# Shor's quantum error correction code with unknown basis

$\newcommand{\ket}[1]{\lvert #1 \rangle}$I've met a problem in quantum secret sharing which involves the use of a quantum error-correction code. (let's make it simple to be the 9-qubit Shor code)

In the Shor code, Alice encodes $\ket{0}$ as $(\ket{000}+\ket{111})^3$, and $\ket{1}$ as $(\ket{000}-\ket{111})^3$, so the code can fight against 1-qubit flip or phase error, and recover the original qubit.

However, what I'm wondering is:

1. What becomes of the 9 qubits after measurement? Do they become the original 1 logical qubit? Or do they stay as 9 entangled qubits?

2. What would happen if Bob, the one who decodes the code, actually has a slightly different orthogonal basis than the original one due to imperfect knowledge or experimental error, e.g. Bob is measuring with $\ket{0'}=\cos(x)\ket{0}+\sin(x)\ket{1}$ and $\ket{1'}=-\sin(x)\ket{0}+\cos(x)\ket{1}$ but $x$ is small? (Or, let's say he is really unlucky and mistook Alice's $\ket{0}$ and $\ket{1}$ basis entirely, so he uses $\ket{+}$ and $\ket{-}$ basis, then Bell basis for him would become $\ket{+++} + \ket{---}$

When $x$ is small, would he still be able to decode out the original qubit using the "wrong" Bell basis? And would there be some way to calculate the error probability?

• This might also be on-topic at Theoretical Computer Science. (Don't crosspost, but consider flagging for migration if you do not receive the answers you seek here) – ACuriousMind Apr 22 '15 at 15:17
• @ACuriousMind Thank you very much for the comment and also your kind editing! If indeed the question doesn't receive answers after some time, may I ask how can I flag for migration? Should I copy the question and mark this question "migrated", or would there be administrators manually switching the questions if they see a request? Sorry for such basic questions about the working of the forum...but I'm really inexperienced about Stack Exchange, and just starting to appreciate its power :) (I actually just saw Professor Shor himself answering questions in the theoretical CS forum!) – Mike Wong Apr 22 '15 at 15:28
• I think we should wait a bit before migrating though. I think the answer to (1) is that they become 1 logical qubit, and with (2) is that we would have some error in the final qubit every now and then. – Manishearth Apr 24 '15 at 2:48
• The answer to (1) depends on the details of what you're calling "measurement". You can correct it in place, so they stay 9 entangled qubits, or you can decode it (with error correction) to the original logical qubit. In both cases, the details of how you would do it using one- and two-qubit operations are quite a bit more complicated than a usual physics measurement operator, although conceptually you can think of it as a standard projection operator on the $2^9$-dimensional Hilbert space. – Peter Shor May 5 '15 at 12:56
• For your second question, you really need to do a calculation to get the exact results. This shouldn't be too cumbersome, but I'm not going to take the time. But qualitatively, if Bob has a small enough mistake in the basis, you can think of this as a small error in each qubit, so Bob will get the logical qubit with a much smaller error. If Bob uses the complementarity basis, my guess is that he gets a completely random qubit, although you would need to do calculation to be completely sure. – Peter Shor May 5 '15 at 13:21

## Q1 : Fate of the 9 qubits

There is no (or 2) answer to your question, since it is implementation dependent :

1. If your implementation of the code is ideal, and you have a way to directly measure the syndromes, the 9 qubits are projected on the global 9 entangled qubits state. It is the case, for example, if the code is used in a fault-tolerant implementation of quantum computing.
2. If the implementation is more realistic, the measurement of the 8 syndromes is performed in 2 steps : 1st you apply a global unitary (through a quantum circuit) on the 9 qubits, then you measure 8 qubits, destroying them in the problem. You have then a single qubit, to which you can apply a unitary(depending on the measurement result) to get the single original qubit.

## Q2: Effect of misalignment

The misalignment behaves like an error, whether $x$ is small or $π/4$ (as in your $\lvert±\rangle$ example). What you describe correspond to applying the operator $\cos x I +\sin x XZ$ to each qubit. Measuring the syndrome projects the state to one with a given number of errors, but as long as there is less than 1 X error and 1 Z error, the code is enough to correct it.

This condition happens with probability $$(\cos^2 x)^9 + 9 (\cos^2 x)^8\sin^2 x=\cos^{16}x(1+8\sin^2x)\simeq 1-64x^4$$ the last approximation being valid for small $x$.

If the rotation is by $\pi/4$, the error rate is 50% and your code will not be able to correct it.