Assume that a density matrix is given in its eigenbasis as $$\rho = \sum_{k}\lambda_k |k \rangle \langle k|.$$ On page 19 of this paper, it states that the Quantum Fisher Information is given as $$F_{Q}[\rho,A] = 2 \sum_{k,l}\frac{(\lambda_k-\lambda_l)^2}{\lambda_k + \lambda_l}|\langle k|A|l\rangle|^2$$
Question: I might be missing something simple but can anyone see why, as stated in the paper, for pure states it follows that $$F_{Q}[\rho,A] = 4(\Delta A)^2?$$
Thanks for any assistance.
Suppose $\lambda_0 = 1$ and the rest are $0$.
$$ F_Q [\rho,A] = 2 \sum_{k,l} \frac{(\lambda_k-\lambda_l)^2}{\lambda_k + \lambda_l} | \langle k |A| l \rangle |^2\\ = 2 \sum_{k=0,l \neq 0} \frac{(1-0)^2}{1 + 0} | \langle 0 |A| l \rangle |^2 + 2 \sum_{k\neq 0,l = 0} \frac{(0-1)^2}{0 + 1} | \langle k |A| 0 \rangle |^2\\ = 4 \sum_{l \neq 0} | \langle l |A| 0 \rangle |^2\\ = 4 \sum_{l \neq 0} \langle 0 |A| l \rangle \langle l |A| 0 \rangle\\ = 4 \langle 0 |A \bigg( \sum_{l \neq 0} | l \rangle \langle l | \bigg) A| 0 \rangle\\ = 4 \langle 0 |A \bigg( \mathbb{I} - |0\rangle\langle 0| \bigg) A| 0 \rangle\\ $$
