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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.

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  • $\begingroup$ Should the entropy not be zero for pure states? (And the entropy of the mixture of states simply be $S = - \alpha \ln \alpha - (1 - \alpha) \ln (1 - \alpha) $ by definition?). $\endgroup$
    – Sebastian Riese
    Commented Jul 13, 2015 at 11:53
  • $\begingroup$ @SebastianRiese State being pure has nothing to do with zero or nonzero entropy. It has to do whether it gives a unique value for all the observables (pure state) or probability distributions for their values (mixed state). For example a pure state for a macroscopical description of a common fluid is given by the fields of density, velocity and temperature. Obviously such description implies the fluid has a nonzero entropy. A mixed state might be of interest in turbulence research when you don't know the fluid velocity at a point, but rather assume there is a probability distribution for it. $\endgroup$
    – Yrogirg
    Commented Jul 13, 2015 at 12:34
  • $\begingroup$ So it is not a pure microstate, but rather a "pure macrostate". Did not meet that nomenclature before, my mistake. $\endgroup$
    – Sebastian Riese
    Commented Jul 13, 2015 at 12:40
  • $\begingroup$ @Yrogirg so you have a system that can now exhibit two micro-states $\omega_1$ and $\omega_2$ with different weights $\alpha$ and $1-\alpha$. So that for certain time of observation $T$ it exhibits $\omega_1$ a fraction $\alpha T$ of the time and exhibits $\omega_2$ a fraction $(1 - \alpha)T$ of the time. Is this coherent with your question? $\endgroup$
    – rmhleo
    Commented Jul 15, 2015 at 9:38
  • $\begingroup$ @rmhleo I can't admit that this would be the only interpretation, but I'd say that yes, one can take it, provided the states do not change in time. $\endgroup$
    – Yrogirg
    Commented Jul 15, 2015 at 12:16

1 Answer 1

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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.

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  • $\begingroup$ Yes, indeed I have found the formula of the same structure when a mixed state was a generalised canonical ensemble in the book Beyond equilibrium thermodynamics. However, what I really would like to know is how it is derived. What is the rationale behind it? $\endgroup$
    – Yrogirg
    Commented Jul 17, 2015 at 9:53
  • $\begingroup$ @Yrogirg: I augmented my answer. Oettinger's book that you mention is nice. $\endgroup$
    – Arnold Neumaier
    Commented Jul 17, 2015 at 12:40
  • $\begingroup$ It seems for me that it is actually $\langle -R \log \frac{\rho}{\exp(S/R)} \rangle$. How to get this? This is not obvious at all for me, otherwise I wouldn't be asking the question. Is this derivation available anywhere in the literature? Probably not physics textbooks, but the ones on information theory. $\endgroup$
    – Yrogirg
    Commented Jul 17, 2015 at 13:07
  • $\begingroup$ @Yrogirg: Your question is very unusual, and I haven't seen it anywhere before. So there is perhaps no derivation anywhere. Perhaps I'll find the time for a detailed derivation later. My formula is guesswork based on the assumption that the result can actually be expressed in terms of the partial entropies only. This might be true under some additional conditions only. Maybe there is a connection with ideal mixtures in chemical thermodynamics? $\endgroup$
    – Arnold Neumaier
    Commented Jul 17, 2015 at 13:35
  • $\begingroup$ @Yrogirg: I added to my answer a caveat concerning my proposed formula. $\endgroup$
    – Arnold Neumaier
    Commented Jul 21, 2015 at 16:47

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