Rényi entropy becomes von Neumann entropy

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SUMMARY

The discussion centers on the transition from Rényi entropy to von Neumann entropy as alpha approaches one. Specifically, the equation (2.41) diverges to infinity when alpha approaches one, leading to the expression (2.42). The key insight is that the limit of the logarithm of the trace of the density matrix raised to the alpha power, ## \lim_{\alpha \to 1} \log(Tr \rho^\alpha) ##, equals zero, resulting in an indeterminate form ## \frac{0}{0} ##. To resolve this, L'Hôpital's rule must be applied to differentiate the numerator and denominator with respect to alpha before taking the limit.

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  • Understanding of Rényi entropy and von Neumann entropy concepts
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  • Proficiency in applying L'Hôpital's rule for limit evaluation
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Lapidus
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In holographic entanglement entropy notes like here, they let alpha go to one in (2.41) and get (2.42). But (2.41) goes towards infinity, when doing that! Can someone explain how alpha --> 1 will make (2.41) into (2.42)? Thank you!

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You should keep in mind that ## Tr \rho=1 ##. So ## \displaystyle \lim_{\alpha \to 1} \log(Tr \rho^\alpha)=0##. So ## \displaystyle \lim_{\alpha \to 1}S_\alpha(\rho)=\frac 0 0 ## and is indeterminate. To calculate it, you need to use L'Hopital's rule and differentiate the numerator and denominator w.r.t. ## \alpha ## and then take the limit.
 
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