Quantum Computing Asked by mbuchberger1967 on November 27, 2020
I am referring to the article Advanced Topics in Quantum Information Theory exercise 4 and to the MS Quantum Kata MS-Quantum-Kata which describes a solution for the Mermin–Peres Magic Square Game.
I could follow the article and do all the maths for myself until the end of the article.
Unfortunately the article ends with
… Following the hint, we consider two maximally entangled states shared between Alice and Bob:$$|Ψ〉_{AB}=1/√2(|00〉_{A_1B_1}+|11〉_{A_1B_1})⊗1/√2(|00〉_{A_2B_2}+|11〉_{A_2B_2})$$
It is then easy to check by using $$1/√2(|00〉+|11〉) =1/√2(|phi^x_0phi^x_0〉+|phi^x_1phi^x_1〉) =1/√2(|phi^y_0phi^y_1〉+|phi^y_1phi^y_0〉)$$ where $|0,1〉$ are the eigenvectors of $σ_z$ and where $|φ^{x,y}_j〉$ denote the eigenvectors of $σ_x$ and $σ_y$ as in the solution to the last exercise (with eigenvalues ±1 each), that the compatibility condition is indeed satisfied.
Unfortunately for me it is not easy to check, I spent multiple nights to try this, but I don’t get it done.
Thanks a lot for any help or hints.
Not an answer yet, but to long for comment: Alice and Bob share two entangled pairs. Now look only at Alice's. The last line gives the state that only one of'em has: $$frac1{sqrt 2}(|00〉+|11〉) =frac1{sqrt 2}(|phi^x_0phi^x_0〉+|phi^x_1phi^x_1〉) =frac1{sqrt 2}(|phi^y_0phi^y_1〉+|phi^y_1phi^y_0〉).$$ You can see that when you trace out Bob. Now all hermitian measurement operator in one row/column commute, which means that they share eigenvectors. So after the collapse at the first measurement, subsequent measurement along communting operators don't alter the state anymore, since they are already eigenstates. I think you have to check that the measured value is the same at the coincident site...
Answered by draks ... on November 27, 2020
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