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.


1 Answer 1


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!

  • $\begingroup$ Thanks for this very helpful answer! I was wondering if you proved the second equality and not the first? (..or am I missing something!?..) $\endgroup$
    – user6818
    Aug 11, 2013 at 3:52
  • $\begingroup$ You're right I skipped the first equality; I'll try to think about this soon and get back to you with a proof of the first (unless you beat me to it!) $\endgroup$ Aug 11, 2013 at 4:36
  • $\begingroup$ Any luck in getting the entanglement entropy from Renyi entropy? :) $\endgroup$
    – user6818
    Sep 2, 2013 at 20:58
  • $\begingroup$ @user6818 Ah man not yet; I haven't had time to look at this in detail yet, but I promise I haven't forgotten :). $\endgroup$ Sep 3, 2013 at 6:10
  • $\begingroup$ Any luck in getting that proof? $\endgroup$
    – user6818
    Oct 15, 2013 at 19:36

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