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# 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, ifAliceandBobboth uncover the same coin, the outcomes are perfectly correlated. We know it always works, we've checked it a million times.

Bobreasons:

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

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

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

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

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!

AliceandBobrepeat 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?Bellassumed 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

AandBto 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)