Deriving entanglement entropy from Renyi entropy My questions are based on this paper - http://arxiv.org/abs/0905.4013

*

*Firstly I want to know as to whether some assumptions are needed about the relationship between the systems $A$ and $B$ for the Hilbert space to factor as tensor products as ("assumed"?) on page 3?
I mean assume the more common reverse scenario - if you are given a system C and you decide to call some part of it as $A$ and the rest as $B$ then does it automatically mean that the Hilbert space of C factors between A and B? (...that doesn't intuitively feel to be true...then what exactly is the assumption being made here?..)


*Secondly given the definition of $S_A$ and $S^{(n)}_A$ as in equations 2 and 3 how does this claimed equality follow that, $S_A = \lim _{n \rightarrow 1} S^{(n)}_A = - \lim_{n \rightarrow 1} \frac{\partial \rho^n_A }{\partial n }$
I am unable to see the proof of the above 2 equalities.
It would be great if someone could help.
 A: 
I want to know as to whether some assumptions are needed about the relationship between the systems $A$ and $B$ for the Hilbert space to factor as tensor products

Here's the general idea behind the factorization in this context.  Disclaimer: this won't be completely rigorous (as with many computations in field theory).  
Consider, for the sake of orienting ourselves, a classical mechanical system with two, independent configuration degrees of freedom $q_A$ and $q_B$.  When one quantizes such a system, one assigns a Hilbert space to each independent degree of freedom, say $\mathcal H_A$ and $\mathcal H_B$, and the Hilbert space for the whole system is the tensor product $\mathcal H_A\otimes\mathcal H_B$.
Now consider some classical theory of a fields on a manifold $M$.  There is an infinite number of degrees of freedom for such a system, one for each point on the manifold, since in order to specify a classical configuration, one would need to specify the value of the fields $\phi$ at every point $x$ on the manifold.  When one quantizes, one therefore assigns a Hilbert space $\mathcal H_x$ to each of these classical degrees of freedom, and the total Hilbert space is the tensor product
$$
  \bigotimes_{x\in M}\mathcal H_x
$$
Now, let's say that I partition my manifold into two regions $M_A$ and $M_B$, namely $M = M_A\cup M_B$ and $M_A\cap M_B = \emptyset$, then notice that the Hilbert space of the system factors as follows:
$$
  \bigotimes_{x_\in M} = \left(\bigotimes_{x\in M_A}\mathcal H_x\right)\otimes\left(\bigotimes_{x\in M_B}\mathcal H_x\right)
$$
The first factor is what one might call $\mathcal H_A$, the Hilbert space corresponding to all degrees of freedom assigned to points in region $M_A$, and similarly for the second factor.

how does this claimed equality follow that, $S_A = lim _{n \rightarrow 1} S^{(n)}_A = - lim _{n \rightarrow 1} \frac{\partial \rho^n_A }{\partial n }$

Let a density matrix $\rho$ with eigenvalues $\lambda_k$ be given.  Its associated von-Neumann entropy is defined by
\begin{align}
    S(\rho)
    &= -\sum_k\lambda_k\ln\lambda_k
\end{align}
Now notice that
\begin{align}
    \frac{d}{da}x^a
    &= \frac{d}{da}e^{\ln x^a}
    =\frac{d}{da}e^{a\ln x}
    =e^{a\ln x}\ln x
    =e^{\ln x^a}\ln x
    =x^a\ln x
\end{align}
and therefore
\begin{align}
    \lim_{a\to 1}\frac{d}{da} x^a = x\ln x
\end{align}
It follows that
\begin{align}
    S(\rho)
    &= -\sum_k\lim_{a\to 1}\frac{d}{da}(\lambda_k)^a
    = -\lim_{a\to 1}\frac{d}{da}\sum_k(\lambda_k)^a.
\end{align}
Now, let $n$ be a positive integer, and define a function $f:\mathbb{Z}^+\to\mathbb{C}$ by
\begin{align}
    f(n) = \sum_k(\lambda_k)^n.
\end{align}
I claim, without proof, that $f$ can be analytically continued so that its domain includes all complex numbers $a$ with $\Re [a]>1$.  Let us call this analytic continuation $f_c$.  I claim, further, without proof that an explicit formula for this analytic continuation is obtained by simply replacing the $n$ with $a$;
\begin{align}
    f_c(a) = \sum_k(\lambda_k)^a.
\end{align}
We can now write the entropy as
\begin{align}
    S(\rho) = -\lim_{a\to 1^+}\frac{d}{da}f_c(a)
\end{align}
For $a =n$ where $n$ is a positive integer, notice that $f$ is simply the trace of $\rho^n$;
\begin{align}
    f(n)
    &= \mathrm{tr}(\rho^n)
\end{align}
If we denote the analytic continuation of $\mathrm{tr}(\rho^n)$ in the variable $n$ to complex values $a$ with $\Re [a]>1$ by $\mathrm{tr}(\rho^a)$, then we obtain the desired formula
\begin{align}
    \boxed{S(\rho) = -\lim_{a\to 1^+}\frac{d}{da}\mathrm{tr}(\rho^a)}.
\end{align}
Addendum. (October 15, 2013)  Here is the proof of the first equality as promised from months ago :).  I use the same notation as above.  Then notice that
\begin{align}
  \lim_{a\to 1^+} f_c(a) &= \mathrm{tr}\rho = 1 \\
  \lim_{a\to 1^+} f_c'(a) &= -S(\rho)
\end{align}
This first equality follows from the fact that density operators are traceless, and the second equality is essentially the second equality I proved from before.  Now, using these two facts, we compute;
\begin{align}
  \lim_{a\to 1^+}\frac{1}{1-a} \ln f_c(a)
  &= \lim_{a\to 1^+} \frac{1}{1-a} \ln(f_c(1) + f_c'(1)(a-1) + O(a-1)^2) \\ 
  &= \lim_{a\to 1^+} \frac{1}{1-a}\ln(1+S(\rho)(1-a)+ O(a-1)^2) \\
  &= \lim_{a\to 1^+} \frac{1}{1-a}(S(\rho)(1-a) + O(1-a)) \\
  &= S(\rho)
\end{align}
as desired!
