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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?

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  • $\begingroup$ 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

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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.

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  • $\begingroup$ 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! $\endgroup$
    – Ian Gershon Teixeira
    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 $. $\endgroup$
    – Ian Gershon Teixeira
    Aug 2, 2022 at 15:40
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    $\begingroup$ 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. $\endgroup$
    – Ian Gershon Teixeira
    Aug 2, 2022 at 16:36
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    $\begingroup$ 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. $\endgroup$
    – Ian Gershon Teixeira
    Oct 21, 2022 at 21:40

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