# ghz

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```					                    GHZ-a quantum game

June 8, 2007

1    The game
The gambles, Alice, Bob and Charlie, play against the house the following
game. The house asks one of two questions, question X or question Y and
they have to give an answer that is ±. They can not communicate during the
game but they are allowed to preplan and make any strategy they choose. If
they are all asked the question x they win if
xa xb xc = −1                             (1.1)
If one of them is asked the question x and the other the question y they win
if
xa yb yc = 1,                              (1.2)
ya xb yc = 1,                              (1.3)
ya yb xc = 1.                              (1.4)
These are the rules of the game. Can they win? There is no classical strategy
to win this game. Namely, they can not pre-decide what Alice, Bob and
Charlie will give as an answer because there is no solution to the equations
above. In fact, if you multiply the left hand side of the 4 equations that xa
appears as a square, and so does ya and so for all others so the left hand side
is +. But the product of the rhs is a minus. This shows there is no solution.

2    The quantum oracle
Now there is a quantum strategy to win the game. They prepare a quantum
state in a special superposition, and then Alice Bob and Charlie each go to a

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separate room with his qubit. the state is the GHZ state (after Grienberger,
Horn, Zeilinger) and is

|GHZ = |111 − |000                            (2.1)

It is readily veriﬁed that it is an eigenstate of Xa ⊗ Xb ⊗ Xc with eigenvalue
−1
Xa ⊗ Xb ⊗ Xc |GHZ = − |GHZ                        (2.2)
and also that it is an eigenstate of all the other questions with eigenvalue
+1, e.g.
Xa ⊗ Yb ⊗ Yc |GHZ = |GHZ                        (2.3)
This means that if they all measure the operator associate with the question,
tell the house the result, (but not each other,) they win.

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