In the last century physicists have developed quantum mechanics, a theory required to explain observations made at the atomic level. Quantum mechanics is a very successful theory: It can precisely predict the colours emitted by atoms and give an explanation for the properties of the elements. It also describes a property discovered in quantum system called "entanglement", but in doing so it forces us to reject locality or reality.

Our common experience of the world is local and real. It is taught to us by experience and expressed with statements such as: "Touch the cup of tea to feel if it's warm" or "The gold is in the safe."

The scenario of a

Because of the strange properties of entanglement, however, the complementary predictions made by the model disagree with the results of the quantum experiment. The only explanation for the discrepancy is that locality or reality must be rejected.

This is inspired by the papers from Mermin[1] and Bell[2].

Upon receiving several postcards, I observe the following. The postcards always have three flaps labeled 1, 2 and 3 and a date written on the postmark. When a flap is lifted a single word is exposed which can be either "heads" or "tails". However, the postcards have a special mechanism: Opening a flap makes ink flow behind the other two and covers the other words. So although I am free to choose which flap to lift, I can only read one of the three words.

Having figured out so much, I decide to start recording my data in a systematic way. Upon receiving the postcard, I pick a random number from 1 to 3 and lift the flap labeled with that number. On the first week I choose the number 1 and lift the corresponding flap; the word "heads" appears. The next week, I choose 3 and the word "heads" appears. The following postcard has "tails" underneath flap 1. After two months my list of flap-numbers and words is: 1H, 3H, 1T, 1H, 2H, 2T, 3H, 3H.

I am intrigued by the behaviour of our friend and wonder what he writes on

Being an investigative person, you combine our lists to make word-pairs with my data on the left and yours on the right:

3H3H <- flap 3 lifted on both postcards, same words exposed: "heads" and "heads"

1T3H

1H1H <- flap 1 lifted on both postcards, same words exposed: "heads" and "heads"

2H1T

2T2T <- flap 2 lifted on both postcards, same words exposed: "tails" and "tails"

3H1H

3H2T

This continues for several years until we each have received over 400 such postcards. We combine our lists to get the word-pairs shown below in Table 1. Although the words "heads" and "tails" seem to be randomly distributed, word-pairs with the same flap-numbers always have the same words! They are highlighted in red to make them stand out.

TABLE 1. Partial list of 432 word-pairs
from the postcards received every week. The green
rectangle highlights my

first entries. The red entries are those for which we randomly chose the same flap-numbers.

first entries. The red entries are those for which we randomly chose the same flap-numbers.

This is an example of a

For example, the theoretical probability of winning a coin toss if you bet "heads" is calculated as follows. The number of desired outcomes is 1 since there is only one outcome that can make you win. The total number of possible outcomes is 2 since the coin has a total of two sides, "heads" or "tails". The probability of winning is 1÷2 = 50%.

Another example: what is the probability of throwing "4" or "5" with a single die? The number of desired outcomes is 2 since there is one "4" and one "5" on the die. The total number of possible outcomes is 6 since a die has six sides. The probability of throwing a "4" or a "5" is 2÷6 = 33.3%.

In an actual experiment, the number of occurrence of a desired outcome must be counted on a large number of measurements to obtain a good

After eight years of receiving postcards from our (boring) friend, we make a statistical analysis of the entries in Table 1.

A) For this first analysis we only look at

Although this analysis is overwhelming, the important thing to retain is that the estimated probabilities are all consistent with a probability of 50%.

Counting the entries in my list where I lifted flap 1: 1H is listed 78 times, 1T is listed 67 times, for a total of 145 words. Therefore the estimated probability of occurrence for "heads" is 78÷145 = 54% and for "tails" is 67÷145 = 46%.

The same procedure gives for my other entries: 2H = 50%, 2T = 50%, 3H = 43%, 3T = 57%, and independently of the flap I chose H = 49%, and T =51%. For your entries: 1H = 49%, 1T = 51%, 2H = 56%, 2T = 44%, 3H = 40%, 3T = 60%, H = 48%, and T =52%.

