I've been reading this blog post about surface code and it says
There are four non-equivalent types of loops: the trivial ones (stabilizers), the horizontal ones ($X_1$ operator), the vertical ones ($X_2$ operator), and those two at the same time ($X_1 X_2$ operator). Therefore, the surface code encodes $k=2$ logical qubits.
I was wondering about the reasoning behind this conclusion.
In my following reasoning I will separate the why and the what.
For the question of why there are two logical qubits, you do not have to worry about loops. It's simply number of physical qubits minus number of independent stabilizer generators.
For the question of what the logical qubits are (or rather their corresponding logical X- and Z-operators), the loops are essential. The loops are the logical X- and Z-operators.
The number of logical qubits $l$ in any stabilizer code is the number of physical qubits $n$ minus the number of independent stabilizer generators $k$ of the code, so: $l = n - k$
Having a $L \times L$ grid (as in the blog post) means that you have $n = 2L^2$ edges (which correspond to physical qubits), $L^2$ vertices (which correspond to Z-stabilizers) and $L^2$ plaquettes (which correspond to X-stabilizers)
The $L^2$ Z-stabilizers are not independent. You can choose one of them to be the product of all others. The same goes for the X-stabilizers. Hence, we have $L^2 - 1$ independent Z-stabilizers and $L^2 - 1$ independent X-stabilizers.
We gather that $n = L^2$ and $k = 2(L^2 - 1)$ and thus $l = n - k = 2$
In any stabilizer code, the logical qubits can be identified via pairs of logical X- and Z-operators. And they must meet these conditions:
The non-contractable loops that the author lays out in the section Loops on the surface code are the logical X- and Z-operators.
You can verify that they meet the above described conditions.
Remember: identifying the qubits via their logical X and Z is a key concept in any stabilizer code.
