I don't understand how it's being proven that error correction can be applied only to X, Z noises and this solves all errors?

Does this have to be with this set being universal? (Z is exactly like phase matrix or rotation matrix.)

Bonus question: When recovering a state in Shors code does it matter if I apply X, Z or Z, X?

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    $\begingroup$ Any 2x2 matrix can be written as a linear combination of I,X,Z and ZX. Moreover, those 4 matrices form an orthogonal basis in matrix space of 2x2 matrices under Hilbert–Schmidt inner product. $\endgroup$ – Danylo Y Mar 9 '19 at 8:33

The reason people focus on X and Z errors is because

  1. X and Z are super simple. You can propagate the errors through common gates like CNOT and H without any trouble. A product of X and Z errors on various qubits before a series of Clifford gates is equivalent to a (different) product of X and Z errors on various qubits after the gates. And error correcting codes are almost all made up of Clifford gates.
  2. You can still get upper bounds on how bad more flexible real quantum errors will be. An accidental 5 degree rotation around the Z axis of a qubit is no worse than a $\sin^2(5^\circ)$ chance of a Z error on that qubit.
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  • $\begingroup$ Two questions: (1) CNOT isn't included in this set only B I understand that you can build it from Z+X, no? (2) Why when recovering we always apply XZ and not ZX? Especially that you can't know which error happened first $\endgroup$ – bilanush Mar 9 '19 at 10:05
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    $\begingroup$ You can't build a CNOT out of X and Z, because it's a two-qubit entangling gate, however during the process of error correction you should not ever apply any gate between your qubits that lead to a CNOT between data qubits, and you projectively measure your stabilizers, which projects the errors onto the Pauli basis. XZ and ZX are the same up to a phase, so it doesn't matter. People just say XZ because that's the natural order. $\endgroup$ – Dripto Debroy Mar 9 '19 at 20:45

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