Please refer this question to understand why I consider the freebit picture important.

In short, it is conjectured, that for certain real systems the most complete physical description possible involves Knightian uncertainty so that some qubits in nature actually "freebits".

So I decided to calculate entropy of a system that involves Knightian uncertainty.

Lets define c-freebit (classical freebit) as a two-level system whose Knightian uncertainty is maximal.

We know that entropy of a two-level system depends on the probabilities of the respective levels, if the probability of the state 0 is $p_0= x$, then the entropy (in natural units) is:

$$S= -\sum_{i=0}^1 p_i \ln p_i = -x \ln x - (1 - x) \ln (1 - x)$$

So if $x=1/2$ then $S=\ln 2$ nat, equal to 1 bit.

But when we introduce the Knigtian uncertainty $x\in[a,b]$, the total entropy becomes $$S=\int_a^b \frac{-(1-x) \ln (1-x)-x \ln (x)}{b-a} \, dx$$

For the maximum case, $a=0$, $b=1$ we get $S=\frac12$ nat.

This is remarkable, because we got a rational fraction of natural units of information! The nat, which plays no role in probablistic description of anything suddenly appears in the evidential description of the Dempster–Shafer formalism!

We obtained that a c-freebit contains 1/2 nat of information!

Given this result I wonder whether something similar happens in quantum world. Can anybody please provide the calculation for entropy of a quantum freebit?

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    $\begingroup$ I don't understand why you're calculating the entropy S by averaging over all x in the interval [0,1]. If you really take seriously that this is Knightian uncertainty, then you ONLY know that x is in [a,b]: you don't even have probabilistic information beyond that, and you certainly don't know that x is uniform. So, not knowing where your formula came from, it's hard to say whether it's "just a neat mathematical coincidence" or whether it has some operational meaning. $\endgroup$ Feb 6 '14 at 15:54
  • $\begingroup$ @Scott Aaronson I just assume that x has uniform probability distribution on the maximum allowed interval. This is the most simple model of freebit. Of course if we consider physical applications we may need take some more complicated distribution. The presented here model seems to represent a system where probability (rather than real state) is taken as a physical quantity and there is no other information about the system. $\endgroup$
    – Anixx
    Feb 6 '14 at 16:03
  • $\begingroup$ @Scott Aaronson for example, if we take a planck scale, a particle contained in planck length seems to represent a freebit. This is because not only the exact position of particle is uncertain, but also the maximum of the particle's probability! This is because it is impossible to measure units smaller than planck scale. So in my impression, one particle in a planck scale interval is just 1/2 nat. This seems somewhat connected with the calculations for the entropy of a black hole, where a planck unit area cantains 1/4 nats of information. $\endgroup$
    – Anixx
    Feb 6 '14 at 16:21
  • $\begingroup$ @Scott Aaronson see also physics.stackexchange.com/questions/98001/… and physics.stackexchange.com/questions/98152/… $\endgroup$
    – Anixx
    Jul 18 '19 at 15:37
  • $\begingroup$ I agree with Scott first comment. See my answer here math.stackexchange.com/a/3298795/312 $\endgroup$
    – leonbloy
    Jul 20 '19 at 15:32

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