Is this a valid stabilizer state?

Question

Suppose I have a code where the codewords are such that there are an equal number of $0$s and $1$s. For example, if $n=6$, codewords are $000111$, $101010$, $110001$ and so on as long as there are three $0$s and three $1$s.

Is it possible to write stabilizers for the quantum version of this code? That is, what are the stabilizers of the code with codewords $\vert 000111\rangle$, $\vert 101010\rangle$, $\vert 110001\rangle$, and so on.

Or alternatively, is the question itself ill-posed and if so, why?

Answer 1

"Classical" stabilizer codes, meaning stabilizer codes where the code space is spanned by a bunch of standard basis states, are equivalent to classical linear codes. So the answer is no because the classical code described here is not a linear code: it does not include the all-zero string $000000,$ for instance. That is to say, this is not a stabilizer code.

Answer 2

If you wanna make a CSS code with such constraints, then you must be able to construct such a code from two valid codes $C_1,C_2\subseteq\mathbb{F}_2^n$. In the case of a length-$n$ code, your constraint is a subset of a family called $\frac{n}{2}-$divisible codes. If we pick two codewords $c,d\in C_1$, then $c+d\in C_1$. Let's see why your example would not work: let $c = 000111$ and $d = 010101$ be two 3-divisible codewords, then $c+d = 010010$ is not 3-divisible.

But your code is in general not possible since any valid code contains the all-zero codeword 0. If you say that any non-trivial codeword is $\frac{n}{2}$-divisible, you can constuct such a code if $n=4k$ for some $k\in\mathbb{N}$, since this will preserve the parity of codewords under addition. (If a code contains one single codeword, then yes $n=2k$.)

Example: let $n=8$ and $C_2 \subseteq C_1 \subseteq \mathbb{F}_2^8$ with generators $G_2 = \langle 00001111 \rangle$ and $G_2 = G_1 + \langle 00111100\rangle$.