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I was reading the CSS ( Steane Code) from the Nielsen & Chuang book. It asked in Ex. 10.27 to prove that: suppose $C_1$ and $C_2$ are $[n,k_1]$ and $[n,k_2]$classical linear codes such that $C_2\subset C_1$ and $C_1$ and $C_2^\perp$ both correct $t$ errors. Codes defined by $$|x+C_2\rangle\equiv \dfrac{1}{\sqrt{|C_2|}}\sum_{y\in C_2}(-1)^{u.y}|x+y+v\rangle $$ and parametrized by $u$ and $v$ are equivalent to $\mathrm{CSS}(C_1, C_2)$ in the sense that they have the same error-correcting properties.

My attempt for this was let the corrupted state be for the bit flip case: $$\dfrac{1}{\sqrt{|C_2|}}\sum_{y\in C_2}(-1)^{u.y}|x+y+v+e_1\rangle$$ now proceeding on the lines of the code $\mathrm{CSS}(C_1,C_2)$, I apply the Parity matrix $H_1$ for $C_1$, on the ancilla to obtain $$ \dfrac{1}{\sqrt{|C_2|}}\sum_{y\in C_2}(-1)^{u.y}|x+y+v+e_1\rangle|H_1e_1\rangle$$ where $H_1(x+y+v)=0$. so i get the position of the flipped qubit by inspecting the position where $1$ occurs.

Now for the phase flip case here is my try $$\dfrac{1}{\sqrt{|C_2|}}\sum_{y\in C_2}(-1)^{u.y}(-1)^{(x+y+v).e_2}|x+y+v\rangle $$ now I apply the Hadamard gate on the qubit to obtain $$\dfrac{1}{\sqrt{|C_2|2^n}}\sum_z\sum_{y\in C_2}(-1)^{u.y}(-1)^{(x+y+v).e_2}(-1)^{(x+y+v).z}|z\rangle $$$$= \dfrac{1}{\sqrt{|C_2|}2^n}\sum_z\sum_{y\in C_2}(-1)^{u.y}(-1)^{(x+y+v).(e_2+z)}|z\rangle$$ Now let $e_2+z=z'$, we get $$ \dfrac{1}{\sqrt{|C_2|2^n}}\sum_z\sum_{y\in C_2}(-1)^{u.y}(-1)^{(x+y+v).z'}|z'+e\rangle$$, proceeding from here the final step that I got was $$\dfrac{1}{\sqrt{2^n/|C_2|}}\sum_{z'+u\in C_2^{\perp}}(-1)^{(x+v)z'}|z'+e_2\rangle$$, Now is this correct, if so how do I proceed, and what should be the answer?

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I am not following all of the calculations in your post (for one thing because your first displayed equation does not mention $C_1$), but I know why the goal of the exercise is true. In fact it is true for any code at all, not just a CSS code. If $\mathcal{C} \subseteq \mathcal{H}_\text{qubit}^{\otimes n}$ is any code, then you get an equivalent code $\mathcal{C}'$ if you apply a separate unitary operator to each of the $n$ qubits. After all, the definition of the error properties does not depend on a choice of basis for each for the $n$ qubits; who is to say whether the code "is" $\mathcal{C}$ or $\mathcal{C}'$ before you have chosen qubit bases.

In the case at hand, translation by your vector $x$ is equivalent to applying a flip operator $X$ in each position $k$ with $x_k = 1$. So this is just applying independent unitaries (either $X$ or $I$) for each of the qubits. In fact you can check the error properties in a particularly explicit way, since $X$ is a Pauli operator. Applying $X$ to the code in some position amounts to replacing the $Y$ and $Z$ errors with $-Y$ and $-Z$ and keeping $X$ just itself; you can see this from whether they commute or anti-commute.

There is a nifty corollary that applies to all additive codes, not just CSS codes. Namely, an additive code is normally defined by demanding that a set of commuting parity check operators all report the value $1$. However, you get an equivalent code if you use any mutual eigenspace of the parity check operators. In fact, if $\mathcal{C} \subseteq \mathcal{H}_\text{qubit}^{\otimes n}$ is an additive code, then the outer Hilbert space $\mathcal{H}_\text{qubit}^{\otimes n}$ is partitioned into equivalent codes.

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We can proceed by applying the parity check matrix $H_2$ for $C_2^\perp$ on the ancilla, which would become $\vert H_2(z'+e_2)\rangle =\vert H_2(z'+u + e_2 -u)\rangle =\vert H_2(e_2-u)\rangle$. Measuring the ancilla gives us $H_2(e_2-u)$. As $u$ is known. we then know $H_2 e_2$, and in turn $e_2$. After correcting the now bit flip $e_2$, we have: $$\dfrac{1}{\sqrt{2^n/\vert C_2\vert}}\sum_{z'+u\in C_2^{\perp}}(-1)^{(x+v)z'}|z'\rangle \label{eq:1}\tag{$\star$}$$ We then re-apply Hadamard gates to each qubit of $\eqref{eq:1}$, which is equivalent to applying Hadamard gates to the state on the final line of the question $$\dfrac{1}{\sqrt{2^n/\vert C_2\vert}}\sum_{z'+u\in C_2^{\perp}}(-1)^{(x+v)z'}|z'+e_2\rangle$$ with $e_2=0$. Hadamard gates are self-inverse, so we get back $$\dfrac{1}{\sqrt{|C_2|}}\sum_{y\in C_2}(-1)^{u\cdot y}(-1)^{(x+y+v)\cdot 0}\vert x+y+v\rangle $$ which is the original encoded state: $$\dfrac{1}{\sqrt{|C_2|}}\sum_{y\in C_2}(-1)^{u\cdot y}\vert x+y+v\rangle $$

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