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I'm reading a book (Quantum Information Meets Quantum Matter) that introduced the bit flip superoperator $$\mathcal{E}_{BF}(\rho) = (1-p)\rho + pX \rho X.$$ For a three qubit state $$|\psi \rangle = \alpha |000\rangle + \beta |111 \rangle$$ the book says that the receiver sees the output state as $$\rho = \mathcal{E}_{BF}^{\otimes 3}(|\psi \rangle \langle\psi|) = (1-p)^3 |\psi \rangle \langle\psi| + (1-p)^2 p X_1 |\psi \rangle \langle\psi| X_1 + \cdot \cdot \cdot$$

The coefficients all make sense to me, but I don't quite understand how only $|\psi \rangle \langle\psi|$ ends up in each term. It seems to me that $$\mathcal{E}_{BF}^{\otimes 3}(|\psi \rangle \langle\psi|) = \mathcal{E}_{BF}(|\psi \rangle \langle\psi|) \otimes \mathcal{E}_{BF}(|\psi \rangle \langle\psi|) \otimes \mathcal{E}_{BF}(|\psi \rangle \langle\psi|)$$ So wouldn't you have factors like $|\psi \rangle \langle\psi| \otimes |\psi \rangle \langle\psi| \otimes |\psi \rangle \langle\psi|$ in the result? It seems like they are saying that $|\psi \rangle \langle\psi| \otimes |\psi \rangle \langle\psi| = |\psi \rangle \langle\psi|$, which I don't think is true.

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    $\begingroup$ A cubit is about 44 cm. You mean a qubit! $\endgroup$
    – march
    Jun 17 at 16:57
  • $\begingroup$ I think without the language from the book (or the book reference at least), it's hard to figure this out. It's possible that they mean the same thing by $\mathcal{E}_{BF}^{\otimes 3}$ and $\mathcal{E}_{BF}$, since, after all, the input state is already a three-qubit state. In the surrounding context, is there any mention of there for some reason being nine qubits? - because that's what would be necessary in order to interpret it the way that you have. I suspect (but obviously can't be sure) that they mean the same thing by $\mathcal{E}_{BF}^{\otimes 3}$ and $\mathcal{E}_{BF}$. $\endgroup$
    – march
    Jun 17 at 22:10
  • $\begingroup$ Then again, the result does have factors of $(1-p)^3$, etc. So maybe $|\psi\rangle \langle \psi|$ in the final expression is just a short-hand for the product of three copies of that state. Again, it's hard to know without more surrounding information from the book that you are reading. $\endgroup$
    – march
    Jun 17 at 22:12
  • $\begingroup$ Can you be more specific about the reference? There's a draft version of that book on the ArXiV, and if you give chapter and section (and maybe even page), it would be helpful for potential answerers to find the relevant section of the book. $\endgroup$
    – march
    Jun 17 at 22:18
  • $\begingroup$ It is in chapter 3. Pg 61 of the text (77 of the pdf) $\endgroup$
    – roshoka
    Jun 17 at 22:20

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The operator $\mathcal{E}_{BF}$ is by definition a single-qubit operator, and so we construct an operator that flips the bit (with noise) of each qubit in a three-qubit system by taking the tensor product of three copies of the operator, i.e., $\mathcal{E}_{BF}^{\otimes 3}$ is an operator that acts on a 3-qubit system. Thus, $$ \mathcal{E}_{BF}^{\otimes 3} = \left( (1-p)\hat{I}_1(\cdot)+\hat{X}_1(\cdot)\hat{X}_1 \right) \otimes \left( (1-p)\hat{I}_2(\cdot)+\hat{X}_2(\cdot)\hat{X}_2 \right) \otimes \left( (1-p)\hat{I}_3(\cdot)+\hat{X}_3(\cdot)\hat{X}_3 \right)\,, $$ where the $(\cdot)$ notation means to "put the argument of the function here", and $\hat{I}$ is an identify operator. The first factor in this product acts on the first element in the tensor product of three qubit states, i.e., on the $n_1$ part of $\lvert n_1 n_2 n_3\rangle\langle n_1n_2n_3\rvert$, the second factor acts on the second part, etc.

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  • $\begingroup$ The argument in each $(\cdot)$ would be $|n_1 n_2 n_3 \rangle \langle n_1 n_2 n_3 |$, right? Or is it the individual $|n_i \rangle \langle n_i |$'s? $\endgroup$
    – roshoka
    Jun 17 at 22:35
  • $\begingroup$ Yep, and the operators shown only on each qubit individually, in order to evaluate all those terms in Eq. 3.6 after plugging in the explicit state, use linearity to separate each piece into two terms (one for the 000 state and one for the 111) state, and then the operator acts on the corresponding piece. $\endgroup$
    – march
    Jun 17 at 22:38
  • $\begingroup$ okay that makes sense. I suspected it was a notation issue on my end. thanks $\endgroup$
    – roshoka
    Jun 17 at 22:44

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