# A2/A Story

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A2

Quantum correlation
Introduction

Lektionen

Abb. 1 (a) A lottery ticket (b) Opening any field on one side will instantaneously blacken the opposing field.
Abb. 2 The four ticket areas and the coincidence as measured by Bob. The correlation marked in red is the one he wants to estimate.
Abb. 3 All four possible combinations for loosing tickets with FL=X and FR=O.

It was a sunny day when Bob - a physicist - decided to take a trip to the fair that happened to be in town. He rode the roller coasters, had some cotton candy and was pretty much enjoying himself, when he passed a scratch-off lottery tent that caught his attention. In bright letters above the tent, a sign said that "Honest Eves Lottery" will be the first lottery to allow the customers to check their chances of winning beforehand.

Bob was surprised, as he had always thought of such lotteries as scams, ripping the people of their money in a quiet obvious fashion. The basic concept of a scratch-off lottery was that there are two kinds of lottery tickets - wins and losses - of which a customer draws one at random from a big pot of tickets. The ratio between wins and losses, however, is set by the owner of the lottery. In an extreme case, he could simply put only losses in the pot and hope that no customer would recognize this. If, on the other hand, the potential customer of the lottery would know about his odds, he could calculate whether the bet was fair, and decide whether he wants to take the risk. At this point, Bob decided to enter the lottery tent.

The rules of the lottery

Bob asks the lady at the counter to explain the rules of the lottery to him: Each ticket has two scratch areas on the front. Each are will reveal a symbol, which is either an “X” or an “O” and the ticket is a win, if both symbols on the front coincide and a loss if they do not coincide. One ticket costs 1€, while a winning ticket gets the customer 2€ in return. From this Bob concludes, that if there were an equal number of wins and losses in the pot, the lottery would be fair. That is that in the long run, neither the lottery owner, nor the player would make any profit.

Remembering the sign outside the tent, Bob asks the owner about how this special scratch-off lottery allows the players to test their chances. She explains to him, that in order for the players to test the statistics of the tickets, each ticket has been equipped with two additional fields on the back of the ticket (see Fig. 1(a)). The rules are then as follows: The player draws one ticket at random from the bowl. Then he must decide: does he wants to buy the ticket or does he wants to use the ticket for testing. If he wants to buy, he opens the two fields on the front (owing 1€ to the owner by doing so) and finds out, if he holds a winning or a losing ticket. If he wants to use the ticket for testing, he does not pay anything, but he may not open the two fields on the front (otherwise he has to pay). This means he may open one field on the front and one field on the back, or the two fields on the back. Additionally, he may not open two opposing fields. In fact, the tickets have been produced in a way, that if a field is opened, the corresponding field on the opposite side of the ticket will turn black. For instance, of Bob chooses to open the left field on the back side, the left field on the front side will turn black (see Fig. 1 (b)).

How can Bob estimate his chances?

Now Bob starts to think: He is not allowed to test the correlation he is interested in (the ratio of winning tickets to losing tickets in the pot) directly, but has to estimate it using the other correlations. Suppose, we name the fields on the front side FL and FR (for “front-left” and “front-right”, respectively) and the fields on the back BL and BR. The correlation he wants to estimate is the probability of two equal symbols on the front side, so P(FL=FR) (where we use the symbol P for probability). The rules of the lottery do not allow him to measure this directly, but he may determine the correlations P(FL=BR), P(BL=BR) and P(BL=FR).

But how does the knowledge of these correlations help him?

The simplest, and most advantageous, case for him would be, if he would always observe equal symbols in his three test. In this case P(FL=BR)=1, so in every instance the symbol FL is equal to the symbol FR. If this also holds for the other two probabilities, it follows that FL=BR, BL=BR and BL=FR in every instance. But this can only be true, if the chain of equalities hold FL=BL=BR=FR. This would imply that all the tickets in the lottery would be winning tickets (as FL=FR), so Bob does not expect this to happen.

How would the estimation be, if the observed correlations on the test cases are not perfect, but only close to maximal?

To do this, Bob notices, that the probability of finding a winning ticket and the probability of finding ticket sum to one, as each ticket is either a win or a loss. This means that P(FL=FR) = 1 - P(FL\neqFR), so to estimate the number of wins he can likewise estimate the number of losses. For a losing ticket it holds that LF \neq FR, so the chain of equalities FL=BL=BR=FR has to be broken at least once. So, if he knows the probability of the three equalities to break, he can directly fin an upper bound for the probability of the chain to break: P(FL \neq FR) \leq P(FL \neq BR) + P(BL \neq BR) + P(BR \neq FR).

One should note that this estimation is actually optimal. A simple way to see this is to consider all possible loosing tickets and to see that the inequality hold for each one. From this it follows that it holds for all combinations of tickets in the lottery bowl. In Fig. 3 the four loosing tickets with the combination FL=X and FR=O are depicted, where the respective correlation on which non-equality is observed is indicated. The other four loosing tickets are similar, if one changes every X of O and vice versa.

Bob applies the estimation

So, Bob begins to draw tickets and determines the probabilities for the three test-combinations. To his surprise, his tests reveal that the probability for inequality in these cases is roughly 15%. If he applies the inequality, he can infer that that the probability that the fields on the front do not coincide is less than 45%. But then the probability of finding a winning ticket was to be at least 55%. Bob is surprised to see that this lottery is not only fair, but actually gives an advantage to the player. On average, he expects to make a profit of 0.10ct. in every round, so he decides to spend his remaining money on lottery tickets. As he opens the tickets, however, he becomes more and more worried as most of his tickets are actually losses - he gets winning tickets with a probability of only 15%. So, poor Bob loses most of his money to Eve and has to end his day at the fair earlier as planned.

On his way home, Bob tries to find out, where his reasoning went wrong. First he rechecks the derivation of the inequality and comes up with a caveat: While the estimation is mathematically correct, it relies on an assumption he made – namely that Eves lottery tickets behave like ordinary lottery tickets in the sense, that the value of every field is predetermined, and opening a field will only reveal the value that had been sitting there all along.

But what if the tickets contain active elements? In this case, the value of any field could be determined right in the moment he starts to scratch the field open. Then a signal would travel to the other fields and change the probabilities there. With such an active process, the observed correlations would be explainable.

Feeling pretty smart, Bob returns to the fair the next day to confront Eve with his reasoning and request his money back (maybe with an extra allowance for not telling anybody her trick). To his surprise, Eve is not impressed at all. She says that there is no signal travelling between the left and the right side of the ticket, and that she is willing to make a new bet with him on this. She will sell Bob the whole lottery pot and will give him ten-times his money back if he is able to confirm his claim. Bob agrees, takes the pot and leaves the tent to call a good friend of his, Alice. They meet and decide on a way to check if there is any signal travelling between the sides of the card. To do so, they cut the cards in half, where one side is given to Alice and the other side to Bob. They agree on a time at which each ticket should be opened and they will decide at random whether the front or the back should be considered. Any signal that would travel between the half tickets would travel at most at the speed of light, so Alice goes on a trip to the far side of the world to ensure that no signal from Bob could reach her in time when they open the fields. After meeting back at the fair they compare their findings. Surprisingly for them (but not for Eve) the observed correlations do not change even if the halves of the tickets are spatially separated. Now, Bob is out of ideas. He returns to the lottery and begs Eve to reveal her secret to him. Eve gently smiles, and points him towards the nearest quantum mechanics lecture, as the lottery tickets can be realized using quantum mechanical states.