Quantum Mechanics and Hidden Variables

Contents

Question

Does Quantum Mechanics requires Hidden Variables ?

Description

If you want to answer this question you of course first must have a clear definition of:
  1. What is Quantum Mechanics.
  2. What are Hidden Variables.
  3. What is required in order to find an answer on this question.
To start with the third we are going to do an experiment, not one, but many.

Answer part 1 - dice # 1

In order to answer this question .
Consider a dice (dice # 1). A dice has 6 sides. Lets do an experiment, throw the dice 600 times and count how many times we throw each value.
The outcome will be, for each of the values, close to 100.
Next we can do a little bit of mathematics and make the following statements.:

Answer part 2 - dice # 2

Next I give you a new dice (dice # 2) and I ask you the question: Does this dice behave the same as # 1.
In order to answer this question you say: lets do the same experiment as with dice #1 and compare the results.
The results are the same, So your answer is: YES.

Answer part 3 - dice # 3

Next I give you a new dice (dice # 3) and I ask you the question: Does this dice behave the same as # 1.
You again do an experiment and the result is for the values from 1 to 6: 120, 80, 100, 100, 100, 100.
There is one possible conclusion: dice #3 is different as dice #1 and #2. What is this difference ?. This is also called a Hidden Variable.
The first clue what is the cause comes from investigating the dice: side 1 and side 2 are opposite each other.
The second clue comes from splitting the dice in 8 equal parts: All the 4 parts adjacent to side 1 have the same weight and all the 4 parts adjacent to side 2 have the same weight. However the weight of the first is more than the second four, meaning that the centre of mass is closer to side 1.
This explains the preference of the value of 1 compared to 2.

Answer part 4 - 3 coins

In answer part 1 to 3 we used a dice. The chance of a certain value is 1/6 to get a certain value.
Now we are going to study a different chance: 1/8.
In order to have a chance of 1/8 we need three coins. The sides of the three coins we label: (A,a), (B,b) and (C,c)
The chance of getting any combination, for Example p(A,B,C) or p(a,b,c) is 1/8. The letter p stands for probability or chance.
With probabilities simple mathematics is possible:
p(A,B) = p(A,B,C) + p(A,B,c) = 1/4
p(A,C) + p(B,c) = p(A,B,C) + p(A,b,C) + p(A,B,c) + p(a,B,c) = 1/2
p(A,C) + p(B,c) = p(A,B) + p(A,b,C) + p(a,B,c) = 1/2
p(A,C) + p(B,c) >= p(A,B)
p(A,C) + p(B,C) >= p(A,B)
The last equation is called The Bell Inequality

The following table demonstrates this inequality by the author using three coins. 

   Throw   p(A,B)   p(A,C)  p(B,C)   p(a,b)  p(a,c)  p(b,c)
 1 a,B,C                       1
 2 a,B,c                                        1
 3 A,b,c                                                1
 4 a,B,C                       1   
 5 A,B,C      1        1       1
 6 a,b,c                                1       1       1  
 7 A,b,C               1              
 8 a,B,C                       1
 9 a,B,C                       1 
10 a,b,C                                1
11 a,B,c                                        1
12 A,b,c                                                1
13 a,b,c                                1       1       1 
14 A,b,C                       1        
15 A,B,c      1                    
16 A,b,c                                                1
   Total      2        2       5        3       4       5
   Result     2   <   (2   +   5)       3   <  (4    +  5)

Answer part 5 - Quantum Mechanics

Quatum Mechanics can be used to predict the result for similar experiments.
However with quantum mechanics the equation is different:
p(A,B) => p(A,C) + p(B,C)
The details of this calculation are not known by the author. See Reflection 2 - General 25 May 2018

Answer part 6 - Entangled Photons

To test which of the two equations is right photons, or better pairs of photons are used.

Each photon has a direction in space. A direction is described by three coordinates:A,B and C
With a simple setup (positron, electron collision) it is possible to create pairs of entangled photons: If photon 1 is A,B,C then then photon 2 is a,b,c
Using an analyzer it is possible to measure one coordinate.

The following table describes the results of a hypothetical test.
The column p(A) shows if a photon in the A direction is tested. The same for p(B) and p(C)
The column p(A,A) shows the correlation if both for photon 1 and photon 2 the A direction is measured. The results shows that the two photons are strongly negative correlated in the A direction.
The column p(A,B) shows the correlation if for photon 1 the A direction and photon 2 the B direction is measured. The expected value is 0 meaning no correlation. The results show a small correlation between each photon pair in the A and B direction.
The column p(A,C) shows the correlation if for photon 1 the A direction and photon 2 the B direction is measured. The results show a small correlation between each photon pair in the A and C direction. 

