Consider an $[[n, k, d]]$ quantum CSS code with $C^{\perp} \subset C$. If I consider the state, \begin{equation} \frac{1}{\sqrt{|C^{\perp}|}} \sum_{x \in C^{\perp}} |x\rangle; \end{equation} is this a logical state of the CSS code? For instance, is it the logical $|+\rangle^{\otimes k}$ state?
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In short, the first state you provided is a logical codeword of the CSS code, but the $|+\rangle^{\otimes k}$ is not because it is a logical superposition over $C$ and not $C^\perp$.
The state you provided follows the general form of a codeword: $$\frac{1}{\sqrt{|C^\perp|}}{\sum_{x\in C^\perp}{|u+x\rangle}}$$ (where $u \in C$) where in this case $u=0$ represents the logical zero state ($|0\rangle_L$). The logical $|+\rangle^{\otimes k}_L$ is a equal superpostion of all logical basis states: $$|+\rangle^{\otimes k} = \frac{1}{\sqrt{2^k}}\sum_{u \in C/C^\perp}{|\bar{u}\rangle}$$ and plugging in for $|\bar{u}\rangle$ we have: $$|+\rangle^{\otimes k}_L = \frac{1}{\sqrt{2^k|C^\perp|}}\sum_{u \in C/C^\perp}\sum_{x\in C^\perp}{|u+x\rangle} = \frac{1}{\sqrt{|C|}}{\sum_{v\in C}{|v\rangle}}$$ because $|C/C^\perp| = 2^k$ and $C= \sqcup_{u \in C/C^\perp}{(u + C^\perp)}$. Which means it is not a logical codeword of the code.
