CST matched 15

63 CDT CDL unmatched

64 CDT CDT matched 16

 

You have misunderstood the claim. The claim isn't about two coins. The claim is about one coin. The correct listing is this:

 

GSL

GST

GDL

GDT

CSL

CST

CDL

CDT

 

Then the two "properties" match if either both are the first type (Gold, Shiny, or Large) or both are of the second type (Copper, Dull, or Tiny). As such, the matches between A and C are:

 

GSL match

GST

GDL match

GDT

CSL

CST match

CDL

CDT match

[Lazarus]

Uncool said:

You have misunderstood the claim. The claim isn't about two coins. The claim is about one coin. The correct listing is this:

 

GSL

GST

GDL

GDT

CSL

CST

CDL

CDT

 

Then the two "properties" match if either both are the first type (Gold, Shiny, or Large) or both are of the second type (Copper, Dull, or Tiny). As such, the matches between A and C are:

 

GSL match

GST

GDL match

GDT

CSL

CST match

CDL

CDT match

 

-------------------------------

 

Lazarus said:

P(A,C) is the probability that both coins will be gold and large Not that the coins will be gold, silver and large.

Thst should read gold, shiny and large. Not gold, silver and large.

[uncool]

Uncool said:

You have misunderstood the claim. The claim isn't about two coins. The claim is about one coin. The correct listing is this:

 

GSL

GST

GDL

GDT

CSL

CST

CDL

CDT

 

Then the two "properties" match if either both are the first type (Gold, Shiny, or Large) or both are of the second type (Copper, Dull, or Tiny). As such, the matches between A and C are:

 

GSL match

GST

GDL match

GDT

CSL

CST match

CDL

CDT match

 

-------------------------------

 

Lazarus said:

P(A,C) is the probability that both coins will be gold and large Not that the coins will be gold, silver and large.

Thst should read gold, shiny and large. Not gold, silver and large.

That's not how P(A, C) was meant by Lorenzo Maccone. From his paper:

 

"Bell’s inequality refers to the correlation among measurement outcomes of the properties: call Psame(A, B) the probability that the properties A of the first object and B of the second are the same: A and B are both 0 (the first coin is gold and the second is shiny) or they are both 1 (the first is copper and the second is dull)."

 

You only measure one property of each coin. I'd gotten it wrong earlier myself, but the paper is quite clear that you only measure one property of each coin, and make exactly one comparison.

Edited December 25, 2015 by uncool

[Lazarus]

From Lorenzo Maccone’s paper:

Bell's inequality refers to the correlation among measurement outcomes of the properties: call Psame(A;B) the probability that the properties A of the first object and B of the second are the same: A and B are both 0 (the first coin is gold and the second is shiny) or they are both 1 (the first is copper and the second is dull). For example, Psame(A;B) = ½ tells me that with 50% chance A = B (namely they are both 0 or both 1). Since the two coins have equal counterfactual properties, this also implies that with 50% chance I get two gold shiny coins or two copper dull coins. Note that the fact that the two coins have the same properties means that Psame(A;A) = Psame(B;B) = Psame(C;C) = 1: if one is made of gold, also the other one will be, or if one is made of copper, also the other one will be, etc.

 

Sure sounds like Lorenzo is talking about 2 coins and that P(A,B) means the probability that the coins match both the gold/copper and shiny/dull conditions, P(A,C) means the probability that the coins match both the gold/copper and large/tiny conditions and P(B,C) means the probability that the coins match both the shiny/dull and large/tiny conditions. Is that wrong?

[uncool]

Ah. There's why I thought there was one coin earlier; it's an equivalent situation. Namely, also from the paper:

 

"Suppose we have two identical objects, namely they have the same properties."

 

The coins are assumed to have the same properties - if one is gold, then so is the other. If one is copper, then so is the other. If one is shiny, then so is the other. As such, there are only 8 possibilities - both are gold, shiny, and large, etc.

[Lazarus]

You are right. The error I made was not taking in to account that the coins were to have identical properties. That does leave me with a bit of a dilemma understanding how this is a proof of Bell’s Theorem. Since both gold/gold and copper/copper are matched, also both shiny/shiny and dull/dull are matched, P(A.B) = 1. Similarly, P(A,C) =1 and P(B,C) = 1. That means P(A.B) + P(A,C) + P(B,C) = 3. I fail to see how certainty + certainty + certainty = 3 * certainty proves Bell’s Theorem.

[swansont]

You are right. The error I made was not taking in to account that the coins were to have identical properties. That does leave me with a bit of a dilemma understanding how this is a proof of Bell’s Theorem. Since both gold/gold and copper/copper are matched, also both shiny/shiny and dull/dull are matched, P(A.B) = 1. Similarly, P(A,C) =1 and P(B,C) = 1. That means P(A.B) + P(A,C) + P(B,C) = 3. I fail to see how certainty + certainty + certainty = 3 * certainty proves Bell’s Theorem.

 

You have to compare this to the QM measurement. But your statement is wrong: P(A,B) = 1 does not derive from the matching. It is a statement about the distribution of the coins. YOU are saying that e.g. gold coins are shiny, and copper ones are dull. That is not an assumption inherent to the experiment. It's a condition YOU have placed on it. If the properties are arbitrarily decided, then P(A,B) = 1/2

 

What the paper says is that no matter what correlation you have between those properties, the sum of those 3 properties is not smaller than 1. But if you treat this as a QM system, the sum of those probabilities will be < 1

[uncool]

You are right. The error I made was not taking in to account that the coins were to have identical properties. That does leave me with a bit of a dilemma understanding how this is a proof of Bell’s Theorem. Since both gold/gold and copper/copper are matched, also both shiny/shiny and dull/dull are matched, P(A.B) = 1. Similarly, P(A,C) =1 and P(B,C) = 1. That means P(A.B) + P(A,C) + P(B,C) = 3. I fail to see how certainty + certainty + certainty = 3 * certainty proves Bell’s Theorem.

