Are the preparations of a logical state in a surface code random and do we need to update the stabilizers after preparation?

Question

Reading Fowler et al, there is something in regards to the intialisation of the surface code in a logical state (ie a state of a logical operator) that is confusing me.

I can think of two examples that illustrate my problem. The first is restricted to planar code. The initialisation of a logical $|0\rangle_{L}$ or $|+\rangle_{L}$ is performed by taking all the data qubits, initialising them as either $|0\rangle$ or $|+\rangle$, and then turning on the stabilisers.

In the case of $|0\rangle^{\otimes n}$, $\hat{Z_{L}} |0\rangle^{\otimes n} = |0\rangle^{\otimes n}$, showing it to be a +1 eigenstate. From here, initialisation proceeds by turning on all the stabilizers. The $\hat{Z}$ stabilizers shouldn't affect this state. So only the $X$ stabilisers will. However, $\hat{Z_{L}}$ commutes with all of them. So this means I can expand it as a linear combination of both +1 and -1 eigenvectors of the $\hat{X}$ stabilizers, correct? However, this means that, while an projection by the $X$ stabilizers will give a +1 eigenstate of $\hat{Z_{L}}$, only the projections

$$\frac{(I+\hat{X_{s}^{i}})}{2}$$

will give me a +1 eigenstate of $\hat{X_{s}^{i}}$, where $\hat{X_{s}^{i}}$ is a stabilizer that measures in the $X$ basis to identify $Z$ errors.

Looking at Fowler et al, in the section where they show how to initialize a +1 $\hat{Z_{L}}$ eigenstate, on page 17, they also say that after measuring the qubits initialized in a state of $\hat{Z_{L}}$ with 3 $\hat{X_{s}^{i}}$ stabilizers, we end up with an eigenstate of all the stabilizers. However, like before, the outcome being a +1 or -1 eigenstate of these $\hat{X_{s}^{i}}$ is random.

Since this is the case, what do we do here? We can't just use the -1 result and update the 3 stabilizers, or in the case of the planar code, however may gave a -1 result. This would result in a change of stabilizers that could affect other parity measurements, especially in the case of the surface code. For example, if I took the stabilizer being data qubits 1 and 2 in Fig.14 of Fowler et al, and changed it to $-\hat{X}$, wouldn't this affect the measurement results? Or is it the case that now it just means that the surrounding qubits are taken to a state that is a simultaneous +1 eigenstate of this new operator and the old ones.

Or do we just take this as a failed initialization attempt and try again?

Answer 1

TL;DR: Initializing data qubits in a physical $X$ (respectively, $Z$) eigenstate and switching on the stabilizers reliably$^1$ prepares a logical $X$ (respectively, $Z$) eigenstate.

Syndrome

Suppose we measure all $n^2-1$ stabilizers of the planar distance-$n$ surface code on an $n\times n$ square block of data qubits. The outcome is a bit string $s\in\{0,1\}^{n^2-1}$ called the syndrome. For convenience, assume that $n=2k+1$, so that $n^2-1=4k(k+1)$. We can then write the syndrome as \begin{align} s=s_1s_2\ldots s_{n^2-1}=x_1x_2\ldots x_{2k(k+1)}z_1z_2\ldots z_{2k(k+1)}\tag1 \end{align} where we implicitly split the syndrome into the $X$ and $Z$ sectors.

Syndrome decomposition of physical Hilbert space

The post-measurement state is the simultaneous eigenvector of the group generated by \begin{align} (-1)^{x_j}X_j\quad(-1)^{z_j}Z_j\tag2 \end{align} where $X_j$ and $Z_j$ denote the well-known stabilizer generators of the planar surface code with Pauli type $X$ and $Z$ respectively. Thus, we have $2^{n^2-1}$ different possible stabilizer groups $G_s$ indexed by all possible syndrome bit strings. Each $G_s$ stabilizes a two-dimensional subspace $\mathcal{H}_s$ of the full Hilbert space $\mathcal{H}$ of $n^2$ data qubits. Moreover, \begin{align} \mathcal{H}=\bigoplus_{s\in\{0,1\}^{n^2-1}}\mathcal{H}_s.\tag3 \end{align}

Evolution of the stabilizer group

Typically, we think of a logical state as an element of the $\mathcal{H}_0$, but we can choose$^2$ as our stabilizer group any $G_s$ and hence as our logical subspace any $\mathcal{H}_s$. In fact, we can consider the action of errors on the quantum state as shuttling the logical state from one $\mathcal{H}_s$ to another$^3$. We could of course apply active corrections to restore $s=0$, but there really is no reason$^4$ to do this. Instead, we let the classical control system keep track of $s$ and take it into account when necessary. Thus, we can think of the stabilizer group as evolving \begin{align} G_{s_1}\to G_{s_2}\to \ldots\tag4 \end{align} throughout the computation due to effects of noise.

Returning now to the question of initialization, when the initial syndrome measurement fails to yield the zero bit string the evolution in $(4)$ begins with $s_1\ne 0$. Once again, it doesn't matter what $s_1$ is as long as we keep track of it in the control system. Even if $s_1=0$ initially, the noise will quickly turn it into a non-zero bit string anyway.

The bottom line is: for the purposes of error correction, the subspaces $\mathcal{H}_s$ are equivalent. As long as we know in which $\mathcal{H}_s$ our quantum information resides, there is no reason to insist it reside in $\mathcal{H}_0$ specifically.

Appendix: Note on decoding

Finally, note that in decoding we are typically interested in the difference between $s_k$ and $s_{k-1}$, i.e. we subtract them modulo two. Therefore, in the absence of noise, where $s_k=s_1$ for all syndrome measurement rounds $k$, the fact that the initial $s_1$ is non-zero does not matter at all. In this case, the difference $s_k-s_{k-1}=0$ anyway.

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$^1$ Assuming enough error correction rounds.

$^2$ The subspace $\mathcal{H}_0$ does stand out from among $\mathcal{H}_s$ when we describe the surface code as the degenerate ground state of a certain Hamiltonian. However, we can define a slightly modified Hamiltonian that recovers such a description for any other $\mathcal{H}_s$.

$^3$ And occasionally changing the logical state in the subspace, but that is for a decoder to detect.

$^4$ In fact, there are reasons not to do this, since active error correction leads to a longer error correction round with a more complicated quantum circuit that presents more opportunities for errors.