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Calculating measurement probabilities for the CHSH game

  1. Mar 24, 2015 #1
    Hi All,

    First, the context for this question can be found in Wilde's Quantum Shannon Theory text on the arXiv, specifically the section starting at the bottom of page 98, entitled "Entanglement in the CHSH Game". My particular question relates to Exercise 3.5.11 on page 100 of the same document... I am working through this book independently and have moved well beyond this point, but frustratingly still can't solve the seemingly simple exercise 3.5.11, and I think it's because I'm missing something basic about how to calculate probabilities of the different outcomes!

    My understanding: Alice and Bob share a maximally entangled Bell state, |φ+> = 1/sqrt(2) [|00> + |11>]. We are given Alice and Bob's measurement strategies for each possible scenario and asked to show that their winning probability is cos2(pi/8), the well-known result of the chsh experiment... If Alice gets x=0 from the ref, she will measure the Z operator (phase flip in the computational basis) and return the result as "a", if she gets x=1 she will measure the X operator (a NOT operation in the computational basis) and return the result as "a". For Bob, upon receiving y=0 he will measure 1/sqrt(2) [ X+Z ] and for y=1 he will measure 1/sqrt(2) [ Z - X ], returning the result as "b" in each case.

    Alice and Bob "win" if aXORb = xANDy, where x and y are 0 or 1, uniformly chosen by the ref. It's straightforward to see that this only holds when Alice and Bob each return a 0 or each return a 1 for three of the four cases. For the final case, corresponding to 1AND1 we need Alice and Bob to return opposite pairs, (0,1) or (1,0).

    My approach (definitely wrong) has been to take each of the four cases, form the tensor product in terms of computational basis operators (not matrices), and separate the resulting operator into measurement operators that result in the different possible outcomes. For example, if Alice and Bob both receive x=y=0, the global measurement operator is

    Z⊗(1/√2)(X+Z) = (1/√2)(|0><0| - |1><1|)⊗(|1><0| + |0><1| + |0><0| - |1><1|)
    = (1/√2) [ (|00><00| + |00><01|) + (|11><11| - |11><10|) + (|01><00| - |01><01|) - (|10><10| + |10><11|)]
    = (1/√2) [ M1 + M2 + M3 - M4]

    It's easy to check that these measurement operators, while not projectors, "resolve" the Bell basis:

    ΣMiMi = IBell

    So I thought that, to find the probability that Alice and Bob collectively return 00 or 11 would just be the sum of the probabilities for M1 and M2 above, namely <φ+|M1+>, and for M2, but this gives the mixed result of 1/2...

    Should I instead be performing local measurements only, as in en. 4.123 of pg 127 of the document? This approach seems to get me closer, but still gives trouble.

    Thanks for taking the time!
     
    Last edited: Mar 24, 2015
  2. jcsd
  3. Mar 30, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Mar 30, 2015 #3

    Physics Monkey

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    The most direct way I know to obtain this answer is just to carefully write out all the projection operators on a case by case basis.

    For example, if x=y=0 so that (x AND y) = 0 then the win conditions are a=b=0 or a=b=1 so that (a XOR b) = 0. Since x=y=0 Alice and Bob are measuring Z and (Z+X) respectively. a=0 corresponds to Z=1 and a=1 corresponds to Z=-1 and similarly with Bob.

    The projector onto the subspace consisting of a=b=0 and a=b=1 is given by $$ P_{x=0,y=0} = \left(\frac{1-Z}{2}\right)\left(\frac{1 - \frac{Z+X}{\sqrt{2}}}{2}\right) + \left(\frac{1+Z}{2}\right)\left(\frac{1 + \frac{Z+X}{\sqrt{2}}}{2}\right). $$

    The probability to win with x=y=0 is then $$ p_{x=0,y=0} = \langle \Phi^+ | P_{x=0,y=0}|\Phi^+\rangle.$$

    Working out the analogous projectors for the other combinations of x and y and taking the expectation value gives the probabilities of a win [itex] p_{x,y}[/itex].

    The total probability of winning is then $$ \frac{1}{4} \sum_{x,y} p_{x,y} $$ because each x and y occurs probability 1/2. If you do everything right you should get the desired answered (I just checked it myself). As a computational hint, it helps to first add up all the projectors before taking the expectation value. Symmetry is also very helpful.

    Hope this helps.
     
  5. Mar 30, 2015 #4

    Strilanc

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    I'm not sure if you already know this or not or are making this mistake or not.

    One of the mistakes I made when learning this stuff was confusing the operator of an observable with the operation you would place in a quantum logic circuit to evaluate the observation. For example, if you want to measure the observable defined by X, you use an H gate instead of an X gate because that's the correct change of basis:

    Code (Text):
    Measuring the X axis:

    input state --H--•--H--> "collapsed" state
                     |
              0 -----⊕-----> measurement result
     
     
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