What is the entropy of a mixed state in classical physics? Consider a classical system which admits certain macroscopic level of description. It is known, that for two pure states $\omega_1$ and $\omega_2$ on this level of description the entropy of the system is $S_1$ and $S_2$ accordingly.
Next, let us consider a mixed state:
$$
\omega_\text{m} = \alpha \omega_1 + (1-\alpha) \omega_2
$$
What is the entropy of $\omega_\text{m}$ in terms of $S_1$ and $S_2$?
I think there should be nothing special regarding the sum of greater number of pure states, so I chose the simplest case of two pure states. If my assumption is not true, please indicate.
For the used notion of states see an entry in n-Lab on observables and states.
 A: In macroscopic units it should be 
$$S=-R\alpha \log(\alpha e^{-S_1/R})-R(1-\alpha)\log\Big(1-\alpha)e^{-S_2/R}\Big)
\\=\alpha \Big(S_1-R\log\alpha\Big)+(1-\alpha)\Big(S_2-R\log(1-\alpha)\Big),$$
where $R$ is the universal gas constant. In the pure case, this reduces to the textbook formula.
But such a formula cannot be true in general. The general formula is
$$S=\langle-R\log\rho\rangle=-R\ Tr (\rho\log\rho),$$ 
where the trace is over the microstates. Now represent $\rho$ as a mixture of two independent distributions. If system $k=1,2$ is in the classically pure but quantum mixed state $\rho_k$ then the mixed classical state has density matrix $\rho=\alpha\rho_1+(1-\alpha)\rho_2$, If one only knows the entropies of the states $k=1,2$ then $S_k=-Tr\ \rho_k\log\rho_k$ and 
$$S=-Tr\ \rho\log\rho=-Tr\Big(\alpha\rho_1+(1-\alpha)\rho_2\Big)\log\rho
=-\alpha\ Tr\rho_1\log\rho-(1-\alpha)\ Tr\ \rho_2\log\rho,$$
which cannot be simplified further without making approximations or assumptions. 
The formula given is the most reasonable ''simplification'' independent of other data about the partial states.
