I'm currently doing a project on two sided Ads-Schwarzschild black holes in the context of Ads/CFT. I want to show that the entanglement entropy between the two CFTs corresponds approximately to the Hawking-Bekenstein black hole entropy.

Hawking-Bekenstein entropy

Let's do this for the 3D case (also known as the BTZ black hole) to keep it simple. The metric of the BTZ black hole is $$ \text{d} s^2 = -f(r)\text{d} t^2 + f(r)^{-1}\text{d} r^2 + r^2\text{d} \phi^2 $$ with $f(r)=k^2(r^2-\mu^2)$ and $\mu^2=\dfrac{8G_nM}{k^2}$. In this case, the horizon area is given by $$ A_h = 2\pi r_h = 2\pi \frac{2\sqrt{2G_nM}}{k} $$ And the black hole entropy is $S_{bh} = \frac{A_h}{4 G}$. The black hole temperature is given by $\frac{\kappa}{2\pi}$ with $\kappa$ the surface gravity. If you compute $\kappa$ (which involves computing Christoffel symbols) and then substitute it into $T_h$, you find $ S_{bh} = 2M/T$ (where $M$ is the black hole mass.)

Entanglement entropy

On the CFT side, such a black hole corresponds to a thermal state of the two CFTs: $$ \mid \psi (t)\rangle = \frac{1}{\sqrt{Z}} \sum_n e^{-\beta E_n/2} e^{-2iE_nt} \mid n \rangle_1 \otimes \mid n \rangle_2 $$ where $ Z = \sum_n e^{-\beta E_n/2} $. After tracing out one of the CFTs we find a reduced density matrix $$ \rho_1 = \text{Tr}_2 \mid\psi(t)\rangle \langle \psi (t) \mid = \frac{1}{Z}\sum_n e^{-\beta E_n}\mid n \rangle \langle n \mid. $$ From this we can compute the entanglement entropy between the two CFTs: $$ S_{ent} = - \text{Tr}(\rho_1 \log \rho_1) = \frac{1}{Z} \sum_n e^{-\beta E_n} (\beta E_n + \log Z). $$

Linking the two

As far as I understand, one should be able to find that these two entropies are approximately equal, but I don't really see how to obtain this. Can anyone point out how I should do this?

edit: I think that the link between the two relies on the properties of super-Yang-Mill’s CFT that needs to be plugged in to obtain the $E_n$. This may be a bit beyond my reach since I don't really have any background in CFT yet.


2 Answers 2


You are right - your problem lies in a lack of knowledge of the CFT side. This is quite complicated, so I'll only be able to give you hints and references. A full answer would easily fill 50 pages!

We are looking at $(2+1)$ dimensional gravity dual to a $(1+1)$ dimensional CFT. Our problem is now to calculate entanglement entropy in the appropriate CFT. Note that this isn't really a "super-Yang-Mills" theory. In fact, we don't have to be exactly sure of the theory at all, because it turns out that conformal symmetry tells us enough to determine the entropy up to some factor.

The next step is to go to the literature, more precisely this paper by Cardy and Calabrese. They calculate the entanglement entropy associated to a defined area $A$ of the $1+1$ dimensional spacetime in which the theory lives. More precisely they calculate a measure of entanglement between everything living inside $A$ and everything living outside $A$.

They proceed by using the replica trick (nice review here) on a lattice QFT which says that

$$ S_A = -\textrm{Tr} (\rho_A \log \rho_A) = -\lim_{n\to 1}\frac{\partial}{\partial n} \textrm{Tr}\rho_A^n $$

where the density matrix $\rho_A$ is given more or less by your formula for $\rho_1$ in the question.

Why does this help? Well it removes the complicated expression inside the $\textrm{Tr}$ and replaces it with one that can be calculated more easily. In fact it can be calculated just from conformal symmetry using a Riemann surface intepretation. Skimming down to equation $(16)$ in the paper above will give you the gist.

After all that hard work the answer comes out to be

$$S_A = c\log(l/a)$$

where $c$ is the central charge of the CFT, $l$ is the diameter of our region $A$, and $a$ is the lattice spacing.

At first glance this hardly looks like the Bekenstein-Hawking formula! In fact, there's a good reason for this. The Bekenstein-Hawking result is effectively semiclassical, why the entanglement entropy is fully quantum. So we should expect to get $S_{BH}$ as the first term in an expansion of $S_A$ perhaps.

Actually realising this is a somewhat technical exercise in matching parameters on each side of the AdS/CFT correspondence. For details, you'll need to immerse yourself in this paper by Cadoni and Mellis. The expansion of which I spoke is equation $(41)$.

Best of luck with your project - I assume it's a Masters dissertation? This topic can be quite daunting at first, but there's a lot of interesting maths and physics. And it really doesn't matter if you don't understand everything - nobody does!

  • $\begingroup$ Thanks for the answer! It was not for my master thesis but for some smaller project, so this was a bet beyond the scope of it. $\endgroup$
    – Lagrangian
    Commented Jun 2, 2015 at 15:42
  • $\begingroup$ I think Cardy and Calabrese's result shown above is the 'area law' on 1 dimensional system, S_BH is on 2 dimensional. So they are different. If you compute the area law in 2 dimension, then it is S_BH (the leading term) as explained by holographic entanglement entropy. $\endgroup$
    – XXDD
    Commented Feb 19, 2016 at 2:26
  • $\begingroup$ This answer is ridiculously good. $\endgroup$
    – user224659
    Commented Oct 18, 2020 at 13:10

There is a nice concise, 2022 summary of this found in this paper by Zurek (Eqn 6). That is, although the horizon of a black hole passes through an empty region of spacetime, it has an entropy associated with it:

\begin{equation} S_{BH} = \frac{A}{4G} = S_{ent} \end{equation}

where the subscripts denote Bekenstein-Hawking (BH) and entanglement (ent), and the two entropies are equated.

While the second equality is known only to be true in certain systems (e.g. AdS/CFT), where the BH entropy exactly gives the entanglement entropy of the boundary quantum field theory, for reasons expanded in the Zurek paper, there is evidence that this relationship is generic to horizons.


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