# Calculating bit flip and phase error using local operations for GHZ state

Suppose each qubit of $$\text{GHZ}_3$$ State is distribute to $$3$$ different parties at different locations through a noisy quantum channel. So each qubit can possibly go through: a bit flip error, a phase error or a combination of these two.

Is there some operation that can be performed locally by each parties to determine the errors in the qubit?

Basically I wanted to generalize the idea for Bell state to GHZ state as given here. Clearly in the case of Bell state when Alice sends other half to Bob, there exists measurements that can be performed localy to determine bit flip and phase error. That is (see here for detail): \begin{aligned} &\Pi_{\mathrm{bf}}=\frac{1}{2}\left(\mathrm{id} \otimes \mathrm{id}-\sigma_{z} \otimes \sigma_{z}\right) \\ &\Pi_{\mathrm{pe}}=\frac{1}{2}\left(\mathrm{id} \otimes \mathrm{id}-\sigma_{x} \otimes \sigma_{x}\right) \end{aligned}

I am interested in estimating amount of error in GHZ state using local operations in order to apply the error correcting codes and to calculate the fidelity.

I think the place you're starting from is misleading. If Alice and Bob share a Bell pair, and one qubit could have had an error, then, yes, they would like to measure observables $$X\otimes X$$ and $$Z\otimes Z$$ to detect the error. In principle, you can measure the values of these two simultaneously because they commute.
However, if you're trying to measure $$Z\otimes Z$$ on two qubits that are spatially separated, you either need
• to use some extra entanglement (which you don't have) to implement the $$Z\otimes Z$$ measurement directly, or
• to measure both qubits in the $$Z$$ basis. This is a bad thing because it completely destroys your state. (To see this differently, you're measuring the observables $$Z\otimes I$$ and $$I\otimes Z$$. These do not commute with $$X\otimes X$$). Hence, you can detect one of the types of error, but your entanglement is completely gone.