# In quantum error correction, what does an "arbitrary error that yields an un-normalized state" mean?

This is from page 434 of Nielsen and Chuang:

. Supposing the state of the encoded qubit is |ψ⟩ before the noise acts, then after the noise has acted the state is E(|ψ⟩⟨ψ|). To analyze the effects of error-correction it’s easiest to focus on the effect error-correction has on a single term in this sum, say Ei|ψ⟩⟨ψ|E†i . As an operator on the first qubit alone Ei may be expanded as a linear combination of the identity, the bit flip, the phase flip, and the combined bit and phase flip.The (un-normalized) quantum state Ei|ψ⟩ can thus be written as a superposition of four terms, |ψ⟩, X1|ψ⟩, Z1|ψ⟩, X1Z1|ψ⟩.

Just before, this is stated:

Indeed, the Shor code protects against much more than just bit and phase flip errors on a single qubit – we now show that it protects against completely arbitrary errors, provided they only affect a single qubit! The error can be tiny – a rotation about the z axis of the Bloch sphere by π/263 radians, say – or it can be an apparently disastrous error like removing the qubit entirely and replacing it with garbage!

I understand as to how it's possible for any error with can be defined as an arbitrary gate acting on a single qubit in the system to be corrected, but what does an arbitrary error that yields an un-normalized quantum state even mean? How is it possible for a qubit's resultant state to be un-normalized?

• The text does not say that the error yields an unnormalized state. They are taking one term from the sum $\sum_i E_i |\psi\rangle \langle \psi | E_i^\dagger$ and analyzing it. Obviously, the entire sum is a normalized density matrix, but each term in it is not. Commented Jun 21, 2023 at 6:17
• @AbdullahKhalid This clears it up perfectly. Commented Jun 21, 2023 at 8:10