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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