7
$\begingroup$

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": $$S_A=\lim_{n\to 1}\frac{tr_A \rho_A^n-1}{1-n}=-\frac{\partial}{\partial n}\log tr_A\rho^n_A|_{n=1}.$$ Could anyone explain why $$-tr_A\rho_A Log\rho_A=-\frac{\partial}{\partial n}log tr_A\rho^n_A|_{n=1}~?$$

$\endgroup$
0

1 Answer 1

6
$\begingroup$

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 $$ - \frac{\partial }{ \partial n} \log \text{tr} \rho_A^n \bigg|_{n=1} = - \sum_i \lambda_i \log \lambda_i = - \text{tr} \rho_A \log \rho_A $$

$\endgroup$
2
  • $\begingroup$ Why $- \frac{\partial }{ \partial n} \log \text{tr} \rho_A^n \bigg|_{n=1} = - \frac{\sum_i \lambda_i^n \log \lambda_i }{\sum_i \lambda_i^n } \bigg|_{n=1}$ is true? $\endgroup$ Commented Dec 30, 2020 at 9:04
  • $\begingroup$ @RathindraNathDas - Well, different the first equation w.r.t. $n$ and it should follow. $\endgroup$
    – Prahar
    Commented Dec 30, 2020 at 23:10

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