Simplifying equation for two qubit syndrome extraction code

Question

In the paper Quantum Error Correction: An Introductory Guide (arXiv), the author gives the following formula for a simple two qubit code (Eq. 19 in the paper).

$$ E|\psi\rangle_L|0\rangle_A \xrightarrow{\text{syndrome extraction}} \frac{1}{2}(\mathbb{I}_1\mathbb{I}_2 + Z_1Z_2) E|\psi\rangle_L|0\rangle_A + \frac{1}{2}(\mathbb{I}_1\mathbb{I}_2 - Z_1Z_2) E|\psi\rangle_L|1\rangle_A $$

For reference, the circuit corresponding to this code is:

Here $E \in \{\mathbb{I}, X_1, X_2, X_1X_2\}$ is an error gate. Now, after giving that equation, the paper says

Now, consider the case where $E = X_1$ so that the logical state occupies the error space $E|\psi\rangle_L \in \mathcal{F}$. In this scenario, it can be seen that the first term in equation (19) goes to zero. [...] Considering the other error patterns, we see that if the logical state is in the codespace (i.e., if $E = \{\mathbb{I}, X_1X_2\}$) then the ancilla is measured as ‘0’.

$\mathcal{F}$ is the error space, which contains the states $|01\rangle$ and $|10\rangle$. By making $|\psi\rangle_L = \alpha|0\rangle_L + \beta|1\rangle_L$ and working out the equation with $E = X_1$ and $E = X_2$ I can see how the first term cancels and we are left with $|1\rangle_A$ in the ancilla qubit. Similarly with $E = X_1X_2$ and $E = \mathbb{I}$ I can see how we are left with $|0\rangle_A$ in the ancilla qubit.

As you can see, I needed to expand the logical qubit to see how the code is able to measure the error. However, I feel like it is implied in the paper that you can figure this out by just substituting $E$ and leaving $|\psi\rangle_L$ as it is. How can I show this? For example, in the case that $E=X_1$, how does the expression

$$ \frac{1}{2}(\mathbb{I}_1\mathbb{I}_2 + Z_1Z_2) X_1|\psi\rangle_L|0\rangle_A + \frac{1}{2}(\mathbb{I}_1\mathbb{I}_2 - Z_1Z_2) X_1|\psi\rangle_L|1\rangle_A $$

simplify such that the first term disappears, without making $|\psi\rangle_L = \alpha|0\rangle_L + \beta|1\rangle_L$? I tried doing it with matrix multiplication but the term didn't disappear.

Answer 1

If $E|\psi_L\rangle\in\mathcal{F}$, then it can be written as a linear combination of the states $|01\rangle,|10\rangle$, so that: \begin{eqnarray} E|\psi_L\rangle=\gamma|01\rangle+\delta|10\rangle. \end{eqnarray} This would correspond to the cases where $E\in\{X_1,X_2\}$. Now for the first term we have a sum of the identity in the $(1,2)$ subspace, and the product $Z_1Z_2$ acting over the state $E|\psi_L\rangle$. You can see that the term $Z_1 Z_2$ will transform the state to $Z_1 Z_2E|\psi_L\rangle=-E|\psi_L\rangle$, because the linear combination above picks opposite values of the $Z$ components, thus cancelling identically with the identity operation. More concretely: \begin{eqnarray} (\mathbb{I}_1\mathbb{I}_2 + Z_1Z_2)E|\psi_L\rangle=E|\psi_L\rangle-E|\psi_L\rangle=0. \end{eqnarray} Note that $(\mathbb{I}_1\mathbb{I}_2 + Z_1Z_2)$ acts only on the $(1,2)$ subspaces, but not in the ancilla $A$.