When using a quantum channel to transmit classical information, we consider an ensemble $\mathcal{E} = \{(\rho_x, p(x))\}$ consisting of states $\rho_x$ labelled with a symbol $x$ from a finite alphabet $\Sigma$, each of which is associated with a probability $p(x)$. With just this ensemble we can compute things like Holevo $\chi$ or entropy or relative entropy. If we further include a communication protocol (sender Alice tries to communicate $x$ to Bob by transmitting $\rho_x$ through channel $\mathcal{N}_{A\rightarrow B}$ with probability $p_X(x)$) we have additional quantities we can compute like channel capacity and Holevo information.

Now define the ensemble with respect to a continuous probability distribution $p(x)$ ($\Sigma$ is no longer finite), e.g. the classical-quantum state associated with $\mathcal{E}$ becomes $$ \sigma_{XB} = \int_\Sigma dx p(x) |x\rangle \langle x | \otimes \rho_B^x, \tag{1} $$ and the state of the ensemble becomes $\rho = \text{Tr}_B (\sigma_{XB})$. Entropy $H(\rho)$ remains well-defined, and an equation for the Holevo quantity like \begin{align} \chi(\mathcal{E}) &:= H(\rho) - \int_\Sigma dx p(x) H( \rho_x) \tag{2} \end{align} doesn't seem wrong in any obvious way. On the other hand, Shannon entropy seems to fall apart for probability densities, and $\chi$ is implicitly describing a scenario in which a continuous variable $x$ will be measured.

But some other quantities seem sketchy, e.g. conditional min-entropy seems well defined but its interpretation in terms of optimal measurements feels off. E.g. trying to adapt the operational interpretation that the conditional min-entropy maximizes the state identification probability gives something like \begin{equation} 2^{-H_{min}(X|B)} = \max_{\{ \Lambda_B^x\}} \int_\Sigma dx p(x) \text{Tr}(\Lambda_B^x \rho_B^x) \tag{3} \end{equation} where the maximization is over all POVMs associating each element of $\Sigma$ with a positive operator in $A$. I do not have a good feel for whether this is a reasonable approach; I am again concerned with the idea that Bob is extracting a continuously-valued variable on his end in some way that would blow up the mutual information between his measurement and $X$ to infinity.


Which information-theoretic quantities retain their operational meaning when we substitute a discrete distribution with a continuous one? References are appreciated

  • $\begingroup$ I think you're conflating a continuous mixture of wuantum states with a continuous random variable classically. They're not analagous in this context. The Shannon enteopy loses some of its properties when you try to extend it to a continuous distribution. Similarly, so does the von Neumann entropy for infinite dimensional systems. If you can continuously mix quantum states then you can continuously mix discrete distributions $\endgroup$
    – Rammus
    Commented Apr 8, 2023 at 6:45
  • 1
    $\begingroup$ Yes I agree with most of that. The last equation above is where I'm confused: Alice prepares a set of finite-dimensional states distributed like $p(x)$, then Bob (tries to) define a measurement for discriminating these states and calls that classical measurement outcome $M$. Now $I(X:M)$ is the mutual information between two continuous classical random variables distributed like $p(x)$ and $p(m)$. My question is whether this causes any issues in interpreting capacities or min-entropies or etc. $\endgroup$
    – forky40
    Commented Apr 8, 2023 at 17:15
  • $\begingroup$ Ok, yes. I've not thought about it carefully but I'd imagine you're right here. Its not obvious you can take the usual quantities and just assume they apply to this case because X and M continuous in that setting. $\endgroup$
    – Rammus
    Commented Apr 8, 2023 at 21:08

1 Answer 1


I've found a partial answer for the case of conditional min-entropy, due to Ref. [1] (Appendix IV.B):

Consider a fixed ensemble $\{(\rho_B(x), p(x))\}_{x \in \Sigma}$, where $p(x)$ is a probability density and each $\rho_B(x)$ is finite dimensional state on system $B$, and let $\{\Lambda(x)\}$ be a collection of bounded, positive operators on $B$ satisfying $\int_{\Sigma} dx \Lambda(x) = I$. Then, they introduce a "window function" $w_\epsilon(z) = \mathbb{I}\{|z| \leq \epsilon\}$ and define an optimal measurement success probability as $$ \eta^* := \sup_{\{\Lambda(x)\}} \text{Tr}\left( \int dx p(x) \rho_B(x) \int_\Sigma dx' \Lambda(x') w_\epsilon(x-x') \right). \tag{1} $$

With max-relative entropy $D_{max}(X \Vert Y) :=\log \lVert Y^{-1/2} X Y^{-1/2} \rVert$ defined for operators $X,Y\geq 0$, the conditional min-entropy is defined \begin{align} H_{min}(A|B)_\rho := - \inf_{\sigma_B \in D(\mathcal{H}_B)} D_{max}(\rho_{AB} \Vert \mathbb{I}_A \otimes \sigma_B). \tag{2} \end{align} Then, they claim the desired equality, $$ \eta^* = 2^{-H_{min}(A|B)_Z}, \tag{3} $$ where $Z$ is an operator on $\mathcal{H}_{AB}$ defined as $$ Z_{AB} := \int_{\Sigma} dx |x\rangle \langle x|_A \otimes \left(\int_{\Sigma}dx' p(x') \rho_B(x') w_\epsilon(x-x')\right). \tag{4} $$ The operators $\{|x\rangle\}_{x \in \Sigma}$ are a collection of unbounded linear forms on $\mathcal{H}_A$, satisfying $\langle x | x'\rangle = \delta(x - x')$ (the Dirac delta). So, this almost recovers Eq. (3) from the question, with the exception of the window operator $w_\epsilon$ being used. We can then identify the expression in parentheses above with a positive operator on $B$, and then factor out its trace to recover Eq. (1) from the question: in that case $H_{min}$ plays exactly the role we're looking for.

The reason why this is a partial answer is that I still have some concerns:

  • I'm not sure if the above prescription is general, whether every operator of the form Eq. (1) in the question could be recovered by some windowed ensemble, but its possible that the ones that can't be recovered aren't important.
  • $Z_{AB}$ doesn't seem to actually be a state on $AB$ - $\text{Tr}_B(Z_{AB})$ will not be compact in general.
  • Its tempting to replace $w_\epsilon(z)$ with a Dirac delta $\delta(z)$ to recover Eq. (3) exactly, but this might lead to some problems. If this doesn't work, I think that suggests certain requirements about what ensembles $\rho(x)$ we can consider.

[1] Johannes Jakob Meyer, et. al. "Quantum metrology in the finite-sample regime." https://arxiv.org/abs/2307.06370


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.