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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?

Ian Gershon Teixeira
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1 Answers1

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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.

Danylo Y
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