# Why is dual-rail encoding called an error-detecting code for amplitude damping?

Exercise 8.23 : Suppose that a single qubit state is represented by using two qubits, as $$|\psi\rangle=a|01\rangle+b|10\rangle$$. Show that $$\mathcal{E}_{AD}\otimes\mathcal{E}_{AD}$$ applied to this state gives a process which can be described by the operation elements $$E_0^{dr}=\sqrt{1-\gamma}I$$ and $$E_1^{dr}=\sqrt{\gamma}[|00\rangle\langle01|+|00\rangle\langle10|]$$, i.e., either nothing $$(E_0^{dr})$$ happens to the qubit, or the qubit is transformed $$(E_1^{dr})$$ into the state $$|00\rangle$$, which is orthogonal to $$|\psi\rangle$$. This is a simple error-detection code and is also the basis for the robustness of the ‘dual-rail’ qubit discussed in Section 7.4

Here $$\mathcal{E}_{AD}$$ represents amplitude damping such that $$\mathcal{E}_{AD}(\rho)=E_0\rho E_0^\dagger+E_1\rho E_1^\dagger$$ where $$E_0=\begin{bmatrix}1&0\\0&\sqrt{1-\gamma}\end{bmatrix}=|0\rangle\langle 0|+\sqrt{1-\gamma}|1\rangle\langle1|$$ and $$E_1=\begin{bmatrix}0&\sqrt{\gamma}\\0&0\end{bmatrix}=\sqrt{\gamma}|0\rangle\langle1|$$, And $$\mathcal{E}(\begin{bmatrix}a&b\\b^*&c\end{bmatrix})=\begin{bmatrix}1-(1-\gamma)(1-a)&b\sqrt{1-\gamma}\\b^*\sqrt{1-\gamma}&c({1-\gamma})\end{bmatrix}$$.

My Attempt

\begin{align} &(\mathcal{E}_{AD}\otimes \mathcal{E}_{AD})(\rho\otimes\sigma)=\mathcal{E}_{AD}(\rho)\otimes \mathcal{E}_{AD}(\sigma)=(E_0\rho E_0^\dagger+E_1\rho E_1^\dagger)\otimes(E_0\sigma E_0^\dagger+E_1\sigma E_1^\dagger)\\ &=E_0\rho E_0^\dagger\otimes E_0\sigma E_0^\dagger+E_0\rho E_0^\dagger\otimes E_1\sigma E_1^\dagger+E_1\rho E_1^\dagger\otimes E_0\sigma E_0^\dagger+E_1\rho E_1^\dagger\otimes E_1\sigma E_1^\dagger\\ &=(E_0\otimes E_0)(\rho\otimes\sigma)(E_0\otimes E_0)^\dagger+(E_0\otimes E_1)(\rho\otimes\sigma)(E_0\otimes E_1)^\dagger+(E_1\otimes E_0)(\rho\otimes\sigma)(E_1\otimes E_0)^\dagger+(E_1\otimes E_1)(\rho\otimes\sigma)(E_1\otimes E_1)^\dagger \end{align} Let $$|0\rangle_D=|01\rangle$$ and $$|1\rangle_D=|10\rangle$$ $$(E_0\otimes E_0)|i\rangle_D=\sqrt{1-\gamma}|i\rangle_D\\ (E_0\otimes E_1)|01\rangle=\sqrt{\gamma}|00\rangle\quad\&\quad(E_1\otimes E_0)|10\rangle=\sqrt{\gamma}|00\rangle\\ (E_0\otimes E_1)|10\rangle=0\quad\&\quad(E_1\otimes E_0)|01\rangle=0\\ (E_1\otimes E_1)|i\rangle_D=0$$

\begin{align} &|\psi\rangle\langle\psi|=(a|01\rangle+b|10\rangle)(a^*\langle01|+b^*\langle 10|)=|a|^2|01\rangle\langle01|+ab^*|01\rangle\langle10|+ba^*|10\rangle\langle01|+|b|^2|10\rangle\langle10|\\\\ \end{align} \begin{align} (\mathcal{E}_{AD}\otimes \mathcal{E}_{AD})(|\psi\rangle\langle\psi|)&=(1-\gamma)(|\psi\rangle\langle\psi|)+\gamma(|a|^2+|b|^2)(|00\rangle\langle 00|)\\ &=(1-\gamma)(|\psi\rangle\langle\psi|)+\gamma|00\rangle\langle 00| \tag{1}\label{eq1} \end{align} \begin{align} E_0^{dr}(|\psi\rangle\langle\psi|)E_0^{dr}{}^\dagger+E_1^{dr}(|\psi\rangle\langle\psi|)E_1^{dr}{}^\dagger=(1-\gamma)(|\psi\rangle\langle\psi|)+\gamma|00\rangle\langle 00| \tag{2}\label{eq2} \end{align}

I think if my math is correct this completes the proof of the first statement.

What is the logic behind calling this dual-rail encoding, an error-detecting code for the amplitude-damping channel?

• I am asking you a question instead of answering yours. I have a different result from your Equation 2, so I failed to prove the first statement. My calculation is $$\because E_1^{dr}(a|01\rangle+b|10\rangle)=\sqrt{r}(a+b)|00\rangle$$ $$\therefore E_1^{dr}|\psi\rangle\langle\psi|E_1^{dr\dagger}=r|a+b|^2|00\rangle\langle00|=r(1+a^*b+b^*a)|00\rangle\langle00|$$ Did I do wrong somewhere? What are your detailed steps? Sorry that I do not have enough reputations to post this in a comment. Aug 6 at 3:36

Equation 1 misses the Kraus operator on the right. The results are ok. You may obtain the same results by considering E acting on a single qubit: $$E_0|0> =|0>,\tag{1}$$ $$E_0|1> =\sqrt{1-\gamma}|1>, E_1|0> =|0>, E_1|1> =\sqrt{\gamma}|0>.`\tag{2}$$ And using the mixed product:
$$(A\otimes B)(C\otimes D)=AC\otimes DB.\tag{3}$$