Paris 2009 paper on Quantum Estimation. From eq. 12 to eq. 16

Question

In the paper "Quantum estimation for quantum technology", by Matteo Paris (2009), one is concerned with estimating a parameter $\lambda$ encoded in a quantum state $\rho_\lambda = \sum_n \rho_n |\psi_n\rangle \langle \psi_n|$ - where both eigenvalue $\rho_n$ and eigenstate $|\psi_n\rangle$ may depend on $\lambda$.

In eq. $(12)$ of the paper, the Symmetric Logarithmic Derivative $L_\lambda$ is written as: $$ L_\lambda = 2\sum_{nm} \frac{\langle \psi_m| \partial_\lambda \rho_\lambda |\psi_n\rangle}{\rho_n + \rho_m} |\psi_m\rangle \langle \psi_n| \tag{12}\label{12}$$

Some paragraphs ahead the paper reads:

In order to separate the two contribution to the QFI we explicitly evaluate $\partial_\lambda \rho_\lambda$ $$ \partial_\lambda \rho_\lambda = \sum_{p} \partial_\lambda \rho_p |\psi_p\rangle \langle \psi_p| + \rho_p |\partial_\lambda \psi_p\rangle \langle \psi_p| + \rho_p |\psi_p\rangle \langle \partial_\lambda \psi_p| $$ Since $\langle \psi_n|\psi_m\rangle = \delta_{nm}$ we have $\partial_\lambda \langle \psi_n|\psi_m\rangle \equiv \langle \partial_\lambda \psi_n|\psi_m\rangle + \langle \psi_n| \partial_\lambda\psi_m\rangle = 0$ and therefore $$\text{Re} \langle\partial_\lambda\psi_n|\psi_m\rangle =0 \qquad \langle\partial_\lambda\psi_n|\psi_m\rangle=-\langle \psi_n| \partial_\lambda\psi_m\rangle=0 \tag{15}\label{15}$$ Using Eq. $(15)$ and the above identities we have

$$ L_\lambda = \sum_{p} \frac{\partial_\lambda \rho_p}{\rho_p} |\psi_p\rangle \langle \psi_p| + 2\sum_{n\neq m} \frac{ \rho_n - \rho_m}{\rho_n + \rho_m} \langle\psi_m|\partial_\lambda \psi_n\rangle|\psi_m\rangle \langle \psi_n| \tag{16}\label{16}$$

My Questions

1. Shouldn't it be only $\text{Re} \langle\partial_\lambda\psi_p|\psi_p\rangle =0$?

As far as I see I cannot tell that $\langle \partial_\lambda \psi_n|\psi_m\rangle$ is the conjugate of $\langle \psi_n| \partial_\lambda\psi_m\rangle$. Also, how exactly did we use the previous expression in going from \eqref{12} to \eqref{16}?

2. Shouldn't it be just $\langle\partial_\lambda\psi_n|\psi_m\rangle=-\langle \psi_n| \partial_\lambda\psi_m\rangle$? If the previous were null we would have no second term in \eqref{16}!

Answer 1

The first question is correct: consider unitary evolution with generator $H_\lambda$ such that $|\partial_\lambda \psi_n\rangle=iH_\lambda |\psi_n\rangle$ for all $n$. The states $|\psi_n\rangle$ need not be orthonormal or eigenstates of $H_\lambda$. We can easily choose a pair with $\langle\psi_n|\psi_m\rangle\equiv z\neq 0$ and consider $|\psi_n\rangle$ to be some eigenstate of $H_\lambda$ with real eigenvalue $E_n$. Then $\mathrm{Re}\langle \partial_\lambda\psi_n|\psi_m\rangle=E_n\mathrm{Re}(-i z)$, which certainly does not need to vanish.

The middle question is answered by substituting in the expansion; this should be done by the reader. The important thing to remember is that the latter terms are of the form $\langle\psi_m|\partial_\lambda\psi_n\rangle$, not $\langle\partial_\lambda\psi_m|\psi_n\rangle$, so that is where the minus sign comes from when doing the substitution.

As for question 2, yes that is what it should be; you have spotted the error and fixed it!