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?


1 Answer 1


You're missing a couple of steps in the process! Note that we don't just use Pauli errors because they're nice. All errors can be expressed in terms of a linear combination of Pauli errors. (And, fair warning, you need all 3 Pauli operators to describe amplitude damping on a single qubit, so the 3-qubit repetition code won't be enough to protect it.)

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.

  • $\begingroup$ I see. To my understanding, amplitude damping can be expressed as a combination of X, Y, and Z errors via the twirling approximation, but the error itself is not Pauli (as in this paper: arxiv.org/pdf/1404.3747.pdf, Section IV.A). Is this correct or not? $\endgroup$
    – dankpit
    Aug 21, 2023 at 15:18
  • $\begingroup$ 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. $\endgroup$
    – DaftWullie
    Aug 22, 2023 at 1:02
  • $\begingroup$ Another way to think about it is that the stabilizer measurements project arbitrary noise into Pauli noise. Following DaftWullie's answer with a full CSS code (say a vanilla surface code), the inclusion of the X type stabilizers helps project the state into |+>_L up to some Pauli noise. Since a bit flip code only protects against X errors, more complex noise (like amplitude damping) does not break down as nicely, as you can see above. $\endgroup$
    – Ian
    Aug 22, 2023 at 10:44

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