If all entries are taken together, the estimated probability of occurrence for

B) For this second analysis we look at the combined

Word-pairs having the same words are listed 139 times, word-pairs having different words are

The estimated probability of occurrence of the

A)

i- A postcard has three flaps, each one hides a single word: "heads" or "tails".

ii- Lifting a flap exposes a word and releases ink that covers the other two, making them unreadable.

iii- Our friend sends pairs of postcards simultaneously via Canada Post, one for you the other for me. A postmark with the sent date is printed on the postcards.

iv- It takes at least two days to deliver the mail, never more than three. This is the fastest way to communicate between each other.

v- A random number is chosen just before lifting the flap labeled with that number. This is done in order to sample the words without any bias.

vi- For every postcard I take note of the date, the chosen flap-number and the exposed word. You follow the same procedure.

vii- There is no other common agreement between any of you, me and our friend.

viii- There is no communication between any of us except for:

- the postcards sent to us by our friend,

- the lists sent to each other.

ix- Our lists are combined into word-pairs for each pair of postcards sent on the same date.

B)

i- One of only two possibilities, "heads" or "tails", is exposed when a flap is lifted. We never see an ink covered word.

ii- The occurrence of the words is random, with no discernible pattern.

iii- The words have a near equal chance of being used, independently of which flap is lifted. This is the result of the statistical analysis (2A).

C)

i- Word-pairs with the same flap-number always have the same words. This is the result of the statistical analysis (2B).

D)

i- Our friend flips a coin to generate the random words on the postcards. The statistical analysis (2A) is consistent with the 50% probability expected from a coin toss.

ii- The word exposed when I lift a flap tells me that a the same word is written in your postcard under the same-numbered flap,

(What Einstein called "element of reality": the word is determined even if nobody looks. "The moon is there when nobody looks.")

iii- Since I am free to choose which flap to lift, all three words on your postcard must already be determined when the postcard is sent.

This is because it is potentially possible for me to predict which word will be written under any of your flaps.

iv- You can deduce the same thing about the words written in my postcard.

v- Transmission of information is impossible at a speed faster than Canada Post can send a message.

("Locality" is assumed: a message cannot influence the other at another location faster than the maximum speed of transmission of messages.)

E)

i- In order to explain observation (3C-i), our friend

1 1H - 2H - 3H

ii- Each template has the same probability of being chosen. This is necessary in order to explain (2A).

iii- The postcard cannot be changed during transit (e.g. by a criminal tampering with the mail). If the criminal opened a flap, we would sometimes expose an ink covered word (which does not happen according to 3B-i). Even if a new postcard was forwarded by the criminal, they could only read one word from off the original and not know which other two words to fill in (because of 3B-ii they cannot predict the words written on the postcards). Forwarding a new postcard would destroy the perfect correlation (2B). Even if an eavesdropper saw which flap I opened, there would not be enough time for them (because of 3A-iv) to mail that information to an accomplice who could intercept your postcard and modify it.

We now want to calculate the probability of obtaining the

This is obtained by taking the template table and reducing it to the case where I lift flap 1. We don't need to know what is behind the other flaps:

1 1H

From (E-ii) we know that each template has the same probability of being chosen. If you lift flap 2, we get the same words for templates 1, 2, 7 and 8 out of eight templates: that is 4÷8 = 50% of the time. If you lift flap 3, we get the same words for templates 1, 3, 6 and 8 out of eight templates: that is also 50% of the time. The same probability of 50% is also obtained if the table is reduced to the case where I lift flap 2 or 3. (Left as an exercise!)

B)

Given that we accept 3A), 3B), 3C) and 3D), we predict that if we lift different flaps the probability of getting the same words in a word-pair is 50%.

The following table lists the results obtained from a statistical analysis of word-pairs with different flap-numbers taken from Table 1.

