# Difference between min/max-entropies and the von Neumann entropy

Consider the (smooth) min-entropy, max-entropy and von Neumann entropy of a given density operator $$\rho_A$$. Does a small gap between $$H_{\max(\min)}(A)_\rho$$ and $$H(A)_\rho$$ implies a small gap between $$H_{\min(\max)}(A)_\rho$$ and $$H(A)_\rho$$? Put in other words, does a small gap between the von Neumann entropy and either min- or max-entropy implies a nearly flat spectrum of $$\rho_A$$?

I'm not sure what you mean exactly with "small gap", but you can easily build examples where $$H(A)$$ and $$H_{\rm max}(A)$$ are "maximally different". For example, $$\rho = \begin{pmatrix}1-\epsilon & 0 \\ 0 &\epsilon\end{pmatrix}$$ for small $$\epsilon>0$$, has $$H(\rho)\simeq 0$$, but $$H_{\rm max}(\rho)=\log|\operatorname{supp}(\rho)|=1$$. You can build similar examples in any dimension. In these cases $$H(\rho)\simeq H_{\rm min}(\rho)$$ but $$H_{\rm max}(\rho)$$ is rather different.

On the other hand, to have $$H(\rho)\simeq H_{\rm max}(\rho)$$ it seems you'd need a more or less balanced nonzero eigenvalues, which would then also make $$H_{\rm min}(\rho)\simeq H(\rho)$$.

Here's a plot of the three quantities for the possible distributions over two outcomes: We can also do the same for three-outcome distributions, obtaining essentially the same result: This is, however, significantly harder to parse without being able to rotate the figure. One way to see that the relations between $$H,H_{\rm min},H_{\rm max}$$ we had before remains here, is to show only the difference between $$H$$ and $$H_{\rm min}$$: This clearly shows that $$H\simeq H_{\rm max}$$ whenever we have a balanced distribution, be it $$P\simeq (1/3,1/3,1/3)$$, or things like $$P\simeq (1/2,1/2,0)$$. But nearby this point, we also have $$H\simeq H_{\rm min}$$.

• Yes, this is basically what I mean by gap. I see your point. It seems a bit strange to me that there's this asymmetry between them. Sep 18, 2022 at 22:03
• There's no asymmetry, it is possible to also make Hmin and H arbitrarily far apart whilst Hmax and H are close. The comment on this in the answer is a little misleading. Sep 19, 2022 at 6:43
• @Rammus I suppose you mean with distributions with more outcomes? It doesn't seem like it's possible for distributions with few outcomes. I guess you'd need a mostly balanced distribution over many outcomes, with a single outcome associated with a much larger probability? Do you have an explicit example?
– glS
Sep 19, 2022 at 7:40
• Yes, you need more outcomes to amplify the gap, but the example is exactly what you say. Large probability on one and flat on the rest. I think I wrote it out explicitly on a recent question of yours but I can't check right now. Sep 19, 2022 at 7:52
• @Rammus if you're referring to the example in this answer, that is, essentially, $p_k=\delta/N$ for $k=1,...,N$ and $p_0=1-\delta$, then we have there $H(p)=H(\delta) + \delta \log N$, $H_{\rm max}(p)=\log_2 (N+1)$, and $H_{\rm min}(p)=\log(1/(1-\delta))$. So $H_{\rm min}\sim 0$, $H_{\rm max}\sim \log N$, and $H\sim \delta \log N$. I don't know that I'd say that in this examples $H$ is "close" to $H_{\rm max}$. They both increase with $\log N$, but they remain quite distant, especially for small $\delta$
– glS
Sep 19, 2022 at 15:29