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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.

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Suppose $\lambda_0 = 1$ and the rest are $0$.

$$ F_Q [\rho,A] = 2 \sum_{k,l} \frac{(\lambda_k-\lambda_l)^2}{\lambda_k + \lambda_l} | \langle k |A| l \rangle |^2\\ = 2 \sum_{k=0,l \neq 0} \frac{(1-0)^2}{1 + 0} | \langle 0 |A| l \rangle |^2 + 2 \sum_{k\neq 0,l = 0} \frac{(0-1)^2}{0 + 1} | \langle k |A| 0 \rangle |^2\\ = 4 \sum_{l \neq 0} | \langle l |A| 0 \rangle |^2\\ = 4 \sum_{l \neq 0} \langle 0 |A| l \rangle \langle l |A| 0 \rangle\\ = 4 \langle 0 |A \bigg( \sum_{l \neq 0} | l \rangle \langle l | \bigg) A| 0 \rangle\\ = 4 \langle 0 |A \bigg( \mathbb{I} - |0\rangle\langle 0| \bigg) A| 0 \rangle\\ $$

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  • $\begingroup$ How do you know that $$ \langle 0 |A \bigg( \mathbb{I} - |0\rangle\langle 0| \bigg) A| 0 \rangle = (\Delta A)^2? $$ $\endgroup$ – John Doe Jan 18 at 12:02
  • $\begingroup$ Write the two terms. One is an expectation value of A^2 while the other is expectation value of A squared. $\endgroup$ – AHusain Jan 18 at 17:52
  • $\begingroup$ Yes that's correct, I overlooked that we are interested only in $|0\rangle$ since it is a pure state. $\endgroup$ – John Doe Jan 18 at 18:03

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