I don't understand the lattice splitting procedure for surface code

I am trying to understand the lattice splitting as described in Horsman paper. In short: I don't understand why measuring the data qubit to split the surface doesn't change the stabilizer results.

The tiny green dots are measurement qubits measuring $$Z$$ stabilizers (on the center of the plaquettes), or $$X$$ stabilizers (on each node).

The wide dots are the data qubits.

At time $$0$$ we have a unique surface. At time $$0^+$$, we are measuring in the $$X$$-basis the pink "wide" dots representing some the data qubits at the row in the middle.

My two questions

In the paper it is said:

Firstly, we can see that none of the joint Z operators changes at all: measuring out the qubits removes a row of face plaquettes from the error correction entirely, and leaves the surrounding face plaquettes untouched

1. Why is that true? If I measure the pink data qubit belonging to the black plaquette with a Pauli $$X$$ measurement, the black plaquette will have its value changed. Indeed the $$X$$ Pauli I applied touches one edge of the black plaquette and hence anti-commutes with it. The same reasoning applies for any plaquette below. The upper and bottom edges of the red plaquette are being touched by a Pauli-$$X$$ measurement. Their product might commute with the plaquette but I am not measuring their product but each Pauli individually (which is equivalent to measuring the product of the Pauli, which would commute with the plaquette, followed by one Pauli applied on one edge, which would anticommute with the plaquette).

2. Assuming there is no problem, what happens after this $$X$$ measurements? Do I stop including the pink qubits in any stabilizer measurement?

• Thanks for the answer. I think I understand your point. Just a check: we are actually measuring the data qubits in order to keep error tracking, is that correct? As we are going to remove one "feet" to the $X$ stab at the boundary, their syndrome might change when we remove the boundary. Those $X$ measurement allow us to keep the tracking of the errors, this is why it is necessary. Would you agree? Jan 27 at 18:48