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This question pertains to the real-time classical processing required while running the surface code.

Consider two logical patches $P$, $P'$. Let's assume that after $d$ rounds, a merge operation is performed on the two patches to produce a merged patch $PP'$. Considering we have $d$ rounds worth of syndromes immediately before the merge operation is performed, does the decoder need to decode errors for both $P$ and $P'$ before the merge operation can start?

I have the same question for splits. Let's assume that after $d$ rounds, a split operation is performed on a patch $PP'$ to produce $P$, $P'$. In this case, does the decoder need to decode $d$ rounds worth of syndromes before the split operation is performed?

Additional clarification: When I say the decoder needs to decode errors before a merge/split operation is performed, I imply that the patches will remain idle until decoding is complete after which the lattice surgery operation is performed.

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The decoder doesn't need to be "caught up" in order to start or stop lattice surgery. You just charge forward and merge/split the patches, and the decoder will figure out the corrections later. The corrections are Pauli gates which can be handled entirely within the classical control system; you don't need to do anything different on the quantum computer so it doesn't have to wait.

This is generically true about most stabilizer codes doing Clifford operations. The decoder can lag arbitrarily far behind without compromising accuracy. Only non-Clifford operations force the decoder to catch up, in order to decide whether or not a fixup Clifford gate occurs on the quantum computer. Even then, it only has to catch up to the point where the non-Clifford gate happened (as opposed to catching up to the present). We call the average delay until the decoder catches up to the non-Clifford operation, in order to decide on the fixup, the "reaction time" of the system. Clifford gate speed is independent of the reaction time.

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  • $\begingroup$ Thanks for the answer! >>> Even then, it only has to catch up to the point where the non-Clifford gate happened (as opposed to catching up to the present). <<< I'm having trouble understanding this. Wouldn't the non-clifford gate to which the decoder has to catch up to also be the latest instruction being executed? $\endgroup$ Feb 1 at 22:36
  • $\begingroup$ @controlfreak It might be the latest logical instruction, but the physical instructions can't stop happening while the decoder catches up or else the system will fall apart. If things go badly you can end up in a state where the logical qubits are idling for a long time, waiting for the decoder, but the physical stabilizer measurements never ever stop. But the logical qubits don't idle until the decoder catches up to the present; once it catches up to where the last measurement happened then you can decide what they need to do next and move on. $\endgroup$ Feb 1 at 23:16

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