# Why is the quantum Fisher information for pure states $F_Q[\rho,A]=4(\Delta A)^2$?

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 $$0$$.
$$F_Q [\rho,A] = 2 \sum_{k,l} \frac{(\lambda_k-\lambda_l)^2}{\lambda_k + \lambda_l} | \langle k |A| l \rangle |^2\\ = 2 \sum_{k=0,l \neq 0} \frac{(1-0)^2}{1 + 0} | \langle 0 |A| l \rangle |^2 + 2 \sum_{k\neq 0,l = 0} \frac{(0-1)^2}{0 + 1} | \langle k |A| 0 \rangle |^2\\ = 4 \sum_{l \neq 0} | \langle l |A| 0 \rangle |^2\\ = 4 \sum_{l \neq 0} \langle 0 |A| l \rangle \langle l |A| 0 \rangle\\ = 4 \langle 0 |A \bigg( \sum_{l \neq 0} | l \rangle \langle l | \bigg) A| 0 \rangle\\ = 4 \langle 0 |A \bigg( \mathbb{I} - |0\rangle\langle 0| \bigg) A| 0 \rangle\\$$
• How do you know that $$\langle 0 |A \bigg( \mathbb{I} - |0\rangle\langle 0| \bigg) A| 0 \rangle = (\Delta A)^2?$$ Jan 18, 2020 at 12:02
• Yes that's correct, I overlooked that we are interested only in $|0\rangle$ since it is a pure state. Jan 18, 2020 at 18:03