This paper is a paper in 2012 and cited by a lot of papers. And there does not exist comment in arxiv or error statement in PRA. But when I reading this paper, I think the right part of the eq(23) should be $2M+(N-M)(N-M+2)$ instead of $M+(N-M)(N-M+2)$. Is there someone who happens to read this paper, so he or she can tell me if this place is truly an error or just my misunderstanding?
My reasoning based on two methods, the first method is to get the contradiction, the second method is to show my proof.
Contradiction: Right below the eq(24), he stated:

Any state that violates Eq.(23) has fewer than $M$ unentangled particles.

And the state below Eq.(29) shows that the state $\mid 1\rangle \bigotimes\mid D_{N-1}^{N/2-1}\rangle$ has Quantum Fisher Information $(\frac{N^2}{2}+\frac{1}{2},\frac{N^2}{2}+\frac{1}{2},0)$,corresponding to three Quantum Firsher Information components, so the total Quantum Fisher Information is $N^2+1$, which violate the $M+(N-M)(N-M+2)$ with $M=1$. But we know $\mid 1\rangle \bigotimes\mid D_{N-1}^{N/2-1}\rangle$ is not the state which has fewer than $M(M=1)$ unentangled particles.

My Proof: Just like the skill he used in the first equation in Eq.(20), i.e.,$$\sum_{l=x,y,z}(\Delta J_l)^2_{\mid\psi_{k-producible}\rangle}=\sum_m\sum_{l=x,y,z}(\Delta J_l)^2_{\mid\psi_{m}\rangle}.$$ That is, change the total sum to the product part sum, $\mid\psi_{k-producible}\rangle=\mid\psi_1^{(N_1)}\rangle\bigotimes\mid\psi_2^{(N_2)}\rangle\bigotimes...$(his Eq.(18)) .
So what I do here, is combine the Observation 1, i.e., $\sum_{l=x,y,z}F_Q[\rho,J_l]\le 2N$ and observation 2, i.e., $\sum_{l=x,y,z}F_Q[\rho,J_l]\le N(N+2)$, change the variable $N$ in the first inequality($2N$) to $M$, and the variable $N$ in the second inequality($N(N+2)$) to $N-M$, we get the inequality in Eq.(23), but just $2M$ instead of $M$.

  • 1
    $\begingroup$ Maybe it would be best if you included the relevant details needed to answer the question in the question. And also your derivation of the alternative result. $\endgroup$
    – Rammus
    Commented Apr 22, 2021 at 7:31
  • $\begingroup$ I've added more details to my question. It seems my derivation is right? $\endgroup$
    – narip
    Commented Apr 22, 2021 at 9:05

1 Answer 1


Your conclusion appears correct to me. It seems that Eq.(23), modified with your proposed change to the RHS, can be verified by combining Eq.(3), for the upper bound on the $M$ unentangled particles, with Eq.(5), for the upper bound on the $N-M$ entangled particles, using the convexity asserted in Section II.

With respect to the $M$ unentangled particles, the result in the paper would seem to require the use of Eq.(4), which relates to information associated with a single arbitrary component of $J$, instead of Eq.(3), which relates to the sum over components of $J$, to represent the information associated with the sum over the components of $J$. This appears to be a mistake.


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