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In Quantum Statistical Mechanics we have the density operator $\rho$ acting on state space $\mathcal{E}$, which gives information about probabilities regarding mixed states. In that setting we define the mean value of the observable $A\in \mathcal{L}(\mathcal{E})$ by

$$\langle A\rangle=\operatorname{tr}(\rho A).$$

With this we define the entropy as

$$S=-k_B\langle \ln \rho\rangle = -k_B\operatorname{tr}(\rho \ln \rho).$$

Now I can't see the intuition behind this. I mean, even the traditional $S = k_B \ln \Omega$ always seemed quite hard for me to grasp, so that saying that "this is the quantum way of saying that" would not help very much here.

Is there some physical intuition that leads us to define the quantity $S$ using that equation? If there is, what is the physical intuition behind this definition?

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closed as unclear what you're asking by ACuriousMind, Jon Custer, user36790, honeste_vivere, Cosmas Zachos Sep 24 '16 at 13:40

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  • $\begingroup$ Maybe it would be helpful, as motivation, to understand why Shannon defined informational entropy in a similar way: en.wikipedia.org/wiki/… $\endgroup$ – Rococo Sep 21 '16 at 22:55
  • $\begingroup$ Followed by: physics.stackexchange.com/q/263197 $\endgroup$ – Rococo Sep 21 '16 at 22:55
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    $\begingroup$ "I mean, even the traditional $S=k_b \lnΩ$ always seemed quite hard for me to grasp, so that saying that "this is the quantum way of saying that" would not help very much here." In that case, this question is unanswerable without first explaining why $S = k_b \ln \Omega$ makes sense, so this question is sort of doomed to failure. Perhaps you should first ask why $S = k_b \ln \Omega$ makes sense. As written, this post is unanswerable. $\endgroup$ – DanielSank Sep 21 '16 at 22:58
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    $\begingroup$ To be clear, I'd love to answer this post, but I'd also like to respect our site best practices and keep questions as focused as possible. $\endgroup$ – DanielSank Sep 21 '16 at 23:04
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    $\begingroup$ @user1620696 Daniel is saying you have to understand the standard version first to even try to understand the quantum version. Ask about $~\ln\Omega$ $\endgroup$ – Bob Bee Sep 22 '16 at 0:13