2
$\begingroup$

You may be familiar with "Klein's inequality"; one form of it is $$ -\operatorname{tr}(\rho \log \sigma) + \operatorname{tr}(\rho \log \rho) \ge 0, $$ stating that relative entropy is nonnegative (where $\rho$ and $\sigma$ are density matrices). There are various other versions of it - some which are equivalent, and some which go by the same name but are inequivalent / more general. A rather general one is $$ \operatorname{tr}(f(A)-f(B)-(A-B)f'(B)) \ge 0 $$ ($f$ convex, such as $f(t) = t \log t$), which is what https://en.wikipedia.org/wiki/Trace_inequality calls Klein's inequality. I'm curious about the origins of the Klein inequality (or perhaps I should say, inequalities, plural).

I see Nielsen and Chuang, for example, cite a paper, O. Klein, Zur quantenmechanischen Begründung des zweiten Hauptsatzes der Wärmelehre, Zeitschrift für Physik A 72 (1931), 767 – 775. However - and maybe I'm just bad at reading German or the notation - I don't think I actually recognize "Klein's inequality" in there.

Did Oskar Klein prove what's now called Klein's inequality? Or if he didn't, then who did, and when?

$\endgroup$

1 Answer 1

2
$\begingroup$

As far as I can tell Klein never published the proof but the proof is rather attributed to him via word of mouth.

In the paper Mean Entropy of States in Quantum‐Statistical Mechanics (free version available here), a form of Klein's inequality is given in Lemma 1. There is a footnote attached to the lemma which states the following:

This lemma, together with its proof, was communicated to one of us (D.W.R.) by Professor R.Jost, who attributed it to O. Klein. If $\mathrm{Tr}[A] = \mathrm{Tr}[B]=1$, this lemma is a particular case of Theorem 1 of H. Umegaki, Kodai Math. Sem. Rep. 14, 59 (1962).

I believe the professor who attributed the inequality to Klein is Res Jost who was working in Zürich at the time which is close to D.W.R. who was in Geneva.

$\endgroup$

Your Answer

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

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