High fidelity implies low entropy

Question

I was reading the security proof of QKD by Lo-Chau. One of their main idea is to bound the entropy of the eavesdropper if the singlet states are of high fidelity. Here I summarized the first lemma:

Let singlet state be $|\psi\rangle = (|01\rangle - |10\rangle)\sqrt{2}$, then if the fidelity of $\langle \psi|^{\otimes n} \rho | \psi \rangle^{\otimes n} > 1 - \delta$, with $\delta <<1$, then the von Neumann entropy $S(\rho) < - (1-\delta) \log_2(1-\delta)- \delta \log_2 \frac{\delta}{2^{2n}-1}$.

In the proof, it is mentioned that if the the fidelity is bounded, then the largest eigenvalue of the density matrix $\rho$ must be larger than $1-\delta$. Then the entropy of $\rho$ is bounded by a diagonal density matrix of $\rho_0$, with diagonal entries of $\{1-\delta, \frac{\delta}{2^{2n}-1} , \dots, \frac{\delta}{2^{2n}-1}\}$

My question: Why does the bounded fidelity implies the largest eigenvalue must be larger than $1-\delta$? I think what the author means is that this eigenvalue correspond to eigenstate of $| \psi \rangle \langle \psi|$, and the remaining eigenstates has uniform eigenvalues of $\frac{\delta}{2^{2n}-1}$. But if this eigenvalue does not correspond to a singlet eigenstate, is it possible to make a density matrix with eigenvalues less than $1-\delta$, but still give fidelity higher than $1-\delta$?

Answer 1

This is easiest to see if you diagonalise the state \begin{align} \rho = \sum_j p_j |\phi_j \rangle\langle \phi_j|, \end{align} where the $|\phi_j\rangle$ are orthogonal and normalized, and the $p_j$ are non-negative and sum to $1$. Taking the fidelity with any pure state (say $|\psi\rangle$) we obtain \begin{align} \langle{\psi}|\rho|\psi\rangle &= \sum_j p_j \langle\psi |\phi_j\rangle\langle \phi_j|\psi\rangle \\ &= \sum_j p_j \lvert\langle \phi_j|\psi\rangle \rvert^2, \end{align} the fidelity is a convex combination of the eigenvalues, $p_j$, of $\rho$ with weights given by the numbers $\lvert\langle \phi_j|\psi\rangle \rvert^2$ (which are non-negative and sum to $1$ so this is a true convex combination), it therefore can't be larger than the larger than the largest eigenvalue.

For a slightly more formal argument let $p_\text{max} = 1-\delta$ be the largest eigenvalue of $\rho$ then clearly \begin{align} \langle{\psi}|\rho|\psi\rangle = \sum_j p_j \lvert\langle \phi_j|\psi\rangle \rvert^2 &\leq \sum_j p_\text{max} \lvert\langle \phi_j|\psi\rangle \rvert^2 = p_\text{max} \sum_j \lvert\langle \phi_j|\psi\rangle \rvert^2 = p_\text{max}. \end{align}