The Second Mystery: Nonlocality

Abstract

In this chapter, we shall discuss the second “impossible thing” to believe before or after breakfast or the second quantum mystery: the existence of instantaneous actions at a distance in Nature. But what does “action” mean here? If there are such actions, do they allow an instantaneous transfer of matter? An instantaneous transfer of energy? An instantaneous transfer of messages? An instantaneous transfer of information? Does their existence contradict the idea that “nothing goes faster than light” (which is supposed to be a consequence of the theory of relativity)? Besides, if quantum mechanics shows that the mind acts directly on matter, do such actions at a distance justify telepathy? We have to discuss each of these points carefully and slowly.

Appendices

Appendices

7.A Nonlocality in Newton’s Theory

Newton’s theory of gravitation has also a nonlocal aspect, but which is different from the one discussed in Sect. 7.3.

Two of the most famous laws of classical physics are due to Newton. Suppose we have two bodies, labelled 1 and 2 whose masses are denoted $M_1$ and $M_2$. Then, they attract each other through a gravitational force proportional to the product $M_1\times M_2$ of their masses and proportional to the inverse of the square of their distance d. The second law is that “the force is equal to the mass times the acceleration”, the acceleration being the rate at which the velocity changes. So, if we change the position of body 2, we change the value of the distance d between the two bodies in that equation, and therefore the value of the acceleration of body 1. If the acceleration changes, this changes the velocity of that body, and if one changes its velocity, one changes the position of the body.¹⁷

This makes actions at a distance possible: since the gravitational force depends on the distribution of matter in the Universe, changing that distribution, say by waving my arm, instantaneously affects the motion of all other bodies in the Universe (of course, the effect is minuscule, but we give this simple example to illustrate the principle). That action at a distance has properties 1 and 2 of Sect. 7.3, since it is instantaneous and acts arbitrarily far.

But it does not have the other properties, 3, 4 and 5 of Sect. 7.3: it does not have property 3 because its effect decreases with the distance, since the effect is proportional to the inverse of the square of the distance d. Besides, it affects all bodies at a given distance equally: there is nothing special in body 2 except its distance with respect to body 1. In other words, unlike what happens with the boxes, it is not individuated, so it does not have property 4.

On the other hand, it can in principle be used to transmit messages (so it does not have property 5 of Sect. 7.3): if I decide to choose, at every minute, to wave my arm or not to wave it, one can use that choice of movements to encode a sequence of zeros and ones and, assuming that the gravitational effect due to that movement can be detected, one can transmit a message instantaneously and arbitrarily far (of course, the further away one tries to transmit it, the harder the detection).

Note that all this refers to Newton’s theory. There have been no experiments performed in this framework that could prove that gravitational forces really act instantaneously or at least at speeds faster than the speed of light, and this is a major difference with respect to the situation in quantum mechanics.

7.B Proof of Bell’s Theorem in Sect. 7.4

As we said in Sect. 7.4.1, there are three questions numbered 1, 2, and 3, and two answers Yes and No. If the answers are given in advance, there are $2^3=8$ possibilities:

$$ \begin{array}{ccc} 1 &{} \qquad 2 &{} \qquad 3 \\ { Yes} &{} \qquad { Yes} &{} \qquad { Yes} \\ { Yes} &{} \qquad { Yes} &{} \qquad { No} \\ { Yes} &{} \qquad { No} &{} \qquad { Yes} \\ { Yes} &{} \qquad { No} &{} \qquad { No} \\ { No} &{} \qquad { Yes} &{} \qquad { Yes} \\ { No} &{} \qquad { Yes} &{} \qquad { No} \\ { No} &{} \qquad { No} &{} \qquad { Yes} \\ { No} &{} \qquad { No} &{} \qquad { No} \end{array}$$

So Alice and Bob could agree, for example to always answer “Yes” to the first question, “No” to the second one and also “No” to the third (let’s call that the YNN strategy). Or they could follow each of the strategies YYN, NYN, and NNN one third of the time. Or they could choose their answers at random among the eight possibilities.

In any case, in each situation there are at least two questions with the same answer. Therefore,

$$\begin{aligned} \begin{array}{l} \text{ Frequency } \text{(answer } \text{ to } \text{1 }\ = \text {answer to 2)}\\ +\ \text{ Frequency } \text{(answer } \text{ to } \text{2 }\ = \text {answer to 3)}\\ +\ \text{ Frequency } \text{(answer } \text{ to } \text{3 }\ = \text {answer to 1)}\ \ge 1\;, \end{array} \end{aligned}$$

(7.4)

since at least one of the identities: answer to 1 $=$ answer to 2, answer to 2 $=$ answer to 3, answer to 3 $=$ answer to 1, holds in every run of the experiment.

But if

$$ \begin{array}{l} \text{ Frequency } \text{(answer } \text{ to } \text{1 } = \text {answer to 2)}\\ =\ \text{ Frequency } \text{(answer } \text{ to } \text{2 } = \text {answer to 3)}\\ =\ \text{ Frequency } \text{(answer } \text{ to } \text{3 } = \text {answer to 1)} = 1/4\;, \end{array} $$

we get $\frac{3}{4} \ge 1$, which is a contradiction.$\blacksquare$

The inequality like (7.4), saying that a sum of frequencies are greater than or equal to 1, is an example of a Bell inequality , i.e., an inequality which is a logical consequence of the assumption of pre-existing values, but which is violated by quantum predictions.

7.C How to Encode Secret Messages?

Suppose that Alice and Bob possess a common key k, namely a random sequence of 0’s and 1’s, and that they want to use that sequence in order to encode a message m which is also a sequence of 0’s and 1’s, but a non random one (messages that have a meaning for us are not random) in such a way that the result looks random.

They can do that by adding “in binary addition” the message to be sent m and the sequence k; binary addition means that one adds each of the corresponding symbols in the two sequences according to the following rules:

• $0+0=0$,

• $0+1=1$,

• $1+0=1$,

• $1+1=0$.

It looks like those rules are the ordinary ones except for the last one where we seemingly made a mistake: $1+1=2$! But these are just rules that we define to be the valid ones if we deal with only two numbers, 0 and 1. There is no symbol 2 here, by definition.

For example, if the message to be sent is $m=01101010$ and the common sequence or key is $k=11011001$, we have $m+k=10110010$ and that is the sequence which is sent by Alice to Bob through an open channel (and can in principle be obtained by the spy).

Since Bob also has the sequence k, it is very easy for him to obtain the original message m. Indeed, with our binary rules, we have $0+0=0$ and $1+1=0$. So, adding twice the same message amounts to adding 0 everywhere, which means that one does not change anything. So, one has $m+k+k=m$. Just to check, add $k=11011001$ to $m+k=10110010$, with the binary rules, and you will obtain back $m=01101010$.

Now, if the sequence k is sufficiently long and sufficiently random one can use it to encode any given message (we just gave one with eight symbols to illustrate the method) and the regularities in the message m will disappear because of the randomness in k.

To check that the sequence $m+k$ will be as random as k, no matter what m is, just consider the least random m one can imagine: $m= 111111111\dots$ that is every symbol of m is 1. Then, with the rules of binary addition, in the sequence $m+k$ one will simply have each symbol 0 in k replaced by 1 and each symbol 1 in k replaced by 0. But if the sequence k is random, the new sequence $m+k$ will also be random.