On page 566, it states that using $S(\rho^{'})-S(\rho,\varepsilon) \ge S(\rho)$ and combining this with $S(\rho) \ge S(\rho^{'})-S(\rho,\varepsilon))$, we get $S(\rho^{'})=S(\rho)-S(\rho,\varepsilon)$. I can see no way in which this case holds, as by subbing this back in you'd end up getting either $S(\rho)-2S(\rho,\varepsilon) \ge S(\rho)$ or $S(\rho) \ge S(\rho)-2S(\rho,\varepsilon)$

The first inequality should never hold, with equality only when $S(\rho,\varepsilon)=0$, and the second should hold, but contradicts the first.

This is meant to be a way to show the necessity of $S(\rho)=S(\rho^{'})-S(\rho,\varepsilon)$

If feel like it should actually be $S(\rho^{'})=S(\rho)+S(\rho,\varepsilon))$, as subbing this back in gives equality in the two cases I stated, which is what you want to show. But I am assuming here that the book is right and I am wrong, so where am I going wrong?


1 Answer 1


Combining those two equations gives

$$S(\rho')-S(\rho,\epsilon)\geq S(\rho)\geq S(\rho')-S(\rho,\epsilon)$$

and since the left and right hand sides are the same, the inequalities must be equalities, which indeed gives $S(\rho')=S(\rho)+S(\rho,\epsilon)$, or equivalently $S(\rho)=S(\rho')-S(\rho,\epsilon)$. It's quite possible this was a typo where they mixed up the $'$ symbol. In my copy of the book Equations 12.132 and 12.136 have it in the right order. It looks like they are trying to prove that Equation 12.132 holds, so I'm fairly confident this is a typo.

  • $\begingroup$ Thanks, I though I was going insane and missing something obvious there. Really frustrating when that happens in a textbook :/ $\endgroup$ Aug 11, 2020 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.