Because P(A, B) isn't testing whether the two coins match in two properties. It's testing whether the first coin has property 1 and the second property 2, or the first doesn't have property 1 and the second doesn't have property 2. As said before: there are 8 possibilities, and we get a "match" for the following:

 

GSL match

GST

GDL match

GDT

CSL

CST match

CDL

CDT match

[Lazarus]

Again, you are correct. I was erroneously using criteria that is different from Lorenzo’s conditions. Now you, Lorenzo, I and my computer all agree that 50 % is the correct result. My computer did an experiment by doing 100,000 iterations of randomly selecting one of the 8 possibilities and counting the results. It came up with 50062 zeros and 49938 ones for P(A,B) , 49960 zeros and 50039 ones for P(A,C) , 50128 zeros and 49872 ones for P(B,C). So both logically and experimentally 50 % is correct.

 

Lorenzo said that Quantum Theory gives a result of 25 %. It is also claimed that Quantum theory is always verified experimentally.

 

How is that conflict resolved?

[Strange]

Again, you are correct. I was erroneously using criteria that is different from Lorenzo’s conditions. Now you, Lorenzo, I and my computer all agree that 50 % is the correct result. My computer did an experiment by doing 100,000 iterations of randomly selecting one of the 8 possibilities and counting the results. It came up with 50062 zeros and 49938 ones for P(A,B) , 49960 zeros and 50039 ones for P(A,C) , 50128 zeros and 49872 ones for P(B,C). So both logically and experimentally 50 % is correct.

 

Lorenzo said that Quantum Theory gives a result of 25 %. It is also claimed that Quantum theory is always verified experimentally.

 

How is that conflict resolved?

 

The experimental test confirm that quantum theory is correct and, as expected, cannot be explained by local hidden variables. There is no conflict. This is exactly what is expected of quantum theory.

[imatfaal]

Again, you are correct. I was erroneously using criteria that is different from Lorenzo’s conditions. Now you, Lorenzo, I and my computer all agree that 50 % is the correct result. My computer did an experiment by doing 100,000 iterations of randomly selecting one of the 8 possibilities and counting the results. It came up with 50062 zeros and 49938 ones for P(A,B) , 49960 zeros and 50039 ones for P(A,C) , 50128 zeros and 49872 ones for P(B,C). So both logically and experimentally 50 % is correct.

 

Lorenzo said that Quantum Theory gives a result of 25 %. It is also claimed that Quantum theory is always verified experimentally.

 

How is that conflict resolved?

 

IT ISN'T. That's the whole point It was a difference between QM and EPR - and differences can be tested

 

And with one bound (well a few more than one) you have discovered the beauty of Bell's Inequality - by a simple mathematical inequality and through showing that the hidden variable hypothesis (of Einstein, Podolsky and Rosen) versus the quantum mechanics hypothesis MUST give different answers (because of the linkage to that inequality) he showed that a test was possible which would distinguish between QM and Hidden Variables. It then fell to other guys to work out how to actually test this - when Aspect and many others later found ways to experimentally confirm the idea the results were clearly in the QM rather than the EPR space

[Lazarus]

It was considerate of everyone to continue correcting my misinterpretations until things became clearer. I am not dogmatic but I still have a small problem with this. It appears to me that the classic 50 % can be experimentally justified, also. At least with the coins.

[swansont]

It was considerate of everyone to continue correcting my misinterpretations until things became clearer. I am not dogmatic but I still have a small problem with this. It appears to me that the classic 50 % can be experimentally justified, also. At least with the coins.

It's not a matter of justfying it. The classical probability is different from the quantum one. It looks like you have yet to analyze the quantum problem for the coins.

[Lazarus]

It's not a matter of justfying it. The classical probability is different from the quantum one. It looks like you have yet to analyze the quantum problem for the coins.

 

My concern is not in the computation of either the Classic or Quantum result. It is that the Quantum result is verified by experiment. The different Classic result is verified by experiment. I can’t be both ways.

[Strange]

 

My concern is not in the computation of either the Classic or Quantum result. It is that the Quantum result is verified by experiment. The different Classic result is verified by experiment. I can’t be both ways.

 

Why not?

[Lazarus]

 

Why not?

 

 

2 + 2 = 5 and 2 + 2 = 6. If there is any logic left in the world, one or both have to be wrong.

[Strange]

There are many ways in which you get different results from quantum theory than you do for classical theories. The quantum world does not follow classical rules, and there is no reason to expect it to.

 

If you don't like the way this universe works, you are free to find another one.

[Lazarus]

Both Quantum and Cllassic are trying to solve the same problem. If there are different experimental results from the same experiment something is seriously wrong with the experiment.

[Strange]

Both Quantum and Cllassic are trying to solve the same problem. If there are different experimental results from the same experiment something is seriously wrong with the experiment.

 

They are measuring different things. In the classical world the nature of the coins in the boxes are fixed before you measure it (they are "real" values or "hidden variables"). In the quantum world, the variables (e.g. spin) are not determined until you measure them (i.e. no hidden variables).

 

Each of these conditions produce different results. When we experiment with real coins we get the classical result. When we experiment with electrons we get the quantum result.