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2

I will attempt to provide some insight regarding your first question. For starters, both quantum surface codes and quantum block codes are stabilizer codes, which means that although they are significantly different in terms of their construction and utility, they still share some common ground. With regard to which code family is more promising, I believe ...

0

I have added the layout from the paper. A Z error on Db will fire Xb and Xa. A Z error on Dc will fire Xa. Thus these two are distinguishable. If a X error occurs on Dc this will fire Zb. This can be corrected by applying X on Dc. If a X error occurs on Db this will also fire Zb. It is also corrected by applying X on Dc. At the end Db and Dc have been ...

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Distinguishing $X$ and $Z$ errors is easy. $X$ errors anti-commute with the $Z$-type stabilizers, and so when you perform a measurement of those parity checks, you get and answer '1'. Similarly, $Z$ errors give you a '1' answer only on the $X$-type parity checks. Also, note that, in the bulk (i.e. not on the edges), you never get a '1' on only one weight-4 ...

3

For $|0\rangle$, $|1\rangle$, $|+\rangle$, and $|-\rangle$ you do transversal initialization (initialize all physical qubits to the desired state, then turn on the stabilizers). For $|i\rangle$ and $|-i\rangle$ you do topological initialization using twists. For $T|+\rangle$ states and other states with nice state distillation protocols, you do noisy low ...

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I've not read the cited paper, so I don't know how this corresponds to anything that they say, but one way that I would think about it is, if I have an unknown qubit state stored on a single qubit, how do I copy this onto a surface code already initialised in logical 0? Now, if it weren't logical qubits, we can easily write down a circuit that would ...

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Each vertex has a physical data qubit. But what exactly do the operators (green circles) represent? Is there an X and Z operator on every vertex site, or only some of them (Figure 1b)? The circles on, e.g. the bottom-left of Fig. 1, show you how to describe each of the stabilizers. So, for every mustard yellow square in the top-left diagram, there is a ...

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You can initialize a qubit to any arbitrary state by gate $U3$ (abbreviation used on IBM Q): $$U3(\theta,\phi,\lambda)= \begin{pmatrix} \cos(\theta/2) & -\mathrm{e}^{i\lambda} \sin(\theta/2) \\ \mathrm{e}^{i\phi}\sin(\theta/2) & \mathrm{e}^{i(\lambda+\phi)} \cos(\theta/2) \end{pmatrix}$$ It is also possible to prepare any multi-qubit quantum ...

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