1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.