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.