# Question about Nielson & Chuang Problem 9.2

I am working on the following problem from the book "Quantum Computation and Quantum Information" by Nielsen and Chuang.

Problem 9.2: Let $$\mathcal{E}$$ be a trace-preserving quantum operation. Show that for each $$\rho$$ there is a set of operation elements $$\{E_i\}$$ for $$\mathcal{E}$$ such that $$F(ρ, \mathcal{E}) = |\operatorname{tr}(\rho E_1)|^2 .\tag{1}$$

Some background:

The $$F(\rho,\mathcal{E})$$ here is the entanglement fidelity. In addition, for quantum operation $$\mathcal{E}$$ such that $$\mathcal{E}(\rho)=\sum_kE_k\rho E_k^{\dagger}$$, we have $$F(\rho,\mathcal{E})=\sum_k|\operatorname{tr}(\rho E_k)|^2$$.

My attempt:

Given $$\rho$$ and a set of operation elements $$\{E_k\}_{k=1}^K$$ for a trace-preserving quantum operation $$\mathcal{E}$$, I'd like to find another set of operation elements $$\{F_j\}$$ such that

$$F(ρ, \mathcal{E}) =\sum_k|\operatorname{tr}(\rho E_k)|^2= |\operatorname{tr}(\rho F_1)|^2 \tag{2}.$$

In particular, I consider $$\{E_k\}$$ such that $$\operatorname{tr}(E_k^{\dagger}E_l)\propto\delta_{kl}$$. (As the entanglement fidelity is independent of the choice of $$\{E_k\}$$ and such $$\{E_k\}$$ can always be found).

I then tried with $$F_1=\sum_kE_k/\operatorname{tr}(\rho E_k)\cdot\sqrt{F(\rho,\mathcal{E})}/K$$, which gives me the desired property $$(2)$$. But then I find it hard to come up with the remaining $$F_j$$s so that it forms a set of operation elements for $$\mathcal{E}$$.

I have also checked various solutions to this book, but none solve this problem. Much appreciate any hints or suggestions.

I have a sketch of ideas, but haven't buttoned up all the details into a proof so use it as a set of hints.

Intuition: Kraus operators have a unitary freedom, thus you can get the next set of Kraus operators $$\{F_i\}$$ by applying a unitary transformation on the original Kraus operators $$\{E_i\}$$. I think you can always find a unitary that "rotates" the Kraus operators relative to a density operator $$\rho$$ so that only one of the Kraus operators has non-zero expectation value for $$\rho$$.

A bit more details:

1. If the channel is unitary (i.e. $$K=1$$), then we are done, as there is only a single Kraus operator
2. For $$K>1$$, we are looking for a unitary $$U$$, which generates the set of $$\{F_i\}$$ from $$\{E_i\}$$ based on:

$$F_i = \sum_j^K U_{ij} E_j$$

1. Interestingly, this means that the expectation value of $$\rho$$ for the Kraus operators also goes through a unitary transformation because of the linearity of the trace:

$$\operatorname{tr}(F_i \rho) = \sum_j^K U_{ij} \operatorname{tr}(E_j \rho)$$

1. Denote the vector of expectation values of $$\rho$$ relative to $$\{E_i\}$$ and $$\{F_i\}$$ with $$|\rho\rangle\rangle_E$$ and $$|\rho\rangle\rangle_F$$, respectively, then:

$$|\rho\rangle\rangle_F = U|\rho\rangle\rangle_E$$

The final step would be to show that there is always a unitary $$U$$ that can rotate $$|\rho\rangle\rangle_E$$ in a way that $$(|\rho\rangle\rangle_F)_1 \neq 0$$ and $$(|\rho\rangle\rangle_F)_{j} = 0$$ for all $$j>1$$.

I think this is true as one can always rotate a vector into an arbitrary axis where all other axes take the 0 value.

• Many thanks for your help! That's a much more systematic way to solve this problem. Regarding my assumption that $\{E_i\}$ is trace-orthonormal, I think it is perhaps unnecessary, as we don't need to expand $\rho$. As long as we can rotate the vector $(\operatorname{tr}(E_j \rho))$ is good enough. As a side note, I also came across a proof sketch for this problem in Lemma 2 of this paper: ieeexplore.ieee.org/abstract/document/850671
– DJD
Nov 22, 2023 at 14:12
• Nice find on that lemma - the ideas are similar there - Kraus operators' unitary freedom comes from the unitary freedom of picking a purification. Also, I agree that the orthonormality of the Kraus ops is not required, I removed the assumption, in the end, the orthonormality (of the vector's basis) comes from the environment's basis. Please accept the answer if you found it helpful. Thanks! Nov 22, 2023 at 15:47