Replica trick for calculating Entanglement Entropy?
Let $\lambda_i$ be the eigenvalues of $\rho_A$. Then $$ \log \text{tr} \rho_A^n = \log \left( \sum_i \lambda_i^n \right) $$ Now, let is differentiate w.r.t. $n$. We find $$  \frac{\partial }{ \partial n} \log \text{tr} \rho_A^n \bigg_{n=1} =  \frac{\sum_i \lambda_i^n \log \lambda_i }{\sum_i \lambda_i^n } \bigg_{n=1} =  \frac{ \sum_i \lambda_i \log \lambda_i }{\sum_i \lambda_i } $$ Now since $\text{tr}_A \rho_A = 1 \implies \sum_i \lambda_i = 1$. Thus, we find $$  \frac{\partial }{ \partial n} \log \text{tr} \rho_A^n \bigg_{n=1} =  \sum_i \lambda_i \log \lambda_i =  \text{tr} \rho_A \log \rho_A $$
Nahc
Updated on January 05, 2021Comments

Nahc almost 3 years
This is probably a simple question. Von Neumann entropy is defined to be $$S_A=tr_A\rho_A \log\rho_A.$$ And it's said that it can be calculate from the "Replica trick": $$S_A=\lim_{n\to 1}\frac{tr_A \rho_A^n1}{1n}=\frac{\partial}{\partial n}\log tr_A\rho^n_A_{n=1}.$$ Could anyone explain why $$tr_A\rho_A Log\rho_A=\frac{\partial}{\partial n}log tr_A\rho^n_A_{n=1}~?$$

Rathindra Nath Das almost 3 yearsWhy $ \frac{\partial }{ \partial n} \log \text{tr} \rho_A^n \bigg_{n=1} =  \frac{\sum_i \lambda_i^n \log \lambda_i }{\sum_i \lambda_i^n } \bigg_{n=1}$ is true?

Prahar almost 3 years@RathindraNathDas  Well, different the first equation w.r.t. $n$ and it should follow.