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If I consider the 3-dimensional structures from https://arxiv.org/abs/1905.08916, these would form a nice 3-dimensional structure for surface code layouts, except I'm not sure what the equivalence rules are. For example, it seems like basic deformations aren't necessarily allowed, since we could deform a CNOT into something that is essentially a lattice surgery merge then split (which as far as I can tell, is not equivalent to a CNOT). Has there been a paper (or is it obvious?) about the equivalence rules for these 3-d diagrams?

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I'm not sure what the equivalence rules are

I gave a talk about this at one point. The slides are here, and define the flows of the lattice surgery constructions as well as the ZX analogues:

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since we could deform a CNOT into something that is essentially a lattice surgery merge then split (which as far as I can tell, is not equivalent to a CNOT)

It's still a CNOT. Anything that maps $X_c \rightarrow X_cX_t$, $X_t \rightarrow X_t$, $Z_t \rightarrow Z_cZ_t$, and $Z_c \rightarrow Z_c$ is a CNOT and deforming or bending the pipes won't change how the parity sheets map input ports to output ports. All topological deformations are allowed. Beyond that there are lots of additional non-topological rewrites that are allowed, corresponding to ZX calculus rewrites.

I recommend reading:

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  • $\begingroup$ If I have two qubits and lattice merge them, then I have a single logical qubit (i.e., given the stabilizers I'm measuring, there is only one logical X operator). Dimensionality alone seems enough to conclude that nothing I can do after this point can be equivalent to a CNOT. Is there something different being done with the stabilizers that avoids this issue? $\endgroup$
    – Sam Jaques
    Commented May 10 at 15:00
  • $\begingroup$ @SamJaques Oh, I see what's confusing you. If you deform away the ancilla qubit, you will find that the bridge connecting the two logical qubits ends up with a quarter turn in it. So it won't correspond to a merge, and you won't be able to fit it into O(d^2) spacetime.. The quarter-turn bridge corresponds to a CNOT. $\endgroup$ Commented May 10 at 16:18
  • $\begingroup$ Oh, a lattice surgery CNOT is a rough merge and a smooth merge! So you can't do it straight across. $\endgroup$
    – Sam Jaques
    Commented May 10 at 17:16

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