2
$\begingroup$

I see on this chapter that, when noise applies to an even number of entangled qubits, such noise can be considered as operating only on half of those qubits. This is interesting, but I'm not sure if I'm well interpreting the model. I suppose that the qubits must be fully entangled, and, maybe, an even amount as well. Is that right?

EDIT: I read that there's the assumption that half of the state is perfect. However, I still think that for some state, e.g. the bell states, this kind of correction is possible.

$\endgroup$

1 Answer 1

1
$\begingroup$

Consider the Bell pair $|\Phi^+\rangle_{\mathcal{AB}}$. By ignoring global phases, the possible errors are $\{I,X,Z,XZ\}^{\otimes 2}$. Many of these operators have no effect -- again, up to a global phase -- to $|\Phi^+\rangle_{\mathcal{AB}}$. Namely: $X^{\otimes 2}, Z^{\otimes 2}, (XZ)^{\otimes 2}$, for similar reasons, also combinations can be considered as an error occurring only on, say, system $\mathcal{A}$.

In fact, assume that $Z_{\mathcal{A}}\otimes (XZ)_{\mathcal{B}}$ happens. Since $Z^{\otimes2}|\Phi^+\rangle_{\mathcal{AB}} = I^{\otimes2}|\Phi^+\rangle_{\mathcal{AB}}$, this is equivalent to $X_{\mathcal{B}}$, which, in turn, is equivalent to $X_{\mathcal{A}}$.

The above reasoning can be applied to the remaining cases. However, this result doesn't imply that the error is detectable and correctable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.