# How do error-correction codes deal with decoherence?

This is a bit of a fundamental question. Normally, in QEC studies, an error model with Pauli errors is used because it is easy to simulate. But I was considering the following scenario:

Suppose you have a bit-flip code (as an example) and your qubit is in the $$|+\rangle_L$$ state (that is $$\frac{|000\rangle + |111\rangle}{\sqrt{2}}$$). If we use amplitude damping to model decoherence errors, then it is possible that we end up with an error on the second qubit:

$$|\psi\rangle = \frac{|000\rangle + |101\rangle}{\sqrt{2}}$$

If we were to apply the $$ZZI$$ and $$IZZ$$ stabilizers of the code, then we would get:

$$ZZI|\psi\rangle = \frac{|000\rangle - |101\rangle}{\sqrt{2}}$$

and similarly for the $$IZZ$$ stabilizer. We will either measure +1 or -1 here as $$|\psi\rangle$$ is not an eigenstate of the stabilizers: now we cannot recover $$|+\rangle_L$$. However, if our error was a simple $$X$$ error on the second qubit:

$$|\phi\rangle = \frac{|010\rangle + |101\rangle}{\sqrt{2}}$$

Then, recovery is straightforward as $$|\phi\rangle$$ is in the -1 eigenspace of both stabilizers.

$$|\psi\rangle$$ seems like an illegal state that completely destroys the logical state. Is this a limitation of such classical codes? Would a code such as the surface code be capable of handling such an error?

Now, let's be a bit more careful with the starting point - what does amplitude damping noise look like? You should describe it with 2 Kraus operators, $$\begin{bmatrix} 1 & 0 \\ 0 & \sqrt{1-\gamma} \end{bmatrix},\begin{bmatrix} 0 & \sqrt{\gamma} \\ 0 & 0 \end{bmatrix}.$$ Sure, you might split it up as choosing to look at the action of one or the other, but neither creates the evolution you specified: \begin{align*} \frac{1}{\sqrt{2}}(|000\rangle+\sqrt{1-\gamma}|111\rangle) \\ \frac{1}{\sqrt{2}}\sqrt{\gamma}|101\rangle \end{align*}
Next, when you measure a syndrome, that's not the same as just multiplying by the stabilizer. Instead, you should introduce an ancilla for each syndrome which records the $$\pm 1$$ outcome: \begin{align*} \frac{1}{\sqrt{2}}(|000\rangle+\sqrt{1-\gamma}|111\rangle)|00\rangle \\ \frac{1}{\sqrt{2}}\sqrt{\gamma}|101\rangle|11\rangle \end{align*} It is those syndromes that you measure to determine the correction. The second case will spot that you need to apply $$X$$ on the second qubit, yielding \begin{align*} \frac{1}{\sqrt{2}}(|000\rangle+\sqrt{1-\gamma}|111\rangle)|00\rangle \\ \frac{1}{\sqrt{2}}\sqrt{\gamma}|111\rangle|11\rangle \end{align*} As I say, this is not fully corrected, because there's also a bit of $$Z$$ error floating around which you've not been able to detect/correct. But, if you could, you would always end up with corrected state.
• It is correct that it is not Pauli in itself. But its action can be decomposed as Paulis (because Paulis form a basis for $2\times 2$ matrices. Twirling doesn't come into it. Aug 22, 2023 at 1:02