I have read some articles about Shor's code (e.g. this one). It is said that Shor's code can correct a single-qubit error. What about two qubit errors? Three qubit errors? It confused me a lot...

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    $\begingroup$ +1 and welcome to the community! We hope to see much more of you here in the future! As for the close vote and downvote, they were by someone else, but I suspect that you might your questions would get a more positive response if you used complete sentences, rather than sentence fragments like "Three qubit errors?", or the problem is that you talk about "single errors" and "single-qubit errors" like they're the same thing. A single 3-qubit error is still a single error. Anyway, hopefully the answer by Adam was useful! $\endgroup$ Jan 25 '21 at 23:30
  • $\begingroup$ Thank you so much for your useful suggestions! $\endgroup$ Jan 29 '21 at 18:01

We can combine weaker codes to obtain stronger codes using concatenation (see this paper or chapter X B in this paper).

Suppose we have a quantum error correcting code that encodes a logical qubit in $n$ physical qubits in a way that enables us to correct any $t$ errors. We can use this code to encode $n$ logical qubits in $n^2$ physical qubits and then we can add a second level of encoding to encode a second-level logical qubit into the $n$ first-level logical qubits.

How many errors can the resulting two-level code correct? For an error to occur at the second level, at least $t+1$ first-level encoded qubits must suffer an error. For an error to occur in any first level encoded qubit, at least $t+1$ physical qubits must suffer an error. Thus, for an error to occur at the second level, there must be at least $(t+1)^2$ physical errors. Consequently, the two-level concatenated code can correct any

$$ (t+1)^2 - 1 = t^2 + 2t $$

physical errors. For example, the two level Shor's code can correct any three physical errors.

Concatenation can be continued to any number of levels and if the physical error rate is low enough it allows us to bring the logical error rate below any desired target value. This last result is known as the threshold theorem.

  • $\begingroup$ Thank you for your reply. So if we have two qubits errors, we need to combine two 9-qubits shor's code as a whole system? $\endgroup$ Jan 26 '21 at 5:38
  • $\begingroup$ Yes, if you'd like to be able to recover from errors on any two physical qubits, then one way is to use two-level concatenated Shor's code. Note that this is excessive in the sense that it can actually recover from errors on any three physical qubits. $\endgroup$ Jan 26 '21 at 5:54
  • $\begingroup$ Thanks for your reply! So if I just have a 9 qubits Shor's code (only), I cannot correct 2 qubits errors. But if one is phase error and the other one is flip error, the 9 qubits shor's code can complete. Isn't it? $\endgroup$ Jan 26 '21 at 7:34
  • $\begingroup$ Yes, if one error is a bit-flip and the other is a phase-flip then we can correct them using the 9-qubit Shor's code. However, if the first error is a bit-flip on the first qubit and the second error is a bit-flip on the second qubit, then we can't correct them. The point is that with 9 qubit Shor's code we cannot correct arbitrary two-qubit errors - we have to get lucky. However, we have a guarantee that we can correct arbitrary $t=1$ qubit errors. Similarly, with two levels of concatenation we have a guarantee that we can correct arbitrary $t=3$ qubit errors - no need for luck :-) $\endgroup$ Jan 26 '21 at 16:20
  • $\begingroup$ Note also that sometimes you can get really lucky and multiple errors make no change to the quantum state at all. For example, a phase-flip on the first and a phase-flip on the second qubit together do nothing at all in 9-qubit Shor's code. $\endgroup$ Jan 26 '21 at 16:27

Shor's code is specifically designed to correct all possible single-qubit errors. There are some two-qubit errors that it can correct for (e.g. Pauli $X$ on one qubit and Pauli $Z$ on another). You will also be able to find specific combinations of multiple-qubit errors for which it can correct, but those will be the exception rather than the rule. Instead, if you want to protect against two-qubit errors, you employ different strategies - either code concatenation, or just find a different error correcting code with a larger distance.

  • $\begingroup$ Thank you so much! That means if qubit 1 has an X error and qubit 2 has a Z error, the shor's code can correct it. Isn't it? $\endgroup$ Jan 29 '21 at 17:58
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    $\begingroup$ Yes. Any X error together with any Z error (including X and Z errors on the same qubit, i.e. a Y error) can be corrected in the 9-qubit Shor's code. $\endgroup$ Jan 29 '21 at 21:45

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