# What are the antisymmetric terms in $\sigma_{mn}$ in the expression for the Fisher information?

Given a parameter-dependent density operator $$\hat\rho^\lambda$$ and its spectral decomposition $$\{\rho_m^\lambda, |\psi_n^\lambda\rangle\}$$, Eq. $$(17)$$ from this review shows that one can compute its quantum Fisher information (QFI) as \begin{align} H(\lambda)&:=H_C(\lambda) +H_Q(\lambda) \\ &=\sum_n\frac{(\partial_\lambda\rho_n^\lambda)^2}{\rho_n^\lambda} + \left(2\sum_{n\ne m}\sigma_{mn}\left|\langle \psi_m^\lambda|\partial_\lambda\psi_n^\lambda\rangle\right|^2\right) \end{align} where the matrix $$\sigma_{mn}$$ is rather mysteriously defined as $$\sigma_{mn}:=\frac{(\rho_n^\lambda-\rho_m^\lambda)^2}{\rho_m^\lambda+\rho_m^\lambda}+\text{any antisymmetric terms}.$$ When the eigenvectors do not depend on $$\lambda$$, then $$H(\lambda)=H_C(\lambda)$$, which is just the classical FI associated to the distribution of the eigenvalues.

I have two questions regarding this expression, including one that is perhaps a little vague.

1. What are these antisymmetric terms, and why aren't they written explicitly?
2. Suppose a state $$\hat\rho^\lambda$$ is such that $$H_Q(\lambda)=0$$, either because the eigenvectors do not depend on $$\lambda$$ or because the scalar product vanishes, as is the case for the state $$p_\lambda|\Psi_\lambda\rangle\langle \Psi_\lambda|+(1-p_\lambda)|0\rangle\langle 0|$$, where $$|0\rangle$$ is the empty state with no photons and $$|\Psi_\lambda\rangle\propto \int d\mathbf r \ A_\lambda(\mathbf r)\hat a^\dagger(\mathbf r)|0\rangle$$ is a general one-photon state. What does it mean for a state to have purely 'classical' Fisher information? Do we need, for instance, entanglement between two or more photons in order for $$H_Q$$ to be nonzero?

Cross-posted on physics.SE

Antisymmetric means that $$\sigma_{nm}=-\sigma_{mn}$$. Since the sum ranges over all values of $$m$$ and $$n$$, adding an antisymmetric term adds something proportional to $$|\langle \psi_m^\lambda|\partial_\lambda\psi_n^\lambda|^2-|\langle \psi_n^\lambda|\partial_\lambda\psi_m^\lambda|^2.$$ Now, these two quantities are equal in magnitude and so their difference is zero, as can be seen by taking the derivative \begin{align}0=\partial_\lambda \delta_{mn}=\partial_\lambda \langle \psi_m^\lambda|\psi_n^\lambda\rangle= \langle \partial_\lambda\psi_m^\lambda|\psi_n^\lambda\rangle+\langle \psi_m^\lambda|\partial_\lambda\psi_n^\lambda\rangle \\ \Rightarrow \langle \partial_\lambda\psi_m^\lambda|\psi_n^\lambda\rangle=-\langle \psi_m^\lambda|\partial_\lambda\psi_n^\lambda\rangle\\ \Rightarrow |\langle \partial_\lambda\psi_m^\lambda|\psi_n^\lambda\rangle|=|\langle \psi_m^\lambda|\partial_\lambda\psi_n^\lambda\rangle|. \end{align} (We also need that $$|\langle \partial_\lambda\psi_m^\lambda|\psi_n^\lambda\rangle|=|\langle \psi_n^\lambda|\partial_\lambda\psi_m^\lambda\rangle|$$.) Since any antisymmetric term adds 0 to the Fisher information, we can include as many extra antisymmetric terms that we like (obviously the easiest thing to do is to not include any antisymmetric term).
For a state to have purely "classical" Fisher information, it can still be as quantum mechanical as you like. I can write $$\rho(\lambda)=\sum_k p_k(\lambda) |\psi_k\rangle\langle \psi_k|$$ with basis states $$|\psi_k\rangle$$ that are as entangled as I like. Even a boring state like $$\rho=|\psi\rangle\langle \psi|$$ that is independent from the parameter can be a maximally entangled pure state in any dimension! The question about having "quantum" additions to the Fisher information is not so much about the state's nonclassical properties; it instead depends on how the parameter dependence is encoded in the state. As you have seen, this simply means that the eigenbasis of the state must change with the parameter, otherwise there are only classical contributions to the Fisher information.