State injection in a surface code

Question

I am reading this paper on lattice surgery by Dominic Horsman et al, and am struggling to understand a simple method they put forward in their Figure 8(a)(b)(c) to inject an arbitrary state into a planar surface.

The protocol they put forward is very simple: initialize all physical qubits to $|0\rangle$ except for one distance-3 line of qubits. For that line of qubits, initialize each one of them to the arbitrary state to be injected $|\psi\rangle=\alpha|0\rangle + \beta|1\rangle$. And then this surface is "stabilized" (I imagine this means 1 round of concurrent measurements of all X- and Z-type stabilizers), yielding a distance-3 logical qubit in that arbitrary logical state $|\psi\rangle_\text{L}=\alpha|000\rangle + \beta|111\rangle$.

I know that in surface code we operate on quiescent states, like described in Fowler's review paper. Am I correct that this simple state injection is viable, because $|0\rangle$ transversal initialization and $|1\rangle$ transversal initialization gives the two orthogonal logical state $|0\rangle_L$ and $|1\rangle_L$ after the code is stabilized, and so a superposition initialization of a line of qubit works as intended?

Could anyone help me understand why such a simple initialization of qubit does the job? Thanks.

Answer 1

I believe your reasoning is correct.

To put it in other words: you define the logical operators $Z_L$ and $X_L$ as a string of Pauli $Z$ and $X$ in the usual way for surface code (i.e., a row from left to right or a ladder from up to bottom, respectively).

Now, notice that you don't start with all three qubits in the desired state, but rather only one of them is initialize in $\alpha|0\rangle + \beta|1\rangle$, whereas the other two are initialized in $|0\rangle$. Then, you perform a CNOT gate, resulting in the state $\alpha|000\rangle + \beta|111\rangle$.

You now want to show that it is actually equivalent to $\alpha|0_L\rangle + \beta|1_L\rangle$. Indeed, after you measure all the stabilizers, any row of $ZZZ$ from left to right will act only on one of these qubits (say, the top one), such that it transforms $\alpha|000\rangle + \beta|111\rangle \to \alpha|000\rangle - \beta|111\rangle = \alpha|0_L\rangle - \beta|1_L\rangle$, which is exactly how $Z_L$ should act. Similarly, any ladder of $XXX$ will act on all three qubits, such that it transforms $\alpha|000\rangle + \beta|111\rangle \to \alpha|111\rangle + \beta|000\rangle = \alpha|1_L\rangle + \beta|0_L\rangle$.

Because the stabilizers commute with these operations, and they act on the state as you want them to act, you are in the desired state.

Notice, however, that the expansion step might have some problems. See here Surface code expansion in the lattice surgery context.