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

Table of contents
- Solution i

Solution

The two-sided eternal black hole is dual to the thermofield double state

∣ Ψ_{β} ⟩ = \frac{1}{Z ( β )} \sum_{n} e^{- β E_{n} /2} ∣ n ⟩_{L}^{*} ∣ n ⟩_{R} .

Tracing out the left CFT gives

ρ_{R} = Tr_{L} ∣ Ψ_{β} ⟩ ⟨ Ψ_{β} ∣ = \frac{e ^{- β H_{R}}}{Z ( β )} .

Thus the geometric period

β

is the boundary inverse temperature, and

S_{BH}

is the Von Neumann entropy

S_{R} = - Tr (ρ_{R} lo g ρ_{R})

, equivalently the entanglement entropy between the two CFTs.

The modular Hamiltonian of this thermal state is

K_{R} = - lo g ρ_{R} = β H_{R} + lo g Z .

For any first-order state variation with

Tr δ ρ = 0

,

δ S_{R} = - Tr (δ ρ lo g ρ_{R}) = Tr (δ ρ K_{R}) = β Tr (δ ρ H_{R}) = β δ ⟨ E ⟩_{ρ} .

This is the first law of entanglement entropy.

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