Given two pure quantum state $\rho=|\psi_\rho\rangle\langle\psi_\rho\mid$ and $\sigma=\mid\psi_\sigma\rangle\langle\psi_\sigma\mid$ ($\rho\neq\sigma$). We know that the fidelity between quantum states reduces to a closed form expression i.e., the squared overlap between the states: $$F(\rho,\sigma)=\operatorname{Tr}\left(\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}\right)^2=|\langle\psi_\rho|\psi_\sigma\rangle|^2.$$

Does the Quantum Relative Entropy (QRS) also reduce to a closed-form expression? The QRS for $\rho\neq\sigma$ is given by $$ S(\rho\|\sigma)=-\operatorname{Tr}(\rho\log\sigma)-S(\rho) $$ Where $S(\rho)$ is the Von Neumann entropy of $\rho$ and it is equal to zero as $\rho$ is pure. So for $\rho\neq\sigma$ the above expression is non vanishing. Is there a way to express this in terms of the overlap between the states?


1 Answer 1


Well, if you go back to the definition of quantum relative entropy you'll notice it is not defined for all states. Or more precisely, it is defined as $$S(\rho\|\sigma) = \operatorname{Tr}(\rho\log\rho)-\operatorname{Tr}(\rho\log\sigma) \equiv -S(\rho) - \operatorname{Tr}(\rho\log\sigma)$$ only if $\operatorname{im}(\rho)\subseteq \operatorname{im}(\sigma)$. In all other cases, there are problems in handling $\operatorname{Tr}(\rho\log\sigma)$, and one then just usually defines $S(\rho\|\sigma)=\infty$ whenever $\operatorname{im}(\rho)\subseteq \operatorname{im}(\sigma)$ does not hold.

$\newcommand{\ketbra}[1]{|#1\rangle\!\langle #1|}$The case of pure states brings this aspect into the spotlight: if $\rho\equiv \ketbra\psi$ and $\sigma\equiv \ketbra\phi$, then $\operatorname{im}(\rho)\subseteq\operatorname{im}(\sigma)$ iff $\rho=\sigma$. In which case $S(\rho\|\sigma)=0$. Otherwise it is infinity.

It might also be interesting to note that you can write the quantum relative entropy between generic positive semidefinite operators $P,Q$ in a way that makes the relations with their spectrum more explicit: if $P=\sum_j \lambda_j(P) \ketbra{x_j}$ and $Q=\sum_k \lambda_k(Q) \ketbra{y_k}$, assuming the condition about the images is respected, you have $$S(P\|Q) = \sum_{jk} |\langle x_j|y_k\rangle|^2 \, \lambda_j(P)(\log\lambda_j(P) - \log\lambda_k(Q)).$$ In the case of pure states you'll have a single nonzero eigenvalue equal to $1$ for both states, and in the case the eigenstates are also equal you get back the previous result. See e.g. Watrous' book (in particular, section 5.2.1) for further details on this.

  • $\begingroup$ If I consider a one parameter family of pure quantum states they have a non zero Quantum Fisher Information (QFI) matrix. For two infinitesimally related pure states the second order term in the expansion of QRS should reproduce QFI for pure states. But this requires a non zero QRS. I am bit puzzled. Is the QFI is also zero? $\endgroup$
    – m1rohit
    Nov 4, 2022 at 11:59
  • $\begingroup$ I just realised essentially the same question was asked before, see quantumcomputing.stackexchange.com/a/14158/55. I'm not sure about the QFI tbh. Maybe the gist is that when you compute infinitesimal displacements of the state (which I think you need to compute what you're saying) those are not rank-1 matrices anymore and therefore you get nontrivial values from there? You can ask that as a separate question if you want $\endgroup$
    – glS
    Nov 4, 2022 at 12:49

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