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.


1 Answer 1


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.

  • 1
    $\begingroup$ 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 $\endgroup$
    – DJD
    Nov 22, 2023 at 14:12
  • $\begingroup$ 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! $\endgroup$ Nov 22, 2023 at 15:47

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