  photon 1  p(A)     photon 2    p(A)   p(A,A)    p(B)    p(A,B)   p(C)   p(A,C)
 1 a,B,C     -1        A,b,c       1       -1      -1        1      -1       1 
 2 a,B,c     -1        A,b,C       1       -1      -1        1       1      -1
 3 A,b,c      1        a,B,C      -1       -1       1        1       1       1
 4 a,B,C     -1        A,b,c       1       -1      -1        1      -1       1
 5 A,B,C      1        a,b,c      -1       -1      -1       -1      -1      -1
 6 a,b,c     -1        A,B,C       1       -1       1       -1       1      -1
 7 A,b,C      1        a,B,c      -1       -1       1       -1      -1      -1          
 8 a,B,C     -1        A,b,c       1       -1      -1        1      -1       1
 9 a,B,C     -1        A,b,c       1       -1      -1        1      -1       1
10 a,b,C     -1        A,B,c       1       -1       1       -1      -1       1
11 a,B,c     -1        A,b,C       1       -1      -1        1       1      -1
12 A,b,c      1        a,B,C      -1       -1       1        1       1       1
13 a,b,c     -1        A,B,C       1       -1       1       -1       1      -1
14 A,b,C      1        a,B,c      -1       -1       1        1      -1      -1
15 A,B,c      1        a,b,C      -1       -1      -1       -1       1       1              
16 A,b,c      1        a,B,C      -1       -1       1        1       1       1
  Total      -2                    2      -16       0        4       0       2

Answer part 7 - Random Photons

To test which of the two equations is right photons, or better pairs of photons are used.

Each photon has a direction in space. A direction is described by three coordinates:A,B and C
With a simple setup (positron, electron collision) it is possible to create pairs of photons: In this particular case there is no relation between photon 1 and photon 2 Using an analyzer it is possible to measure one coordinate.

The following table describes the results of a hypothetical test.
The column p(A) shows if a photon in the A direction is tested. The same for p(B) and p(C)
The column p(A,A) shows the correlation if both for photon 1 and photon 2 the A direction is measured. The results shows that the two photons are not correlated in the A direction.
The column p(A,B) shows the correlation if for photon 1 the A direction and photon 2 the B direction is measured. The expected value is 0 meaning no correlation. The results show a small correlation between each photon pair in the A and B direction.
The column p(A,C) shows the correlation if for photon 1 the A direction and photon 2 the B direction is measured. The results show no correlation between each photon pair in the A and C direction. 

  photon 1  p(A)     photon 2    p(A)   p(A,A)    p(B)    p(A,B)   p(C)   p(A,C)
 1 a,B,C     -1        A,b,C       1       -1      -1        1       1      -1 
 2 a,B,c     -1        a,b,c      -1        1      -1        1      -1       1
 3 A,b,c      1        a,b,C      -1       -1      -1       -1       1       1
 4 a,B,C     -1        a,b,C      -1        1      -1        1      -1       1
 5 A,B,C      1        a,B,c      -1       -1       1        1      -1      -1
 6 a,b,c     -1        a,B,C      -1        1       1       -1       1      -1
 7 A,b,C      1        a,B,c      -1       -1       1        1      -1      -1          
 8 a,B,C     -1        a,b,C      -1        1      -1        1       1      -1
 9 a,B,C     -1        A,B,C       1       -1       1       -1       1      -1
10 a,b,C     -1        a,b,C      -1        1      -1        1       1      -1
11 a,B,c     -1        a,B,c      -1        1       1       -1      -1       1
12 A,b,c      1        a,b,C      -1       -1      -1       -1       1       1
13 a,b,c     -1        A,b,C       1       -1      -1        1       1      -1
14 A,b,C      1        A,B,c       1        1       1        1      -1      -1
15 A,B,c      1        a,B,C      -1       -1       1        1       1       1              
16 A,b,c      1        a,b,c      -1       -1      -1       -1       1       1
  Total      -2                   -8       -2      -2        4       4      -1
The overall result is that when the photons are random there is no correlation measured.

Quantum Randi Challenge

The Quantum Randi Challenge is a contest to prove that Quantum Mechanics is wrong. (I think)
For a critical evaluation about this challenge read this: Critical Evaluation QRC
The authors of this Challenge do not seem to understand that this whole Challenge is more a disfavour than a favour to propagate the understanding of science and physics.

Reflection 1 - General

The opinion of the author is that the above hypothetical test does show anything special. At least not that quantum mechanics is better compared to clasical knowledge.
The author does not exclude that single photons behave "differently". However in order to make any statement more eleborate experiments should be done and the results should be made publically available.
What the author also asthonised is that the calculations of quantum mechanics are not described in detail.

For some there are experiments done which show that clasical knowledge is not complete. Those experiments apperently show that there is a relation between the A and B component. The explanation is that faster than light communication is involved because the measurement of one analyzer influences the results of the other analyzer. Unfortunate no detail about those test is available, neither if this influence is a function between the distance of the analyzer and the source (point of collision) of the photons, nor which analyzer influences which analyzer.

Reflection 2 - Entanglement 25 May 2018

In Answer part 5 - Quantum Mechanics both the inequality theorem and Quantum Mechanics is discussed.
The inequality equation is: p(A,B) => p(A,C) + p(B,C)
The corresponding Quantum Mechanics equation: p(A,B) >= p(A,C) + p(B,C)
This information comes Scientific American. See ScientificAm November 1979 - Bernard d'Espagnat . As mentioned in the article the details are not discussed.

The issue is that there are certain experiments with agree with the inequality equation. See Answer part 4 - 3 coins above. In that case 3 coins are tested, each simulating the +X,+Y and +Z axis.

The type of experiments used to demonstrate Quantum Mechanics involve entanglement. In that case there is a certain correlation between the observations when the same axis are measured in both directions. That does not exist in case the inequality theorem is tested using coins.

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Original Version: 29 June 1996
Modified: 3 October 2013
Modified: 25 May 2018

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