The two-sided eternal black hole is dual to the <thermofield double state>
$$
|\Psi_\beta\rangle
=\frac1{\sqrt{Z(\beta)}}
\sum_n e^{-\beta E_n/2}|n\rangle_L^*|n\rangle_R.
$$
Tracing out the left CFT gives
$$
\rho_R=\operatorname{Tr}_L|\Psi_\beta\rangle\langle\Psi_\beta|
=\frac{e^{-\beta H_R}}{Z(\beta)}.
$$
Thus the geometric period $\beta$ is the boundary inverse temperature, and $S_{\rm BH}$ is the <Von Neumann entropy> $S_R=-\operatorname{Tr}(\rho_R\log\rho_R)$, equivalently the entanglement entropy between the two CFTs.

The <modular Hamiltonian> of this thermal state is
$$
K_R=-\log\rho_R=\beta H_R+\log Z.
$$
For any first-order state variation with $\operatorname{Tr}\delta\rho=0$,
$$
\delta S_R
=-\operatorname{Tr}(\delta\rho\log\rho_R)
=\operatorname{Tr}(\delta\rho K_R)
=\boxed{\beta\operatorname{Tr}(\delta\rho H_R)
=\beta\,\delta\langle E\rangle_\rho.}
$$
This is the <first law of entanglement entropy>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 354 / 2 / c / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-354/2/c/i/solution