I am trying to understand why we need to put the boundary data qubits in the ground state $|0\rangle$ for a rough merging procedure.
From my current understanding, if no errors occurred (perfect qubits) it would be unnecessary. The point of doing it is only in order to preserve fault tolerance.
In this image (taken from Horsman's paper), we have two surfaces (left and right) containing a single logical qubit. I call the left and right logical states $|\psi_L\rangle=\alpha|0\rangle+\beta|1\rangle$ and $|\psi_R\rangle=\alpha'|0\rangle+\beta'|1\rangle$. Then we will merge those two surfaces following the two steps:
- Put all the pink data qubits in $|0\rangle$
- Treat all the qubits as if we had a single surface. In practice it means that we include the pink data qubits in the $Z$ stabilizer measurements, and we measure the $X$ stabilizers at the boundary. To ensure fault-tolerance we then measure the stabilizers for $d$ clock cycles ($d$ being the code distance).
When doing step 2, we see that the product of the $X$ stabilizer measurement will be equal to $X_L X_R$ (the product of the logical Pauli of the two surfaces). Based on that, and after some calculations and some conventions taken, we can show that the resulting surface will have a quantum state being
$$|\Psi\rangle=\alpha |\psi_R\rangle + \beta X (-1)^M |\psi_R \rangle$$
Where $M$ is the eigenvalue of $X_L X_R$.
In the absence of any errors, for me it is unnecessary to put the intermediate pink data qubits in $|0\rangle$ (hence step 1). Indeed, we don't need step 1 to see that $X_L^1 X_L^2$ is being measured. And the reasoning yielding a final state being $|\Psi\rangle$ doesn't need to know that the pink qubits were initially in $|0\rangle$ (at least from my understanding, please correct me if you believe I am wrong).
For this reason, I guess that step 1 is here, to ensure fault tolerance. I would like to understand precisely why it is necessary.
While I am here focused on merging procedure, the philosophy behind my question is more general. It is very frequent to have to put qubits in some specific state before turning on or off stabilizer measurements and I really struggle to understand why. A related question has been asked here (but I don't get the answer there either).