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Measuring in a basis other than the computational (Z) basis will generally destroy any entanglement present in the GHZ state, as the entanglement is defined with respect to specific basis states. Why?

(FYI, if I measure GHZ in z state, it will give me a probability consisting of results of 111 and 000 only which is correct. But if I measure it different basis such as X, I will get also probability results of 010, 100, etc. When I do the same in bell state, it gave me perfect probability results, in all basis measurements)

glS
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Rayhan Kabir
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1 Answers1

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The two-qubit GHZ state (Bell state) case is special: any measurement in a basis $$ \cos\theta X+\sin\theta Z $$ will return perfectly correlated results between the two parties.

In the case of a larger GHZ state, measuring one qubit in the Z is the only single-qubit measurement that returns a fully separable state of the other qubits (and hence has completely predictable outcomes when you measure in the corresponding bases, which happen to all be the $Z$ basis). Any other measurement that you make will return an entangled state, which means there's at least two different possible outcomes when you make single-qubit measurements.

DaftWullie
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