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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.

enter image description here

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?

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Why [doesn't the data measurement change the boundary stabilizers]? 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.

The basis of the data measurement has to match the type of the boundary you are introducing. The boundary type is important because it results in all the opposite-type stabilizers that the data measurements touch not being used anymore in the configuration after the data measurements. Yes, you anticommute with them. You destroy them. But you don't need them anymore.

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

That's right. You start doing the surface code cycle for two separated qubits, completely ignoring that they used to be one patch.

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  • $\begingroup$ 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? $\endgroup$
    – Marco Fellous-Asiani
    Commented Jan 27, 2022 at 18:48
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    $\begingroup$ Yes, the X data qubit measurements are necessary in order to track Z type errors. Those Z type errors may be piercing the XX parity sheet shared between the two logical qubits which is anchored to the spacelike boundary formed by the measurements so they have to be decoded. The ZZ parity sheet is instead anchored to the spatial boundaries, so X errors near the time boundaries don't have to be solved quite as exactly. $\endgroup$
    – Craig Gidney
    Commented Jan 27, 2022 at 19:15
  • $\begingroup$ Hmm ok. I dont have a full understanding and how to correct acknowledging the time dimension (for measurement qubit errors) yet so I dont understand exactly but thanks. $\endgroup$
    – Marco Fellous-Asiani
    Commented Jan 27, 2022 at 20:22

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