3
$\begingroup$

A lot of well known codes (5 qubit code, 7 qubit Steane code, 9 qubit Shor code) have logical zero and local one which can be written as (a global scalar times) a linear combination of computational basis kets with only $ \pm 1 $ as coefficients.

Is it true that for any stabilizer code there always exists a choice of logical code words that can be written as (a global scalar times) a linear combination of computational basis kets where every coefficient is $ \pm 1, \pm i $?

Here is my reasoning for why this should be true :

Suppose that we have a stabilizer code with stabilizer $ S $. Then we can form a projector onto the codespace by $$ P=\sum_{g \in S} g $$ We can take the image of the computational basis under this projector $ P $. For any given $ g $ and any computational basis ket $ v $ then $$ gv=\pm v' $$ for some other computational basis ket $ v' $ (or possibly $ \pm i $ if $ g $ has some factors of $ Y $). Adding all this up then $ Pv $ must be some $ \mathbb{Z}[i] $ linear combination of the computational basis kets. Is there always a way to factor out a global scalar and just get $ \pm 1,\pm i $ relative coefficients?

$\endgroup$
1
  • $\begingroup$ I think this can even be generalized : the entire encoding matrix has entries from +1,-1,0 (and maybe an overall normalizing scalar). You can take $k$ rows of encoding matrix as basis for codespace. Maybe an even larger generalization : any matrix derived from a tableau will also have this form. I don't have a proof but I did notice the pattern. A possible path to a proof could be looking at the form of the clifford gates involved in building the encoder $\endgroup$
    – unknown
    Aug 1, 2022 at 16:42

1 Answer 1

5
$\begingroup$

Yes, any stabilizer codespace is a linear span of stabilizer states.

Any stabilizer state has a power of 2 non-zero amplitudes, which are $\pm 1, \pm i$ up to a common factor. See Eq. (2) in https://arxiv.org/abs/0811.0898

There are other interesting properties, for example, the number of imaginary amplitudes is either 0 or half the number of non-zero amplitudes, as claimed in corollary 2 in https://arxiv.org/abs/1711.07848. Beware, thought, part (iv) of that corollary seems incorrect.

$\endgroup$
6
  • $\begingroup$ Very elegant to observe that you can always pick stabilizer codewords to be stabilizer states stabilized by the $ k $ generators of the code stabilizer $ S $ and $ n-k $ of the $ \pm \overline{Z}_i $ logical Z operators from $ N(S)/S $ ( $ \pm $ depending on whether you want 0 o 1 for the ith logical qubit). Then corollary 2 of thm 9 (which is really appendix A of the other paper they cite arxiv.org/abs/0811.0898) completes the picture by listing facts about stabilizer states like their $ \pm 1, \pm i $ coefficients. Interesting answer and a great reference! $\endgroup$ Aug 2, 2022 at 15:15
  • $\begingroup$ Do you know of any code that uses coefficients other than $ \pm1 $ in an essential way? The class of stabilizer states where exactly half the nonzero entries are $ \pm i $ seems to indicate that such codes exist. But perhaps any code with some complex coefficients is just equivalent by a local unitary to a stabilizer code with all $ \pm 1 $ coefficients? Bottom line: I'd love to see any example code you know that uses $ \pm i $ coefficients. If not then I'll just ask another question about $ \pm i $ versus just $ \pm 1 $. $\endgroup$ Aug 2, 2022 at 15:40
  • 2
    $\begingroup$ Also am I missing something or is part (iv) of Thm 9 Corollary 2 just totally not true (even with footnote h)? They don't provide a proof or really any explanation so I can't try to follow their logic. But it seems like the stabilizer state for \begin{align*} &XZZXI \\ &IXZZX \\ &XIXZZ \\ &ZXIXZ \\ &ZZZZZ \end{align*} is a clear counterexample. This stabilizer state is the logical 0 for the 5 qubit code en.wikipedia.org/wiki/Five-qubit_error_correcting_code. It has 6 $1$s and 10 $-1$s. 6 and 10 are not powers of 2. You only use part (ii) so your answer is fine. Just an observation. $\endgroup$ Aug 2, 2022 at 16:36
  • 1
    $\begingroup$ Yeah, part (iv) seems incorrect, there are counterexamples in their own table 2 (if we ignore the footnote h). I've edited the answer. I don't know about specific examples you are looking for. Look at that table 2 at first. $\endgroup$
    – Danylo Y
    Aug 2, 2022 at 18:37
  • $\begingroup$ An error in Thm 9 of arxiv.org/pdf/1711.07848.pdf misplaces a factor of 2. The proper text is $ i^{(c^Ty+2y^TQy)} $. That means that the linear functional $ c^T $ controls the presence of $ i $ (which explains/ proves that part (iii) of corollary 2 is correct since the kernel of a linear functional is either everything or it is a codimension 1 subspace) while the quadratic part controls the sign $ i^{2y^TQy}=(-1)^{y^TQy} $. There's no reason for the kernel of a quadratic form to be a power of 2 (i.e. no reason for it to be a binary linear subspace), so that's why (iv) is nonsense. $\endgroup$ Oct 21, 2022 at 21:40

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.