Context:

Three coins on the table. Each is either heads or tails. You can uncover any one of the three coins, revealing whether it is heads or tails but then you choose two the other two coins disappear --- you'll never know whetehr those two other coins are heads or tails.

I'm a bit confused by Preskill's description of a Bell experiment [$$\dagger$$]. He describes a game, using correlated coins. Donald sitting in Denver manufactures three pairs of correlated coins. He then sends one set of coins (containing coin from each correlated pair) to Alice in Pasadena and another set of coins to Bob in Waterloo. Now according to the rules of the game: if Alice or Bob uncover any one coin from their set (to check if it's head or tails), the other two coins instantaneously disappear (and we can't ever know the states of those disappeared coins). Also, if Alice and Bob uncover the same number coin from their respective set (say if both choose coin number 2), they always observe the same side (i.e. either both get heads or both get tails).

There are many sets of coins, identically prepared by Donald (in Denver). For each of the three coins, in Pasadena or Waterloo, the probability is $$\frac{1}{2}$$ that the coin is heads or tails. But, if Alice and Bob both uncover the same coin, the outcomes are perfectly correlated. We know it always works, we've checked it a million times.

Bob reasons:

• We know the correlation is always perfect.
• And surely what Alice does in Pasadena exerts no influence on what Bob finds when he uncovers a coin in Waterloo.
• So, in effect, Alice and Bob, working together can learn the outcome when any two of the coins are uncovered in Waterloo.

Bell reasons:

$$\sum_{x,y,z \in \{H,T\}} P(x,y,z)= 1$$

$$P(1,2)_{\text{same}} = P(HHH) + P(HHT) + P(TTH) + P(TTT),$$ $$P(2,3)_{\text{same}} = P(HHH) + P(THH) + P(HTT) + P(TTT),$$ $$P(1,3)_{\text{same}} = P(HHH) + P(HTH) + P(THT) + P(TTT).$$

$$\therefore P(1,2)_{\text{same}} + P(2,3)_{\text{same}} + P(1,3)_{\text{same}} = 1 + 2P(HHH) + 2P(TTT) \geq 1$$

I do understand that the sum of these three probabilities is greater than one because there are some constraints already involved and that the three cases: $$(1,2)_{\text{same}}, (2,3)_{\text{same}}$$ and $$(1,3)_{\text{same}}$$ are not mutually exclusive; like if we uncover all three coins at least two have to be the same. So naturally, there's some redundancy leading to a sum of probabilities that is greater than one!

Alice and Bob repeat the measurement a million times, and found ...[$$\dagger\dagger$$] $$P(1,2)_{\text{same}} = P(2,3)_{\text{same}} = P(1,3)_{\text{same}} = \frac{1}{4}$$

How could Bell's prediction be wrong? Bell assumed the probability distribution describes our ignorance about the actual state of the coins under the black covers, and there is no "action at a distance" between Pasadena and Waterloo. The lesson:

• Don't reason about counterfactuals ("I found H when I uncovered 1; I would have found either H or T if I had uncovered 2 instead, I just don't know which."). When the measurements are incompatible, then if we do measurement 1 we can't speak about what would have happened if we had done measurement 2 instead.

• Quantum randomness is not due to ignorance. Rather, it is intrinsic, occuring even when we have the most complete knowledge that Nature will allow.

Note that the quantum correlations do not allow A and B to send signals to one another.

I believe Preskill's actually referring to shared pairs of Bell states between Alice and Bob here rather than the coin setup. I'll try to clarify the difference between the two setups.

The Classical Coin Game:

If we were to consider the setup using classical coins as described in the question, the probabilities would have turned out to be considerably different. It is given that the same numbered coins on Alice's and Bob's side always turn out to have the same side. Now, the important point here is that "coins" can't exist in superposition, unlike qubits! Their only states of existence are either heads or tails. So according to the correlation described in the question if we can know the state of all the coins on Alice's side we already know the the state of all the coins on Bob's side! That is,

$$P(1,2)_{\text{same}} = P(HHH) + P(HHT) + P(TTH) + P(TTT)$$ $$\implies P(1,2)_{\text{same}}= \frac{1}{2}.\frac{1}{2}.\frac{1}{2} + \frac{1}{2}.\frac{1}{2}.\frac{1}{2} + \frac{1}{2}.\frac{1}{2}.\frac{1}{2} + \frac{1}{2}.\frac{1}{2}.\frac{1}{2} = \frac{1}{2}$$

Similarly, $$P(2,3)_{\text{same}}$$ and $$P(1,3)_{\text{same}}$$ are also $$\frac{1}{2}$$! As you can see, the answer isn't $$\frac{1}{4}$$ in this setup.

