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I was going through the security proof of the BB84 protocol by Dr. Ramona Wolf. I am having trouble following the equation (used at 8:53) for the fidelity,

\begin{align*} F(\rho_{ABE},\left|\phi\right\rangle ^{\otimes m}\otimes\sigma_{E})= & F(\rho_{AB},\left|\phi\right\rangle ^{\otimes m})\,, \end{align*} where it has been claimed that accroding to Uhlmann's theorem there is a state $\sigma_{E}$, which is a system of Eve, for which the above relation is true. $\rho_{ABE}$ is a purification of $\rho_{AB}$.

1.How do derive the relation using Uhlmann's Theorem?

2.If the equation holds true for only particular state $\sigma_{E}$ of Eve's system, What gaurantees that Eve would not end up with a different state on her part of the system? What I have understood is that in the ideal case the state of the whole system can be described by \begin{align*} \rho_{ABE}= & \rho_{UU}\otimes\sigma_{E}\,, \end{align*} where $\rho_{UU}=\sum\frac{1}{\left|U\right|}\left|U\right\rangle \left\langle U\right|\otimes\left|U\right\rangle \left\langle U\right|$, i.e there are no correlations between the System of Alice and Bob and of Eve, with no restriction on the actual state of Eve's system.

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The relation in 1. is essentially a statement of Uhlmann's theorem. Ulmann's theorem states: Given two density matrices $\rho$ and $\sigma$, we have that the fidelity $F(\rho, \sigma)$ is given by

$$F(\rho, \sigma) = \max_{|\psi_\rho\rangle, |\phi_\sigma\rangle} F(|\psi_\rho\rangle, |\phi_\sigma\rangle)$$

Where $|\psi_\rho\rangle$ and $|\phi_\sigma\rangle$ are purifications of $\rho$ and $\sigma$ respectively.

For most purifications, the fidelity between the purified states is not the maximum. But there must be some choice in which that fidelity is maximized. In this case, let $\rho_{ABE}$ be that purification for $\rho_{AB}$, and since $|\phi\rangle^{\otimes m}$ is already pure, we can choose $\sigma_E$ to be any state.

To answer 2., we want to argue a "no-go" result. That is, we want to show that in the ideal case, Eve gets cannot gain eavesdrop, and in all other cases, Eve performs even worse.

In this case, results from the fact that if Eve manages to choose the best possible state $\sigma_E$, then we have the relationship $F(\rho_{ABE}, |\phi\rangle^{\otimes m} \otimes \sigma_E) = F(\rho_{AB}, |\phi\rangle^{\otimes m})$, but in all other cases, the relationship is $\leq$, rather than equality.

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  • $\begingroup$ Thank you! That really helped a lot. $\endgroup$ Mar 9 at 20:45
  • $\begingroup$ @HypnoticZebra please consider accepting the answer if this has resolved your query! $\endgroup$
    – FDGod
    Mar 10 at 6:11

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