Rényi entropy becomes von Neumann entropy

Lapidus · Mar 19, 2017

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!

Text page with equation 2.41, the Rényi entropy S_α(ρ) = 1/(1−α) log(Tr ρ^α) for 0 < α ≤ ∞, and equation 2.42, the von Neumann entropy S(ρ) = −Tr(ρ log ρ) = −Σ_k λ_k log λ_k.

ShayanJ · Mar 19, 2017

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.

Rényi entropy becomes von Neumann entropy

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