The Quantum Game:

Here Alice and Bob share three $$|\Phi^+\rangle$$ Bell pairs. I'll denote Alice's qubits with $$A$$ (numbered $$1,2,3$$) and Bob's qubits with $$B$$ (numbered $$1,2,3$$). The overall state is:

$$|\Phi^+\rangle \otimes |\Phi^+\rangle \otimes |\Phi^+\rangle$$ $$\small{= \frac{1}{\sqrt{2}}(|0\rangle_{A1}|0\rangle_{B1} + |1\rangle_{A1} |1\rangle_{B1}) \otimes \frac{1}{\sqrt{2}}(|0\rangle_{A2}|0\rangle_{B2} + |1\rangle_{A2} |1\rangle_{B2}) \otimes \frac{1}{\sqrt{2}}(|0\rangle_{A3}|0\rangle_{B3} + |1\rangle_{A3} |1\rangle_{B3})}$$

From here, in this quantum case too:

$$P(1,2)_{\text{same}} = P(\text{A1} = |0\rangle \ \cap \ \text{B2} = |0\rangle) + P(\text{A1} = |1\rangle \ \cap \ \text{B2} = |1\rangle)$$ $$= (\frac{1}{\sqrt 2})^2.(\frac{1}{\sqrt 2})^2 + (\frac{1}{\sqrt 2})^2.(\frac{1}{\sqrt 2})^2 = \frac{1}{2}$$

Question:

So I'm not quite sure how Preskill came up with the probabilities $$P(1,2)_{\text{same}} = P(2,3)_{\text{same}} = P(1,3)_{\text{same}} = \frac{1}{4}$$ in either the quantum case or the classical case. Could someone clarify? I did go through the proof for $$P(X=Y) = \frac{1}{4}$$ in the light polarization experiment, but in that experiment $$P(X=Y)$$ means given photon passed X, the probability that it passes Y is. There they didn't mention anything about Bell pairs, and so I can't quite relate the two experiments. Here the situation is rather different; we want to know what is the probability of the two qubits $$\text{A1}$$ and $$\text{B2}$$ having the same state after consecutive or simultaneous measurements? And for this question I don't quite see how the answer is $$\frac{1}{4}$$.

[$$\dagger$$]: John Preskill - Introduction to Quantum Information (Part 1) - CSSQI 2012 (timestamp included)

[$$\dagger \dagger$$]: John Preskill - Introduction to Quantum Information (Part 2) - CSSQI 2012 (timestamp included)

1 Answer

I do understand that the sum of these three probabilities is greater than one because there are some constraints already involved; like if we uncover all three coins at least two have to be the same. So naturally, there's some redundancy leading to a sum of probabilities that is greater than one!

I would say the explanation is simpler than that. Remember that the rule for adding probabilities and them summing to 1 only applies if (i) the set of events is mutually exclusive and (ii) covers everything that could possibly happen. Now, the set of things that are added together do include everything that can happen because, as you said, at least two coins must come out the same. However, they're not mutually exclusive. It can be that 1 and 2 get the same answer and 1 and 3 get the same answer. These are the cases TTT and HHH.

I believe Preskill's actually referring to shared pairs of Bell states between Alice and Bob here rather than the coin setup.

Implicitly, yes. But that's not important. The philosophy is important. Bell predicts something. You do an experiment of some form. The experimental details are irrelevant, but the outcome is different from the prediction. Why? One of the assumptions of the model must be incorrect.

So I'm not quite sure how Preskill came up with the probabilities

As I tried to say above, that is not necessarily the point. He could have picked anything arbitrarily to make his point.

If I were trying to build a model, I'd build it a bit differently to yours. I'd give Alice and Bob a single entangled state $$|\Psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$ Alice and Bob each choose from one of 3 possible measurement bases, $$Z,\quad \frac{-Z+\sqrt{3}X}{2},\quad \frac{-Z-\sqrt{3}X}{2}.$$ When a measurement is performed, it gives answers $$\pm 1$$ which we correspond with the heads and tails outcomes of the coins. You can verify that for each basis $$\sigma$$, $$\langle\Psi|\sigma\otimes\sigma|\Psi\rangle=1,$$ meaning that is both parties choose the same measurement (aka coin), they get the same result. You can also see that the individual outcomes are 50:50 by evaluating $$\langle\Psi|\sigma\otimes\mathbb{I}|\Psi\rangle=0.$$

If we take two different measurements $$\sigma$$ and $$\tau$$, then $$\langle\Psi|\sigma\otimes\tau|\Psi\rangle=-\frac12.$$ Since this is effectively evaluating $$P(\text{same})-P(\text{different})$$ with $$P(\text{same})+P(\text{different})=1$$, you can see that $$P(\text{same})=\frac14$$.

How is this different from what you suggested? Well one of the important things is that the analogy should be clearer. With measurements on the same qubit, you can only measure one of them, and you cannot simultaneously measure the outcomes, i.e. the other two coins disappear. In what you were measuring, it was still in principle possible to perform all 3 measurements because the measurement operators would have commuted.