4
$\begingroup$

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}?

  1. 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}!
$\endgroup$
2
  • 1
    $\begingroup$ Please verify my edit: I assume there is a typo in the question, as the result quoted in the paper says $\mathrm{Re}\langle \partial_\lambda\psi_n|\psi_m\rangle=0,$ not that $\mathrm{Re}\langle \psi_n|\psi_m\rangle=0$ (we know the latter to just be $\delta_{nm}$). $\endgroup$ Nov 22, 2022 at 1:01
  • $\begingroup$ @QuantumMechanic Thanks for the edit and for the answer. My middle question referred only the knowledge of $\text{Re} \langle\partial_\lambda\psi_p|\psi_p\rangle =0$, it seems to me that it is not used in going from (12) to (16). It seems to me that $\langle\partial_\lambda\psi_n|\psi_m\rangle=-\langle \psi_n| \partial_\lambda\psi_m\rangle$ suffices. $\endgroup$
    – G Frazao
    Nov 22, 2022 at 11:51

1 Answer 1

3
$\begingroup$

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!

$\endgroup$
2
  • 1
    $\begingroup$ the $|\psi_n\rangle$ are orthogonal since they are eigenstates of the Hermitian $\rho$, by the Spectral Theorem. Continuing your line of thought with $|\partial_\lambda \psi_n\rangle=iH_\lambda |\psi_n\rangle$ we have $\mathrm{Re}\langle \partial_\lambda\psi_n|\psi_m\rangle=\mathrm{Re}(-i \langle \psi_n|H_\lambda|\psi_m\rangle)=\mathrm{Re}(-i H_{\lambda,nm})$. Now $H_{\lambda,nn} \in \mathbb{R}$, but in general $H_{\lambda,nm}$ might be a fully entitled complex number for $n \neq m$. Thanks again for your time and explanation. $\endgroup$
    – G Frazao
    Nov 22, 2022 at 12:28
  • $\begingroup$ @GFrazao this is an important point, thank you. The proof is then slightly harder, because one can imagine that there is always the freedom to redefine the phases of the eigenstates. We have to count the number of independent phases in some matrix $H_{\lambda,nm}$, which are $N(N-1)/2$ for rank $N$ matrices $\rho$, versus the $N$ eigenstates defining $N-1$ relative phases. For $N>2$ one cannot always find a set of eigenstates of $\rho$ for a given, arbitrary $H_\lambda$ such that $H_{\lambda,nm}\in\mathbb{R}$ . $\endgroup$ Nov 29, 2022 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.