Words |
Flap 1 |
Flap 2 |
Flap 3 |

Flap 1 |
100% are the
same ( or 1T1T)1H1H |
55% are the same ( or 1T2T)1H2H |
61% are the same ( or 1T3T)1H3H |

Flap 2 |
54% are the same ( or 2T1T)2H1H |
100% are the
same ( or 2T2T)2H2H |
45% are the same ( or 2T3T)2H3H |

Flap 3 |
55% are the same ( or 3T1T)3H1H |
49% are the same ( or 3T2T)3H2H |
100% are the
same ( or 3T3T)3H3H |

TABLE 3: Statistical properties of 432
word-pairs obtained with a classical experiment. Green
boxes: probability

of getting the same words for the 293 word-pairs with different flap-numbers.

of getting the same words for the 293 word-pairs with different flap-numbers.

The analysis is therefore compatible with the predicted probability of 50%.

A real quantum experiment can be made using a system with a single Source and two Detectors. The Source emits a pair of simultaneous light-pulses in opposite directions: one light-pulse toward Detector A and the other toward Detector B (see Figure A).

The detectors measure the polarization of the light-pulses using an analyzer and two photodetectors. When a photodetector detects a light-pulse, it amplifies the signal to produce an electrical output large enough to be easily measured. The analyzer redirects the light-pulses according to their polarization: If a light-pulse goes through the analyzer, the photodetector labeled "heads" generates an output; if a light-pulse is reflected by the analyzer, the photodetector labeled "tails" generates an output. It is possible to produce light-pulses which will only trigger one photodetector at a time. Either "heads" or "tails" will produce an output at "H" or "T" respectively, never both at the same time. Hence the name "photon" is used because the indivisible quantized light-pulse can't be split: it will either go straight into photodetector "heads" or be reflected into photodetector "tails".

FIGURE A:
Simplified schematics of an experimental
setup. Two Detectors measure the polarization
of the received light.

The Source produces simultaneous photon-pairs.

The Source produces simultaneous photon-pairs.

After the photon-pair is emitted, the Detectors are rotated individually about the axis of the light beam. The entire assembly analyzer-photodetectors is rotated. The Detectors are set randomly to any of three positions:

Setting 1: aligned with the vertical at 0°,

Setting 2: rotated by 60° with respect to the vertical, or,

Setting 3: rotated by 120° with respect to the vertical.

Figure B shows an example with Detector A rotated by 60° and Detector B aligned with the vertical at 0°.

FIGURE B: Configuration with
Detector A at Setting 2 and Detector B at Setting 1

Such a setup which uses the above concepts has been built Ref. [3].

The analysis of output-pairs gives the following results:

A) The probability for producing an output "heads" is 50% and the probability of obtaining "tails" is 50%.

B) The probability of the same outputs for an output-pair is near 100% with the same detector setting.

This analysis reveals that

In this case, the model as described in (4) makes the same prediction for the quantum experiment:

When the output-pairs from the quantum experiment Ref. [3] are analyzed, the following results are obtained which can be summarized in the following table:

Outputs |
Setting 1 |
Setting 2 |
Setting 3 |

Setting 1 |
>99% are
the same |
~25% are the same |
~25% are the same |

Setting 2 |
~25% are the same |
>99% are the same | ~25% are the same |

Setting 3 |
~25% are the same |
~25 % are the same |
>99% are the same |

TABLE 4: Statistical properties of measured
occurrence of output-pairs. The results

obtained with the quantum experiment for different settings are not

compatible with the 50% predicted by the classical theory.

obtained with the quantum experiment for different settings are not

compatible with the 50% predicted by the classical theory.

It is

Entanglement describes the statistical property resulting from only

Entanglement, as described by quantum mechanics, is a solution which predicts experimental results. However, it rejects "reality" (Einstein would not reject elements of reality!) and "locality" (Einstein called this spooky action at a distance!) This tells us that the behaviour of the quantum world is far from our common experience.

[2] John Bell, "On the Einstein Podolsky Rosen Paradox," Physics. 1 (3), 195–200 (1964).

[3] Alain Aspect

[4] Xiao-song Ma

Updated 2018-2-17

© Louis Marmet 2018