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

enter image description here

Where $E \in \{\mathbb{I}, X_1, X_2, X_1X_2\}$ 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.

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