Fake Stabilizer States

Question

Consider an $ n $ qubit stabilizer state $ |\psi> $ with stabilizer group $ S $. Then the projector onto this one dimensional stabilizer subspace is given by $$ \sum_{g \in S} g $$ $ |\psi> $ can always be obtained by applying the projector to some computational basis ket $ |b> $ ($ b $ is some length $ n $ bit string). $$ |\psi>= \sum_{g \in S} g |b> $$ Every $ g \in S $ can be written $ g=\alpha_g g_Xg_Z $ where $ g_X $ is a tensor products of $ X $s and $ I $s and $ g_Z $ is a tensor product of $ Z $s and $ I $s and $ \alpha_g $ has one of the four values $$ \alpha_g= \pm 1, \pm i $$ $$ \sum_{g \in S} g | b> =\sum_{g \in S} \alpha_g g_X g_Z | b >=\sum_{g \in S} (-1)^{g_Z\cdot b} \alpha_g g_X | b>= \sum_{g \in S} (-1)^{g_Z\cdot b} \alpha_g | g_X +b> $$ where $ |g_X> $ is the computational basis ket for the bit string corresponding to $ g_X $ (each $ I $ is a $ 0 $ each $ X $ is a $ 1 $). The $ g_X $ that show up this way are exactly the span of the $ g_X $ for the stabilizer generators. So the bit strings in the $ |g_X> $ will form a binary vector space. And thus the $ |g_X+b> $ will be exactly an affine translate of a binary vector space by the fixed vector $ b $ (In particular, the size of the support must be $ 2^{dim(C)} $ since that is the size of a binary vector space $ C $, so this is a proof the the fact of part ii of Thm 9 Corollary 2 discussed in Stabilizer codes and 1,-1 coefficients that the support is always a power of $ 2 $ )

From this we can conclude that every stabilizer state is a superposition over the $ 2^{dim(C)} $ computational basis kets of some affine binary vector space $ C $ with all coefficients taking one of the values $ \pm1, \pm i $.

Indeed this question What are nontrivial examples of stabilizer codes whose codewords have some $\pm i$ coefficients? says that every stabilizer state is equivalent (by Pauli operators) to a stabilizer state with just $ \pm 1 $ coefficients (no $ \pm i $ coefficients)

This leads me to ask: If $ |\psi> $ is a uniform superposition over a an affine binary vector space $ C $ with all coefficients $ \pm 1 $ then must it be the case that $ | \psi> $ is a stabilizer state?

Call a state which is of that form but is not a stabilizer state a "fake stabilizer state"

Question: Do fake stabilizer states exist?

Note: Every single qubit fake stabilizer state is an actual stabilizer state. Up to global scalar the six possible single qubit fake stabilizer states (including the ones with $ i $ here for the sake of completeness) are $$ |0>\;,\;|1>\;,\;|0>+|1>\;,\;|0>-|1>\;,\;|0>+i|1>\;,\;|0>-i|1> $$ These are stabilizer states for
$$ Z,-Z,X,-X,Y,-Y $$ respectively (where here we use standard convention $ Y=iXZ $)

Edit: I looked into this a bit more and it seems that an $ n $ qubit state is a stabilizer state if and only if it is of the form $$ \frac{1}{\sqrt{2^k}}\sum_{u \in \mathbb{F}_2^k} i^{\ell(y)} (-1)^{q(y)} |y=Ru+t \rangle $$ for some vector $ t \in \mathbb{F}_2^n $, $ n \times k $ binary matrix $ R $, linear function $ \ell $ and quadratic function $ q $. This result is originally theorem 5 of

https://arxiv.org/pdf/quant-ph/0304125.pdf

but is given in a slightly more digestible form in the appendix of

https://arxiv.org/pdf/0811.0898.pdf

So the essence of the counterexample given in the answer is that the distribution of $ +1,-1 $ coefficients does not correspond to any quadratic function $ \mathbb{F}_2^3 \to \mathbb{F}_2 $.

Answer 1

The answer is yes, they do exist.

This answer is based on the assumed validity of https://arxiv.org/abs/1711.07848 that has an exhaustive list of 3 qubit stabilizer states in the appendix.

Here are two 3 qubit examples not contained in that table: 1--1-111 and 111-1--1 (using the shorthand notation in the paper, 1 means +1 coefficient for the given basis state, - means -1). Expressed as a sum:

$$ \frac{1}{2 \sqrt{2}}( \sum_{x\notin S} |x\rangle - \sum_{x\in S} |x\rangle) $$ , where choices for $S$ corresponding to the two states above are $\{x \in \mathbb{F}_2^3 | wt(x) = 1\}$ and $\{x \in \mathbb{F}_2^3 | wt(x) = 2\}$.

---

Credit goes to my colleague, Andrew Nemec, who should be on SE but didn't feel like creating an account yet.