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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

This is given as Exercise 8.23, Page 381, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang

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?

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  • $\begingroup$ 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. $\endgroup$
    – Jintao Yu
    Aug 6, 2023 at 3:36

2 Answers 2

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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:

$$ (A\otimes B)(C\otimes D)=AC\otimes DB.\tag{3}$$

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I don't think the maths reasoning is really rigorous here (neither it is from the physics point of view), even if you get the right answer \begin{alignat*}{1} \varepsilon_{AD}\otimes \varepsilon_{AD}(\rho) &=\gamma|0\rangle\langle 0|+(1-\gamma)\rho \end{alignat*}

For instance I don't think it even make sense to write things like

$(E_1\otimes E_1)(\rho\otimes\sigma)(E_1\otimes E_1)^\dagger$

I see what you mean from the previous lines though.

More fundamentally, you first establish a formula to compute $\varepsilon_{AD}\otimes \varepsilon_{AD}$ on a product state $\rho\otimes\sigma$;can you explain why it is right to apply it to the entangled state $|\psi\rangle \langle\psi|$ ?

Besides I agree with Jintao Yu comment, the Kraus operators are not correct; we have $E_1^{dr}|\psi\rangle\langle\psi|E_1^{dr\dagger}=\gamma(1+a^*b+b^*a)|00\rangle\langle00|$

Please check out my solution here https://github.com/pptacher/quantum/blob/master/ch8_quantum_noise_quantum_operations/ch8_quantum_noise_quantum_operations.pdf

To answer the question "Why is dual-rail encoding called an error-detecting code for amplitude damping?":

The output state is a mixed state:

  • with probability $\gamma$, the state is projected to $|00\rangle$, orthogonal to $|\psi\rangle$.
  • with probability $1-\gamma$, state is unchanged.

Since $|00\rangle$ is orthogonal to $|\psi\rangle$, you can detect amplitude dampling errors with measurement operators: \begin{alignat*}{1} M_0 & = & |00\rangle\langle 00|\qquad\text{orthogonal projector on }\mathrm{span}\{|00\rangle\} \\ M_1 & = & |01\rangle\langle 01|+|10\rangle\langle 10|+|11\rangle\langle 11|\qquad\text{orthogonal projector on }\mathrm{span}\{|01\rangle,|10\rangle,|11\rangle\} \\ \end{alignat*}

  • If the state decayed to $|00\rangle$, then with probability $1$ the result of the mesurement will be $|00\rangle$.
  • Otherwise, with probability $1$ the result of the measurement will be the original $|\psi\rangle$.
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