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Consider a rotated surface code. Let the surface code have $Z$ stabilizers along the top and bottom boundary and $X$ stabilizers along the left and right boundary.

If I initialize all the physical qubits in $\vert 0\rangle$ and then turn on the stabilizers, I have a logical state - let's call this state $\vert 0\rangle_L$. This is just an arbitrary choice, since I could have called it $\vert 1\rangle_L$ or $\vert +\rangle_L$ or anything else I like.

Now, is it still up to me to decide what is the logical $X$ operator of this code? We choose usually a row of physical $X$ operators to be defined as $X_L$. Given that I have called the surface code's logical state $\vert 0\rangle_L$, do I still have a second freedom of choice in what I define to be $X_L$?

Obviously, once I define $X_L$, the operator that anticommutes with it and commutes with all stabilizers is $Z_L$.

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Almost...

Yes, the stabilizers of a code define a space. You are nominally free to pick any pair of states within that to be logical 0 and 1, and hence define logical Z.

You then have freedom to define the logical X to be anything that commutes with the stabilizers and anticommutes with logical Z. So the choice is not entirely arbitrary, but it's partially constrained. For a single qubit, you can visualise the Bloch sphere. I can pick any axis through the sphere to define Z. The equator around that axis gives you all the possible logical Xs.

So, why do I say almost? You're not quite finished. Once you've selected logical X and logical Z, there are two operators that anticommute with both. You have a further choice of which you want to be +Y and which -Y. In particular, that means that your statement

Obviously, once I define $X_L$, the operator that anticommutes with it and commutes with all stabilizers is $Z_L$.

is not so obvious. There are many operators that anticommute with $X_L$ and commute with all the stabilizers (all linear combinations $\cos\theta Z_L+\sin\theta Y_L$).

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  • $\begingroup$ Thanks! I had two questions: About the first paragraph, can I really pick any two states (that are not identical and are $+1$ eigenstates of all stabilizers) to be $\vert 0\rangle_L$ and $\vert 1\rangle_L$ respectively? Or are there more restrictions? I was doing it the other way i.e. first pick a specific stabilizer state to be logical $\vert 0\rangle$, then define logical $X$ (which is an arbitrary operator that commutes with all stabilizers but not a product of stabilizers itself) and go from there to determine logical $\vert 1\rangle$. Is my procedure also correct? $\endgroup$
    – user29393
    Commented Jul 9 at 13:47
  • $\begingroup$ Ah yes, your procedure also works so long as you ensure that logical X maps your logical 0 to an orthogonal state. But the Z logical is already implicitly defined by the selection of the state. $\endgroup$
    – DaftWullie
    Commented Jul 9 at 13:59
  • $\begingroup$ There are no further restrictions that are mathematically imposed on you for selecting the logical states, although many authors that want to develop further results will typically assume that logical x and z are tensor products of paulis. $\endgroup$
    – DaftWullie
    Commented Jul 9 at 14:01
  • $\begingroup$ Thank you for the clarifications $\endgroup$
    – user29393
    Commented Jul 9 at 17:28

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