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?

  • $\begingroup$ What do you mean by "physical $|1\rangle$ probability"? $\endgroup$
    – Yunzhe
    Commented Jun 17 at 5:19
  • $\begingroup$ 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... $\endgroup$ Commented Jun 17 at 10:07


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