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Being not at all a specialist of the subject my question is maybe naive, I apologize if it's the case. The question is the following: There are many different version of entropy functionals around (Boltzmann, Tsallis and many others...). Usually they satisfy the first 3 SHannon-Khinchin axioms but the 4th, due to the uniquess theorem of Shannon-Khinchin, is some sort of twisted version of composability like $$S(AB) = S(A) + S(B) + (some other term) .$$ (In case of Shannon entropy we know that $$S(AB) = S(A) + S(B)).$$ There are also various notions of composability around (see e.g. G. Tempesta). MY question is the following:

IS it physically interesting or even reasonable (do we have examples of physical situations?) to have a "sub-composable" entropy S in the sens that $$S(AB) \leq S(A) + S(B)~?$$

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  • $\begingroup$ I assume your expression "<=" means "≤", which can be written in Mathjax as $\leq$. Is that correct? $\endgroup$
    – lsusr
    Jun 11, 2019 at 19:03
  • $\begingroup$ yes, sorry to not have used mathjax. $\endgroup$ Jun 11, 2019 at 19:06

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I'm not an expert in this field either, but I'd hazard a guess that quantum entanglement can be interpreted as a sub-composable entropy.

The entropy of two entangled particles is lower than the additive entropy of each particle on its own. This might be cheating the spirit of your question, though, since one half of a pair of entangled particles is not a complete system by itself. But $S(AB)<S(A)+S(B)$ implies that $S(A)$ and $S(B)$ share information in some way. It's hard to imagine how $A$ and $B$ can share information without being different parts of the same system.

The entropy of an entangled system is called its von Neumann entropy, which comes from the definition of Shannon entropy. In this sense, a quantum mechanical state obeys the relation $S(AB)\leq S(A)+S(B)$. In particular, an entangled state obeys the relation $S(AB)<S(A)+S(B)$ and an unentangled system obeys $S(AB)=S(A)+S(B)$.

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  • $\begingroup$ Great thanks, very interesting. I will have a look at this notion of entanglement entropy $\endgroup$ Jun 11, 2019 at 19:05

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