# $\langle Z \rangle_L$ in the Distance Two Surface Code

In an experimental realization of the distance 2 surface code, the codewords are: $$|0\rangle_L = \frac{1}{\sqrt{2}} (|0000\rangle + |1111\rangle), |1\rangle_L = \frac{1}{\sqrt{2}} (|0101\rangle + |1010\rangle),$$ and $$Z_L = Z_1 Z_2$$ (or equivalently $$Z_L = Z_3Z_4$$). In the paper, Fig 5 (a) is a plot of the $$Z_L$$ expectation value for $$|0\rangle_L$$ and $$|1\rangle_L$$ over repeated error correction cycles, which corresponds to the physical $$|1\rangle$$ probability, i.e., the plot shows physical $$|1\rangle$$ probability goes to zero, $$\langle Z_L \rangle$$ goes to zero for both $$|0\rangle_L$$ and $$|1\rangle_L$$. I'm struggling to see why this is the case. If the physical $$|1\rangle$$ probability is 0, will $$\langle Z_L \rangle$$ not just be 1?

• What do you mean by "physical $|1\rangle$ probability"? Commented Jun 17 at 5:19
• This is just taken verbatim from the (right) y-axis of the plot. I interpreted it to mean the probability the physical qubits in the logical qubit haven't decayed from the $|1\rangle$ state, which may be incorrect and where my confusion lies... Commented Jun 17 at 10:07